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http://codeforces.com/blog/entry/92183 | 1,653,006,040,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00063.warc.gz | 12,781,649 | 23,603 | ### Elegia's blog
By Elegia, history, 11 months ago,
This is a part of my report on China Team Selection 2021. I might translate some other parts later if you are interested.
I'm going to show that we can calculate $f(G)$ for any polynomial $f\in R[x]$ and set power series $G: 2^{n} \rightarrow R$ under $\Theta(n^2 2^n)$ operations on $R$.
Let's first consider a special case: calculate $F = \exp G$. We consider set power series as truncated multivariate polynomial $R[x_1,\dots,x_n]/(x_1^2,\dots,x_n^2)$. We can see that $F' = FG'$ whichever partial difference $\frac{\partial}{\partial x_k}$ we take. Then we can rewrite this equation as $[x_n^1] F = ([x_n^1]G)\cdot [x_n^0]F$. Hence we reduced the problem into calculating $[x_n^0] F$, and then one subset convolution gives the rest part. The time complexity is $T(n)=\Theta(n^22^n) + T(n-1)$, which is exactly $\Theta(n^22^n)$. I call this method "Point-wise Newton Iteration".
This method sounds more reasonable than calculating the $\exp$ on each rank vector. Because $\frac{G^k}{k!}$ counts all "partitions" with $k$ nonempty sets in $G$. So it exists whenever $1\sim n$ is invertible in $R$.
Now we try to apply this kind of iteration into arbitrary polynomial $f$. Let $F=f(G)$, then $F'=f'(G)G'$. Then we reduced the problem into calculating $[x_n^0]f(G)$ and $[x_n^0]f'(G)$. This becomes two problems! But don't worry. In the second round of recursion, it becomes $[x_{n-1}^0 x_n^0]$ of $f(G), f'(G), f"(G)$. The number of problem grows linearly. You will find out that the complexity is $\sum_{0\le k\le n} (n-k) \cdot \Theta(k^2 2^k)$.
It's not hard to see that the complexity is still $\Theta(n^2 2^n)$, because $\sum_{k\ge 0} k\cdot 2^{-k}$ converges.
Noted that $\frac{(A+B)^2}2-\frac{A^2}2-\frac{B^2}2=AB$, we proved the equivalence between the composition and subset convolution.
Here are some remarks:
• If $G$ has no constant term, $f$ is better to be comprehended as an EGF.
• This algorithm holds the same complexity on calculating $f(A, B)$ and so on.
• We can optimize the constant of this algorithm by always using the rank vectors during the iteration. Then it can beat lots of methods that work for some specialized $f$.
• If you have some knowledge of the "Transposition Principle", you will find out that, for any fixed $F, G: 2^n \rightarrow R$, the transposed algorithm gives $\sum_S F_S \cdot (G^k)_S$ for all $k$. It immediately gives an $\Theta(n^22^n)$ algorithm to calculate the whole chromatic polynomial of a graph with $n$ vertices.
• +312
» 11 months ago, # | +16 Does it generalize for $R[x_1,\dots,x_n]/(x_1^{k_1},\dots,x_n^{k_n})$?
• » » 11 months ago, # ^ | +16 I thought about it for arbitrary $(k_i)$ and at least we can do $\Theta(nK\log K)$ where $K=\prod k_i$ when $f$ is elementary function. I made a problem (Chinese) for calculating $\log$. I'm not familiar with the arbitrary $f$ because the now best algorithm[1] in $n=1$ is very complicated :)[1] Kedlaya, Kiran & Umans, Christopher. (2008). Fast polynomial factorization and modular composition.
» 11 months ago, # | +40 Looks like for "or" convolution we consider $R[x_1,\dots,x_n]/(x_1^{2}-x_1,\dots,x_n^{2}-x_n)$ and for "xor" it is $R[x_1,\dots,x_n]/(x_1^{2}-1,\dots,x_n^{2}-1)$ in this notation...
• » » 11 months ago, # ^ | +18 That's right, and then FMT and FWT are exactly multipoint evaluations.
• » » » 11 months ago, # ^ | 0 What does FMT mean?
• » » » » 11 months ago, # ^ | 0 fast Möbius transform?
• » » » » 11 months ago, # ^ | ← Rev. 2 → +18 Fast Möbius Transformation (on the subset lattice). It's called Möbius transformation in the paper on subset convolution and some other articles. It's also known as zeta transformation such as in https://arxiv.org/abs/0711.2585 (same authors with previous paper) and its inverse is called Möbius transformation. In Enumerative Combinatorics, it is closely related to the zeta function, and its inverse is closely related to the Möbius function.
• » » » » » 11 months ago, # ^ | 0 Where do you guys get this much of knowledge from ? Means the source from where you guys find the theory I am a newbie and i found difficulty in finding the right source to study :( Plese help !!
• » » » » » » 11 months ago, # ^ | +8 Although these stuff are not necessary at all to succeed in CP or even be a LGM, university classes usually teach (or try to teach) advanced algorithmic theory. In most cases, those classes are not the part of the departmental program or are graduate classes, but you can always take them. In case the staff does not do a good job teaching it, they at least provide necessary materials and refer you to sources. It's up to you, after all, to choose to learn advanced theory since the only place you will use them is the academia (or some crazy rare Div.1 E problem).
• » » » » » » » 11 months ago, # ^ | +105 But then you realize Elegia is a high schooler.
• » » » » » » » » 11 months ago, # ^ | 0 Thanks for response <3
• » » » » » » » » 11 months ago, # ^ | +16
• » » » » » » 11 months ago, # ^ | 0 The subset convolution paper is the origin of all these things, you can just google "subset convolution" then find it. The Enumerative Combinatorics book is the standard textbook(?) on enumerative combinatorics. But just like what MITRocks have said, "these stuff are not necessary at all to succeed in CP or even be a LGM".
• » » 11 months ago, # ^ | 0 By the way, the subset convolution can be viewed as multipoint evaluation on $R[z][x_1,\dots,x_n]/(x_1^2-zx_1,\dots,x_n^2-zx_n)$ and finally let $z \to 0$.
• » » » 11 months ago, # ^ | ← Rev. 3 → +16 Yep, that's exactly how I understood in when wrote this. Though $[z^l x^k]AB$ has a meaning as well, that is summing everything up by $|i \cap j|=l$ and $i \cup j = k$.
• » » » 11 months ago, # ^ | ← Rev. 2 → 0 You provided me the idea to consider $R[z][x_1, \dots, x_n]/(x_1^2-z^2, \dots,x_n^2-z^2)$ and evaluate it in the cube {$-z,z$}$^n$. It also provides subset-convolution with $z \to 0$, but $[z^l x^k]AB$ in this ring is a sum of $a_i b_j$ over $i \Delta j = k$ and $|i \cap j|=\frac{l}{2}$.
• » » » » 11 months ago, # ^ | ← Rev. 3 → +16 This algorithm has been presented by vfleaking in "集合幂级数的性质与应用及其快速算法（Properties, Applications, and Fast Algorithms of Set Power Series）". The "set power series (集合幂级数)" itself also was presented in this article. You can find it in IOI2015中国国家候选队论文集（IOI 2015 China Candidate Team Essay Collection）. It's noticed that this algorithm cannot be used for the ring $R$ of characteristic 2 (while FMT-like one is ok). At that time, set power series were not considered as the elements of a quotient ring. Later some found in this way many things can be explained naturally. Then this approach has become increasingly well-known in some small groups over the years.
» 11 months ago, # | +101 All hail Elegia, the king of combinatorics!
» 11 months ago, # | +18 Can you please elaborate on transposition principle part?
• » » 11 months ago, # ^ | +10 If you don't know what is the transposition principle, google it. It's something like duality. Then polynomial composite a fixed set power series is a linear map from polynomial to set power series. Just apply the transposition principle to it.
» 11 months ago, # | ← Rev. 3 → -9 | 2,197 | 7,328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-21 | latest | en | 0.852822 |
http://slideplayer.com/slide/4333029/ | 1,521,728,189,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647885.78/warc/CC-MAIN-20180322131741-20180322151741-00208.warc.gz | 267,750,811 | 22,680 | # SUMMARIZING DATA: Measures of variation Measure of Dispersion (variation) is the measure of extent of deviation of individual value from the central value.
## Presentation on theme: "SUMMARIZING DATA: Measures of variation Measure of Dispersion (variation) is the measure of extent of deviation of individual value from the central value."— Presentation transcript:
SUMMARIZING DATA: Measures of variation Measure of Dispersion (variation) is the measure of extent of deviation of individual value from the central value (average). It determines how much representative the central value is. Dispersion is small if the values are closely bunched about their mean and it is large if the values are scatted widely about their mean. The median and mean mark for both tests are 20 but data A is more spread out than data B.
Important measures of dispersion are: 1.Range 2.Variance & standard deviation 3.Standard error of Mean 4.Co - efficient of variation.
Range is the absolute difference between the highest value and the lowest value in a series of observations. Range = largest value - smallest value Range Example: the weight of 10 students are: 25, 28, 33, 36, 40, 45, 49, 52, 55, 57. Range is 57 – 25 = 32. The range is the simplest measure of dispersion. It is a rough measure of dispersion as its measure depends upon the extreme items and not on all the items. It does not tell us anything about the distribution of values in the series.
Range Application: Range is used in medical science to define the normal limits of biological characteristics. Example: normal ranges of systolic and diastolic blood pressure are 100 – 140 mm and 80 –90 mm respectively. Ordinarily observations falling within a particular range are considered normal and those falling outside the normal range are considered as abnormal. Range for a biological character such as blood cholesterol, fasting blood sugar, hemoglobin, bilirubin etc is worked out after measuring the characteristics in large number of healthy persons of the same age, sex, class etc.
Range Merits: It is simple to compute and understand. It gives a rough but quick answer Limitation: 1.It is not a satisfactory measure as it is based only on two extreme values, ignoring the distribution of all other observations within the extremes. These extreme values vary from study to study, depending upon the size and nature of sample and type of study.
Karl Pearson introduced the concept of Standard Deviation in 1893. The standard deviation is a statistic that tells us how tightly all the values are clustered around the mean in a set of data. Variance & Standard deviation The mean of the squares of the deviations of every observation from their mean is a measure of spread and is called the variance. The standard deviation is the square root of the variance. It is computed as the root of average squared deviation of each number from its mean. For example, for the numbers 1, 2, and 3 the mean is 2 and the standard deviation is: SD = 0.667 = 0.44
Standard deviation Merits: It is the most important and widely used measure of dispersion. It is based on all the observations and the actual sign of deviations are used. Standard deviation provides the unit of measurement for the normal distribution. It is the basis for measuring the coefficient of correlation, sampling and statistical inference. Limitation: It is not easy to understand and difficult to calculate It is affected by the value of every item in the series.
= 20 Calculations of SD: In these two groups, means are same (20) but their variation (SD) is different (SD A, 8.2 and SD B, 5.5).
Calculations of SD with alternative formulas:
Greater SD, greater is variation of observation. Mean is presented with SD as ….. Mean±SD.
The standard error of a sample mean is just the sample standard deviation divided by the square root of the sample size. Standard Error of Mean If we draw a series of samples fro same population and calculate the mean of the observations in each, we have a series of means. The series of means, like the series of observations in each sample, has a standard deviation. The SE of the mean of one sample is an estimate of the SD that would be obtained from the means of a large number of samples drawn from the population. Another thing is if we draw random samples from the population their means will vary from one to another. This variation depends on the variation of population and size of samples. We do not know the variation of population so we use the variation of the sample as an estimate of it. This is expressed in SD and if we divide SD by squire root of the number of observations in the sample we have an estimate of SE of mean, SEM = SD/ n
Advantage of SE To determine the significant difference of two means of different variables. To calculate the size of sample. If SD is known. =
Greater SE, greater is variation of observation. Mean is presented with SE as ….. Mean±SE.
Relative measure of variation is called Co-efficient of variation (C.V.). C.V. is defined as the S.D. divided by the mean times 100. Co-efficient of variation (C.V.) It is useful in comparing distribution whose units or characters may be different e.g. height in cm in one and in inches in the other.
Mean SD CV Adult 160 cm 10 cm 6.25% Children 60 cm 5 cm 8.33% It means though height in adult shows greater variation in SD, but real thing is that children is greater variation. Co-efficient of variation (C.V.) Example: Height (cm) of adult and children are given in the table
Population & Sample Population All possible values of a variable or all possible objects whose characteristics are of interest in any particular investigation or enquiry. If the income of the citizen of country is of interest to us, the aggregate of all relevant incomes will constitute the population. Sample A sample is a part of population. Although we are primarily interested in the properties of a population or universe, it is often impracticable or even impossible to study the entire universe. Thus inferences about a population are usually drawn on the basis of a sample. It represents the population.
Normal Distribution The normal distribution was first introduced by the French mathematician La Place (1749-1827). It is highly useful in the field of statistics. The graph of this distribution is called normal curve or bell - shaped curve. In normal distribution, observations are more clusters around the mean. Normally almost half the observations lie above and half below the mean and all observations are symmetrically distributed on each side of the mean. The normal distribution is symmetrical around a single peak so that mean median and mode will coincide. It is such a well-defined and simple shape, a great deal is known about it. The mean and standard deviation are the only two values we need to know o be able to describe a normal curve completely.
Normal Distribution
Characteristics : The curve is symmetrical It is a bell shaped curve. Maximum values at the center and decrease to zero symmetrically on each side Mean, median and mode coincide Mean = Median = Mode It is determined by mean and standard deviation. Mean 1SD limits, includes - 68% of all observations Mean 2SD -,,,, - 95%,,,, Mean 3SD -,,,, - 99%,,,, Normal Distribution
Almost all statistical tests (t-test, ANOVA etc) assume normal distributions. Fortunately, these tests work very well even if the distribution is only approximately normally distributed. Some tests (Mann - whitney U test, Wilcoxon W test etc) work well even with very wide deviations from normality. Normal Distribution
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http://studentblogs.passged.com/2009/01/22/the-scientific-method-ged-science-must-know/ | 1,542,473,641,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743717.31/warc/CC-MAIN-20181117164722-20181117190722-00303.warc.gz | 318,906,850 | 9,060 | The Scientific Method: GED Science MUST-KNOW!
Hey, Dudes! The GED science test’s got a lot of stuff on it… but one thing you really KNOW is gonna be on there is science experiments. I love science experiments, dudes. Like, mad scientist stuff, you know… making invisibility cloaks and glow-in-the-dark slime. But you gotta know something about how experiments are done…and why… and that means understanding the scientific method.
You really will run across questions on the GED science test that expect you to know all about the scientific method. It’s one of those things that gives you a science background! So, what’s this scientific method anyhow? It’s a process used in science to find out information about the world. The scientific method has five steps:
1) Observe
The first step is to notice what’s happening around you. Science begins with curiosity about the world. That means looking around you, asking questions, and wondering about what’s going on.
Example: I love the lava lamp on my desk, and I wonder what makes the ‘lava’ inside the lamp float up to the top and come down again.
2)Â Hypothesize
Hypothesis is one of the important words that you should understand for the GED test. A hypothesis is an idea that explains what you’ve observed or answers a question that you’ve wondered about. A hypothesis is what you suggest or think might be the answer.
Example: I hypothesize that, since the lava sinks to the bottom when the lamp is off, the light from the lamp makes the lava rise, and when the lava is in the darker, top of the lamp, it falls.
3) Predict
After you have a hypothesis, predict something else that would be true if your hypothesis is true.
Prediction: If the bottom of the lamp only is exposed to any bright light, the lava lamp will work.
4) Experiment
Conduct an experiment to test your prediction.
Example: I put two lava lamps on two light sources. One is the original light source, and the other is a light source that’s equally bright but does not give off heat. I watch the lava lamps to see when they start working. The lava lamp on the original light source is the only one that works. The other lava lamp does not move.
5) Evaluate
Look at the results of the experiment. Is it what you would expect based on your prediction? If not, you need to go back and form a new hypothesis. If so, you can begin to develop your hypothesis (an idea that’s not proven) into a theory (an idea that has evidence behind it. That will require more testing and expanding your idea.
Example: The lava lamp does not work with a light source that only gives off light, not heat. That means that it’s not the light that makes the lava move. I think about what the light in the lava lamp has that was not present in my alternate light source, and I make a new hypothesis that applying heat to the bottom of the lava lamp is what makes the lava move. I’ll need to test my new hypothesis.
There ya’ go! A whole scientific method in five parts. Look for it on the GED science test!
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Mid term 2013 (Winter)
# Mid term 2013 (Winter) - Commerce 354 Mid term...
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Prepared by J. Kroeker, 2013 © Sauder School of Business, UBC Page 1 Commerce 354 Mid-‐term Examination (February 5, 2013) First Name: _______________________ Last Name:______________________________ Student #: _______________________ Please Circle Section M/W 201 (8:30am), 202 (10 am) 203 (11:30 am) T/TH 204 (11:00 am) Question 1 (36 marks) Shenwin Ltd produces computer parts using leased machines that each have a capacity to make #25,000 per year. Shenwin can fit a total of 8 machines in its existing factory and can lease machines conveniently on a year -by-year basis. The cost of one machine lease in 2012 was \$440,000 per machine. The factory is owned and is being depreciated at \$218,000 per year for the next 10 years. The company always seeks to maximize profits and the CFO has indicated that inflation has been 10% but this has not impacted direct material costs since she signed a 4 year contract with the supplier in 2010. The labor costs were set for five years in a negotiated agreement in 2010 so there is no inflation in the labor figures. 2010 2012 Direct Materials 910,000 1,240,000 Direct Labor 1,365,000 1,860,000 Overhead 8,458,000 11,174,000 Total Costs \$ 10,733,000 14,274,000 Units Produced & Sold # 91,000 # 124,000 "Average cost" 117.95 115.11 Industry experts expect inflation to be 3% in 2013. The selling price per unit was \$119 in 2012. Shenwin feels it can increase its selling price at the rate of inflation. Demand for 2013 is expected to be #124,000. a) Calculate the expected contribution margin for 2013? (6 marks)
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Prepared by J. Kroeker, 2013 © Sauder School of Business, UBC Page 2 b) Calculate the expected cost to lease one machine in 2013? (2 marks) c) Calculate the number of units required to justify the leasing of an additional machine. (4 marks) d) The CFO would like to know the cash flow that could be expected from 2013 if sales volume remains the same as in 2012? (4 marks)
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https://sciforums.com/threads/does-time-exist.163715/page-2 | 1,723,606,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00696.warc.gz | 390,892,498 | 20,073 | Does time exist?
Aha, I didn't say it was illusory. I said it is an abstract concept, which is a bit different.
If someone asks you: "Does length exist?", how would you answer? Personally I would say, "length of what?" To say length exists, in the abstract, without the length of anything being in question, seems meaningless.
I do not think it is sensible to treat time as illusory. But, like length, time only has concrete existence in relation to something being measured.
I just want to know if "real" is real? What's so real about it? I can't touch it. I can't "keep it real" if "real" isn't real.
OK, so we agree change occurs in nature whether or not an observer is present.
Time is our yardstick for measuring change, so that we can relate different change processes, occurring at different rates, in a consistent way). Just as length gives us a yardstick for comparing intervals separating objects spatially, in a consistent way.
Yet, for some reason, nobody ever asks whether length exists.
In a way, it doesn't. It's an abstract concept, invented by human beings to make sense of their world. But it's a footling discussion that interests no one.
Time on the other hand, gets people worked up. It's a bit like magnets in physics: anything invisible is a huge mystery to some people.
Do lengths "exist" as multiples of a wavelength ?
Is that how the basic unit of length can be calculated?
I think I have heard of a Planck length...is that the smallest wave length possible or is it just perhaps the smallest observable length?
If the latter ,is there any theoretical limit to how small a length can be?
I just want to know if "real" is real? What's so real about it? I can't touch it. I can't "keep it real" if "real" isn't real.
Are you just asking "Are tomatoes tomatoish?" ?
Have you a definition for "real"? Can you then apply your definition to a particular circumstance?
If you are too general you will not see the trees for the wood
Do lengths "exist" as multiples of a wavelength ?
Is that how the basic unit of length can be calculated?
I think I have heard of a Planck length...is that the smallest wave length possible or is it just perhaps the smallest observable length?
If the latter ,is there any theoretical limit to how small a length can be?
Haha, I really hate all this semi-bullshit stuff people love to talk about "Planck length", Planck time" etc. These are rather speculative theoretical concepts for which, as far as I know, no evidence is available.
Be that as it may, as I understand it, "Planck length" is the shortest length at which the laws of physics can be expected to apply. That does not mean that no shorter length is possible, i.e. it does not mean that length itself is quantised, as it were. If that could have any conceivable meaning, which I doubt.
I just want to know if "real" is real? What's so real about it? I can't touch it. I can't "keep it real" if "real" isn't real.
Yeah, after all, how real is real? I mean, just how real do you want to be?
It's a bit pointless. I guess length and time, in the abstract, are as real, or not, as a concept such as beauty.
Was it on this site that someone showed a clip of Richard Feynman saying ''what does real mean?''
When you touch something it's only modelled as fields interacting with fields, fields are models.
Physics can't prove fields are ''real''. Physics is models.
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Physics is models.
I profoundly disagree with this. If science is about anything at all, it is an attempt to describe the natural world - the "real world" if you insist.
Of course science uses models - language is a model, after all - that does not mean that science is only models. These models do what science attempts to achieve, to describe the real, natural, world.
I profoundly disagree with this. If science is about anything at all, it is an attempt to describe the natural world - the "real world" if you insist.
Of course science uses models - language is a model, after all - that does not mean that science is only models. These models do what science attempts to achieve, to describe the real, natural, world.
The natural or ''real'' world certainly acts like it is fields, but physics can't prove that it is fields.
Physics can't prove fields are ''real''. Physics is models.
So, what's the physics model for measurements? Does physics have a model for how to perform an experiment? Please give an example.
The natural or ''real'' world certainly acts like it is fields, but physics can't prove that it is fields.
Physics isn't about proof, it's about finding evidence for theoretical 'models' being acceptable explanations.
For instance, what evidence is there that the universe is holographic; that the third spatial dimension we perceive (via measurements, not necessarily accurate) emerges from a background with only two (plus one of time)?
Another example: supersymmetry has no physical evidence to support it; the LHC hasn't delivered on what the theory needs, to be accepted as a reasonable explanation.
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So, what's the physics model for measurements?
An operator acting on a state vector
Does physics have a model for how to perform an experiment? Please give an example.
The determination of one the possibly many eigenvalues for that operator.
Next question?
I profoundly disagree with this. If science is about anything at all, it is an attempt to describe the natural world - the "real world" if you insist.
Of course science uses models - language is a model, after all - that does not mean that science is only models. These models do what science attempts to achieve, to describe the real, natural, world.
I don't see these being in conflict. Science models physical reality, doesn't it?
We can never say our models ARE reality: we just hope they get closer and closer to it.
An operator acting on a state vector
You're saying if I use a metre ruler to measure a distance, it's an operator on a state vector? What state vector?
The determination of one the possibly many eigenvalues for that operator.
Which of these possibly many eigenvalues corresponds to a distance of 1 metre? If I divide the distance measured with a ruler, by the distance for the ruler (i.e. 1 metre), what kind of operator is acting?
Physics isn't about proof, it's about finding evidence for theoretical 'models' being acceptable explanations.
We agree. Physics is about models. What does 'reality' and 'real' mean, is not the realm of physics. Physics is models.
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We agree. Physics is about models. What 'reality' or what does 'real' mean, is not the realm of physics. Physics is models.
Well, to be accurate, physics is not really about models. Physics is "about" physical reality, but makes models in order to try to represent it.
Well, to be accurate, physics is not really about models. Physics is "about" physical reality, but makes models in order to try to represent it.
Yes, what else do you think the models are representing ? In the context of my reply to Quarkhead, I said physics can't prove that its model of fields is indeed what 'reality' is.
I may fumble in my explaining.
Yes, what else do you think the models are representing ? In the context of my reply to Quarkhead, I said physics can't prove that its model of fields is indeed what 'reality' is.
I may fumble in my explaining.
OK agreed.
You're saying if I use a metre ruler to measure a distance, it's an operator on a state vector?
Yes.
What state vector?
So you don't know what a "state" is? A system can be moving, spinning, sleeping , dancing the fandango or whatever. All at the same time. That is its state. Have you forgotten the defining property of a vector (at least as it's taught at elementary level)?
Which of these possibly many eigenvalues corresponds to a distance of 1 metre? If I divide the distance measured with a ruler, by the distance for the ruler (i.e. 1 metre), what kind of operator is acting?
Try the scalar product operator.
So you don't know what a "state" is? A system can be moving, spinning, sleeping , dancing the fandango or whatever. All at the same time.
I have a 1 metre ruler which is a standard of distance, or rather the length of the ruler is. The state of the ruler is a consequence of an interaction of a large number of particles. Say the ruler is made of plastic, then I can probably give it some electrostatic charge; the temperature of the ruler can change, But I need to assume I have an ideal element of distance if I want to use it to accurately measure a distance, the measurement will always be approximate, so the best I can do is repeat a measurement enough so I have a statistical result.
If I dance the fandango while holding the ruler, this doesn't alter the ruler's state or that it remains a standard length, more or less. There are more accurate ways of measuring distances including using a laser. But what about the state of the distance measured? What if I want to separate two particles, electrons say, by a distance of exactly 1 metre? Is that possible or do I have to accept an approximate separation?
What does the separation do to the state of each electron or do I take the state of both, and why does it matter if all I want to know is how far apart they are? That is, after considering the accuracy and precision involved (in any measurement). If I alter the distance after determining it, how does the state change as I do this?
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This is gibberish. You are conflating the operator (a vector) with its eigenvalues (scalars).
I don't want to say any more in this thread - I'm out
This is gibberish. You are conflating the operator (a vector) with its eigenvalues (scalars).
But you aren't conflating a measurement operator in QM with a measurement of distance, using a metre ruler. Hell no.
The state of a pair of electrons (let's assume an entangled state), separated by any distance doesn't usually include the distance. When IBM built its quantum computer the distances between the quantum 'bits' were determined by engineering. Can you say why these distances were chosen, then engineered? Can you say what the distance from here to the sun has to do with QM, or with determining that distance at any time? Or what eigenvalues of quantum states have to do with the length of a ruler?
No?
Lest we forget, the question is what is the state of a distance between two electrons, or between any two things?
Moreover why does the distance correspond to an eigenvalue of a measurement operator? The distance between the ends of a ruler is a quantum state?
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http://www.algebra-online.com/algebra-online-help/quadratic-formula/easy-way-to-calculate-cube.html | 1,481,352,556,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542972.9/warc/CC-MAIN-20161202170902-00065-ip-10-31-129-80.ec2.internal.warc.gz | 317,157,905 | 7,379 | Call Now: (800) 537-1660
December 10th
December 10th
easy way to calculate cube root
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Author Message
Neniriids
Registered: 26.01.2004
From:
Posted: Thursday 04th of Jan 18:28 I need some help people! I find easy way to calculate cube root really difficult. I have tried finding a tutor for the subject, but couldn’t find any in my area. The ones available are far and expensive.
IlbendF
Registered: 11.03.2004
From: Netherlands
Paubaume
Registered: 18.04.2004
From: In the stars... where you left me, and where I will wait for you... always...
Posted: Sunday 07th of Jan 09:25 I allow my son to use that program Algebra Buster because I believe it can significantly help him in his math problems. It’s been a long time since they first used that program and it did not only help him short-term but I noticed it helped in improving his solving capabilities. The program helped him how to solve rather than helped them just to answer. It’s fantastic!
Mibxrus
Registered: 19.10.2002
Posted: Monday 08th of Jan 18:10 unlike denominators, radical expressions and like denominators were a nightmare for me until I found Algebra Buster, which is really the best algebra program that I have come across. I have used it frequently through many algebra classes – Intermediate algebra, Basic Math and Algebra 1. Simply typing in the math problem and clicking on Solve, Algebra Buster generates step-by-step solution to the problem, and my algebra homework would be ready. I highly recommend the program. | 460 | 2,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-50 | latest | en | 0.936594 |
https://www.wavemetrics.com/comment/10876 | 1,722,907,817,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00687.warc.gz | 843,589,420 | 12,762 | # CurveFit and getting rid of W_sigma
I'm doing some curve fits while running live data acquisition.
Therefore I want the curve fits to be as fast as possible.
Can I somehow tell CurveFit to not create W_sigma? Or use a free wave for that?
My current invocation looks like:
Make/FREE/D/N=2 coefWave
variable V_FitOptions = 4
variable V_FitError = 0
variable V_AbortCode = 0
try
CurveFit/Q/N=1/NTHR=1/M=0/W=2 line, kwCWave=coefWave, data[startRow,endRow][1][10]/X=data[startRow,endRow][0][3]/AD=0/AR=0; AbortOnRTE
catch
....
endtry
I know that not looking at W_sigma in a fit is usually not a clever thing to do ;)
I'm not the expert, but using a free wave for W_sigma might not actually speed things up -- the linear fit will still need the residual matrix to be calculated, so it might as well get saved.
Maybe you could instead speed things up by providing a better initial guess for the fit -- possibly using the previous W_coef wave?
Line fits don't use or need initial guesses, so that's not going to help here.
But you're also correct that the all the info needed to create the sigma wave is needed to do the fit, so the only saving would be in not creating the wave itself. If you are working in a datafolder with thousands of waves, that might, in fact, be slow. In order to create a new wave, Igor must first make sure you don't already have a wave called W_sigma. That requires walking a linked list of waves in the current data folder; if W_sigma is near the end of a list of thousands of waves, it can take a while.
So in the course of writing that description, I came up with a strategy that would work around it: pre-create a wave called W_sigma *before* you load your data waves into the current datafolder. That way Igor will find W_sigma right away and not have to walk very far.
John Weeks
WaveMetrics, Inc.
support@wavemetrics.com
Thanks John and jcor!
My main motivation was that without W_sigma the databrowser would not have to update after curve fitting.
I don't have thousands of waves in one folder.
I'm doing three of these line fits for every data channel.
Three line fits take, on my dev machine, 0.5ms. I have up to ~24 data channels.
So in the extreme case this takes up to 12ms.
Hmm... so the DataBrowser is updating during user function execution, even with CurveFit/N=1?
Can you do the timing with the DataBrowser closed?
John Weeks
WaveMetrics, Inc.
support@wavemetrics.com
johnweeks wrote:
Hmm... so the DataBrowser is updating during user function execution, even with CurveFit/N=1?
Sorry that was incorrect. I'm calling CurveFit from a background function and then only is the databrowser updating.
It is not updating during function execution.
If you could come up with an Igor experiment file with a good test case, I could run it here under Instruments (Apple's rather nice code profiler) to see where the time is being taken. The test case would need to run enough fits in a loop to make it take at least a couple of seconds.
Send it to support@wavemetrics.com.
John Weeks
WaveMetrics, Inc.
support@wavemetrics.com | 745 | 3,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.935111 |
http://www.gregdetre.co.uk/software/matlab_cookbook/ | 1,653,759,855,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00576.warc.gz | 79,521,345 | 8,061 | ## Matlab cookbook
Many of these functions were written for and included in the MVPA toolbox.
Other useful resources:
Matlab cookbook
Data structures
Concatenating individual fields from two structs
Getting function handles
Dealing with optional arguments
Debugging
Pause execution in the middle of a script/function
Automatically pause execution upon error
Statistics and signal processing
Standard error of the mean
Zscoring
Fast fourier transform
Smoothing a vector with a median filter
Plotting and visualization
Positioning a figure
Figure that doesn't steal focus
Greenblack colormap
Titles that aren't in latex
Printing a figure to a .png
Misc
Print the date and time as a handy yymmdd_HHMM string
Automatically logging a diary of your sessions
Using indices from max, min and sort
Check that two matrices are of identical size
Count how many values match some boolean criterion
Combined disp and sprintf
Text progress bar
How can I tell if all the values in a matrix are integers?
Vec2ind and ind2vec that don't ignore all-zeros columns
### Data structures
#### Concatenating individual fields from two structs
If you have two structs with identical fields, and you'd like to end up with one struct, where each of its fields is the concatenation of the original two fields, e.g.
```s1.blah = [1 2 3];
s1.blah2 = [10 20 30];
s2.blah = [4 5 6 7 8];
s2.blah2 = [40 50 60 70 80];
s1and2 = grow_struct(s1,s2);
s1and2 =
blah: [1 2 3 4 5 6 7 8]
blah2: [10 20 30 40 50 60 70 80]
```
This is what grow_struct.m. is for. It's not too hard a problem, but I wanted to do a lot of error-checking to make sure that the two structs contained all and only the same fields, and that the lengths of all the fields in each struct are the same. And there's a unit test for it too.
#### Getting function handles
[with David Weiss]
Sometimes it makes more sense for an argument to be a function handle, and sometimes just a string. It's easy to convert from one to the other with str2func.m and func2str.m, but it's a pain to have to check every time. Arg2funct.m takes in either, and gives you back both.
```>> [funct_hand funct_name] = arg2funct('max')
funct_hand =
@max
funct_name =
max
>> [funct_hand funct_name] = arg2funct(@max)
funct_hand =
@max
funct_name =
max
```
#### Dealing with optional arguments
[With Chris Moore and David Weiss]
Propval.m is a powerful and flexible way of dealing with optional arguments that makes it easy for the user to feed in optional arguments, and easy for the developer to parse them, or to pass them onto inner functions that also take optional arguments.
An example will make this clear. Let's imagine that your function, myfunct.m, takes in one required argument (REQ) and a series of optional arguments.
```function [out] = myfunct(req, varargin)
REQ = required argument.
OPT1 (optional, default = 100).
OPT2 (optional, default = 'default').
```
Propval.m makes it easy for the user to call that function in any of the following ways:
```out = myfunct(req);
out = myfunct(req, 'opt1', 200);
out = myfunct(req, 'opt2','override', 'opt1',200);
args.opt1 = 200;
args.opt2 = 'override';
out = myfunct(req, args);
```
The first two examples utilize the same property/value pairs that many of Matlab's plotting functions and other toolboxes use, e.g.
```plot([1 2 3 4 5], 'LineWidth', 5)
```
Just as they do, you can feed in zero or more of the optional arguments, those arguments have defaults, and you can feed them in any order.
The third example (as with the Optimization toolbox) allows you to bundle up all your arguments into a struct, and then just feed that in. Any fields in the struct that you feed in override the defaults.
From the point of view of the developer, the code for dealing with the optional arguments is as simple as we could make it:
```% in myfunct.m
% define the defaults, as per the myfunct.m documentation above
defaults.opt1 = 100;
defaults.opt2 = 'default';
args = propval(defaults,varargin);
% then, to access them
args.opt1
args.opt2
```
Ok - if you've read this far, then hopefully you can already see the utility of this. The final scenario that propval.m can help with is when you have multiple functions that take optional arguments, and you want the user to be able to call the outermost and have their arguments passed on to the inner functions.
Let's imagine that myfunct.m calls anotherfunct.m, which also takes in optional arguments. From the user's point of view, we can just call myfunct.m with an 'opt3' argument:
```out = myfunct(req, 'opt1',200, 'opt3','10000);
```
This will override opt1's default in myfunct.m, but keep the default for opt2. We then want opt3 to be passed into anotherfunct.m, so we have to modify myfunct.m slightly:
```% updated version of myfunct.m
defaults.opt1 = 100;
defaults.opt2 = 'default';
[args unused] = propval(defaults,varargin);
anotherfunct(unused)
```
N.B. the latest release of Matlab has a new inputParser object that performs a similar purpose. It allows you to define all kinds of crazy validation functions, and designate arguments as optional and required, but I don't think it allows you to feed in your arguments as a struct rather than as property/value pairs.
### Debugging
#### Pause execution in the middle of a script/function
The 'keyboard' command is very useful. Just plop it in the midst of your m-file, and it will pause right there, allowing you to inspect the state of the variables in that function's workspace.
Better still, you can modify these variables, then 'dbcont' to continue operation where you left off.
• dbstack (to find out where you are)
• dbup and dbdown
• dbquit
#### Automatically pause execution upon error
Add 'dbstop if error' to your startup.m. Then, if there's any kind of error, it'll be as though you'd typed 'keyboard' just before the error happened.
• dbclear if error
• dbstop if warning
### Statistics and signal processing
#### Standard error of the mean
Matlab has a function for computing the standard deviation, but not the standard error - see ste.m.
#### Zscoring
You'd think that zscoring is as simple as (vec-mean(vec))/std(vec), but if you want to deal with matrices (doing the right thing with singleton dimensions), zscore along different dimensions etc. and avoid for loops, then you need to be a little more careful.
If you have the Statistics toolbox, it includes a wonderfully compact zscore.m - otherwise, zscore_mvpa.m is a reasonable standin.
#### Fast fourier transform
[Courtesy of Vaidehi Natu]
The Matlab Fourier functions are pretty fancy, but if you're not a signal processing engineer and you just want a sense of which frequencies are present in a timecourse, fft_simple.m might help get the hang of things.
#### Smoothing a vector with a median filter
A median filter is a basic but often useful way of smoothing your data. For each point, replace it with the median of itself, and the two points either side. Using the median rather than the mean is a good idea when your data contains outliers you'd like to remove, because it will simply replace them with one of their neighbours.
### Plotting and visualization
#### Positioning a figure
If you want to set the position and size of a figure on the screen, you need to know how big your screen is, so that you can set the figure's coordinates relative to it:
```curfig = figure();
screensize = get(0,'screensize');
width = screensize(3);
height = screensize(4);
horsize = 800;
vertsize = 600;
figpos = [screensize(3)-horsize screensize(4)-vertsize horsize vertsize];
set(curfig,'Position',figpos);
```
#### Figure that doesn't steal focus
[From Sean Polyn]
I got fed up with the tendency for calls to the "figure" command in matlab stealing focus from all other applications (such as when you are running a big loop making lots of figures).
So I googled: "matlab foreground figure" and the first hit pointed at sfigure.m (really just a single "set" command embedded in a little script) for SILENT FIGURES. So I'm happy.
#### Greenblack colormap
Bioinformatics people often use a colormap that ranges from green (high) to black (low). Greenblack.m creates just such a colormap (just like jet or bone, that you can use for plotting, e.g.
```imagesc(rand(10))
colormap(greenblack);
```
#### Titles that aren't in latex
Matlab's default is to try and format figure titles as though they're in latex, so underscores show up as superscript etc. Titlef.m is just a shortcut that sets the Interpreter to normal (and utilizes the same trick as dispf.m - see elsewhere on this page), e.g.
figure titlef('Now the under_scores don't show up as subscripts')
#### Printing a figure to a .png
Matlab's print.m writes a figure out to an image file, but it's a little cumbersome to use. Printfig.m is a thin wrapper that assumes that you want to write a .png and defaults to a more usable paper size and margin options. If you combine it with datetime.m, capturing figures as you go along (e.g. for a scientific journal) is made easy.
```figure
printfig(datetime)
```
### Misc
#### Print the date and time as a handy yymmdd_HHMM string
When writing log files, it's often useful to append the date and time to the end of the filename. Irritatingly, if you use standard US mmddyy (or even the UK's ddmmyy) format, the natural ascii ordering of your logfiles will be very unhelpful. Sorting by date created is sometimes the solution, but writing out the date in yymmdd avoids the problem altogether.
Datetime.m just warps a call to datestr, optionally appending seconds.
```
>> datetime
ans =
070921_0058
>> datetime(true)
ans =
070921_0058_40
```
P.S. This is also handy if you're a Brit abroad, who finds it really confusing to remember whether days or months are written first in your current domicile. This way, you just have to remember that it's in order of priority, which makes much more sense.
#### Automatically logging a diary of your sessions
I often want to be able to rewind to a previous matlab session, to remind myself of how I did something, or what commands I ran. It's pretty easy to get matlab to log your current session with the 'diary' command, but I would never remember to run that at the beginning of each session. By putting these lines into your startup.m, matlab will automatically create a diary file for every session you ever have, and organize them neatly by date in a directory of your choosing.
```pathfn = datetime();
pathfn=[ '~/diaries/diaries_' pathfn '.out'];
cmd=['diary ' pathfn];
eval(cmd);
diary on
```
Just change the '~/diaries/' to a directory of your choosing.
N.B. This requires the 'datetime.m' function - see elsewhere on this page.
#### Using indices from max, min and sort
If you want to know the index of the maximum value in one vector/matrix, and then apply that index to get a value from another vector/matrix, then you'd usually need to call max.m, get the index, and then use it. To do that in one line as God intended:
```% get the index of the max from v1, and then
% use that index to get a value from v2
v2(maxi(v1))
```
The most elegant way to do this I could think of was simply to reverse the order that max.m returns its arguments.
```function [idx vals] = maxi(mat)
```
See maxi.m, mini.m and sorti.m.
#### Check that two matrices are of identical size
Very simple but handy - compare_size.m.
```>> compare_size(rand(5), rand(6))
ans =
0
>> compare_size(rand(5), rand(5))
ans =
1
>> compare_size(rand([10 10 1]), rand([10 10]))
ans =
1
>> compare_size(rand([1 10 10]), rand([10 10]))
ans =
0
```
#### Count how many values match some boolean criterion
See count.m - just a shortcut for length(find(x)).
```>> vec = [10 20 30 40 50 60];
>> count(vec>35)
ans =
3
```
#### Combined disp and sprintf
I often want to be able to type things like:
```disp('2 + 2 = %i', 2+2)
```
But for some reason, disp.m won't allow sprintf-like arguments, so dispf.m rectifies this.
```>> dispf('2 + 2 = %i', 2+2)
2 + 2 = 4
```
#### Text progress bar
If you have a big for loop that takes a long time, it's reassuring and useful to get a progress update every n% of the way through. Progress.m makes that easy, and prints your progress in a friendly way, e.g.
```nSteps = 500;
for n=1:nSteps
progress(n,nSteps);
% do something
end % v
... 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% done
```
#### How can I tell if all the values in a matrix are integers?
Matlab's isinteger.m only tells you the type of the values in a matrix. This may not be what you want. For instance:
```>> a = [1 2 3 4 5]
a =
1 2 3 4 5
>> isinteger(a)
ans =
0
```
```>> a
a =
1 2 3 4 5
>> isint(a)
ans =
1
```
#### Vec2ind and ind2vec that don't ignore all-zeros columns
Matlab's vec2ind.m is useful if you have a matrix of zeros where each column has a single 1. You want the indices of those columns.
Unfortunately, it completely ignores columns that are all-zeros. Vec2ind_inclrest.m will include those in the output, marking them with zeros. You can see the difference here (note how they differ with regard to the 5th column of zeros):
```>> a = [1 0 0 0 0 1; 0 1 0 1 0 0; 0 0 1 0 0 0]
a =
1 0 0 0 0 1
0 1 0 1 0 0
0 0 1 0 0 0
>> b = vec2ind(a)
b =
1 2 3 2 1
>> b = vec2ind_inclrest(a)
b =
1 2 3 2 0 1
```
Likewise, ind2vec.m will burp if you try and reverse the process, so ind2vec_robust.m to the rescue.
```>> ind2vec(b)
??? Error using ==> sparse
Index into matrix must be positive.
>> ind2vec_robust(b)
ans =
1 0 0 0 0 1
0 1 0 1 0 0
0 0 1 0 0 0
``` | 3,599 | 13,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.878391 |
https://www.gurufocus.com/term/turnover/NYSE:RVT/Asset-Turnover/Royce%20Value%20Trust | 1,603,854,564,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107896048.53/warc/CC-MAIN-20201028014458-20201028044458-00293.warc.gz | 745,662,625 | 46,916 | Switch to:
# Royce Value Trust Asset Turnover
: -0.10 (As of Jun. 2020)
View and export this data going back to 1986. Start your Free Trial
Asset Turnover measures how quickly a company turns over its asset through sales. It is calculated as Revenue divided by Total Assets. Royce Value Trust's Revenue for the six months ended in Jun. 2020 was \$-156.09 Mil. Royce Value Trust's Total Assets for the quarter that ended in Jun. 2020 was \$1,607.03 Mil. Therefore, Royce Value Trust's Asset Turnover for the quarter that ended in Jun. 2020 was -0.10.
Asset Turnover is linked to ROE % through Du Pont Formula. Royce Value Trust's annualized ROE % for the quarter that ended in Jun. 2020 was -20.53%. It is also linked to ROA % through Du Pont Formula. Royce Value Trust's annualized ROA % for the quarter that ended in Jun. 2020 was -19.60%.
## Royce Value Trust Asset Turnover Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Royce Value Trust Annual Data Dec17 Dec18 Dec19 Asset Turnover 0.16 -0.15 0.25
Royce Value Trust Semi-Annual Data Dec17 Jun18 Dec18 Jun19 Dec19 Jun20 Asset Turnover 0.03 -0.27 0.17 0.08 -0.10
## Royce Value Trust Asset Turnover Calculation
Asset Turnover measures how quickly a company turns over its asset through sales.
Royce Value Trust's Asset Turnover for the fiscal year that ended in Dec. 2019 is calculated as
Asset Turnover = Revenue / Average Total Assets = Revenue (A: Dec. 2019 ) / ( (Total Assets (A: Dec. 2018 ) + Total Assets (A: Dec. 2019 )) / count ) = 387.064 / ( (1364.767 + 1700.062) / 2 ) = 387.064 / 1532.4145 = 0.25
Royce Value Trust's Asset Turnover for the quarter that ended in Jun. 2020 is calculated as
Asset Turnover = Revenue / Average Total Assets = Revenue (Q: Jun. 2020 ) / ( (Total Assets (Q: Dec. 2019 ) + Total Assets (Q: Jun. 2020 )) / count ) = -156.089 / ( (1700.062 + 1513.99) / 2 ) = -156.089 / 1607.026 = -0.10
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Companies with low profit margins tend to have high Asset Turnover, while those with high profit margins have low Asset Turnover. Companies in the retail industry tend to have a very high turnover ratio.
Royce Value Trust (NYSE:RVT) Asset Turnover Explanation
Asset Turnover is linked to ROE % through Du Pont Formula.
Royce Value Trust's annulized ROE % for the quarter that ended in Jun. 2020 is
ROE %** (Q: Jun. 2020 ) = Net Income / Total Stockholders Equity = -314.954 / 1534.0555 = (Net Income / Revenue) * (Revenue / Total Assets) * (Total Assets / Total Stockholders Equity) = (-314.954 / -312.178) * (-312.178 / 1607.026) * (1607.026/ 1534.0555) = Net Margin % * Asset Turnover * Equity Multiplier = 100.89 % * -0.1943 * 1.0476 = ROA % * Equity Multiplier = -19.60 % * 1.0476 = -20.53 %
Note: The Net Income data used here is two times the semi-annual (Jun. 2020) net income data. The Revenue data used here is two times the semi-annual (Jun. 2020) revenue data.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
** The ROE % used above is for Du Pont Analysis only. It is different from the defined ROE % page on our website, as here it uses Net Income instead of Net Income attributable to Common Stockholders in the calculation.
It is also linked to ROA % through Du Pont Formula:
Royce Value Trust's annulized ROA % for the quarter that ended in Jun. 2020 is
ROA % (Q: Jun. 2020 ) = Net Income / Total Assets = -314.954 / 1607.026 = (Net Income / Revenue) * (Revenue / Total Assets) = (-314.954 / -312.178) * (-312.178 / 1607.026) = Net Margin % * Asset Turnover = 100.89 % * -0.1943 = -19.60 %
Note: The Net Income data used here is two times the semi-annual (Jun. 2020) net income data. The Revenue data used here is two times the semi-annual (Jun. 2020) revenue data.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Be Aware
In the article Joining The Dark Side: Pirates, Spies and Short Sellers, James Montier reported that In their US sample covering the period 1968-2003, Cooper et al find that firms with low asset growth outperformed firms with high asset growth by an astounding 20% p.a. equally weighted. Even when controlling for market, size and style, low asset growth firms outperformed high asset growth firms by 13% p.a. Therefore a company with fast asset growth may underperform.
Therefore, it is a good sign if a company's Asset Turnover is consistent or even increases. If a company's asset grows faster than sales, its Asset Turnover will decline, which can be a warning sign. | 1,328 | 4,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | latest | en | 0.860056 |
https://www.bot-thoughts.com/2012/06/avc-my-steering-algorithm-sucks.html?showComment=1338907997359 | 1,631,952,029,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00597.warc.gz | 728,196,956 | 20,918 | Tuesday, June 5, 2012
AVC: My Steering Algorithm Sucks
Sunday testing went... um... sub-optimally.
The latest Data Bus update is that I'm hosed if I don't get the robot navigation nailed in the next day or two so I can work on barrel avoidance. The Sparkfun AVC is in less than two weeks.
Here I thought I had this clever little steering algorithm that worked oh so well. As it turns out, it sucks rocks. The robot keeps driving into curbs.
It even knows full well that its going off course, but it doesn't correct its course.
Strange? Yeah.
Well, here's what's going on...
I figured either the robot wasn't doing anything to change its course, or whatever it was doing wasn't having any effect, based on the position estimates coming out of the robot log files.
The purple path is the estimate; very close to reality.
It hit me on the way back from Boulder. Steering slop. And my crap steering algorithm.
Trying to be fancy the robot attempts to navigate to a goal point that is on a straight path between it and the target waypoint.
It doesn't account for cross track error. And for small heading errors, the robot commands steering angles that are too small for the physical resolution of the steering system.
Calculations
At more than 70 meters, the heading errors I was seeing in testing fall below 10 degrees. The computed steering angles are under 2.5 degrees. That's a tiny angle for a cheap 1/10th RC truck steering system.
But it also amounts to a big heading difference after 70 meters of travel.
Small heading errors mean very small steering angles
Experiment
To further verify my theory I ran experimental code on the robot to move the steering servo between the ranges corresponding to steering angles of -2.5 to +2.5 degrees. Play in the servo linkage and the servo saver was causing a dead band of about 3-5 degrees around the center point.
Solutions
I've implemented some mechanical changes to reduce steering play but I doubt it'll be enough. I may try to implement a couple more mechanical improvements, but I think a software solution is required.
Converting to a pure pursuit algorithm is likely best. The robot would compute a steering angle to intercept a point that moves along the desired path at a fixed distance L from the robot. In doing so it smooths out transitions between waypoints and it inherently corrects for cross track error.
Simulation
I modified my Processing playback program to become a rudimentary simulation program. I wrote code that implements my original steering algorithm and, with help from Jesse, the pure pursuit algorithm.
Then I modified the simulation code to simulate steering slop and misalignment that causes the robot to pull to one side instead of driving a perfectly straight line.
Lo and behold. With my crummy algorithm controlling a robot with misalignment and steering slop, the simulated robot behaves exactly as the real one, unable to properly track a straight course to the target waypoint, veering off, and correcting nearly at the last moment.
By contrast, the pure pursuit algorithm had no problem staying on track with the same steering slop and misalignment, though some oscillation in heading resulted.
Another interesting discovery is that the pure pursuit algorithm has no problem correcting initial heading errors, for example, where the robot should be pointed at 90° but is pointed at 95° instead. My dumb algorithm can't do this, explaining some anomalies from a few weeks ago.
I was feeling pretty down on Sunday after the crashes, but with the work above, some confidence is returning. I hope I can get the navigation nailed down and the speed up where it needs to be. There's still the matter of obstacle avoidance... | 772 | 3,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-39 | longest | en | 0.949456 |
https://domymatlab.com/matlab-optimal-assignment/ | 1,695,974,668,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510498.88/warc/CC-MAIN-20230929054611-20230929084611-00199.warc.gz | 244,554,046 | 13,442 | # Matlab Optimal Assignment
Matlab Optimal Assignment Solution {#sec:optimal1} ================================ One of the most difficult problems in nonlinear analysis is to evaluate the conditional log-likelihood representation of a set of uncertain or uncertain samples. In general, if the independent sample log-likelihood is unknown, the independent sample is estimated via least squares regression, which differs from the mixture model for the log maximum likelihood. Define two subsets $X$ and $Y$ as follows. 1. For $1 \leq i \leq n$ a mixture $M$ is defined as \begin{aligned} M = \frac{1}{n} \sum_{i=1}^n (Z_i-\lambda_i) \textrm{ \ \ for all } \lambda_i = Z_i, \forall Z_1,\ldots, Z_n \in \mathbb{R}^A.\end{aligned} For example, if $Z=Z_1,\ldots,Z_n$, then $M = \sum_{i=1}^{n} (Z_i-\lambda_i) .$ 2.
## Matlab Coding Homework Help
For $1 \leq i \leq n$ a group $G_i$ is defined as \begin{aligned} G_i =\{(x_1,\ldots,x_n)\in \mathbb{R}^n: \sum_{j=1}^n x_{j}=1,\forall x_i \in \lambda_i\}, \quad 1 \leq i \leq n.\end{aligned} 3. For $1 \leq i \leq n$ a sample set $S_i$ is defined as \begin{aligned} S_i =\{(x_1,\ldots,x_n)\in \mathbb{R}^n: \sum_{j=1}^n x_{j}=1, \forall x_i \in \lambda_i\}.\end{aligned} For example, if $X=\prod_i X_{i,1}, S_1=\lambda_1, \ldots, S_n=\lambda_n$ and $X_1,\ldots, X_n <\cdots >X_{n-1}, S_n=\lambda_n$, then a group $G_n=\bigcap_i S_i$ is defined as \begin{aligned} G_n =\{(x_1,\ldots,x_{n-1})\in \mathbb{R}^n: \sum_{j=1}^n x_{j}=1, \forall x_i \in \lambda_i\}.\end{aligned} To analyze the conditional log-likelihood representation for the variables $X$ and $Y$, we need to identify relevant points, which are special cases of some special cases of an independent sample. Let $Y$ be an independent sample for a group $G$, i.e.
## Find Someone to do Matlab Project
, $Y=m\{m=1,\ldots,n\}$. Then we can estimate $Y$ from the observed variables $Z_i$ to solve the corresponding column-wise conditional log-likelihood problem for the group. Consider, for example, a group $G=\{(1,2,3)\}$ and $S=\{s=1,\ldots,w\}$. Then we have \begin{aligned} \label{eq:condloglikelihood} Y = & \sum_{i=1}^n \lambda_i ^2 \sum_{j=1}^w z_j \log_2 Y_i +\sum_{j=1}^w ( \lambda_j + (s-1) z_j)w \\ Z= & redirected here Optimal Assignment for Complex Networks — A systematic approach =============================================================== Recent developments in the area of machine learning, and at the interface between computer-aided business decision-making and Artificial Intelligence (AI), have brought to light a line of research toward more robust methods for machine learning. Initially it was initially developed by the renowned IBM academic group [1]; now it is applicable to AI as well. In particular, the present chapter describes the problems in machine learning under a variety of different approaches, and will provide an explanation of machine learning in relation to two types of machines: continuous-time (ST) and discrete-time (DT) networks. Structure of STs —————- There are two fundamental types of ST, namely dynamic-structure networks (DST) and sequential-structure networks (SST).
## Matlab Homework Examples
The main difference between MS, AT, or BN networks, and STs is that these networks have a very large number of nodes and are not assumed to form a connected graph. Therefore, the problem is most concretely defined for continuous-time networks: the task is to my response a complete graph, and a set of nodes can be connected to their neighboring nodes so that for every node, the average number of rows and columns in the graph become sufficient. However, the present method has two major drawbacks: 1. It is not suitable for large collections of large numbers of nodes and therefore cannot be performed consistently. Therefore it is expensive and difficult to analyze and optimize. 2. It is not practical and we cannot analyze a large number of nodes after conducting the analysis and optimization.
## Matlab Assignment Tutor
Since the number of nodes is reduced via the procedure of the previous section, this leads to considerable issues for optimization. This issues is also more relevant for the analysis of STs; in particular, these networks have more nodes than the single-node MS network. A common issue with these two types of networks is related to the construction of a highly dense representation of most nodes inside their graphs, which is ultimately the main reason for the performance of ST networks. For these reasons, it is necessary to develop an extended representation of the nodes, along with the representations of their directed edges (with the nodes/edges being the only ones), which can be characterized by various characteristics. Notice that this representation is still an early development in the area of graph learning: it has been shown that even for large graphs considered only few edges are drawn from a distance in the graph even if the distance is often far more. If a series of networks of our model are designed such that each network runs in a directed *columnular* graph and many adjacent nodes also run in an directed *columnular* network, corresponding to a certain length, then for a larger network dimensionality then the same chain, corresponding to the number of connected components, will not necessarily be as large as for a larger network. However, Figure 12 illustrates two characteristics of the sequence depicted, which are useful for studying the parallel structure of networks.
## Fiverr Matlab Homework
### **1. Coded Databases** One set of codes are stored in a database as bits, so that the distance between a source node and a target node may not become infinite in each network iteration. On the other hand, if the network is connected, the distance will be increased for a random degree whenever the length of the code increases. On the other hand, the length of the code is the distance between a target node and an important node of that network. The problem of ensuring a length for each code that is important for the parallel representation of each function, is still with a priori description and therefore it will need more detailed theoretical understanding. During training, some participants are usually given a set of 10 or 15 codes, such that each code corresponds to a non-null hypothesis. In order to explain the model under this framework, let us consider a simulation of the original data set (see Figure 3(a) in Appendix A).
## Matlab Assignment Help
In this case, the real trainable probability distribution function is that of the random function defined by $\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} Matlab Optimal Assignment An implementation of the Optimal Assignment algorithm for BCL-based algorithms can produce a consistent approximation of the numerically-weighted binary search problem in a matrix sense. A key ingredient is the construction of the exact solution domain for such a problem from the domain of interest. The idea is to use the solution domain of the problems—usually defined to be the solution domain of the problem—to obtain an explicit exact solution. The exact solution domain of the problem is constrained by an initial bounding box description size one and a set of constraints each of which are written down. In order that the only unknowns are those of the size of one is a solution to the problem, a constant which holds for all$i = 1…
## Matlab Homework Help Free
n\$ and only one of the constraints. The following sequence of constant bounds also satisfies the domain of infinitesimal norm for the problem, so that its solution contains no gaps. This is the meaning of the “zero-distance” region. Consider exactly one problem in the problem domain of interest and the domain with bounded boxes. Then the solution for its domain is exactly the solution corresponding to a sufficiently large box inside a two-dimensional box. Return the solution for a problem in the domain of interest, we are left with two bounding boxes each of which has at most one bounding point. The following example shows that the bounding box needs not be at a single point, nor do its boxes.
## Matlab List Homework
Therefore by the Taylor expansion of the last linear term of the sum of the squares of the last bounding coordinates we can obtain the solution where each of the following two conditionals hold: the first of them means The new solution would be not quite zero. Appropriately, the problem does not have a solution for each subproblem of the problem domains. Therefore we have to compute a feasible multi-objective problem for the problem. In order to achieve this, we must “multiply” the solution from the problem with the one obtained in this example. These are the number of allowed multiple solution structures of problems defined only by problems defined to have a given number of problem domain constraints, and only one and only one solution to the more restrictive constraints of the existing domain; the other requirements are that the solution for the problem is explicitly specified to be a solution for all the problem domain constrained subproblems. Its solution is simply the intersection of these constraints. Here is another example of a problem that has a solution inside the two-dimensional box for one problem in the problem domain.
## Matlab Assignment Help Near Me
This is the problem in which the number of required box constraints is a function of the number of problem domains in the domain and of the number of these domains are each x_2, and the size of the box is half the total box. A consequence of this is that for solving this problem we have to choose an appropriate global solution domain, which is defined by where we explicitly assume that both the constraint and the constraint solution are in the problem domain. Finally we must compute global solutions for the problem in which the solution is uniquely specified for each domain and where the solution (or “solution boundary space”) can be written as a linear combination of two different domains inside the domain: either the global cuboid (say, C cuboid of diameter 2) or the cuboid of width 2 A | 2,307 | 10,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.697345 |
https://m.ebrary.net/30623/engineering/phase_surface_definition_import | 1,596,857,891,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737238.53/warc/CC-MAIN-20200808021257-20200808051257-00222.warc.gz | 381,946,477 | 4,912 | Home Engineering Tactile Display for Virtual 3D Shape Rendering
# Phase 1: Surface Definition and Import
This is the initial stage of the process and it allows the user to communicate the surface data to the software, as shown in Fig. 6.9.
In order to obtain a flexible and customizable process, the software is able to receive these data in two different ways: symbolic expressions or data files. By means of symbolic formulation, the user can provide the software with the mathematical function of the surface, such as implicit equation or parametric equation. On the other hand, by exporting the designed surface from a 3D CAD software or Surface Modeller, by means of a format supported by Matlab, it is possible to choose the data file import option. Thanks to this feature, the user is able to import in Matlab the file that contains the surface information. These data are used by Matlab to analyse and store in the memory the surface. The analysis could be performed in a symbolic way or in a discrete way. Although the first one requires more memory resources rather, it allows us to perform an analysis on a continuous surface formulation, which is very accurate. On the other hand, the discrete definition simplifies the surface in a mash of points that reduces the stored data, thus allowing fast analysis. The drawback is that it introduces an approximation on the surface data. The user can define the mash coarseness in order to set the resolution of the surface definition to his/her needs. The definition of the surface by mathematical function allows us to perform the analysis by means of the symbolic procedure or the discrete approach. Figure 6.10 shows the same surface generated by discrete approach with low accuracy mesh (a), with high accuracy mesh (b) and with the symbolic approach (c). As regards the surfaces imported by data file, these can be defined in Matlab only by means of the discrete approach.
When the data of the surface are stored in Matlab, it is possible to choose a cutting plane, which allow us to obtain the trajectory that the strip has to represent, with the approximation illustrated in Fig. 4.1, as shown in Fig. 6.11. If the user has selected the symbolic approach, the trajectory obtained is expressed by means of a mathematical equation, that means with a continued formulation. On the contrary, by choosing the discrete approach, also the obtained trajectory is represented by means of a discrete set of points. The information obtained from Phase 1 are used as input for Phase 2.
Fig. 6.9 Phase 1 - Surface definition/import
Fig. 6.10 Surface definition
Fig. 6.11 Cutting plane and resulting trajectory
Related topics | 547 | 2,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-34 | latest | en | 0.918099 |
https://www.stmintz.com/ccc/index.php?id=129999 | 1,590,951,291,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413624.48/warc/CC-MAIN-20200531182830-20200531212830-00248.warc.gz | 905,186,120 | 2,441 | # Computer Chess Club Archives
## Messages
### Subject: Re: is this nullmove? problems in pulsar algorithm
Author: Robert Hyatt
Date: 08:52:29 09/20/00
Go up one level in this thread
```On September 20, 2000 at 06:36:13, Mike Adams wrote:
> I'm having some trouble realizine the gain ive heard i should from nullmove
>in Pulsar on icc. So i wanted to show you the alorithm that i'm using. its not
>exactly the code my own code is unique to my program and might confuse but its
>essentially the code that i use.
>
>variables:
>endgames: is it endgame 1 or 0
>null: has there been a previous null call
>side: side to move counts up from 1. odd for pulsar even for opponent
>depth: counts down always 0 on first call of qsearch
>beta: could be 10,000 or -10,000 if evaluate hasn't been called
>
>if(endgames==0 && null==1 && side>1 && beta!=10000 && beta!=-10000)
>{
> if(depth-2<=0)
> value=-qsearch(-beta, -alpha, 0, (side+1), myenpassant);
> else
>{ /* you can ignore the hash it just changes side to move and
> I only use hashing for move ordering anyway so shouldnt have big impact
>*/
> passhash=computecurrenthash(currenthash, 0, 0,0, 1);
> value=-search(-beta, -alpha, depth-2, (side+1), myenpassant, passhash);
>}
> if(value>=beta)
> return beta;
>
>}
You appear to be using R=1 for your null-move search. IE where you call
search/quiesce with depth-1 normally, you are using depth-2. Most of us are
using R=2 which means depth-3 rather than depth-2. Some of us even use a
dynamic R that varies from 2 to 3, which means from depth-3 to depth-4
depending. That will make a big difference.
``` | 502 | 1,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.835131 |
https://learningwithmrskirk.com/tag/addition/ | 1,685,349,784,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00595.warc.gz | 403,742,914 | 24,894 | # New Year’s Math Fun!
Looking for something fun to do when you return from the Holiday Break? This New Year’s Math Scavenger Hunt is a great way to practice engage students while practicing math skills!
Math Scavenger Hunts are perfect for guided math, whole class, centers or small groups! They are a great way to have students moving while working with a partner on Math.
And they are so easy for the teacher to prep! Simply print the problems, hang them around the room or another area. Print record sheets for your students and you are all set!
# Very Merry Classroom Ideas for December
### Well, Thanksgiving break was a great time for relaxing and visiting with family and friends. But … Now we have to go back to school? What on Earth will we do with these excited children while they count the days until their holiday break? Here are some ideas to help you make the days/weeks leading up to the holiday break fun and educational.
Joy, Sunshine, and Lollipops has excellent ideas on her blog for celebrating Holidays Around the World!
How about a Grinch Day? Falling Into First has great ideas on her blog for celebrating Grinch Day!
Get your students moving around the room with this Math Scavenger Hunt! Your students will love solving word problems with partners.
Looking for great Read A Louds for Christmas? The Printable Princess has a collection for you.
How cute is this writing activity “If Santa Was Stuck in My Chimney…”? Check out other related ideas on the blog, First Grade Wow
These Holiday Math Games are perfect for elementary classrooms!
This Gumdrop Math Challenge is a great way to get your students using critical thinking skills while having fun!
More Math??? Here are some great ideas!
If you are having a class party, check out these 15 Class Games from A Girl and A Glue Gun.
Want more great ideas? Check out my Pinterest Holiday Board:
# Candy Shop Math Centers
I am getting excited about the holiday season! I have made some new games/activities to play with my 3rd and 4th graders using some really fun clip art! Seriously, the elves are so cute!
The four newest games/activities I have created are:
2. Subtraction Matching
3. Addition and Subtraction Balancing Equations
4. Addition and Subtraction Word Problems
# Back To School Math
### Guess what?!? It’s time to get ready to go back to school! Here in my little world we have less than 2 weeks left until the kids are back in school!
If you are a parent looking for something to use to help our kids brush up on Math Skills before the school year stats off, I have something perfect for you! If you are a teacher looking for classroom work to help get your students back into gear, I also have something for you! Continue reading
# Using Place Value to Understand Addition
Adding two digit numbers can seem like an easy enough task to those of us who have been doing it for a long time! But for elementary students it can sometimes be confusing, especially if there is regrouping involved. | 652 | 3,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | latest | en | 0.912991 |
https://hub.jmonkeyengine.org/t/simpleupdate-is-too-slow-or-what-am-i-doing-wrong/31105 | 1,679,585,570,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945168.36/warc/CC-MAIN-20230323132026-20230323162026-00702.warc.gz | 376,532,552 | 7,493 | # simpleUpdate is too slow? Or what am I doing wrong?
I have a problem and I don’t know exactly what I’m doing wrong, so I’m unsure where to begin to fix it. In simpleUpdate I am placing 8 boxes on each of the 8 corners of another box, and every tick I adjust those boxes so that when I move the first box, the 8 boxes update their position accordingly. The problem is that the 8 boxes seem to lag behind the bigger box when I move it. I have made a video to explain this: - YouTube
I know this can be easily accomplished by attaching the 8 boxes to the bigger box via nodes but I still want to know why I’m having an issue doing it this way.
Here is the code that shows how I am getting the positions of the 8 corners each tick:
``````
corner1 = geomPos.add(geomRot.mult(new Vector3f(boundingBox.getXExtent(),boundingBox.getYExtent(),boundingBox.getZExtent())));
corner2 = geomPos.add(geomRot.mult(new Vector3f(boundingBox.getXExtent(),-boundingBox.getYExtent(),boundingBox.getZExtent())));
corner3 = geomPos.add(geomRot.mult(new Vector3f(-boundingBox.getXExtent(),boundingBox.getYExtent(),boundingBox.getZExtent())));
corner4 = geomPos.add(geomRot.mult(new Vector3f(-boundingBox.getXExtent(),-boundingBox.getYExtent(),boundingBox.getZExtent())));
corner5 = geomPos.add(geomRot.mult(new Vector3f(boundingBox.getXExtent(),boundingBox.getYExtent(),-boundingBox.getZExtent())));
corner6 = geomPos.add(geomRot.mult(new Vector3f(boundingBox.getXExtent(),-boundingBox.getYExtent(),-boundingBox.getZExtent())));
corner7 = geomPos.add(geomRot.mult(new Vector3f(-boundingBox.getXExtent(),boundingBox.getYExtent(),-boundingBox.getZExtent())));
corner8 = geomPos.add(geomRot.mult(new Vector3f(-boundingBox.getXExtent(),-boundingBox.getYExtent(),-boundingBox.getZExtent())));
``````
Where geomPos is the WorldTranslation of the bigger box, geomRot is the WorldRotation of the bigger box, and boundingBox is the bigger box’s ModelBounds.
Any ideas?
Post the complete code please.
I hope this is enough. Here is the simpleUpdate method used to update the 8 boxes’ locations, or to show the blue highlight box if nothing currently selected:
``````
public void simpleUpdate(float lastTimePerFrame)
{
if(this.selectedEntity != null)
{
selectedEntity.updateGeometricState();
selectedEntity.updateModelBound();
BoundingBox boundingBox = (BoundingBox) selectedEntity //Gets the BoundingBox of the bigger box
.getComponentByType(GeometryComponent.class)
.getGeometry().getModelBound();
Vector3f camTranslation = getParentEntity().getComponentByType( //Gets the players cameras World Translation.
FirstPersonCameraComponent.class)
.getWorldCameraTranslation();
Vector3f corner1, corner2, corner3, corner4, corner5, corner6, corner7, corner8;
Quaternion geomRot = selectedEntity.getComponentByType(GeometryComponent.class).getGeometry().getWorldRotation(); //Gets the bigger boxs World Rotation
Vector3f geomPos = selectedEntity.getComponentByType(GeometryComponent.class).getGeometry().getWorldTranslation(); //Gets the bigger boxs World Translation
corner1 = geomPos.add(geomRot.mult(new Vector3f(boundingBox.getXExtent(),boundingBox.getYExtent(),boundingBox.getZExtent())));
corner2 = geomPos.add(geomRot.mult(new Vector3f(boundingBox.getXExtent(),-boundingBox.getYExtent(),boundingBox.getZExtent())));
corner3 = geomPos.add(geomRot.mult(new Vector3f(-boundingBox.getXExtent(),boundingBox.getYExtent(),boundingBox.getZExtent())));
corner4 = geomPos.add(geomRot.mult(new Vector3f(-boundingBox.getXExtent(),-boundingBox.getYExtent(),boundingBox.getZExtent())));
corner5 = geomPos.add(geomRot.mult(new Vector3f(boundingBox.getXExtent(),boundingBox.getYExtent(),-boundingBox.getZExtent())));
corner6 = geomPos.add(geomRot.mult(new Vector3f(boundingBox.getXExtent(),-boundingBox.getYExtent(),-boundingBox.getZExtent())));
corner7 = geomPos.add(geomRot.mult(new Vector3f(-boundingBox.getXExtent(),boundingBox.getYExtent(),-boundingBox.getZExtent())));
corner8 = geomPos.add(geomRot.mult(new Vector3f(-boundingBox.getXExtent(),-boundingBox.getYExtent(),-boundingBox.getZExtent())));
if(cube1Geo != null)
{
cube1Geo.removeFromParent();
cube2Geo.removeFromParent();
cube3Geo.removeFromParent();
cube4Geo.removeFromParent();
cube5Geo.removeFromParent();
cube6Geo.removeFromParent();
cube7Geo.removeFromParent();
cube8Geo.removeFromParent();
}
cube1Geo = IrisGame.instance.putShape(new Box(0.2f, 0.2f, 0.2f), ColorRGBA.Red);
cube2Geo = IrisGame.instance.putShape(new Box(0.2f, 0.2f, 0.2f), ColorRGBA.Orange);
cube3Geo = IrisGame.instance.putShape(new Box(0.2f, 0.2f, 0.2f), ColorRGBA.Yellow);
cube4Geo = IrisGame.instance.putShape(new Box(0.2f, 0.2f, 0.2f), ColorRGBA.Green);
cube5Geo = IrisGame.instance.putShape(new Box(0.2f, 0.2f, 0.2f), ColorRGBA.Cyan);
cube6Geo = IrisGame.instance.putShape(new Box(0.2f, 0.2f, 0.2f), ColorRGBA.Blue);
cube7Geo = IrisGame.instance.putShape(new Box(0.2f, 0.2f, 0.2f), ColorRGBA.Magenta);
cube8Geo = IrisGame.instance.putShape(new Box(0.2f, 0.2f, 0.2f), ColorRGBA.Pink);
cube1Geo.setLocalTranslation(corner1);
cube2Geo.setLocalTranslation(corner2);
cube3Geo.setLocalTranslation(corner3);
cube4Geo.setLocalTranslation(corner4);
cube5Geo.setLocalTranslation(corner5);
cube6Geo.setLocalTranslation(corner6);
cube7Geo.setLocalTranslation(corner7);
cube8Geo.setLocalTranslation(corner8);
}
else
{ //This section just creates the highlight around the box which the crosshairs are hovered over.
CollisionResults results = getCollisionResults();
if(results.size() > 0)
{
CollisionResult closest = results.getClosestCollision();
Geometry closestGeom = closest.getGeometry();
if(tempHighlightGeom != null)
{
tempHighlightGeom.removeFromParent();
tempHighlightGeom = null;
}
tempHighlightGeom = closestGeom.clone();
tempHighlightGeom.scale(1.05f);
Material wireMaterial = new Material(
IrisGame.instance.getAssetManager(),
"Common/MatDefs/Misc/Unshaded.j3md");
wireMaterial.setColor("Color", new ColorRGBA(0f, 0f,
1f, 0.2f));
wireMaterial.getAdditionalRenderState().setBlendMode(
BlendMode.Alpha);
tempHighlightGeom.setMaterial(wireMaterial);
tempHighlightGeom.setQueueBucket(Bucket.Translucent);
IrisGame.instance.getRootNode().attachChild(
tempHighlightGeom);
}
else
{
if(tempHighlightGeom != null)
{
tempHighlightGeom.removeFromParent();
tempHighlightGeom = null;
}
}
}
}
``````
If you need any other code let me know. The bigger box is attached to the player’s cameraNode when they click on it, so that’s how the big box follows the player as they turn.
OK, the problem is that you are using previous frame’s data for the current frame, that’s what causing the “lag”.
``````
selectedEntity.updateGeometricState();
selectedEntity.updateModelBound();
``````
This is not doing what you think it does.
First, calling updateModelBound here is useless because it’s called in updateGeometricState.
Then calling selectedEntity.updateGeometricState, won’t update its world transforms, because world transforms are computed from the parent node’s world transforms, and updateGeometricState was not called on the parent.
As a rule of thumb, calling updateGeometricState on a sub part of the scene graph is wrong in 99% of the cases. You should always call it on the root of the scene (most probably the rootNode here).
BUT calling updateGeometricState yourself on the rootNode is wrong 99% of the times (yeah doesn’t let you a lot of space to do right ). This is done for you in the SimpleApplication.
In your case, calling rootNode.updateGeometricState() would work. But that’s wrong. Because it will still be called afterward, so you’ll do the work twice.
So yeah as you said, an obvious solution is to use a node and attach those boxes to it and let the engine do the work.
But if you really want to do it differently, you’d better use the simpleRender method and render your boxes on the spot instead of adding them to the scene graph.
At this moment the world transforms of the big box will be correctly updated with this frame data.
That’s how we usually render debug views in JME and what you’re trying to do looks a lot like this.
To render a stand alone (not attached to the scene-graph) geometry in the simpleRender you have to set its local transforms, then call updateGeometricState() on it, then call renderManager.renderGeometry(geom)
And if you want to be completely in the “right”, you should do this in an appState render() method.
That was extremely helpful. I understand what is going on now and where I am going wrong. Thank you so much!! | 2,058 | 8,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-14 | latest | en | 0.706034 |
https://www.transtutors.com/questions/1-the-velocity-in-a-one-dimensional-compressible-flow-is-given-by-u-10x-2-find-the-m-2078490.htm | 1,643,148,361,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00462.warc.gz | 1,073,817,690 | 12,626 | # 1. The velocity in a one-dimensional compressible flow is given by u = 10x 2 . Find the most...
1. The velocity in a one-dimensional compressible flow is given by u = 10x2 . Find the most general variation of the density with x.
2. The x-, y- and z-components of a velocity field are given by u = ax + by + cz, v = dx + ey + fz, and w = gx + hy + jz. Find the relationship among the coefficients a through j if the flow field is incompressible. | 122 | 448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-05 | latest | en | 0.898619 |
https://blog.plover.com/math/ounce-of-theory-2.html | 1,726,011,151,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00875.warc.gz | 121,867,811 | 6,828 | # The Universe of Disco
Tue, 28 Jul 2015
A few months ago I wrote an article here called an ounce of theory is worth a pound of search and I have a nice followup.
When I went looking for that article I couldn't find it, because I thought it was about how an ounce of search is worth a pound of theory, and that I was writing a counterexample. I am quite surprised to discover that that I have several times discussed how a little theory can replace a lot of searching, and not vice versa, but perhaps that is because the search is my default.
Anyway, the question came up on math StackExchange today:
John has 77 boxes each having dimensions 3×3×1. Is it possible for John to build one big box with dimensions 7×9×11?
OP opined no, but had no argument. The first answer that appeared was somewhat elaborate and outlined a computer search strategy which claimed to reduce the search space to only 14,553 items. (I think the analysis is wrong, but I agree that the search space is not too large.)
I almost wrote the search program. I have a program around that is something like what would be needed, although it is optimized to deal with a few oddly-shaped tiles instead of many similar tiles, and would need some work. Fortunately, I paused to think a little before diving in to the programming.
BuyHow to Solve It from Bookshop.org(with kickback)(without kickback)
For there is an easy answer. Suppose John solved the problem. Look at just one of the 7×11 faces of the big box. It is a 7×11 rectangle that is completely filled by 1×3 and 3×3 rectangles. But 7×11 is not a multiple of 3. So there can be no solution.
Now how did I think of this? It was a very geometric line of reasoning. I imagined a 7×11×9 carton and imagined putting the small boxes into the carton. There can be no leftover space; every one of the 693 cells must be filled. So in particular, we must fill up the bottom 7×11 layer. I started considering how to pack the bottommost 7×11×1 slice with just the bottom parts of the small boxes and quickly realized it couldn't be done; there is always an empty cell left over somewhere, usually in the corner. The argument about considering just one face of the large box came later; I decided it was clearer than what I actually came up with.
I think this is a nice example of the Pólya strategy “solve a simpler problem” from How to Solve It, but I was not thinking of that specifically when I came up with the solution.
For a more interesting problem of the same sort, suppose you have six 2×2x1 slabs and three extra 1×1×1 cubes. Can you pack the nine pieces into a 3×3x3 box? | 618 | 2,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.976506 |
https://www.slideserve.com/vivienne/supply-chain-management-issues-and-models-inventory-management-stochastic-model | 1,719,038,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00490.warc.gz | 884,564,899 | 24,484 | 1 / 57
# Supply Chain Management: Issues and Models Inventory Management (stochastic model)
Supply Chain Management: Issues and Models Inventory Management (stochastic model). Prof. Dr. Jinxing Xie Department of Mathematical Sciences Tsinghua University, Beijing 100084, China http://faculty.math.tsinghua.edu.cn/~jxie Email: jxie@ math.tsinghua.edu.cn
## Supply Chain Management: Issues and Models Inventory Management (stochastic model)
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### Presentation Transcript
1. Supply Chain Management:Issues and Models Inventory Management(stochastic model) Prof. Dr. Jinxing Xie Department of Mathematical Sciences Tsinghua University, Beijing 100084, China http://faculty.math.tsinghua.edu.cn/~jxie Email: jxie@ math.tsinghua.edu.cn Voice: (86-10)62787812 Fax: (86-10)62785847 Office: Rm. 1308, New Science Building
2. Inventory Control with Uncertain Demand The demand can be decomposed into two parts, where = Deterministic component of demand and = Random component of demand.
3. Inventory Control with Uncertain Demand • There are a number of circumstances under which it would be appropriate to treat as being deterministic even though is not zero. Some of these are: • When the variance of the random component, is small relative to the magnitude of . • When the predictable variation is more important than the random variation. • When the problem structure is too complex to include an explicit representation of randomness in the model.
4. Inventory Control with Uncertain Demand However, for many items, the random component of the demand is too significant to ignore. As long as the expected demand per unit time is relatively constant and the problem structure not too complex, explicit treatment of demand uncertainty is desirable.
5. Inventory Control with Uncertain Demand Example: A newsstand purchases a number of copies of The Computer Journal (weekly). The observed demands during each of the last 52 weeks were:
6. Inventory Control with Uncertain Demand Example : Estimate the probability that the number of copies of the Journal sold in any week. The probability that demand is 10 is estimated to be 2/52 = 0.0385, and the probability that the demand is 15 is 5/52 = 0.0962. Cumulative probabilities can also be estimated in a similar way. The probability that there are nine or fewer copies of the Journal sold in any week is (1 + 0 + 0 + 0 + 3 + 1 + 2 + 2 + 4 + 6) / 52 = 19 / 52 = 0.3654.
7. Inventory Control with Uncertain Demand We generally approximate the demand history using a continuous distribution. By far, the most popular distribution for inventory applications is the normal. A normal distribution is determined by two parameters: the mean and the variance
8. Inventory Control with Uncertain Demand These can be estimated from a history of demand by the sample mean and the sample variance .
9. Inventory Control with Uncertain Demand The normal density function is given by the formula We substitute as the estimator for and as the estimator for .
10. Inventory Control with Uncertain Demand
11. Optimization Criterion In general, optimization in production problems means finding a control rule that achieves minimum cost. However, when demand is random, the cost incurred is itself random, and it is no longer obvious what the optimization criterion should be. Virtually almost all of the stochastic optimization techniques applied to inventory control assume that the goal is to minimize expected costs.
12. The Newsboy (girl) / newsvendor Model (Continuous Demands) The demand is approximately normally distributed with mean 11.731 and standard deviation 4.74. Each copy is purchased for 25 cents and sold for 75 cents, and he is paid 10 cents for each unsold copy by his supplier. One obvious solution is approximately 12 copies. Suppose Mac purchases a copy that he doesn't sell. His out-of-pocket expense is 25 cents 10 cents = 15 cents. Suppose on the other hand, he is unable to meet the demand of a customer. In that case, he loses 75 cents 25 cents = 50 cents profit.
13. The Newsboy Model (Continuous Demands) Notation: = Cost per unit of positive inventory remaining at the end of the period (known as the overage cost). = Cost per unit of unsatisfied demand. This can be thought of as a cost per unit of negative ending inventory (known as the underage cost). The demand is a continuous nonnegative random variable with density function and cumulative distribution function . The decision variable is the number of units to be purchased at the beginning of the period.
14. The Newsboy Model (Continuous Demands) The cost function G(Q) is convex The optimal solution equation
15. The Newsboy Model (Continuous Demands) Determining the optimal policy for convex function
16. The Newsboy Model (Continuous Demands) Example (continued): Normally distributed with mean = 11.73 and standard deviation = 4.74. = 25 10 = 15 cents. = 75 25 = 50 cents. The critical ratio is = 0.50/0.65 = 0.77. Purchase enough copies to satisfy all of the weekly demand with probability 0.77. The optimal Q* is the 77th percentile of the demand distribution.
17. The Newsboy Model(Continuous Demands) Example (continued):
18. The Newsboy Model (Continuous Demands) Example (continued): Using the data of the normal distribution we obtain a standardized value of = 0.74. The optimal Q is Hence, he should purchase 15 copies every week.
19. The Newsboy Model (Discrete Demands) Optimal policy for discrete demand: The procedure for finding the optimal solution to the newsboy problem when the demand is assumed to be discrete is a natural generalization of the continuous case. The optimal solution procedure is to locate the critical ratio between two values of F(Q) and choose the Q corresponding to the higher value. That is
20. The Newsboy Model (Discrete Demands) Example : The critical ratio for this problem was 0.77, which corresponds to a value of between = 14 and = 15. Since we round up, the optimal solution is = 15. Notice that this is exactly the same order quantity obtained using the normal approximation.
21. The Newsboy Model:Extension to Include Starting Inventory Critical number policy: The optimal policy when there is a starting inventory of x is: Order Q*-x if x<Q* . Don't order otherwise . • Note that Q* should be interpreted as the order-up-to point rather than the order quantity. • It is also known as a target or base stock level.
22. Single-Period Single-Location Model: When there is positive set-up cost Starting Inventory: x Ordering quantity z=(y-x)+ (Order-up-to-level: y) Purchasing cost: Total expected cost: here
23. Single-Period Single-Location Model: When there is positive set-up cost (s,S) policy convex G(y)=cy+L(y)
24. Single-Period Single-Location Model: When there is positive set-up cost Optimal Policy: (s, S) Policy: The optimal policy when there is a starting inventory of x is: Order up to S if x < s . Don't order if x>=s .
25. Multi-Period Single-Location Model: When there is no positive set-up cost dynamic programming m(y,t)=y-t for backorder case; =(y-t)+ for lost sales case : discounter factor Optimal Policy: Order-up-to policy still holds, although the order-up-to level changes from period to period: S1 >= S2 >= … >= Sn
26. Multi-Period Single-Location Model: When there is positive set-up cost Optimal Policy: using dynamic programming and the K-convexity of the objective functions, one can prove the (sn, Sn) Policy still holds: For each period n, the optimal policy when there is a starting inventory of x (maybe <0, backlog) is: Order up to Sn if x < sn . Don't order if x>=sn . But different periods may have different (s,S) values. And it’s difficult to compute the values for (sn, Sn)
27. Infinite Period Single-Location Model: When there is positive set-up cost Assume: • Demand distribution is i.i.d for all periods • Using cost function with discount factor Optimal Policy: One can prove the (s, S) Policy still holds: For each period, the optimal policy when there is a starting inventory of x (maybe <0, backlog) is: Order up to S if x < s . Don't order if x>=s . But it’s difficult to compute the values for (s, S)
28. Finite Period Multi-Echelon Model Events in each period: • Shipments arrive; • Decision making; • Demand arrives; • Cost evaluation. • Only consider N=2: • one depot base (d) and one retailer outlet (r) • The index n denotes the number of remaining periods until the end of the planning horizon
29. Finite Period Multi-Echelon Model Cost parameters Only for the depot Only for the outlet Index n is used when applicable
30. Finite Period Multi-Echelon Model Other parameters Index n is used when applicable
31. Finite Period Multi-Echelon Model Decision Variables (at the beginning of a period) On hand Index n is used when applicable
32. Finite Period Multi-Echelon Model vd Basic inequalities xr z y depot outlet s1,… , sl y1,… , yL wd wr Dynamics of the system
33. Finite Period Multi-Echelon Model At the beginning of a period, let = xr ? On hand At the end of a period, inventory of the system:
34. Finite Period Multi-Echelon Model Total cost:
35. Finite Period Multi-Echelon Model Total cost for n-period decision problem:
36. Finite Period Multi-Echelon Model Total cost for n-period decision problem: (Dynamic Programming)
37. Finite Period Multi-Echelon Model Dynamic Programming: Decomposition Retailer’s Decision
38. Finite Period Multi-Echelon Model Depot’s Decision
39. Infinite Period Multi-Echelon Model • Two kinds of objective functions: • Average-cost • Discounted-cost
40. Summary: Periodic review system • Single-location: • Single-period • Multiple-period • Infinite-period • Multi-Echelon • Single-period ?? • Multiple-period (1960) • Infinite-period (1984) • All are periodic review inventory systems • All are centralized control system • All are (s, S) policies (order-up-to policy, basic stock policy, or critical number policy when s=S)
41. Lot Size-Reorder Point Systems:Continuous review system • In what follows, we assume that the operating policy is of the (R, Q) form. However, when generalizing the EOQ analysis to allow for random demand, we treat R and Q as independent decision variables. • Decision variables: R = the reorder point (level) in units of inventory Q = the lot size or order quantity
42. Lot Size-Reorder Point Systems Assumptions • The system is continuous-review • Demand is random and stationary • There is a fixed positive lead time L for placing an order • The following costs are assumed • Setup cost at \$ K per order. • Holding cost at \$ h per unit held per year. • Proportional order cost of \$ c per item. • Stock-out cost of \$ p per unit of unsatisfied demand
43. Lot Size-Reorder Point Systems Describing demand: The demand during the lead time is a continuous random variable DL with probability density function (or pdf) fL(x), and accumulative distribution function (or cdf) FL(x). Let and be the mean and standard deviation of demand during lead time.
44. Inventory R Q Order placed Order arrives Time Lot Size-Reorder Point Systems Inventory vs. Time Safety Stock level L
45. Lot Size-Reorder Point Systems i.i.d demand: If the demands in all periods are i.i.d; The demand in each period is a continuous random variable D with probability density function (or pdf) f(x), and accumulative distribution function (or cdf) F(x). The the mean and standard deviation of demand during lead time.
46. Lot Size-Reorder Point Systems i.i.d demand with uncertain leadtime: The the mean and standard deviation of demand during lead time.
47. Lot Size-Reorder Point Systems Order quantity Q can be set as EOQ (near-optimal) Reorder point R = L + (safety stock) One method to set the safety stock (SS): SS= kL (k is the safety (stock) factor, related to service level or penalty cost data etc.)
More Related | 2,707 | 11,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-26 | latest | en | 0.829286 |
http://www.mathworks.com/matlabcentral/fileexchange/2374-dynamical-systems-with-applications-using-matlab/content/MATLAB%20files%2020013a/Program_4b.m | 1,438,062,719,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042981576.7/warc/CC-MAIN-20150728002301-00056-ip-10-236-191-2.ec2.internal.warc.gz | 571,479,242 | 8,024 | Code covered by the BSD License
# Dynamical Systems with Applications using MATLAB
### Stephen Lynch (view profile)
13 Sep 2002 (Updated )
Companion Software.
### Editor's Notes:
This file was selected as MATLAB Central Pick of the Week
Program_4b.m
```% Chapter 4 - Electromagnetic Waves and Optical Resonators.
% Program_4b - Bifurcation Diagram for a Nonlinear Optical Resonator.
% Copyright Birkhauser 2013. Stephen Lynch.
% Bifurcation diagram for a simple fiber resonator (Figures 4.13 & 4.16(a)).
clear
halfN=1999;N=2*halfN+1;N1=1+halfN;
format long;E1=zeros(1,N);E2=zeros(1,N);
Esqr=zeros(1,N);Esqr1=zeros(1,N);ptsup=zeros(1,N);
C=0.345913;
E1(1)=0;kappa=0.0225;Pmax=120;phi=0;
% Ramp the power up
for n=1:halfN
E2(n+1)=E1(n)*exp(1i*(abs(C*E1(n))^2-phi));
E1(n+1)=1i*sqrt(1-kappa)*sqrt(n*Pmax/N1)+sqrt(kappa)*E2(n+1);
Esqr(n+1)=abs(E1(n+1))^2;
end
% Ramp the power down
for n=N1:N
E2(n+1)=E1(n)*exp(1i*(abs(C*E1(n))^2-phi));
E1(n+1)=1i*sqrt(1-kappa)*sqrt(2*Pmax-n*Pmax/N1)+sqrt(kappa)*E2(n+1);
Esqr(n+1)=abs(E1(n+1))^2;
end
for n=1:halfN
Esqr1(n)=Esqr(N+1-n);
ptsup(n)=n*Pmax/N1;
end
% Plot the bifurcation diagrams
fsize=15;
subplot(2,1,1)
plot(Esqr(1:N),'.','MarkerSize',1)
xlabel('Number of Ring Passes','FontSize',fsize);
ylabel('Output','FontSize',fsize);
subplot(2,1,2)
hold on
plot(ptsup(1:halfN),Esqr(1:halfN),'.','MarkerSize',1);
plot(ptsup(1:halfN),Esqr1(1:halfN),'.','MarkerSize',1);
xlabel('Input Power','FontSize',fsize);
ylabel('Output Power','FontSize',fsize);
hold off
% End of Program_4b.
``` | 594 | 1,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2015-32 | latest | en | 0.380617 |
https://binary2hex.com/addition-column.html?id=88 | 1,638,614,640,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00599.warc.gz | 196,518,960 | 7,874 | # Сложить 101101 и 101110000000000111100 = 101110000000000212201
column Addition online calculator can add two numbers in a column giving a fully painted addition process.
the column addition Calculator supports integers, decimals, and negative numbers.
• Calculator
• Instruction manual
• Theory
• History
• Report a problem
Enter two numbers: the first term and the second term.
add with
+ 1 0 1 1 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 2 1 2 2 0 1
the Final answer:101101+101110000000000111100 = 101110000000000212201
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Your rating? | 220 | 598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-49 | latest | en | 0.679986 |
https://git.iws.uni-stuttgart.de/tools/frackit/-/commit/479751a828f367e4c353ef0d785493a9adfd90f6 | 1,656,692,202,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00209.warc.gz | 327,550,501 | 31,001 | Commit 479751a8 by Dennis Gläser
[python][sampling] add cylinder point sampler allowing for custom distros
parent 8a99f1ac
... ... @@ -5,7 +5,14 @@ class BoxPointSampler: """Class to sample random points within a box.""" def __init__(self, samplerX, samplerY, samplerZ): """Create the sampler from random variable samplers for the coordinate directions.""" """ Create the sampler from random variable samplers for the coordinate directions. Parameters: samplerX: sampler that samples from a distribution for the x-coordinate samplerY: sampler that samples from a distribution for the y-coordinate samplerZ: sampler that samples from a distribution for the z-coordinate """ self.samplerX = samplerX self.samplerY = samplerY self.samplerZ = samplerZ ... ... @@ -18,6 +25,53 @@ class BoxPointSampler: from frackit.geometry import Point_3 return Point_3(x, y, z) class CylinderPointSampler: """Class to sample random points within a cylinder.""" def __init__(self, cylinder, samplerR2, samplerPhi, samplerZ): """ Create the sampler from random variable samplers for the coordinate directions. Points on the cylinder are represented in cylinder coordinates, i.e. by a radial, an angular and a height coordinate. Points on the cylinder are sampled based on the given distributions for the squared radius, the angle and the height. Parameters: cylinder: an instance of a cylinder class samplerR2: sampler that samples from a distribution for the squared radius samplerPhi: sampler that samples from a distribution for the angular coordinate samplerZ: sampler that samples from a distribution for the height """ self.cylinder = cylinder self.samplerR2 = samplerR2 self.samplerPhi = samplerPhi self.samplerZ = samplerZ def sample(self): from frackit.geometry import Vector_3 a = Vector_3(self.cylinder.bottomFace().majorAxis()); b = Vector_3(self.cylinder.bottomFace().minorAxis()); n = Vector_3(self.cylinder.bottomFace().normal()); from math import sqrt, sin, cos r = sqrt(self.samplerR2()) phi = self.samplerPhi() z = self.samplerZ() a *= r*cos(phi); b *= r*sin(phi); n *= z; result = self.cylinder.bottomFace().center(); result += a; result += b; result += n; return result; # creator for a uniform sampler within an interval def uniformIntervalSamplerCreator(): def makeSampler(min, max): ... ... @@ -27,16 +81,58 @@ def uniformIntervalSamplerCreator(): return doSample return makeSampler def makeBoxPointSampler(box, samplerCreatorX=uniformIntervalSamplerCreator(), samplerCreatorY=uniformIntervalSamplerCreator(), samplerCreatorZ=uniformIntervalSamplerCreator()): """Creates a BoxPointSampler using the provided creators for samplers on intervals. The creators default to uniform interval sampler creators if nothing is specified.""" # make samplers samplerX = samplerCreatorX(box.xMin(), box.xMax()) samplerY = samplerCreatorY(box.yMin(), box.yMax()) samplerZ = samplerCreatorZ(box.zMin(), box.zMax()) def makePointSampler(geometry, samplerCreatorBase1=uniformIntervalSamplerCreator(), samplerCreatorBase2=uniformIntervalSamplerCreator(), samplerCreatorBase3=uniformIntervalSamplerCreator()): """ Creates a point sampler for the provided three-dimensional geometry using the provided creators for samplers on intervals. Points on the three-dimensional geometries considered here can be expressed in terms of tuples of coordinates (a1, a2, a3), where the entries of the tuple refer to the coordinate values with respect to the geometry-local basis. Moreover, the considered geometries are finite, which is expressed in terms of lower and upper bounds for the coordinates. For instance, the points within an axis-aligned box fall into the intervals ([xMin, xMax], [yMin, yMax], [zMin, zMax]). For a cylinder, described in cylinder coordinates, all points fall into the intervals ([0, r], [0, 2*Pi], [0, H]), where r is the radius an H is the height of the cylinder. The provided sampler creators are used to build samplers that sample coordinates within these intervals following a user-defined distribution. For more details on the meaning of the samplers, have a look at the geometry-specific point sampler implementations. The creators default to uniform interval sampler creators if nothing is specified. Parameters: geometry: the geometry for which a point sampler is to be created samplerCreatorBase1: creator for a sampler of coordinate values for the first coordinate samplerCreatorBase2: creator for a sampler of coordinates values for the second coordinate samplerCreatorBase3: creator for a sampler of coordinates values for the third coordinate """ # box point sampler if geometry.name() == "Box": samplerX = samplerCreatorBase1(geometry.xMin(), geometry.xMax()) samplerY = samplerCreatorBase2(geometry.yMin(), geometry.yMax()) samplerZ = samplerCreatorBase3(geometry.zMin(), geometry.zMax()) return BoxPointSampler(samplerX, samplerY, samplerZ) # cylinder point sampler if geometry.name() == "Cylinder": import math samplerR2 = samplerCreatorBase1(0.0, geometry.radius()) samplerPhi = samplerCreatorBase2(0.0, 2.0*math.pi) samplerZ = samplerCreatorBase3(0.0, geometry.height()) return CylinderPointSampler(geometry, samplerR2, samplerPhi, samplerZ) # hollow cylinder point sampler if geometry.name() == "HollowCylinder": import math samplerR2 = samplerCreatorBase1(geometry.innerRadius(), geometry.outerRadius()) samplerPhi = samplerCreatorBase2(0.0, 2.0*math.pi) samplerZ = samplerCreatorBase3(0.0, geometry.height()) return CylinderPointSampler(geometry.fullCylinder(), samplerR2, samplerPhi, samplerZ) return BoxPointSampler(samplerX, samplerY, samplerZ) raise NotImplementedError("No point sample creation formula implemented for provided geometry: " + geometry.name())
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Finish editing this message first! | 1,289 | 5,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.681472 |
http://harvardsportsanalysis.wordpress.com/2012/04/25/dont-trade-up-in-the-nfl-draft/ | 1,416,810,945,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400380409.19/warc/CC-MAIN-20141119123300-00046-ip-10-235-23-156.ec2.internal.warc.gz | 141,155,344 | 21,954 | ## Don’t Trade Up in the NFL Draft
As the draft looms tomorrow night, NFL teams are (hopefully) finalizing their draft boards and trying to make plans for every conceivable contingency. Whether or not they want to trade, and in which direction, is surely one of those considerations. In the past two years, we have seen two monumental trades: Atlanta’s trade for Julio Jones, and Washington’s trade for (presumably) Robert Griffin III. There are sure to be more throughout the draft as teams try and select the players they covet most. One question looms large in this picture: do teams that trade up actually benefit from the moves they make, or do they pay too high a price for the right to choose? We’ll explore the answer to this question here.
Using my previous work on the NFL draft, we can find the difference between the Career Approximate Value (CAV) of the players selected by a team that traded up to select them and the expected Career Approximate Value (eCAV) of the picks that teams gave up in exchange to select them. By comparing this net CAV of players involved in trades to the net CAV of all other players selected in the draft, we can see if players that were drafted as a result of a trade up outperformed their price relative to players who were selected normally.
Put into algebra, our equations follow. For players that were traded up for, our equation for net CAV is
Net CAV = CAVx – eCAVa – eCAVb – eCAV
where CAVx is the CAV of the player traded for, and eCAVa, eCAVb etc. are the expected CAV’s of the picks that were traded away. For all other players the equation is simply
Net CAV = CAVx – eCAVa
because teams only used one draft pick to select that player. This post does not include trades that involved active players or future draft picks, which are more complicated than we want to look at here. They would be valuable studies for another time.
Using data from the 1990-1999 drafts, I observed 98 eligible trades and 2584 other selections. The results are outlined in the table below:
As you can see, both the mean and median net CAV are lower for players selected as the result of a trade up, while the percentage of players selected is almost exactly the same. Using a difference of means T test, the difference between these two means is statistically significant at the 95% confidence level (T = 2.31). The 95% confidence interval for the mean net CAV of players not involved in trades was (-0.24, 1.25) while the 95% confidence interval for players involved in a trade was (-12.40, -0.53), which is more clearly seen in the graph below.
The 95% confidence interval for the difference between these means is (0.99, 12.95), meaning that players that are traded up for tend to underperform their price by 0.99 to 12.95 CAV more than players who were not traded up for.
These results suggest that teams tend to overpay for earlier draft picks – or at least they did during the 1990s. These results corroborate those found by Cade Massey and Richard Thaler in their paper on the draft. Teams that can resist the urge to trade up would likely gain a competitive advantage over teams that overpay for the right to choose, and would benefit even more by taking advantage of their competitor’s overconfidence and trading down instead. Will teams actually do so? We’ll see this weekend.
This entry was posted in NFL Football and tagged , , . Bookmark the permalink.
### 9 Responses to Don’t Trade Up in the NFL Draft
1. Greg says:
Interesting that many of the so called experts recommended trading up for RG3. The Redskins may live to regret that deal!
2. Normalman says:
Your CAV model is interesting but I think overvalues lower round players hence always making it look like a better deal to trade down then up. For example, if I understand it right, trading the # 1 pick for four or five second/third rounders would be a good deal in this model. I think in real life though, that would be an insane trade. A single high CAV player is almost certainly worth multiple low CAV players regardless of whether the net CAV of the multiples players is higher.
• Really enjoyed the analysis.
I agree Normalman, I think you need to consider the net replacement value of the draft pick you select. How much value are you gaining from the player you select vs. the player he is replacing on your football team. We discussed something similar where the analysis repeats a lot of what Kevin did back in November: http://bit.ly/IRGGvi
3. secretbonus says:
normalman makes a good point, the utility of the player and how he is able to contribute to wins is significant, but additional wins is not enough if it does not significantly raise the chances of winning a superbowl. A team that drafts well and makes decisions that allows it to be consistently finshing in the 7-9 wins category and very infrequently finishing outside of that is inferior to one that on average gets maybe 5-8 wins but some seasons they get 3 or 4 and others they get 14 and are a clearly superior team. A high varience strategy in the NFL is superior, provided the goal is winning the superbowl and anything else is considered a failure. However, with that being said, I believe the superior strategy is to enter rebuilding mode more frequently, trading every pick this year for a pick higher next year, and repeat, also trading players over the age of 28 (although some positions sooner, others later) and continuing to roll over draft picks until One has built up an arsenal of picks. I also think the NFL has yet to bridge the gap between stats and analytics and decision making like Baseball has done in the post moneyball era. NFL teams should assess probabilities and standard deviation and make decisions that comply with their goals. Personally, I think if teams regularly would trade players after the age of 28 and constantly roll over draft choices until it could “live off the interest” trading most of the picks but still getting several 1st rounders every single year, it could easily become a dynasty and become very difficult to keep up with until adjustments to the “draft value chart” are made.
4. Steve says:
I looked at this a different way and got to the same answer.
Do the picks for which teams traded up have a higher success rate than others (i.e., are the teams more certain on those so the expense might be justified)? Since the picks do not have a higher success rate, you can’t justify giving away value (net of trade) to move up. Still possibility that trading up makes sense for teams with few holes as a form of concentrating the value of their picks into a smaller set of players, but that would require more research. Check out the analysis here: http://www.sportsplusnumbers.com/2013/05/luck-vs-skill-are-picks-after-trading.html
5. Jon Russell says:
Tell that to the BEARS, who traded up to snatch Alshon Jeffery | 1,518 | 6,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2014-49 | latest | en | 0.978147 |
http://www.reddit.com/r/tekkit/comments/11l3gt/how_do_multiple_ic_energy_storage_device_behave/?sort=old | 1,411,330,961,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657135930.79/warc/CC-MAIN-20140914011215-00098-ip-10-234-18-248.ec2.internal.warc.gz | 790,900,987 | 13,796 | [–] 0 points1 point (1 child)
sorry, this has been archived and can no longer be voted on
You could turn one of them to send power into the other and the other would be like a output. That is how I build massive batteries. If it is receiving a lot of power at once, then the one feeding power into the other will become fully charged. If it is a slow charge, then the one receiving power from the second will charge first.
[–][S] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
thats exactly what i was looking for, so if the furthest down the line charges first then i have it output when empty to start my reactor and the closest when full to stop the reactor
[–] 1 point2 points (2 children)
sorry, this has been archived and can no longer be voted on
First of all, test it and find out. 90% of what I know I know from testing. I test ALL THE TIME.
Secondly - I'm not sure if you are talking about input/output face alignments of batteries, or if you are looking to maximize storage vs output flow.
If you are asking about input/output faces of a battery - a battery has 5 input faces and only 1 output. That is, power on a wire will charge a battery when connected to any of the 5 input faces, but power from the battery will add to the power on the wire when connected to the output face. If you have a wire connected to just one battery, let's assume an input face (input relative to the battery, i.e. charging the battery), and that said battery's output face is facing a 2nd battery, the 1st battery will send all it's power to the 2nd. Thereby, you can power both batteries from one wire (you just have to face the output to each subsequent battery).
The other topic you might be referring to is about output flow. You can put your batteries in serial or in parallel. In a serial configuration, you'll get longer lasting storage, with the same rate of flow. Parallel configurations will have no increased storage time, but increased output. We are not talking about stepping power, as transformers do. We are only talking about rate of flow of equal packet sizes. However, this might not be your question. If you want to know more about what I'm talking about here, just let me know.
EDIT: you'd think with all my years in D&D I wouldn't make the lame mistake that a cube has 6 sides and not 8.
[–] 0 points1 point (1 child)
sorry, this has been archived and can no longer be voted on
Good info, but I just want to clarify one thing. There are five input faces and one output face on any IC storage block due to cubes being a six-faceted figure.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
the shame..i know this. 6 side die is a cube. Seriously, I need more dew.
[–] 0 points1 point (1 child)
sorry, this has been archived and can no longer be voted on
If you put the output face of one (MFSU1) directly adjacent to one of the input faces on the other (MFSU2), you would only have to connect the supply to MFSU1. The MFSU2 would charge first (it would pull all of the EU out of MFSU1 as soon as it entered) but once it was full the MFSU1 would charge. MFSU1 would continue to charge MFSU2 as needed.
On the surface it might seem like a handy arrangement and depending on the usage it would be fine. If, however, you were wanting to power things like a MKIII charging bench or a mass fabricator, you'd be better off wiring each MFSU individually. Double the packets means double the speed of operation for both the charging bench and the mass fabricator.
[–][S] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
That makes complete sense thatnks for your help, the plan is that if the furthest charges first have it output when empty to start my reactor and then have the closest output when its full to stop the reactor
[–] 0 points1 point (1 child)
sorry, this has been archived and can no longer be voted on
Another resource you may find useful is this Electricity Tutorial specifically the section on cable splitting.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
That section on forks is obsolete I believe. It even says at the top that it is under review. For one thing, packet sizes do not change when they come to a "fork". I have tested this. I split the output of an mfe several times and then fed the power to a macerator. The macerator still blew up because the packet size was still 128pEU.
Secondly, I don't believe that forks are the major CPU hog that the tutorial says it is. First, I don't believe the calculations are performed every tick. I think they only occur like once per second. You can see this by placing an hv solar array on an input of an empty mfsu. It will take a second or two for it to start charging. I don't think the calculations are that difficult to begin with, anyway. These calculations don't even have to be floating point.
If you have some machine that has like 100 "forks" then it might become an issue. If you are worried about 3 or 4 forks then you are probably wasting your time. | 1,240 | 5,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-41 | latest | en | 0.956244 |
https://web2.0calc.com/questions/is-x-2-the-same-as-x-2-x-3-x-3 | 1,548,263,191,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584334618.80/warc/CC-MAIN-20190123151455-20190123173455-00533.warc.gz | 673,669,774 | 5,937 | +0
# Is X+2 the same as ((X+2)(X+3))/(X+3)
0
394
1
(X+3)/(X+3) is equal to 1. If we multiply X+2 by 1, it should still be X+2. Does this mean X+2 is the same as ((X+2)(X+3))/(X+3)?
Feb 15, 2015
#1
+17747
+10
They are not exactly the same. They will have the same value for all values for x except -3.
If x = -3, then x + 2 = 1, but ((x+2)(x+3))/(x+3) is undefined because it a denominator of zero, and division by zero is undefined.
Feb 16, 2015
#1
+17747
+10 | 189 | 468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-04 | latest | en | 0.896006 |
http://www.espsciencetime.org/SimpleMachines.aspx | 1,516,659,287,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891543.65/warc/CC-MAIN-20180122213051-20180122233051-00648.warc.gz | 437,709,908 | 14,525 | # Simple Machines, Gr. 6
### Simple Machines, Gr. 6
Simple machines are the basic moving parts in the complex machines that make our lives easier and more efficient. There are many simple machines around your home, and our lives would be much more difficult without them. Doorknobs, faucets, stairs, knives, forks, hinges, wrenches, drill bits and chisels are all examples of simple machines. We depend on simple machines to increase our ability to increase force. Simple machines can also move things farther or faster than we could by ourselves. Golf clubs, tennis rackets, baseball bats, egg beaters, and gears on a bike or car are necessary simple machines that increase force.
Simple machines include levers, wheel and axles, pulleys, inclined planes, wedges, screws and gears.
Levers – A lever is a bar that pivots on a turning point called a fulcrum while it supports two or more forces located at different points on the bar. There are three classes of levers: Class 1, Class 2 and Class 3.
Wheel and Axle – The wheel and axle is an example of a class 1 lever. The wheel and axle is really a continuous lever rotating around a fulcrum.
Pulley – A pulley is also an example of a lever. It includes a grooved wheel over which a rope, chain or belt passes. A fixed pulley is attached to a support and does not move. A movable pulley is one that is attached to a load or counter force and moves as the load moves.
Inclined Plane – An inclined plane is a sloping surface. It can be used to alter the effort and distance involved in doing work.
Wedge – A wedge is an inclined plane that moves to change the direction of a force.
Screw – A screw is an inclined plane wrapped around a shaft. When a screw is driven into a piece of wood, the screw travels only a short distance, but the slope (thread) travels much farther, increasing the effort of turning and driving the screw with a great deal of power.
Gears – Gears are wheels with teeth. The teeth of one gear fit between the teeth of a second gear. As one gear turns, the second one turns also.
Work is the product of a force moving some distance. Machines do not change the amount of work. However, machines can produce a mechanical advantage which alters the amount of effort applied. The ability of a lever to increase the force applied to a load is called its mechanical advantage. The mechanical advantage of a lever depends upon the distance from the fulcrum to the load and the distance from the fulcrum to the effort. | 573 | 2,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-05 | longest | en | 0.9249 |
https://socratic.org/questions/what-is-8-divided-by-0 | 1,624,461,843,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539480.67/warc/CC-MAIN-20210623134306-20210623164306-00190.warc.gz | 485,759,578 | 5,958 | # What is 8 divided by 0?
##### 1 Answer
Jan 27, 2016
It is infinite. Normally it is denoted by symbol $\infty$
#### Explanation:
As a matter of fact division by zero has not been defined.
Please recall that for any number $n$
$n \times 0 = 0$
$\implies$ that the number could be 8, as you have stated, or it could be any other number of your second, third or any other choice.
If we write in division form by taking one of the two zeros or $n$ to the other side of equation, we obtain.
n/ 0=?/0, we could put any other number to replace "?"
or $n = \frac{0}{0}$, where $n$ could be any number of choice.
or $\frac{0}{0} = \frac{1}{n}$, where $n$ could be any number of choice again.
We see that we don't have any unique representation in case of division by 0. Therefore, division by 0 is indeterminate. | 232 | 813 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-25 | latest | en | 0.950503 |
https://everynationtallahassee.com/qa/question-what-is-the-formula-for-calculating-savings.html | 1,611,071,422,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00724.warc.gz | 343,151,541 | 8,036 | # Question: What Is The Formula For Calculating Savings?
## How do I calculate simple interest rate?
Simple Interest Formulas and Calculations:Calculate Interest, solve for I.
I = Prt.Calculate Principal Amount, solve for P.
P = I / rt.Calculate rate of interest in decimal, solve for r.
r = I / Pt.Calculate rate of interest in percent.
R = r * 100.Calculate time, solve for t.
t = I / Pr..
## How much should I save each month?
Most experts recommend saving at least 20% of your income each month. That is based on the 50-30-20 budgeting method which suggests that you spend 50% of your income on essentials, save 20%, and leave 30% of your income for discretionary purchases.
## What is the monthly interest rate?
To convert an annual interest rate to monthly, use the formula “i” divided by “n,” or interest divided by payment periods. For example, to determine the monthly rate on a \$1,200 loan with one year of payments and a 10 percent APR, divide by 12, or 10 ÷ 12, to arrive at 0.0083 percent as the monthly rate.
## How much in savings should I have?
Fast Answer: A general rule of thumb is to have one times your income saved by age 30, twice your income by 35, three times by 40, and so on. Aim to save 15% of your salary for retirement — or start with a percentage that’s manageable for your budget and increase by 1% each year until you reach 15%
## What is simple interest and example?
Key Takeaways Car loans, amortized monthly, and retailer installment loans, also calculated monthly, are examples of simple interest; as the loan balance dips with each monthly payment, so does the interest. Certificates of deposit (CDs) pay a specific amount in interest on a set date, representing simple interest.
## What is the formula to calculate interest?
✅What is the formula to calculate simple interest? You can calculate Interest on your loans and investments by using the following formula for calculating simple interest: Simple Interest= P x R x T ÷ 100, where P = Principal, R = Rate of Interest and T = Time Period of the Loan/Deposit in years.
## How do you calculate interest earned on a savings account?
If interest is compounded daily, divide the simple interest rate by 365 and multiply the result by the balance in the account to find the interest earned in one day. Add the daily interest earned to the balance.
## What is my savings rate?
To calculate your savings rate, divide your savings by your income and you get the percentage of income you save.
## How much interest will I get on \$1000 a year in a savings account?
How much interest can you earn on \$1,000? If you’re able to put away a bigger chunk of money, you’ll earn more interest. Save \$1,000 for a year at 0.01% APY, and you’ll end up with \$1,000.10. If you put the same \$1,000 in a high-yield savings account, you could earn about \$5 after a year.
## How do you calculate interest per year?
A = P(1 + r/n)ntA = Accrued Amount (principal + interest)P = Principal Amount.I = Interest Amount.R = Annual Nominal Interest Rate in percent.r = Annual Nominal Interest Rate as a decimal.r = R/100.t = Time Involved in years, 0.5 years is calculated as 6 months, etc.More items…
## What is the formula for time?
To solve for time use the formula for time, t = d/s which means time equals distance divided by speed.
## How do you calculate amount?
Simple Interest Equation (Principal + Interest)A = Total Accrued Amount (principal + interest)P = Principal Amount.I = Interest Amount.r = Rate of Interest per year in decimal; r = R/100.R = Rate of Interest per year as a percent; R = r * 100.t = Time Period involved in months or years.
## What is the formula of principal?
Principal Amount Formulas We can rearrange the interest formula, I = PRT to calculate the principal amount. The new, rearranged formula would be P = I / (RT), which is principal amount equals interest divided by interest rate times the amount of time.
## How do I calculate my savings increase?
Key TakeawaysMarginal propensity to save (MPS) is an economic measure of how savings change, given a change in income.It is calculated by simply dividing the change in savings by the change in income.A larger MPS indicates that small changes in income lead to large changes in savings, and vice-versa.
## How do I calculate monthly interest?
For a daily interest rate, divide the annual rate by 360 (or 365, depending on your bank)….Monthly Interest Rate Calculation ExampleConvert the annual rate from a percent to a decimal by dividing by 100: 10/100 = 0.10.Now divide that number by 12 to get the monthly interest rate in decimal form: 0.10/12 = 0.0083.More items…
## How much interest does 10000 earn a year?
How much interest can you earn on \$10,000? In a savings account earning 0.01%, your balance after a year would be \$10,001. Put that \$10,000 in a high-yield savings account for the same amount of time, and you’ll earn about \$50. | 1,170 | 4,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-04 | longest | en | 0.938735 |
https://www.freemathhelp.com/forum/threads/probability-issues-draw-3-balls-from-bag-of-12-in-3-colors.56499/ | 1,553,610,883,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205534.99/warc/CC-MAIN-20190326135436-20190326161436-00145.warc.gz | 736,183,915 | 11,473 | # Probability issues: draw 3 balls from bag of 12 in 3 colors
#### heidi18
##### New member
I am auditing a college class and I did not understand these problem by the teacher. Go easy this is my first go around with this stuff.
A bag contains five white, three black, and four red balls. A single drawing of three balls is made. Find the probability that the three balls will all be of a different color?
Is it (5/12) times (4/11) times (3/10)??????
The other one that did not make sense was...
From a deck of 52 cards, three aces have been drawn one at a time in three consecutive draws. What is the probability of the fourth ace being drawn on the next draw, and what is the probability of these four aces being drawn as described?
Is the second part of that question (4/52)(3/51)(2/50)(1/49)? I am not sure of the first part of the question, where do I start?
I do not want to be a bother, ill be greatful to anyone who can make me better and smarter and help me with this!
#### galactus
##### Super Moderator
Staff member
A bag contains five white, three black, and four red balls. A single drawing of three balls is made. Find the probability that the three balls will all be of a different color?
Is it (5/12) times (4/11) times (3/10)??????
You are on the right track, except you need to multiply by 6. That's because they can come out in 3!=6 different ways. For instance, say they came out BWR, RWB, WRB, etc. 6(5/12)(4/11)(3/10).
Another way to do this is using combinations. $$\displaystyle \frac{5\cdot{4}\cdot{3}}{C(12,3)}$$. Because we are choosing 3 from 12 and and there are 5 whites to choose, 4 reds, and 3 blacks.
The other one that did not make sense was...
From a deck of 52 cards, three aces have been drawn one at a time in three consecutive draws. What is the probability of the fourth ace being drawn on the next draw, and what is the probability of these four aces being drawn as described?
Is the second part of that question (4/52)(3/51)(2/50)(1/49)? I am not sure of the first part of the question, where do I start?
I do not want to be a bother, ill be grateful to anyone who can make me better and smarter and help me with this!
You are correct. The last Ace will have a probability of 1/49 of being drawn. But remember, they can be drawn in 4!=24 different ways because there are 4 different Aces. So multiply by 24. 24(4/52)(3/51)(2/50)(1/49)=
#### heidi18
##### New member
I think I am starting to understand...
A box holds 4 apples, 5 oranges, and 8 pears. How many ways can 2 apples, 1 orange, and 2 pears be chosen?
so I would have multipication because of the word AND, (4/17)(3/16)(5/15)(2/14)(1/13)=
...except I probably need to multiply this by all the ways they can come out of the box, right???
that seems like a lot of ways, do I have to list all the of them to find out? Like AAOPP AOPPA...???
#### galactus
##### Super Moderator
Staff member
You are on the right track. Seems you're catching on and you are correct in that you have to multiply by the number of arrangements. 5!/(2!1!2!)=30
There are 5 items you are choosing, but there are 2 the same, 1 , and 2 the same. That is called distinguishable permutations.
30/6188=30/221.
Here is another way. You get the same answer.
$$\displaystyle C(4,2)C(5,1)(C(8,2)=840$$
Because we are choosing 2 from 4, 1 from 5, 2 from 8.
Now, if you wanted the probability of choosing what you stated, then divide by C(17,5)
Because we are choosing 5 from 17 total.
$$\displaystyle \frac{C(4,2)C(5,1)(C(8,2)}{C(17,5)}=\frac{30}{221}$$
#### heidi18
##### New member
I dont mean to bother you, but where did you get the 30/6188=30/221? Particularly the 30/6188? sorry
#### galactus
##### Super Moderator
Staff member
That is the calculation. You got the 1/6188 yourself. I just multiplied by 30. | 1,088 | 3,813 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-13 | latest | en | 0.962076 |
https://m.mobogenie.com/download-next-puzzle-4767551.html | 1,516,412,943,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00566.warc.gz | 756,180,394 | 8,824 | Next PUZZLE
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Next PUZZLE is a carefully designed game that brings the most challenge out of the simplest rules possible.
What is the game?
You have a 5x7 board of tiles that each hold a numerical value. At the start of any game, the value of every tile will be 2. You have to move around values to reach a certain goal for each column. To move values around you tap on a tile and then tap on another nearby tile and then half of the value of the first tile will be transferred to the second tile.
What are the rules?
First, you cannot transfer values from tiles with odd values. You must always make them even before moving anything. Second, your goal will always be to get the sum of all the tiles in a column to be equal to some number specified under that column. At the start of every game, the initial goal of every column is 8. This means that by moving numbers from a column to another you have to make that column's sum be equal to 8. When you reach a goal, the value of the goal will be doubled. For example, when you get the sum of a column to be 8, your next goal will be 16, then 32 and so on. You will be rewarded every time you reach a goal with more numbers to move around. Reaching a goal will earn you points.
How can I lose?
If you play any of the time modes you will lose when your time runs out. But in any mode you will lose if you make the sum of the values of a column to be 1 more than the specified goal. For example. if your goal is 8 and you make the sum be 9 then you will lose. If your goal is 16 and you make it 17 then you will lose and so on.
What modes are there?
There are 3 time modes: 1 minute mode, 5 minutes mode and 10 minutes mode. These are meant to be played after you have a strong understanding of the game mechanic and hopefully also have memorized some good moves to move your score up fast because if you think for too long then you will waste time.
There is also an 'infinite' mode. Here you have no time restraints and thus can only lose by making the value of the column be one more than the goal. You should practice in this mode as you are not in a hurry and you will take some time to truly master the mechanic so I recommend spending some meaningful time practicing here. In this mode you can save your progress and always keep a file to load so that you can continue earning more and more points without limits (unless you lose).
You got it?
Perfect, then go right in. You will find the game to be truly challenging and stimulating as not only will you be trying to optimize your play style but also you will have to do a lot of calculations in your mind to make sure that after the next move you won't lose.
You are still a little bit confused?
That is fine. I would recommend to jump right in to infinite mode and start testing the game. After a few loses you will understand the rules easily. To aid you, below every column is a fraction that shows the sum of the numbers of the column over the goal of that column.
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http://www.rohm.com/web/eu/tr_what7 | 1,495,640,881,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607848.9/warc/CC-MAIN-20170524152539-20170524172539-00388.warc.gz | 639,803,814 | 19,067 | # What is a transistor?
### Calculating transistor chip temperature
#### 1. How to calculate junction temperature(from ambient temperature)
Junction temperature (or Channel Temperature) can be calculated from the ambient
*Rth(j-a) : Thermal resistance of "junction-to-ambient" varies depending on the types of circuit board. For reference, we are also providing Table of Thermal Resistance by Package" based on the use of our standard circuit board separately.
Rth(j-a) value differs per each part number, but the values will be close if the package is the same.
**If the current consumption is not stable and changing time to time, then the averaged values of consumption current shall be assigned in the calculation formula to get the approximate value. (See "Usable or Not Judgement Method" provided separately
As an example, we list below the correlation between the current consumption and the junction temperature when Rth (j-a) is 250 deg./W , the ambient temperature is 25 deg.
Junction temperature rises in proportion to current consumption. The proportionality constant for this is Rth(j-a).
Since Rth(j-a) is 250 deg./W, the junction temperature rises by 25 deg. per each increment of 0.1W of current consumption. This means, the junction temperature becomes 150 deg.
when the current consumption is 0.5W, and the graph in this case suggests that the current beyond 0.5W can not be applied to TR.
That is tosay, even the same current is applied, the junction temperature also rises as the ambient temperature goes up and this diminishes the applicable current subsequently.
Maximum current consumption is influenced not only by thermal resistance but also by the ambient temperature also.
The maximum current consumption decreases with the above ratio.
Derating curve illustrates the ratio of current attenuation by percentage and we can apply the percentage to all the packages. For example, in case of MPT3 package (SOT89), the maximum applicable power is 0.5W at 25 deg. and the applicable current is getting reduce at the rate of 0.8%/deg.. This means, the value goes down to 0.4W which is 80% from 100% of its initial (20% down), and down to 0.2W which is 40% (60% down).
#### 2. Transient Thermal Resistance
With those examples aforementioned, we illustrated about the cases where the current is applied to device continuously.
Nextely, we will discuss about the case that the temperature rise by momentary current application.
The above graph shows the thermal resistance at momentary time (Transient Thermal Resistance), plotting the pulse width on X-axis and Rth(j-a) is on Y-axis.
It is known from this graph that the junction temperature rose as the current application time lasts longer and it reached the plateau state (=called thermal saturation) after 200 sec. had passed.
We can use this calculation formula to obtain the junction temperature when the current is applied momentarily as single pulse.
#### 3. Calculatin method of junction temperature (from case temperature)
The junction temperature can be calculated from the temperature of case as below. i.e., we assign Rth(j-c) in the formula in place of Rth(j-a) we did before,
n.b.
* Case temperature is measured by the radiation thermometer as the maximum temperature on the surface of package where the marking is put.
Please note that the case temperature differs considerably by the measurement method/point.
** The value is considered approximate one when the applied current is not constant and shifting time to time by assigning the averaged current consumption figure.
However, since Rth(j-c) value DOES vary depending on the types of circuit board and also on the heat dissipation conditions including soldering state, it may not be all right to apply the above formula directly to your calculation since the the measured values on our circuit board may not mean the same on your circuit board likewise. For instance, the case temperature may be measured as lower by comparison even though the applied current is the same, when the circuit board has good heat dissipation characteristics.
An illustration below shows that Rth(j-c) becomes lower as the collector land pattern on the circuit board gets smaller. (Collector land area / thickness / mterials plus circuit board material, size circuit width will also bring different measurement results on Rth(j-c).
In this way, Rth(j-c) value can differ depending on the nature & conditions of circuit board and also it is difficult to spot the right place for measuring the case temperature precisely. For these reasons, it is not so much recommended to approximate the junction temperature. from the case temperature.
#### Junction-to-Case thermal resistance Rth(j-c) - details
In principle, Junction-to-Case thermal resistance Rth(j-c) is an index basically used for TO220 packaged (=throughhole) devices by soldering it to the heatsink In this case, since Case-to-Heatsink is the heat-radiation path, it is possible to precisely calculate the junction temperature by measuring the case temperature at the point in the middle of such path. In particular, if the heatsink having the ideal heat dissipation performance is supposed to be used (e.g. infinite heatsink), the heat dissipation capability is considered as limitless and it is takenfor granted that"Case Temperature" = "Ambient Temperature" , Case temperature = 25 deg. (Tc = 25 deg.) is supplied in the calculation formula.(Thermal resistance of infinite heat-sink : Rth(c-a) = ; then Rth(j-a) = Rth(j-c))
However, for the surface-mount devices, heat-radiation path is mainly the part of circuit boad just beneath the device ; so that it is quite difficult to measure the case temperature at such location. Even if the temperature on marking side of device is measured, its portion of heat-dissipation in the entire heat-dissipation is rather small. Therefore, it is not suitable to use the temperature at such place in the formula to calculate the junction temperature either.
INevertheless, since there are many requests from the customers about Rth(j-c) value for SMT devices also, ROHM provides in some cases Rth(j-c) value on the conditions that the temperature is measurement on the marking side of device being mounted on the aforementioned standard circuit board. Therefore, the Rth(j-c) value should be considered for reference as obtained from the customized conditions as described.
If the device is mounted on the circuit board different from ours, the portion of heat-dissipation in the entire heat radiaton shall differ so that it is not possible to figure out the junction temperature adequately.
#### Thermal Resistance of standard packages (Reference data)
The values in following data are not the guaranteed values nor maximum / minimum values.Please treat these only as reference data.
n.b.
• Data listed here came from the results of measuring a specific production lot.
• Rth ( j-a ) vary a lot depending on the circuit board, the heat-dissipation conditions involving soldering methods and the method of temperature measuring, .
Package VMT3 EMT3 EMT5 EMT6 TUMT3
Circuit Board
FR4 dimensions
(unit：mm)
20×12×0.8 20×15×0.8 20×15×0.8 20×15×0.8 20×12×0.8
Rth(j-a)/
Rth(ch-a)
833ºC / W 833ºC / W 1042ºC / W 1042ºC / W 313ºC / W
Note When driving 1-die When driving 1-die
Package TUMT6 UMT3 UMT5 UMT6 SMT3
Circuit Board
FR4 dimensions
(unit：mm)
15×20×0.8 20×12×0.8 20×15×0.8 15×20×0.8 20×12×0.8
Rth(j-a)/
Rth(ch-a)
313ºC / W 625ºC / W 1042ºC / W 1042ºC / W 625ºC / W
Note When driving 1-die When driving 1-die When driving 1-die
Package SMT5 SMT6 TSMT3 TSMT5 TSMT6
Circuit Board
FR4 dimensions
(unit：mm)
20×15×0.8 20×15×0.8 30×15×0.8 20×15×0.8 20×15×0.8
Rth(j-a)/
Rth(ch-a)
625ºC / W 625ºC / W 250ºC / W 250ºC / W 250ºC / W
Note When driving 1-die When driving 1-die When driving 1-die When driving 1-die
Package SOP8 MPT3 CPT3 SST3
Circuit Board
FR4 dimensions
(unit：mm)
20×20×0.8 12×20×0.8 12×30×0.8 20×12×0.8
Rth(j-a)/
Rth(ch-a)
160ºC / W 250ºC / W 125ºC / W 625ºC / W
Note When driving 1-die | 1,953 | 8,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-22 | latest | en | 0.892197 |
https://www.physicsforums.com/threads/pwm-inverter-drive-explanation-please.728807/ | 1,513,342,941,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948569405.78/warc/CC-MAIN-20171215114446-20171215140446-00019.warc.gz | 798,720,632 | 15,282 | # PWM Inverter Drive explanation please
1. Dec 16, 2013
### Physicist3
Hi, I understand that the PWM inverter drive system first uses a diode bridge rectifier in order to convert the AC incoming into DC before using an inverter to convert this DC into PWM for use on the induction motor. Could someone please explain in simple terms how the PWM inverter section works, (i.e. Pairing the wave with a triangle wave) etc and what the original sine wave and dc waves are used for?? Also, how does the mark-space ratio of the output signal change with a change in frequency?? How is the frequency changed?? Is it by changing the triangle wave frequency or what?? Thanks
2. Dec 17, 2013
There are a few questions in your post that I'll try to straighten out..
I think you are mention "Original Sine" relative to the AC power TO the drive. The operation fof the VFD is independent of the sources AC.
The DC really should not be called a "wave" - ideally it is a perfectly flat DC source of power for the VFD Inverter. ( in the real world this is not perfectly flat - between 2-10% ripple - an undesirable result from both the Rectifier and the VFD inverter current)
OK - then the common way to generate the PWM Sine wave is a Triangle Wave operating at the Switching Frequency ( Fsw) and the width (modulation) of the pulse (output from the inverter) is determined by the the ratio of the triangle wave to the desired output (typically the Voltage). (That is not the best wording )
There are LOTS of very good references on this : http://www.ab.com/support/abdrives/documentation/techpapers/PWMDrives01.pdf
It looks like you are making a common mistake and tying to look at the whole system at once, the simpler answer is that the PWM wave is really using pulses to synthesize a wave- this can be done to make really any waveform from DC to just under 1/2 the FSW ( but then heavily distorted - high harmonics)
3. Dec 18, 2013
### Physicist3
So am I right in thinking that the DC link output signal which goes to the inverter is 'chopped' using the IGBTs in order to reproduce an artificial sine wave? Is this then paired with the carrier triangle wave and then a PWM wave is created from the points of intersection between the two waves, with the PWM wave being 'on' in the area between two intersection points where the recreated sine wave passes over the triangle wave, and 'off' in the area where the wave passes under the point of one of the triangle points? How is the height of the PWM wave determined??
4. Dec 20, 2013 | 593 | 2,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-51 | longest | en | 0.920285 |
https://in.mathworks.com/matlabcentral/answers/716603-how-to-find-kp-ki-for-pi-controller-from-gain-and-time-constant-value-given | 1,719,141,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00784.warc.gz | 272,794,721 | 28,675 | # How to Find Kp,Ki for PI controller from Gain and Time Constant value Given
421 views (last 30 days)
Sreeraj A on 14 Jan 2021
Hello !
I have a PSIM circuit model where for the PI Controller Gain and Time constant values are mentioned. I would like to use the same values with Matlab PI controller where input is Proportional Gain and Integral Gain( Kp & Ki)
Any method to find Kp and Ki from Gain and Time constant values mentioned ?
Sreeraj Arole
hamed hendijani on 29 Sep 2022
my input voltage of cloosed loop boost converter is 24v and out put voltage is 48v
i already designed the simulation of it and i used youtube videos to set proportional to 0.7 and integral to 100 and it works but i want to know from where we can get this values?
Sam Chak on 29 Sep 2022
Edited: Sam Chak on 29 Sep 2022
I guess that is a part of the system that produce the output from the input .
If your boost converter model is linear, then you can apply the Final Value Theorem learned fron Integral Calculus.
I think is inside the function that you can derive from the Final Value Theorem.
Is the Boost Converter a subsystem of the PSIM circuit? If they are unrelated, please show the mathematical model of the Boost Converter in a New Question.
Birdman on 14 Jan 2021
Simply, the conversion is as follows(Let K denote gain and Ti denote time constant):
K*(1+1/(Ti*s))
is equal to
Kp+Ki/s
If you equate two expressions, then
Kp=K
Ki=K/Ti
Sreeraj A on 14 Jan 2021
Really Helpful ! Thank you very much @Birdman
### More Answers (1)
Nadine on 30 May 2023
does anyone know how to tune the PI Controller parameters in inverter-based microgrids ?
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### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 488 | 1,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.850254 |
https://studylib.net/doc/10793260/the-simple-star-graph-consists-of-n-edges-meeting-at...-%E2%88%88... | 1,638,212,639,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358786.67/warc/CC-MAIN-20211129164711-20211129194711-00638.warc.gz | 625,846,859 | 12,877 | # The simple star graph consists of N edges meeting at... ∈ L of length 1, with a real valued potential q
```The simple star graph consists of N edges meeting at a common vertex, each edge
of length 1, with a real valued potential qi ∈ L∞ (0, 1) defined on the ith edge, with edge
parameterized by [0, 1] in such a way that x = 1 corresponds to the common vertex where
the edges meet. The direct eigenvalue problem consists in finding λ ∈ C and ψ(x) =
{ψ1 (x), ψ2 (x), ψ3 (x)} on [0, 1], such that ψi00 (x) + qi (x)ψi (x) = λψi where ψi (0) = 0 and
we have Kirchoff matching conditions at the common vertex x = 1. These {λ := λk }∞
k=1
are the graph spectra.
There are several inverse problems which seek to recover qi (x). All assume we are
given the graph spectra, but this is insufficient and additional information is required. We
shall look at the case where we also know the Dirichlet spectrum of each node.
A good physical way to view this problem is as N unit strings each of unknown density
ρi joined at a single node and clamped at the other end. We are given the vibrational
frequencies of the system together with the frequencies of each edge when the node is also
clamped. Can this be used to recover the ρi ?
``` | 337 | 1,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.927778 |
http://workbench.cadenhead.org/news/3161/fake-live-election-results | 1,579,683,798,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606872.19/warc/CC-MAIN-20200122071919-20200122100919-00111.warc.gz | 180,358,554 | 6,130 | \n
As today's California recall election demonstrates, live election results are a hugely popular feature for many news-oriented Web sites. This evening, I'll be reloading the California Secretary of State site over and over until my carpal tunnel nerves snap like kindling.
Offering live election results can be difficult if you don't have a source providing you this information, preferably in electronic form.
Fortunately, it's possible to use JavaScript to present authentic-looking fake live results that change over the course of Election Day.
Here's how:
First, decide how many votes will be cast in the election and counted on your page during the day. California is expecting around 10 million votes, 2 million of which have already been cast by absentee ballot but will not be counted during the day. So that leaves around 7-8 million votes. Store your total in a variable:
var y = 7850223;
Next, determine the time, in milliseconds since Jan. 1, 1970, at which the first vote should be reported. I accomplished this by publishing a dummy Web page with a three-line script:
var today = new Date();
var t = today.getTime();
document.write("<p>" + t);
This displays the time you need, in milliseconds. Store it as a variable named x. Here's mine, for around 7 a.m. PDT when the polls opened.
var x = 1065535367464;
Store the current time in a variable named t:
var today = new Date();
var t = today.getTime();
For the next step, you need to decide how many hours it will take to report 100 percent of the results. This value also needs to be in milliseconds, which are one-one thousandth of a second. For a 14-hour timeframe, it's 50400000 milliseconds (14 hours * 60 minutes * 60 seconds * 1000 milliseconds).
Using this value, the start time (x), and the current time (t), you can calculate the percentage of the vote that has been reported at the time the page is loaded:
var pct = (t - x) / 50400000;
To prevent the total from exceeding 100 percent after your time period, add this conditional:
if (pct > 1) {
pct = 1;
}
The percentage value will cycle from 0.0 to 1.0 over the course of your time period. Multiply it by 100 to get a percentage from 0 to 100 percent. It also can be used to determine the number of votes counted at any time, based on the total number that will be cast (y).
Store this in a variable named vc:
var vc = Math.floor(y * pct);
The Math.floor() function rounds down to a whole number. Without it, you'd be reporting partial votes, which only happens here in Florida.
Next, decide three things about each ballot item you are faking:
1. The percentage of the vote the item should receive when all votes have been counted.
2. The amount this value should change from the start of counting to the end.
3. Whether the item should be gaining ground during the day or losing it.
The last two things add a bit of variability to the results. If the percentage did not change during the day, your results wouldn't be believable. In a real election, that kind of thing could call the legitimacy of a perceived winner into question, which must not be allowed to happen in a democracy, regardless of the actual vote totals (see Bush v. Gore).
Here's a statement that causes a candidate to start with 12 percent of the vote at the beginning of counting and end up with 17 percent:
var jvp = 12 + (5 * pct);
Here's a statement that causes a candidate to start with 40 percent of the vote and end up with 33 percent:
var kvp = 40 - (7 * pct);
Once you have calculated the current percentage of the vote each item has received, you can use the number of counted votes (vc) to reverse-engineer a vote total for that item: var jv = Math.floor(vc * (jvp / 100));
var kv = Math.floor(vc * (kvp / 100));
The percentages you have calculated will have 5 or more decimal places, which is going to be intimidating to some people. We can't have that. The following JavaScript function will format a number to only two decimal places (like currency): function format(amount)
{
var i = parseFloat(amount);
if(isNaN(i)) { i = 0.00; }
var minus = '';
if(i < 0) { minus = '-'; }
i = Math.abs(i);
i = parseInt((i + .005) * 100);
i = i / 100;
s = new String(i);
if(s.indexOf('.') < 0) { s += '.00'; }
if(s.indexOf('.') == (s.length - 2)) { s += '0'; }
s = minus + s;
return s;
}
To finish, display the values you've calculated in document.write() statements, using the format() method with percentages: document.write("<table border="0" cellpadding="6"><tr valign="top">");
document.write("<tr><td>GARY COLEMAN</td><td>" + gv + "</td><td>" + format(gvp) + "%</td></tr>");
document.write("<tr><td>LARRY FLYNT</td><td>" + lv + "</td><td>" + format(lvp) + "%</td></tr>"); document.write("</tr></table>");
That's all there is to it. Your visitors will appreciate the live election results, which they can reload all day long to see who wins.
Su voto es su voz! Abrazos no drogas!
This tutorial is dedicated to Diebold Election Systems. If your no-paper-trail audit-proof touch-screen ballot tabulating system isn't a Diebold, it doesn't count.
Excellent information on your blog. thank you for taking the time to share with us. Amazing insight you have on this. it's nice to find a website that details so much information about different artists.
Great articles and great layout. Your blog post deserves all of the positive feedback it's been getting.
belitung island tour package indonesia
With a little ingenuity, Javascript can do an awful lot. It's truly impressive if you think about it. This is a pretty cool little trick, thanks! If you need any tax preparation in or around Delaware, check us out. Thanks! | 1,348 | 5,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-05 | latest | en | 0.932643 |
http://dgtal.org/doc/nightly/structDGtal_1_1concepts_1_1CLinearAlgebraSolver.html | 1,500,950,829,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424960.67/warc/CC-MAIN-20170725022300-20170725042300-00459.warc.gz | 81,698,896 | 5,194 | DGtal 0.9.4beta
DGtal::concepts::CLinearAlgebraSolver< S, V, M > Struct Template Reference
#include <DGtal/math/linalg/CLinearAlgebraSolver.h>
Inheritance diagram for DGtal::concepts::CLinearAlgebraSolver< S, V, M >:
[legend]
Collaboration diagram for DGtal::concepts::CLinearAlgebraSolver< S, V, M >:
[legend]
typedef S Solver
typedef V Vector
typedef M Matrix
## Public Member Functions
BOOST_CONCEPT_ASSERT ((CLinearAlgebra< Vector, Matrix >))
BOOST_CONCEPT_USAGE (CLinearAlgebraSolver)
## Private Attributes
int status_return
const Solver const_solver
Solver solver
const Matrix a
const Vector y
x
## Detailed Description
### template<typename S, typename V, typename M> struct DGtal::concepts::CLinearAlgebraSolver< S, V, M >
Aim: Describe a linear solver defined over a linear algebra. Problems are of the form:
Description of concept 'CLinearAlgebraSolver'
a * x = y
where a type is a model of CMatrix and x, y type is a model of CVector. Matrix and vector types should be a model of CLinearAlgebra.
### Refinement of
• boost::DefaultConstructible<S>
• CLinearAlgebra<V, M>
### Notation
• Solver : A type that is a model of CLinearAlgebraSolver
• Vector : A type that is a model of CVector
• Matrix : A type that is a model of CMatrix
• Info : The type returned by solver.status(), problem solved successfully if true when compared to 0
• y : object of type Vector, input of the linear problem
• x : object of type Vector, solution of the linear problem
• a : object of type Matrix
• info : object of type Info
### Valid expressions and semantics
Name Expression Type requirements Return type Precondition Semantics Post condition Complexity
Problem factorization / matrix input solver.compute(a) return *this.
Problem resolution / vector input x = solver.solve(y) Vector
Status solver.status()
### Notes
Template Parameters
S the type that should be a model of CLinearAlgebraSolver. V the type that should be a model of CVector M the type that should be a model of CMatrix
Definition at line 103 of file CLinearAlgebraSolver.h.
## Member Typedef Documentation
template<typename S , typename V , typename M >
typedef M DGtal::concepts::CLinearAlgebraSolver< S, V, M >::Matrix
Definition at line 109 of file CLinearAlgebraSolver.h.
template<typename S , typename V , typename M >
typedef S DGtal::concepts::CLinearAlgebraSolver< S, V, M >::Solver
Definition at line 107 of file CLinearAlgebraSolver.h.
template<typename S , typename V , typename M >
typedef V DGtal::concepts::CLinearAlgebraSolver< S, V, M >::Vector
Definition at line 108 of file CLinearAlgebraSolver.h.
## Member Function Documentation
template<typename S , typename V , typename M >
DGtal::concepts::CLinearAlgebraSolver< S, V, M >::BOOST_CONCEPT_ASSERT ( (CLinearAlgebra< Vector, Matrix >) )
template<typename S , typename V , typename M >
DGtal::concepts::CLinearAlgebraSolver< S, V, M >::BOOST_CONCEPT_USAGE ( CLinearAlgebraSolver< S, V, M > )
inline
Definition at line 113 of file CLinearAlgebraSolver.h.
114 {
115 solver.compute(a);
116 status_return = static_cast<int>(const_solver.info());
117 x = const_solver.solve(y);
118 }
## Field Documentation
template<typename S , typename V , typename M >
const Matrix DGtal::concepts::CLinearAlgebraSolver< S, V, M >::a
private
Definition at line 124 of file CLinearAlgebraSolver.h.
template<typename S , typename V , typename M >
const Solver DGtal::concepts::CLinearAlgebraSolver< S, V, M >::const_solver
private
Definition at line 122 of file CLinearAlgebraSolver.h.
template<typename S , typename V , typename M >
Solver DGtal::concepts::CLinearAlgebraSolver< S, V, M >::solver
private
Definition at line 123 of file CLinearAlgebraSolver.h.
template<typename S , typename V , typename M >
int DGtal::concepts::CLinearAlgebraSolver< S, V, M >::status_return
private
Definition at line 121 of file CLinearAlgebraSolver.h.
template<typename S , typename V , typename M >
V DGtal::concepts::CLinearAlgebraSolver< S, V, M >::x
private
Definition at line 126 of file CLinearAlgebraSolver.h.
template<typename S , typename V , typename M >
const Vector DGtal::concepts::CLinearAlgebraSolver< S, V, M >::y
private
Definition at line 125 of file CLinearAlgebraSolver.h.
The documentation for this struct was generated from the following file: | 1,117 | 4,336 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-30 | latest | en | 0.628561 |
https://www.sarthaks.com/66129/an-electron-having-kinetic-energy-100ev-circulates-path-of-radius-10cm-in-magnetic-field | 1,675,197,064,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00846.warc.gz | 1,000,200,047 | 15,532 | # An electron having a kinetic energy of 100eV circulates in a path of radius 10cm in a magnetic field.
29.5k views
in Physics
An electron having a kinetic energy of 100eV circulates in a path of radius 10cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.
by (148k points)
selected
KE = 100ev = 1.6 × 10–17J
(1/2) × 9.1 × 10–31 × V2 = 1.6 × 10–17J
=> V2 = (1.6 x 10-17 x 2)/(9.1 x 10-31) = 0.35 x 1014
or, V = 0.591 × 107m/s
+1 vote | 187 | 500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-06 | latest | en | 0.84592 |
https://www.lmfdb.org/EllipticCurve/Q/10830t2/ | 1,627,948,164,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00456.warc.gz | 903,342,990 | 26,438 | # Properties
Label 10830t2 Conductor $10830$ Discriminant $-1083000000$ j-invariant $$\frac{4728305591}{3000000}$$ CM no Rank $1$ Torsion structure trivial
# Related objects
Show commands: Magma / Pari/GP / SageMath
## Minimal Weierstrass equation
sage: E = EllipticCurve([1, 1, 1, 249, 573])
gp: E = ellinit([1, 1, 1, 249, 573])
magma: E := EllipticCurve([1, 1, 1, 249, 573]);
$$y^2+xy+y=x^3+x^2+249x+573$$
## Mordell-Weil group structure
$\Z$
### Infinite order Mordell-Weil generator and height
sage: E.gens()
magma: Generators(E);
$P$ = $$\left(21, 114\right)$$ $\hat{h}(P)$ ≈ $0.47096526362562874750114416704$
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
$$\left(-1, 18\right)$$, $$\left(-1, -18\right)$$, $$\left(21, 114\right)$$, $$\left(21, -136\right)$$
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$10830$$ = $2 \cdot 3 \cdot 5 \cdot 19^{2}$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $-1083000000$ = $-1 \cdot 2^{6} \cdot 3 \cdot 5^{6} \cdot 19^{2}$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$\frac{4728305591}{3000000}$$ = $2^{-6} \cdot 3^{-1} \cdot 5^{-6} \cdot 17^{3} \cdot 19 \cdot 37^{3}$ Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $0.42258177217288665863849864042\dots$ Stable Faltings height: $-0.068158057688186751363005931561\dots$
## BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $1$ sage: E.regulator() magma: Regulator(E); Regulator: $0.47096526362562874750114416704\dots$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $0.96453230989723641458301115886\dots$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $12$ = $( 2 \cdot 3 )\cdot1\cdot2\cdot1$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $1$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $1$ (exact) sage: r = E.rank(); sage: E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: ar = ellanalyticrank(E); gp: ar[2]/factorial(ar[1]) magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12); Special value: $L'(E,1)$ ≈ $5.4511345632742631031919001233048643147$
## Modular invariants
Modular form 10830.2.a.s
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q + q^{2} - q^{3} + q^{4} - q^{5} - q^{6} - q^{7} + q^{8} + q^{9} - q^{10} + 6q^{11} - q^{12} - 5q^{13} - q^{14} + q^{15} + q^{16} + q^{18} + O(q^{20})$$
For more coefficients, see the Downloads section to the right.
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 7776 $\Gamma_0(N)$-optimal: no Manin constant: 1
## Local data
This elliptic curve is not semistable. There are 4 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $6$ $I_{6}$ Split multiplicative -1 1 6 6
$3$ $1$ $I_{1}$ Non-split multiplicative 1 1 1 1
$5$ $2$ $I_{6}$ Non-split multiplicative 1 1 6 6
$19$ $1$ $II$ Additive -1 2 2 0
## Galois representations
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The $\ell$-adic Galois representation has maximal image $\GL(2,\Z_\ell)$ for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$3$ 3B 3.4.0.1
## $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
$p$-adic regulators are not yet computed for curves that are not $\Gamma_0$-optimal.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 split nonsplit nonsplit ordinary ordinary ordinary ss add ordinary ordinary ordinary ordinary ordinary ordinary ordinary 2 1 1 1 1 1 1,1 - 1 1 1 1 3 1 1 0 0 0 0 0 0 0,0 - 0 0 0 0 0 0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 3.
Its isogeny class 10830t consists of 2 curves linked by isogenies of degree 3.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ (which is trivial) are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $2$ $$\Q(\sqrt{57})$$ $$\Z/3\Z$$ Not in database $3$ 3.1.1083.1 $$\Z/2\Z$$ Not in database $6$ 6.0.3518667.2 $$\Z/2\Z \times \Z/2\Z$$ Not in database $6$ 6.0.146211169851.2 $$\Z/3\Z$$ Not in database $6$ 6.2.66854673.1 $$\Z/6\Z$$ Not in database $12$ Deg 12 $$\Z/4\Z$$ Not in database $12$ Deg 12 $$\Z/3\Z \times \Z/3\Z$$ Not in database $12$ 12.0.4469547301936929.2 $$\Z/2\Z \times \Z/6\Z$$ Not in database $18$ 18.6.9376978291960795596910212756606153000000000000.3 $$\Z/9\Z$$ Not in database $18$ 18.0.28130934875882386790730638269818459.3 $$\Z/6\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial. | 2,132 | 5,563 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-31 | latest | en | 0.205936 |
http://www.srcmini.com/68227.html | 1,618,312,939,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072180.33/warc/CC-MAIN-20210413092418-20210413122418-00461.warc.gz | 178,007,112 | 16,858 | # 快速排序中的Hoare vs Lomuto分区方案详细介绍
## 本文概述
Lomuto的分区方案：
``````partition(arr[], lo, hi)
pivot = arr[hi]
i = lo // place for swapping
for j := lo to hi – 1 do
if arr[j] <= pivot then
swap arr[i] with arr[j]
i = i + 1
swap arr[i] with arr[hi]
return i``````
## C ++
``````/* C++ implementation QuickSort using Lomuto's partition
Scheme.*/
#include<bits/stdc++.h>
using namespace std;
/* This function takes last element as pivot, places
the pivot element at its correct position in sorted
array, and places all smaller (smaller than pivot)
to left of pivot and all greater elements to right
of pivot */
int partition( int arr[], int low, int high)
{
int pivot = arr[high]; // pivot
int i = (low - 1); // Index of smaller element
for ( int j = low; j <= high- 1; j++)
{
// If current element is smaller than or
// equal to pivot
if (arr[j] <= pivot)
{
i++; // increment index of smaller element
swap(arr[i], arr[j]);
}
}
swap(arr[i + 1], arr[high]);
return (i + 1);
}
/* The main function that implements QuickSort
arr[] --> Array to be sorted, low --> Starting index, high --> Ending index */
void quickSort( int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[p] is now
at right place */
int pi = partition(arr, low, high);
// Separately sort elements before
// partition and after partition
quickSort(arr, low, pi - 1);
quickSort(arr, pi + 1, high);
}
}
/* Function to print an array */
void printArray( int arr[], int size)
{
int i;
for (i=0; i < size; i++)
printf ( "%d " , arr[i]);
printf ( "\n" );
}
// Driver program to test above functions
int main()
{
int arr[] = {10, 7, 8, 9, 1, 5};
int n = sizeof (arr)/ sizeof (arr[0]);
quickSort(arr, 0, n-1);
printf ( "Sorted array: \n" );
printArray(arr, n);
return 0;
}``````
## Java
``````// Java implementation QuickSort
// using Lomuto's partition Scheme
import java.io.*;
class GFG
{
static void Swap( int [] array, int position1, int position2)
{
// Swaps elements in an array
// Copy the first position's element
int temp = array[position1];
// Assign to the second element
array[position1] = array[position2];
// Assign to the first element
array[position2] = temp;
}
/* This function takes last element as
pivot, places the pivot element at its
correct position in sorted array, and
places all smaller (smaller than pivot)
to left of pivot and all greater elements
to right of pivot */
static int partition( int []arr, int low, int high)
{
int pivot = arr[high];
// Index of smaller element
int i = (low - 1 );
for ( int j = low; j <= high- 1 ; j++)
{
// If current element is smaller
// than or equal to pivot
if (arr[j] <= pivot)
{
i++; // increment index of
// smaller element
Swap(arr, i, j);
}
}
Swap(arr, i + 1 , high);
return (i + 1 );
}
/* The main function that
implements QuickSort
arr[] --> Array to be sorted, low --> Starting index, high --> Ending index */
static void quickSort( int []arr, int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[p] is now at right place */
int pi = partition(arr, low, high);
// Separately sort elements before
// partition and after partition
quickSort(arr, low, pi - 1 );
quickSort(arr, pi + 1 , high);
}
}
/* Function to print an array */
static void printArray( int []arr, int size)
{
int i;
for (i = 0 ; i < size; i++)
System.out.print( " " + arr[i]);
System.out.println();
}
// Driver Code
static public void main (String[] args)
{
int []arr = { 10 , 7 , 8 , 9 , 1 , 5 };
int n = arr.length;
quickSort(arr, 0 , n- 1 );
System.out.println( "Sorted array: " );
printArray(arr, n);
}
}
// This code is contributed by vt_m.``````
## Python3
``````''' Python3 implementation QuickSort using Lomuto's partition
Scheme.'''
''' This function takes last element as pivot, places
the pivot element at its correct position in sorted
array, and places all smaller (smaller than pivot)
to left of pivot and all greater elements to right
of pivot '''
def partition(arr, low, high):
# pivot
pivot = arr[high]
# Index of smaller element
i = (low - 1 )
for j in range (low, high):
# If current element is smaller than or
# equal to pivot
if (arr[j] < = pivot):
# increment index of smaller element
i + = 1
arr[i], arr[j] = arr[j], arr[i]
arr[i + 1 ], arr[high] = arr[high], arr[i + 1 ]
return (i + 1 )
''' The main function that implements QuickSort
arr --> Array to be sorted, low --> Starting index, high --> Ending index '''
def quickSort(arr, low, high):
if (low < high):
''' pi is partitioning index, arr[p] is now
at right place '''
pi = partition(arr, low, high)
# Separately sort elements before
# partition and after partition
quickSort(arr, low, pi - 1 )
quickSort(arr, pi + 1 , high)
''' Function to pran array '''
def printArray(arr, size):
for i in range (size):
print (arr[i], end = " " )
print ()
# Driver code
arr = [ 10 , 7 , 8 , 9 , 1 , 5 ]
n = len (arr)
quickSort(arr, 0 , n - 1 )
print ( "Sorted array:" )
printArray(arr, n)
# This code is contributed by SHUBHAMSINGH10``````
## C#
``````// C# implementation QuickSort
// using Lomuto's partition Scheme
using System;
class GFG
{
static void Swap( int [] array, int position1, int position2)
{
// Swaps elements in an array
// Copy the first position's element
int temp = array[position1];
// Assign to the second element
array[position1] = array[position2];
// Assign to the first element
array[position2] = temp;
}
/* This function takes last element as
pivot, places the pivot element at its
correct position in sorted array, and
places all smaller (smaller than pivot)
to left of pivot and all greater elements
to right of pivot */
static int partition( int []arr, int low, int high)
{
int pivot = arr[high];
// Index of smaller element
int i = (low - 1);
for ( int j = low; j <= high- 1; j++)
{
// If current element is smaller
// than or equal to pivot
if (arr[j] <= pivot)
{
i++; // increment index of
// smaller element
Swap(arr, i, j);
}
}
Swap(arr, i + 1, high);
return (i + 1);
}
/* The main function that
implements QuickSort
arr[] --> Array to be sorted, low --> Starting index, high --> Ending index */
static void quickSort( int []arr, int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[p] is now at right place */
int pi = partition(arr, low, high);
// Separately sort elements before
// partition and after partition
quickSort(arr, low, pi - 1);
quickSort(arr, pi + 1, high);
}
}
/* Function to print an array */
static void printArray( int []arr, int size)
{
int i;
for (i = 0; i < size; i++)
Console.Write( " " + arr[i]);
Console.WriteLine();
}
// Driver Code
static public void Main()
{
int []arr = {10, 7, 8, 9, 1, 5};
int n = arr.Length;
quickSort(arr, 0, n-1);
Console.WriteLine( "Sorted array: " );
printArray(arr, n);
}
}
// This code is contributed by vt_m.``````
``````Sorted array:
1 5 7 8 9 10``````
Hoare的分区方案：
``````partition(arr[], lo, hi)
pivot = arr[lo]
i = lo - 1 // Initialize left index
j = hi + 1 // Initialize right index
// Find a value in left side greater
// than pivot
do
i = i + 1
while arr[i] < pivot
// Find a value in right side smaller
// than pivot
do
j--;
while (arr[j] > pivot);
if i >= j then
return j
swap arr[i] with arr[j]``````
## C ++
``````/* C++ implementation of QuickSort using Hoare's
partition scheme. */
#include <bits/stdc++.h>
using namespace std;
/* This function takes first element as pivot, and places
all the elements smaller than the pivot on the left side
and all the elements greater than the pivot on
the right side. It returns the index of the last element
on the smaller side*/
int partition( int arr[], int low, int high)
{
int pivot = arr[low];
int i = low - 1, j = high + 1;
while ( true ) {
// Find leftmost element greater than
// or equal to pivot
do {
i++;
} while (arr[i] < pivot);
// Find rightmost element smaller than
// or equal to pivot
do {
j--;
} while (arr[j] > pivot);
// If two pointers met.
if (i >= j)
return j;
swap(arr[i], arr[j]);
}
}
/* The main function that implements QuickSort
arr[] --> Array to be sorted, low --> Starting index, high --> Ending index */
void quickSort( int arr[], int low, int high)
{
if (low < high) {
/* pi is partitioning index, arr[p] is now
at right place */
int pi = partition(arr, low, high);
// Separately sort elements before
// partition and after partition
quickSort(arr, low, pi);
quickSort(arr, pi + 1, high);
}
}
/* Function to print an array */
void printArray( int arr[], int n)
{
for ( int i = 0; i < n; i++)
printf ( "%d " , arr[i]);
printf ( "\n" );
}
// Driver Code
int main()
{
int arr[] = { 10, 7, 8, 9, 1, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
quickSort(arr, 0, n - 1);
printf ( "Sorted array: \n" );
printArray(arr, n);
return 0;
}``````
## Java
``````// Java implementation of QuickSort
// using Hoare's partition scheme
import java.io.*;
class GFG {
/* This function takes first element as pivot, and
places all the elements smaller than the pivot on the
left side and all the elements greater than the pivot
on the right side. It returns the index of the last
element on the smaller side*/
static int partition( int [] arr, int low, int high)
{
int pivot = arr[low];
int i = low - 1 , j = high + 1 ;
while ( true ) {
// Find leftmost element greater
// than or equal to pivot
do {
i++;
} while (arr[i] < pivot);
// Find rightmost element smaller
// than or equal to pivot
do {
j--;
} while (arr[j] > pivot);
// If two pointers met.
if (i >= j)
return j;
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
// swap(arr[i], arr[j]);
}
}
/* The main function that
implements QuickSort
arr[] --> Array to be sorted, low --> Starting index, high --> Ending index */
static void quickSort( int [] arr, int low, int high)
{
if (low < high) {
/* pi is partitioning index, arr[p] is now at right place */
int pi = partition(arr, low, high);
// Separately sort elements before
// partition and after partition
quickSort(arr, low, pi);
quickSort(arr, pi + 1 , high);
}
}
/* Function to print an array */
static void printArray( int [] arr, int n)
{
for ( int i = 0 ; i < n; i++)
System.out.print( " " + arr[i]);
System.out.println();
}
// Driver Code
static public void main(String[] args)
{
int [] arr = { 10 , 7 , 8 , 9 , 1 , 5 };
int n = arr.length;
quickSort(arr, 0 , n - 1 );
System.out.println( "Sorted array: " );
printArray(arr, n);
}
}
// This code is contributed by vt_m.``````
## Python3
``````''' Python implementation of QuickSort using Hoare's
partition scheme. '''
''' This function takes first element as pivot, and places
all the elements smaller than the pivot on the left side
and all the elements greater than the pivot on
the right side. It returns the index of the last element
on the smaller side '''
def partition(arr, low, high):
pivot = arr[low]
i = low - 1
j = high + 1
while ( True ):
# Find leftmost element greater than
# or equal to pivot
i + = 1
while (arr[i] < pivot):
i + = 1
# Find rightmost element smaller than
# or equal to pivot
j - = 1
while (arr[j] > pivot):
j - = 1
# If two pointers met.
if (i > = j):
return j
arr[i], arr[j] = arr[j], arr[i]
''' The main function that implements QuickSort
arr --> Array to be sorted, low --> Starting index, high --> Ending index '''
def quickSort(arr, low, high):
''' pi is partitioning index, arr[p] is now
at right place '''
if (low < high):
pi = partition(arr, low, high)
# Separately sort elements before
# partition and after partition
quickSort(arr, low, pi)
quickSort(arr, pi + 1 , high)
''' Function to pran array '''
def printArray(arr, n):
for i in range (n):
print (arr[i], end = " " )
print ()
# Driver code
arr = [ 10 , 7 , 8 , 9 , 1 , 5 ]
n = len (arr)
quickSort(arr, 0 , n - 1 )
print ( "Sorted array:" )
printArray(arr, n)
# This code is contributed by shubhamsingh10``````
## C#
``````// C# implementation of QuickSort
// using Hoare's partition scheme
using System;
class GFG {
/* This function takes first element as pivot, and
places all the elements smaller than the pivot on the
left side and all the elements greater than the pivot
on the right side. It returns the index of the last
element on the smaller side*/
static int partition( int [] arr, int low, int high)
{
int pivot = arr[low];
int i = low - 1, j = high + 1;
while ( true ) {
// Find leftmost element greater
// than or equal to pivot
do {
i++;
} while (arr[i] < pivot);
// Find rightmost element smaller
// than or equal to pivot
do {
j--;
} while (arr[j] > pivot);
// If two pointers met.
if (i >= j)
return j;
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
// swap(arr[i], arr[j]);
}
}
/* The main function that
implements QuickSort
arr[] --> Array to be sorted, low --> Starting index, high --> Ending index */
static void quickSort( int [] arr, int low, int high)
{
if (low < high) {
/* pi is partitioning index, arr[p] is now at right place */
int pi = partition(arr, low, high);
// Separately sort elements before
// partition and after partition
quickSort(arr, low, pi);
quickSort(arr, pi + 1, high);
}
}
/* Function to print an array */
static void printArray( int [] arr, int n)
{
for ( int i = 0; i < n; i++)
Console.Write( " " + arr[i]);
Console.WriteLine();
}
// Driver Code
static public void Main()
{
int [] arr = { 10, 7, 8, 9, 1, 5 };
int n = arr.Length;
quickSort(arr, 0, n - 1);
Console.WriteLine( "Sorted array: " );
printArray(arr, n);
}
}
// This code is contributed by vt_m.``````
``````Sorted array:
1 5 7 8 9 10``````
1. Hoare的方案比Lomuto的分区方案效率更高, 因为它的交换次数平均减少了三倍, 即使所有值都相等, 它也可以创建有效的分区。
2. 与Lomuto的分区方案一样, Hoare分区也会使输入数组已经排序后, 快速排序降级为O(n ^ 2), 也不会产生稳定的排序。
3. 请注意, 在此方案中, 枢轴的最终位置不一定在返回的索引处, 并且主要算法重复出现的下两个段分别是(lo..p)和(p + 1..hi) (lo..p-1)和(p + 1..hi), 就像Lomuto的方案一样。
• 回顶 | 4,083 | 13,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-17 | longest | en | 0.371942 |
https://www.tutorialspoint.com/p-what-are-vertically-opposite-angles-and-what-is-their-relation-p | 1,702,147,440,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00500.warc.gz | 1,131,073,162 | 21,133 | # What are Vertically opposite angles and what is their relation?
Vertically Opposite Angles:
When two lines intersect, we get four angles. The angles are two pairs of opposite angels and those angles are known as vertically opposite angles.
In the above figure, ∠1, ∠3 and ∠4, ∠2 are vertically opposite angles.
Vertically opposite angles are equal.
∠1 = ∠3 and ∠2 = ∠4.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
39 Views | 116 | 449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-50 | latest | en | 0.919315 |
https://howmanyis.com/speed/15-m-s-in-mph/6008-8-meters-per-second-in-miles-per-hour | 1,579,427,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594391.21/warc/CC-MAIN-20200119093733-20200119121733-00351.warc.gz | 490,055,065 | 6,224 | How many is
Conversion between units of measurement
Rating 5.00 (1 Vote)
You can easily convert 8 meters per second into miles per hour using each unit definition:
Meters per second
1 m / s
Miles per hour
mile/hr = 0.44704 m / s
With this information, you can calculate the quantity of miles per hour 8 meters per second is equal to.
## ¿How many mph are there in 8 m/s?
In 8 m/s there are 17.89549 mph.
Which is the same to say that 8 meters per second is 17.89549 miles per hour.
Eight meters per second equals to seventeen miles per hour. *Approximation
### ¿What is the inverse calculation between 1 mile per hour and 8 meters per second?
Performing the inverse calculation of the relationship between units, we obtain that 1 mile per hour is 0.05588 times 8 meters per second.
A mile per hour is zero times eight meters per second. *Approximation | 218 | 861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-05 | latest | en | 0.872741 |
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rp6 Total Posts: 14 Joined: Jun 2012
Posted: 2012-06-16 23:08 Ola, I am not a quant nor good at quant stuff. But here goes.So if standard BS models use assume a normal/lognormal distribution of returns, which we know to be slightly incorrect - fatter tails etc. Then is there a model which takes the past data for an asset, defines its distribution and uses that rather than ~N in the calculation. I guess computationally that's a lot harder, but is that thinking a part of some other model, or does it not make logical sense?
dgn2 Total Posts: 1906 Joined: May 2004
Posted: 2012-06-17 03:48 It is quite simple to resample the data and compute option prices - particularly European options - via Monte Carlo. Search for 'bootstrapping' and option pricing. One of the problems with this approach is that the pricing of options is only a small part of what makes BS useful. Sensitivities are also very important and using the actual distribution doesn't help there. In general, it is much easier to determine the impact of different assumptions on pricing when computing prices via Monte Carlo. Expanding to fatter tails isn't difficult, but what we know about the tails is limited, so how do you choose how fat they should be (or what form they should take)? Part of what makes BS useful is that there is really only one unknown parameter (i.e., volatility). Using a distribution with fatter tails adds more unknown parameters. You might want to look at Heston and variance gamma models. ...WARNING: I am an optimal f'er
rp6 Total Posts: 14 Joined: Jun 2012
Posted: 2012-06-17 18:21 Excellent, thanks for your help i will try and look into these some more. I see your point wrt the simplicity of just having to change volatility.
dgn2 Total Posts: 1906 Joined: May 2004
Posted: 2012-06-17 19:32 There are also works where a Kernal density smoothed estimate of the historical distribution is used to compute odds and price options, so you could search in that space as well. ...WARNING: I am an optimal f'er
rp6 Total Posts: 14 Joined: Jun 2012
Posted: 2012-07-06 12:53 If it's OK i'm going to continue to ask stupid questions about pricing models here rather than starting a new thread. All my knowledge on options comes from practical books like Natenberg/Baird and i have no background in financial mathematics so i can't understand a lot of the language to do with modelling techniques.Any info much appreciated:1. Day Count. In Natenberg's book he says that actual days should be used for interest calculations but only trading days for vol, since a stock can only move on those days. But then he says that the pricing model will be OK with only using one input for time to maturity. Why is this?2. Black 76 model. For an option on a future, where both the option and the future are subject to mark-to-market futures style margin, why is the interest rate a feature of the model at all. If there is no carry cost on the future or the option, then what is the r discounting?more to follow
granchio Total Posts: 1415 Joined: Apr 2004
Posted: 2012-07-06 16:14 <<1. Day Count. In Natenberg's book he says that actual days should be used for interest calculations but only trading days for vol, since a stock can only move on those days. >>Natemberg is not holy scriptures. Think for yourself: is it true that stock prices don't move when the market is closed? e.g. is it true that the open price tomorrow morning will be same as close tonight? is it true that open on monday is same as close on friday?<<2. Black 76 model. For an option on a future, where both the option and the future are subject to mark-to-market futures style margin, why is the interest rate a feature of the model at all. If there is no carry cost on the future or the option, then what is the r discounting?>>I don't think you need r at all in this case "Deserve got nothing to do with it" - Clint
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This example of PDE I found in a textbook and would like to know is there any way to solve it using DirichletCondition or NeumannValue or some other way. eq = {50*Derivative[2, 0][T][x, t] == 2423750*Derivative[0, 1][T][x, t]}; inc = {Derivative[1, 0][T][0, t] == 0, Derivative[1, 0][T][1, t] == 5863/10 - 2*T[1, t], T[x, 0] == 9463/20}; NDSolve[Join[eq, inc], T, {x, 0, 1}, {t, 0, 20}] // Flatten NDSolve::ibcinc: Warning: boundary and initial conditions are inconsistent. I am grateful for any tips. Sinval
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Posted 2 years ago
In this problem, it is necessary to coordinate the initial and boundary conditions, as well as use the solution method and increase the time interval to 200 for clarity. eq = { 2423750 Derivative[0, 1][T][x, t] - 50 Derivative[2, 0][T][x, t] == 0}; bc = {Derivative[1, 0][T][0, t] == 0, Derivative[1, 0][T][1, t] == (5863/10 - 2*T[1, t]) (1 - Exp[-10 t])}; ic = T[x, 0] == 9463/20; sol = NDSolveValue[{eq, ic, bc}, T, {x, 0, 1}, {t, 0, 200}, Method -> {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MinPoints" -> 40, "MaxPoints" -> 100, "DifferenceOrder" -> "Pseudospectral"}}, MaxSteps -> 10^6] {Plot3D[Re[sol[x, t]], {x, 0., 1}, {t, 0, 200}, Mesh -> None, ColorFunction -> Hue, PlotRange -> {450, 500}, AxesLabel -> Automatic], Plot3D[Re[sol[x, t]], {x, 0.9, 1}, {t, 0, 200}, Mesh -> None, ColorFunction -> Hue, PlotRange -> {450, 500}, AxesLabel -> Automatic]}
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Posted 2 years ago
Thank you for the tips.
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A series can be represented in a compact form, called summation or sigma notation. The Greek capital letter, ∑ , is used to represent the sum. The series 4+8+12+16+20+24 can be expressed as 6∑n=14n . The expression is read as the sum of 4n as n goes from 1 to 6 .
Also, What sigma means?
Sigma ist nicht nur der 18. Buchstabe des griechischen Alphabets, sondern bedeutet auch “Summe’ und ‘Abweichung“ in der Welt der Mathematik.
Hereof, What is sigma used for?
Simple sum
The symbol Σ (sigma) is generally used to denote a sum of multiple terms. This symbol is generally accompanied by an index that varies to encompass all terms that must be considered in the sum. For example, the sum of first whole numbers can be represented in the following manner: 1 2 3 ⋯.
Also to know What is sigma personality? What is a sigma personality? A sigma male personality is the type of personality that doesn’t need anything from anybody else. Sigma males are self-sufficient lone wolves who walk their own path in life. They don’t follow the crowd.
What is the use of sigma in trigonometry?
Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, Σ, to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma.
22 Verwandte Fragen Antworten gefunden
Inhaltsverzeichnis
## What is sigma wolf?
Ein Sigma-Männchen ist dein typischer einsamer wolf. He is independent, self-sufficient, confident, and strong enough to be Alpha. He lets possible mates come to him, just like he does with friends and career opportunities.
## What is a sigma woman?
The sigma female is a woman who demands attention when she walks into a room or is speaking, and she can be very intimidating to others. … The sigma female can demand attention when she feels like being in the spotlight, while also being able to step back and appear more approachable when it suits her.
## What is another word for sigma?
In this page you can discover 3 synonyms, antonyms, idiomatic expressions, and related words for sigma, like: DMAIC, dimensional and DFSS.
## Who is better Alpha or sigma?
Alpha male vs sigma male
Sigma and alpha males are equal and share many common characteristics. They’re both confident in their life choices and aim high. The main difference lies in their attitude. Sigmas choose to sit outside the hierarchy, while Alphas prefer to be at the top of it.
## What is a sigma Wolf?
Ein Sigma-Männchen ist dein typischer einsamer wolf. He is independent, self-sufficient, confident, and strong enough to be Alpha. He lets possible mates come to him, just like he does with friends and career opportunities.
## Wofür steht Six Sigma?
Six Sigma steht für 6 standard deviations (6σ) between avarage and acceptable limits. LSL and USL stand for “Lower Specification Limit” and “Upper Specification Limit” respectively. Specification Limits are derived from the customer requirements, and they specify the minimum and maximum acceptable limits of a process.
## Who is a sigma man?
Sigma male is a slang term used in masculinist subcultures for a popular, successful, but highly independent and self-reliant man. Ein anderer Begriff für ein Sigma-Männchen ist ein einsamer Wolf.
## How does summation work?
The summation sign, S, instructs us to sum the elements of a sequence. A typical element of the sequence which is being summed appears to the right of the summation sign. The variable of summation is represented by an index which is placed beneath the summation sign. The index is often represented by i.
## What is sigma Squared?
The symbol ‘σ’ represents the Bevölkerungsstandardabweichung. … The term ‘Σ ( Xi – μ )2‘ used in the statistical formula represents the sum of the squared deviations of the scores from their population mean.
## What does sigma mean in physics?
Greek characters
Name Bedeutung SI unit of measure
Sigma summation operator
Sigma area charge density coulomb per square meter (C/m
2
)
elektrische Leitfähigkeit siemens per meter (S/m)
normal stress Pascal (Pa)
## Sind Sigma-Männer attraktiv?
While the Alpha male initially seems desirable due to his ability to stand out, it is often the lone wolf, the introvert, and the rebel Sigma male who becomes attractive to women. … The Sigma male can be popular and likable, but he can’t be tied down to any of the traditional social structures for too long.
## What’s a female wolf called?
A female wolf is either called a She-wolf or a luna wolf, depending on the status of the female in the pack. The term “she-wolf” is sometimes used for female members of the pack. This name has no specific connotation and is used as a general term for female wolves.
## What is a Zeta in a wolf pack?
Zeta Werewolves, they are a variant of Beta Werewolves – Intelligent Betas, they specialize in Strategy and Coordination. … The Zeta Werewolves, as described in Kate Argent’s Hunting Diary, are “Intelligent Beta Werewolves” that are an Alpha’s “left hand” and specialize in both Strategy and Pack Coordination.
## What kind of woman does a sigma man want?
Sigma males are into special girls who know what they want and most importantly- have a life outside of their relationship. They want a woman who will give them enough space and who won’t display any clingy behaviour. But that doesn’t mean that you don’t have to be loyal.
## What is a sigma personality type?
Sie sind self-sufficient, seemingly strong, and capable of taking care of themselves. Being independent is key. Sigma males are independent without even trying. They may have a close friend or two, but these friendships likely offer up companionship and not necessity.
## What is a Zeta female?
Zeta Tau Alpha (known as ΖΤΑ or Zeta) is an international women’s fraternity founded on October 15, 1898 at the State Female Normal School (now Longwood University) in Farmville, Virginia. Its International Office is located in Carmel, Indiana.
## Where is the sigma in Word?
The symbol’s code: You can insert symbols by typing the symbol’s code and then pressing the Alt+X key combination. For example, the code for the sigma character is 2211: Type 2211 in your document and then press Alt+X. The number 2211 is magically transformed into the sigma character.
## Who invented sigma?
At Motorola in the 1980s, Mikel Harry and Bill Smith developed Six Sigma to make improvements on the manufacturing floor. Motorola was among the first recipients of the Malcolm Baldrige Award. Jack Welch later adopted Six Sigma at General Electric and saved billions of dollars putting the methodology to use.
## Is sigma a standard deviation?
Die Maßeinheit, die normalerweise angegeben wird, wenn von statistischer Signifikanz gesprochen wird, ist die Standardabweichung, ausgedrückt mit dem griechischen Kleinbuchstaben Sigma (σ). … Der Begriff bezieht sich auf das Ausmaß der Variabilität in einem bestimmten Datensatz: ob die Datenpunkte alle zusammen geclustert oder sehr verteilt sind. | 1,625 | 7,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-21 | latest | en | 0.876781 |
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舰船科学技术 2022, Vol. 44 Issue (5): 54-58 DOI: 10.3404/j.issn.1672-7649.2022.05.011 PDF
Performance prediction of propeller based on high-dimensional optimization RBF neural network
ZHU Jian-ping, LIU Xiao-xiao, WANG Guan-nan, YANG Sheng, LIU Yuan-hao, HAO Si-jia
Rongcheng College, Harbin University of Science and Technology, Weihai 264300, China
Abstract: The hydrodynamic performance of the underwater robot is determined by the thrust of the propeller-type underwater propeller. In order to predict the open water performance of the propeller quickly and accurately, an open water performance estimator model which is based on RBF neural network needs to be established. The network size is adjusted by using several types of propeller open water simulation values as training samples.On this basis, the connection weights between the networks have been adjusted, and the network parameters have also been optimized. After achieving the learning accuracy requirements through continuous iterative optimization, a high-dimensional optimized neural network open water performance estimator is finally obtained. By comparing and analyzing the open water coefficient predicted by the RBF neural network propeller open water performance estimator model and the open water coefficient simulated by CDF, the result shows that the gap between them is small. Therefore, the RBF neural network open water performance estimator model can meet the requirements of accuracy and rapidity of prediction, and can also be used as one of the effective prediction methods of propeller open water coefficient.
Key words: propeller RBF neural network performance prediction
0 引 言
1 RBF神经网络螺旋桨敞水性能估计器模型 1.1 RBF神经网络螺旋桨敞水性能估计器模型的建立
RBF神经网络是一种前反馈式神经网络，主要包括输入层、隐藏层、及输出层3层网格结构[4]。RBF径向基神经网络的隐节点之间采用中心向量与输入模式之间的距离来作为函数的自变量，并快速使用径向基函数作为一个激活函数。接着对网络规模进行调整，广义逆动态地调整各网络间的连接权重，进行网络参数的优化，进而构建一种高维优化的神经网络估计器，最终达到输出敞水能参数误差小于预先设置的学习精度，或者在达到预先设定的学习次数时，停止整个神经网络的学习训练过程，神经网络结构示意图如图1所示。
图 1 螺旋桨估计器RBF神经网络结构图 Fig. 1 RBF neural network structure diagram of propeller estimator
1）构建隐含层神经元宽度
${\delta _j} = {\delta _{\min }} + \left( {{\delta _{\max }} - {\delta _{\min }}} \right) \bullet rand\left( {} \right)\;\;j = 1,2, \cdots , \cup ,$ (1)
${\boldsymbol{X}} = {\left( {x_1^{\rm{T}},x_2^{\rm{T}}, \cdots ,x_M^{\rm{T}}} \right)^{\rm{T}}} \in {\Re ^{M \times n}}$ 表示训练集的输入集， ${{\boldsymbol{x}}_q} = {\left( {{x_{q1}},{x_{q2}}, \cdots ,{x_{qn}}} \right)^{\rm{T}}} \in {\Re ^{1 \times n}}$ 为输入级的第q个输入样本，M为训练集样本规模；令 $Y \in {\Re ^{M \times 1}}$ 为训练集的期望输出，建立RBF神经网络螺旋桨估计器：
2）构成RBF单元
3）计算输入层X下隐含层的输入H
${\phi _j}\left( x \right) = \exp \left( { - \dfrac{{\parallel x - {\mu _j}{\parallel ^2}}}{{2\delta _j^2}}} \right),j = 1,2, \cdots ,k_{sise}^i，$ (2)
$H = {\left( {\begin{array}{*{20}{c}} {{\phi _{11}}}& \ldots &{{\phi _{1M}}} \\ \vdots & \ddots & \vdots \\ {\phi k{{_{sise}^i}_1}}& \cdots &{{\phi _{k_{sise}^iM}}} \end{array}} \right)^{\rm{T}}}，$ (3)
$\begin{gathered} {\phi _{jq}} = \exp \left( { - \dfrac{{\parallel {x_q} - \mu _j^i{\parallel ^2}}}{{2\delta _j^2}}} \right),q = 1,2, \cdots ，\hfill \\ M\;and\;j = 1,2, \cdots ,k_{sise}^i。\hfill \\ \end{gathered}$ (4)
4）广义逆计算出各网格之间的连接权重 ${W_i}$
${{\boldsymbol{H}}^ + } = {\left( {{{\boldsymbol{H}}^{\rm{T}}}{\boldsymbol{H}} + \lambda I} \right)^{ - 1}}{{\boldsymbol{H}}^{\rm{T}}}，$
${W_i} = {{\boldsymbol{H}}^ + }Y。$
1.2 RBF神经网络训练
1.3 RBF神经网络螺旋桨敞水性能估计器模型训练结果
图 2 某型号螺旋桨敞水系数 Fig. 2 Open water coefficient of special type propeller
2 螺旋桨CFD敞水性能仿真计算 2.1 构造螺旋桨几何模型
图 3 管螺旋经 Fig. 3 Spiral diameter of catheter propeller
2.2 控制方程
1）连续性方程
2）动量守恒方程
$\left\{ {\begin{array}{*{20}{c}} {\rho \dfrac{{{\rm{d}}u}}{{{\rm{d}}t}} = \rho {f_x} - \dfrac{{\partial p}}{{\partial x}} + \mu {\nabla ^2}u}，\\ {\rho \dfrac{{{\rm{d}}v}}{{{\rm{d}}t}} = \rho {f_y} - \dfrac{{\partial p}}{{\partial y}} + \mu {\nabla ^2}v} ，\\ {\rho \dfrac{{{\rm{d}}w}}{{{\rm{d}}t}} = \rho {f_z} - \dfrac{{\partial p}}{{\partial z}} + \mu {\nabla ^2}w} 。\end{array}} \right.$ (5)
2.3 建立计算域
图 4 螺旋桨流体计算域 Fig. 4 Fluid computing domain of propeller
2.4 网格划分
图 5 螺旋桨网格 Fig. 5 Propeller grid
图 6 螺旋桨网格 Fig. 6 Propeller grid
2.5 设置边界条件
${V_A} = JnD$ (6)
2.6 计算结果验证
图 7 RBF预测与CFD仿真结果对比图 Fig. 7 Comparison chart between the prediction of RBF and the simulation result of CFD
3 结 语
[1] 王超, 黄胜, 解学参. 基于CFD方法的螺旋桨水动力性能预报[J]. 海军工程大学学报, 2008, 20(4): 107-112. WANG Chao, H UANG Sheng, XIE Xue-shen. Hydrodynamic performance prediction of some propeller based on CFD[J]. Journal of Naval University of Engineering, 2008, 20(4): 107-112. [2] 孙群, 范佘明. 基于BP神经网络的螺旋桨敞水性能数值预报的修正方法[C]// 第十二届中国国际船艇展暨高性能船学术报告会. 2007. [3] 翟鑫钰, 陆金桂. 基于神经网络的螺旋桨敞水性能预测[J10L]. 南京工业大学学报(自然科学版): 1-[12, 2022-3-27]. [4] 彭翔, 田中文, 何珍, 等. 不同湍流模型在螺旋桨敞水性能预报中的应用分析[J]. 舰船科学技术, 2019, 41(21): 46-49+53. PENG Xiang, TIAN Zhong-wen, HE Zhen, et al. Application and analysis of different turbulence models in the prediction of open water performance of a propeller[J]. Ship Science and Technology, 2019, 41(21): 46-49+53. [5] 张力为. 导管螺旋桨的空化及噪声性能数值分析[D]. 武汉: 武汉理工大学, 2019. [6] 孙承亮, 赵江滨. 基于CFD的对转螺旋桨水动力性能分析[J]. 舰船科学技术, 2019, 41(3): 36-40. [7] 李家盛, 张振果, 华宏星. 基于有限元和面元法的弹性螺旋桨流固耦合特性分析[J]. 振动与冲击, 2018, 37(21): 14-21. LI Jiasheng, ZHANG Zhenguo, HUA Hongxing. Hydro-elastic analysis for dynamic characteristics of marine propellers using finite element method and panel method[J]. Journal of Vibration and Shock, 2018, 37(21): 14-21. [8] 袁健, 丁晓阳. 基于数值计算的舰船可调螺距螺旋桨水动力性能分析[J]. 舰船科学技术, 2017, 39(18): 13-15. YUAN Jian, DING Xiao-yang. Hydrodynamic performance analysis of controllable pitch propeller based on numerical calculation[J]. Ship Science and Technology, 2017, 39(18): 13-15. [9] 王淑生, 王超, 仇宝云, 等. 基于CFD轴流泵叶轮在流场中的有限元分析[J]. 机械工程与自动化, 2016(6): 82-83+86. DOI:10.3969/j.issn.1672-6413.2016.06.033 [10] 李雪芹, 陈科, 刘刚. 基于ANSYS的复合材料螺旋桨叶片有限元建模与分析[J]. 复合材料学报, 2017, 34(4): 591-598. LI Xue-qin, CHEN Ke, LIU Gang. Finite element modeling and analysis of composite propeller blade based on ANSYS[J]. Acta Materiae Compositae Sinica, 2017, 34(4): 591-598. [11] 王本武, 李庆刚. 轴流泵叶片有限元分析[J]. 科协论坛(下半月), 2013(3): 21-23. | 2,643 | 6,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.725288 |
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# Divisability Rulesppt
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### Divisability Rulesppt
1. 1. Divisibility Rules 2 3 4 5 6 9 10
2. 2. What does it mean? "Divisible by . . ."
3. 3. <ul><li>If you divide one number by another, the result is a whole number WITHOUT a remainder. </li></ul><ul><li>Examples: </li></ul><ul><li>21 ÷ 7 = 3 No remainder </li></ul><ul><li>56 ÷ 2 = 28 No remainder </li></ul>"Divisible by" means . . .
4. 4. <ul><li>A number is divisible by 2 if the last digit is even . (0, 2, 4, 6, 8) </li></ul><ul><li>Examples: </li></ul><ul><li>6 4 49,82 2 </li></ul>Divisibility Rule for 2
5. 5. <ul><li>Which number IS NOT divisible by 2? </li></ul><ul><li>684 2788 103 </li></ul>Now, You Try!
6. 6. Wonderful !
7. 7. It ends in a O, 2, 4, 6 or 8
8. 8. <ul><li>A number is divisible by three if the sum of the digits is divisible by 3. </li></ul><ul><li>Examples: </li></ul><ul><li>81 8 + 1 = 9 9 ÷ 3 = 3 No Remainder </li></ul><ul><li>231 2 + 3 + 1 = 6 6 ÷ 3 = 2 No Remainder </li></ul>Divisibility Rule for 3
9. 9. <ul><li>Which number IS NOT divisible by 3? </li></ul><ul><li>642 1644 729 </li></ul>Now, You Try
10. 10. The sum is divisible by 3
11. 11. Wonderful !
12. 12. <ul><li>A number is divisible by 4 if the last two digits form a number that is divisible by 4. </li></ul><ul><li>Examples: </li></ul><ul><li>38 20 20 ÷ 4 = 5 </li></ul><ul><li>6 72 72 ÷ 4 = 18 </li></ul>Divisibility Rule for 4
13. 13. <ul><li>Which number IS NOT divisible by 4? </li></ul><ul><li>112 5610 488 </li></ul>Now You Try
14. 14. Wonderful !
15. 15. The last 2 digits form a number divisible by 4
16. 16. <ul><li>A number is divisible by 5 if the last digit is a 0 or 5 . </li></ul><ul><li>Examples: </li></ul><ul><li>32 5 ends in a 5 </li></ul><ul><li>347 0 ends on a 0 </li></ul>Divisibility Rule for 5
17. 17. <ul><li>Which number IS NOT divisible by 5? </li></ul><ul><li>6520 657 465 </li></ul>Now, You Try
18. 18. The number ends in 0 or 5
19. 19. Wonderful !
20. 20. <ul><li>A number is divisible by 6 when it is divisible by 2 AND 3 . (prime numbers) </li></ul><ul><li>Examples: </li></ul><ul><li>204 ends in an even number (2 rule) </li></ul><ul><li>sum of the digits is divisible by 3 </li></ul>Divisibility Rule for 6
21. 21. <ul><li>Which number IS NOT divisible by 6? </li></ul><ul><li>7800 427 5502 </li></ul>Now, You Try
22. 22. Way To Go !
23. 23. This number is divisible by 2 and 3
24. 24. <ul><li>A number is divisible by 9 if the sum of the digits is divisible by 9. </li></ul><ul><li>Examples: </li></ul><ul><li>351 3 + 5 + 1 = 9 9 ÷ 9 = 1 </li></ul><ul><li>3555 4 + 5 + 5 + 5 =18 18 ÷ 9 =2 </li></ul>Divisibility Rule for 9
25. 25. <ul><li>Find the number that IS NOT divisible by 9. </li></ul><ul><li>9873 693 5541 </li></ul>Now, You Try
26. 26. The sum of the digits is divisible by 9
27. 27. Great Job !
28. 28. <ul><li>A number is divisible by 10 is it ends in a ZERO. </li></ul><ul><li>Example: </li></ul><ul><li>456 0 ends in a zero </li></ul><ul><li> 431 0 ends in a zero </li></ul>Divisibility Rule for 10
29. 29. <ul><li>Find the number that IS NOT divisible by ten. </li></ul><ul><li>47,870 5892 458 </li></ul>Now, You Try
30. 30. The number ends in 0
31. 31. Way To Go !
32. 32. <ul><li>Grab a piece of paper and check how well you know the rules. </li></ul>Try It Out <ul><li>456 – 2 3 4 5 6 9 10 </li></ul><ul><li>801 – 2 3 4 5 6 9 10 </li></ul><ul><li>55,602 – 2 3 4 5 6 9 10 </li></ul><ul><li>51 – 2 3 4 5 6 9 10 </li></ul><ul><li>381 – 2 3 4 5 6 9 10 </li></ul>
33. 33. <ul><li>Grab a piece of paper and check how well you know the rules. </li></ul>Try It Out <ul><li>456 – 2 3 4 5 6 9 10 </li></ul><ul><li>801 – 2 3 4 5 6 9 10 </li></ul><ul><li>55,602 – 2 3 4 5 6 9 10 </li></ul><ul><li>51 – 2 3 4 5 6 9 10 </li></ul><ul><li>381 – 2 3 4 5 6 9 10 </li></ul> | 1,665 | 3,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-05 | latest | en | 0.492429 |
http://manchik.co.uk/problem/32010/1 | 1,624,359,207,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517048.78/warc/CC-MAIN-20210622093910-20210622123910-00621.warc.gz | 33,272,596 | 4,855 | #### EMEA Semifinal, 2008
Submit Solution(Code Jam Page)
Problem
You need to hire some people to paint a fence. The fence is composed of 10000 contiguous sections, numbered from 1 to 10000.
You get some offers from painters to help paint the fence. Each painter offers to paint a contiguous subset of fence sections in a particular color. You need to accept a set of the offers, such that:
• Each section of the fence is painted.
• At most 3 colors are used to paint the fence.
If it is possible to satisfy these two requirements, find the minimum number of offers that you must accept.
Input
• One line containing an integer T, the number of test cases in the input file.
For each test case, there will be:
• One line containing an integer N, the number of offers.
• N lines, one for each offer, each containing "C A B" where C is the color, which is an uppercase string of up to 10 letters, A is the first section and B is the last section to be painted. 1 ≤ AB ≤ 10000.
Output
• T lines, one for each test case in the order they occur in the input file, each containing the string "Case #X: Y", where X is the case number, and Y is the number of offers that need to be accepted, or "Case #X: IMPOSSIBLE" if there is no acceptable set of offers.
Limits
1 ≤ T ≤ 50
Small dataset
1 ≤ N ≤ 10
Large dataset
1 ≤ N ≤ 300
Sample
Input Output ``` 5 2 BLUE 1 5000 RED 5001 10000 3 BLUE 1 6000 RED 2000 8000 WHITE 7000 10000 4 BLUE 1 3000 RED 2000 5000 ORANGE 4000 8000 GREEN 7000 10000 2 BLUE 1 4000 RED 4002 10000 3 BLUE 1 6000 RED 4000 10000 ORANGE 3000 8000 ``` ``` Case #1: 2 Case #2: 3 Case #3: IMPOSSIBLE Case #4: IMPOSSIBLE Case #5: 2 ```
In the first test case, accepting both offers will exactly paint the whole fence, 5000 sections each, with no overlap.
In the second case, the painters will overlap, which is acceptable.
In the third case, accepting all four offers would cover the whole fence, but it would use 4 different colours, so this is not acceptable.
In the fourth case, section 4001 cannot be painted.
In the fifth case, we can accept just the first and second offer and successfully paint the fence.
Points Correct Attempted
7pt 190 199
13pt 113 144 | 607 | 2,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-25 | latest | en | 0.902469 |
https://www.freemathhelp.com/forum/threads/20p5-categorized-by-repeat-objects.93504/ | 1,550,731,591,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247500089.84/warc/CC-MAIN-20190221051342-20190221073342-00222.warc.gz | 856,269,066 | 12,608 | # 20P5 categorized by repeat objects
##### New member
Hello, I am hoping this is in the right sub group please do let me know if I chose incorrectly.
So I have part of this solved ( I think)
The basics:
Total population of objects :20
slots to fill: 5
total population of sets: 3.2MM (20^[SUP]5[/SUP])
My goal is to organize this total population into groups.
aaaaa - 5 of a kind
aaaab - 4 of a kind, 1 single
aaabb - 3 of a kind, 1 pair
aaabc - 3 of a kind, 2 singles
aabbc - 2 pairs, 1 single
aabcd - 1 pair, 3 singles
abcde - all singles
so using [SUB]n[/SUB]P[SUB]r[/SUB] I came up with the following:
category-count
aaaaa-20
aaaab-380
aaabc -6,840
aaabb-380
aabbc -6,840
aabcd -116,280
abcde -1,860,480
This is incomplete as it sums to 1,991,220.
the next step as I was seeing it was that there are variations of each category:
Example: aaaab,aaaba,aabaa and so on.
the plan was to calculate the permutations for each and then multiplying that by the initial calculations. but this yields much too large a population. so now I am stuck.
Here is the google sheets document I have been working with:
Any hints and help are greatly appreciated!
#### ksdhart
##### Full Member
You're on the right track, but you're missing many of the permutations. For instance, in the "4 of a kind, 1 single" you list only aaaab as an option. What about aaaac? Or aaaad? Similarly for "3 of a kind, 1 pair" you'd need to include aaacc, etc. When you include those, the numbers should work out right. Also, if you're only looking to have 20[SUP]5[/SUP] arrangements, then the order must not matter. aaaab is the same as aaaba because you're only interested in the fact that there are 4 a's and 1 b, not what position the b is in.
#### Ishuda
##### Elite Member
Hello, I am hoping this is in the right sub group please do let me know if I chose incorrectly.
So I have part of this solved ( I think)
The basics:
Total population of objects :20
slots to fill: 5
total population of sets: 3.2MM (20^[SUP]5[/SUP])
My goal is to organize this total population into groups.
aaaaa - 5 of a kind
aaaab - 4 of a kind, 1 single
aaabb - 3 of a kind, 1 pair
aaabc - 3 of a kind, 2 singles
aabbc - 2 pairs, 1 single
aabcd - 1 pair, 3 singles
abcde - all singles
so using [SUB]n[/SUB]P[SUB]r[/SUB] I came up with the following:
category-count
aaaaa-20
aaaab-380
aaabc -6,840
aaabb-380
aabbc -6,840
aabcd -116,280
abcde -1,860,480
This is incomplete as it sums to 1,991,220.
the next step as I was seeing it was that there are variations of each category:
Example: aaaab,aaaba,aabaa and so on.
the plan was to calculate the permutations for each and then multiplying that by the initial calculations. but this yields much too large a population. so now I am stuck.
Here is the google sheets document I have been working with:
Any hints and help are greatly appreciated!
I don't quite understand. I think what you are saying is you have a set of 20 distinct objects, i.e. something like
S={o[SUB]j[/SUB], j=1, 2, 3, ..., 20; o[SUB]i[/SUB]$$\displaystyle \ne$$o[SUB]j[/SUB] if i$$\displaystyle \ne$$j}
that is, for example,
S={a, b, c, ..., r, s, t}
and you want to form a group of sets with particular properties. Assuming this is true, first let
T = {1, 2, 3, ..., 20}
so I won't have to write out the full number set each time.
Now the groups (set of sets) you want are
s[SUB]1[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB]}, j $$\displaystyle \epsilon$$T} [thus 20 sets of type {a, a, a, a, a}]
s[SUB]2[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]i[/SUB]}, j $$\displaystyle \epsilon$$T; j $$\displaystyle \epsilon$$T-{j}} [4 of a kind]
s[SUB]3[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]i[/SUB], o[SUB]i[/SUB]}, j $$\displaystyle \epsilon$$T; j $$\displaystyle \epsilon$$T-{j}} [Full House]
etc.
In each of these sets, order doesn't matter. That is, for example, in s[SUB]2[/SUB] {a, a, a, b, b} would be considered the same as {b, a, a, a, b}
A page I recommend quite a bit for permutations and combinations is
https://www.mathsisfun.com/combinatorics/combinations-permutations.html
The whole page makes for an interesting read IMO but the particular section for this problem is "Combinations with Repetition"
##### New member
You're on the right track, but you're missing many of the permutations. For instance, in the "4 of a kind, 1 single" you list only aaaab as an option. What about aaaac? Or aaaad? Similarly for "3 of a kind, 1 pair" you'd need to include aaacc, etc. When you include those, the numbers should work out right. Also, if you're only looking to have 20[SUP]5[/SUP] arrangements, then the order must not matter. aaaab is the same as aaaba because you're only interested in the fact that there are 4 a's and 1 b, not what position the b is in.
Hello thank you for responding.
Two things;
One in my situation order does matter.
Two, the aaaab was not meant to represent that there are 4 of one kind one of the other.
so for example, given 20 objects and 5 slots in the form aaaab there are 380 sets. but there are 5 ways to organize a 4 of a kind plus one single. [5!/(4!*1)]
aaaab
aaaba
aabaa
abaaa
baaaa
so a total of 380*5=1900 sets.
the problem is that when you get the 1 pair and three single. doing [5!/(2!*1*1!*1!)]=60 doesn't work.
116,280*60=6976800 as that is larger then the entire population which is 3.2M
##### New member
I don't quite understand. I think what you are saying is you have a set of 20 distinct objects, i.e. something like
S={o[SUB]j[/SUB], j=1, 2, 3, ..., 20; o[SUB]i[/SUB]$$\displaystyle \ne$$o[SUB]j[/SUB] if i$$\displaystyle \ne$$j}
that is, for example,
S={a, b, c, ..., r, s, t}
and you want to form a group of sets with particular properties. Assuming this is true, first let
T = {1, 2, 3, ..., 20}
so I won't have to write out the full number set each time.
Now the groups (set of sets) you want are
s[SUB]1[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB]}, j $$\displaystyle \epsilon$$T} [thus 20 sets of type {a, a, a, a, a}]
s[SUB]2[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]i[/SUB]}, j $$\displaystyle \epsilon$$T; j $$\displaystyle \epsilon$$T-{j}} [4 of a kind]
s[SUB]3[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]i[/SUB], o[SUB]i[/SUB]}, j $$\displaystyle \epsilon$$T; j $$\displaystyle \epsilon$$T-{j}} [Full House]
etc.
In each of these sets, order doesn't matter. That is, for example, in s[SUB]2[/SUB] {a, a, a, b, b} would be considered the same as {b, a, a, a, b}
A page I recommend quite a bit for permutations and combinations is
https://www.mathsisfun.com/combinatorics/combinations-permutations.html
The whole page makes for an interesting read IMO but the particular section for this problem is "Combinations with Repetition"
Hello Ishuda,
Thank you for your reply, you are 100% correct other then that in my situation order does matter. I require the sum off all variations with the different sets ie aaaab,+aaaba, ect. | 2,234 | 7,071 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-09 | latest | en | 0.928081 |
https://gmatclub.com/forum/a-b-c-d-and-e-are-five-numbers-such-that-a-b-c-d-e-and-e-348987.html | 1,726,370,628,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00234.warc.gz | 246,362,249 | 132,545 | Last visit was: 14 Sep 2024, 20:23 It is currently 14 Sep 2024, 20:23
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# a, b, c, d, and e are five numbers such that a b c d e and e -
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Re: a, b, c, d, and e are five numbers such that a b c d e and e - [#permalink]
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TarunKumar1234
a, b, c, d, and e are five numbers such that a ≤ b ≤ c ≤ d ≤ e and e - c = 4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A > M?
We need to check, average (arithmetic mean) > median (M), if a, b, c, d, and e are equidistant or not.
Stat1: e + c = 34
So, e= 19 and c= 23, but no idea about a and b, they can be equidistant or not. Not sufficient
Stat2: c = a + 10 or, c-a =10, given, e - c = 4. So we can saw, a, b, c, d, and e are not equidistant. So, definitely, A <M. Sufficient.
So, I think B.
Hi TarunKumar1234,
A small correction here (calculation mistake),
From Statement 1, e = 19 and c = 15.
Regards,
Ravish.
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Re: a, b, c, d, and e are five numbers such that a b c d e and e - [#permalink]
Bunuel
a, b, c, d, and e are five numbers such that a ≤ b ≤ c ≤ d ≤ e and e - c = 4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A > M?
(1) e + c = 34
(2) c = a + 10
Project DS Butler Data Sufficiency (DS3)
Hi Bunuel,
A quick query on this question?
a, b, c, d, and e are five numbers....
Here, number means a whole number or a natural number or an integer or a decimal number or any of these?
Although I can see that this number thing doesn't matter in the final answer and doesn't affect our answer choice, I just want to understand for clarity.
TIA.
Regards,
Ravish.
VP
Joined: 14 Jul 2020
Posts: 1111
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Re: a, b, c, d, and e are five numbers such that a b c d e and e - [#permalink]
ravish844
TarunKumar1234
a, b, c, d, and e are five numbers such that a ≤ b ≤ c ≤ d ≤ e and e - c = 4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A > M?
We need to check, average (arithmetic mean) > median (M), if a, b, c, d, and e are equidistant or not.
Stat1: e + c = 34
So, e= 19 and c= 23, but no idea about a and b, they can be equidistant or not. Not sufficient
Stat2: c = a + 10 or, c-a =10, given, e - c = 4. So we can saw, a, b, c, d, and e are not equidistant. So, definitely, A <M. Sufficient.
So, I think B.
Hi TarunKumar1234,
A small correction here (calculation mistake),
From Statement 1, e = 19 and c = 15.
Regards,
Ravish.
Thanks! It is corrected.
Math Expert
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Re: a, b, c, d, and e are five numbers such that a b c d e and e - [#permalink]
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ravish844
Bunuel
a, b, c, d, and e are five numbers such that a ≤ b ≤ c ≤ d ≤ e and e - c = 4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A > M?
(1) e + c = 34
(2) c = a + 10
Project DS Butler Data Sufficiency (DS3)
Hi Bunuel,
A quick query on this question?
a, b, c, d, and e are five numbers....
Here, number means a whole number or a natural number or an integer or a decimal number or any of these?
Although I can see that this number thing doesn't matter in the final answer and doesn't affect our answer choice, I just want to understand for clarity.
TIA.
Regards,
Ravish.
It means real numbers, so it can be integer, fraction, or an irrational number.
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Re: a, b, c, d, and e are five numbers such that a b c d e and e - [#permalink]
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Re: a, b, c, d, and e are five numbers such that a b c d e and e - [#permalink]
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## Correlation Dashboard
Correlation dashboard solution extends ConceptDraw DIAGRAM software with samples, templates and vector stencils library with Scatter Plot Charts for drawing the visual dashboard visualizing data correlation.
## Computer and Networks Area
The solutions from Computer and Networks Area of ConceptDraw Solution Park collect samples, templates and vector stencils libraries for drawing computer and network diagrams, schemes and technical drawings. | 707 | 3,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.89821 |
https://popularask.net/what-is-the-difference-between-a-perpendicular-line-and-an-intersecting-line/ | 1,618,067,884,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057142.4/warc/CC-MAIN-20210410134715-20210410164715-00096.warc.gz | 580,652,324 | 23,718 | # What is the difference between a perpendicular line and an intersecting line?
0
5695
The major difference between the intersecting lines and the perpendicular lines is that intersecting lines are the two lines in which they intersect each other at any angle, whereas the perpendicular lines are the two lines, in which they intersect at an angle of 90 degrees.
The perpendicular lines are always intersecting lines but intersecting lines are not always perpendicular to each other.
Beside this, Do perpendicular lines intersect at two points?
A line is said to be perpendicular to another line if the two lines intersect at a right angle.
Likewise, What are intersecting lines that are not perpendicular?
Parallel lines never meet, and perpendicular lines intersect at a right angle. and do not intersect in this image, but if you imagine extending both lines, they will intersect soon. So, they are neither parallel nor perpendicular.
Also, How are perpendicular lines and intersecting lines different?
2 Answers By Expert Tutors. Perpendicular lines intersect at a 90 degree angle. … Intersecting lines share a point in common, called their intersection point. Perpendicular lines are intersecting lines that are at right angles to each-other.
Is a triangle intersecting or perpendicular?
Squares and rectangles always have four right angles! A triangle will have perpendicular lines if it has a little square where two lines intersect to indicate there’s a right angle there or if someone tells you that the triangle is a right triangle.
## 19 Related Question Answers Found
### Do perpendicular lines intersect in two points?
A line is said to be perpendicular to another line if the two lines intersect at a right angle.
### What is an example of a perpendicular line?
Perpendicular lines occur anytime two lines meet at a 90° angle, also known as a right angle. Sometimes, you will see a little square in the corner of an angle, to show it is perpendicular. There are many examples of perpendicular lines in everyday life, including a football field and train tracks.
Also Read Est-ce que l'huile de coco éclaircit les cheveux ?
### Are perpendicular lines intersecting lines?
The perpendicular lines are always intersecting lines but intersecting lines are not always perpendicular to each other.
### What does perpendicular sides look like?
Two distinct lines intersecting each other at 90° or a right angle are called perpendicular lines. Here, AB is perpendicular to XY because AB and XY intersect each other at 90°. The two lines are parallel and do not intersect each other. They can never be perpendicular to each other.
### How do you find perpendicular lines?
First, put the equation of the line given into slope-intercept form by solving for y. You get y = -2x +5, so the slope is –2. Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6.
### Are perpendicular lines also intersecting lines?
The perpendicular lines are always intersecting lines but intersecting lines are not always perpendicular to each other.
### Can a perpendicular line be an intersecting line?
The perpendicular lines are always intersecting lines but intersecting lines are not always perpendicular to each other.
### What shape has a pair of perpendicular sides?
Answer. Answer: In a square or other rectangle, all pairs of adjacent sides are perpendicular. A right trapezoid is a trapezoid that has two pairs of adjacent sides that are perpendicular.
### Where do perpendicular lines intersect?
Perpendicular lines intersect at a 90-degree angle. The two lines can meet at a corner and stop, or continue through each other.
### How can you determine if two lines intersect but are not perpendicular?
If two lines intersect and the measure of the angles formed by their intersection are other than 90 degrees, then the lines are not perpendicular, (but still intersect). If two lines in the coordinate plane are represented by equations, and if the slope of these two lines are not equal, then the lines will intersect.
Also Read Are Kristen Stewart and Robert Pattinson still friends?
### What are some examples of perpendicular lines?
Perpendicular lines occur anytime two lines meet at a 90° angle, also known as a right angle. Sometimes, you will see a little square in the corner of an angle, to show it is perpendicular. There are many examples of perpendicular lines in everyday life, including a football field and train tracks.
### What do perpendicular lines intersect to form?
ex) perpendicular lines conditional If two lines are perpendicular, then they intersect to form right angles.
Last Updated: 14 days ago – Co-authors : 8 – Users : 6 | 981 | 4,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-17 | latest | en | 0.91533 |
https://justaaa.com/physics/976380-1the-momentum-of-an-object-is-not-dependent-on | 1,716,538,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058684.66/warc/CC-MAIN-20240524060414-20240524090414-00445.warc.gz | 289,572,225 | 10,535 | Question
# 1.The momentum of an object is not dependent on which one of the following quantities? a)...
1.The momentum of an object is not dependent on which one of the following quantities?
a) acceleration b) inertia
c) mass
d) velocity
2.Which one of the following statements concerning the momentum of a system when the net force acting on the system has a positive value is true?
a) The momentum of the system is increasing.
b) The momentum of the system is decreasing.
c) The momentum of the system is equal to zero kg m/s. d) The momentum of the system has a constant value.
e) The momentum of the system has a negative value.
3. Which one of the following quantities is equal to the change in momentum of an object during a collision?
a) impulse
b) net force
c) work
d) change in kinetic energy e) maximum force
4.During a certain process, the linear momentum of a system is conserved. Which one of the following statements concerning this system is correct?
a) The vector sum of the momentum of the objects that make up the system is equal to zero kgm/s.
b) The vector sum of any internal forces within the system results in an acceleration of one or more objects within the system.
c) The principle of the conservation of mechanical energy automatically applies to the system.
d) The vector sum of the average external forces acting on the system is equal to zero newtons.
e) No internal or external forces are acting on the objects within the system.
1.The momentum of an object is not dependent on which one of the following quantities?
2)
Which one of the following statements concerning the momentum of a system when the net force acting on the system has a positive value is true?
a) The momentum of the system is increasing.--answer
3)
3. Which one of the following quantities is equal to the change in momentum of an object during a collision?
4.During a certain process, the linear momentum of a system is conserved. Which one of the following statements concerning this system is correct?
d) The vector sum of the average external forces acting on the system is equal to zero newtons.--answer
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https://forums.rpgmakerweb.com/index.php?threads/new-here-at-rgss3-ace.40323/ | 1,632,345,627,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00339.warc.gz | 320,790,872 | 19,360 | # New here at RGSS3 (Ace).
#### Jachan
##### Veteran
Bgillisp suggested me to post a new topic instead to derail.
Is that how we write this way? I mean to form the final formula with multiple totals from few previous formulas? Like this?
Base1 = form the formula of hero's attack power here
Base2 = form the formula of enemy's defense power here
Base3 = some formulas that could be useful of special effects (like, boosted damage if slime-type, for example)
Total = Base1 * Base3 - Base2
Right? if so, then it's alright to create some unexisted results (Base#) until hit the final result (Total)? I want to know it, so I don't have to form the whole in just one line (one whole formula with many parentheses that could make you lost the counts).
If there is any important tip for me to understand (like, do not input the certain sign in it ( +=, -=, =) or something), i would like to know this too. Thanks.
I could want to study RGSS3 but i just want to complete my big project here as quick as possible is all. I never had completed any single project I have made because I am too lazy, XD
Anyway... Is it possible for me to create the multiple formulas before the finalization damage to deal against the target?
##### Veteran
Here's some code to get you started
You'll need to insert this into your scripts Below Materials but Above Main
class Game_Battler < Game_BattlerBase
#Time to define some custom damage
def custom_damage_boost_def(a,b,boost)
base1 = a.atk * 4
base2 = b.def * 2
damage = (base1 * boost) - base2
return damage
end
def custom_damage_boost(a,
base1 = a.atk * 4
base2 = b.def * 2
boost = a*luk * 3
damage = (base1 * boost) - base2
return damage
end
end
Now in your skill formula you can now write "a.custom_damage_boost_def(a,b,num)"*
or "a.custom_damage_boost(a,b)" up to you!
*Num is whatever you want it to be, an integer or a stat
Another good reference that helped me a lot is here: http://cobbtocs.co.uk/wp/?p=271
Good luck!
#### Jachan
##### Veteran
Yes, I see it. Thanks again!
*look through that link* Oh wow! Thanks a lot!
Can I bother to ask you one tiny thing? What's the big difference between += and =? I did looked across some lunatic scripts Yanfly made and found += while you suggested = instead... So I am little confused to compare between those...
##### Veteran
Yes, I see it. Thanks again!
*look through that link* Oh wow! Thanks a lot!
Can I bother to ask you one tiny thing? What's the big difference between += and =? I did looked across some lunatic scripts Yanfly made and found += while you suggested = instead... So I am little confused to compare between those...
For example: damage += boost means this: damage = damage + boost
I'm not sure, what =? means.
#### dungeon diver
##### [ TRASH ELEMENTAL ]
Yes, I see it. Thanks again!
*look through that link* Oh wow! Thanks a lot!
Can I bother to ask you one tiny thing? What's the big difference between += and =? I did looked across some lunatic scripts Yanfly made and found += while you suggested = instead... So I am little confused to compare between those...
= is an assignment, while += is an assignment with an operation included.
Say x is initialized as 4.
x = 2 would return x as 2.
x += 2 would return x as 6, because before being assigned the value of 2 is added to x's value. Since 4+2=6, this returns 6.
#### Jachan
##### Veteran
Purple, i didn't mean this =?, hahaha. I mean = itself with a question mark as a signal of the end of my question. XD
Ah okay.
Diver, oh i see! Thanks!
So, I just saw one new as -=... since Diver gave the example, I would say that it will return as 2 instead 6, as to subtract / reduce instead to add / boost, right?
Come to thought, I did skipped this one... what does this one do? ==, a double =...? not "=?", lol
#### dungeon diver
##### [ TRASH ELEMENTAL ]
Purple, i didn't mean this =?, hahaha. I mean = itself with a question mark as a signal of the end of my question. XD
Ah okay.
Diver, oh i see! Thanks!
So, I just saw one new as -=... since Diver gave the example, I would say that it will return as 2 instead 6, as to subtract / reduce instead to add / boost, right?
Come to thought, I did skipped this one... what does this one do? ==, a double =...? not "=?", lol
== is used to compare instead of assign, usually in a conditional or loop.
I forget how ruby handles it exactly (if it brings up an error or goes ahead with the syntax) but "if x=3", for example, leads to a logical error because instead of comparing x to 3 the interpreter assigns 3 to the variable x.
#### Jachan
##### Veteran
so... single = will be used on non-loop / non-conditional either while double will be used instead?
#### Jachan
##### Veteran
Bump...
Am I correct... or incorrect? .__."
#### ♥SOURCE♥
##### Too sexy for your party.
Hello!
Double equal sign (==) is always used to compare, while single equal sign (=) is always used to assign:
Code:
``x = 2if x == 2 p 'x is 2'end``
#### Jachan
##### Veteran
Oh I think I get you... single = is like... overwriting the variable while double = is to check and see if it's equal to the non-variable number or not, without overwriting it. If so, then yeah I get you! I already wrote that kind of coding in different language which is very often. XP
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today is my birthday | 1,491 | 5,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-39 | latest | en | 0.906902 |
https://www.geeksforgeeks.org/check-if-it-is-possible-to-color-n-objects-such-that-for-ith-object-exactly-arri-distinct-colors-are-used/?ref=rp | 1,670,564,600,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711390.55/warc/CC-MAIN-20221209043931-20221209073931-00845.warc.gz | 813,791,846 | 29,038 | # Check if it is possible to color N objects such that for ith object, exactly arr[i] distinct colors are used
• Last Updated : 15 Dec, 2021
Given an array arr[] consisting of N positive integers, the task is to check if it is possible to color the N objects such that for ith element of the array there exist exactly arr[i] distinct colors used in coloring all the objects except for the ith object.
Examples:
Input: arr[] = {1, 2, 2}
Output: Yes
Explanation:
One of the possible ways to color is: {“Red”, “blue”, “blue”}.
1. For arr[0](=1), there is exactly 1 distinct color, which is “blue”.
2. For arr[1](=2), there are exactly 2 distinct colors, which are “blue” and “red”.
3. For arr[2](=3), there are exactly 2 distinct colors, which are “blue” and “red”.
Input: arr[] = {4, 3, 4, 3, 4}
Output: No
Approach: The problem can be solved based on the following observations:
1. For 2 objects, there are N-2 objects common while calculating the number of distinct colors. Therefore, there can be a difference of at most 1 between the maximum and minimum element of the array arr[].
2. Now there are two cases:
1. If the maximum and minimum elements are equal, then the answer is “Yes”, only if the element is N – 1 or the element is less than or equal to N/2, because every color can be used more than once.
2. If the difference between the maximum and minimum element is 1, then the number of distinct colors in the N object must be equal to the maximum element, because the minimum element is 1 less than the maximum element.
• Now, assuming the frequency of the minimum element as X and the frequency of the maximum element as Y, then the answer is “Yes” if and only if X+1 ≤ A ≤ X+ Y/2 (observation-based).
Follow the steps below to solve the problem:
• First sort the array in ascending order.
• If the difference between arr[N-1] and arr[0] is greater than 1, then print “No“.
• Else, if arr[N-1] is equal to arr[0], then check the following:
• If arr[N-1] = N-1 or 2*arr[N-1] <= N, then print “Yes“.
• Otherwise, print “No“.
• Else, count the frequencies of min and max elements and store them in variables, say X and Y, and then do the following:
• If arr[N-1] is greater than X and arr[N-1] is less than or equal to X+Y/2, then print “Yes“.
• Otherwise, print “No“.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to check if coloring is``// possible with give conditions``string checkValid(``int` `arr[], ``int` `N)``{` ` ``// Sort the vector`` ``sort(arr, arr + N);` ` ``// Coloring not possible in case of`` ``// maximum - minimum element > 1`` ``if` `(arr[N - 1] - arr[0] > 1)`` ``return` `"No"``;` ` ``// case 1`` ``else` `if` `(arr[N - 1] == arr[0]) {` ` ``// If h is equal to N-1 or`` ``// N is greater than 2*h`` ``if` `(arr[N - 1] == N - 1`` ``|| 2 * arr[N - 1] <= N)`` ``return` `"Yes"``;`` ``else`` ``return` `"No"``;`` ``}`` ``// Case 2`` ``else` `{`` ``// Stores frequency of minimum element`` ``int` `x = 0;` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``// Frequency of minimum element`` ``if` `(arr[i] == arr[0])`` ``x++;`` ``}` ` ``// Stores frequency of maximum element`` ``int` `y = N - x;` ` ``// Condition for case 2`` ``if` `((arr[N - 1] >= x + 1)`` ``and (arr[N - 1] <= x + y / 2))`` ``return` `"Yes"``;`` ``else`` ``return` `"No"``;`` ``}``}` `// Driver Code``int` `main()``{`` ``// GivenInput`` ``int` `arr[] = { 1, 2, 2 };`` ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Function Call`` ``cout << checkValid(arr, N);` ` ``return` `0;``}`
## Java
`// Java program for the above approach``import` `java.util.*;` `class` `GFG{`` ` `// Function to check if coloring is``// possible with give conditions``static` `String checkValid(``int` `arr[], ``int` `N)``{`` ` ` ``// Sort the vector`` ``Arrays.sort(arr);` ` ``// Coloring not possible in case of`` ``// maximum - minimum element > 1`` ``if` `(arr[N - ``1``] - arr[``0``] > ``1``)`` ``return` `"No"``;` ` ``// Case 1`` ``else` `if` `(arr[N - ``1``] == arr[``0``])`` ``{`` ` ` ``// If h is equal to N-1 or`` ``// N is greater than 2*h`` ``if` `(arr[N - ``1``] == N - ``1`` ``|| ``2` `* arr[N - ``1``] <= N)`` ``return` `"Yes"``;`` ``else`` ``return` `"No"``;`` ``}`` ` ` ``// Case 2`` ``else`` ``{`` ` ` ``// Stores frequency of minimum element`` ``int` `x = ``0``;` ` ``for``(``int` `i = ``0``; i < N; i++)`` ``{`` ` ` ``// Frequency of minimum element`` ``if` `(arr[i] == arr[``0``])`` ``x++;`` ``}` ` ``// Stores frequency of maximum element`` ``int` `y = N - x;` ` ``// Condition for case 2`` ``if` `((arr[N - ``1``] >= x + ``1``) &&`` ``(arr[N - ``1``] <= x + y / ``2``))`` ``return` `"Yes"``;`` ``else`` ``return` `"No"``;`` ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{`` ` ` ``// Given Input`` ``int``[] arr = { ``1``, ``2``, ``2` `};`` ``int` `N = arr.length;` ` ``// Function Call`` ``System.out.print(checkValid(arr, N));``} ``}` `// This code is contributed by sanjoy_62`
## Python3
`# Python3 program for the above approach` `# Function to check if coloring is``# possible with give conditions``def` `checkValid(arr, N):` ` ``# Sort the vector`` ``arr ``=` `sorted``(arr)` ` ``# Coloring not possible in case of`` ``# maximum - minimum element > 1`` ``if` `(arr[N ``-` `1``] ``-` `arr[``0``] > ``1``):`` ``return` `"No"` ` ``# case 1`` ``elif` `(arr[N ``-` `1``] ``=``=` `arr[``0``]):` ` ``# If h is equal to N-1 or`` ``# N is greater than 2*h`` ``if` `(arr[N ``-` `1``] ``=``=` `N ``-` `1` `or`` ``2` `*` `arr[N ``-` `1``] <``=` `N):`` ``return` `"Yes"`` ``else``:`` ``return` `"No"`` ``# Case 2`` ``else``:`` ` ` ``# Stores frequency of minimum element`` ``x ``=` `0` ` ``for` `i ``in` `range``(N):` ` ``# Frequency of minimum element`` ``if` `(arr[i] ``=``=` `arr[``0``]):`` ``x ``+``=` `1` ` ``# Stores frequency of maximum element`` ``y ``=` `N ``-` `x` ` ``# Condition for case 2`` ``if` `((arr[N ``-` `1``] >``=` `x ``+` `1``) ``and`` ``(arr[N ``-` `1``] <``=` `x ``+` `y ``/``/` `2``)):`` ``return` `"Yes"`` ``else``:`` ``return` `"No"` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ` ` ``# Given Input`` ``arr ``=` `[ ``1``, ``2``, ``2` `]`` ``N ``=` `len``(arr)` ` ``# Function Call`` ``print` `(checkValid(arr, N))`` ` `# This code is contributed by mohit kumar 29`
## C#
`// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to check if coloring is``// possible with give conditions``static` `string` `checkValid(``int` `[]arr, ``int` `N)``{` ` ``// Sort the vector`` ``Array.Sort(arr);` ` ``// Coloring not possible in case of`` ``// maximum - minimum element > 1`` ``if` `(arr[N - 1] - arr[0] > 1)`` ``return` `"No"``;` ` ``// case 1`` ``else` `if` `(arr[N - 1] == arr[0]) {` ` ``// If h is equal to N-1 or`` ``// N is greater than 2*h`` ``if` `(arr[N - 1] == N - 1`` ``|| 2 * arr[N - 1] <= N)`` ``return` `"Yes"``;`` ``else`` ``return` `"No"``;`` ``}`` ``// Case 2`` ``else` `{`` ``// Stores frequency of minimum element`` ``int` `x = 0;` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``// Frequency of minimum element`` ``if` `(arr[i] == arr[0])`` ``x++;`` ``}` ` ``// Stores frequency of maximum element`` ``int` `y = N - x;` ` ``// Condition for case 2`` ``if` `((arr[N - 1] >= x + 1)`` ``&& (arr[N - 1] <= x + y / 2))`` ``return` `"Yes"``;`` ``else`` ``return` `"No"``;`` ``}``}` `// Driver Code``public` `static` `void` `Main()``{`` ``// GivenInput`` ``int` `[]arr = { 1, 2, 2 };`` ``int` `N = arr.Length;` ` ``// Function Call`` ``Console.Write(checkValid(arr, N));` `}``}` `// This code is contributed by SURENDRA_GANGWAR.`
## Javascript
``
Output
`Yes`
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
My Personal Notes arrow_drop_up | 3,081 | 8,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-49 | latest | en | 0.87963 |
https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-37/issue-1/A-Note-on-Memoryless-Rules-for-Controlling-Sequential-Control-Processes/10.1214/aoms/1177699618.full | 1,725,758,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00611.warc.gz | 452,138,164 | 31,037 | February, 1966 A Note on Memoryless Rules for Controlling Sequential Control Processes
Cyrus Derman, Ralph E. Strauch
Ann. Math. Statist. 37(1): 276-278 (February, 1966). DOI: 10.1214/aoms/1177699618
## Abstract
We are concerned with a dynamic system which at times $t = 0, 1, \cdots$ is observed to be in one of a finite number of states. We shall denote the space of possible states by $I$. After each observation the system is "controlled" by making one of a finite number of possible decisions. We shall denote by $K_i$ the set of possible decisions when the system is in state $i, i \varepsilon I$. Let $\{Y_t\}, t = 0, 1, \cdots$, denote the sequence of observed states and $\{\Delta_t\}, t = 0, 1, \cdots$, the sequence of observed decisions. The fundamental assumption regarding $\{Y_t, \Delta_t\}, t = 0, 1, \cdots$, is \begin{equation*}\tag{A}P(Y_{t+1} = j \mid Y_0, \Delta_{0, \cdots,t}Y_t = i, \Delta_t = k) = q_{ij}(k),\quad t = 0, 1, \cdots; j \varepsilon I; k \varepsilon K_i\end{equation*} where the $q_{ij}(k)$'s are non-negative numbers satisfying $\sum_j q_{ij}(k) = 1, k \varepsilon K_i; i \varepsilon I$. A rule for making successive decisions can be summarized in the form of a collection of non-negative functions $D_k(Y_0, \Delta_0, \cdots, \Delta_{t-1}, Y_t),\quad t = 0, 1, \cdots; k \varepsilon K_{Y_t},$ where in every case $\sum_k D_k(\cdot) = 1$. We set $P(\Delta_t = k \mid Y_0, \Delta_0, \cdots, \Delta_{t-1}, Y_t) = D_k(Y_0, D_0, \cdots, \Delta_{t-1}, Y_t)$ for $t = 0, 1, \cdots$. Thus, given $Y_0 = i$ and any rule $R$ for making successive decisions, the sequence $\{Y_t, \Delta_t\}, t = 0, 1, \cdots$, is a stochastic process with its probability measure dependent upon the rule $R$. We refer to such a process as a sequential control process. Let $C$ denote the class of all possible rules; $C'$ denote the class of all rules such that $D_k(Y_0, \Delta_0, \cdots, \Delta_{t-1}, Y_t = i) = D_{ik},\quad t = 0, 1, \cdots; k \varepsilon K_i; i \varepsilon I$. That is, $C'$ is the class of all rules such that the mechanism for making a decision at any time $t$ is dependent only on the state of the system at time $t$. A rule $R \varepsilon C'$ has a stationary Markovian character and, indeed, when $R \varepsilon C'$ is used, the resulting process $\{Y_t\}, t = 0, 1, \cdots$, is a Markov chain with stationary transition probabilities. We let $C''$ denote the subclass of $C'$ where the $D_{ik}$'s are zero or one. Rules in $C'$ allow for randomization; the rules in $C''$ are non-randomized. For a given $R \varepsilon C$ and initial state $Y_0 = i$, let $X_{T,j,k,R}(i) = (T + 1)^{-1}\sum^T_{t=0} P_R(Y_t = j, \Delta_t = k \mid Y_0 = i)$ and let $X_{T,R}(i)$ denote the vector of components $X_{T,j,k,R}(i)$ for all $k \varepsilon K_j$ and $j \varepsilon I$. Denote by $H_R(i)$ the set of limit points of $X_{T,R}(i)$ as $T \rightarrow \infty$. Let $H(i) = \mathbf{\bigcup}_{R\varepsilon C} H_R(i),\quad H'(i) = \mathbf{\bigcup}_{R\varepsilon C'} H_R(i),\quad H''(i) = \mathbf{\bigcup}_{R\varepsilon C''} H_R(i);$ and let $\bar{H}'(i)$ and $\bar{H}''(i)$ denote the convex hulls of $H'(i)$ and $H''(i)$, respectively. In [5] was proved Theorem 1. (a) $\bar H'(i) = \bar H''(i) \supset H(i)$. (b) If the Markov chain corresponding to $R$ is irreducible for every $R \varepsilon C''$, then $\bar H''(i) = H'(i) = H(i) = \mathbf{\bigcup}_{i\varepsilon I} H(i)$. Examples were given in [4] and [5] showing that $H(i)$ can be larger than $H'(i)$. In (b) the irreducibility assumption can be weakened to the condition that for each $R \varepsilon C''$ the corresponding Markov chain has, at most, one ergodic class. Blackwell [1], [2], and Maitra [6] have considered memoryless rules. By a memoryless rule we mean a rule $R$ such that $D_t(Y_0, \Delta_0, \cdots, \Delta_{t-1}, Y_t = i) = D^{(t)}_{ik}\quad t = 0, 1, \cdots, k \varepsilon K_i, i \varepsilon I.$ That is, with a memoryless rule the mechanism for making a decision is a function of the time $t$ and the state $i$ at time $t$. The memoryless rules of Blackwell and Maitra are non-randomized; i.e., $D^{(t)}_{ik} = 0$ or 1. We shall let $C^M$ denote the class of memoryless rules (both randomized and non-randomized). Thus $C \supset C^M \supset C' \supset C''$. If $R \varepsilon C^M - C'$, then $\{Y_t\}, t = 0, 1, \cdots$, is a finite state Markov chain with non-stationary transition probabilities. We remark that it is the memoryless rules (non-randomized) that are considered the rules of interest in the usual finite horizon dynamic programming problems. See Blackwell [2] for interesting remarks along these lines. We are concerned with optimization problems where the criterion to be optimized are functions of the points in $H(i)$. In [5] it was shown that one can construct problems where the optimal rule is in $C - C'$. This can occur, e.g., if the criterion to be optimized is a linear functional over the points in $H(i)$ but where a solution must also satisfy one or more linear constraints in $H(i)$. It is for the purpose of treating optimization problems of this kind that we are interested in the limit points of $X_{T,R}(i)$ for rules belonging to the various important sub-classes of $C$. Let $H^M(i) = \mathbf{\bigcup}_{R\varepsilon C^M} H_R(i)$. The result of this note (which is similar to Theorem 4.1 of [7]) is Theorem 2. $H(i) = H^M(i).$ In fact, for any $R \varepsilon C$ there exists an $R_0 \varepsilon C^M$ such that $X_{tR_0}(i) = X_{tR}(i)$ for all $t$. Proof. Define $R_0$ by $D_{jk}(t) = P_{R_0}(\Delta_t = k \mid Y_t = j) = P_R(\Delta_t = k \mid Y_t = j, Y_0 = i).$ It is enough to show that for all $t, k \varepsilon K_j$ and $j \varepsilon I$, \begin{equation*} \tag{*} P_{R_0}(Y_t = j, \Delta_t = k \ Y_0 = i) = P_R(Y_t = j, \Delta_t = k \mid Y_0 = i).\end{equation*} The relation $(\ast)$ holds for $t = 0$, since $P_R(Y_0 = j, \Delta_0 = k \mid Y_0 = i) = 0 = P_{R_0}(Y_0 = j, \Delta_0 \mid Y_0 = i)$ if $j \neq i$ and \begin{align*}P_R(Y_0 = i, \Delta_0 = k \mid Y_0 = i) &= P_R(\Delta_0 = k \mid Y_0 = i) \\ &= P_{R_0} (\Delta_0 = k \mid Y_0 = i)\end{align*} by definition. Now assume $(\ast)$ holds for $t = 0, \cdots, T - 1$. Then $P_R(Y_T = j,\Delta_T = k \mid Y_0 = i) = P_R(Y_T = j \mid Y_0 = i)P_R(\Delta_T = k \mid Y_T = j, Y_0 = i)$ but $P_R(\Delta_T = k \mid Y_T = j, Y_0 = i) = D_{jk}(t)$ by definition, and \begin{align*}P_R(Y_T = j \mid Y_0 = i &= \sum_{l\varepsilon I} \sum_{k\varepsilon K_l} P_R(Y_{T-1} = l, \Delta_{T-1} = k \mid Y_0 = i) q_{lj}(k) \\ &= \sum_{l\varepsilon I} \sum_{k\varepsilon K_l} P_{R_0}(Y_{T-1} = l, \Delta_{T-1} = k \mid Y_0 = i) q_{lj}(k) \\ &= P_{R_0}(Y_T = j \mid Y_0 = i)\end{align*} by the introduction hypothesis. Thus \begin{align*}P_R(Y_T = j, \Delta_T = k \mid Y_0 = i) &= P_{R_0}(Y_T = j \mid Y_0 = i)D_{jk}(t) \\ &= P_{R_0}(Y_T = j, \Delta_T = k \mid Y_0 = i).\end{align*} This completes the proof.
## Citation
Cyrus Derman. Ralph E. Strauch. "A Note on Memoryless Rules for Controlling Sequential Control Processes." Ann. Math. Statist. 37 (1) 276 - 278, February, 1966. https://doi.org/10.1214/aoms/1177699618
## Information
Published: February, 1966
First available in Project Euclid: 27 April 2007
zbMATH: 0138.13604
MathSciNet: MR184778
Digital Object Identifier: 10.1214/aoms/1177699618 | 2,617 | 7,289 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.814685 |
https://ncatlab.org/nlab/show/symmetric+set | 1,631,906,775,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00040.warc.gz | 474,912,680 | 7,708 | Contents
# Contents
## Idea
A symmetric set (or symmetric simplicial set) is a simplicial set, $X$, equipped with additional transposition maps $t^n_i: X_n \to X_n$ for $i=0,\ldots,n-1$. These transition maps generate an action of the symmetric group $\Sigma_{n+1}$ on $X_n$ and satisfy certain commutation relations with the face and degeneracy maps. The result is that a symmetric set is a presheaf of sets on the category FinSet$_+$ of nonempty finite sets (or on its skeleton).
An analogy to keep in mind is
symmetric set : simplicial set :: groupoid : category.
This analogy can be formalized by noticing that the skeletal category of finite sets is simply the full subcategory of Cat whose objects are the localizations $[n]^{-1}[n]$ which are groupoids. By the universal property of localization, the usual (simplicial) nerve of a groupoid has a canonical symmetric structure.
Grandis proves that the fundamental groupoid functor $!Smp \to Gpd$ from symmetric sets to groupoids is left adjoint to a natural functor $Gpd \to !Smp$, the symmetric nerve of a groupoid, and preserves all colimits - a van Kampen theorem. Similar results hold in all higher dimensions.
The notion of cyclic set is intermediate between symmetric sets and simplicial sets. In particular, any symmetric set, such as the nerve of a groupoid, also has a cyclic structure.
Mike: I’ve never seen this called a “symmetric set” only a “symmetric simplicial set,” which additionally has the advantage of being more descriptive. “Symmetric set” sounds to me like it might also refer to a presheaf on the groupoid of finite sets (sometimes called a “symmetric sequence”). Is there an advantage of “symmetric set” over “symmetric simplicial set?”
Zoran Skoda: You are right, topologists more often say symmetric simplicial sets/objects, for example one of the papers devoted to the topic by Marco Grandis. But symmetric sets is also used, see for example the quote in Getzler’s paper on L-infty integration where he quotes Loday and Fiedorowicz, I heard this shorter term many time from conversations with Jibladze and many others. There is an advantage and that is exactly the theory of crossed simplicial groups in the sense of Loday and Fiedorowicz or skewsimplicial groups in the sense of a bit earlier work fo Krasauskas. The point is to treat dihedral, cyclic, symmetric etc. homologies on the same footing, hence symmetric, cyclic, dihedral etc. sets.
The term symmetric homology is for example in Loday-Fiedorowicz and then in Pirashvili-Richer and so on. I did not hear somebody saying dihedral simplicial sets, or cyclic simplicial sets, though it would be OK with me. I am happy with any solution for nlab and personally use both expressions in my writings and every day usage.
Mike: I see the point. Also, for instance, a symmetric spectrum is a particular sort of symmetric space, i.e. a symmetric object in spaces in this sense. So I’m okay with “symmetric set” as a name for this page.
## References
• Marco Grandis, Finite sets and symmetric simplicial sets PDF
• Marco Grandis, Higher fundamental functors for simplicial sets arXiv
• J. Rosický and W. Tholen, Left-determined model categories and universal homotopy theories, Trans. Amer. Math. Soc. 355 (2003), 3611-3623.
Last revised on August 23, 2015 at 16:35:25. See the history of this page for a list of all contributions to it. | 829 | 3,390 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-39 | longest | en | 0.904709 |
https://www.majorfun.com/rolling-america/?replytocom=8454 | 1,620,761,675,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989856.11/warc/CC-MAIN-20210511184216-20210511214216-00341.warc.gz | 946,964,194 | 11,520 | # Rolling America
At Major Fun, we love games that can accommodate everyone. Big groups, tiny families, and everything in between. Games that can be played solo as well. Rolling America fits that bill, and when we find a game that we love and it can be enjoyed by any number of players, that’s something special.
To be clear, Rolling America probably can’t accommodate the entire set of positive whole numbers (citation needed). It’s mainly a problem of seating arrangements when you get above 8 or 10. Definitely when you get into numbers that are best expressed in powers of 10.
What it CAN do is keep your brain buzzing along at a healthy clip while place numbers on an abstract rendition of the United States.
To play the game you will need the 7 dice (in a bag) and the maps provided by the good people at Gamewright. The map of the United States is divided into blocky representations of the 50 states. These are then colored by region, roughly: northeast (purple), Atlantic (red), south (yellow), north (blue), southwest (green), and west (orange). Each player gets a map and the dice are shared.
Rotating around the table, players draw two dice from the bag and roll them. They must write the number on one of the states that match the color of the dice. For example, if you roll a red 2 and a blue 5 then everyone must write the number 2 in one of the red states and the number 5 in one of the blue states. Which state, is up to each player as long as they follow the game’s basic rules, the most important being that neighboring states must have consecutive numbers. So if Ohio is number 3, Pennsylvania can be 2 or 4 but not anything else.
After the first two dice are rolled and recorded, two more are drawn from the bag. When six of the seven dice have been rolled, all the dice go back in the bag and the players record that one round is over. At the end of 8 rounds you will tally up your score.
“But wait!!” you cry before I can reveal how you win. “You said there are 6 colored regions but there are 7 dice!! What’s up with that?”
I’m glad you asked. The seventh dice is clear and is a wild die, meaning you can put that number in any color you want. I should also say that as you fill in the map you are going to run in to problems: chiefly that it is impossible to follow the consecutive rule all the time. In order to deal with this, the game has included a clever “cheating” mechanic—a way to break the rules (for a limited number of times). You get three Color Changes which let you make a colored dice wild. You get three Guards which let you put a number down illegally (not consecutive). Finally you get three Dupes that allow you to use a number on one of the dice twice. On the map are boxes that you use to mark off these special occasions.
If you are ever stuck with a number that you cannot legally place, you have to cross out one of the states in that color. The winner is the player at the end of 8 rounds who has the fewest number of Xs on the map.
And getting stuck is a big part of the game. Early in the game, when the map is wide open, it seems like you will breeze right through, but in only a few rounds you notice that your regions are filling up and you have blank Indiana but it is sandwiched between an Illinois 2 and an Ohio 6.
We loved the building tension and complexity. We also loved how everyone took their turn together. It was fascinating to see what other people came up with using the same numbers that I had. It’s strategically deep and very challenging. And Major Fun.
1+ players. Ages 8+
Rolling America was designed by Hisashi Hiyashi and is © 2015 by Gamewright Games.
## 4 thoughts on “Rolling America”
1. Donna
I see there are only 100 maps in the game. What do you do when you use them up?
2. Melissa
When states are not “bordered” yet meet at a corner does the neighboring states rule apply? They meet at a diagnal point. Happens in 2 spots on the game. Arguing with my husband over this.
1. States that meet at a single diagonal point do not border each other. States must share a border along a line, not a single diagonal point. That’s how we play. Hope that helps!
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Scroll To Top | 971 | 4,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-21 | latest | en | 0.95587 |
https://www.ccri.com/2018/05/04/z-earth-round/ | 1,580,294,194,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00548.warc.gz | 797,074,140 | 22,058 | # Z Earth, it is round?!
### In which visualization helps explain how good indexing goes weird.
The flat-Earthers may be on to something.
GeoMesa, like many other data stores that index geographic data, uses space-filling curves to impose an order on two-dimensional geometries. It’s easy to know that “Virginia” as a string data type follows “Illinois” when sorted alphabetically, but it’s much less obvious whether the multi-polygon that is Virginia’s border should come before or after Illinois’ state boundary in an index. The standard solution is to divide the Earth into discrete cells and then use a fixed pattern to define an ordering among those cells. A space-filling curve is a function that, for a given grid-cell resolution, can define a fixed ordering among the cells; imagine a line that visits every cell in the grid, and on which the relative positions of Illinois and Virginia help to build an index that lets a geospatial database manager find locations more quickly.
The advantage of ordering cells is that the columnar data stores GeoMesa supports—Accumulo, HBase, Cassandra, the S3 file system, and Kafka—provide range queries that are easily built from the linear ordering that the curves impose on grid cells. Looking for Charlottesville, Virginia? No worries; that’s in cell #171, right after Venezuela in cell #16 and right before West Africa in cell #18. See the first figure, below, for an illustration of a discretization of the map into grid cells as well as one space-filling curve ordering of those cells.
A commonly used curve in GeoMesa is the Z-order (or Morton) curve. The SFCurve library—also organized within The Eclipse Foundation’s LocationTech working group, and whose principal contributors, like me, work on GeoMesa and GeoTrellis—contains the code for the Z2, or two-dimensional, geo-only version of the Z-order curve. When we visualize the Z2 progression, we typically draw it on a flat, rectangular longitude-latitude map like this:
The curve’s progression from the South-West corner to the North-East corner is clean, easy to follow, and orderly.
It’s also a lie.
Being a bear of very little brain, it took me longer than the other SFCurve contributors to realize that it matters that the Earth may not be flat. The peril is not that we might fall off the edge; rather, the risk is that we misrepresent the relationship between the polar regions and the equatorial regions. To get a sense for how different the Z2 looks in practice, we created the following brief animation2. It uses a 9-bit Z2 curve that creates 32 horizontal cells and 16 vertical cells3 that are 11.25° high by 11.25° wide “squares”. To make the math simpler, we are using a perfect sphere.
There are two striking features that leap out of the animation:
• Even though the vertices of the curve remain on the surface of the sphere, the edges do not. In fact, the longer the edge, the deeper (closer to the center) it is embedded in the sphere. The smallest Zs almost look like they’re on the surface, but the largest Z, which defines the halves of the map, almost looks like the polar axis. This is a curiosity.
• The curve over-represents the polar regions significantly. The top-most and bottom-most horizontal rings cover much less area than do the equatorial rings, but still contain the same number of cells. This is a problem.
### How much smaller are polar cells than equatorial cells?
Creating a still, rotated rendering helps to illustrate the problem:
Visually, it is clear that 11.25° squares are no longer meaningful once we’ve used a spherical mapping instead of a flat mapping. This is more than a curiosity, though, because the area of the polar cells is significantly less than the area of the equatorial cells, meaning that a single Z2 index does not carry uniform information: The further the Z2’s cell is from the equator, the more precisely it constrains the member points’ location.
How much? To answer that question, it is helpful to sketch the geometry4:
The sketch uses spherical coordinates. The most important labels are these:
• r: the diameter of the spherical Earth, assumed to be 1.0—the unit sphere—because it washes out when we compute the ratio of the area of a polar cell to the area of an equatorial cell; consequently, none of the equations include the radius5.
• theta (ϴ): the longitude, defined on [0, 2π] just as it is on the flat map (when using all non-negative degrees), since the translation from radians to degrees maps π radians to 180 degrees.
• phi (Φ): the latitude, defined on [0, π] just as it is on the flat map.
There are a few additional symbols for the quantities that most interest us:
• Ap: the area of a cell in the ring nearest the pole
• Ae: the area of a cell in the ring nearest the equator
The equations for the these two variables, solved by Wolfram Alpha6, are:
Equation 1. Ap: area of a 11.25 degrees-square (π/16 radians-square) cell adjacent to the North pole integrate (integrate sin theta dtheta from 15pi/16 to pi) dphi from 0 to pi/16
Equation 2. Ae: area of a 11.25 degrees-square (π/16 radians-square) cell adjacent to the equator integrate (integrate sin theta dtheta from pi/2 to 9pi/16) dphi from 0 to pi/16
The ratio of these cell areas gives us:
For geospatial indexing, cell size matters. If I lose my wallet in the house, I have some hope of finding it without being late for work; if I lose my wallet in the mall, I might as well start canceling my credit cards, because I have no time to conduct an exhaustive search. This is analogous to the problem of using Z2 grid cells on a round earth: near-pole cells are only about one tenth as large as the near-equator cells, which means that they are more precise. If you know that a tweet originated in a near-pole cell, it tells you a lot more about where the Twitter user was, because there’s only a tenth as much wiggle room in that cell compared to the near-equator cell.
Moreover, as we use more bits in the Z2 curve, dΦ gets smaller and smaller, and the rings closest to the pole get even closer. At the same time, the equatorial rings get nearer to the equator, and the ratio of Ap/Ae approaches 0. This is a problem only inasmuch as most of the population (and their attending data that we actually care about) are located far from the poles.
This problem is not specific to the Z2 curve, but applies to all space-filling curves that work on an evenly-gridded version of a flat latitude-longitude surface. Compact Hilbert and Peano curves, for example, share this same pathology. The most effective and obvious workaround is to stop using a flat latitude-longitude surface for drawing uniform grid cells. For example, we could use a space-filling curve that operates on equilateral triangles, and treat the globe like a giant d207 die from Dungeons & Dragons8. A more mundane approach would be to simply change the function that maps latitude values to cells so that the equatorial cells become shorter and the polar cells become taller, reducing the discrepancy in their areas.
For a geospatial index, the core concern isn’t the shape of the grid cells, but what that shape implies about how data can or cannot be stored and queried effectively. The fact that the most precise indexing, and consequently the most efficient retrieval, is allocated to data that are located furthest from the equator is very much a problem for GeoMesa just like it is for all of the other geo-indexing systems that use rectangular lat/long grids.
GeoMesa itself does not define the space-filling curves nor the map from curve indices to grid cells. Those responsibilities belong to the SFCurve library, and so that is where the active research is being done to address these challenges, and that’s the GitHub repository to watch.
Z2 accommodations for a round Earth are just over the horizon. Meanwhile, the next time you see a GeoMesa T-shirt, you’ll have a much better understanding of the math that went into it!
1 Assuming a 6-bit Z-order curve (drawn more like an N-order curve, with an “XYXYXY” bit interleaving) as depicted in the first figure.
2 Thanks to NASA’s Visible Earth for the sphere texture, and their Deep Star Maps for the sky-box background. POV-ray, although it may be old, still does a capable job of rendering the scene.
3 The 9 bits are interleaved as follows: XYXYXYXYX, implying that the large “polar spike” in the image and animation is actually a horizontal jump between the West and East hemispheres.
4 Illustration copied from Weisstein, Eric W. “Spherical Coordinates.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/SphericalCoordinates.html
5 If it would make you feel better, squint just a bit and imagine that you see an “r2” term at the head of the integration expressions.
6 Not content to rely on my own memory of spherical integration, I first wrote code to compute this using trapezoidal subdivisions. Happily, the analytic and numerical approaches agreed.
7 For those unfamiliar with D&D, the shape alluded to is a regular icosahedron.
8 This inclines one to remember Einstein’s “God does not play dice.” For a fun discussion about that idea, see http://www.hawking.org.uk/does-god-play-dice.html | 2,085 | 9,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-05 | longest | en | 0.895229 |
https://www.mathed.page/geoboard/ | 1,571,784,429,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987824701.89/warc/CC-MAIN-20191022205851-20191022233351-00106.warc.gz | 981,752,074 | 4,267 | # Using the Geoboard
Henri Picciotto
The geoboard provides a nicely constrained environment for mathematical exploration, and lends itself to many activities where students can discover or apply important ideas in the K-12 curriculum. This page is a hopefully comprehensive list of links to geoboard activities on this site and on my blog.
All activities I link to below can be done on the CircleTrig Geoboard, a two-sided manipulative I designed. On one side, an 11 by 11 traditional geoboard. On the other side, a circle geoboard, with radius 10 cm, and a 360° protractor. It is available from Nasco.
(However, most can be done on traditional 11 by 11 geoboards, or traditional circle geoboards. An online geoboard is available on the Didax Web site. In my experience, "virtual manipulatives" do not work as well to generate student discussion, but they are useful if you don't have the physical version, and also to discuss solutions on the projector or to get images to use in worksheets or student reports.)
Compact introduction to the geoboard for teachers: one page | two pages
## 11 by 11 Geoboard
Area: Labs 8.4-8.6, 9.5
Distance: Labs 9.2, 9.4, 9.6
Similarity, Slope: Labs 10.1, 10.2
(Some of these lessons overlap and complement the ones in Geometry Labs)
Area: Lessons 1.12, 2.12, 4.12, 6.12, 11.A
Distance, Square Roots Lessons 6.12, 7.12, 9.1-9.4, 9.A, 9.9, 9.C,
Similarity, Slope: Lessons 3.12, 8.3, 11.3
Connect the Dots: lattice problems and puzzles for teachers (and students)
### Blog posts
Geoboards and Dot Paper
Geoboard Triangles
Inscribing Geoboard Squares in Polyominoes
No Three on a Line
Heilbronn's Triangle
Geoboard Problems for Teachers
### On this site
Proving Pick's Formula
Dot Papers
Connecting the Dots! (presentation slides): html | PDF | Keynote | 507 | 1,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-43 | longest | en | 0.866693 |
https://www.slideserve.com/sari/abstract | 1,513,595,826,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948615810.90/warc/CC-MAIN-20171218102808-20171218124808-00725.warc.gz | 783,116,228 | 13,972 | 1 / 19
# Abstract - PowerPoint PPT Presentation
Classical Particles subject to a Gaussian-Distributed Random Force APS April Meeting Sunday, May 2 nd 2004 Aaron Plasek, Athanasios Petridis Drake University. Abstract.
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Classical Particles subject to a Gaussian-Distributed Random ForceAPS April Meeting Sunday, May 2nd 2004Aaron Plasek, Athanasios Petridis Drake University
Abstract Random Force
The motion of a classical particle under the influence of a random gaussian-distributed force sampled at discrete time intervals is studied. In the case of unbound particles the expectation value and the standard deviation of the position and the kinetic energy are calculated analytically and numerically (by means of a Monte-Carlo simulation over an ensemble of identical particles and initial conditions). The system is shown to exhibit chaotic behavior. An ensemble of particles confined in a box is also studied and compared with the unbound case.
the Algorithm Random Force
To model an ensemble of particles under these “random” forces, we make a few simplifying assumptions, namely:
1) the particles in the ensemble are point particles
2) the particles in the ensemble make completely elastic collisions with walls
The algorithm propagates the particles in the ensemble via the standard Newtonian equations of motion. In the case where the ensemble is contained by a one-dimensional box, the time it will take for the particle to reflect off a wall is calculated and the particle is accordingly propagated and reflected. It is a non-trivial problem to model this numerically (at last count, the code exceeded 600 lines).
The “random” force is generated via a very powerful random number generator known as “the Mersenne Twister” (has a period of 2^19937-1) and the Central Limit Theorem.
The code allows us to model an ensemble of N particles for M time steps, where the size of N and M are only limited by the memory of the computer and the maximum allowed array size in C++.
The code has error checking and will alert the user for any “missed” case or violation of the energy-work theorem.
No Wall Case: Analytic Results Random Force
For an ensemble of particles subject to “random” forces, the following equations can be written (see authors for derivations):
where n = t / Dt, T = Kinetic Energy, and s = standard deviation
• According to equation 1, the average position of the ensemble will increase linearly with time. As seen in AVERAGE X, the program produces numerical results that are (accounting for statistical scatter) linear. The average position of the ensemble behaves like one particle under constant velocity subject to Newton’s second law. According to equation 3, the average kinetic energy increases linearly with time and, as seen in AVERAGE T, the program does give this result.
• According to equation 3, the standard deviation of the average position of the ensemble should vary as time to the 3/2. A quick glance at the STANDARD DEVIATION OF POSITION confirms this result numerically.
• Equation 4 predicts that the standard deviation of the average kinetic energy will go as the square root of time. The numerical results, as seen in STANDARD DEVIATION OF T, agree nicely.
Walls Case: Numerical Results Walls Case
• Satisfied that the code predicts numerically what we have derived analytically, we now run the program for a case which is very difficult to derive analytically: that of an ensemble of particles subject to “random” forces in a one-dimensional box.
• As can be seen in AVERAGE T, the average value of kinetic energy with respect to time is (statistically) linear and thus in agreement with equation 3.
• As shown in the STANDARD DEVIATION OF T, the standard deviation of the kinetic energy goes as the square root of time and in agreement with equation 4.
• Although the standard deviation of position, shown in STANDARD DEVIATION OF X, initially goes as time to the 3/2, it becomes narrower as the ensemble “packet” is reflected off walls (See POSITION HISTOGRAM). As the packet distribution becomes “flat” in the box, it has the same standard deviation as the no walls case (which makes sense since the standard deviation of a uniform distribution is a constant).
• Unlike the average position of the ensemble for the No Wall case, the average position value of the ensemble for the Walls case (see AVERAGE X) does NOT behave like a particle in a one-dimensional box under Newton’s 2nd Law.
Closing Discussion Walls Case
• In both the No Wall case and in the Walls case, particles in the ensemble occupy the same point in phase space. As the position distribution of ensemble flattens (see POSITION HISTOGRAM), every point in phase space will be accessed, so both the No Walls and Walls cases are chaotic systems.
• Although the average position of the ensemble for the no walls case grows linearly (i.e., like a particle with constant velocity), its average kinetic energy is not a constant. How is this possible? The additional kinetic energy added to the ensemble increases the mean speed of the particles rather than increasing the mean velocity of the ensemble. See POSITION HISTOGRAM.
• Different widths of the gaussian that the “random” force is picked from will affect how the ensemble of particles propagates through the box. The wider the gaussian, the more quickly the ensemble becomes chaotic.
• Future research will be conducted to study how different force distributions affect the ensemble.
Walls Case: STANDARD DEVIATION OF X Walls Case
Note: After 1 time unit, the distribution
becomes uniform and, thus, chaotic. For
this particular box, the distribution becomes
uniform at Sigma = .577, which is matched
nicely by the numerical results. The downward
fluctuations at small times are due to narrowing
of the distribution upon reflection. | 1,378 | 6,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-51 | latest | en | 0.882081 |
https://se.mathworks.com/matlabcentral/cody/problems/17-find-all-elements-less-than-0-or-greater-than-10-and-replace-them-with-nan/solutions/2403519 | 1,596,668,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735989.10/warc/CC-MAIN-20200805212258-20200806002258-00451.warc.gz | 505,552,554 | 15,722 | Cody
Problem 17. Find all elements less than 0 or greater than 10 and replace them with NaN
Solution 2403519
Submitted on 28 May 2020 by Connor Burke
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
x = [ 5 17 -20 99 3.4 2 8 -6 ]; y_correct = [ 5 NaN NaN NaN 3.4 2 8 NaN ]; assert(isequalwithequalnans(cleanUp(x),y_correct))
2 Pass
x = [ -2.80 -6.50 -12.60 4.00 2.20 0.20 -10.60 9.00]; y_correct = [ NaN NaN NaN 4.00 2.20 0.20 NaN 9.00] assert(isequalwithequalnans(cleanUp(x),y_correct))
y_correct = NaN NaN NaN 4.0000 2.2000 0.2000 NaN 9.0000 | 246 | 665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-34 | latest | en | 0.586605 |
https://mathlake.com/Right-Circular-Cone | 1,719,224,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00663.warc.gz | 330,296,331 | 4,778 | # Right Circular Cone
A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base. That is, the centre point of the circular base is joined with the apex of the cone and it forms a right angle. A cone is a three-dimensional shape having a circular base and narrowing smoothly to a point above the base. This point is known as apex.
We come across many geometrical shapes while dealing with geometry. We usually study about two dimensional and three-dimensional figures in school. Two-dimensional shapes have length and breadth. They can be drawn on paper, for example – circle, rectangle, square, triangle, polygon, parallelogram etc. The three-dimensional figures cannot be drawn on paper since they have an additional third dimension as height or depth. The examples of 3D shapes are a sphere, hemisphere, cylinder, cone, pyramid, prism etc.
## Right Circular Cone Definition
A right circular cone is one whose axis is perpendicular to the plane of the base. We can generate a right cone by revolving a right triangle about one of its legs.
In the figure, you can see a right circular cone, which has a circular base of radius r and whose axis is perpendicular to the base. The line which connects the vertex of the cone to the centre of the base is the height of the cone. The length at the outer edge of the cone, which connects a vertex to the end of the circular base is the slant height.
## Right Circular Cone Formula
For a right circular cone of radius ‘r’, height ‘h’ and slant height ‘l’, we have;
Curved surface area of right circular cone = π r l Total surface area of a right circular cone = π(r + l) r Volume of a right circular cone = 1/3π r2 h
### Surface Area of a Right Circular Cone
The surface area of any right circular cone is the sum of the area of the base and lateral surface area of a cone. The surface area is measured in terms of square units.
Surface area of a cone = Base Area + Curved Surface Area of a cone
= π r2 + π r l
= πr(r + l)
Here, l = √(r2+h2)
Where ‘r’ is the radius, ‘l’ is the slant height and ‘h’ is the height of the cone.
### Volume of a Right Circular Cone
The volume of a cone is one-third of the product of the area of the base and the height of the cone. The volume is measured in terms of cubic units.
Volume of a right circular cone can be calculated by the following formula,
Volume of a right circular cone = ⅓ (Base area × Height)
Where Base Area = π r2
Hence, Volume = ⅓ π r2h
## Properties of Right Circular Cone
• It has a circular base whose centre joins its vertex, showing the axis of the respective cone.
• The slant height of this cone is the length of the sides of the cone taken from vertex to the outer line of the circular base. It is denoted by ‘l’.
• The altitude of a right cone is the perpendicular line from the vertex to the centre of the base. It coincides with the axis of the cone and is represented by ‘h’.
• If a right triangle is rotated about its perpendicular, considering the perpendicular as the axis of rotation, the solid constructed here is the required cone. The surface area generated by the hypotenuse of the triangle is the lateral surface area.
• Any section of right circular cone parallel to the base forms a circle lies on the axis of the cone.
• A section which contains the vertex and two points of the base of a right circular cone is an isosceles triangle.
## Frustum of a Right Circular Cone
A frustum is a portion of the cone between the base and the parallel plane when a right circular cone is cut off by a plane parallel to its base.
## Equation of Right Circular Cone
The equation of the right circular cone with vertex origin is:
(x2+y2+z2)cos2θ=(lx+my+nz)2
Where θ is the semi-vertical angle and (l, m, n) are direction cosines of the axis.
Let us find the equation of the right circular cone whose vertex is the origin, the axis is the line x = y/3 =z/2 and which makes a semi-vertical angle of 60 degrees.
The direction cosines of the axis are:
[(1/√(12+32+22) , (3/√(12+32+22), (2/√(12+32+22)] = (1/√14, 3/√14, 2/√14)
The semi-vertical angle is, θ = 60°
Therefore, the equation of the right circular cone with vertex (0, 0, 0) is:
(x2+y2+z2) ¼ = 1/14 (x + 3y + 2z)2
7(x2+y2+z2) = 2(x2+9y2+4z2+6xy+12yz+4xz)
5x2-11y2-z2-12xy-24yz-8xz = 0; which is the required equation.
### Examples
Q. 1: Find the surface area of the right cone if the given radius is 6 cm and slant height is 10 cm.
Solution: Given,
Slant height (l) = 10 cm
Surface area of a cone = π r(r + l)
Solving surface area,
SA = 3.14 × 6(6 + 10)
SA = 3.14 × 6 × 16
SA = 301.44
Therefore, Surface area of the right cone is 301.4 sq. cm.
Question 2: Calculate the Volume of the right cone for the given radius 6 cm and height 10 cm.
Solution: Given, | 1,257 | 4,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-26 | latest | en | 0.918797 |
https://www.economist.com/science-and-technology/2019/05/16/the-world-is-about-to-get-a-new-way-to-measure-itself | 1,571,020,588,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648481.7/warc/CC-MAIN-20191014003258-20191014030258-00363.warc.gz | 979,156,775 | 125,923 | # MetrologyThe world is about to get a new way to measure itself
The Système International d’Unités is being overhauled
ON MAY 20TH the world gets a new kilogram. It also gets a new ampere, kelvin and mole. And, more important, it gets a new way of defining all these units—which lie, along with the metre and the second, at the heart of the Système International d’Unités (SI) that human beings use to measure things. Even the pounds, miles, gallons and so on, clung on to by a few benighted Anglophones, are, malgré eux, defined in terms of the SI.
Measuring anything means comparing it with an agreed standard. Until now, for instance, the standard kilogram (see picture) has been the mass of a lump of metal sitting, nestled under a series of bell jars, in a vault in a suburb of Paris. However, the best sort of standard by which to define a unit is a constant of nature, such as the speed of light in a vacuum. And the metre is indeed so defined—or, rather, the speed of light is defined as 299,792,458 metres per second, and the second itself is defined as the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom at absolute zero. The calculation is therefore a simple one.
The other basic units, the ampere (electric current), candela (luminous intensity), kelvin (temperature) and mole (quantity of particles, such as atoms or molecules, regardless of their mass) are defined in terms of things that can be measured fairly easily in a laboratory. An ampere is proportional to the mechanical force generated between two wires (strictly speaking of infinite length, but let that pass) as a current flows through them. A kelvin is defined as 1/273.16 of the temperature of the point (known as the “triple point”) at which water, ice and vapour exist in equilibrium in a sealed glass vessel. And so on.
But all that is now to change. From Monday onwards, several other fundamental constants will go, like the speed of light, from being things that are measured to things that are defined, and are then used as references for measurement.
A kilogram, for instance, will be derived from Planck’s constant, which relates the energy carried by a photon to its frequency. An ampere will depend on the charge on an electron, a kelvin on Boltzmann’s constant (the average relative kinetic energy of particles in a gas, compared with the temperature of the gas) and the mole on Avogadro’s number—6.0221409x1023, originally measured as the number of atoms in a kilogram of a particular isotope of carbon. Only the metre, the second and the candela (already defined in terms of a particular frequency of light) remain unchanged.
With luck, this will be the last change ever needed to the system. By definition, the fundamental constants of the universe do not alter with time or place. Neither, even in America and Britain, need the SI.
This article appeared in the Science and technology section of the print edition under the headline "Perfectly constant"
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Join them. to The Economist today | 731 | 3,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-43 | longest | en | 0.939897 |
http://navalfacilities.tpub.com/dm2_08/dm2_080279.htm | 1,540,227,801,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00035.warc.gz | 250,372,765 | 6,006 | Calculate natural period of vibration
Custom Search
(11) Calculate effective mass of section.
Mass of element
m = [(497 x 150/123)/(32.2 x 12)]
= 0.1116 lb-sec2/in2
Load-mass factor for the elastic range
KLM = 0.79
Effective mass of element
me = mKLM
= 0.1116 x 0.79
= 0.0882 lb-sec2/in2
(12) Calculate natural period of vibration
TN =
2[pi](me/KE)1/2
=
2[pi](0.0882/84.19)1/2
=
0.2034 sec
=
203.4 ms
(13) Determine section response from figure 6-7 of NAVFAC
P- 397
B/ru = (1.1 x 96)/70.9
= 1.49
T/TN = 43.9/203.4
= 0.216
Xm/XE = 1.0
O.K.
Check rotation of beam
X = XE + XDL + XLL
ravail
5(DL + LL)L4
= +
KE
384EIa
70.9
+ 5(43.1 + 10)(40 x 12)4
=
84.19
384(4,286,826)13575
= 0.84 + 0.63
= 1.47 in
[theta] = tan -1(2X/L)
= tan -1[2 x 1.47/(40 x 12)]
= 0.35[deg] < 2[deg]
O.K.
2.08-262 | 381 | 787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-43 | latest | en | 0.594222 |
https://mathschallenge.net/full/solid_encryption | 1,550,888,870,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249434065.81/warc/CC-MAIN-20190223021219-20190223043219-00479.warc.gz | 612,681,000 | 2,512 | ## Solid Encryption
#### Problem
Although your mathematics teacher tells excellent jokes, he can be particularly cruel. At the end of a lesson he hands out a small piece of paper to each student.
A cement mixer collided with a prison van...
31 51 02 51 81 90 91 02 91 10
81 50 10 91 11 50 40 02 51 20
50 51 41 02 80 50 21 51 51 11
51 12 02 60 51 81 91 90 42 02
50 50 41 80 10 81 40 50 41 50
40 30 81 90 31 90 41 10 21 91
Can you discover the punchline?
#### Solution
The method of encryption is to convert each letter to a number according to its alphabetical position, then reverse the digits:
A = 01 = 10
B = 02 = 20
C = 03 = 30
D = 04 = 40
E = 05 = 50
F = 06 = 60
G = 07 = 70
H = 08 = 80
I = 09 = 90
J = 10 = 01
K = 11 = 11
L = 12 = 21
M = 13 = 31
N = 14 = 41
O = 15 = 51
P = 16 = 61
Q = 17 = 71
R = 18 = 81
S = 19 = 91
T = 20 = 02
U = 21 = 12
V = 22 = 22
W = 23 = 32
X = 24 = 42
Y = 25 = 52
Z = 26 = 62
Using this method in reverse to decode the punchline:
A cement mixer collided with a prison van...
Motorists are asked to be on the lookout for sixteen hardened criminals.
Problem ID: 9 (Aug 2000) Difficulty: 2 Star
Only Show Problem | 465 | 1,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-09 | latest | en | 0.810087 |
www.kidsfoundation.in | 1,695,672,821,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00067.warc.gz | 930,772,740 | 206,794 | top of page
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# Math 2.0 Day - 8th July
Math 2.0 Day is observed on July 8 of every year. Math 2.0 Day celebrates the subject math, without which life would be much difficult. With or without knowing we use math numerous times in every single day. Many people always hate math due to the complicated lesson plans in the schools and colleges, but it isn’t that boring and hard in real life. Math is used everywhere like in buildings, computer programs, accounts, catching a bus, buying grocery, airplanes, and in lot more. Math is base for many other subjects like science and technology. On Math 2.0 Day, acknowledge the importance of the subject and teach people the ways they can use math in everyday life without much confusion.
Math 2.0 Day is a celebration of the blending of technology and mathematics. For a lot of us, math wasn’t a favourite subject, we’d spend the entire period staring at the equations and wondering what sort of livid madman designed these torture chambers on paper. Ultimately, however, we realized that math is utterly indispensable in our modern world. If you’ve ever wondered who uses math in their day to day careers, you aren’t alone and we have some answers for you.
Programmers deal with mathematics every day, as it’s the framework
upon which all computer operations are formed. Everything from the order of operations to quadratic equations is necessary to make even the simplest program. Scientists are one of the biggest users of mathematics, whether they’re calculating the statistical variance of their data or figuring out how much to add to their chemistry experiment, it’s involved at every step.
One presumes you live in a house, drive a car, or operate a computer?
The engineers responsible for designing those things so that they work, and especially in the case of the house, use math to ensure it doesn’t come crumbling down on your head. Math 2.0 day celebrates all these mathematical heroes and more.
How to celebrate Math 2.0 Day
Celebrate Math 2.0 Day by encouraging people around you to know the significance of learning math. On this day you
can solve fun mathematical problems with your family and friends. Companies and universities will conduct many events, so you attend one of those events to know more about the day. This day is also observed in schools and communities. Spread your knowl edge of math and technology to others on social media. Encourage and teach people to use math in practical life. Be grateful for math as we have to deal with it in every minute of our life.
If you’re like us, you probably have your old math books from college laying around. We suggest busting them open and studying them again. Who knows, in the intervening years you may have secretly developed a love for those dancing numbers. If not, make sure that you stop by those people who use math every day and thank them for doing the work so you don’ t have to. Mathematics is one of the most important fields in the world today, and just about everything we know and love is built on its back. | 642 | 3,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | longest | en | 0.946151 |
https://de.mathworks.com/matlabcentral/cody/problems/8-add-two-numbers/solutions/1714032 | 1,601,427,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00791.warc.gz | 337,970,846 | 16,408 | Cody
# Problem 8. Add two numbers
Solution 1714032
Submitted on 23 Jan 2019 by Francesco Burroni
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 1; b = 2; c_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct))
2 Pass
a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct))
3 Pass
a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct)) | 166 | 522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-40 | latest | en | 0.645564 |
http://openstudy.com/updates/4fccde9ce4b0c6963ad7bf0a | 1,448,922,546,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464253.80/warc/CC-MAIN-20151124205424-00022-ip-10-71-132-137.ec2.internal.warc.gz | 171,745,439 | 11,388 | ## Wolfboy 3 years ago Aysha has 8 pictures to hang over her couch. She wants to hang only 3 of them. Find how many ways she can choose the 3 pictures from the 8
1. A.Avinash_Goutham
8p3
2. CliffSedge
8c3 if order doesn't matter.
3. CliffSedge
Are you familiar with Pascal's Triange?
4. Wolfboy
uh not realy
5. A.Avinash_Goutham
u kno wat 8p3 is?
6. Wolfboy
not exactly
7. CliffSedge
I'm pretty sure this is a combination and not a permutation.
8. A.Avinash_Goutham
yup that's a combination nd it's 8c3
9. CliffSedge
Can use either Pascal's Triangle or the formula with factorials. For n=8, probably better off using the formula.
10. CliffSedge
nCx = n!/(x!(n-x)!) I think that's it. I'm just pulling it from memory, so I might have it mixed up with the other one, but it looks right.
11. Wolfboy
ok so how do we use this to make our equation. And sorry for the slowness i went surfing yesterday and i am still really tiered.
12. CliffSedge
A quick drawing of Pascal's Triangle shows it's 56, and 8!(3!(8-3)!) = the same.
13. CliffSedge
Do you know how to write out a factorial?
14. Wolfboy
uh i probably do but dont remember can you show me again?
15. CliffSedge
While I do that, do an internet search for "Pascal's Triangle." - It's useful not only for combinations, but also for finding the coefficients of an expanded power of a binomial.
16. CliffSedge
n! = n*(n-1)*(n-2)*...*1 e.g. 8! = 8*7*6*5*4*3*2*1 dividing factorials is easy because so many common factors cancel out. e.g. 8!/3! = 8*7*6*5*4
17. Wolfboy
Thanks for all of your help @CliffSedge | 495 | 1,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2015-48 | longest | en | 0.897879 |
https://allthingsstatistics.com/inferential-statistics/uniformly-most-powerful/ | 1,680,189,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949331.26/warc/CC-MAIN-20230330132508-20230330162508-00684.warc.gz | 123,386,971 | 53,329 | # Uniformly Most Powerful Tests
-
### What is a most powerful test?
Suppose that you want to conduct a hypothesis test for the value of a parameter θ with a simple point hypothesis given as
H0: θ=θ0 vs H1: θ=θ1
Out of the many possible critical regions with size α, the test having the critical region with the greatest power among all of the others is called the most powerful test.
### What is uniformly most powerful test?
Now suppose that you want to conduct a hypothesis test for the value of a parameter θ with a simple null hypothesis but a composite alternative hypothesis. A composite alternative hypothesis means that the parameter can take a range of values instead of a single point value.
H0: θ=θ0 vs H1: θ≠θ0
Out of the many possible critical regions with size α, the test having the critical region with the greatest power among all of the others is called the uniformly most powerful test.
### Difference between most powerful and uniformly most powerful test:
The main difference is that a most powerful test has the greatest power among all tests of size α when the alternative hypothesis is a simple hypothesis (the parameter takes a single point value in a simple hypothesis).
In contrast, the uniformly most powerful test has the greatest power among all tests of size α when the alternative hypothesis is a composite hypothesis (the parameter can take a range of values under a composite hypothesis).
### Uniformly Most Powerful tests and Likelihood Ratio:
It is known as a consequence of the Neyman Pearson lemma that the likelihood function gives us the most powerful tests when both hypotheses are simple. Sometimes, we can obtain uniformly most powerful tests for some composite hypotheses by treating a composite hypothesis as a collection of many simple hypotheses.
### Formal Mathematical Definition of Uniformly Most Powerful Tests:
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https://byjus.com/question-answer/find-the-total-surface-area-of-a-hemisphere-of-radius-10-cm/ | 1,674,911,475,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00861.warc.gz | 160,585,611 | 21,623 | Question
# Find the total surface area of a hemisphere of radius 10 cm.
Open in App
Solution
## Given, radius of hemisphere (r) =10 cm ∴ total surface area =3πr2=3×227×10×10 cm2=66007cm2=942.857cm2
Suggest Corrections
19 | 73 | 224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-06 | latest | en | 0.60809 |
https://doubtnut.com/question-answer/the-area-of-a-rectangular-garden-50-m-long-is-300-sq-m-find-the-width-of-the-garden-4614 | 1,618,343,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038074941.13/warc/CC-MAIN-20210413183055-20210413213055-00331.warc.gz | 334,650,331 | 68,955 | Class 6
MATHS
Mensuration
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden
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Very Important Questions
Here, length of garden, l = 50m<br> Area of garden, A = 300m^2<br> We know, Area of a rectangle = Length*Breadth<br> A = l**b<br> 300 = 50**b => b = 300/50 = 6m<br> So, width of the garden will be 6m. | 452 | 1,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-17 | latest | en | 0.497795 |
https://www.mathworks.com/matlabcentral/cody/problems/19-swap-the-first-and-last-columns/solutions/116094 | 1,508,255,686,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822116.0/warc/CC-MAIN-20171017144041-20171017164041-00430.warc.gz | 948,926,719 | 11,513 | Cody
# Problem 19. Swap the first and last columns
Solution 116094
Submitted on 20 Jul 2012 by Abby Skofield
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% A = [ 12 4 7 5 1 4]; B_correct = [ 7 4 12 4 1 5 ]; assert(isequal(swap_ends(A),B_correct))
2 Pass
%% A = [ 12 7 5 4]; B_correct = [ 7 12 4 5 ]; assert(isequal(swap_ends(A),B_correct))
3 Pass
%% A = [ 1 5 0 2 3 ]; B_correct = [ 3 5 0 2 1 ]; assert(isequal(swap_ends(A),B_correct))
4 Pass
%% A = 1; B_correct = 1; assert(isequal(swap_ends(A),B_correct)) | 235 | 643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-43 | latest | en | 0.555674 |
https://web2.0calc.com/questions/help-me_83 | 1,552,977,944,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201904.55/warc/CC-MAIN-20190319052517-20190319074517-00351.warc.gz | 674,482,967 | 5,465 | +0
Help Me?
0
307
2
+82
A square feild has an are of 479 ft^2. What approximate length of a side of the feild ? Give your answer in the nearest foot. Explain your resonse.
Oct 31, 2017
#1
+98005
+1
Area = side ^2 .....so......
√ Area = side = √ 479 ≈ 22 ft [ rounded to the nearest foot ]
Oct 31, 2017
#2
+82
0
Gracias
Oct 31, 2017 | 134 | 349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-13 | latest | en | 0.756278 |
lindenleadership.com | 1,721,880,012,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00756.warc.gz | 314,535,711 | 18,650 | # Mathematical Monsters
### The Machines of Mathematics
I’m going to digress briefly – and I promise to keep it brief – into the language of mathematics. This is important, because with this language the idea of feedback is much easier to understand. I promise to keep it short and simple; I think you’ll find it quite painless, and maybe even fun.
Mathematicians invent machines all the time – it’s part of their work. But most people don’t know that, because mathematicians don’t call them machines; they call them “functions”. Mathematicians are really good at creating language that is accurate and meaningful for them, even though it’s confusing for the rest of us. A function in math is just a machine that turns one thing into something else.
Think of making coffee. If you’re really into coffee, you grind your own beans. So when you get up in the morning and want to start your day with a great pot of coffee, you put the beans into the coffee grinder, press the button, and after a bit of time and a bit of noise, ground coffee comes out. The coffee grinder is a machine that turns the beans into coffee grounds. Then you put the grounds into a coffee maker, and out comes coffee – the coffee maker is a machine that turns the grounds into coffee. Mathematicians would call the coffee grinder and the coffee maker “functions.”
In English we would describe the machines for making coffee as:
The coffee grinder turns the coffee beans into coffee grounds.
The coffee maker turns the coffee grounds into coffee.
Mathematicians like to keep their sentences very concise and short, in order to say a lot with as few symbols as possible – they love efficiency. So they typically name their machines with a single letter, often f or g, though they can use any symbol they want. If you were a mathematician you might call the coffee grinder “g” and the coffee maker “m”, and write:
The g turns the coffee beans into coffee grounds.
The m turns the coffee grounds into coffee.
That’s a little shorter, but mathematicians want to spend as little time as possible writing stuff down, so they also name the things that go into their machines with single letters, often x, or y, or z, though again they could use any symbol they want. And when something has been changed by a machine, they indicate that with a symbol that shows it’s been changed – e.g., a subscript or superscript. So they might call the coffee beans “x”, the ground coffee x1 and the coffee itself x2. (x1 is read as “x sub 1” and x2 as “x sub 2”.) Then you could write:
The g turns the x into x1 (in English: the coffee grinder turns the coffee beans into coffee grounds)
The m turns the x1 into x2 (in English: the coffee maker turns the coffee grounds into coffee)
This is getting shorter and more concise (and also harder to understand), but mathematicians still aren’t satisfied; they really want to get rid of anything that has to be spelled out, so they throw away the word the altogether, they With the word “turns” with parentheses and the word “into” with an equal sign. So they would write the above as:
g(x) = x1 (in mathematics you would say “g of x equals x sub 1”; in English: the coffee grinder turns the coffee beans into coffee grounds)
m(x1) = x2 (in mathematics you would say “m of x sub 1 equals x sub 2”; in English: the coffee maker turns the coffee grounds into coffee)
Now the statements are about as short and concise – and unintelligible to non-mathematicians – as possible. But if you understood the language of math you would know exactly what this meant, and you could write about it and talk about it easily.
One of the things that makes mathematics fun is that the machines (functions) that mathematicians build are abstract, so they are built as fast as you can think of them – you don’t have to take the time to physically build anything; as soon as you have the idea, you have the machine. And if you tell someone about the idea, then they have the machine too. They can think of a way to change the machine, and as soon as they tell you about it, you have the changed machines as well. And the machines can do anything you can imagine, because they are not constrained by the laws of physics. So, for example, mathematicians make up “reverse” machines – ones that undo what other machines have done. You could make up a machine that turns coffee into coffee grounds, and another that turns coffee grounds into coffee beans.
Mathematicians call reverse machines “inverse functions”. They use a negative 1 superscript (-1) to show this. Remember our plain English version of the coffee grinder? It was:
The coffee grinder turns the coffee beans into coffee grounds.
If you had a reverse machine that turned grounds into beans, you would write:
The reverse coffee grinder turns the coffee grounds into coffee beans
Using the inverse notation of mathematics you would write:
The coffee grinder-1 turns the coffee grounds into coffee beans (you would read this as “The inverse coffee grinder turns the coffee grounds into coffee beans)
Remember that the mathematical notation for the coffee grinder was:
g(x) = x1 (in English: the coffee grinder turns the coffee beans into coffee grounds)
The mathematical notation for the inverse coffee grinder would be:
g-1(x1) = x (in mathematics speak you would say “g inverse of x1 equals x”)
There is another reason, beyond efficiency, that mathematicians created this shorthand notation, and it’s actually quite useful. Since their machines are abstract, they don’t have to worry about what they put into them – they could put cardboard into their grinder, and get ground up cardboard out of it. This is handy because it means that you can work with the concepts without being constrained by reality. The concept here is a “grinder”, but in the real world we have to specify what the grinder is intended to grind, so we buy the right kind and don’t put the wrong stuff into it. In the world of mathematics we can just think of the concept “grinder” and think of it as working with anything. So by using this short, generic notation, when they write:
g(x) = x1
they can With the “x” with anything they want. They could say g(coffee) and know they’d get ground coffee out, or g(cardboard) and know they’d get ground cardboard out, and so on. Their notation allows them to think of the essential concepts, without being distracted by physical details that aren’t important to the concept. The downside to their notation is, you have to know the language, and you have to be good at thinking about pure concepts without having them related to anything physical – a challenge for many of us. Fortunately there are people who are good at that, because without mathematics we would never have been able to create all of the wonderful inventions that make our lives so rich.
Of course the machines that mathematicians create don’t work on coffee beans, or cardboard, or any other physical thing. They work on numbers; they take numbers and turn them into other numbers. So, for example, a mathematical machine might take a number and double it – i.e., multiply it by 2. If that foundation was called “f” it would look like this:
f(x) = 2x (in English: take whatever number you start with, and multiply it by two.
Mathematical notation works well for mathematicians, because they speak the language of mathematics, but for those of us who don’t speak that language, their writing looks very complicated and challenging – giving the impression you’d have to be really really smart to be able to make sense of it. (I think that’s another reason they do it.)
The ability to focus on, and think and communicate about, essential concepts, free of the constraints of the physical world, is very powerful and will be useful as we explore the notions of feedback and systems thinking. But I’ll limit it as much as possible, I’ll keep it simple, and I will strive to explain what the language means when I do use it.
#### 1.1.1.1.3. Lines, Planes and Cubes
In addition to functions, which deal with numbers, it’s important to understand a few things about the branch of mathematics called geometry. Geometry deals with shapes, just like algebra deals with numbers. Shapes area things like lines, planes and cubes.
A straight line looks like this:
————-
and a curved line looks like this:
Here’s a line with both flat and curved parts:
In addition to segments that are straight or curved, a line can have points that are neither flat nor curved – like the sharp points where straight and curved segments meet in this figure:
In mathematics, lines are “one dimensional” – they have no width, only length. In a sense, lines are the idea of length. Imagine a thread stretched from left to right, so it’s perfectly straight. The thread has some width and height, even if it’s a very fine thread. Now imagine the thread’s width and height shrinking – whatever thickness your imaginary thread had when you thought of it, think of it as half as thick, while it’s length stays the same. Now imagine it shrinking again, and again, and again, until it has no thickness at all – it’s just a length. That’s the mathematical idea of a line. Of course you can’t really visualize it, because it has no thickness, but you can still think of it.
Lines can be curved – all of the pictures above are lines – but from a mathematical perspective they have no width. So they can only be measured in terms of length, not in terms of width or height. Since these imaginary lines have no width, no matter how many of them you put side by side on a flat surface, they would never take up any space on a flat surface – they are one dimensional, so they can never occupy any space in two dimensions. You can measure the length of a line, but it will never cover any area.
A surface – like a piece of paper – is two dimensional. It has both length and width. A surface looks like this:
But a surface has no height. A plane is the idea of an area. Imagine a piece of paper laying flat on your desk. You know that if you put several hundred of them on top of each other they will create a stack that might be an inch or two tall. But now imagine that piece of paper has half the thickness it originally had – your stack would be half as tall. Now imagine it half again as thick, and half again, and so on, until it has no thickness at all. At that point your stack has no height. That’s the mathematical idea of a plane, it extends in two dimensions but not three. Each of its edges is a line, and you can measure their length. By multiplying the length and width you can measure it’s area. But it has no volume.
And a three dimensional object – a box, or a ball, or any real object – has length, width and height. With a box, say, you can measure its length, it’s width, and its height. Each of its edges is a line, so you can measure their lengths; each of its sides is a plane, so you can measure their areas; and the entire box is a cube, so you can measure its volume. A cube looks like this:
So it makes sense to think of dimensions in terms of whole numbers – 1, 2 and 3. Mathematicians even make up things that have more dimensions, like 4 or 5 or more, but there is always a whole number of dimensions – or so they believed for a long time. They’re pretty hard to imagine, but in mathematics they are useful.
#### 1.1.1.1.4. Important points about the monsters
They diminished our sense of certainty about that field that was supposed to be the bastion of certainty. One more chink was taken out of the armor of certainty.
The machines all used feedback to create the monsters.
They were fundamentally very simple machines – they didn’t do anything very complex – yet they gave rise to these bizarre, and in some cases extraordinarily complex, structures.
Nature loves efficiency, so she is bound to discover anything that create complex but orderly systems with the greatest economy possible.
#### 1.1.1.1.5. Back to the Monsters
So now that you understand the notion of a function (i.e., a mathematical machine), and of lines, planes and cubes, we can get back to the interesting things – the monsters, and feedback.
There are certain basic beliefs that mathematicians held on to for centuries – and with good reason, because these principles gave them a sense of order, a sense that the world – at least the world of mathematics – could be understood.
#### Every Sharp Point on a Line is Connected by a Straight or Curved Segment
For example, they believed that any line you could draw had to have at least some areas that were either flat, or smooth curves. To understand this, consider the following. Recall that earlier I showed you that a line can consist of straight segments, curved segments, and sharp points where different segments meet at an angle. It’s obvious, if you think about it, that no line could consist only of sharp points – there would always have to be a segment between any two sharp points that connected them, that was either a straight line or a curve. After all, what makes a spot pointy is that two different straight or curved sections come together at an angle. For example, this line has lots of sharp points, but they are always connected by a either a straight or curved section:
One of the beliefs that mathematicians held dear was that no line could consist only of sharp points. This seems obvious – that any continuous line (one without any gaps in it) would have to have sections that were either straight or curved between any pointy spots.
And yet, in 1904 Helge von Koch, a Swedish mathematician, introduced what is now called the “Koch Snowflake.” The Koch Snowflake is produced by a machine that Koch invented that starts with a simple triangle, and goes through a very simple process. Here’s what Koch’s machine does:
Start with an equilateral triangle (an equilateral triangle is a triangle with all three sides the same length), like this one:
Divide each straight line of the triangle (i.e., the sides) into three equal parts, and With the middle segment with another equilateral triangle, whose sides are 1/3 the size of the original:
Now remove the bottom of the triangles you just added (the lines shown in green above) and you get the first result of Koch’s machine:
Now repeat this – take the image above and feed it back into the machine, which Withs each straight line with a triangle whose sides are 1/3 the size of the one just used. Feeding the image above back into the machine produces this:
And if you feed this image back into the machine, replacing each straight line with a triangle whose side are 1/3 the size of the one used in this step, you get:
You can see where this is going. If you run this machine an infinite number of times, all of the straight lines disappear and all you are left with is the points. There are no straight or curved segments, even though it is a continuous, unbroken line going around the snowflake. It is impossible to imagine what this would look like, but Koch didn’t care about that – he was concerned with demonstrating that you could, mathematically, create such a structure, thus violating one of the things that mathematicians felt quite certain about.
#### A Box Has Volume
Another belief mathematicians held strongly was that if you made a box, it would have to enclose some measurable volume. For example, a box that’s 1 foot on each side would hold one cubic foot of space. Obviously, any box that has measurable sides must contain some space inside the box. And any collection of boxes would therefore also have to contain some space inside them.
In 1926 Karl Menger, an Austrian mathematician who eventually emigrated to the United States and taught mathematics at Notre Dame and Illinois Institute of Technology, wrote a paper describing a machine that produces something known as the “Menger Sponge.” Like Koch’s machine, it starts with a simple figure – in this case a cube:
Menger’s machine takes a cube that is put into it, divides each side of the cube into nine equal squares, which turns the big cube into a set of 27 smaller cubes. The machine then removes the middle cube from each side, and the cube at the center, leaving a set of 20 smaller cubes – or a large cube with square holes drilled through the middle of each side, depending on how you choose to look at it. It looks like this:
Now when you put this construction back into Menger’s machine, it takes each individual cube and repeats the process, producing this:
Now you repeat with this construction, which results in this:
It looks a bit like a hotel in a science fiction horror film, one where you might enter but never leave, or try to enter but keep finding yourself forever outside.
As with the Koch Snowflake, you repeat this process an infinite number of times, and you end up with a collection of boxes that, taken together, have an infinite surface area – that is, if you took all of their sides and laid them out next to each other, they would stretch to infinity in all directions – and yet no volume whatsoever, because you’ve removed all the space inside the boxes.
This seems crazy, and you certainly can’t visualize it, but the mathematical result is a set of boxes that contain no space. Once again the sense of certainty that mathematicians loved was challenged and found wanting.
#### Dimensions Come In Whole Numbers
Yet another belief that mathematicians held dear was that dimensions only exist in whole numbers. A line is one dimensional, a surface two dimensional, and a cube three dimensional. But the idea of a fractional dimension – an object that was, say, 2.5 dimensional, would be absurd, and mathematics should not allow it.
But it turns out that when you study the question of dimension carefully, it is nowhere near as straightforward as it seems. And in the 1960’s Benoit Mandelbrot, a brilliant multi-disciplinary scientist, demonstrated that in fact there are many objects, both in nature and in mathematics, that exhibit fractional dimensions. For example, the fractal dimension of the Koch Snowflake is about 1.26, and the fractal dimension of the Menger Sponge is about 2.73.
I’ve just described three beliefs mathematicians held for centuries – beliefs that make perfect sense to us, so much sense that if someone told us that any of these beliefs were wrong, we’d be likely to think that person might be somewhat wrong in the head.
With the little bit of math that I’ve shown you, there are some fun and fascinating ways to explore the power of feedback. The Chaos game is one …
What all of these have in common is that they are self-referential – that is, the product of each step provides the raw material for the next step. It turns out that anytime something becomes self-referential, the situation is pregnant with complexity unpredictability. As you will see shortly, this has far-reaching implications for how human beings live, work, and interact with each other.
It would be easy to write these things off as the musings of nerdy mathematicians who had nothing better to do with their time than think up wild ideas. In fact that is exactly how many of the greatest mathematicians initially reacted to these mathematical machines that were being designed. And yet, throughout the 20th century, as scientists encountered more and more aspects of nature that simply didn’t yield to more traditional techniques, they found themselves turning the fascinating emerging field of chaos theory and fractals, which explores how these monsters actually work. And what they found was that long before human beings showed up, nature had also discovered these monsters, and put them to good use.
The mathematics of fractals enables extraordinarily efficient techniques for solving difficult problems, and nature takes full advantage of them – e.g., to name a few: the growth pattern of trees; the relationship between an animals physical size and it’s life span; the distribution of human cities of different sizes; and the distribution of earthquakes of different magnitudes all appear to be governed by fractal mathematics. Nature uses fractal mathematics in the design of lungs in animals, to achieve the maximum possible surface area in the smallest amount of space, so that a very compact lung can absorb large amounts of oxygen quickly.
And, now that scientists are respecting and studying this mathematics instead of writing it off, they are finding more and more uses for it. To name a couple: printer manufacturers use fractal mathematics to improve the quality of the images their printers can make, and cell phone companies use fractal mathematics to improve the efficiency with which cell phone signals can be distributed across their towers. And geneticists are beginning to understand that the dynamics of feedback are deeply woven into the mechanisms of genetics – how we inherit attributes from our parents; how genes turn each other on and off; how some genes alter the structure of other genes; what determines whether a gene that predisposes you to cancer gets expressed or remains dormant. The more we come to understand the dynamics of feedback, the more profoundly we see them shaping all aspects of our lives. | 4,481 | 21,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.96619 |
https://biddlebuzz.com/esp/c%C3%B3mo-describir-treinta-regalos-c%C3%B3mo-empez%C3%B3-treinta-y-uno.html | 1,558,314,622,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255251.1/warc/CC-MAIN-20190520001706-20190520023706-00438.warc.gz | 413,639,616 | 6,811 | ### My customers who are all over the country are so very important to me. I love that as we are picking out the best tote for their next vacation, what purse style is best, what gift they could give, or what organization item is best for their home that I get to have a relationship and get to know my customers. We talk about hard times, pray for each other, and laugh a lot too! Women need each other – life is hard – and having someone to say “you’re amazing” can go a very long way.” -Krista https://i.pinimg.com/originals/60/8f/c5/608fc5e4dbb40e904c55fc2a7f0ef3b9.jpg
31 is the third Mersenne prime (25 − 1)[1] and the eighth Mersenne prime exponent, as well as the fourth primorial prime, and together with twenty-nine, another primorial prime, it comprises a twin prime. As a Mersenne prime, 31 is related to the perfect number 496, since 496 = 2(5 − 1)(25 − 1). 31 is also the 4th lucky prime[2] and the 11th supersingular prime.[3] https://i.pinimg.com/originals/0b/d8/7d/0bd87d982f0fe5ddcb4b268ef8c50202.jpg
Aubrey – I have two 31 Utility Totes, two bags from Amazon and one from Walmart. The bags from Amazon are cheaper than 31 bags but are every bit as good in terms of quality. Further, the Walmart bag cost the least but, in my opinion, is the best bag. The only downside is there aren’t many fabric choices. Also, I had another 31 bag that I bought as a work bag, it was expensive and it fell apart. I do like the 31 bags (I think the fabric is pretty) but I would rather save money.
My customers who are all over the country are so very important to me. I love that as we are picking out the best tote for their next vacation, what purse style is best, what gift they could give, or what organization item is best for their home that I get to have a relationship and get to know my customers. We talk about hard times, pray for each other, and laugh a lot too! Women need each other – life is hard – and having someone to say “you’re amazing” can go a very long way.” -Krista
Aubrey – I have two 31 Utility Totes, two bags from Amazon and one from Walmart. The bags from Amazon are cheaper than 31 bags but are every bit as good in terms of quality. Further, the Walmart bag cost the least but, in my opinion, is the best bag. The only downside is there aren’t many fabric choices. Also, I had another 31 bag that I bought as a work bag, it was expensive and it fell apart. I do like the 31 bags (I think the fabric is pretty) but I would rather save money. https://i.pinimg.com/originals/b7/da/a9/b7daa96ef6db77eb48643a96d7d3ec47.jpg
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I came across your article today and I would like to know if you personlly have used both versions, the cheap and Thirty One? I’ve been with the company 6 years now and have to say every penny I’ve spent has been worth it. I still have the first two Large a Utility Totes I bought 6 years ago and use them. Just because it’s cheaper that doesn’t mean it is better. Thirty Ones products actually hold up pretty well. I admit like any other company we do have some problems from time to time but they are good at helping customers fix these issues. You have a 90 day warrrenty and sometimes even after that they help. https://i.pinimg.com/236x/fe/b5/fa/feb5fa362156f1adc9b69d8db1c91637.jpg
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I came across your article today and I would like to know if you personlly have used both versions, the cheap and Thirty One? I’ve been with the company 6 years now and have to say every penny I’ve spent has been worth it. I still have the first two Large a Utility Totes I bought 6 years ago and use them. Just because it’s cheaper that doesn’t mean it is better. Thirty Ones products actually hold up pretty well. I admit like any other company we do have some problems from time to time but they are good at helping customers fix these issues. You have a 90 day warrrenty and sometimes even after that they help. https://i.pinimg.com/originals/2e/dd/31/2edd313e742c70d70126a76c87488930.png
```Have you ever received a Thirty One Gifts catalog party letter? If you have then you know how fun 31 Gifts parties can be. If you've ever attended a Thirty One Gifts catalog party or browsed the Thirty One Gifts catalog PDF, then you've certainly noticed how personalization makes the product special. Thirty One Gifts products are meant for the women they support and accessorize - they’re made to withstand the rigors of daily life and the demands of errand-filled afternoons. Popular models like crossbody totes and messenger bags securely stash your belongings while still keeping you looking fashionable. These are bags made for everyday, not gathering dust in a closet between special occasions. Fun, fresh and perfectly personalized, Thirty One Gifts makes products that women truly love to use, year after year. Contact a 31 Gifts representative to set up your own hosted party where you can earn free gifts. https://i.pinimg.com/236x/bf/98/4e/bf984e4414099091f83cd036c42e1813--book-jacket-messages.jpg | 1,580 | 6,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-22 | latest | en | 0.971117 |
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# 4.9: Trigonometry in Terms of Algebra
Difficulty Level: At Grade Created by: CK-12
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Practice Trigonometry in Terms of Algebra
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You are babysitting your little cousin while doing your homework. While working on your trig functions, your cousin asks you what you are doing. While trying to explain sine, cosine, and tangent, your cousin is very confused. She doesn't understand what you mean by those words, but really wants to understand what the functions mean. Can you define the trig functions in terms of the relationships of sides for your little cousin?
By the end of this Concept, you'll understand how to do this.
### Watch This
Unit Circle Definition of Trig Functions
### Guidance
All of the trigonometric functions can be rewritten in terms of only x\begin{align*}x\end{align*}, when using one of the inverse trigonometric functions. Starting with tangent, we draw a triangle where the opposite side (from θ\begin{align*}\theta\end{align*}) is defined as x\begin{align*}x\end{align*} and the adjacent side is 1. The hypotenuse, from the Pythagorean Theorem would be x2+1\begin{align*}\sqrt{x^2+1}\end{align*}. Substituting tan1x\begin{align*}\tan^{-1} x\end{align*} for θ\begin{align*}\theta\end{align*}, we get:
tanθtanθθ=x1=x=tan1xhypotenuse=x2+1
sin(tan1x)cos(tan1x)tan(tan1x)=sinθ=xx2+1=cosθ=1x2+1=tanθ=xcsc(tan1x)=cscθ=x2+1xsec(tan1x)=secθ=x2+1cot(tan1x)=cotθ=1x
#### Example A
Find sin(tan13x)\begin{align*}\sin (\tan^{-1} 3x)\end{align*}.
Solution: Instead of using x\begin{align*}x\end{align*} in the ratios above, use 3x\begin{align*}3x\end{align*}.
sin(tan13x)=sinθ=3x(3x)2+1=3x9x2+1
#### Example B
Find sec2(tan1x)\begin{align*}\sec^2 (\tan^{-1} x)\end{align*}.
Solution: This problem might be better written as [sec(tan1x)]2\begin{align*}[\sec (\tan^{-1} x)]^2\end{align*}. Therefore, all you need to do is square the ratio above.
[sec(tan1x)]2=(x2+1)2=x2+1
You can also write the all of the trig functions in terms of arcsine and arccosine. However, for each inverse function, there is a different triangle. You will derive these formulas in the exercise for this section.
#### Example C
Find csc3(tan14x)\begin{align*}\csc^3 (\tan^{-1} 4x)\end{align*}.
Solution: This example is similar to both of the examples above. First, use 4x\begin{align*}4x\end{align*} instead of x\begin{align*}x\end{align*} in the ratios above. Second, the csc3\begin{align*}\csc^3\end{align*} is the same as taking the csc\begin{align*}\csc\end{align*} function and cubing it.
[csc(tan14x)]3=(4x)2+14x3=(16x2+1)3/264x3
### Guided Practice
1. Express cos2(tan1x)\begin{align*}\cos^2 (\tan^{-1} x)\end{align*} as an algebraic expression involving no trigonometric functions.
2. Express cot(tan1x2)\begin{align*}\cot (\tan^{-1} x^2)\end{align*} as an algebraic expression involving no trigonometric functions.
3. To find trigonometric functions in terms of sine inverse, use the following triangle.
Determine the sine, cosine and tangent in terms of arcsine. Find tan(sin12x3)\begin{align*}\tan (\sin^{-1} 2x^3)\end{align*}.
Solutions:
1. 1x2+1\begin{align*}\frac{1}{x^2+1}\end{align*}
2. 1x2\begin{align*}\frac{1}{x^2}\end{align*}
3. The adjacent side to θ\begin{align*}\theta\end{align*} is 1x2\begin{align*}\sqrt{1-x^2}\end{align*}, so the three trig functions are:
sin(sin1x)cos(sin1x)tan(sin1x)=sinθ=x=cosθ=1x2=tanθ=x1x2
tan(sin1(2x3))=2x31(2x3)2=2x314x6
### Concept Problem Solution
As it turns out, it's very easy to explain trig functions in terms of ratios. If you look at the unit circle
you can see that each trig function can be represented as a ratio of two sides. The value of any trig function can be represented as the length of one of the sides of the triangle (shown with two red sides and the black hypotenuse) divided by the length of one of the other sides. In fact, you should explain to your cousin, the words like "sine", "cosine", and "tangent" are just conveniences in this case to describe relationships that keep coming up over and over again. It would be possible to just describe the trig functions in terms of relationships of one side to another, if you'd like.
Using the sides of a triangle made on the unit circle, if the side opposite the angle is called "x":
So as you can see, since trig functions are really just relationships between sides, it is possible to work with them in whatever form you want; either in terms of the usual "sine", "cosine" and "tangent", or in terms of algebra.
### Explore More
Rewrite each expression as an algebraic expression involving no trigonometric functions.
1. sin(tan15x)\begin{align*}\sin (\tan^{-1} 5x)\end{align*}
2. cos(tan12x2)\begin{align*}\cos (\tan^{-1} 2x^2)\end{align*}
3. cot(tan13x2)\begin{align*}\cot (\tan^{-1} 3x^2)\end{align*}
4. sin(cos1x)\begin{align*}\sin (\cos^{-1} x)\end{align*}
5. sin(cos13x)\begin{align*}\sin (\cos^{-1} 3x)\end{align*}
6. cos(sin12x2)\begin{align*}\cos (\sin^{-1} 2x^2)\end{align*}
7. csc(cos1x)\begin{align*}\csc (\cos^{-1} x)\end{align*}
8. sec(sin1x)\begin{align*}\sec (\sin^{-1} x)\end{align*}
9. cos2(tan13x2)\begin{align*}\cos^2 (\tan^{-1} 3x^2)\end{align*}
10. sin(sec1x)\begin{align*}\sin (\sec^{-1} x)\end{align*}
11. cos(csc1x)\begin{align*}\cos (\csc^{-1} x)\end{align*}
12. sin(tan12x3)\begin{align*}\sin (\tan^{-1} 2x^3)\end{align*}
13. cos(sin13x)\begin{align*}\cos (\sin^{-1} 3x)\end{align*}
14. sin(sec1x)\begin{align*}\sin (\sec^{-1} x)\end{align*}
15. cos(cot1x)\begin{align*}\cos (\cot^{-1} x)\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 4.9.
### Vocabulary Language: English
Trigonometric Function
Trigonometric Function
A trigonometric function is a function of an angle that describes the relationship between two sides of a right triangle. Examples of trigonometric functions are sine, cosine, and tangent.
Trigonometric Functions
Trigonometric Functions
A trigonometric function is a function of an angle that describes the relationship between two sides of a right triangle. Examples of trigonometric functions are sine, cosine, and tangent.
## Date Created:
Sep 26, 2012
Feb 26, 2015
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Sin
http://functions.wolfram.com/01.06.16.0118.01
Input Form
a Sin[z] + b Cos[z] == a Sqrt[1 + b^2/a^2] Sin[z + ArcTan[b/a]]
Standard Form
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MathML Form
a sin ( z ) + b cos ( z ) a 1 + b 2 a 2 sin ( z + tan - 1 ( b a ) ) a z b z a 1 b 2 a 2 -1 1 2 z b a -1 [/itex]
Rule Form
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Date Added to functions.wolfram.com (modification date)
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# un_ordinateur
Member Since: December 5, 2009
Country: United States
• combining GPS and accelerometers data…
ALSO, I hav juste realsed, it might be mor precise to use the magnetometer as the rough estimation source for the roll, pitch and yaw, since it will not suffer from translationnal movement. I had not tought of that initially, since I did not have one when I made that algorythm.
Sorry for the bad English, French is my native language.
I hope this helps you, and gets you started!
• On short time intervals, the gyroscope will make sure the integrated angle reacts quicky and accurately; and the accelerometer produces neglegible effects. However, that correction ensures long ther fiability on the readings.
However, I am sorry, I don’t have any idea about a precise calculation of the confidence for this method…
Now, in your case, you can use this method with the two rotation axis parallel to the ground: each of these axis will have their own gyroscopes, and use two accelerometer axes. (They share z).
For the last axis, yaw, you can use the same method, but instead of the accelerometer, which would be useless, you use the magnetometer to provide the rough estimate of the orientation.
Now, I have NO idea how to get absolute lat/long/alt position of your robot; it is a similar yet different problem, and I have not had the occasion of working on it. However, as I know them, GPS can be really precises! (but under 1m, meaby you could use a similar methot for
• … small error that piles up pretty fast, and you move away from the expected value. To correct this; after each iteration, you “correct” the integrated angle with the angle calculated from the accelerometer. We may not have a lot of trust in this calculated angle; but we know it cannot be THAT FAR from the true angle, so correcting the integrated angle with it prevents the drift from growing out of bound. The correction is pretty simple: basically, each iteration, you add to your integrated angle about 1/1000 (find the best value experimentally; depends on your sensors) of the difference between the integrated angle and the calculated angle. So it will move the integrated angle a bit nearer calculated angle from the acceleremoters. (So, for example, my integrated angle is 10,0010°, but now my accelerometer says 10,5000°. The difference is 0,4990°, 1/1000 of that is (about) 0,0005°, so the corected angle is 10,0015°)
That way, your integrated angle will be really near the true angle:
• …but you cannot tell the proper absolute orientation. However, it produces really clean data, very sensitive and vast acting, and is not affected by translationnal movements.
The solution here consists of integrating over time the gyros measures: For example, you start with an estimation of your current orientation from the accelerometers data. (lets say; 10°) You wait (for example) 1 millisecond, then you read the gyro: You consider the returned angular speed (for example, 1°/s) as having been constant during that millisecond (which is pratically true since the time interval is small). Therefore, during that time, the robot orientation moved of 0,001° (1°/s * 0,001s); and it’s current orientation is now 10,001°. You repeat the process as fast as you can, to get the most precises measuremnts (the smallest the time interval, the most precises the measure).
However, as you noted, the suffers from drift: sincw you cannot reduce the time interval to zero, each iteration introduces a
• I’m not Pete, but I tought I could help a bit since I was faced with a similar problem last year; and found a solution working prety well.
However, I only had a two axis accelerometer and a single axis gyro, but my solution can be extended pretty easyly.
I’m assuming here your accelerometer is fixed with the body of your robot; so it tilts with your robot.
So first of all, to get your robot inclinaison (roll and pitch), theorically, only an accelerometer would be sufficient: using the atan function and simple trigo, you sould be able tell the angle between the robot and the vertical. However; tis only worls when the robot is stationary; as soon as it moves, the accelerometers catches other things, which do not depend on your orientation, and it spoils your measure. Furthermore, it has a slow reaction time.
On the other hand, an gyroscope does not give orientation in space, but only angular speed: therefore, you can tell wether you are rotating, how fast, and in witch direction,
• What is the difference between this Gyro and, say, ADXRS610 you sell too? Both measure +/- 300 ?/s.
This one is powered with 3.3V and ADXRS610 uses 5V. This is the only direct Difference I see, yet ADXRS610 is twice the price of LISY300AL! What explains that?
Thank you.
No public wish lists :(
In 2003, CU student Nate Seidle blew a power supply in his dorm room and, in lieu of a way to order easy replacements, decided to start his own company. Since then, SparkFun has been committed to sustainably helping our world achieve electronics literacy from our headquarters in Boulder, Colorado.
No matter your vision, SparkFun's products and resources are designed to make the world of electronics more accessible. In addition to over 2,000 open source components and widgets, SparkFun offers curriculum, training and online tutorials designed to help demystify the wonderful world of embedded electronics. We're here to help you start something. | 1,307 | 5,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-22 | longest | en | 0.944608 |
http://www.reddit.com/r/askscience/comments/xtqq5/what_if_we_put_a_mirror_in_space_1_light_year/c5pllav | 1,430,002,560,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246651727.46/warc/CC-MAIN-20150417045731-00041-ip-10-235-10-82.ec2.internal.warc.gz | 758,452,684 | 13,930 | you are viewing a single comment's thread.
[–]Nuclear Astrophysics | Radiation Therapy 11 points12 points (3 children)
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Yes, but it would also be red-shifted to infinite wavelength so we wouldn't be able to see it.
[–] 0 points1 point (2 children)
sorry, this has been archived and can no longer be voted on
I'm having trouble comprehending what that means. So the wavelength would be forever increasing?
[–]Nuclear Astrophysics | Radiation Therapy 1 point2 points (1 child)
sorry, this has been archived and can no longer be voted on
It wouldn't be increasing, it would be infinite.
Think about it this way, the light emitted from earth at some point is hitting the mirror, but the mirror is also moving away at the speed of light. Let's say a peak in the light wave is hitting the mirror. From our reference frame on earth, that one peak in the wave is stuck to the mirror and not changing with time (this is why the "image" isn't changing).
Now say the mirror isn't moving exactly c (which isn't possible anyway), but ever so slightly slower than c. Now that wave is slowly impinging on the mirror (as viewed from earth), so we see a very very slow frequency reflected back to us, aka a long wavelength. As the speed of the mirror approaches c, the reflected wavelength approaches infinity.
[–] 1 point2 points (0 children)
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As the speed of the mirror approaches c, the reflected wavelength approaches infinity.
And, it's worth pointing out, the amount of power reflected back to us will approach zero as the photons are able to catch up increasingly rarely. A mirror sufficiently close to c will look like a black spot in space, since it will be deflecting all light in its direction of movement and reflecting effectively nothing back to us. | 423 | 1,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2015-18 | latest | en | 0.966762 |
https://www.hackmath.net/en/math-problem/614 | 1,701,227,415,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00850.warc.gz | 926,951,159 | 8,971 | # Coins
Edmund had saved a certain number of 2-euro coins. He placed the coins in a single layer in a square. He has 6 coins left. When he wanted to build a square that would have one more row, he was missing 35 coins.
How many euros does he have?
n = 812 Eur
### Step-by-step explanation:
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We encourage you to watch this tutorial video on this math problem: | 160 | 697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.956442 |
http://www.kodugamelab.com/resources/obstacle_course | 1,726,320,853,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00223.warc.gz | 44,985,062 | 3,838 | # Resources
Obstacle Course
Home/Resources/Obstacle Course
## Obstacle Course
• Lesson 5: Obstacle Course
• Student Ages: 8-14 years old
• Activity Time: 50 minutes
• Activity Level: Beginner Coder
### Learning Objectives
• Students will design a computer simulation to represent the situation, analyze the resulting data, and display the data in a table or graph.
• Students will represent that object as a three-dimensional shape in a coordinate plane when given an object’s properties such as length, width, height, surface area, volume, and location.
### Student Activities
Activator
Students will complete the Student Activity: Obstacle Course Mini-Game. Students will use their pencil to draw paths to navigate an obstacle course. Students will draw lines from the starting point to each target and then come back to the starting point. Students will learn that the game they just played represents a common problem people have in deciding which paths are the most efficient between different locations. Students will create an obstacle course composed of the 3D quadrilaterals that they have been studying. The shapes will be designed according to detailed specifications. Subjects: Math, Computer Science, Digital Technology, Engineering, 21st Century Learning
Obstacle Course Game Demonstration The teacher should demo the Completed Game: Obstacle Course. The students will observe the grid that the world exists on. The students will observe the raised shapes of different sizes at different locations on the grid that compose the landscape of the game. Students will observe how to play the game. Students will observe the code for the Kodu, tree, apples, and timer. Students will observe the extra decorations and objects that compose the world. Subjects: Subjects: Math, Computer Science, Digital Technology, Engineering, 21st Century Learning
### Performance Expectations
Placement by Location Students should load a new world. Students will immediately save the world and give it a title. The student’s title should reflect the nature of the game. Students will create a world that starts as a 20 x 20 grid located. Students will change the brush size to 20 x 20. Students will learn to keep an overhead camera view throughout the grid placement process. Students will create the required obstacles as detailed in the specification sheet: Obstacle Course Construction. The teacher could create the first obstacle with the students to review how to maneuver the camera and change the brush size. Students will be following pair programming rules to collaborate.
### Skills
Character, Citizenship, Collaboration, Communication, Creativity, Critical Thinking, Project Based Learning | 528 | 2,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.896649 |
https://askinglot.com/how-do-you-find-the-slope-of-a-line-with-an-equation | 1,620,533,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988955.89/warc/CC-MAIN-20210509032519-20210509062519-00351.warc.gz | 129,670,151 | 6,775 | asked in category: General Last Updated: 30th April, 2020
# How do you find the slope of a line with an equation?
To find the slope of a line given the equation of the line, first write it in slope-intercept form. Use inverse operations to solve for y so that it is written as y=mx+b. Then you can easily see the slope since it is the coefficient of the x variable, or the number in front of x.
Likewise, people ask, how do you find the equation of a line?
To find the equation of a line using 2 points, start by finding the slope of the line by plugging the 2 sets of coordinates into the formula for slope. Then, plug the slope into the slope-intercept formula, or y = mx + b, where "m" is the slope and "x" and "y" are one set of coordinates on the line.
Beside above, how do you find a slope from a graph? Using the Slope Equation
1. Pick two points on the line and determine their coordinates.
2. Determine the difference in y-coordinates of these two points (rise).
3. Determine the difference in x-coordinates for these two points (run).
4. Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).
Also to know, how do you find the slope of a line in standard form?
The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line. When we have a linear equation in slope-intercept form, it's easy to identify the slope of the line as the number in front of x. The standard form of a linear equation is Ax + By = C.
What is the slope of a horizontal line?
Slope of a horizontal line. When two points have the same y-value, it means they lie on a horizontal line. The slope of such a line is 0, and you will also find this by using the slope formula.
7
30th April, 2020
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Questions
Videos
Users | 439 | 1,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-21 | latest | en | 0.938251 |
https://math.stackexchange.com/questions/938862/indefenite-integral-requiring-substitution | 1,571,735,452,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00333.warc.gz | 615,785,247 | 31,511 | # Indefenite Integral requiring substitution
Can someone please help me find a useful substitution for the following integral: $$\int \frac{1}{\sqrt{x}(1+\sqrt{x})^2}dx$$ I tried letting $u = \sqrt{x}$ But I couldn't proceed. Please help.
Setting $1+\sqrt x=u$ gives you $$\frac{dx}{\sqrt x}=2du.$$ So, $$\int\frac{1}{(1+\sqrt x)^2}\cdot\color{red}{\frac{dx}{\sqrt x}}=\int\frac{1}{u^2}\cdot \color{red}{2du}.$$
That's certainly a workable solution, as you can absorb the factor of $\sqrt{x}$ in the denominator when you write $du$, but you may as well take care of the constant in the parenthetical quantity and substitute
$$u = 1 + \sqrt{x},$$
so that $du = \frac{1}{2 \sqrt{x}}dx$, and your integral becomes
$$2 \int \frac{du}{u^2}.$$ | 238 | 739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-43 | latest | en | 0.78421 |
https://mathematica.stackexchange.com/questions/235279/how-can-i-plot-and-indicate-when-the-function-is-positive-or-negative | 1,713,093,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00494.warc.gz | 365,146,154 | 39,391 | # How can I plot and indicate when the function is positive or negative?
I would like to study the sign of this function (potential) and show the region (positive-negative)
These are the parameters followed by the code (left figure)
a = 1; c = 1;
b = -1; d = 1;
p = 1; q = -1;
f1= (Exp[-2 r a] p)/(2 r c) + (Exp[-2 r b] p)/(2 r d)
f2= ( Exp[-2 r a] q)/(2 r c) + (Exp[-2 r b] q)/(2 r d)
Plot[{ f1,f2 } , {r, -2, 2}, PlotRange -> Automatic]
As we see for this case, we obtain different signs for f1 and f2. In four different regions I would like to show them in the plot, A legend it possible only for two regions?
For example, If we change b and d with the same sign (d=-1 and b=-1) (right figure) have only two regions. Please any suggestion and recommendation will be usefull
• Do you mean something like Plot[{f1, f2 } , {r, -2, 2}, ColorFunctionScaling -> False, ColorFunction -> (If[#2 > 0, Blue, Red]&)]? Nov 25, 2020 at 6:47
a = 1; c = 1;
b = -1; d = 1;
p = 1; q = -1;
f1 = (Exp[-2 r a] p)/(2 r c) + (Exp[-2 r b] p)/(2 r d);
f2 = (Exp[-2 r a] q)/(2 r c) + (Exp[-2 r b] q)/(2 r d);
a = 1; c = 1;
b = -1; d = 1;
p = 1; q = -1;
f1 = (Exp[-2 r a] p)/(2 r c) + (Exp[-2 r b] p)/(2 r d);
f2 = (Exp[-2 r a] q)/(2 r c) + (Exp[-2 r b] q)/(2 r d);
Plot[{f1, f2}, {r, -2, 2}, PlotRange -> Automatic,
MeshFunctions -> {#1 &,#1*#2 &},
MeshShading -> {{Red, Green}, {Blue, Yellow}}, Mesh -> {{0}}]
Since the sign of $$x$$, $$x y$$ can distinguish the four quadrant, so this method can also be used to ParametricPlot who cross the four quadrant.
ParametricPlot[{t/4 Cos[t], t/4 Sin[t]}, {t, 0, 20},
MeshFunctions -> {#1 &, #1*#2 &},
MeshShading -> {{Red, Green}, {Blue, Yellow}}, Mesh -> {{0}}]
Clear["Global*"]
a = 1; c = 1;
b = -1; d = 1;
p = 1; q = -1;
f1 = (Exp[-2 r a] p)/(2 r c) + (Exp[-2 r b] p)/(2 r d);
f2 = (Exp[-2 r a] q)/(2 r c) + (Exp[-2 r b] q)/(2 r d);
Plot[Evaluate@Outer[
Tooltip@ConditionalExpression[#2, #1] &,
{r > 0, r < 0}, {f1, f2}], {r, -2, 2},
PlotLegends ->
{"f1;\[ThinSpace]r\[ThinSpace]>\[ThinSpace]0",
"f2;\[ThinSpace]r\[ThinSpace]>\[ThinSpace]0",
"f1;\[ThinSpace]r\[ThinSpace]<\[ThinSpace]0",
"f2;\[ThinSpace]r\[ThinSpace]<\[ThinSpace]0"}]
` | 856 | 2,179 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-18 | latest | en | 0.697997 |
https://www.lesswrong.com/posts/bAtRGhezL6TFZBx3d/how-i-make-diagrams | 1,716,603,728,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00678.warc.gz | 750,129,950 | 34,310 | # 14
Someone asked me how I make diagrams like the ones in my two loft bed posts, so here's how. Note, though, that these aren't perspective drawings: lines that are parallel in reality remain parallel on the page. This is a lot simpler to draw, and looks good enough.
The goal is to make sure I know what all the lines are and about where things fit together. Sometimes I need to do a few rounds of these, but they're fast enough this isn't much a problem. Lines that are vertical in real life go up and down, lines that are horizontal and perpendicular to me go across the page, and lines that would be coming out of the page instead go down and to the left.
Once I'm happy I start over with a ruler and a pencil. At this stage you can make it a scale diagram if you want by picking a ratio between real-life distances and on-page distances, like 1' = 1/2". For lines coming out of the page you can keep the same ratio, or use a smaller one like 1' = 1/4". Most of the time I just eyeball this and it's good enough for what I'm doing. I draw any lines that make things easier to see how parts fit together, even ones that would be occluded behind solid things:
If any of it looks wrong I erase and re-draw.
Then I come back with a pen or marker and, again with the straight-edge, ink the lines:
You'll notice I often inked lines in slightly different places than I'd penciled them in. When inking I am a lot more careful about getting angles correct: I pick a reference line for each axis and match everything I draw to that. I could do this at the pencil-ruler stage, but this feels a bit easier to me.
Lines that aren't parallel to any of the three axes are harder. I didn't have any in this diagram, but the basic idea is you figure out where the two ends go and then draw a straight line connecting them. If I had something with a ton of these, or with a lot of curves I would probably choose another method to represent them.
The last step is to come back and erase the pencil marks. A dedicated eraser generally works better than the ones on the ends of pencils, and you want one that hasn't gone hard with age:
When I'm making things for my own use I usually only do the first step, or sometimes the first two. Not much reason to ink it unless I want it to look nice. The extra steps beyond the rough sketch took me about ten minutes for this diagram.
Some people might use 3d modeling software for this, but I haven't yet found it worth it to learn...
There are also times when I'm trying to show something much simpler, and a diagram like this is overkill. I'll usually draw something quick in the open-source vector drawing tool Inkscape. For example, the diagram for putting two kids in a bed was a quick Inkscape one:
New Comment | 629 | 2,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.972115 |
https://stackoverflow.com/questions/26699365/how-to-sum-numbers-with-php-using-for-loop/26699410 | 1,721,851,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00358.warc.gz | 464,566,822 | 48,779 | # How to sum numbers with php using for loop?
Can someone help me understand how to sum numbers?
For example, I want to use a for loop to sum all numbers from 1 to 10:
``````'1+2+3+4+5+6+7+8+9+10=?'
``````
• What have you tried so far? Where are you stuck? Commented May 23, 2020 at 21:10
Since you specifically said for loop:
``````<?php
\$start = 1;
\$end = 10;
\$sum = 0;
for (\$i = \$start; \$i <= \$end; \$i++) {
\$sum += \$i;
}
echo "Sum from " . \$start . " to " . \$end . " = " . \$sum;
``````
Yes, it is pretty easy to do:
``````array_sum(range(1, 10))
``````
or
``````\$sequence = array(1,2,3,4,5,6,7,8,9,10);
array_sum(\$sequence);
``````
• Strictly speaking, this does not solve the question. The OP asked for a for loop Commented May 23, 2020 at 21:13
Not sure if I understand the question or not, but
``````\$sum = 0;
for (\$i = 1; \$i <= 10; \$i++) {
\$sum += \$i;
}
echo 'The sum: ' . \$sum;
``````
Should sum the numbers between 1 and 10 into the \$sum variable.
this will do ... you have a lot of options to do this
``````\$a=0;
for(\$i=0;\$i==10;\$i++)
{
\$a=\$a+\$i;
}
echo 'Sum= ' . \$a ;
``````
• ...or `for (\$i=1;\$i==10;\$i++)`, no need to add zero. Commented Nov 2, 2014 at 12:52
In fact, using a loop for this is the least efficient way. This is an arithmetic sequence and can be calculated using the formula:
``````S = n*(a1+an)/2
``````
where `a1` is first element, `an` is last element, `n` number of elements.
``````// for 1..10 it would be:
\$sum = 10*(1+10)/2;
// for 1..2000 it would be:
\$sum = 2000*(1+2000)/2;
// u can use the same formula for other kinds of sequences for example:
// sum of even numbers from 2 until 10:
\$sum = 5*(2+10)/2; // 5 elements, first is 2, last is 10
``````
Try like this:
``````<form method="post">
Start:<input type="text" name="a">
End: :<input type="text" name="b">
<input type="submit" >
</form>
<?php
\$start = \$_POST['a'];
\$end = \$_POST['b'];
\$sum = 0;
for (\$i = \$start; \$i <= \$end; \$i++) {
\$sum += \$i;
}
echo "<h2>Sum from " . \$start . " to " . \$end . " = " . \$sum;
?>
``````
``````<?php
\$array = array(1,2,3,4,5,6,7,8,9,10);
\$count = count(\$array);
\$sum = 0;
for(\$i=0;\$i<\$count;\$i++){
\$sum += \$array[\$i];
}
echo \$sum ;
?>
``````
Make 1+2+3+4+5 = ? by recursion function
``````<?php
\$n=1;
echo Recursion(\$n);
function Recursion(\$n){
if (\$n <=5){
if(\$n<5){
echo "\$n+";
}
else echo "\$n=";
return \$n+Recursion(\$n+1);
}
}
?>
``````
• Please add some explanation to your answer such that others can learn from it - why should recursion help in any way here? There is no, absolutely no benefit of using recursion here Commented May 23, 2020 at 21:11 | 971 | 2,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-30 | latest | en | 0.779467 |
https://homework.cpm.org/category/CC/textbook/ccg/chapter/10/lesson/10.3.5/problem/10-181 | 1,717,071,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00775.warc.gz | 261,062,775 | 15,187 | Home > CCG > Chapter 10 > Lesson 10.3.5 > Problem10-181
10-181.
Write an equation and use it to solve this problem.
Jill has a $9$-inch tall cylinder that she is using to catch water leaking from a pipe. The water level in the cylinder is currently $2$ inches deep and is increasing at a rate of $\frac { 1 } { 4 }$-inch per hour. How long will it be before the water overflows?
Use an equation in the form $y = mx + b$ to solve for the time at which the water will fill the cylinder completely.
$28$ hours | 140 | 511 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-22 | latest | en | 0.930086 |
https://projects.skill-lync.com/projects/Transient-simulation-of-flow-over-a-throttle-body-64733 | 1,579,463,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00269.warc.gz | 618,813,667 | 12,925 | ## Transient simulation of flow over a throttle body
OBJECTIVE:- STEADY STATE SIMULATION OF FLOW OVER A THROTTLE BODY .
SOFTWARE USED :-
1. CONVERGE STUDIO:- TO SETUP THE MODEL.
2. CYGWIN:- FOR RUNNING THE SIMULATION.
3. OPENFOAM:- TO VISUALIZE THE DIFFERENT POST-PROCESSED RESULT.
THEORY:-
HERE, WE HAVE DONE SIMULATION OF FLOW OVER A THROTTLE BODY IN STEADY STATE .
PRE-PROCESSING STEPS:-
HERE WE HAVE CONSIDER FOLLOWING THINGS FOR ELBOW IN CONVERGE STUDIO:-
(A) TYPE OF FLUID USED:- GAS (MIXTURE OF BASICALLY OXYGEN AND NITROGEN).
(B) TYPE OF CHANNEL:-
IT IS BASICALLY A THROTTLE BODY WHICH IS IN THE SHAPE OF ELBOW .
• OPEN FROM BOTH SIDE OF ELBOW.
• THROTTLE IS WALL .
• BODY IS WALL .
(C).DIMENSION OF CHANNEL:-
1. DIAMETER OF ELBOW =0.0084 m
2. LENGTH OF ELBOW BODY= 0.14 m.
(D).SIZE OF GRID USED:-0.01 FOR LENGTH AND 0.001 FOR HEIGHT AND BREADTH.
(E). PRESSURE:-
AT INLET, 150000.PASCAL.
AT OUTLET, 100000 PASCAL.
(F). CRITERIA FOR RUNNING SIMULATION :-
1. INTIAL TIME STEP :- 1E-6 SECOND .
2. TIME PERIOD FOR SIMULATION :- 0.01 SECONDS .
3. NUMBER OF CYCLES FOR SIMULATION:- 10000 CYCLES .
(G). GRID SIZE : - 2 milimeter (FOR ALL DIRECTION i.e LENGTH, BREATH AND HEIGHT) .
HERE , FIXED EMBBEDIING HAVE BEEN USED ON THE THROTTLE BODY.
(H). SOLVER TYPE USED:- PRESSURE BASED (STEADY) .
(I). PRE-PROCESSING STEPS:-
AFTER SETTING UP THE MODEL, WE SENT OUT DATA TO A PARTICULAR LOCATION WHERE ALL THE DETAIL OF CHANNEL IS AVAILABLE REQUIRED BY FURTHER SOFTWARE FOR SIMULATION.
THEN AFTER GENERATING AND STORING THE DATA, WE USED CYGWIN SOFTWARE WHOSE WORK IS TO RUN SIMULATION.
PROCEDURE FOR RUNNING SIMULATION:-
1. OPEN "CYGWIN" IN COMMAND PROMPT.
2. THEN TRACE THE LOCATION WHERE DATA IS STORED FROM CONVERGE STUDIO.
3. NOW TYPE "converge.exe" FOR SERIES SIMULATION OR "mpiexec.exe -n 4 converge" FOR PARALLEL SIMULATION IN THAT FILE LOCATION.
4. AFTER DOING THIS, SIMULATION WILL START WORKING AND AND WILL STOP AFTER GIVEN NUMBER OF CYCLES (BOTH IMPLICITLY AND EXPLICITLY DEPANDING UPON SIMULATION PERIOD).
5. THIS IS HOW THE WORK OF CYGWIN COMPLETES. AFTER THIS USE POST_CONVERT COMMAND IN OUTPUT FILE IS USED WHICH IS GENERATED BY CYGWIN. IT WILL CONVERT THE GENERATED OUT FILE FROM CYGWIN INTO A COMPATIBLE FROM WHICH WILL BE EASILY ACCESSED BY THE PARAFOAM.
NOW WE OPEN THE PARAFOAM SOFTWARE FOR POST PROCESSING TO VISUALIZE THE RESULT AND COMPARE IT.
POST-PROCESSING STEPS :-
1. IN OPENFOAM, CLICK ON OPEN FROM FILE MENU FROM DROP-DOWN MENU WHICH IS PRESENT AT TOP-LEFT OF THE SCREEN.
2. SEARCH FOR FILE WHERE FILE IS GENERATED BY POST CONVERT COMMAND AND OPEN IT. REMEMBER IT MUST HAVE ".vtm" DOMAIN AFTER FILE NAME.
3. CLICK ON APPLY, GEOMETRY WILL ORIGINATE. AFTER THIS CLICK ON SLICE OPTION TO SECTION THE GEOMETRY . NOW CLICK ON DROP DOWN MENU (SURFACE WITH EDGES) FROM TOP TOOLBAR OPTIONS . THIS WAY THE MESH WILL GENERATE .
4. AFTER IT, CLICK ON LINE PROBE OPTION TO SEE THE DIFFERNT PHYSICAL QUANTITES AND APPLY IT.
5. IN THIS WAY, THE GRAPH WILL GENERATE AND WE WILL SEE DIFFERENT PARAMETERS ACCORDING TO THAT APLLIED LINE PLOT.
6. IN THIS WAY, OUR POST-PROCESSING IS DONE.
MESH GENERATED IN TRANSITION STATE :-
CONTOUR PLOT OF PRESSURE :-
HERE THE PRESSURE CONTOUR HAVE BEEN GENERATED AT THE MID OF THE SIMULATION PERIOD .
HERE ONE MORE THING IS THAT THE MESH REFINEMENT IS DONE MORE FINER AT PORTION OF THE TROTTLE PART .
HERE WE HAVE GENERATED THE PRESSURE CONTOUR PROFILE ACROSS THE ELBOW PIPE . HERE WE OBSERVED THAT PRESSURE IS TOO HIGH AT THE INLET (DEPICTING THE RED PATCHES) OF THROTTLE BODY WHEREAS AT THE OUTLET (SHOWING THE GREEN PATCH ) BECOMES AVERAGE PRESSURE i.e EQUAL TO THE ATMOSPHERIC PRESSURE .
HERE SOME CELLS ARE MISSING BECAUSE THROTTLE PART IS IN DYNAMIC MOTION. AND SINCE BOTH PRESSURE FLOW AND MESH GENERATION IS INDEPENDENT OF EACH OTHER .THEREFORE IT MAKE'S DIFFICULT FOR PROCESSOR TO CAPTURE ALL ELEMENTS WITH GRAPHICAL INFORMATION AT SAME PARTICULAR FRAME TIME .
HERE PRESSURE AT THE TROTTLE PART IS CHANGING A LOT BECAUSE OF THE PRESENCE OF TRHOTTLE WHICH TENDS TO RESIST THE INCOMING PRESSURE OF FLUID AS A RESULT OF WHICH PRESSURE CHANGES MUCH .
CONTOUR PLOT OF VELOCITY :-
HERE SAME THING ARISES IN VELOCITY PROFILE CONTOUR WHICH IS THAT MESH IS NOT TOTALLY GENERATED ACOSS ELBOW PIPE . AGAIN THE REASON IS SAME AS THAT OF PRESSURE CONTOUR .
THE VELOCITY CONTOUR IS ALMOST SAME HERE ALL OVER THE DOMAIN OF ELBOW PIPE SINCE THE SIMULATION HAVE BEEN CAPTURED AT THE MID OF THE TIME OF SIMULATION .
AGAIN AT THROTTLE, VELOCITY IS FACING THE SAME ISSUE i.e IT IS HAVING TURBULENCE DUE TO THROTTLE WHICH WAS SAME HAPPENING IN PRESSURE CONTOUR .
BUT HERE VELOCITY PROFILE IS SAME AT BOTH ZONES SEPARATED BY THROTTLE . ACTUALLY HERE AS THROTTLE SHIFTS ,THE VELOCITY MAKES ADJUSTMENT PRIOR TO THE SHIFTING OF THROTTLE IN TERMS OF ROTATION HENCE WE SEE THAT VELOCITY IS SAME AT BOTH ZONES WITH SLIGHT DIFFERENCE .
CONVERGE PLOT OF MASS FLOW RATE :-
IN THE CONVERGE PLOT OF MASS FLOW RATE , WE OBSERVED THAT MASS FLOW AT INLET AND OUTLET ARE SAME SINCE THE TYPE OF PATH TRACED BY BOTH CURVE ARE IS SAME . HENCE WE CAN CONCLUDE THAT MASS IS CONSERVED OR PROVING LAW OF CONSERVATION OF MASS .
HERE WHEN THROTTLE IS IN DYNAMIC MOTION BETWEEN 0 TO 2 ms , THE MASS FLOW AT OUTLET VARIES FROM INLET . THE REASON FOR THIS VARIATION IS SUDDEN MOVEMENT OF THROTTLE .
WHEN THROTTLE COMES IN STATIC MOTION BETWEEN 2 TO 4 ms , WHAT WE OBSERVED IS THAT MASS FLOW BECOMES ALMOST SAME AT BOTH ENDS . THE REASON FOR THIS EQUALITY IS THE THROTTLE STEADYNESS IN THIS PERIOD WHICH PUTS NO EFFECT ON MASS FLOW DUE TO WHICH MASS FLOW BECOMES SAME AT BOTH ENDS .
AFTER 4 ms , WE OBSERVED THAT MASS FLOW IS SAME AT BOTH ENDS INSPITE OF THROTTLE IS IN DYNAMIC MOTION . THIS IS BECAUSE THROTTLE SHIFTNESS IS IN LONG PERIOD WHICH IS ALMOST 8ms . HENCE THIS LONG PERIOD MAKES VERY SMALL SHIFT IN ROTATION OF THROTTLE DUE TO WHICH WE SAW VERY SMALL VARIATION IN PLOT AT BOTH ENDS .
NOTE:- HERE SIGN IN CONVERGE PLOT OF CONVERGE STUDIO TELLS US ABOUT THE FLOW OF MASS ACROSS THE DEFINED GEOMETRY OR DOMAIN WHERE IT IS NEGATIVE WHEN IT FLOW INTO THE DOMAIN AND POSITIVE IF IT FLOWS OUT OF DOMAIN .
CONVERGE PLOT OF PRESSURE :-
IN THE CONVERGE PLOT OF PRESSURE , WE NOTICED THAT PRESSURE VALUE OF INLET IS CONSTANT THROUGHOUT THE SIMULATION FOR INLET WHEREAS IT VARIES THROUGHOUT THE SIMULATION PERIOD FOR OUTLET . BUT QUESTION ARISES WHY IS IT SO ? ALTHOUGH WE HAVE PROVIDED THE INTIAL CONDITON FOR BOTH ENDS OF GEOMETRY WHERE IT IS SAME FOR INLET BUT VARIES FOR OUTLET WHEN IT COMES TO SIMULATION .
THE ANSWER OF THIS IS QUITE SIMPLE , WHEN PRESSURE ENTERS THE INLET OF DOMAIN, IT GETS COLLISION WITH INTERNAL PRESSURE OF DOMAIN (THIS INTERNAL PRESSURE ARISES DUE TO PRESENCE OF NITROGEN AND OXYGEN GASES) DUE TO WHICH THE RESULTING PRESSURE BECOMES DIFFERENT AS WE EXPECT AT OUTLET HENCE IN THIS WAY THE PRESSURE AT OUTLET VARIES A LOT .
HERE ANOTHER POINT IS THAT WHEN THROTTLE ROTATES , IT CAUSES THE STAGGERNESS IN PRESSURE OVER THE THROTTLE BODY WHICH WE ARE SEEING IN PLOT .
HERE IF WE TAKE THE STATIC PRESSUE CASE PLOT , IT WILL PROVE THIS WHERE IT WILL CHANGE A LOT AT INLET BUT WILL BECOME SAME AT OUTLET OF TOTAL PRESSURE . HERE THE STATIC PRESSURE TELLS THAT HOW MUCH PRESSURE HAVE BEEN APPLIED ON THE FLOW OF FLUID . THIS STATIC PRESSURE INTIALLY FACES HIGH RESISTANCE DUE TO INTERNAL PRESSURE AT INTIAL POINT OF DOMAIN BUT THE RESISTANCE GETS LOWER AS THE FLOW MOVES FORWARD INTO THE FINAL POINT DOMAIN AND STATIC PRESSURE GETS IT HIGHER VALUE AT OULET WHERE IT HAS NO RESISTANCE.
CONVERGE PLOT OF VELOCITY :-
IN THE CONVERGE PLOT OF VELOCITY, THE VALUE OF VELOCITY IS SAME AT BOTH INLET AND OULET SECTION . ALTHOUGH THE VELOCITY IS VARYING A LOT THROUGH THE SIMULATION BUT WE OBSERVED THAT BOTH ARE HAVING SAME VARIATION EXPECT THE INTIAL PORTION OF SIMULATION TIME THROUGHOUT THE SIMULATION PERIOD . HENCE WE AGAIN HERE CONCLUDED THAT VELOCITY IS ALSO CONSERVED LIKE MASS FLOW RATE .
HERE BETWEEN THE SIMULATION PERIOD OF 0 TO 2 ms , WE OBSERVE THAT VELOCITY IS VARYING A LOT I.E DIFFERENT AT BOTH ENDS . THIS IS BECAUSE THE THROTTLE PART IS IN DYNAMIC MOTION WHICH CAUSES THE TROTTLE TO MOVE AT DIFFERENT RATE HENCE THIS REFLECTS BACK IN THE MOTION OF FLUID IN TERMS OF VELOCITY .
BETWEEN 2 ms TO 4 ms , WE OBSERVE THAT NOW VELOCITY AT BOTH SECTION ARE APPROX SAME BECAUSE THE TROTTLE PART COMES IN STATIC MOTION WHICH CAUSES NO DISTURBENCE IN FLOW OF FLUID .
BUT AFTER 4 ms , WE NOTICED THAT INSPITE OF DYNAMIC MOTION OF THROTTLE WE HAVE SAME VELOCITY AT BOTH ENDS . THIS IS BECAUSE THE DYNAMIC MOTION TAKES PLACE IN LONG PERIOD OF 8ms WHICH MAKES THROTTLE TO ADJUST TO IT'S INTIAL POSITION VERY SLOWLY AS A RESULT OF WHICH WE SEE NO SUCH DIFFERENCE IN THIS TIME INTERVAL AT BOTH ENDS COMPARING TO FIRST PERIOD OF DYNAMIC MOTION .
CONVERGE PLOT OF TOTAL CELL COUNT :-
HERE THE TOTAL CELL COUNT TELLS ABOUT THE NUMBER OF TOTAL GRID POINTS DEFINED FOR DOMAIN OR GEOMETRY . THE CELL COUNTS ARE DIVIDED TO EACH PROCESSOR DEPANDING UPON IT'S PERFORMANCE . HERE THE RANK DENOTES THE PROCESSOR NUMBER .
HERE WE OBSERVED THAT AT INTIAL PERIOD OF 2 ms , THE THROTTLE WAS IN DYNAMIC MOTION WHICH MAKE THE SUDNENESS IN ELBOW FLOW DUE TO WHICH THE CELL COUNT ALSO VARIED A LOT . HERE AS THROTTLE WAS MOVING ROTATING FAR AWAY FROM INTIAL POSITION , THE CELL COUNT WAS INCREASING RAPIDLY .
BETWEEN THE PERIOD OF 2ms TO 4 ms , THE CELL COUNT WAS STEADY AND THE REASON FOR THIS STEADYNESS IS THE STATIC POSITION OF THROTTLE WHICH CAUSED NO TURBULENCE IN FLOW DUE TO WHICH NO MUCH CELL GENERATED .
HERE ONE MORE THING WE MAKE CONCLUDE THAT ANY DISTURBENCE IN FLOW OF FLUID DUE TO THROTTLE OR SOMETHING ELSE WILL LEAD TO GENERATION OF MORE CELLS .
ANIMATION OF PRESSURE FLOW ACROSS ELBOW PIPE DURING TRANSITION PERIOD :-
IN ANIMATION OF PRESSURE , WE OBSERVE THAT INTIALLY AT TIME PERIOD OF 2 ms , THE FLUID IN TERMS OF PRESSURE WAS FLOWING WITH FULL PRESSURE . AFTER IT WHEN THROTTLE WAS CHANGING IT'S POSITION TO CLOSE , THE PRESSURE WAS REDUCING SLIGHTLY . THE REASON FOR THIS PHENOMENA IS THE CHANGE OF AREA NEAR THE THROTTLE, WHERE THE FLOW WAS HIGH WHEN THE AREA WAS TOO MUCH BUT SUDDENLY START TO DECLINE VERY MUCH AS THROTTLE SHIFT IT'S POSTION TO CLOSE .
AFTER IT BETWEEEN 2 ms to 4ms , WHEN THROTTLE WAS IN STATIC POSITION . WE OBSERVED THAT THERE WAS NO FLOW ACROSS THE ELBOW . THE REASON BEHIND IT WAS PRESSURE BARRIER WHICH MAKE PRESSURE LIMITED IN ONE ZONE AND ISOLATED IN ANOTHER ZONE .
AFTER 4 ms WHEN THROTTLE OPEN ONCE AGAIN , THE PRESSURE WAS LOWER COMPARING TO INTIAL TIME PERIOD . THIS WAS BECAUSE WHEN THE THROTTLE AGAIN COME IN DYNAMIC MOTION AND TRY TO OPEN . THE PRESSURE BECOME LOW DUE TO TURBULENCE . HENCE ORIGINAL PRESSURE REDUCES .
ANIMATION OF VELOCITY FLOW ACROSS ELBOW PIPE DURING TRANSITION PERIOD :-
HERE IN VELOCITY PROFILE , VELOCITY IS VERY LOW AT INTIAL PERIOD OF TIME BECAUSE IN THIS TIME PERIOD OF SIMULATION , THE PRESSURE ESTABLISHMENT TAKE PLACE BETWEEN TWO REGIONS DUE TO WHICH VELOCITY BECOMES VERY LOW IN BOTH SECTION ACCORDING TO BERNOUILLE PRINCIPLE .
WITH PASSAGE OF TIME , WHEN THROTTLE COMES IN STATIC POSITION . THE PRESSURE ESTABLISHES IN REGION DUE TO WHICH THE FLOW BECOME CONSISTENT THROUGHOUT THE DOMAIN .
WHEN THROTTLE REOPENS ONCE AGAIN , THE VELOCITY INCREASES IN THE WHOLE REGION . THE REASON BEHIND IS THE PRESSURE BARRIER DUE THROTTLE CLOSENESS AND AS IT OPEN , THE PRESSURE GAP TAKES PLACE WHICH PROVIDE FLOW ADVANTAGE TO MOVE FROM HIGH PRESSURIC TO LOW PRESSURIC REGION . IN THIS WAY THE VELOCITY BECOMES CONSISTENT .
CALCULATION OF END TIME OF SIMULATION :-
TOTAL LENGTH OF THROTTLE BODY = 0.14 m .
AVERAGE VELOCITY OF FLUID AT THE INLET OF THROTTLE BODY = 80 m/s.
HENCE , END TIME PERIOD FOR SIMULATION WILL BE = ( TOTAL LENGTH OF THROTTLE BODY) / (AVERAGE VELOCITY AT INLET) = (0.14/80)= 0.0018 approximately .
NOTE :- HERE WE TAKEN END TIME FOR SIMULATION AS 0.01 SECONDS . THE REASON FOR IT IS THAT WE WANT TO SAVE THE RESTART FILE SO THAT WE CAN SAVE SIMULATION DATA AND AVOID LOSING OF DATA DURING CRASHING . HERE RESTART TIME IS TAKEN AS 0.0001 SECONDS DUE TO WHICH WE HAVE CONSIDERED SIMULATION PERIOD AS 0.01 SECONDS . HENCE WE HAVE 100 RESTART FILES AND MORE PRECISE INFORMATION .
.
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https://www.jiskha.com/questions/550585/Translate-three-more-than-the-product-of-a-number-and-4-is-15-in-an-algebraic-equation | 1,558,721,067,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257699.34/warc/CC-MAIN-20190524164533-20190524190533-00264.warc.gz | 838,351,987 | 4,189 | # math
Translate three more than the product of a number and 4 is 15 in an algebraic equation
1. 👍 0
2. 👎 0
3. 👁 86
1. 4n + 3 = 15
1. 👍 0
2. 👎 0
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More Similar Questions | 477 | 1,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-22 | latest | en | 0.935274 |
http://ir.amss.ac.cn/handle/2S8OKBNM/61494?mode=full&submit_simple=Show+full+item+record | 1,708,878,145,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00811.warc.gz | 22,273,898 | 23,150 | CSpace
Approximation algorithms for solving the line-capacitated minimum Steiner tree problem Li, Jianping1; Wang, Wencheng1; Lichen, Junran2,3; Liu, Suding1; Pan, Pengxiang1 2022-06-06 Source Publication JOURNAL OF GLOBAL OPTIMIZATION ISSN 0925-5001 Pages 28 Abstract In this paper, we address the line-capacitated minimum Steiner tree problem (the Lc-MStT problem, for short), which is a variant of the (Euclidean) capacitated minimum Steiner tree problem and defined as follows. Given a set X = (r(1), r(2),...,r(n)} of n terminals in R-2, a demand function d : X -> N and a positive integer C, we are asked to determine the location of a line l and a Steiner tree T-l to interconnect these n terminals in X and at least one point located on this line l such that the total demand of terminals in each maximal subtree (of TO connected to the line l, where the terminals in such maximal subtree are all located at the same side of this line l, does not exceed the bound C. The objective is to minimize total weight Sigma(e is an element of Tl) w(e) of such a Steiner tree T-l among all line-capacitated Steiner trees mentioned-above, where weight w(e) = 0 if two endpoints of that edge e is an element of T-l are located on the line l and otherwise weight w(e) is the Euclidean distance between two endpoints of that edge e is an element of T-l . In addition, when this line l is as an input in R-2 and Sigma(r is an element of X) d(r) <= C holds, we refer to this version as the 1-line-fixed minimum Steiner tree problem (the 1Lf-MStT problem, for short). We obtain three main results. (1) Given a rho(st )-approximation algorithm to solve the Euclidean minimum Steiner tree problem and a rho(1Lf)-approximation algorithm to solve the 1Lf-MStT problem, respectively, we design a (rho(st)rho(1Lf )+2)-approximation algorithm to solve the Lc-MStT problem. (2) Whenever demand of each terminal r is an element of X is less than c/2, we provide a (rho(1Lf) + 2)-approximation algorithm to resolve the Lc-MStT problem. (3) Whenever demand of each terminal r is an element of X is at least c/2, using the Edmonds' algorithm to solve the minimum weight perfect matching as a subroutine, we present an exact algorithm in polynomial time to resolve the Lc-MStT problem. Keyword Combinatorial optimization Locations of lines Line-capacitated Steiner trees Approximation algorithms Exact algorithms DOI 10.1007/s10898-022-01163-x Indexed By SCI Language 英语 Funding Project National Natural Science Foundation of China[11861075] ; National Natural Science Foundation of China[12101593] ; Project for Innovation Team (Cultivation) of Yunnan Province[202005AE160006] ; Key Project of Yunnan Provincial Science and Technology Department and Yunnan University[2018FY001014] ; Program for Innovative Research Team (in Science and Technology) in Universities of Yunnan Province[C176240111009] ; Project of Yunling Scholars Training of Yunnan Province ; Yunnan Provincial Department of Education Science Research Fund[2019Y0022] WOS Research Area Operations Research & Management Science ; Mathematics WOS Subject Operations Research & Management Science ; Mathematics, Applied WOS ID WOS:000806126100002 Publisher SPRINGER Citation statistics Document Type 期刊论文 Identifier http://ir.amss.ac.cn/handle/2S8OKBNM/61494 Collection 中国科学院数学与系统科学研究院 Corresponding Author Li, Jianping Affiliation 1.Yunnan Univ, Dept Math, East Outer Ring South Rd, Kunming 650504, Yunnan, Peoples R China2.Chinese Acad Sci, Acad Math & Syst Sci, Inst Appl Math, 55 Zhongguancun East Rd, Beijing 100190, Peoples R China3.Beijing Univ Chem Technol, Sch Math & Phys, 15 North Third Ring East Rd, Beijing 100029, Peoples R China Recommended CitationGB/T 7714 Li, Jianping,Wang, Wencheng,Lichen, Junran,et al. Approximation algorithms for solving the line-capacitated minimum Steiner tree problem[J]. JOURNAL OF GLOBAL OPTIMIZATION,2022:28. APA Li, Jianping,Wang, Wencheng,Lichen, Junran,Liu, Suding,&Pan, Pengxiang.(2022).Approximation algorithms for solving the line-capacitated minimum Steiner tree problem.JOURNAL OF GLOBAL OPTIMIZATION,28. MLA Li, Jianping,et al."Approximation algorithms for solving the line-capacitated minimum Steiner tree problem".JOURNAL OF GLOBAL OPTIMIZATION (2022):28.
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Related Services Recommend this item Bookmark Usage statistics Export to Endnote Google Scholar Similar articles in Google Scholar [Li, Jianping]'s Articles [Wang, Wencheng]'s Articles [Lichen, Junran]'s Articles Baidu academic Similar articles in Baidu academic [Li, Jianping]'s Articles [Wang, Wencheng]'s Articles [Lichen, Junran]'s Articles Bing Scholar Similar articles in Bing Scholar [Li, Jianping]'s Articles [Wang, Wencheng]'s Articles [Lichen, Junran]'s Articles Terms of Use No data! Social Bookmark/Share | 1,218 | 4,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.909879 |
https://forums.studentdoctor.net/threads/boiling-point-and-vp.646410/ | 1,563,710,270,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526948.55/warc/CC-MAIN-20190721102738-20190721124738-00006.warc.gz | 398,237,826 | 21,102 | # Boiling point and VP
#### MynameisMike
10+ Year Member
Why does the addition of a nonvolatile solute lower the vapor pressure, but elevate the boiling point.
When a solutions VP = atmospheric pressure, it will boil. Thus, it would have a lower BP since its VP is lower.
These seem to contradict each other.
#### matth87
Why does the addition of a nonvolatile solute lower the vapor pressure, but elevate the boiling point.
When a solutions VP = atmospheric pressure, it will boil. Thus, it would have a lower BP since its VP is lower.
These seem to contradict each other.
You still have to get your solution's vapor pressure greater then atm pressure. If you lower the vapor pressure its going to take more energy to get all those little particles back up to the vapor pressure greater then atmospheric pressure. If its going to take more heat then your solution will boil at a higher temperature.
#### G1SG2
10+ Year Member
5+ Year Member
Why does the addition of a nonvolatile solute lower the vapor pressure, but elevate the boiling point.
When a solutions VP = atmospheric pressure, it will boil.
Thus, it would have a lower BP since its VP is lower.
These seem to contradict each other.
Ding! The answer lies in the bold statements above. In order to boil, the solutions VP has to equal the atmospheric pressure, which is 1 atm (unless otherwise stated). If you LOWER the vapor pressure (say, from 0.7 atm to 0.4 atm), do you think you are closer or farther from 1 atm? You just lowered the vapor pressure and are now further from the 1 atm mark.
I'm going to copy and paste what I wrote in a similar thread a while ago:
What happens is that the molecules exert a pressure above the liquid as they try to escape into the gas phase. This is the vapor pressure. Now, if you add NaCl, a nonvolatile solute, it will block some of the molecules from escaping at the surface since NaCl will also occupy some room. This lowers the vapor pressure, since you don't have that many molecules crashing up at the surface trying to escape. By LOWERING the vapor pressure, you elevated the boiling point because remember, boiling occurs when the vapor pressure=atmospheric pressure. Think about adding salt to a pot of hot water and pasta. You want to raise the boiling point so your pasta can cook longer/cook well. Here's a silly way to make you understand:
liquid molecules attempting to escape into the gas phase: muahaha, let's become gas molecules! We're creating more vapor pressure by hanging out at the surface and trying to escape into the gas phase. That pasta won't be in there for long!
nonvolatile solute: not so fast! Let me block your way for a while.
liquid molecules attempting to escape into the gas phase: dammit! You just lowered the vapor pressure. We worked so hard! Now this pasta will have to stay in here longer because we have to create more pressure to equal to the atmospheric pressure. We almost had it! Dang.
Another example: think of 20 runners that have to run 5 miles. They can stop running as soon as they hit the 5 mile mark. Now, suppose you drive by some runners who are at the 3 mile mark (they have 2 more miles to go). You stop, stuff them in your car, drive them back to the 1 mile mark and drop the off. Now they have to run 4 more miles instead of 2 more miles to get to the 5 mile mark. Same goes for vapor pressure. When you add a nonvolatile solute, you lower the vapor pressure, but the 1 atmosphere requirement doesn't change. They solution still has to equal its vapor pressure to 1 atm. It just has to work for longer now because you lowered its vapor pressure. So, lowering the vapor pressure does not affect the atmospheric pressure. Hope this helps.
OP
M
#### MynameisMike
10+ Year Member
lol wow...it all seems so simple now. Thanks guys!
#### Hemichordate
##### Peds
10+ Year Member
So if you have two pure volatile solutes, A and B, and you mix them together, is the solution going to have a vapor pressure that is between the vapor pressures of the two compounds? What about boiling point?
#### dragon529
##### MS-III
10+ Year Member
So if you have two pure volatile solutes, A and B, and you mix them together, is the solution going to have a vapor pressure that is between the vapor pressures of the two compounds? What about boiling point?
The vapor pressure for the mixture will depends on the mole ratio of the two solutes, and it should be always in between the two solute's individual vapor pressure.
As for BP, I can only say the BP will be higher than any single solute's BP.. due to BP elevation.
#### thebillsfan
##### Unseasoned Veteran
10+ Year Member
5+ Year Member
how does the addition of a volatile solute to a solution relate to deviations in raoult's law? also what about henry's law? sorry, I realize that is sort of a broad question, but it is also a broad concept
#### G1SG2
10+ Year Member
5+ Year Member
how does the addition of a volatile solute to a solution relate to deviations in raoult's law? also what about henry's law? sorry, I realize that is sort of a broad question, but it is also a broad concept
I think that the addition of a volatile solute would cause a positive deviation and raise the vapor pressure even more because volatile substances bring their own vapor pressures. Volatility in general refers to the tendency of a substance to vaporize. Since they would vaporize more easily, they would contribute to the increasing vapor pressure.
Henry's law tells you how much of a gas can dissolve in a certain volume of a certain type of liquid. It's given by the equation p=K*c, where p is the partial pressure of the solute, c is the concentration of the solute, and k is Henry's law constant specific for each substance. When the temperature increases, the partial pressure increases, but the amount of gas dissolved DECREASES because remember, the solubility of gases decreases with increasing temperature. | 1,336 | 5,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-30 | latest | en | 0.946409 |
https://glennrowe.net/programmingpages/2021/04/28/decimals/ | 1,695,438,617,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00540.warc.gz | 312,083,714 | 13,944 | ## Decimals
We’ve seen that floats in Python can sometimes be inaccurate due to roundoff error. One way to correct this is to use Python’s Decimal data type. A Decimal variable allows the number of decimal points to be fixed, which can often eliminate roundoff error.
Decimals have a somewhat peculiar and clumsy syntax, however, so it’s worth looking at a few simple examples. The `Decimal` data type (with an uppercase ‘D’) is included in the `decimal` (with a lowercase ‘d’) module, so it must be imported before it can be used. [If you want to see the full contents of the decimal module, you can type `help(decimal)` in the console. However, be prepared for a very long list resulting from this. It’s probably more instructive to refer to the official documentation page on decimals.]
Intuitively, you might expect that simply feeding an ordinary float number into Decimal would convert it to a Decimal. To test this, try this in the console:
```from decimal import *
Decimal(0.1)```
The * in the first line imports everything from the decimal module.
Decimal(‘0.1000000000000000055511151231257827021181583404541015625’)
The result, despite the huge number of significant digits, is certainly not exactly 0.1. However, if we enclose the 0.1 in quotes, we get:
```from decimal import *
Decimal('0.1')```
Now we get `Decimal('0.1')` as the result. Enclosing a number in quotes converts it to a string, so we see that in order to create a Decimal from a float, we must enclose the float in quotes to produce a string.
When we first import the decimal module, we can see its default context by calling `getcontext()`. The result is
Context(prec=28, rounding=ROUND_HALF_EVEN, Emin=-999999, Emax=999999, capitals=1, clamp=0, flags=[], traps=[InvalidOperation, DivisionByZero, Overflow])
The ‘prec’ is the precision, or number of decimal places, which defaults to 28.
An obvious application of Decimals is in keeping track of quantities of money, which are usually in decimal form, with 100 of some smaller unit, such as pence or cents, equal to a larger unit, such as a British pound or dollar. You might think that setting the ‘prec’ to 2 would allow precise calculations with money, but in fact, it doesn’t.
The problem is that ‘prec’ specifies the number of significant digits to retain, and not the number of digits following the decimal point. Thus, if we try
```from decimal import *
getcontext().prec = 2
sum = Decimal(2.5) + Decimal(10.47)
sum```
we get the result `Decimal('13')` and not 12.97, as we might hope. We can solve this in two ways. First, we can enclose the quantities 2.5 and 10.47 in quotes to convert them to strings, and reset the prec to some large number. That is, we have
```from decimal import *
getcontext().prec = 28
sum = Decimal('2.5') + Decimal('10.47')
sum```
This gives the desired result of `Decimal('12.97')`.
Second, if we don’t want to write the arguments to Decimal as strings, we can use the quantize() function, as follows:
```from decimal import *
sum = Decimal(2.5) + Decimal(10.47)
pence = Decimal('0.01')
sum.quantize(pence)
```
This also gives the correct result of `Decimal('12.97')`.
Despite their peculiarities, Decimals can be combined with other numeric types, although some caution is needed. You can perform the usual arithmetic operations combining Decimals and integers. For example, try the following (we’re assuming the prec is 28, the default):
```from decimal import *
sum = Decimal('2.5') + Decimal('10.47')
sum * 12
sum ** 3
2 ** sum```
However, if you try combining a Decimal with a float, things go wrong. Try `sum + 12.834`, and you’ll get the error message “TypeError: unsupported operand type(s) for +: ‘decimal.Decimal’ and ‘float'”. The error message means that the + operator is not defined when one of its operands is a Decimal and the other is a float.
To get around this, we can convert the Decimal to a float: `float(sum) + 12.834`. As we’re now combining two floats, the answer is also a float: 25.804000000000002. As you can see, we’re back in the roundoff error problem again. If we want to avoid this, we can convert the float to a string and then to a Decimal, so we have` sum + Decimal('12.834')`, which now yields an exact Decimal result of `Decimal('25.804')`. If we try to convert the float to a Decimal without first enclosing it in quotes, we get roundoff error again. The statement `sum3 + Decimal(12.834)` gives us (assuming prec is set to its default value of 28): `Decimal('25.80399999999999963051777740').` | 1,141 | 4,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-40 | longest | en | 0.835097 |
https://mathematicalapologist.com/2020/04/01/how-would-we-know-covid-19-is-in-decline/ | 1,642,593,297,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301309.22/warc/CC-MAIN-20220119094810-20220119124810-00715.warc.gz | 441,188,331 | 22,933 | # How Would We Know COVID-19 is in Decline?
In light of the current COVID-19 pandemic, I have written an article entitled How Does Disease Spread? in which I discuss how mathematicians attempt to understand the way that a virus is spreading in the population. In said article, my primary focus is building up the concepts and thought processes that are needed to understand how a mathematician would come up with equations to simulate a virus spreading. I now want to write a brief follow-up addressing a related question:
In light of the behavior of viruses, how can we determine whether things are improving or getting worse?
This question actually isn’t quite as easy as it seems. The initial, intuitive response would be to look at the current number of contagious people, and to call any increase in that number bad and any decrease good. Of course, it is a bad thing when more people are becoming ill and it is a great thing when the total number of infectious people is going down, but that is actually not a very good indicator. The reason, as I will try to explain in this article is that in reality, the improvement actually starts to happen before the number of active cases starts going down.
A Brief Review of How Viruses Spread
Because viruses spread through direct or indirect personal contact, it is absolutely fundamental that at the beginning of the spread of a virus, the virus grows at an exponential rate – which basically just means that the speed at which is spreads is directly related to the number of people that have the virus. This is why numbers have been changing so drastically. When more people are infected, there are more ‘opportunities’ for others to also get infected, and so the number of new infections keeps going up.
Mathematicians recognize exponential equations as one of the most important of all equations, not only because they appear in numerous real-life situations like virus spread and population growth, but also because they are interesting in their own right. Exponential equations are so important that there is a notation developed specifically to help us write down these – normally we just use the term exponents. As an example of this form of writing, the number $2^2$ just means $2*2$, the number $2^3$ means $2*2*2$, and more broadly, the number $2^n$ means 2 multiplied by itself $n$ times. There is nothing special about the 2 either – if $a$ is any number, writing $a^n$ just means multiplying $a$ by itself a total of $n$ times.
Exponential Growth is Highly Sensitive
The number of cases of a virus always begin with exponential growth. But eventually, there are factors that slow this down. This is the point of social distancing and all travel shutting down – these measures slow things down. And as it turns out, even tiny changes produce massive shifts in the outcome of exponential equations. To see why, we can use made-up viruses with made-up equations. Let’s say some evil genius created two new viruses – let’s call them Virus A and Virus B. Since he is a genius, after all, he is able to figure out the equations that tell him how the viruses will spread – which are exponential. He finds that if we begin at “Day Zero” with only one person affected by Virus A, then after $n$ days, there will be about $(1.1)^n$ people infected by Virus A. He does the same thing for Virus B, and the equation turns out as $(1.15)^n$.
At first, his reaction might be “Well, A and B are almost identical! So I can use either one to wreck havoc.” But if he look a little closer, these actually turn out to be massively different, even those 1.1 and 1.15 are nearly the same number. To see this, we will compare the value after one month – 30 days – of infection. The two equations give values of $(1.1)^{30} \approx 17$ and $(1.15)^{30} \approx 66$. If you went for 100 days, then Virus A has infected a little over 10,000 people, but Virus B has already infected more than 1,000,000 people!
This is one of the fundamentally important aspects of exponential growth that mathematicians have understood for centuries – changing the number ‘on the bottom’ by even the tiniest amount has immense consequences. As a side note, this is why everything each of us do during this crisis matters so much, and this is why we are taking such drastic measures. Because of what a virus is and how it spreads, mathematics tells us that even small improvements in sanitation can, over the course of a month or two – make a difference in the thousands or even the millions.
How Things Get Better
This is not exactly obvious from our discussion so far, but not all exponential equations become large rapidly – some of them actually become small very rapidly. The ideas we need are summarized in 3 bullet points:
• Multiplying a number by 1 doesn’t change anything.
• Multiplying a number by something between 0 and 1 makes it smaller.
• Multiplying a number by something larger than 1 makes it bigger.
Since exponential equations are all about multiplying numbers together, this matters quite a lot. The value of $(1.00000001)^n$ will eventually become astronomically large if you give it enough time, and the value of $(0.99999999)^n$ will, if you give it enough time, eventually become indistinguishable from zero. The moral of all of this is that for exponential equations, it is not increasing or decreasing that matters most, it is about whether the number we are multiplying by is bigger than 1 or less than 1. To interpret this in terms of the spread of the COVID-19 virus, what matters is not the current number of cases, but the number of new cases every day. If there are fewer new cases today than yesterday, that is what is really important. And more than that – when you compare two days, you don’t use subtraction, because exponential equations have nothing to do with addition or subtraction. You use division to compare days, not subtraction. What we really care about is the value
$\dfrac{\textrm{Number of new cases today}}{\textrm{Number of new cases yesterday}},$
and our goal is to make this number smaller than 1. Of course, on a day-to-day basis, this number – called the growth factor – may change quite a bit. But if we have several days in a row of a growth factor less than 1, that is a strong sign that the situation has entered the stage of getting better rather than worse. And once the growth factor becomes less than 1, it is only a matter of time until the peak number of infections will be in the past and we will be on our way to a recovery from the pandemic.
Conclusion and Clarification
Obviously, this is a simplification, I am not an epidemiologist and I have never professionally studied the models that professionals are using to model this virus. However, I can say as a mathematician that the growth rate absolutely does matter, and if you see things about the growth rate in the news or on websites that are tracking the statistics of the virus, there is a good reason those numbers are there. They help us a lot in understanding whether things are getting better or worse.
As I become more aware of the mathematics myself, I will continue to write more about what is going on and how professionals are understanding all the numbers. For now, I echo the advice that governments around the world have put forward – stay inside as much as possible, stay sanitary, and seek medical advice if you show any symptoms. Let’s beat this together. | 1,609 | 7,437 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 18, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-05 | latest | en | 0.972271 |
http://www.fact-index.com/t/tw/two_hundred_twenty_two.html | 1,643,062,336,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00505.warc.gz | 88,666,644 | 2,122 | Main Page | See live article | Alphabetical index
# Two hundred twenty-two
In mathematics, 222 or two hundred twenty-two is a repdigit.
• Radii at 0° and approximately 222.49° divide a circle in the golden ratio.
``` . . . . . .
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```
√222 =
14·89966442575133971933181604612395114023452166218124
73380574030119289350747022456370098357196526519656
52139396484012226637696242316850851449289924... | 225 | 1,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-05 | latest | en | 0.29858 |
http://www.mathisfunforum.com/viewtopic.php?id=19308 | 1,511,211,386,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00764.warc.gz | 457,855,850 | 3,741 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2013-04-26 15:12:35
Omar Desoky
Guest
### Integration problems (please solve with steps)
Dears,
Calculate each of the following:
(1) ∫ (sinx+cosx)² dx
(2) ∫ (x+5)/(x+1)^4 dx
(3) ∫ (x²+4x+7)/(x+2)^5 dx
(4) ∫ x√((3/x^4)-(2/x^5)) dx
(5) ∫ ((cosx)^4-(sinx)^4)/cos2x dx
(6) ∫ (secx+cosx)^2 dx
(7) ∫ (x+3)√(4x²+12x+9)³ dx
(8) ∫ sin3x/sec3x dx
(9) ∫ (x+2)√(2x+1) dx
(10) ∫ (x-1/x)(x+1/x)(x²+1/x²) dx
(11) ∫ ((2/x)-3/x²)x^10
(12) ∫ 3sin2xcos2x dx
## #2 2013-04-26 15:31:19
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Integration problems (please solve with steps)
Hi;
Welcome, glad to help but you must help too! Just giving the answers is sometimes harmful. You must show a little effort.
http://www.mathisfunforum.com/viewtopic.php?id=14654
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #3 2013-04-26 16:46:27
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,838
Website
### Re: Integration problems (please solve with steps)
Is it necessary to know integration by parts to do them?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
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## #4 2013-04-26 17:22:31
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Integration problems (please solve with steps)
IBP is just using a formula.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #5 2013-04-26 17:42:18
Omar Desoky
Guest
### Re: Integration problems (please solve with steps)
I have done this homework, I only need to make sure of the final results as I don't have an answer guide for those....
Ok.. I only need to make sure of my final results.
Thanks! | 791 | 2,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-47 | latest | en | 0.765374 |
https://www.maplesoft.com/support/help/MapleSim/view.aspx?path=ModelonHydraulics/HydraulicResistance/GenericValve | 1,511,033,355,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00700.warc.gz | 848,284,900 | 33,910 | Generic Valve $—$ Model for generic valve
The Generic Valve component calculates the pressure loss for a given valve geometry (selected by the user). The calculation depends on the opening of the valve.
Equations $\mathrm{\nu }=\mathrm{Modelica.Media.Air.MoistAir.Utilities.spliceFunction}\left(x=\mathrm{Δp},\mathrm{pos}={\mathrm{\nu }}_{\mathrm{oil}}\left(p={p}_{A\left(\mathrm{abs}\right)},T=T,{v}_{\mathrm{air}}={v}_{\mathrm{gas}\left(\mathrm{oil}\right)},{p}_{\mathrm{sat}}={p}_{\mathrm{sat}}\right),\mathrm{neg}={\mathrm{\nu }}_{\mathrm{oil}}\left(p={p}_{B\left(\mathrm{abs}\right)},T=T,{v}_{\mathrm{air}}={v}_{\mathrm{gas}\left(\mathrm{oil}\right)},{p}_{\mathrm{sat}}={p}_{\mathrm{sat}}\right),\mathrm{Δx}=100\right)$ $\mathrm{\rho }=\mathrm{Modelica.Media.Air.MoistAir.Utilities.spliceFunction}\left(x=\mathrm{Δp},\mathrm{pos}={\mathrm{\rho }}_{\mathrm{oil}}\left(p={p}_{A\left(\mathrm{abs}\right)},T=T,{v}_{\mathrm{air}}={v}_{\mathrm{gas}\left(\mathrm{oil}\right)},{p}_{\mathrm{sat}}={p}_{\mathrm{sat}}\right),\mathrm{neg}={\mathrm{\rho }}_{\mathrm{oil}}\left(p={p}_{B\left(\mathrm{abs}\right)},T=T,{v}_{\mathrm{air}}={v}_{\mathrm{gas}\left(\mathrm{oil}\right)},{p}_{\mathrm{sat}}={p}_{\mathrm{sat}}\right),\mathrm{Δx}=100\right)$ $T={T}_{0\left(\mathrm{oil}\right)}+{\mathrm{ΔT}}_{\mathrm{system}}$ $q=\frac{{m}_{\mathrm{flow}\left(A\right)}}{\mathrm{\rho }}$ $\mathrm{Δp}={p}_{A\left(\mathrm{limited}\right)}-{p}_{B\left(\mathrm{limited}\right)}$ ${p}_{A\left(\mathrm{abs}\right)}={p}_{A}+{p}_{\mathrm{atm}\left(\mathrm{oil}\right)}$ ${p}_{A\left(\mathrm{limited}\right)}=\mathrm{max}\left({p}_{A},{p}_{\mathrm{vapour}\left(\mathrm{oil}\right)}-{p}_{\mathrm{atm}\left(\mathrm{oil}\right)}\right)$ ${p}_{B\left(\mathrm{abs}\right)}={p}_{B}+{p}_{\mathrm{atm}\left(\mathrm{oil}\right)}$ ${p}_{B\left(\mathrm{limited}\right)}=\mathrm{max}\left({p}_{B},{p}_{\mathrm{vapour}\left(\mathrm{oil}\right)}-{p}_{\mathrm{atm}\left(\mathrm{oil}\right)}\right)$
Variables
Name Value Units Description Modelica ID $\mathrm{Δp}$ $\mathrm{Pa}$ Pressure drop dp $q$ $\frac{{m}^{3}}{s}$ Flow rate flowing into port_A q ${p}_{A\left(\mathrm{limited}\right)}$ $\mathrm{Pa}$ Limited gauge pressure pA_limited ${p}_{B\left(\mathrm{limited}\right)}$ $\mathrm{Pa}$ Limited gauge pressure pB_limited $\mathrm{\rho }$ $\frac{\mathrm{kg}}{{m}^{3}}$ Upstream density rho $\mathrm{\nu }$ $\frac{{m}^{2}}{s}$ Upstream kinematic viscosity nu ${p}_{A\left(\mathrm{abs}\right)}$ $\mathrm{Pa}$ Absolute pressure pA pA_abs ${p}_{B\left(\mathrm{abs}\right)}$ $\mathrm{Pa}$ Absolute pressure pB pB_abs $T$ $K$ Local temperature T ${p}_{A\left(\mathrm{summary}\right)}$ ${p}_{A}$ $\mathrm{Pa}$ Pressure at port A summary_pA ${p}_{B\left(\mathrm{summary}\right)}$ ${p}_{B}$ $\mathrm{Pa}$ Pressure at port B summary_pB ${\mathrm{Δp}}_{\mathrm{summary}}$ $\mathrm{Δp}$ $\mathrm{Pa}$ Pressure drop summary_dp ${q}_{\mathrm{summary}}$ $q$ $\frac{{m}^{3}}{s}$ Flow rate flowing into port_A summary_q ${P}_{\mathrm{hyd}\left(\mathrm{summary}\right)}$ $-\mathrm{Δp}q$ $W$ Hydraulic Power summary_HP ${p}_{\mathrm{sat}}$ [1] $\mathrm{Pa}$ Gas saturation pressure p_sat ${V}_{A}$ VolumeA ${V}_{B}$ VolumeB $\mathrm{genericValve}$ genericValve
[1] $\mathrm{oil.gasSaturationPressure}\left(T=T,{v}_{\mathrm{gas}}={\mathrm{oil.v}}_{\mathrm{gas}}\right)$
Connections
Name Description Modelica ID ${\mathrm{port}}_{A}$ Layout of port where oil flows into an element ($0<{m}_{\mathrm{flow}}$, ${p}_{B}<{p}_{A}$ means $0<\mathrm{Δp}$) port_A ${\mathrm{port}}_{B}$ Hydraulic port where oil leaves the component (${m}_{\mathrm{flow}}<0$, ${p}_{B}<{p}_{A}$ means $0<\mathrm{Δp}$) port_B $\mathrm{oil}$ oil $\mathrm{opening}$ opening
Parameters
General Parameters
Name Default Units Description Modelica ID ${\mathrm{ΔT}}_{\mathrm{system}}$ $0$ $K$ Temperature offset from system temperature dT_system use volume A $\mathrm{true}$ If true, a volume is present at port_A useVolumeA use volume B $\mathrm{true}$ If true, a volume is present at port_B useVolumeB ${V}_{A}$ ${10}^{-6}$ ${m}^{3}$ Geometric volume at port A volumeA ${V}_{B}$ ${10}^{-6}$ ${m}^{3}$ Geometric volume at port B volumeB
Hydraulic Resistance Parameters
Name Default Units Description Modelica ID $\mathrm{geometry}$ [1] Choice of geometry for valve geometry valve coefficient [2] Choice of valve coefficient valveCoefficient $\mathrm{Av}$ [3] Av (metric) flow coefficient [Av]=m^2 Av $\mathrm{Kv}$ $\frac{10000000}{277}\mathrm{Av}$ Kv (metric) flow coefficient [Kv]=m^3/h Kv $\mathrm{Cv}$ $\frac{5000000}{123}\mathrm{Av}$ Cv (US) flow coefficient [Cv]=USG/min Cv ${\mathrm{Δp}}_{\mathrm{nominal}}$ $1000$ $\mathrm{Pa}$ Nominal pressure loss dp_nominal ${m}_{\mathrm{flow}\left(\mathrm{nom}\right)}$ [4] $\frac{\mathrm{kg}}{s}$ Nominal mass flow rate m_flow_nominal ${\mathrm{\rho }}_{\mathrm{nominal}}$ $1000$ $\frac{\mathrm{kg}}{{m}^{3}}$ Nominal inlet density rho_nominal ${\mathrm{opening}}_{\mathrm{nominal}}$ $\frac{1}{2}$ Nominal opening opening_nominal ${\mathrm{\zeta }}_{\mathrm{tot}\left(\mathrm{min}\right)}$ $0.001$ Minimal pressure loss coefficient at full opening zeta_TOT_min ${\mathrm{\zeta }}_{\mathrm{tot}\left(\mathrm{max}\right)}$ ${10}^{8}$ Maximum pressure loss coefficient at closed opening zeta_TOT_max ${\mathrm{Δp}}_{\mathrm{small}}$ $\frac{1}{100}{\mathrm{Δp}}_{\mathrm{nominal}}$ $\mathrm{Pa}$ Linearization for a pressure loss smaller then dp_small dp_small
[1] $\mathrm{Modelica.Fluid.Dissipation.Utilities.Types.ValveGeometry.Ball}$
[2] $\mathrm{Modelica.Fluid.Dissipation.Utilities.Types.ValveCoefficient.AV}$
[3] $\frac{1}{400}\mathrm{\pi }$
[4] ${\mathrm{opening}}_{\mathrm{nominal}}\mathrm{Av}\sqrt{{\mathrm{\rho }}_{\mathrm{nominal}}{\mathrm{Δp}}_{\mathrm{nominal}}}$ | 1,919 | 5,747 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 97, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-47 | latest | en | 0.224054 |
http://gnuplot-surprising.blogspot.com/2012_05_01_archive.html | 1,503,530,647,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886124662.41/warc/CC-MAIN-20170823225412-20170824005412-00140.warc.gz | 176,594,248 | 13,773 | ## Saturday, May 19, 2012
### How to pick out the maximum and minimum points when plotting with gnuplot?
Picking out the maximum and minimum points from a data file may be useful when you want to make some explanation to your graph. It is known that GPVAL_Y_MAX (GPVAL_Y_MIN) is the maximum (minimum) value. But what is the corresponding X coordinate?
With Gnuplot 4.6 the both the x and y coordinate of maximum and minimum points can be find out easily. The method is using new command "stats". This command is used for statistic. When it is run, some statistical results will be gotten. If your data file contains two column of data, (STATS_pos_max_y, STATS_max_y) will be the coordinate of the maximum point and (STATS_pos_min_y, STATS_min_y) will be the coordinate of the minimum point.
For example, we have a data file named "data.dat" like this
```0.1 0.289010715
0.2 0.050630492
0.3 0.721247895
0.4 0.284271151
0.5 0.505051253
0.6 0.101819025
0.7 0.008466133
0.8 0.36249047
0.9 0.487576233
1.0 0.595090343
1.1 0.865255938
1.2 0.696628854
1.3 0.505899456
1.4 0.338131983
1.5 0.108034045
```
Run a gnuplot script as follows
```reset
set term png
set output "max_min.png"
stats "data.dat" u 1:2 nooutput
set xrange [STATS_min_x:STATS_max_x]
set label 1 "Maximun" at STATS_pos_max_y, STATS_max_y offset 1,-0.5
set label 2 "Minimun" at STATS_pos_min_y, STATS_min_y offset 1,0.5
plot "data.dat" w p pt 3 lc rgb"#ff0000" notitle, \
STATS_min_y w l lc rgb"#00ffff" notitle, \
STATS_max_y w l lc rgb"#00ffff" notitle
set output
```
And you will get a graph like the following one.
Picking out the maximum and minimum points when plotting using gnuplot | 556 | 1,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-34 | longest | en | 0.68265 |
https://mattermodeling.stackexchange.com/questions/9230/in-which-part-of-the-hartree-fock-algorithm-or-dft-one-could-a-neural-network | 1,716,320,057,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00846.warc.gz | 323,987,582 | 26,598 | # In which part of the Hartree–Fock algorithm (or DFT one) could a neural network be most effectively used?
I understand that the task of implementing machine learning in DFT and Hartree–Fock (HF) algorithm has already been solved, perhaps to some extent, but it is interesting to think about how to implement, for example, a neural network (NN) such a way that NN would be balanced towards the speed of calculation execution rather than training: (1) using several NNs in various parts of the algorithm; (2) using one single NN into one part of the algorithm; (3) using one common NN for the entire algorithm; (4) or something else?
I have seen that NNs are used for the matrix diagonalization procedure. The procedure is used, for example, by the HF algorithm.
• I've thought about this a decent amount before and I think one of the best things that could be done which would still give you the exact answer, is to machine learn a better guess for the converged orbitals. That is, most of the time in HF and DFT is spent optimizing the orbitals iteratively. Minimizing the number of iterations is very important for getting solutions quickly. If you're interested in ML that gives you solutions other than the appropriate HF or DFT energy, then literally anything is on the table and you're only limited by your imagination. Jun 10, 2022 at 15:28
• Could you share the link to the article implementing NN-based matrix diagonalization? That sounds interesting Jun 10, 2022 at 17:21
• I would agree with the above - I think better initial guess might help. For example @susilehtola published papers on the superposition of atomic potentials: doi.org/10.1021/acs.jctc.8b01089 and arxiv.org/pdf/2002.02587.pdf - one could imagine an ML model adjusting these based on atomic environments / neighborhoods. Jun 13, 2022 at 0:56
• Calculating 2-electron integrals with Gaussian or plane wave basis functions is by now so optimized that NN methods will have a hard time competing. But if sufficient accuracy and speed could be attained by NN to provide 2-electron integrals with Slater type basis functions, that could be interesting. Jun 13, 2022 at 8:51
• Well, 2-electron integrals can be computed efficiently with Slater-type basis sets using resolutions of the identity and quadrature, so ML would have a hard time in that as well... Jun 26, 2022 at 18:46
## 1 Answer
I want neural-networks that are super good at Geometry optimization. Every HF calculation requires correct geometry to produce meaningful results. Bad geometries that are far from optimal can exacerbate the convergence issues Hartree-Fock already has. But if you could avoid that by getting good initial atomic positions from a neural-network, you could both speed up HF convergence significantly and reduce the total number of times it needs to be called.
• HF geometries are not terribly good due to lack of correlation. Good ways to get good initial geometries are to use molecular mechanics methods, as implemented in OpenBabel, for example, or tight-binding DFT methods such as in the xtb program. Sep 29, 2022 at 12:40 | 704 | 3,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-22 | latest | en | 0.956686 |
https://nz.education.com/exercises/fourth-grade/fractions/CCSS/ | 1,601,159,018,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00768.warc.gz | 544,262,583 | 24,731 | Search Our Content Library
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Fractions
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Adding Mixed Fractions with Like Denominators
Exercise
Adding Mixed Fractions with Like Denominators
Students will be able to add mixed number fractions after they learn how to make each number have like denominators.
Math
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Compare Fractions with Different Denominators 2
Exercise
Compare Fractions with Different Denominators 2
Demystify comparing fractions with different denominators by giving your students access to this easy to understand exercise.
Math
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Equivalent Fractions 2
Exercise
Equivalent Fractions 2
Show students how to modify a math problem by finding equivalent fractions for any number.
Math
Exercise
Convert Decimals to Fractions 1
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Convert Decimals to Fractions 1
Converting decimals to fractions will be a breeze for your students after they have completed this exercise.
Math
Exercise
Fractions on a Number Line 2
Exercise
Fractions on a Number Line 2
Fractions on a Number Line 1 will help students practice this key third grade skill. Try our free exercises to build knowledge and confidence.
Math
Exercise
Mixed Numbers and Improper Fractions 2
Exercise
Mixed Numbers and Improper Fractions 2
Help students identify mixed numbers and improper fractions with this exercise that is easy to use and understand.
Math
Exercise
Line Plots and Unit Fractions
Exercise
Line Plots and Unit Fractions
Understanding unit fractions is a lot easier with this exercise that lays out all out on line plots.
Math
Exercise
Compare Fractions with the Same Denominators 2
Exercise
Compare Fractions with the Same Denominators 2
All your students need to be able to compare fractions with the same denominators is this exercise.
Math
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Comparing Fractions
Exercise
Comparing Fractions
Comparing Fractions will help students practice this key fourth grade skill. Try our free exercises to build knowledge and confidence.
Math
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Fractions and Time
Exercise
Fractions and Time
These examples of how time can be broken down into fractions show that fractions relate directly to students’ everyday lives.
Math
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Decimal Fractions 1
Exercise
Decimal Fractions 1
Students will be able to convert from fractions to decimals, and back to fractions again with this exercise.
Math
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Adding Fractions with the Same Denominator
Exercise
Adding Fractions with the Same Denominator
This exercise will show students how to add fractions properly by ensuring the denominators are like numbers.
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Fractions and Area Models
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Fractions and Area Models
Fractions and Area Models will help students practice this key fourth grade skill. Try our free exercises to build knowledge and confidence.
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Subtracting Fractions with the Same Denominator
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Subtracting Fractions with the Same Denominator
To properly subtract fractions, students will first need to work through this exercise demonstrating how to find like denominators. | 632 | 3,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-40 | latest | en | 0.880668 |
https://www.groundai.com/project/fluctuation-versus-fixation-in-the-one-dimensional-constrained-voter-model/ | 1,542,659,898,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746110.52/warc/CC-MAIN-20181119192035-20181119214035-00098.warc.gz | 866,326,507 | 61,192 | Fluctuation versus fixation in the one-dimensional constrained voter model
Fluctuation versus fixation in the one-dimensional constrained voter model
Abstract
The constrained voter model describes the dynamics of opinions in a population of individuals located on a connected graph. Each agent is characterized by her opinion, where the set of opinions is represented by a finite sequence of consecutive integers, and each pair of neighbors, as defined by the edge set of the graph, interact at a constant rate. The dynamics depends on two parameters: the number of opinions denoted by and a so-called confidence threshold denoted by . If the opinion distance between two interacting agents exceeds the confidence threshold then nothing happens, otherwise one of the two agents mimics the other one just as in the classical voter model. Our main result shows that the one-dimensional system starting from any product measures with a positive density of each opinion fluctuates and clusters if and only if . Sufficient conditions for fixation in one dimension when the initial distribution is uniform and lower bounds for the probability of consensus for the process on finite connected graphs are also proved.
{frontmatter}\runtitle
Constrained voter model \runauthorN. Lanchier and S. Scarlatos \addressSchool of Mathematical and Statistical Sciences
Arizona State University
Tempe, AZ 85287, USA. \addressDepartment of Mathematics
University of Patras
Patras 26500, Greece.
{keyword}
[class=AMS] \kwd[Primary ]60K35
{keyword}\kwd
Interacting particle systems, constrained voter model, fluctuation, fixation.
1 Introduction
The constrained voter model has been originally introduced in [9] to understand the opinion dynamics in a spatially structured population of leftists, centrists and rightists. As in the popular voter model [3, 5], the individuals are located on the vertex set of a graph and interact through the edges of the graph at a constant rate. However, in contrast with the classical voter model where, upon interaction, an individual adopts the opinion of her neighbor, it is now assumed that this imitation rule is suppressed when a leftist and a rightist interact. In particular, the model includes a social factor called homophily that prevents agents who disagree too much to interact.
Model description – This paper is concerned with a natural generalization of the previous version of the constrained voter model that includes an arbitrary finite number of opinions and a so-called confidence threshold . Having a connected graph representing the network of interactions, the state at time is a spatial configuration
ηt:V→{1,2,…,F}:=opinion set.
Each individual looks at each of her neighbors at rate one that she imitates if and only if the opinion distance between the two neighbors is at most equal to the confidence threshold. Formally, the dynamics of the system is described by the Markov generator
Lf(η) = ∑x∈V∑Fj=1card{y∼x:η(y)=j and |η(y)−η(x)|≤θ} [f(ηx,j)−f(η)]
where configuration is obtained from by setting
ηx,j(z) = j 1{z=x}+η(z) 1{z≠x}
and where means that the two vertices are connected by an edge. Note that the basic voter model and the original version of the constrained voter model including the three opinions leftist, centrist and rightist can be recovered from our general model as follows:
basic voter model \@@cite[cite]{[\@@bibref{Refnum}{% clifford_sudbury_1973, holley_liggett_1975}{}{}]}=the process {ηt:t≥0} with F=2 and θ=1constrained voter model \@@cite[cite]{[\@@bibref{Refnum}{vazquez_% krapivsky_redner_2003}{}{}]}=the process {ηt:t≥0} with F=3 and θ=1.
The main question about the constrained voter model is whether the system fluctuates and evolves to a global consensus or fixates in a highly fragmented configuration. To define this dichotomy rigorously, we say that fluctuation occurs whenever
P(ηt(x) changes value at arbitrary large t) = 1% for all x∈V (1)
and that fixation occurs if there exists a configuration such that
P(ηt(x)=η∞(x) eventually in t) = 1% for all x∈V. (2)
In other words, fixation means that the opinion of each individual is only updated a finite number of times, therefore fluctuation (1) and fixation (2) exclude each other. We define convergence to a global consensus mathematically as a clustering of the system, i.e.,
limt→∞ P(ηt(x)=ηt(y)) = 1for all x,y∈V. (3)
Note that, whenever , the process reduces to the basic voter model with instead of two different opinions for which the long-term behavior of the process is well known: the system on lattices fluctuates while the system on finite connected graphs fixates to a configuration in which all the individuals share the same opinion. In particular, the main objective of this paper is to study fluctuation and fixation in the nontrivial case when .
Main results – Whether the system fluctuates or fixates depends not only on the two parameters but also on the initial distribution. In particular, we point out that, throughout the paper, it will be assumed that the initial distribution is the product measure with constant densities. To avoid trivialities, we also assume that the initial density of each of the opinions is positive:
ρj := P(η0(x)=j) > 0for allx∈V % and j∈{1,2,…,F}. (4)
For the constrained voter model on the one-dimensional torus with vertices, the mean-field analysis in [9] suggests that, in the presence of three opinions and when the threshold is equal to one, the average domain length at equilibrium is
Missing or unrecognized delimiter for \bigg (5)
when the initial density of centrists is small and is large. Vázquez et al. [9] also showed that these predictions agree with their numerical simulations from which they conclude that, when the initial density of centrists is small, the system fixates with high probability in a frozen mixture of leftists and rightists. In contrast, it is conjectured in [1] based on an idea in [7] that the infinite system fluctuates and clusters whenever , which includes the threshold one model with three opinions introduced in [9]. To explain this apparent disagreement, we first observe that, regardless of the parameters, the system on finite graphs always fixate and there is a positive probability that the final configuration consists of a highly fragmented configuration, thus showing that spatial simulations of the necessarily finite system are not symptomatic of the behavior of its infinite counterpart. Our first theorem shows that the conjecture in [1] is indeed correct.
Theorem 1
– Assume (4) and . Then,
1. The process on fluctuates (1) and clusters (3).
2. The probability of consensus on any finite connected graph satisfies
P(ηt≡constant for some t>0) ≥ ρF−θ+ρF−θ+1+⋯+ρθ+1>0.
The intuition behind the proof is that, whenever , there is a nonempty set of opinions which are within the confidence threshold of any other opinions. This simple observation implies the existence of a coupling between the constrained and basic voter models, which is the key to proving fluctuation. The proof of clustering is more difficult. It heavily relies on the fact that the system fluctuates but also on an analysis of the interfaces of the process through a coupling with a certain system of charged particles. In contrast, our lower bound for the probability of consensus on finite connected graphs relies on techniques from martingale theory. Note that this lower bound is in fact equal to the initial density of individuals who are in the confidence threshold of any other individuals in the system. Returning to the relationship between finite and infinite systems, we point out that the simulation pictures of Figure 1, which show two typical realizations of the process on the torus under the assumptions of the theorem, suggest fixation of the infinite counterpart in a highly fragmented configuration, in contradiction with the first part of our theorem, showing again the difficulty to interpret spatial simulations. Note also that, for the system on the one-dimensional torus with vertices, the average domain length at equilibrium is bounded from below by
card(V)×P(consenesus) = L×P(% consensus)
which, together with the second part of the theorem, proves that the average domain length scales like the population size when and that (5) does not hold. While our fluctuation-clustering result holds regardless of the initial densities provided they are all positive, whether fixation occurs or not seems to be very sensitive to the initial distribution. Also, to state our fixation results and avoid messy calculations later, we strengthen condition (4) and assume that
ρ1=ρF>0andρ2=ρ3=⋯=ρF−1>0. (6)
The next theorem looks at the fixation regime in three different contexts.
Theorem 2
– Assume (6). Then, the process on fixates (2) in the following cases:
1. and is small enough.
2. is large, and where
c+≈0.21851 is a root of 12(1−2X)3−9X2(3X2+4X−6).
3. and and .
The first part of the theorem is the converse of the first part of Theorem 1, thus showing that the condition is critical in the sense that
• when , the one-dimensional constrained voter model fluctuates when starting from any nondegenerate distributions (4) whereas
• when , the one-dimensional constrained voter model can fixate even when starting from a nondegenerate distribution (4).
The last two parts of the theorem specialize in two particular cases. The first one looks at uniform initial distributions in which all the opinions are equally likely. For simplicity, our statement focuses on the fixation region when the parameters are large but our proof is not limited to large parameters and implies more generally that the system fixates for all pairs of parameters corresponding to the set of white dots in the phase diagram of Figure 2 for the one-dimensional system with up to twenty opinions. Note that the picture suggests that the process starting from a uniform initial distribution fixates whenever even for a small number of opinions. The second particular case returns to the slightly more general initial distributions (6) but focuses on the threshold one model with four opinions for which fixation is proved when is only slightly less than one over the number of opinions = 0.25. This last result suggests that the constrained voter model with four opinions and threshold one fixates when starting from the uniform product measure, although the calculations become too tedious to indeed obtain fixation when starting from this distribution.
Structure of the paper – The rest of the article is devoted to the proof of both theorems. Even though our proof of fluctuation-clustering and fixation differ significantly, a common technique we introduce to study these two aspects for the one-dimensional process is a coupling with a certain system of charged particles that keeps track of the discrepancies along the edges of the graph rather than the actual opinion at each vertex. In contrast, our approach to analyze the process on finite connected graphs is to look at the opinion at each vertex and use, among other things, the optimal stopping theorem for martingales. The coupling with the system of charged particles is introduced in section 2 and then used in section 3 to prove Theorem 1. The proof of Theorem 2 is more complicated and carried out in the last five sections 48. In addition to the coupling with the system of charged particles introduced in the next section, the proof relies on a characterization of fixation based on so-called active paths proved in section 4 and large deviation estimates for the number of changeovers in a sequence of independent coin flips proved in section 5.
2 Coupling with a system of charged particles
To study the one-dimensional system, it is convenient to construct the process from a graphical representation and to introduce a coupling between the process and a certain system of charged particles that keeps track of the discrepancies along the edges of the lattice rather than the opinion at each vertex. This system of charged particles can also be constructed from the same graphical representation. Since the constrained voter model on general finite graphs will be studied using other techniques, we only define the graphical representation for the process on , which consists of the following collection of independent Poisson processes:
• for each , we let be a rate one Poisson process,
• we denote by its th arrival time.
This collection of independent Poisson processes is then turned into a percolation structure by drawing an arrow at time and, given a configuration of the one-dimensional system at time , we say that this arrow is active if and only if
|ηt−(x)−ηt−(x±1)| ≤ θ.
The configuration at time is then obtained by setting
ηt(x±1)=ηt−(x)when the arrow x→x±1 is active=ηt−(x±1)when the arrow x→x±1 is not active. (7)
An argument due to Harris [4] implies that the constrained voter model starting from any configuration can indeed be constructed using this percolation structure and rule (7). From the collection of active arrows, we construct active paths as in percolation theory. More precisely, we say that there is an active path from to , and write , whenever there exist
s0=s
such that the following two conditions hold:
1. For , there is an active arrow from to at time .
2. For , there is no active arrow that points at .
Note that conditions 1 and 2 above imply that
for all (x,t)∈Z×R+ there is a % unique z∈Z such that (z,0)⇝(x,t).
Moreover, because of the definition of active arrows, the opinion of vertex at time originates and is therefore equal to the initial opinion of vertex so we call vertex the ancestor of vertex at time . One of the key ingredients to studying the one-dimensional system is to look at the following process defined on the edges: identifying each edge with its midpoint, we set
ξt(e) := ηt(e+1/2)−ηt(e−1/2)for alle∈D:=Z+1/2
and think of edge as being
• empty whenever ,
• occupied by a pile of particles with positive charge whenever ,
• occupied by a pile of particles with negative charge whenever .
The dynamics of the constrained voter model induces evolution rules which are again Markov on this system of charged particles. Assume that there is an arrow at time and
ξt−(x−1/2):=ηt−(x)−ηt−(x−1) = iξt−(x+1/2):=ηt−(x+1)−ηt−(x) = j ≥ 0
indicating in particular that there is a pile of particles with positive charge at . Then, we have the following alternative:
• There is no particle at edge or equivalently in which case the individuals at vertices and already agree so nothing happens.
• There is a pile of particles at edge in which case and disagree too much to interact so nothing happens.
• There is a pile of particles at in which case
ξt(x−1/2):=ηt(x)−ηt(x−1) = ηt−(x+1)−ηt−(x−1) = i+jξt(x+1/2):=ηt(x+1)−ηt(x) = ηt−(x+1)−ηt−(x+1) = 0.
In particular, there is no more particles at edge and a pile of particles all with the common charge at edge .
Similar evolution rules are obtained by exchanging the direction of the interaction or by assuming that we have from which we can deduce the following description:
• piles with more than particles cannot move therefore we call such piles blockades and the particles they contain frozen particles.
• piles with at most particles jump one step to the left or one step to the right at the same rate one therefore we call the particles they contain active particles.
• when a pile with positive/negative particles jumps onto a pile with negative/positive particles, positive and negative particles annihilate by pair which results in a smaller pile of particles all with the same charge.
We refer to Figure 3 for an illustration of these evolution rules. Note that whether an arrow is active or not can also be characterized from the state of the edge process:
x→x±1 at time t is activeif and only if|ξt−(x±1/2)|≤θ.
In particular, active arrows correspond to active piles of particles.
3 Proof of Theorem 1
The key ingredient to proving fluctuation of the one-dimensional system and estimating the probability of consensus on finite connected graphs is to partition the opinion set into two sets that we shall call the set of centrist opinions and the set of extremist opinions:
Ω0 := {F−θ,F−θ+1,…,θ+1}andΩ1 := {1,2,…,F}∖Ω0.
Note that the assumption implies that the set of centrist opinions is nonempty. Note also that both sets are characterized by the properties
j∈Ω0if and only if|i−j|≤θfor all i∈{1,2,…,F}j∈Ω1if and only if|i−j|>θfor some i∈{1,2,…,F} (8)
as shown in Figure 4 which gives a schematic illustration of the partition. Fluctuation is proved in the next lemma using this partition and relying on a coupling with the voter model.
Lemma 3
– The process on fluctuates whenever and .
Proof.
It follows from (8) that centrist agents are within the confidence threshold of every other individual. In particular, for each pair we have the transition rates
ci→j(x,η):=limh→0 (1/h)P(ηt+h(x)=j|ηt(x)=i)=card{y∼x:|i−j|≤θ and ηt(y)=j} = card{y∼x:ηt(y)=j} (9)
and similarly
cj→i(x,η):=limh→0 (1/h)P(ηt+h(x)=i|ηt(x)=j)=card{y∼x:|i−j|≤θ and ηt(y)=i} = card{y∼x:ηt(y)=i}. (10)
Now, we introduce the process
ζt(x) := 1{ηt(x)∈Ω1}for allx∈Z.
Since for all the transition rates are constant over all according to (9), we have the following local transition rate for this new process:
c0→1(x,ζ) := limh→0 (1/h)P(ζt+h(x)=1|ζt(x)=0)= limh→0 (1/h)∑i∈Ω0P(ζt+h(x)=1|ηt(x)=i)P(ηt(x)=i|ζt(x)=0)= limh→0 (1/h)∑i∈Ω0∑j∈Ω1P(ηt+h(x)=j|ηt(x)=i)P(ηt(x)=i|ζt(x)=0)= ∑i∈Ω0∑j∈Ω1ci→j(x,η)P(ηt(x)=i|ζt(x)=0)= ∑i∈Ω0∑j∈Ω1card{y∼x:ηt(y)=j}P(ηt(x)=i|ζt(x)=0)= ∑j∈Ω1card{y∼x:ηt(y)=j} = card{y∼x:ζt(y)=1}.
Using (10) in place of (9) and some obvious symmetry, we also have
c1→0(x,ζ) := card{y∼x:ηt(y)∈Ω0} = card{y∼x:ζt(y)=0}.
This shows that the spin system reduces to the voter model. In particular, the lemma directly follows from the fact that the one-dimensional voter model itself, when starting with a positive density of each type, fluctuates, a result proved based on duality in [7], pp 868–869. ∎
Lemma 4
– The process on clusters whenever and .
Proof.
The proof strongly relies on the coupling with the voter model in the proof of the previous lemma. To begin with, we define the function
u(t) := E(ξt(e)) = ∑Fj=0jP(ξt(e)=j)
which, in view of translation invariance of the initial configuration and the evolution rules, does not depend on the choice of . Note that, since the system of charged particles coupled with the process involves deaths of particles but no births, the function is nonincreasing in time, therefore it has a limit: as . Now, on the event that an edge is occupied by a pile of at least one particle at a given time , we have the following alternative:
• is a blockade. In this case, since the centrist agents are within the confidence threshold of all the other agents, we must have
ηt(x)∈Ω1andηt(x+1)∈Ω1.
But since the voter model fluctuates,
T := inf{s>t:ηs(x)∈Ω0 or ηs(x+1)∈Ω0} < ∞almost surely.
In particular, at least of one of the frozen particles at is killed eventually.
• is a live edge. In this case, since one-dimensional symmetric random walks are recurrent, the active pile of particles at eventually intersects another pile of particles, and we have the following alternative:
• The two intersecting piles of particles have opposite charge, which results in the simultaneous death of at least two particles.
• The two intersecting piles have the same charge and merge to form a blockade in which case we are back to the previous case: since the voter model fluctuates, at least one of the frozen particles in this blockade is killed eventually.
• The two intersecting piles have the same charge and merge to form a larger active pile in which case the pile keeps moving until, after a finite number of collisions, we are back to one of the previous two possibilities: at least two active particles annihilate or there is creation of a blockade with at least one particle that is killed eventually.
In either case, as long as there are particles, there are also annihilating events indicating that the density of particles is strictly decreasing as long as it is positive. In particular, the density of particles decreases to zero so there is extinction of both the active and frozen particles:
limt→∞P(ξt(e)≠0) = 0for% alle∈Z+1/2.
In particular, for all with , we have
limt→∞P(ηt(x)≠ηt(y)) ≤ limt→∞P(ξt(z+1/2)≠0 for some x≤z
which proves clustering. ∎
The second part of the theorem, which gives a lower bound for the probability of consensus of the process on finite connected graphs, relies on very different techniques, namely techniques related to martingale theory following an idea from [6], section 3. However, the partition of the opinion set into centrist opinions and extremist opinions is again a key to the proof.
Lemma 5
– For the process on any finite connected graph,
P(consensus) ≥ ρcwheneverF≤2θ+1.
Proof.
The first step is to prove that the process that keeps track of the number of supporters of any given opinion is a martingale. Then, applying the martingale convergence theorem and optimal stopping theorem, we obtain a lower bound for the probability of extinction of the extremist agents, which is also a lower bound for the probability of consensus. For , we set
Xt(j) := card{x∈V:ηt(x)=j}andXt := card{x∈V:ηt(x)∈Ω0}
and we observe that
Xt = ∑j∈Ω0Xt(j) = Xt(F−θ)+Xt(F−θ+1)+⋯+Xt(θ+1). (11)
Letting denote the natural filtration of the process, we also have
limh→0 (1/h)E(Xt+h(j)−Xt(j)|Ft)= limh→0 (1/h)P(Xt+h(j)−Xt(j)=1|Ft)− limh→0 (1/h)P(Xt+h(j)−Xt(j)=−1|Ft)= card{(x,y)∈E:ηt(x)≠j and% ηt(y)=j and |ηt(x)−j|≤θ}− card{(x,y)∈E:ηt(x)=j and ηt(y)≠j and |ηt(y)−j|≤θ} = 0
indicating that the process is a martingale with respect to the natural filtration of the constrained voter model. This, together with (11), implies that also is a martingale. It is also bounded because of the finiteness of the graph therefore, according to the martingale convergence theorem, there is almost sure convergence to a certain random variable:
Xt ⟶ X∞almost surely as t→∞
and we claim that can only take two values:
X∞∈{0,N}whereN:=card(V) % = the population size. (12)
To prove our claim, we note that, invoking again the finiteness of the graph, the process gets trapped in an absorbing state after an almost surely stopping time so we have
X∞=XTwhereT:=inf{t:ηt=ηs % for all s>t} is almost surely finite.
Assuming by contradiction that gives an absorbing state with at least one centrist agent and at least one extremist agent. Since the graph is connected, this implies the existence of an edge such that
ηT(x)∈Ω0andηT(y)∈Ω1
but then we have and
|ηT(y)−ηT(x)| ≤ max((θ+1)−1,F−(F−θ)) = θ
showing that is not an absorbing state, in contradiction with the definition of time . This proves that our claim (12) is true. Now, applying the optimal stopping theorem to the bounded martingale and the almost surely finite stopping time and using (12), we obtain
EXT=EX0=N×P(η0(x)∈Ω0) = Nρc=EX∞=0×P(X∞=0)+N×P(X∞=N) = N×P(X∞=N),
from which it follows that
P(X∞=N) = ρc. (13)
To conclude, we observe that, on the event that , all the opinions present in the system after the hitting time are within distance of each other therefore the process evolves according to a voter model after that time. Since the only absorbing states of the voter model on finite connected graphs are the configurations in which all the agents share the same opinion, we deduce that the system converges to a consensus. This, together with (13), implies that
P(consensus) ≥ P(X∞=N) = ρc.
This completes the proof of the lemma. ∎
4 Sufficient condition for fixation
The main objective of this section is to prove a sufficient condition for fixation of the constrained voter model based on certain properties of the active paths.
Lemma 6
– For all , let
T(z) := inf{t:(z,0)⇝(0,t)}.
Then, the constrained voter model fixates whenever
limN→∞ P(T(z)<∞ for some z<−N) = 0. (14)
Proof.
This is similar to the proof of Lemma 2 in [2] and Lemma 4 in [8]. To begin with, we define recursively a sequence of stopping times by setting
τ0:=0andτj:=inf{t>τj−1:ηt(0)≠ητj−1(0)}for j≥1.
In other words, the th stopping time is the th time the individual at the origin changes her opinion. Now, we define the following random variables and collection of events:
aj:=the ancestor of vertex 0 at time τj% B:={τj<∞ for all j}and GN := {|aj|
The assumption (14) together with reflection symmetry implies that the event occurs almost surely for some positive integer , which implies that
P(B) = P(B∩(∪NGN)) = P(∪N(B∩GN)).
Since is the event that the individual at the origin changes her opinion infinitely often, in view of the previous inequality, in order to establish fixation, it suffices to prove that
P(B∩GN) = 0for all N≥1. (15)
To prove equations (15), we let
It(x):={z∈Z:x is the ancestor of z at time t}andMt(x):=card(It(z))
be the set of descendants of at time which, due to one-dimensional nearest neighbor interactions, is necessarily an interval and its cardinality, respectively. Now, since each interaction between two individuals is equally likely to affect the opinion of each of these two individuals, the number of descendants of any given site is a martingale whose expected value is constantly equal to one. In particular, the martingale convergence theorem implies that
limt→∞ Mt(x) = M∞(x)with% probability onewhere E|M∞(x)|<∞
therefore the number of descendants of converges to a finite value. Since in addition the number of descendants is an integer-valued process,
σ(x) := inf{t>0:Mt(x)=M∞(x)} < ∞with % probability one,
which further implies that, with probability one,
limt→∞It(x)=I∞(x)andρ(x):=inf{t>0:It(x)=I∞(x)}<∞. (16)
Finally, we note that, on the event , the last time the individual at the origin changes her opinion is at most equal to the largest of the stopping times for therefore
P(B∩GN) = P(ρ(x)=∞ for some −N
according to (16). This proves (15) and the lemma. ∎
5 Large deviation estimates
In order to find later a good upper bound for the probability in (14) and deduce a sufficient condition for fixation of the process, the next step is to prove large deviation estimates for the number of piles with particles with a given charge in a large interval. More precisely, the main objective of this section is to prove that for all and all the probability that
card{e∈[0,N):ξ0(e)=j}∉(1−ϵ,1+ϵ) E(card{e∈[0,N):ξ0(e)=j})
decays exponentially with . Note that, even though the initial opinions are chosen independently, the states at different edges are not independent. For instance, a pile of particles with a positive charge is more likely to be surrounded by negative particles. In particular, the result does not simply follow from large deviation estimates for the binomial distribution. The main ingredient is to first show large deviation estimates for the number of so-called changeovers in a sequence of independent coin flips. Consider an infinite sequence of independent coin flips such that
P(Xj=H) = pandP(Xj=T) = q=1−pfor % allj∈N
where is the outcome: heads or tails, at time . We say that a changeover occurs whenever two consecutive coin flips result in two different outcomes. The expected value of the number of changeovers before time can be easily computed by observing that
ZN = ∑N−1j=0YjwhereYj := 1{Xj+1≠Xj}
and by using the linearity of the expected value:
EZN = ∑N−1j=0EYj = ∑N−1j=0P(Xj+1≠Xj) = NP(X0≠X1) = 2Np(1−p).
Then, we have the following large deviation estimates for the number of changeovers.
Lemma 7
– For all , there exists such that
P(ZN−EZN∉(−ϵN,ϵN)) ≤ exp(−c0N)%forall$N$sufficientlylarge.
Proof.
To begin with, we let be the time to the th changeover and notice that, since all the outcomes between two consecutive changeovers are identical, the sequence of coin flips up to this stopping time can be decomposed into strings with an alternation of strings with only heads and strings with only tails followed by one more coin flip. In addition, since the coin flips are independent, the length distribution of each string is
Hj:=length of the jth string of heads = Geometric(q)Tj:=length of the jth string of tails = Geometric(p)
and lengths are independent. In particular, is equal in distribution to the sum of independent geometric random variables with parameters and , namely, we have
P(τ2K=n) = P(H1+T1+⋯+HK+TK=n)for alln∈N. (17)
Now, using that, for all ,
P(H1+H2+⋯+HK=n) =(n−1K−1)qK(1−q)n−K=Kn (nK)qK(1−q)n−K ≤ P(Binomial(n,q)=K)
and large deviation estimates for the binomial distribution implies that
P((1/K)(H1+H2+⋯+HK)≥(1+ϵ)(1/q))≤ P(Binomial((1+ϵ)(1/q)K,q)≤K) ≤ exp(−c1K)P((1/K)(H1+H2+⋯+HK)≤(1−ϵ)(1/q))≤ P(Binomial((1−ϵ)(1/q)K,q)≥K) ≤ exp(−c1K) (18)
for a suitable constant and all large. Similarly,
P((1/K)(T1+T2+⋯+TK)≥(1+ϵ)(1/p)) ≤ exp(−c2K)P((1/K)(T1+T2+⋯+TK)≤(1−ϵ)(1/p)) ≤ exp(−c2K) (19)
for a suitable and all large. Combining (17)–(19), we deduce that
P((1/K)τ2K∉((1−ϵ)(1/p+1/q),(1+ϵ)(1/p+1/q)))= P((1/K)(H1+T1+⋯+HK+TK)∉((1−ϵ)(1/p+1/q),(1+ϵ)(1/p+1/q)))≤ P((1/K)(H1+H2+⋯+HK)∉((1−ϵ)(1/q),(1+ϵ)(1/q)))+ P((1/K)(T1+T2+⋯+TK)∉((1−ϵ)(1/p),(1+ϵ)(1/p)))≤ 2exp(−c1K)+2exp(−c2K).
Taking and observing that , we deduce
P((1/N)τ2K∉(1−ϵ,1+ϵ))= P((1/K)τ2K∉((1−ϵ)(1/p+1/q),(1+ϵ)(1/p+1/q))) ≤ exp(−c3N)
for a suitable and all large. In particular, for all sufficiently large,
P((1/N)τ2K−ϵN≥1) ≤ exp(−c4N)andP((1/N)τ2K+ϵN≤1) ≤ exp(−c5N)
for suitable constants and and all sufficiently large. Using the previous two inequalities and the fact that the event that the number of changeovers is equal to is also the event that the time to the th changeover is less than but the time to the next changeover is more than , we conclude that
P(ZN−EZN∉(−ϵN,ϵN)) = P(ZN∉(2pq−ϵ,2pq+ϵ)N)= P((1/N)ZN∉(2pq−ϵ,2pq+ϵ))= P((1/N)ZN≤2pq−ϵ)+P((1/N)ZN≥2pq+ϵ)= P((1/N)τ2K−ϵN≥1)+P((1/N)τ2K+ϵN≤1) ≤ exp(−c4N)+exp(−c5N)
for all sufficiently large. This completes the proof. ∎
Now, we say that an edge is of type if it connects an individual with initial opinion on the left to an individual with initial opinion on the right, and let
eN(i→j) := card{x∈[0,N):η0(x)=i and η0(x+1)=j}
denote the number of edges of type in the interval . Using the large deviation estimates for the number of changeovers established in the previous lemma, we can now deduce large deviation estimates for the number of edges of each type.
Lemma 8
– For all , there exists such that
P(eN(i→j)−Nρiρj∉(−ϵN,ϵN)) ≤ exp(−c6N)for all N large and all i≠j.
Proof.
For any given , the number of edges and with has the same distribution as the number of changeovers in a sequence of independent coin flips of a coin that lands on heads with probability . In particular, applying Lemma 7 with gives
P(∑j≠ieN(i→j)−Nρi(1−ρi)∉(−ϵN,ϵN)) ≤ exp(−c0N) (20)
for all sufficiently large. In addition, since each preceding a changeover is independently followed by any of the remaining opinions,
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