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https://byjus.com/mean-median-mode-formula/ | [
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"# Mean Median Mode Formula\n\nThe Mean, Median and Mode are the three measures of central tendency. Mean is the arithmetic average of a data set. This is found by adding the numbers in a data set and dividing by the number of observations in the data set. The median is the middle number in a data set when the numbers are listed in either ascending or descending order. The mode is the value that occurs the most often in a data set and the range is the difference between the highest and lowest values in a data set.\n\nThe Mean\n\n$\\large \\overline{x}=\\frac{\\sum x}{N}$\n\nHere,\n∑ represents the summation\nX represents observations\nN represents the number of observations .\n\nIn the case where the data is presented in a tabular form, the following formula is used to compute the mean\n\nMean = ∑f x / ∑f\n\nWhere ∑f = N\n\nThe Median\n\nIf the total number of observations (n) is an odd number, then the formula is given below:\n\n$\\large Median=\\left(\\frac{n+1}{2}\\right)^{th}observation$\n\nIf the total number of the observations (n) is an even number, then the formula is given below:\n\n$\\large Median=\\frac{\\left(\\frac{n}{2}\\right)^{th}observation+\\left(\\frac{n}{2}+1\\right )^{th}observation}{2}$\n\nConsider the case where the data is continuous and presented in the form of a frequency distribution, the median formula is as follows.\n\nFind the median class, the total count of observations ∑f.\n\nThe median class consists of the class in which (n / 2) is present.\n\n$$\\text { Median }=1+\\left[\\frac{\\frac{\\mathrm{n}}{2}-\\mathrm{c}}{\\mathrm{f}}\\right] \\times \\mathrm{h}$$\n\nHere\n\nl = lesser limit belonging to the median class\n\nc = cumulative frequency value of the class before the median class\n\nf = frequency possessed by the median class\n\nh = size of the class\n\nThe Mode\n\n$\\large The\\;mode\\;is\\;the\\;most\\;frequently\\;occuring\\;observation\\;or\\;value.$\n\nConsider the case where the data is continuous and the value of mode can be computed using the following steps.\n\na] Determine the modal class that is the class possessing the maximum frequency.\n\nb] Calculate the mode using the below formula\n\n$$\\text { Mode }=1+\\left[\\frac{f_{m}-f_{1}}{2 f_{m}-f_{1}-f_{2}}\\right] \\times h$$\n\nl = lesser limit of modal class\n\n$$f_{m}$$ = frequency possessed by the modal class\n\n$$f_{1}$$ = frequency possessed by the class before the modal class\n\n$$f_{2}$$ = frequency possessed by the class after the modal class\n\nh = width of the class\n\nHow Are Mean, Median And Mode Related?\n\nThe 3 estimates of central tendency that is the mean, median and mode are related by the following empirical relationship.\n\n2 Mean + Mode = 3 Median\n\nFor example, if it is required to compute the mean, median and mode of the data that is continuous grouped, then the values of the mean and median can be found using the above formulae. The value of the mode can be found using the empirical formula.\n\nIf the value of the mode is 65 and the median = 61.6, then find the value of the mean.\n\nThe value of the mean can be calculated using the formula,\n\n2 Mean + Mode = 3 Median\n\n2 Mean = (3 × 61.6) – 65\n\n2 Mean = 119.8\n\nMean = 119.8 / 2\n\nMean = 59.9\n\n### Solved Examples\n\nQuestion 1: Find the mean, median, mode, and range for the following list of values:\n\n13, 18, 13, 14, 13, 16, 14, 21, 13\n\nSolution:\n\nGiven data: 13, 18, 13, 14, 13, 16, 14, 21, 13\n\nThe mean is the usual average.\n\nMean = {13 + 18 + 13 + 14 + 13 + 16 + 14 + 21 + 13} / {9} = 15\n\n(Note that the mean is not a value from the original list. This is a common result. You should not assume that your mean will be one of your original numbers.)\n\nThe median is the middle value, so to rewrite the list in ascending order as given below:\n\n13, 13, 13, 13, 14, 14, 16, 18, 21\n\nThere are nine numbers in the list, so the middle one will be\n\n{9 + 1} / {2} = {10} / {2} = 5\n\n= 5th number\n\nHence, the median is 14.\n\nThe mode is the number that is repeated more often than any other, so 13 is the mode.\nThe largest value in the list is 21, and the smallest is 13, so the range is 21 – 13 = 8.\n\nMean = 15\nMedian = 14\nMode = 13\nRange = 8\n\nQuestion 2: The value of the mean of five numbers is observed to be 18. If one number is not included, the mean is 16. Find the number that is excluded.\n\nFrom the question,\n\nThere are 5 observations that mean n = 5.\n\nThe value of the mean = 18\n\nx̄ = 18\n\nx̄ = ∑ x / n\n\n∑ x = 5 * 18 = 90\n\nThe sum of the five observations is 90.\n\nAssume the excluded number to be “a”\n\nThe sum of four observations = 90 – a\n\nMean of 4 observations = (90 – a) / 4\n\n16 = (90 – a) / 4\n\n90 – a = 64\n\na = 26\n\n⇒ The excluded number is 26.\n\n More topics in Mean Median Mode Formula Arithmetic Mean Formula Geometric Mean Formula Harmonic Mean Formula Sample Mean Formula Weighted Mean Formula Effect Size Formula\n\nVery nice\n\n2. Arnav Khare\n\nvery nice app/website\n\n3. Kowshik\n\nSuper\n\n5. Good\n\n6. Abdiweli Ali\n\nDefine mean, median, mode and state formulas for each of them (grouped and ungrouped)? What is range?\n\n7. Mahul Majumder\n\nThank you it was very helpful for me\n\n8. Anush Mannu\n\nThis cleared my doubt. THANKS:)"
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https://forum.ozgrid.com/forum/index.php?thread/60164-worksheet-change-event-intersect-error-91/&postID=560569 | [
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"# Worksheet Change Event Intersect Error 91\n\n• Hi There\nThis is my code:\n\nwhen excution comes to this line\n\nCode\n``If Intersect(Target, Range(\"D6:D6000\")) Is Nothing Or Intersect(Target, Range(\"D6:D6000\")) = \"\" Then``\n\na run time error (91) appears, it tells object variable or with block variable not set.\n\nany idea why ? Thank you\n\n******************\n[SIZE=6]Yours\nh [/SIZE]\n\n• Re: Run Time Error 91\n\ni corrected my code, but still the error msg appear ???\n\n******************\n[SIZE=6]Yours\nh [/SIZE]\n\n• Re: Run Time Error 91\n\nIf the result of the intersect is nothing then the second part of the test will fail.\n\nAlso the second part of the test will not do what you want. It will not check all cells in the rather being empty.\n\n[vba] If Intersect(Target, Range(\"D6:D6000\")) Is Nothing Or _\nApplication.WorksheetFunction.CountA(Range(\"D6:D6000\")) = 0 Then\nIf Intersect(Target, Range(\"F6:F6000\")) Is Nothing Or _\nApplication.WorksheetFunction.CountA(Range(\"F6:F6000\")) = 0 Then\nIf Intersect(Target, Range(\"I6:I6000\")) Is Nothing Or _\nApplication.WorksheetFunction.CountA(Range(\"I6:I6000\")) = 0 Then\nExit Sub\nElse\n\nMsgBox \"Do Something......\"\nEnd If\nEnd If\nEnd If\n[/vba]\n\n[h4]Cheers\nAndy\n[/h4]\n\n• Re: Run Time Error 91\n\ni did understand you quit well, it worked good\nAndy, I donot know what to say to thank you, bt thank you very very much\n\n******************\n[SIZE=6]Yours\nh [/SIZE]\n\n• Re: Run Time Error 91\n\nHi\n\nif i delete a row within the Target, then the code trigered..........\n\nis there a way stop the code excution when i delete a row within the target ?\n\n******************\n[SIZE=6]Yours\nh [/SIZE]\n\n• Re: Worksheet Change Event Intersect Error 91\n\nHelmekki,\n\nCan your code be simplified to this?\n\nCode\n``````Private Sub Worksheet_Change(ByVal Target As Range)\nIf Intersect(Target, Range(\"D6:D6000,F6:F6000,I6:I6000\")) Is Nothing Or _\nApplication.WorksheetFunction.CountA(Range(\"D6:D6000,F6:F6000,I6:I6000\")) = 0 Then\nExit Sub\nElse\nMsgBox \"Do Something......\"\nEnd If\nEnd Sub``````\n• Re: Worksheet Change Event Intersect Error 91\n\nI havnt tested this (But its an idea):\n\n• Re: Worksheet Change Event Intersect Error 91\n\nReafidy , thank you very much it worked fine",
null,
"******************\n[SIZE=6]Yours\nh [/SIZE]\n\n• Re: Worksheet Change Event Intersect Error 91\n\nYour original code was not a Worksheet change event, hence my response that Target\nwas NOT defined ...\n\nCode\n``````Sub T()\netc .............``````\n• Re: Worksheet Change Event Intersect Error 91\n\nhelmekki, if you must move the goal posts at least post to say as much.\n\n• Re: Worksheet Change Event Intersect Error 91\n\nI apologise Ivan..........i shoud have made clear from the start",
null,
"Ok ............it was misorganised posting...............:(\n\nwill take care of such posts .:cool:\n\nThanks\n\n******************\n[SIZE=6]Yours\nh [/SIZE]"
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https://www.embedded.com/design/mcus-processors-and-socs/4460801/Implementing-floating-point-algorithms-in-FPGAs-or-ASICs | [
"Advertisement\n\n# Implementing floating-point algorithms in FPGAs or ASICs\n\nJune 19, 2018\n\nFloating-point is the most preferred data type to ensure high-accuracy calculations for algorithm modeling and simulation. Traditionally, when you want to deploy such floating-point algorithms to FPGA or ASIC hardware, your only choice is to convert every data type in the algorithm to fixed-point to conserve hardware resources and speed up calculations. Converting to fixed-point reduces mathematical precision, and sometimes it can be challenging to strike the right balance between data type word lengths and mathematical accuracy during conversion. For calculations that require high dynamic range or high precision (for example, designs that have feedback loops), fixed-point conversion can consume weeks or months of engineering time. Also, in order to achieve numerical accuracy, a designer has to use large fixed-point word lengths.\n\nIn this article, we will introduce The MathWorks' Native Floating-Point workflow for ASIC/FPGA design, using an IIR filter as an illustration. We will then review the challenges of using fixed-point, and we will compare the area and frequency tradeoffs of using single-precision floating point vs. fixed-point. We will also show how a combination of floating-point and fixed-point can give you much higher accuracy while reducing conversion and implementation time in real-world designs. You will see how modeling directly in floating-point can be important, and how sometimes it can significantly reduce area and improve speed in real-world designs with high dynamic range requirements, contrary to the popular belief that fixed-point is always more efficient compared to floating-point.\n\nNative Floating-Point Implementation: Under the Hood\n\nHDL Coder implements single-precision arithmetic by emulating the underlying math on the FPGA or ASIC resources (Figure 1). The generated logic unpacks the input floating-point signal into sign, exponent, and mantissa -- individual integers that are 1, 8, and 23 bits wide, respectively.",
null,
"Figure 1. How HDL Coder maps a single-precision floating-point multiplication to fixed-point hardware resources (© 1984–2018 The MathWorks, Inc.)\n\nThe generated VHDL or Verilog logic then performs the floating-point calculation (a multiplication in the case shown in Figure 1) by figuring out the sign bit resulting from the input sign bits, the magnitude multiplication, and the addition of exponents and corresponding normalization necessary to compute the result. The last stage of the logic packs the sign, exponent, and mantissa back into a floating-point data type.\n\nTackling Dynamic Range Issues with Fixed-Point Conversion\n\nA simple expression like (1-a)/(1+a), if it needs to be implemented with high dynamic range, can be translated naturally by using single-precision floating-point (Figure 2).",
null,
"Figure 2. Single-precision implementation of (1-a)/(1+a) (© 1984–2018 The MathWorks, Inc.)\n\nHowever, implementing the same equation in fixed-point requires many steps and numerical considerations (Figure 3).",
null,
"Figure 3. Fixed-point implementation of (1-a)/(1+a) (© 1984–2018 The MathWorks, Inc.)\n\nFor example, you must break the division into multiplication and reciprocal, use approximation methods such as Newton-Raphson or LUT (look-up table) for nonlinear reciprocal operation, use different data types to carefully control the bit growth, select the proper numerator and denominator types, and use specific output types and accumulator types for the adders and subtractors.\n\nExploring IIR Implementation Options\n\nLet's look at an infinite impulse response (IIR) filter example. An IIR filter requires high dynamic range calculation with a feedback loop, making it tricky to converge on a fixed-point quantization. Figure 4a shows a test environment comparing three versions of the same IIR filter with a noisy sine wave input. The sine wave has an amplitude of 1, and the added noise increases the amplitude slightly.",
null,
"Figure 4a. Three implementations of an IIR filter with noisy sine wave input (© 1984–2018 The MathWorks, Inc.)\n\nThe first version of the filter is double precision (Figure 4b). The second version is single-precision. The third version is a fixed-point implementation (Figure 4c). This implementation resulted in data types up to 22 bits in word length, with 1 bit allocated for the sign and 21 bits allocated for the fraction. This particular data type leaves 0 bits to represent the integer value, which makes sense given that its range of values will always be between -1 and 1 for the given stimulus. If the design has to work with different input values, that needs to be taken into account during fixed-point quantization.",
null,
"Figure 4b. IIR_filter implementation, shown with double-precision data types (© 1984–2018 The MathWorks, Inc.)",
null,
"Figure 4c. IIR_filter_fixpt implementation, which uses fixed-point data types (© 1984–2018 The MathWorks, Inc.)\n\nThe test environment is set up to compare the results of the single-precision and fixed-point filters with the double-precision filter, which is considered to be the golden reference. In both cases, a loss of precision will yield a certain amount of error. The question is whether that error is within an acceptable tolerance for our application.\n\nWhen we ran Fixed-Point Designer to perform the conversion, we specified an error tolerance of 1%. Figure 5 shows the results of the comparisons. The error for the single-precision version is on the order of 10-8, while the error for the fixed-point data type is on the order of 10-5. This is within the error tolerance we specified. If your application needs higher precision, you may need to increase your fixed-point word lengths.",
null,
"Figure 5. Simulation results comparing the double-precision IIR filter results with the single-precision results (top) and fixed-point results (bottom) (© 1984–2018 The MathWorks, Inc.)\n\nConverging on this quantization takes experience with hardware design, a comprehensive understanding of the possible system inputs, clear accuracy requirements, and some assistance from Fixed-Point Designer. This effort is worthwhile if it helps you shrink your algorithm for production deployment. But what about cases where you need to simply deploy to prototype hardware, or where the accuracy requirements make it difficult to reduce the physical footprint? A solution in these cases is to use single-precision Native Floating-Point.\n\nSimplifying the Process with Native Floating Point\n\nUsing Native Floating-Point has two benefits as follows:\n\n• You don't have to spend time trying to analyze the minimum number of bits needed to maintain sufficient precision for a wide variety of input data.\n• The dynamic range of single-precision floating-point operations scales much more efficiently with a fixed cost of 32 bits.\n\nNow, the design process is much simpler, and you know that with the bits of sign, exponent, and mantissa, you can represent a wide dynamic range of numbers. The table in Figure 6 compares the resource utilization of the floating-point and the fixed-point implementations of the IIR filter using the data type choices shown in Figure 5.",
null,
"Figure 6. Resource usage comparison between the fixed-point and floating-point implementations of the IIR filter (© 1984–2018 The MathWorks, Inc.)\n\nWhen you compare the results obtained from the floating-point and fixed-point implementations, remember that floating-point calculations require more operations than simple fixed-point arithmetic. Using single-precision will result in higher physical resource usage when you deploy to an FPGA or ASIC. If circuit area is a concern, then you will need to trade off higher precision and resource usage. You can also use a combination of floating-point and fixed-point to reduce area while preserving single-precision to achieve high dynamic range in numerically intensive computation islands.\n\n>> Continue reading this article on our sister site, EEWeb: \"Targeting Floating-Point algorithms for FPGA or ASIC deployment.\"\n\n• 08.08.2018"
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http://washingtonart.net/whealton/PFCN.html | [
"Steve Whealton\n\nPrime Factor Configuration Pattern Numbers\n\nn\n\npfcpn(n)\n\nThe fundamental theorem of arithmetic insists that each positive whole number can uniquely be expressed as the product of powers of prime numbers.\n\nWe will use this fundamental theorem to devise a partitioning of all natural numbers into disjoint sets, each set being identifiable by the pattern of the powers of its prime factors.\n\nEach pattern, and thus also each set, will be given a unique numberóthe \"Prime Factor Configuration Pattern Number\" (pfcpn) for that pattern, and it is this pfcpn that will ultimately be used in making visual or musical patterns.\n\nNitty Gritty\n\nWe begin by choosing a few fairly diverse numbers that will be useful as examples:\n\n 2364 873 2222 149 2000 6561 210 12 18 30 60 105\n\nNext, those same numbers are shown along with their prime factors. Each factorization is given in ascending order for the component prime numbers, from left to right.\n\nExponents will be indicated by using the \"^\" symbol. For example, \"2^3\" will be used to signify \"Two cubed,\" or \"Two to the third power.\"\n\n 2364 = 2^2 * 3^1 * 197^1 873 = 3^2 * 97^1 2222 = 2^1 * 11^1 * 101^1 149 = 149^1 2000 = 2^4 * 5^3 6561 = 3^12 210 = 2^1 * 3^1 * 5^1 * 7^1 12 = 2^2 * 3^1 18 = 2*1 * 3^2 30 = 2^1 * 3^1 * 5^1 60 = 2^2 * 3^1 * 5^1 105 = 3^1 * 5^1 * 7^1\n\nOur task is to group the natural numbers together according to the pattern of exponents that appears when numbers are broken down into their prime factors in the manner shown above.\n\nBelow is another chart made from the same numbers. This time, the prime factors are left out, and only the exponents are shown:\n\n 2364 2 1 1 873 2 1 2222 1 1 1 149 1 2000 4 3 6561 12 210 1 1 1 1 12 2 1 18 1 2 30 1 1 1 60 2 1 1 105 1 1 1\n\nThe idea of leaving out the prime factors and focusing only on their exponents is the key idea in this endeavor.\n\nSeveral of our chosen numbers can now be seen to have the same pattern of exponents:\n\n 2364 60 = 2 1 1 873 12 = 2 1 2222 30 105 = 1 1 1\n\nSo 60 and 2364 will end up in the same subset. 12 and 873 will go together into another subset, and 30, 105, and 2222 will belong to yet a third subset.\n\nFrom Patterns to Numbers\n\nIt remains now only to decide how best to label, or enumerate, each subset, and to begin partitioning.\n\nWhat we need is a chart that gives a unique number to each distinct\n\nThe first step is to create a chart that gives a unique number to each distinct prime factor configuration pattern. The pattern takes account of all of the distinct prime factors that are present. The pattern also registers to what powers those prime factors are raised. They appear in order from the smallest to the largest. I call the number given to each pattern a \"prime factor configuration pattern number\" (pfcpn).\n\nFirst, zero and one are unique. Neither is a prime, and neither is a composite number, either. So each of them must be categorized uniquely. The most direct way to do this is to give each of them a class of its own.\n\n pfcpn(0) = 0 pfcpn(1) = 1\n\nThe next number, two, is a prime. The next pfcpn available is also two. So two will be the pfcpn for all primes.\n\n pfcpn(2) = 2 pfcpn(3) = 2 pfcpn(5) = 2 pfcpn(7) = 2 pfcpn(11) = 2 pfcpn(13) = 2 pfcpn(17) = 2 pfcpn(19) = 2 pfcpn(23) = 2 pfcpn(29) = 2\n\nÖ and so on.\n\nThe smallest number not covered above is 4, the smallest square of a prime. The next available pfcpn available is 3. So all squares of primes will belong to the subset designated by the number 3:\n\n pfcpn(4) = 3 pfcpn(9) = 3 pfcpn(25) = 3 pfcpn(49) = 3 pfcpn(121) = 3 pfcpn(169) = 3 pfcpn(289) = 3 pfcpn(361) = 3 pfcpn(529) = 3 pfcpn(841) = 3\n\nÖ etc.\n\nEach new pattern encountered during this ascent through the integers is given the next available number:\n\n pfcpn(0) = 0 pfcpn(1) = 1 pfcpn(2) = 2 pfcpn(4) = 3 pfcpn(6) = 4 pfcpn(8) = 5 pfcpn(12) = 6 pfcpn(16) = 7 pfcpn(18) = 8 pfcpn(24) = 9\n\nIn the chart above, the first 10 pfcpns are indicated, along with the smallest number that exhibits each pattern. These \"smallest\" numbers range from 0, for pfcpn(0), up to 9, for pfcpn(24).\n\nNotice that although both 12 and 18 are the product of one prime squared and another prime present only to the first power, the pfcpn system distinguishes between the two and gives them different numbers.\n\nThe reasoning behind this lies in the fact that with 12, the squared prime (2) is smaller than the prime that is not squared (3), whereas with 18, the prime that is squared (3) is larger than the prime that is not squared (2).\n\nAnother way of creating numbers out of prime factor configuration patterns would be to treat the two classes (as represented by 12 and 18) the same. Such a system would feature fewer distinct patterns. The difference between the two systems is very close to the difference between permutations and combinations.\n\nThe chart shown running down the left-hand side of this page shows the first 129 integers, from 0 up to 128, along with their prime factor configuration pattern numbers.\n\nTo continue the list above 128, one must keep track of each new pattern that is encountered, giving each one the next available pfcpn.\n\nKeeping Track on the Fly\n\nA mnemonic device is useful in working with larger numbers and their prime factor configuration patterns and numbers. The idea is to associate each configuration not so much with its pfcpn itself, but rather with the smallest number that exhibits that pfcpn. This is what the chart immediately above does.\n\nWhen determining the pfcpn for, say, 2364, the first step is to factor 2364. From that factorization, the PFCP exponent pattern 2 1 1 is arrived at by then putting the prime factors, along with their exponents, in numerical order. Then the prime factors themselves are ignored, and only the exponents are considered.\n\nThe next step is to take the 2 1 1 pattern and applying it to the smallest possible set of distinct prime factors; in this case, to 2, 3, and 5. Applying the given pattern to the smallest primes will always result in the smallest number that shows the pfcpn in question:\n\n 2^2 * 3^1 * 5^1 = 60\n\nThis tells us that 60 is the smallest number exhibiting the 2 1 1 pfcpn.\n\nThe above step, with practice, becomes easy to perform mentally. Before long, the 2 1 1 pattern evokes 60 almost automatically.\n\nNow, you need only look up 60 in the PFCP table that you have already created. There, you will find that the pfcpn for 60 is 15. This tells you that the pfcpn for any number (such as 2364) that exhibits the 2 1 1 pattern will also be 15.\n\nAll this may seem like a lot of extra work. But the happy fact is that when you are sifting through large numbers of numbers and examining their prime factor configuration patterns, one after another, this method works. Associating a number like 2364 with 60, once you have figured out that 2 1 1 is the prime factor configuration pattern for 2364, eventually becomes rather easy; at least I find it so.\n\nWhy Bother?\n\nSo what good are these partitions and these index numbers?\n\nMany times, in making visual or musical patterns by mathematical means, it will be necessary to squeeze a huge range of numbers down into a much smaller range. (See squeezing for more on this topic.)\n\nColor index numbers typically lie in the ranges from 0 up to 15 or up to 255. In MIDI, the available pitches are numbered from 0 up to 127.\n\nAlthough pfcpn(22) is encountered at 128, one can speculate that pfcpn(255) will not be met with until a very large number of numbers have been analyzed and their pfcpns assigned.\n\nThe process can indeed continue indefinitely. Creating a master algorithm to figure out pfcpns will require more thinking time and programming time than I have been able to devote to the task, so far.\n\nFor the time being, I keep a pfcpn chart in array form. Whenever I feel like doing some menial labor, I get out my reference books and use them to help me figure out a few more pfcpns, thus slowly enlarging my master list. Then whenever I feel like working with pfcpns, I begin with the array and build images or musical patterns around it.\n\nUnlike the case with simple banding using primes, the visual result of using pfcpns is but subtly different from the visual result of using any of the other systems of disguising and squeezing numbers.\n\nBut once you have enjoyed and coped and struggled with as many visaual and musical patterns as I have, a reliable difference, even a subtle one, will come be to be treasured.\n\n1\n\n1\n\n2\n\n2\n\n3\n\n2\n\n4\n\n3\n\n5\n\n2\n\n6\n\n4\n\n7\n\n2\n\n8\n\n5\n\n9\n\n3\n\n10\n\n4\n\n11\n\n2\n\n12\n\n6\n\n13\n\n2\n\n14\n\n4\n\n15\n\n4\n\n16\n\n7\n\n17\n\n2\n\n18\n\n8\n\n19\n\n2\n\n20\n\n6\n\n21\n\n4\n\n22\n\n4\n\n23\n\n2\n\n24\n\n9\n\n25\n\n3\n\n26\n\n4\n\n27\n\n5\n\n28\n\n6\n\n29\n\n2\n\n30\n\n10\n\n31\n\n2\n\n32\n\n11\n\n33\n\n4\n\n34\n\n4\n\n35\n\n4\n\n36\n\n12\n\n37\n\n2\n\n38\n\n4\n\n39\n\n4\n\n40\n\n9\n\n41\n\n2\n\n42\n\n10\n\n43\n\n2\n\n44\n\n6\n\n45\n\n8\n\n46\n\n4\n\n47\n\n2\n\n48\n\n13\n\n49\n\n3\n\n50\n\n8\n\n51\n\n4\n\n52\n\n6\n\n53\n\n3\n\n54\n\n14\n\n55\n\n4\n\n56\n\n9\n\n57\n\n4\n\n58\n\n4\n\n59\n\n2\n\n60\n\n15\n\n61\n\n2\n\n62\n\n4\n\n63\n\n6\n\n64\n\n16\n\n65\n\n4\n\n66\n\n10\n\n67\n\n2\n\n68\n\n6\n\n69\n\n4\n\n70\n\n10\n\n71\n\n2\n\n72\n\n17\n\n73\n\n2\n\n74\n\n4\n\n75\n\n8\n\n76\n\n6\n\n77\n\n4\n\n78\n\n10\n\n79\n\n2\n\n80\n\n13\n\n81\n\n7\n\n82\n\n4\n\n83\n\n2\n\n84\n\n15\n\n85\n\n4\n\n86\n\n4\n\n87\n\n4\n\n88\n\n9\n\n89\n\n2\n\n90\n\n18\n\n91\n\n4\n\n92\n\n6\n\n93\n\n4\n\n94\n\n4\n\n95\n\n4\n\n96\n\n19\n\n97\n\n2\n\n98\n\n8\n\n99\n\n6\n\n100\n\n12\n\n101\n\n2\n\n102\n\n10\n\n103\n\n2\n\n104\n\n9\n\n105\n\n10\n\n106\n\n4\n\n107\n\n2\n\n108\n\n20\n\n109\n\n2\n\n110\n\n10\n\n111\n\n4\n\n112\n\n13\n\n113\n\n2\n\n114\n\n10\n\n115\n\n4\n\n116\n\n6\n\n117\n\n6\n\n118\n\n14\n\n119\n\n4\n\n120\n\n21\n\n121\n\n3\n\n122\n\n4\n\n123\n\n4\n\n124\n\n6\n\n125\n\n5\n\n126\n\n18\n\n127\n\n2\n\n128\n\n22"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9106855,"math_prob":0.9797174,"size":8704,"snap":"2019-51-2020-05","text_gpt3_token_len":2454,"char_repetition_ratio":0.1637931,"word_repetition_ratio":0.016829534,"special_character_ratio":0.3282399,"punctuation_ratio":0.08342923,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98215,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T21:21:47Z\",\"WARC-Record-ID\":\"<urn:uuid:c420a96b-4315-4207-88ff-3f4c4b8e08de>\",\"Content-Length\":\"74516\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1f452b0-dfd9-4c01-b2b5-684aceadf10d>\",\"WARC-Concurrent-To\":\"<urn:uuid:5375e783-32cc-457c-bd1c-34dcca9e8d8d>\",\"WARC-IP-Address\":\"67.210.126.135\",\"WARC-Target-URI\":\"http://washingtonart.net/whealton/PFCN.html\",\"WARC-Payload-Digest\":\"sha1:BJGTSYQTBAZWYL45CKZ7CJJXITHGDFAM\",\"WARC-Block-Digest\":\"sha1:XE2HVWTM73ACAYRVXQBDUIVY6SLKD6ZO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250591234.15_warc_CC-MAIN-20200117205732-20200117233732-00075.warc.gz\"}"} |
https://numbermatics.com/n/2304/ | [
"# 2304\n\n## 2,304 is an even composite number composed of two prime numbers multiplied together.\n\nWhat does the number 2304 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 27 divisors.\n\n2304 is an even composite number. It is composed of two distinct prime numbers multiplied together. It has a total of twenty-seven divisors.\n\n## Prime factorization of 2304:\n\n### 28 × 32\n\n(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3)\n\nSee below for interesting mathematical facts about the number 2304 from the Numbermatics database.\n\n### Names of 2304\n\n• Cardinal: 2304 can be written as Two thousand, three hundred four.\n\n### Scientific notation\n\n• Scientific notation: 2.304 × 103\n\n### Factors of 2304\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 10\n• Sum of prime factors: 5\n\n### Divisors of 2304\n\n• Number of divisors d(n): 27\n• Complete list of divisors:\n• Sum of all divisors σ(n): 6643\n• Sum of proper divisors (its aliquot sum) s(n): 4339\n• 2304 is an abundant number, because the sum of its proper divisors (4339) is greater than itself. Its abundance is 2035\n\n### Bases of 2304\n\n• Binary: 1001000000002\n• Base-36: 1S0\n\n### Squares and roots of 2304\n\n• 2304 squared (23042) is 5308416\n• 2304 cubed (23043) is 12230590464\n• 2304 is a perfect square number. Its square root is 48\n• The cube root of 2304 is 13.2077089955\n\n### Scales and comparisons\n\nHow big is 2304?\n• 2,304 seconds is equal to 38 minutes, 24 seconds.\n• To count from 1 to 2,304 would take you about thirty-eight minutes.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 2304 cubic inches would be around 1.1 feet tall.\n\n### Recreational maths with 2304\n\n• 2304 backwards is 4032\n• 2304 is a Harshad number.\n• The number of decimal digits it has is: 4\n• The sum of 2304's digits is 9\n• More coming soon!"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8577887,"math_prob":0.9854495,"size":2665,"snap":"2020-34-2020-40","text_gpt3_token_len":718,"char_repetition_ratio":0.11762495,"word_repetition_ratio":0.04954955,"special_character_ratio":0.30806753,"punctuation_ratio":0.16129032,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99456936,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T12:29:07Z\",\"WARC-Record-ID\":\"<urn:uuid:a7a79fb0-7c6b-4961-bba1-0f8e8355d2a3>\",\"Content-Length\":\"18342\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e655899-a46f-4310-ac22-1fba05ef0322>\",\"WARC-Concurrent-To\":\"<urn:uuid:410bba7d-e1b3-44f0-89d0-4786da3e8edb>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/2304/\",\"WARC-Payload-Digest\":\"sha1:ZZO6O4KIDP3GWD6H55O36NJNZAXC6PUX\",\"WARC-Block-Digest\":\"sha1:FRQGVPPENUCEO6EGWEDUSBZCJ3ENQZRN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400279782.77_warc_CC-MAIN-20200927121105-20200927151105-00575.warc.gz\"}"} |
https://www.ibm.com/developerworks/cn/opensource/os-objorient/ | [
"使用 PHP 创建图形的巧妙方法\n\nPHP 内置的制图基本操作与绘图程序非常类似。它们对于绘制图像来说功能非常强大；但是如果您希望自己的图像是一组对象集合时，这就不太适合了。本文将向您展示如何在 PHP 图形库的基础上构建一个面向对象的图形库。您将使用 PHP V5 中提供的面向对象的扩展。\n\n基础知识\n\n图 1. 图形环境和图形对象接口",
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"GraphicsEnvironment 类中保存了图形对象和一组颜色，还包括宽度和高度。saveAsPng 方法负责将当前的图像输出到指定的文件中。\n\nGraphicsObject 是任何图形对象都必须使用的接口。要开始使用这个接口，您所需要做的就是使用 render 方法来画这个对象。它是由一个 Line 类实现的，它利用 4 个坐标：开始和结束的 x 值，开始和结束的 y 值。它还有一个颜色。当调用 render 时，这个对象从 sx,sy 到 ex,ey 画一条由名字指定的颜色的线。\n\n清单 1. 基本的图形库\n<?php\nclass GraphicsEnvironment\n{\npublic \\$width;\npublic \\$height;\npublic \\$gdo;\npublic \\$colors = array();\npublic function __construct( \\$width, \\$height )\n{\n\\$this->width = \\$width;\n\\$this->height = \\$height;\n\\$this->gdo = imagecreatetruecolor( \\$width, \\$height );\n\\$this->addColor( \"white\", 255, 255, 255 );\nimagefilledrectangle( \\$this->gdo, 0, 0,\n\\$width, \\$height,\n\\$this->getColor( \"white\" ) );\n}\npublic function width() { return \\$this->width; }\npublic function height() { return \\$this->height; }\npublic function addColor( \\$name, \\$r, \\$g, \\$b )\n{\n\\$this->colors[ \\$name ] = imagecolorallocate(\n\\$this->gdo,\n\\$r, \\$g, \\$b );\n}\npublic function getGraphicObject()\n{\nreturn \\$this->gdo;\n}\npublic function getColor( \\$name )\n{\nreturn \\$this->colors[ \\$name ];\n}\npublic function saveAsPng( \\$filename )\n{\nimagepng( \\$this->gdo, \\$filename );\n}\n}\nabstract class GraphicsObject\n{\nabstract public function render( \\$ge );\n}\nclass Line extends GraphicsObject\n{\nprivate \\$color;\nprivate \\$sx;\nprivate \\$sy;\nprivate \\$ex;\nprivate \\$ey;\npublic function __construct( \\$color, \\$sx, \\$sy, \\$ex, \\$ey )\n{\n\\$this->color = \\$color;\n\\$this->sx = \\$sx;\n\\$this->sy = \\$sy;\n\\$this->ex = \\$ex;\n\\$this->ey = \\$ey;\n}\npublic function render( \\$ge )\n{\nimageline( \\$ge->getGraphicObject(),\n\\$this->sx, \\$this->sy,\n\\$this->ex, \\$this->ey,\n\\$ge->getColor( \\$this->color ) );\n}\n}\n?>\n\n清单 2. 基本图形库的测试代码\n<?php\nrequire_once( \"glib.php\" );\n\\$ge = new GraphicsEnvironment( 400, 400 );\n\\$ge->addColor( \"black\", 0, 0, 0 );\n\\$ge->addColor( \"red\", 255, 0, 0 );\n\\$ge->addColor( \"green\", 0, 255, 0 );\n\\$ge->addColor( \"blue\", 0, 0, 255 );\n\\$gobjs = array();\n\\$gobjs []= new Line( \"black\", 10, 5, 100, 200 );\n\\$gobjs []= new Line( \"blue\", 200, 150, 390, 380 );\n\\$gobjs []= new Line( \"red\", 60, 40, 10, 300 );\n\\$gobjs []= new Line( \"green\", 5, 390, 390, 10 );\nforeach( \\$gobjs as \\$gobj ) { \\$gobj->render( \\$ge ); }\n\\$ge->saveAsPng( \"test.png\" );\n?>\n\n% php test.php\n%\n\n图 2. 简单的图形对象测试",
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"",
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"添加维数\n\n图 3. 不同 z 值的面",
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"",
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"图 4. 给系统添加另外一维：z 值",
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"",
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"清单 3. 可以处理 z 信息的图形库\n<?php\nclass GraphicsEnvironment\n{\npublic \\$width;\npublic \\$height;\npublic \\$gdo;\npublic \\$colors = array();\npublic function __construct( \\$width, \\$height )\n{\n\\$this->width = \\$width;\n\\$this->height = \\$height;\n\\$this->gdo = imagecreatetruecolor( \\$width, \\$height );\n\\$this->addColor( \"white\", 255, 255, 255 );\nimagefilledrectangle( \\$this->gdo, 0, 0,\n\\$width, \\$height,\n\\$this->getColor( \"white\" ) );\n}\npublic function width() { return \\$this->width; }\npublic function height() { return \\$this->height; }\npublic function addColor( \\$name, \\$r, \\$g, \\$b )\n{\n\\$this->colors[ \\$name ] = imagecolorallocate(\n\\$this->gdo,\n\\$r, \\$g, \\$b );\n}\npublic function getGraphicObject()\n{\nreturn \\$this->gdo;\n}\npublic function getColor( \\$name )\n{\nreturn \\$this->colors[ \\$name ];\n}\npublic function saveAsPng( \\$filename )\n{\nimagepng( \\$this->gdo, \\$filename );\n}\n}\nabstract class GraphicsObject\n{\nabstract public function render( \\$ge );\nabstract public function z();\n}\nabstract class BoxObject extends GraphicsObject\n{\nprotected \\$color;\nprotected \\$sx;\nprotected \\$sy;\nprotected \\$ex;\nprotected \\$ey;\nprotected \\$z;\npublic function __construct( \\$z, \\$color, \\$sx, \\$sy, \\$ex, \\$ey )\n{\n\\$this->z = \\$z;\n\\$this->color = \\$color;\n\\$this->sx = \\$sx;\n\\$this->sy = \\$sy;\n\\$this->ex = \\$ex;\n\\$this->ey = \\$ey;\n}\npublic function z() { return \\$this->z; }\n}\nclass Line extends BoxObject\n{\npublic function render( \\$ge )\n{\nimageline( \\$ge->getGraphicObject(),\n\\$this->sx, \\$this->sy,\n\\$this->ex, \\$this->ey,\n\\$ge->getColor( \\$this->color ) );\n}\n}\nclass Rectangle extends BoxObject\n{\npublic function render( \\$ge )\n{\nimagefilledrectangle( \\$ge->getGraphicObject(),\n\\$this->sx, \\$this->sy,\n\\$this->ex, \\$this->ey,\n\\$ge->getColor( \\$this->color ) );\n}\n}\nclass Oval extends BoxObject\n{\npublic function render( \\$ge )\n{\n\\$w = \\$this->ex - \\$this->sx;\n\\$h = \\$this->ey - \\$this->sy;\nimagefilledellipse( \\$ge->getGraphicObject(),\n\\$this->sx + ( \\$w / 2 ),\n\\$this->sy + ( \\$h / 2 ),\n\\$w, \\$h,\n\\$ge->getColor( \\$this->color ) );\n}\n}\n?>\n\n清单 4. 更新后的测试代码\n<?php\nrequire_once( \"glib.php\" );\nfunction zsort( \\$a, \\$b )\n{\nif ( \\$a->z() < \\$b->z() ) return -1;\nif ( \\$a->z() > \\$b->z() ) return 1;\nreturn 0;\n}\n\\$ge = new GraphicsEnvironment( 400, 400 );\n\\$ge->addColor( \"black\", 0, 0, 0 );\n\\$ge->addColor( \"red\", 255, 0, 0 );\n\\$ge->addColor( \"green\", 0, 255, 0 );\n\\$ge->addColor( \"blue\", 0, 0, 255 );\n\\$gobjs = array();\n\\$gobjs []= new Oval( 100, \"red\", 50, 50, 150, 150 );\n\\$gobjs []= new Rectangle( 200, \"black\", 100, 100, 300, 300 );\nusort( \\$gobjs, \"zsort\" );\nforeach( \\$gobjs as \\$gobj ) { \\$gobj->render( \\$ge ); }\n\\$ge->saveAsPng( \"test.png\" );\n?>\n\n图 5. 红圆在黑方框之后",
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"",
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"\\$gobjs []= new Oval( 200, \"red\", 50, 50, 150, 150 );\n\\$gobjs []= new Rectangle( 100, \"black\", 100, 100, 300, 300 );\n\n图 6. 红圆现在在黑方框上面了",
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"",
null,
"Group 类的代码如清单 5 所示。\n\n清单 5. Group 类\nfunction zsort( \\$a, \\$b )\n{\nif ( \\$a->z() < \\$b->z() ) return -1;\nif ( \\$a->z() > \\$b->z() ) return 1;\nreturn 0;\n}\nclass Group extends GraphicsObject\n{\nprivate \\$z;\nprotected \\$members = array();\npublic function __construct( \\$z )\n{\n\\$this->z = \\$z;\n}\npublic function add( \\$member )\n{\n\\$this->members []= \\$member;\n}\npublic function render( \\$ge )\n{\nusort( \\$this->members, \"zsort\" );\nforeach( \\$this->members as \\$gobj )\n{\n\\$gobj->render( \\$ge );\n}\n}\npublic function z() { return \\$this->z; }\n}\n\nGroup 对象的任务是保持一个对象数组，然后在画图时，逐个对对象zo进行排序和画图。\n\n清单 6. 更新后的测试代码\n<?php\nrequire_once( \"glib.php\" );\n\\$ge = new GraphicsEnvironment( 400, 400 );\n\\$ge->addColor( \"black\", 0, 0, 0 );\n\\$ge->addColor( \"red\", 255, 0, 0 );\n\\$ge->addColor( \"green\", 0, 255, 0 );\n\\$ge->addColor( \"blue\", 0, 0, 255 );\n\\$g1 = new Group( 0 );\n\\$g1->add( new Oval( 200, \"red\", 50, 50, 150, 150 ) );\n\\$g1->add( new Rectangle( 100, \"black\", 100, 100, 300, 300 ) );\n\\$g1->render( \\$ge );\n\\$ge->saveAsPng( \"test.png\" );\n?>\n\n创建 viewport\n\nviewport 是一个人造的坐标系统，可以转换成图像的物理坐标系统。viewport 的扩展可以是您希望的任何东西。例如，x 和 y 轴的起点和终点可以是 -2 和 2，这样 viewport 坐标平面的中心就是 0, 0。这对于三角图形（例如 sin 和 cosine）来说是很好的一个 viewport。或者，这个 viewport 也可以是不对称的，其中 y 值的范围从 -1 到 1，x 值的范围是从 0 到 10,000，这取决于您的需要。\n\n图 7. 所添加的图形环境 viewport 转换",
null,
"",
null,
"清单 7. 具有 viewport 支持的图形库\n<?php\nclass GraphicsEnvironment\n{\npublic \\$width;\npublic \\$height;\npublic \\$gdo;\npublic \\$colors = array();\npublic function __construct( \\$width, \\$height )\n{\n\\$this->width = \\$width;\n\\$this->height = \\$height;\n\\$this->gdo = imagecreatetruecolor( \\$width, \\$height );\n\\$this->addColor( \"white\", 255, 255, 255 );\nimagefilledrectangle( \\$this->gdo, 0, 0,\n\\$width, \\$height,\n\\$this->getColor( \"white\" ) );\n}\npublic function width() { return \\$this->width; }\npublic function height() { return \\$this->height; }\npublic function addColor( \\$name, \\$r, \\$g, \\$b )\n{\n\\$this->colors[ \\$name ] = imagecolorallocate(\n\\$this->gdo,\n\\$r, \\$g, \\$b );\n}\npublic function getGraphicObject()\n{\nreturn \\$this->gdo;\n}\npublic function getColor( \\$name )\n{\nreturn \\$this->colors[ \\$name ];\n}\npublic function saveAsPng( \\$filename )\n{\nimagepng( \\$this->gdo, \\$filename );\n}\n\npublic function tx( \\$x )\n{\nreturn \\$x * \\$this->width;\n}\n\npublic function ty( \\$y )\n{\nreturn \\$y * \\$this->height;\n}\n}\nabstract class GraphicsObject\n{\nabstract public function render( \\$ge );\nabstract public function z();\n}\nfunction zsort( \\$a, \\$b )\n{\nif ( \\$a->z() < \\$b->z() ) return -1;\nif ( \\$a->z() > \\$b->z() ) return 1;\nreturn 0;\n}\nclass Group extends GraphicsObject\n{\nprivate \\$z;\nprotected \\$members = array();\npublic function __construct( \\$z )\n{\n\\$this->z = \\$z;\n}\npublic function add( \\$member )\n{\n\\$this->members []= \\$member;\n}\npublic function render( \\$ge )\n{\nusort( \\$this->members, \"zsort\" );\nforeach( \\$this->members as \\$gobj )\n{\n\\$gobj->render( \\$ge );\n}\n}\npublic function z() { return \\$this->z; }\n}\nabstract class BoxObject extends GraphicsObject\n{\nprotected \\$color;\nprotected \\$sx;\nprotected \\$sy;\nprotected \\$ex;\nprotected \\$ey;\nprotected \\$z;\npublic function __construct( \\$z, \\$color, \\$sx, \\$sy, \\$ex, \\$ey )\n{\n\\$this->z = \\$z;\n\\$this->color = \\$color;\n\\$this->sx = \\$sx;\n\\$this->sy = \\$sy;\n\\$this->ex = \\$ex;\n\\$this->ey = \\$ey;\n}\npublic function render( \\$ge )\n{\n\\$rsx = \\$ge->tx( \\$this->sx );\n\\$rsy = \\$ge->ty( \\$this->sy );\n\\$rex = \\$ge->tx( \\$this->ex );\n\\$rey = \\$ge->ty( \\$this->ey );\n\\$this->draw( \\$rsx, \\$rsy, \\$rex, \\$rey,\n\\$ge->getGraphicObject(),\n\\$ge->getColor( \\$this->color ) );\n}\nabstract public function draw( \\$sx, \\$sy,\n\\$ex, \\$ey, \\$gobj, \\$color );\npublic function z() { return \\$this->z; }\n}\nclass Line extends BoxObject\n{\npublic function draw( \\$sx, \\$sy, \\$ex, \\$ey,\n\\$gobj, \\$color )\n{\nimageline( \\$gobj, \\$sx, \\$sy, \\$ex, \\$ey,\n\\$color );\n}\n}\nclass Rectangle extends BoxObject\n{\npublic function draw( \\$sx, \\$sy, \\$ex, \\$ey,\n\\$gobj, \\$color )\n{\nimagefilledrectangle( \\$gobj, \\$sx, \\$sy,\n\\$ex, \\$ey, \\$color );\n}\n}\nclass Oval extends BoxObject\n{\npublic function draw( \\$sx, \\$sy, \\$ex, \\$ey,\n\\$gobj, \\$color )\n{\n\\$w = \\$ex - \\$sx;\n\\$h = \\$ey - \\$sy;\nimagefilledellipse( \\$gobj,\n\\$sx + ( \\$w / 2 ), \\$sy + ( \\$h / 2 ),\n\\$w, \\$h, \\$color );\n}\n}\n?>\n\nGraphicsEnvironment 类中的 viewport 转换代码是高亮显示的，正如 GraphicsObject 中的 render 代码一样，这会回调图形环境来进行坐标转换的工作。\n\n清单 8. 使用新 viewport 坐标的测试代码\n\\$g1 = new Group( 0 );\n\\$g1->add( new Oval( 200, \"red\", 0.1, 0.1, 0.5, 0.5 ) );\n\\$g1->add( new Rectangle( 100, \"black\", 0.4, 0.4, 0.9, 0.9 ) );\n\n图 8. 具有灵活 viewport 规范的图形环境",
null,
"清单 9. 更新后的 GraphicsEnvironment 代码\nclass GraphicsEnvironment\n{\npublic \\$vsx;\npublic \\$vsy;\npublic \\$vex;\npublic \\$vey;\npublic \\$width;\npublic \\$height;\npublic \\$gdo;\npublic \\$colors = array();\npublic function __construct( \\$width, \\$height,\n\\$vsx, \\$vsy, \\$vex, \\$vey )\n{\n\\$this->vsx = \\$vsx;\n\\$this->vsy = \\$vsy;\n\\$this->vex = \\$vex;\n\\$this->vey = \\$vey;\n\\$this->width = \\$width;\n\\$this->height = \\$height;\n\\$this->gdo = imagecreatetruecolor( \\$width, \\$height );\n\\$this->addColor( \"white\", 255, 255, 255 );\nimagefilledrectangle( \\$this->gdo, 0, 0,\n\\$width, \\$height,\n\\$this->getColor( \"white\" ) );\n}\npublic function width() { return \\$this->width; }\npublic function height() { return \\$this->height; }\npublic function addColor( \\$name, \\$r, \\$g, \\$b )\n{\n\\$this->colors[ \\$name ] = imagecolorallocate(\n\\$this->gdo,\n\\$r, \\$g, \\$b );\n}\npublic function getGraphicObject()\n{\nreturn \\$this->gdo;\n}\npublic function getColor( \\$name )\n{\nreturn \\$this->colors[ \\$name ];\n}\npublic function saveAsPng( \\$filename )\n{\nimagepng( \\$this->gdo, \\$filename );\n}\n\npublic function tx( \\$x )\n{\n\\$r = \\$this->width / ( \\$this->vex - \\$this->vsx );\nreturn ( \\$x - \\$this->vsx ) * \\$r;\n}\n\npublic function ty( \\$y )\n{\n\\$r = \\$this->height / ( \\$this->vey - \\$this->vsy );\nreturn ( \\$y - \\$this->vsy ) * \\$r;\n}\n}\n\n清单 10. viewport 测试代码\n<?php\nrequire_once( \"glib.php\" );\n\\$ge = new GraphicsEnvironment( 400, 400,\n-1000, -1000, 1000, 1000 );\n\\$ge->addColor( \"black\", 0, 0, 0 );\n\\$ge->addColor( \"red\", 255, 0, 0 );\n\\$ge->addColor( \"green\", 0, 255, 0 );\n\\$ge->addColor( \"blue\", 0, 0, 255 );\n\\$g1 = new Group( 0 );\n\\$g1->add( new Oval( 200, \"red\", -800, -800, 0, 0 ) );\n\\$g1->add( new Rectangle( 100, \"black\", -400, -400, 900, 900 ) );\n\\$g1->render( \\$ge );\n\\$ge->saveAsPng( \"test.png\" );\n?>\n\n图 9. viewport 绘制的图像转换为一个 400X400 的图像",
null,
"",
null,
"\\$ge = new GraphicsEnvironment( 400, 200,\n-1000, -1000, 1000, 1000 );\n\n图 10. 图形的 400X200 版本",
null,
"",
null,
"创建 viewport\n\n相关主题\n\n• 您可以参阅本文在 developerWorks 全球站点上的 英文原文\n• PHP.net 是学习有关 PHP 的最新知识的地方。\n• 访问 GD 图形库\n• James Foley 和 Andries van Dam 著的 Computer Graphics: Principles and Practice in C（Addison-Wesley Professional，1995 年），是有关图形的一份很好的资源。\n• Adobe 的 Scalable Vector Graphics 是一个有趣的地方，可以查找有关图形方面的信息。\n• 要学习更多有关 PHP 的内容，请访问 developerWorks 上的 PHP 项目资源\n• 访问 developerWorks 开放源码专区，获得丰富的 how-to 信息、工具和项目更新，以帮助您用开放源码技术进行开发，并与 IBM 产品结合使用。\n• 使用 IBM 试用软件 改进您的下一个开放源码开发项目，可以通过下载或从 DVD 中获得这些软件。\n\n评论\n\nstatic.content.url=http://www.ibm.com/developerworks/js/artrating/\nSITE_ID=10\nZone=Open source, Web development\nArticleID=101587\nArticleTitle=使用 PHP 创建图形的巧妙方法\npublish-date=01042006"
]
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure1.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure2.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure2.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure4.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure6.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure7.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure7.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure8.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure9.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure9.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure10.gif",
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"https://www.ibm.com/developerworks/cn/opensource/os-objorient/figure10.gif",
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.53856444,"math_prob":0.82400763,"size":14690,"snap":"2019-43-2019-47","text_gpt3_token_len":7138,"char_repetition_ratio":0.21088111,"word_repetition_ratio":0.5658466,"special_character_ratio":0.3526889,"punctuation_ratio":0.21921569,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97424406,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,1,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,1,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T14:34:52Z\",\"WARC-Record-ID\":\"<urn:uuid:db7bd020-8557-49fd-bb1d-45143c887483>\",\"Content-Length\":\"60586\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:999da587-fac4-4e2e-84b7-6ca24aa02b6f>\",\"WARC-Concurrent-To\":\"<urn:uuid:b1635adb-3a2f-4964-9693-ac520f2f085d>\",\"WARC-IP-Address\":\"104.118.202.217\",\"WARC-Target-URI\":\"https://www.ibm.com/developerworks/cn/opensource/os-objorient/\",\"WARC-Payload-Digest\":\"sha1:R7V4QNUHZNXHBMVI2KNIOBDJQ26BM7NP\",\"WARC-Block-Digest\":\"sha1:63OYHD7DD7RVEEVLNLFMOU5WB2CSL7RQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987822098.86_warc_CC-MAIN-20191022132135-20191022155635-00350.warc.gz\"}"} |
https://www.andreaperlato.com/theorypost/introduction-to-naive-bayes/ | [
"# Introduction to Naive Bayes\n\nIt is a Probability Classifier. Naïve Bayes is the first algorithm that should be considered for solving Text Classification Problem which involves High Dimensional training Dataset. A few examples are: Sentiment Analysis and Classifying Topics on Social Media.\nIt also refers to the Bayes’ Theorem also known as Bayes’ Law that give us a method to calculate the Conditional Probability: that is the probability of an event, based on previous knowledge available on the events.\n\nConsider for example to have two features: Salary and Age. Some Walks to go at work and other Drive at work.\n\n$P(W a l k s | X)=\\frac{P(X | \\text { Walks }) * P(\\text { Walks })}{P(X)}$\n\nThe X represent the features (Salary or Age) of a specific person.\n\nWhat is the probability that the person goes to work on foot based on X (Salary and Age)? P(Walks) = Prior probability, P(X) = Marginal likelihood, P(X|Walks) = Likelihood. The result of this calculation is P(Walks|X) = Posterior probability. Now, the same have to be made for Drive.",
null,
"Finally, we have to compare P(Walks|X) vs. P(Drives|X), and from there we can decide in which class we have to put our new observatioin.\n\nPrior Probability\nIn our example is the Probability that somebody Walks at work without knowing about hisAge or Salary. Looking at the figure below, we have to calculate the number of red observations and devide it by the overall number.",
null,
"Marginal Likelihood\nWe select a circle around our point (person) that we have to estimate. Then, we calculate the probability of Similar Features (walks, drives) that this person has with the other observations (person). So, Marginal Likelihood says to us what is the likelihood of any new random variable that we have fully inside the circle that we defined.",
null,
"Likelihood\nIt answer the following question: what is the probability of a randomly selected observations to be similar to the observations that we selected by the circle? In our example, this means that we take into consideration only the red points (i.e. people that work at work). We are asking what is the likelihood that a randomly selected data point is someone that exhibit features similar to the people selected in the random circle.",
null,
"Naive Bayes in Machine Learning\nIn a Machine Learning classification problem where there are multiple features and classes, the aim of the Naïve Bayes is to calculate the conditional probability of an object with the feature vector x1, x2, xn. So, it is able to calculate the probability of a particular class Ci given N number of features.\nNaive Bayes is an effective and commonly-used, machine learning classifier. It is a probabilistic classifier that makes classifications using the Maximum A Posteriori decision rule in a Bayesian setting. It can also be represented using a very simple Bayesian network. Naive Bayes classifiers have been especially popular for text classification, and are a traditional solution for problems such as Spam Detection.\n\nAn intuitive explanation for the Maximum A Posteriori Probability MAP is to think probabilities as degrees of belief. For example, how likely are we vote for a candidate depends on our prior belief. We can modify our stand based on the evidence. Our final decision, based on evidence, is the posterior belief, which is what happens after we sifted through the evidence. MPA is simply the maximum posterior belief: after going through all the debates, what is your most likely decision?"
]
| [
null,
"https://www.andreaperlato.com/img/posteriorbayes.png",
null,
"https://www.andreaperlato.com/img/priorprobability.png",
null,
"https://www.andreaperlato.com/img/marginallikelihood.png",
null,
"https://www.andreaperlato.com/img/likeliprob.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9436138,"math_prob":0.96948105,"size":3436,"snap":"2021-43-2021-49","text_gpt3_token_len":732,"char_repetition_ratio":0.13111888,"word_repetition_ratio":0.0,"special_character_ratio":0.20256111,"punctuation_ratio":0.09546166,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9951298,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T22:42:51Z\",\"WARC-Record-ID\":\"<urn:uuid:ea2eacf9-7011-41cb-9e8a-23cdce20299c>\",\"Content-Length\":\"7197\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0f8fb61f-e803-40a4-af44-3aed80c1d201>\",\"WARC-Concurrent-To\":\"<urn:uuid:70247e37-d16d-44d0-869a-bc612ed1aadf>\",\"WARC-IP-Address\":\"52.73.153.209\",\"WARC-Target-URI\":\"https://www.andreaperlato.com/theorypost/introduction-to-naive-bayes/\",\"WARC-Payload-Digest\":\"sha1:ZZ4VULGM4M6VXIZZTEU5H47JSC4QW66S\",\"WARC-Block-Digest\":\"sha1:4PBFYST7WEG6BKFCRGYNAP6ZNMPOSMJC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585183.47_warc_CC-MAIN-20211017210244-20211018000244-00427.warc.gz\"}"} |
https://math.stackexchange.com/questions/87732/different-ways-of-treating-vector-calculus | [
"Different ways of treating vector calculus?\n\nI learned that there are different ways of treating vector calculus: vector fields and differential forms, if I understand correctly. The former is used in calculus, and the latter is in differential geometry. My memory of calculus is vague on the vector calculus part and I have limit knowledge about differential geometry, so I was wondering how these two ways are different and related in a brief? There is no need to refrain from using some terminology in reply, and I can look them up if I am not familiar.\n\nA side question: is multivariate calculus synonym of vector calculus?\n\nThanks for your input!\n\nPS: The book description of Differential Forms: Integration on Manifolds and Stokes's Theorem by Steven H. Weintraub is from which I learned the above:\n\nBook Description:\n\nThis text is one of the first to treat vector calculus using differential forms in place of vector fields and other outdated techniques. Geared towards students taking courses in multivariable calculus, this innovative book aims to make the subject more readily understandable. Differential forms unify and simplify the subject of multivariable calculus, and students who learn the subject as it is presented in this book should come away with a better conceptual understanding of it than those who learn using conventional methods.\n\n• Treats vector calculus using differential forms\n• Presents a very concrete introduction to differential forms\n• Develops Stokess theorem in an easily understandable way\n• Gives well-supported, carefully stated, and thoroughly explained definitions and theorems.\n• Provides glimpses of further topics to entice the interested student\n• MV Calculus and \"Vector Calculus\" generally both mean the same thing: Analysis performed in $\\mathbb{R}^n$. In either case, differential forms and the associated abstract machinery (manifolds, exterior algebra, etc) is a (vast?) generalization of \"Vector Calculus\". The results in vector calculus can be obtained by using $\\mathbb{R}^3$ as the manifold and there is a very famous set of isomorphisms that relate the exterior derivative to the classical vector calculus operations DIV/GRAD/CURL. See the first pages of Bott and Tu's Differential Forms in Algebraic topology for details on this. – ItsNotObvious Dec 2 '11 at 14:19\n\nTrying to put it simply, I'll look at n - dimensional Euclidean space first. Then, differential forms (1 forms to be more precise) are dual to vector fields, that is, a 1 - form $\\omega$ is nothing but a linear function on the space of tangent vectors in p.\nFor differential forms the notion of exterior product or wedge product is rather important, n-forms in n-dimensional space corrsponding to a multiple of the determinant function $\\det$. This makes them volume forms and gives rise to a an integral (as are one forms when restricted to the tangent space of curves) for which particularly nice theorems (like Stokes theorem) are rather elegantly to formulate. k - forms are then natural candidates for volume form on k-dimensional submanifolds"
]
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https://www.tutorialtpoint.net/2023/03/numpy-array-slicing-accessing-array.html | [
"# NumPy Array Slicing - Accessing array elements using index python\n\nNumPy arrays can be sliced using the colon (:) operator, which allows you to extract a portion of the array. The slicing syntax is similar to that of Python lists, but with additional features for multidimensional arrays. Here are some examples of slicing NumPy arrays:\n\n1. We pass slice instead of index like this: `[start:end]`.\n2. We can also define the step, like this: `[start:end:step]`.\n3. If we don't pass start its considered 0\n4. If we don't pass end its considered length of array in that dimension\n5. If we don't pass step its considered 1\n1. Slicing a 1-D array:\n\nYou can slice a 1-D array using the colon (:) operator. For example:\n\n2. Slicing a 2-D array:\n\nYou can slice a 2-D array using the colon (:) operator for each dimension. For example:\n\n3. Slicing with strides:\n\nYou can use the colon (:) operator with a third parameter, called stride, to skip elements in the array. For example:\n\n4. Assigning a sliced array:\n\nYou can assign a sliced array to a new value. For example:\n\nMore\n\nMore\n\nMFCS\nCOA\nPL-CG\nDBMS\nOPERATING SYSTEM\nSOFTWARE ENG\nDSA\nTOC-CD\nARTIFICIAL INT\n\nMore\n\nMore\n\nMore\n\nMore\nTop"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8591009,"math_prob":0.76722616,"size":956,"snap":"2023-40-2023-50","text_gpt3_token_len":222,"char_repetition_ratio":0.12920168,"word_repetition_ratio":0.029069768,"special_character_ratio":0.24267782,"punctuation_ratio":0.14925373,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9653025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T14:19:10Z\",\"WARC-Record-ID\":\"<urn:uuid:9297f981-a7fa-4b01-81e4-d56a7193bbc9>\",\"Content-Length\":\"191888\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c2c9b22-9a61-449f-a7cc-8bc47f2a9493>\",\"WARC-Concurrent-To\":\"<urn:uuid:179c1153-eec9-4fde-90c3-1aba4b700459>\",\"WARC-IP-Address\":\"216.239.34.21\",\"WARC-Target-URI\":\"https://www.tutorialtpoint.net/2023/03/numpy-array-slicing-accessing-array.html\",\"WARC-Payload-Digest\":\"sha1:272YG6ZBB4ZT2FIFHSNO4T5UR2ILH7OV\",\"WARC-Block-Digest\":\"sha1:67MHXZZA24J25EV7WZBQ7EKSH4Y3DMLM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100427.59_warc_CC-MAIN-20231202140407-20231202170407-00502.warc.gz\"}"} |
http://www.kylesconverter.com/illuminance/lumens-per-square-inch-to-kilolux | [
"# Convert Lumens Per Square Inch to Kilolux\n\n### Kyle's Converter > Illuminance > Lumens Per Square Inch > Lumens Per Square Inch to Kilolux\n\n Lumens Per Square Inch (lm/in2) Kilolux (klx) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18\nReverse conversion?\nKilolux to Lumens Per Square Inch\n(or just enter a value in the \"to\" field)\n\nPlease share if you found this tool useful:\n\nUnit Descriptions\n1 Lumen per Square Inch:\n1 Lumen per square inch is equal to 1550.0031000062 Lux (SI unit). A measurement of the number of lumens falling on an area of one square inch. 1 lm/in2 = 1550.0031000062 lx.\n1 Kilolux:\n1 Kilolux is equal to 1000 Lux (SI unit). Lux measures the number of lumens within an area of one square meter. 1 klx = 1000 lx.\n\nConversions Table\n1 Lumens Per Square Inch to Kilolux = 1.5570 Lumens Per Square Inch to Kilolux = 108.5002\n2 Lumens Per Square Inch to Kilolux = 3.180 Lumens Per Square Inch to Kilolux = 124.0002\n3 Lumens Per Square Inch to Kilolux = 4.6590 Lumens Per Square Inch to Kilolux = 139.5003\n4 Lumens Per Square Inch to Kilolux = 6.2100 Lumens Per Square Inch to Kilolux = 155.0003\n5 Lumens Per Square Inch to Kilolux = 7.75200 Lumens Per Square Inch to Kilolux = 310.0006\n6 Lumens Per Square Inch to Kilolux = 9.3300 Lumens Per Square Inch to Kilolux = 465.0009\n7 Lumens Per Square Inch to Kilolux = 10.85400 Lumens Per Square Inch to Kilolux = 620.0012\n8 Lumens Per Square Inch to Kilolux = 12.4500 Lumens Per Square Inch to Kilolux = 775.0016\n9 Lumens Per Square Inch to Kilolux = 13.95600 Lumens Per Square Inch to Kilolux = 930.0019\n10 Lumens Per Square Inch to Kilolux = 15.5800 Lumens Per Square Inch to Kilolux = 1240.0025\n20 Lumens Per Square Inch to Kilolux = 31.0001900 Lumens Per Square Inch to Kilolux = 1395.0028\n30 Lumens Per Square Inch to Kilolux = 46.50011,000 Lumens Per Square Inch to Kilolux = 1550.0031\n40 Lumens Per Square Inch to Kilolux = 62.000110,000 Lumens Per Square Inch to Kilolux = 15500.031\n50 Lumens Per Square Inch to Kilolux = 77.5002100,000 Lumens Per Square Inch to Kilolux = 155000.31\n60 Lumens Per Square Inch to Kilolux = 93.00021,000,000 Lumens Per Square Inch to Kilolux = 1550003.1"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5878475,"math_prob":0.99888164,"size":1466,"snap":"2020-45-2020-50","text_gpt3_token_len":518,"char_repetition_ratio":0.38987687,"word_repetition_ratio":0.33582088,"special_character_ratio":0.41609824,"punctuation_ratio":0.11217949,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9922968,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T19:04:06Z\",\"WARC-Record-ID\":\"<urn:uuid:c5b1e880-1e85-4fae-b5d9-9f6a54b5d0d4>\",\"Content-Length\":\"19548\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f18c1e4-1597-411f-ae0e-67ec31f2d638>\",\"WARC-Concurrent-To\":\"<urn:uuid:c3fb27c5-bc30-4a90-8aa4-29e10dc06447>\",\"WARC-IP-Address\":\"99.84.185.102\",\"WARC-Target-URI\":\"http://www.kylesconverter.com/illuminance/lumens-per-square-inch-to-kilolux\",\"WARC-Payload-Digest\":\"sha1:73V23LXUGS7NK3QBRAIL5CZW3XLTDMER\",\"WARC-Block-Digest\":\"sha1:K5QWOV223BTJOH2YAG3R5XP746LUY3UN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141202590.44_warc_CC-MAIN-20201129184455-20201129214455-00476.warc.gz\"}"} |
https://lists.samba.org/archive/rsync/2005-September/013722.html | [
"# (no subject)\n\n=?GBK?q?=B2=FA=C6=B7=CD=E2=B9=DB=C9=E8=BC=C6=A1=A2=BD=E1=B9=B9=C9=E8=BC=C6?= linwu50 at 163.com\nThu Sep 15 01:15:38 GMT 2005\n\n```HTML attachment scrubbed and removed\n```\n\nMore information about the rsync mailing list"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5933142,"math_prob":0.84953195,"size":552,"snap":"2021-21-2021-25","text_gpt3_token_len":160,"char_repetition_ratio":0.13321167,"word_repetition_ratio":0.65,"special_character_ratio":0.3206522,"punctuation_ratio":0.13761468,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9980511,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T14:27:05Z\",\"WARC-Record-ID\":\"<urn:uuid:9716c519-e0f8-4541-8c84-6211871b5ee6>\",\"Content-Length\":\"2809\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f2cfb688-d55b-456f-b357-5a7b0f561284>\",\"WARC-Concurrent-To\":\"<urn:uuid:39d64b56-ce92-4b93-a3f1-3f8c63c8e6c3>\",\"WARC-IP-Address\":\"144.76.82.148\",\"WARC-Target-URI\":\"https://lists.samba.org/archive/rsync/2005-September/013722.html\",\"WARC-Payload-Digest\":\"sha1:V57FG4RHXUYFVNUHZFVE7SIGGOVSNELN\",\"WARC-Block-Digest\":\"sha1:UXMNJ3FOX4WSN3VXLI4NQJ3WMI5STDSL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487621273.31_warc_CC-MAIN-20210615114909-20210615144909-00255.warc.gz\"}"} |
https://jobscaptain.com/venn-diagram-questions-and-answers-pdf/ | [
"# Venn Diagram Questions and Answers PDF\n\nBy | November 22, 2022",
null,
"Venn diagrams are one of the most popular methods for visualizing relationships between sets of data. In this post, we will take a look at some of the most common Venn diagram questions and answers.\n\nWhen it comes to competitive exams, one of the most popular and important topics is Venn diagram problems and solutions. This is because Venn diagrams are not only used to test your mathematical ability but also your logical and analytical skills.\n\nIn this blog, we will be discussing everything you need to know about Venn diagrams, including what they are, how to use them and some important Venn diagram questions and answers PDF.\n\nContent Index\n\n## What is a Venn Diagram?\n\nA Venn diagram is a graphical representation of sets and their relationships. A Venn diagram consists of a number of overlapping circles, each representing a set.\n\nThe size of the circle corresponds to the cardinality of the set, that is, the number of elements in the set. The overlapping regions of the circles represent the relationships between the sets.\n\n## How to Use a Venn Diagram?\n\nVenn diagrams are used to represent relationships between sets of data. They can be used to represent simple relationships, such as “A is a subset of B”, or more complex relationships, such as “A intersects B but is not a subset of B”.\n\nVenn diagrams can also be used to solve problems. For example, given a set of data, you can use a Venn diagram to find the probability of certain events occurring.\n\nNow that we know what Venn diagrams are and how to use them, let’s take a look at some important Venn diagram questions and answers PDF.\n\nQuestion 1: What is the probability that an event will occur if it is known that the event is a subset of another event?\n\nAnswer: If the event is a subset of another event, then the probability of the event occurring is 1.\n\nQuestion 2: What is the probability that an event will occur if it is known that the event is not a subset of another event?\n\nAnswer: If the event is not a subset of another event, then the probability of the event occurring is 0.\n\nQuestion 3: What is the probability that an event will occur if it is known that the event intersects another event but is not a subset of the other event?\n\nAnswer: If the event intersects another event but is not a subset of the other event, then the probability of the event occurring is 1/2.\n\nQuestion 4: What is the probability that an event will occur if it is known that the event does not intersect another event?\n\nAnswer: If the event does not intersect another event, then the probability of the event occurring is 0."
]
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"https://jobscaptain.com/wp-content/uploads/2022/11/Venn-Diagram-Questions-and-Answers-PDF.png",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9344336,"math_prob":0.9421993,"size":3679,"snap":"2023-40-2023-50","text_gpt3_token_len":781,"char_repetition_ratio":0.18312925,"word_repetition_ratio":0.27429467,"special_character_ratio":0.19951074,"punctuation_ratio":0.080882356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981456,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T02:53:47Z\",\"WARC-Record-ID\":\"<urn:uuid:9d2b1f00-df8b-4fc3-a1ad-d38132bb5100>\",\"Content-Length\":\"61066\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04b14996-5423-4d09-8855-ef152c8ba849>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea23828c-ee4f-4eab-94df-75c79096852c>\",\"WARC-IP-Address\":\"104.21.82.48\",\"WARC-Target-URI\":\"https://jobscaptain.com/venn-diagram-questions-and-answers-pdf/\",\"WARC-Payload-Digest\":\"sha1:QN73ELAY3ZNHZIVQ2ONXE43IOSTSLITC\",\"WARC-Block-Digest\":\"sha1:GHMSK7FHD27JWP4DGRWJBX72ALNDR6HB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510481.79_warc_CC-MAIN-20230929022639-20230929052639-00241.warc.gz\"}"} |
https://iluvsaving.savingadvice.com/fun/ | [
"User Real IP - 3.228.11.9\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => 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Array\n(\n => 110.38.29.246\n)\n\n => Array\n(\n => 14.192.140.199\n)\n\n => Array\n(\n => 24.68.10.106\n)\n\n => Array\n(\n => 47.30.209.179\n)\n\n => Array\n(\n => 106.223.123.21\n)\n\n => Array\n(\n => 103.224.48.30\n)\n\n => Array\n(\n => 104.131.19.173\n)\n\n => Array\n(\n => 119.157.100.206\n)\n\n => Array\n(\n => 103.10.226.73\n)\n\n => Array\n(\n => 162.208.51.163\n)\n\n => Array\n(\n => 47.30.221.227\n)\n\n => Array\n(\n => 119.160.116.210\n)\n\n => Array\n(\n => 198.16.78.43\n)\n\n => Array\n(\n => 39.44.201.151\n)\n\n => Array\n(\n => 71.63.181.84\n)\n\n => Array\n(\n => 14.142.192.218\n)\n\n => Array\n(\n => 39.34.147.178\n)\n\n => Array\n(\n => 111.92.75.25\n)\n\n => Array\n(\n => 45.135.239.58\n)\n\n => Array\n(\n => 14.232.235.1\n)\n\n => Array\n(\n => 49.144.100.155\n)\n\n => Array\n(\n => 62.182.99.33\n)\n\n => Array\n(\n => 104.243.212.187\n)\n\n => Array\n(\n => 59.97.132.214\n)\n\n => Array\n(\n => 47.9.15.179\n)\n\n => Array\n(\n => 39.44.103.186\n)\n\n => Array\n(\n => 183.83.241.132\n)\n\n => Array\n(\n => 103.41.24.180\n)\n\n => Array\n(\n => 104.238.46.39\n)\n\n => Array\n(\n => 103.79.170.78\n)\n\n => Array\n(\n => 59.103.138.81\n)\n\n => Array\n(\n => 106.198.191.146\n)\n\n => Array\n(\n => 106.198.255.122\n)\n\n => Array\n(\n => 47.31.46.37\n)\n\n => Array\n(\n => 109.169.23.76\n)\n\n => Array\n(\n => 103.143.7.55\n)\n\n => Array\n(\n => 49.207.114.52\n)\n\n => Array\n(\n => 198.54.106.250\n)\n\n => Array\n(\n => 39.50.64.18\n)\n\n => Array\n(\n => 222.252.48.132\n)\n\n => Array\n(\n => 42.201.186.53\n)\n\n => Array\n(\n => 115.97.198.95\n)\n\n => Array\n(\n => 93.76.134.244\n)\n\n => Array\n(\n => 122.173.15.189\n)\n\n => Array\n(\n => 39.62.38.29\n)\n\n => Array\n(\n => 103.201.145.254\n)\n\n => Array\n(\n => 111.119.187.23\n)\n\n => Array\n(\n => 157.50.66.33\n)\n\n => Array\n(\n => 157.49.68.163\n)\n\n => Array\n(\n => 103.85.125.215\n)\n\n => Array\n(\n => 103.255.4.16\n)\n\n => Array\n(\n => 223.181.246.206\n)\n\n => Array\n(\n => 39.40.109.226\n)\n\n => Array\n(\n => 43.225.70.157\n)\n\n => Array\n(\n => 103.211.18.168\n)\n\n => Array\n(\n => 137.59.221.60\n)\n\n => Array\n(\n => 103.81.214.63\n)\n\n => Array\n(\n => 39.35.163.2\n)\n\n => Array\n(\n => 106.205.124.39\n)\n\n => Array\n(\n => 209.99.165.216\n)\n\n => Array\n(\n => 103.75.247.187\n)\n\n => Array\n(\n => 157.46.217.41\n)\n\n => Array\n(\n => 75.186.73.80\n)\n\n => Array\n(\n => 212.103.48.153\n)\n\n => Array\n(\n => 47.31.61.167\n)\n\n => Array\n(\n => 119.152.145.131\n)\n\n => Array\n(\n => 171.76.177.244\n)\n\n => Array\n(\n => 103.135.78.50\n)\n\n => Array\n(\n => 103.79.170.75\n)\n\n => Array\n(\n => 105.160.22.74\n)\n\n => Array\n(\n => 47.31.20.153\n)\n\n => Array\n(\n => 42.107.204.65\n)\n\n => Array\n(\n => 49.207.131.35\n)\n\n => Array\n(\n => 92.38.148.61\n)\n\n => Array\n(\n => 183.83.255.206\n)\n\n => Array\n(\n => 107.181.177.131\n)\n\n => Array\n(\n => 39.40.220.157\n)\n\n => Array\n(\n => 39.41.133.176\n)\n\n => Array\n(\n => 103.81.214.61\n)\n\n => Array\n(\n => 223.235.108.46\n)\n\n => Array\n(\n => 171.241.52.118\n)\n\n => Array\n(\n => 39.57.138.47\n)\n\n => Array\n(\n => 106.204.196.172\n)\n\n => Array\n(\n => 39.53.228.40\n)\n\n => Array\n(\n => 185.242.5.99\n)\n\n => Array\n(\n => 103.255.5.96\n)\n\n => Array\n(\n => 157.46.212.120\n)\n\n => Array\n(\n => 107.181.177.138\n)\n\n => Array\n(\n => 47.30.193.65\n)\n\n => Array\n(\n => 39.37.178.33\n)\n\n => Array\n(\n => 157.46.173.29\n)\n\n => Array\n(\n => 39.57.238.211\n)\n\n => Array\n(\n => 157.37.245.113\n)\n\n => Array\n(\n => 47.30.201.138\n)\n\n => Array\n(\n => 106.204.193.108\n)\n\n => Array\n(\n => 212.103.50.212\n)\n\n => Array\n(\n => 58.65.221.187\n)\n\n => Array\n(\n => 178.62.92.29\n)\n\n => Array\n(\n => 111.92.77.166\n)\n\n => Array\n(\n => 47.30.223.158\n)\n\n => Array\n(\n => 103.224.54.83\n)\n\n => Array\n(\n => 119.153.43.22\n)\n\n => Array\n(\n => 223.181.126.251\n)\n\n => Array\n(\n => 39.42.175.202\n)\n\n => Array\n(\n => 103.224.54.190\n)\n\n => Array\n(\n => 49.36.141.210\n)\n\n => Array\n(\n => 5.62.63.218\n)\n\n => Array\n(\n => 39.59.9.18\n)\n\n => Array\n(\n => 111.88.86.45\n)\n\n => Array\n(\n => 178.54.139.5\n)\n\n => Array\n(\n => 116.68.105.241\n)\n\n => Array\n(\n => 119.160.96.187\n)\n\n => Array\n(\n => 182.189.192.103\n)\n\n => Array\n(\n => 119.160.96.143\n)\n\n => Array\n(\n => 110.225.89.98\n)\n\n => Array\n(\n => 169.149.195.134\n)\n\n => Array\n(\n => 103.238.104.54\n)\n\n => Array\n(\n => 47.30.208.142\n)\n\n => Array\n(\n => 157.46.179.209\n)\n\n => Array\n(\n => 223.235.38.119\n)\n\n => Array\n(\n => 42.106.180.165\n)\n\n => Array\n(\n => 154.122.240.239\n)\n\n => Array\n(\n => 106.223.104.191\n)\n\n => Array\n(\n => 111.93.110.218\n)\n\n => Array\n(\n => 182.183.161.171\n)\n\n => Array\n(\n => 157.44.184.211\n)\n\n => Array\n(\n => 157.50.185.193\n)\n\n => Array\n(\n => 117.230.19.194\n)\n\n => Array\n(\n => 162.243.246.160\n)\n\n => Array\n(\n => 106.223.143.53\n)\n\n => Array\n(\n => 39.59.41.15\n)\n\n => Array\n(\n => 106.210.65.42\n)\n\n => Array\n(\n => 180.243.144.208\n)\n\n => Array\n(\n => 116.68.105.22\n)\n\n => Array\n(\n => 115.42.70.46\n)\n\n => Array\n(\n => 99.72.192.148\n)\n\n => Array\n(\n => 182.183.182.48\n)\n\n => Array\n(\n => 171.48.58.97\n)\n\n => Array\n(\n => 37.120.131.188\n)\n\n => Array\n(\n => 117.99.167.177\n)\n\n => Array\n(\n => 111.92.76.210\n)\n\n)\n```\nViewing the 'Fun' Category: MariRDH's Personal Finance Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Category: Fun",
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"# Viewing the 'Fun' Category\n\nMarch 27th, 2008 at 02:19 am\n\nI just noticed that I see a google ad under each post I click on specifically (not if I am looking at the whole blog, but if I click a specific post), and another ad at the top of this page where I am adding a new blog post.\n\nThey do not bother me. I am just wondering - Did these just start or have I not been paying attention? *blush*\n\n## Taxes & Refund Money Dilemma\n\nFebruary 2nd, 2008 at 04:27 am\n\nI have all of our W-2s and 1099s ready to go, but am waiting on a statement from my school (tells how much I paid last year in tuition/fees so I can get a credit), and the statements from SallieMae for both of our student loans (tells how much we paid in interest on our loans and lowers our AGI).\n\nWe will definitely be getting a refund this year so I'd love to get moving and get the taxes filed. I started putting the information we do have into TaxSlayer (free federal and state filing for us this year), and found out that I also have to wait until after February 8th for a certain form that the IRS has not yet released for 2007 (I think it is the form to report the school tuition/fees actually).",
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"I am anxious to receive the money so I can put my plans into action. Between refunds from federal and state taxes, we should have enough to reach \\$6,000 in the EF and have some leftover.\n\nThe leftover amount will probably be about \\$800. Here is where I have been second-guessing myself. I intended to put the money over and above the EF towards the \\$8,800 credit card that is at 0% until Feb 2009, but now I am wondering if I should put it toward a trip to DisneyWorld for the 3 of us.\n\nMy daughter has never been and she will be 10 years old this year. I am feeling sad that we have not been able to take her and, while I know she is still a little girl, she is getting older. We have alot of plans (pay off debt, buy a house, save for her Bat Mitzvah, etc), and if we don't do it this year I am afraid we will not be able to do it for another 4-5 years or more...and then she really will miss the magic of seeing Disney through the eyes of a child. :'(\n\nI guess I will talk it over with my husband and see what he thinks, but his eyes tend to glaze over whenever I try to crunch the numbers with him.\n\n## Nearly tripled my money!\n\nJanuary 12th, 2008 at 09:53 pm\n\nOkay so it was a very small amount, but I was giddy with excitement last night.",
null,
"I took my hubby to Atlantic City to see Don Rickles for his birthday. We had dinner and had a little time to waste before the show, so we put a few dollars in the slot machines. He played (and lost) \\$5 on video poker, then I put in \\$1 on a very strange nickle slot that I could not understand so I lost that. Oops. We both rambled over to another section of nickle slots and put \\$1 apiece into different machines. He bet his whole dollar on one spin and lost. I bet 5 credits (still have no idea how much that actually was...maybe 25 cents) and WON! It kept adding more and more credits until I had a final total of \\$20.65! Seven of the \\$8 we spent was mine, so I nearly tripled my money. I cashed out, took my money and ran.",
null,
"After that we went straight to the show and really laughed and enjoyed ourselves. It was a great night...and I wound up with \\$13.65 to add to my EF.\n\nI also received my check from ViewPoint Forums for \\$34 which will be deposited to my EF Wednesday (my next day off from work).\n\nNew CF Total: \\$1,325.89"
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https://de.maplesoft.com/support/help/maple/view.aspx?path=equation | [
"Equations and Inequalities, =, , , =, , = - Maple Programming Help\n\nHome : Support : Online Help : Programming : Logic : Relations : equation\n\nEquations and Inequalities, =, <>, <, <=, >, >=\n\nDescription\n\n • An equation is represented externally using the binary operator =. An expression which is an equation has two operands, the left-hand side and the right-hand side. The names = and equation are known to the type function.\n • There are three internal data types for inequalities, corresponding to the operators <>, <, and <=. Inequalities involving the operators > and >= are converted to the latter two cases for purposes of representation. An inequality has two operands, the left-hand side and the right-hand side. The names <>, <, <= are known to the type function.\n • Comparisons of numeric values are carried out in the corresponding numeric computation environment. For example, the test 3.141 < 3.142 is evaluated by subtraction in the floating-point environment determined by Digits. Hence, if Digits > 3, this returns true. If Digits <= 3, this test returns false.\n • These operators are viewed as relational operators in a Boolean context or by the evalb function. For more information, see boolean.\n\n • The equation and inequality operators are thread safe as of Maple 15.\n • The Equations and Inequalities, =, <>, <, <=, >, >= command is thread-safe as of Maple 15.\n\nExamples\n\n > $e≔a=b$\n ${e}{≔}{a}{=}{b}$ (1)\n > $\\mathrm{type}\\left(e,'\\mathrm{equation}'\\right)$\n ${\\mathrm{true}}$ (2)\n > $e≔f\\left(x\\right)\n ${e}{≔}{f}{}\\left({x}\\right){<}{g}{}\\left({x}\\right)$ (3)\n > $\\mathrm{type}\\left(e,'\\mathrm{equation}'\\right)$\n ${\\mathrm{false}}$ (4)\n > $\\mathrm{type}\\left(e,\\mathrm{<}\\right)$\n ${\\mathrm{true}}$ (5)\n > $\\mathrm{lhs}\\left(e\\right)$\n ${f}{}\\left({x}\\right)$ (6)\n > $\\mathrm{rhs}\\left(e\\right)$\n ${g}{}\\left({x}\\right)$ (7)\n > $\\mathrm{eqs}≔\\left\\{\\mathrm{a1}=\\mathrm{b1},\\mathrm{a2}=\\mathrm{b2},\\mathrm{a3}=\\mathrm{b3}\\right\\}$\n ${\\mathrm{eqs}}{≔}\\left\\{{\\mathrm{a1}}{=}{\\mathrm{b1}}{,}{\\mathrm{a2}}{=}{\\mathrm{b2}}{,}{\\mathrm{a3}}{=}{\\mathrm{b3}}\\right\\}$ (8)\n > $\\mathrm{map}\\left(\\mathrm{lhs},\\mathrm{eqs}\\right)$\n $\\left\\{{\\mathrm{a1}}{,}{\\mathrm{a2}}{,}{\\mathrm{a3}}\\right\\}$ (9)\n > $a=2$\n ${a}{=}{2}$ (10)\n > $a$\n ${a}$ (11)\n > $\\mathrm{assign}\\left(\\right)$\n > $a$\n ${2}$ (12)"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.88957405,"math_prob":1.0,"size":1359,"snap":"2019-35-2019-39","text_gpt3_token_len":343,"char_repetition_ratio":0.14760147,"word_repetition_ratio":0.2,"special_character_ratio":0.30684328,"punctuation_ratio":0.17826086,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998826,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-19T15:06:38Z\",\"WARC-Record-ID\":\"<urn:uuid:df1f3915-2fa2-428a-9267-88a7ec1a158f>\",\"Content-Length\":\"187138\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b0f6a5d7-d486-4170-95df-27c8897a8fc9>\",\"WARC-Concurrent-To\":\"<urn:uuid:156181e6-269c-492b-9202-7bac99cf3520>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://de.maplesoft.com/support/help/maple/view.aspx?path=equation\",\"WARC-Payload-Digest\":\"sha1:EBYG45W5I7RSRJFFQDLBBFFXYHPHHILI\",\"WARC-Block-Digest\":\"sha1:7NWWURAFKOM5NOZXVG7YOM2MBZH3Q4UK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573533.49_warc_CC-MAIN-20190919142838-20190919164838-00266.warc.gz\"}"} |
https://mathoverflow.net/questions/332135/characterization-of-operator-ranges/332155 | [
"# Characterization of operator ranges\n\nMy question is motivated by the following little proposition:\n\nProposition. For a vector subspace $$V$$ of a Banach space $$(X, \\|\\cdot\\|_X)$$ the following assertions are equivalent:\n\n(i) There exists a Banach space $$Z$$ and a bounded linear operator $$T: Z \\to X$$ with range $$V$$.\n\n(ii) There exists a complete norm $$\\|\\cdot\\|_V$$ on $$V$$ such that the canonical embedding of $$(V, \\|\\cdot\\|_V)$$ into $$(X,\\|\\cdot\\|_X)$$ is continuous.\n\n(See below for a proof.)\n\n$$\\,$$\n\nQuestion. Are (i) and (ii) also equivalent to the following assertion (iii)?\n\n(iii) There exists a bounded linear operator $$S: X \\to X$$ with range $$V$$.\n\n$$\\,$$\n\nProof of the Proposition. Obviously (ii) implies (i), so assume that (i) holds. Let $$\\tilde T: Z / \\ker T \\to X$$ denote the injective operator induced by $$T$$; then $$\\tilde T$$ also has range $$V$$. The inverse $${\\tilde T}^{-1}$$ is a closed linear operator $$X \\supseteq V \\to Z / \\ker T$$, so $$V$$ becomes a Banach space with respect to the graph norm given by $$\\|x\\|_V := \\|x\\|_X + \\|{\\tilde T}^{-1}x\\|_{Z / \\ker T}$$ for all $$x \\in V$$.\n\nRemark. For Hilbert spaces results of this type can, for instance, be found in the paper \"Fillmore and Williams: On Operator Ranges (1971)\". In fact, the above proof is an adaptation of an argument that appears in the proof of Theorem 1.1 of this paper.\n\n## 1 Answer\n\nThe answer to your question is \"No\". It can be seen in the following way: If there exists an operator $$S$$ mentioned in the Question, then, using the standard techniques, one can show that $$V$$ has to be isomorphic to a quotient space of $$X$$. So it remains to show that there exists $$X$$ and an operator range in $$X$$ for which this condition fails. This can be done by using injective nuclear operators with non-closed range from any separable Banach space $$V$$ into $$X$$ and by picking $$V$$ and $$X$$ in such a way that $$V$$ is not a quotient of $$X$$. For example, let $$X$$ be a separable Hilbert space and $$V$$ be a separable Banach space which is not isomorphic to a Hilbert space.\n\nExample 4.12 in Cross, R. W.; Ostrovskiĭ, M. I.; Shevchik, V. V. [Operator ranges in Banach spaces. I. Math. Nachr. 173 (1995), 91–114] is a much stronger example. In the same paper you can find more details on the proof sketched above.\n\n• Thanks a lot for your answer and for the reference to your paper! By the way, is there a Part II of the paper? (I wasn't able to find it.) – Jochen Glueck May 22 '19 at 20:48\n• Unfortunately we were unable to continue collaboration, and part II never appeared. – Mikhail Ostrovskii May 23 '19 at 5:55"
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http://vinzaar.com/solving-systems-of-equations-by-graphing-word-problems.html | [
"# Solving systems of equations by graphing word problems. Solving Systems of Linear Equations by Graphing Examples 2019-01-12\n\nSolving systems of equations by graphing word problems Rating: 9,2/10 581 reviews\n\n## Solving Systems Of Equations By Substitution Worksheets",
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"Perimeter, Rectangle, Length, Width - Word Problem The perimeter of a rectangle is 44 feet and the length is 1 more than twice the width. How many students can a bus carry? The first day of my solving systems by substitution lesson, I see so many heads explode. The slope is -1, so go down one unit and over one unit and draw a point. Solve the given system of equations by graphing. We've found a system of equations: The problem asks how many items sold will give Tammy and Lisa the same amount of money, so we want to find x when the y-values on the two lines are the same. So first, in the red equation, I want to see if my y number is equal to my x number plus 4.\n\nNext\n\n## Solving Systems of Equations Using Algebra Calculator",
null,
"Graph your amount, Y, on the y-axis and your friend's amount, F, on the x-axis it actually doesn't matter which is which as long as you label them correctly. This point works in the first equation. Sample Problem Tammy and Lisa work in retail, in different shops. Why did one line cross the other line? Three times as many student tickets were sold compared to adult tickets. Solving systems of equations by elimination with two variables 2. Next, you need to graph both equations on the same coordinate plane.\n\nNext\n\n## Solving Systems of Equations by Graphing",
null,
"At the end of the day, you pool your money. It's nice to be confident; confidence is attractive. She's not scarfing them at the rate that Loren is, but she could still stand to slow down the assembly line into her mouth a bit. The number of cookies depends on the number of days, so let's have x be the number of days that have passed since Sunday, and y be the number of cookies left at the end of that day. This specific photograph Solving Systems Of Equations by Graphing Worksheet Problem solving Worksheets solving Systems Equations Word Problems posted by means of Samantha Williamson on 2018-11-05 02:46:05. Anyway, we're not in the business of asking questions. Such is the life of an actor.\n\nNext\n\n## Word problems systems of equations and inequalities",
null,
"Solving systems of linear equations by graphing worksheet - Problems 1. Below you can download some free math worksheets and practice. Aside from the hardship letter, preparing the financial worksheet is also one of the most vital part in modifying your home mortgage. The first account pays 7% interest and the second account pays 9% annual interest. Draw a neat line connecting these points.\n\nNext\n\n## Word problems systems of equations and inequalities",
null,
"A pity, because they can tear it up on the dance floor. They'll also never have the same number of roaches infesting their respective hovels. Let's solve the system of equations: Using substitution, we get: However, there's something funny going on here: the lines intersect at , which would be a quarter of a day in the past. The senior class at High School A rented and filled 8 vans and 4 buses with 256 students. Sample Problem Loren and Marisol each bought cookies on Sunday and started eating their cookies the next day. If they don't, we may need to have our negatives developed.\n\nNext\n\n## Systems of Linear Equations Word Problems with Two Lines",
null,
"Establishing a totally totally free calendar is one certain technique to generate income at a fundraiser. To solve a system of equations by graphing, graph each equation and identify the point where the two lines intersect. Solution: Let a be the price of an adult ticket, and let s represent the price of a student ticket. This complete unit is ready to copy! Calculate the speed of the boat in still water and the speed of the river. We need to find the two equations described by the problem. Print the worksheets after all the grades are entered.\n\nNext\n\n## Solving Systems of Equations Using Algebra Calculator",
null,
"Again, we can use substitution, in the same way that Loren and Marisol should have considered substituting celery sticks for cookies. Solving systems of equations with 3 variables by elimination 6. Now we'll move on to some word problems that involve finding where two lines intersect—in other words, solving a system of equations. What that means is that I have to graph both lines and find where they cross. Check the solution by plugging the values into each equations.\n\nNext\n\n## 13 Engaging Ideas for Teaching Systems of Equations",
null,
"It appears to be 0, -3. How many students can a van carry? They're trying to eat as many of them as possible before they need to give cookies up entirely for Lent. After how many days will they have the same number of pennies? It takes it three hours to travel the same distance during the return trip against the same tail wind. Loren took 6 days to eat 30 cookies, while Marisol took 8 days to eat 24 cookies. Past, Present, Future, Age Algebra Word Problem: 3 years ago, Kim was twice as old as John. Again, to make sure you're doing things neatly over long distances, double-check yourself by also drawing in the x-intercept. Word problems can be so difficult, so I absolutely use this activity to help students master them! Sample Problem Lois and Joseph start saving pennies on New Year's Day.\n\nNext\n\n## Word problems systems of equations and inequalities",
null,
"Age Related Algebra Word Problem Sally is twice as old as Julia and Tim is five years older than Julia. That is, we want to find where the lines intersect, which we can do by solving the system of equations. How long will it take for the two trains to meet? Two Trains Traveling In Opposite Directions Word Problem: Two trains are currently 600 miles apart traveling in opposite directions. These are tiny movements on a big line, so draw one more point at the x-intercept to make sure you've got things nicely graphed in the long run. There's no solution to this system of linear equations. For each line, the independent variable x is the number of items sold, and the dependent variable y is the amount of money the seller makes. The following word problems are not notorious for their subtlety.\n\nNext\n\n## System of Equations 3 Variables, Elimination, Substitution, Graphing, Inequalities & Word Problems",
null,
"Here is a list of topics included in this video: 1. The first train is traveling east at 80mph and the second train is traveling west at 40mph. Worksheet will open in a new window. This video contains tons of practice problems and examples to help you prepare for your next algebra test or exam. Upstream Downstream Boat River Distance Rate Time Word Problem: It takes 5 hour for a boat to travel 80 miles downstream and 10 hours for it to travel the same distance upstream against the river current.\n\nNext"
]
| [
null,
"https://www.ixl.com/screenshot/5e719b59ee590c7fc5389ca32099018364ea24c6.png",
null,
"http://www.algebra-class.com/images/System-of-Equations-1.gif",
null,
"https://worksheets-library.com/images2/solving-systems-of-equations-by-graphing-word-problems-worksheet/solving-systems-of-equations-by-graphing-word-problems-worksheet-3.jpg",
null,
"https://www.ixl.com/screenshot/c5fc58a8ecaf7ec840bd160f7f81fc0d15eaba1a.png",
null,
"https://embed-ssl.wistia.com/deliveries/11a722b5435c497a2efa46c1ff47683bab4e8b20.jpg",
null,
"https://www.shmoop.com/images/algebra/alg_syslineq_syseq_narr_graphik_5.png",
null,
"http://locart.club/wp-content/uploads//2018/11/systems-of-equations-word-problems-worksheet-algebra-2-math-systems-of-linear-equations-word-problems-worksheet-answers-doc-solving-systems-equations-by-graphing-worksheet-math-solver-free.jpg",
null,
"https://omegaprzywidz.com/wp-content/uploads/2018/07/solving-systems-of-equations-by-graphing-worksheet-answers-fresh-worksheet-systems-linear-equations-word-problems-worksheet-of-solving-systems-of-equations-by-graphing-worksheet-answers.jpg",
null,
"https://unboy.org/wp-content/uploads/2018/12/systems-of-equations-graphing-worksheet-system-equations-word-problems-worksheet-graphing-new-systems-of-systems-of-equations-graphing-worksheet.jpg",
null,
"http://www.cpalms.org/Uploads/resources/59144/Rubric/36510/graphics/MFAS_ApplesAndPeaches_Image3.jpg",
null
]
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https://socratic.org/questions/how-do-you-solve-8-b-10-4-2b-7 | [
"# How do you solve 8/(b+10)=4/(2b-7)?\n\nApr 15, 2018\n\n$b = 8$\n\n#### Explanation:\n\nStep 1: cross multiply the two fractions\n\n$8 \\left(2 b - 7\\right) = 4 \\left(b + 10\\right)$\n\nStep 2: Use the distributive property on either side of the equations\n\n$16 b - 56 = 4 b + 40$\n\nStep 3: add $56$ to both sides\n\n$16 b - 56 + 56 = 4 b + 40 + 56$\n\n$16 b = 4 b + 96$\n\nStep 4: Subtract $4 b$ on both sides of the equation to isolate the variable\n\n$12 b = 96$\n\nStep 5: divide and simplify\n\n$b = 8$"
]
| [
null
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https://middleburyfoods.com/qa/quick-answer-what-is-e-used-for-in-real-life.html | [
"",
null,
"# Quick Answer: What Is E Used For In Real Life?\n\n## Which is more important PI or E?\n\nIf you draw and measure lots of geometric figures, you might think that π is most important, because it is used to measure perimeters and areas of circular figures.\n\nIf you are a banker, you might think e is the most important.\n\nBanks can use e to calculate continuous interest..\n\n## Why is E so important?\n\nIt turns out the answer is the irrational number e, which is about 2.71828…. Of course, e is more than just any number. It’s one of the most useful mathematical constants. … It’s also an important number in physics, where it shows up in the equations for waves, such as light waves, sound waves, and quantum waves.\n\n## What is E to the infinity?\n\nWhen e is raised to power infinity,it means e is increasing at a very high rate and hence it is tending towards a very large number and hence we say that e raised to the power infinity is infinity. Now… When e is raised to the power negetive infinity , it tends towards a very small number and hence tends to zero.\n\n## What does the weird e mean in math?\n\nIt’s the Greek capital letter Σ sigma. Roughly equivalent to our ‘S’. It stands for ‘sum’. Read this for starters.\n\n## What is e math term?\n\nThe number e, known as Euler’s number, is a mathematical constant approximately equal to 2.71828, and can be characterized in many ways. It is the base of the natural logarithm. It is the limit of (1 + 1/n)n as n approaches infinity, an expression that arises in the study of compound interest.\n\n## What is the most famous number?\n\n10 Famous NumbersThe Greek letter pi represents a value of approximately 3.14159, the ratio between the circumference and diameter of a circle. … e, known as Euler’s number, is approximately 2.71828 and is another nonrepeating, nonterminating number. … 10100 is a Googol. … 0 has nothing to it. … 1 is the first counting number.More items…\n\n## What is the value of E Power 0?\n\n1Value of e to power zero is e is equal to 1.\n\n## What is the significance of E?\n\nIt is often called Euler’s number after Leonhard Euler (pronounced “Oiler”). e is an irrational number (it cannot be written as a simple fraction). e is the base of the Natural Logarithms (invented by John Napier). e is found in many interesting areas, so is worth learning about.\n\n## What is the rarest number?\n\nOther examples of rare numbers are 65, 621770, 281089082, 2022652202, 868591084757, 872546974178 … (Sequence A035519 of OEIS). If we consider palindromic rare numbers, there are infinitely many rare numbers.\n\n2 Answers. These two numbers are not related. At least, they were not related at inception ( π is much-much older, goes back to the beginning of geometry, while e is a relatively young number related to a theory of limits and functional analysis).\n\n## What is E to zero?\n\nAny number raised to zero is one. Zero is neither positive nor negative so the minus sign before it is redundant. e is constant quantity(roughly equal to 2.71) and when raised to the power 0 it results in 1 as the answer.\n\n## What is E in log?\n\nThe natural logarithm of a number is its logarithm to the base of the mathematical constant e, where e is an irrational and transcendental number approximately equal to 2.718281828459. The natural logarithm of x is generally written as ln x, loge x, or sometimes, if the base e is implicit, simply log x.\n\n## Why is e called natural logarithm?\n\nThis natural base of exponential functions is also used as the base for the logarithm functions, thus naming it as the natural logarithm function.\n\n## What do we use E for?\n\ne is the base rate of growth shared by all continually growing processes. e lets you take a simple growth rate (where all change happens at the end of the year) and find the impact of compound, continuous growth, where every nanosecond (or faster) you are growing just a little bit."
]
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null,
"https://mc.yandex.ru/watch/66668599",
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https://www.studyvirus.com/set-27-inequalities-for-sbi-po-and-sbi-clerk-2019-must-go-through-these-questions/ | [
"# Set-27 Inequalities For SBI PO and SBI Clerk 2019 | Must Go Through These Questions",
null,
"Dear Aspirants,\nWe are providing the most important Inequalities for SBI PO 2019, SBI Clerk 2019 and all other competitive bank and insurance exams. These questions have very high chances to be asked in SBI PO 2019, SBI Clerk 2019.\nGet the Best Test Series for SBI PO 2019 at the most affordable price (Based on Real Exam Pattern) – Click Here\nDownload the Best GK Gaming App for Current Affairs and GK (Bank+SSC)– Click here (App No 1) (App No 2)\n\nDirections:(1-5) In each of the following questions some statements are followed by two conclusions numbered I and II. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read the conclusions and then decide which of the given conclusions logically follows from the given statements. Give answer,\n“P\\$Q” means “P is greater than Q”\n“P#Q” means “P is either less than or equal to Q”\n“P%Q” means “P is neither greater nor less than Q”\n“P*Q” means “P is either greater than or equal to Q”\n“P&Q” means “P neither greater than nor equal to Q”\n\n1.\nStatements:\nA%B&C#D, C&E*F%G\nConclusions:\nI. E\\$A\nII. B&G\n\n2.\nStatements:\nI#J&K%L, M\\$N*O#J\nConclusions:\nI. M\\$O\nII. O&L\n\n3.\nStatements:\nP#Q%R&S, T\\$U*R%V\nConclusions:\nI. P&T\nII. P%V\n\n4.\nStatements:\nM#N%O*P, Q&O\\$R%S\nConclusions:\nI. M&R\nII. S*M\n\n5.\nStatements:\nX#Y%Z*A, B&Z\\$C*D\nConclusions:\nI. D&Y\nII. B&X\n\nDirections:(6-10) In each of the following questions some statements are followed by two conclusions numbered I and II. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read the conclusions and then decide which of the given conclusions logically follows from the given statements. Give answer\n\n6.\nStatements:\nA > B < C; D < E > B; D= F ≤ G.\nConclusions:\nI. A > D\nII. A < G\n\n7.\nStatements:\nG≤ F< I; F≤ H< K=L; K≤ M >N.\nConclusions:\nI. G >N\nII.G< M\n\n8.\nStatements:\nP< Q< R=S; S≤ K >M; K≤ L=O >N.\nConclusions:\nI. P< N\nII. N≤ P\n\n9.\nStatements:\nI >J≤ H >L; H< T ≥N; T< M >O.\nConclusions:\nI. J< M\nII. N< M\n\n10.\nStatements:\nW ≤ X > Y < Z; Y< A <C; A> F< G.\nConclusions:\nI. W< C\nII.Y < G\n\nCheck your Answers below:\n\n1. Directions:(1-5) In each of the following questions some statements are followed by two conclusions numbered I and II. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read the conclusions and then decide which of the given conclusions logically follows from the given statements. Give answer,\n“P\\$Q” means “P is greater than Q”\n“P#Q” means “P is either less than or equal to Q”\n“P%Q” means “P is neither greater nor less than Q”\n“P*Q” means “P is either greater than or equal to Q”\n“P&Q” means “P neither greater than nor equal to Q”\n\nStatements:\nA%B&C#D, C&E*F%G\nConclusions:\nI. E\\$A\nII. B&G\n\nAns:2",
null,
"",
null,
"I. E >A (True)\nII. B< G (False)\n\n2. ##### 2. Question\n\nStatements:\nI#J&K%L, M\\$N*O#J\nConclusions:\nI. M\\$O\nII. O&L\n\nAns:1",
null,
"",
null,
"I. M >O (True)\nII. O< L (True)\n\n3. ##### 3. Question\n\nStatements:\nP#Q%R&S, T\\$U*R%V\nConclusions:\nI. P&T\nII. P%V\n\nAns:2",
null,
"",
null,
"I. P< T (True)\nII. P=V (False\n\n4. ##### 4. Question\n\nStatements:\nM#N%O*P, Q&O\\$R%S\nConclusions:\nI. M&R\nII. S*M\n\nAns:4",
null,
"",
null,
"I. M< R (False)\nII.S ≥M (False)\n\n5. ##### 5. Question\n\nStatements:\nX#Y%Z*A, B&Z\\$C*D\nConclusions:\nI. D&Y\nII. B&X\n\nAns:2",
null,
"",
null,
"I. D< Y (True)\nII. B< X (False)\n\n6. ##### 6. Question\n\nDirections:(6-10) In each of the following questions some statements are followed by two conclusions numbered I and II. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read the conclusions and then decide which of the given conclusions logically follows from the given statements. Give answer\n\nStatements:\nA > B < C; D < E > B; D= F ≤ G.\nConclusions:\nI. A > D\nII. A < G\n\nAns:4\nA > B < E > D =F ≤ G\nI. A > D (False)\nII. A < G (False)\n\n7. ##### 7. Question\n\nStatements:\nG≤ F< I; F≤ H< K=L; K≤ M >N.\nConclusions:\nI. G >N\nII.G< M\n\nAns: 2\nG≤ F≤ H< K≤ M >N\nI. G >N (False)\nII. G< M (True)\n\n8. ##### 8. Question\n\nStatements:\nP< Q< R=S; S≤ K >M; K≤ L=O >N.\nConclusions:\nI. P< N\nII. N≤ P\n\nAns:3\nP< Q< R=S≤ K≤ L=O>N\nI. P< N (False)\nII. N≤ P (False)\nBoth conclusions are not follows but here the same variables with all three symbols denotes either conclusions follows.\n\n9. ##### 9. Question\n\nStatements:\nI >J≤ H >L; H< T ≥N; T< M >O.\nConclusions:\nI. J< M\nII. N< M\n\nAns: 5\nJ≤ H< T< M , N≤ T< M\nI. J< M (True)\nII. N< M (True)\n\n10. ##### 10. Question\n\nStatements:\nW ≤ X > Y < Z; Y< A <C; A> F< G.\nConclusions:\nI. W< C\nII.Y < G\n\nAns:4\nI.W< C (False)\nII.Y >G (False)",
null,
"#### About Study Virus\n\nView all posts by Study Virus →\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
]
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null,
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"https://estore.ibpsguide.com/ebook-admin/uploads/image/productimage1529478233386.jpg",
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"https://estore.ibpsguide.com/ebook-admin/uploads/image/productimage1529478452268.jpg",
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"https://estore.ibpsguide.com/ebook-admin/uploads/image/productimage1529478477874.jpg",
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null,
"https://estore.ibpsguide.com/ebook-admin/uploads/image/productimage1529478663317.jpg",
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null,
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null,
"https://estore.ibpsguide.com/ebook-admin/uploads/image/productimage1529478452268.jpg",
null,
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null,
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https://physics.stackexchange.com/questions/599015/using-persistence-length-to-establish-straightness-of-position-data/599024#599024 | [
"# Using Persistence length to establish straightness of position data\n\nIn the following picture I am showing 3D position data for 4 different tracks just to illustrate how the tracks looks like.",
null,
"As we can see that we have all kind of behaviors occurring from helical motion(bottom left) to somewhat Brownian motion. I need to establish a metric through which I can compare these tracks with each other to establish which one is more straighter than the other. All of these tracks have different time resolution. Here is what I have tried so far:\n\nPersistence Length is a concept associated with polymers to calculate bending stiffness. But We can also think of it as a metric to define if track is flexible or not. Here is how I calculate persistence length for these tracks.\n\nWe can assume a bond length for these tracks. We can fit a line for the points which correspond to that particular bond length. Then we can use following formula to calculate the persistence length. $$l_{p} = \\frac{\\sum_{1}^{n}l_{1}.l_{n}}{l} .$$ I have taken this formula from reference 1. My issue is that it works fine in the case if track does not change direction more than 90 degrees. If this change happens in the start of the track, I get negative persistence length which does not make sense to me at all. Following figure shows how it looks in practice for the case: a) when it gives something meaningful. b) when it is negative.",
null,
"I completely understand why I am getting negative value but this made me think that I am not using this concept correctly. If someone can point out what is my mistake here that would be really helpful. Moreover if you have another suggestion to characterize straightness please feel free to point it out.\n\n1: Zhang, J. Z., Peng, X. Y., Liu, S., Jiang, B. P., Ji, S. C., & Shen, X. C. (2019). The Persistence Length of Semiflexible Polymers in Lattice Monte Carlo Simulations. Polymers, 11(2), 295.\n\nThe equation you cited is not the usual definition of the persistence length, $$L_p$$, which is the decay length of the average angle between segments. Perhaps this formula is roughly approximate; I'm not going to bother reading the reference. (It certainly seems odd to privilege segment 1, and not perform any averaging.) Why not calculate a persistence length in the standard way, which would prevent any weird negative lengths?\n• The averaging could be over thermal fluctuations, but could be over segments. For any pair of segments separated by distance $l$, one can calculate $T$, and so have lots of measurements of cos($T$) for the same track. (Roughly, if the total length is 100 units, you'd have 99 values of $T$ for $l=1$, about 50 for $l=2$, etc.) Dec 8 '20 at 15:19"
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"https://i.stack.imgur.com/mthpF.jpg",
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"https://i.stack.imgur.com/tfon1.jpg",
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https://techwhiff.com/learn/a-company-would-like-to-examine-the-linear/199109 | [
"# A company would like to examine the linear relationship between the age and credit score of...\n\n###### Question:",
null,
"A company would like to examine the linear relationship between the age and credit score of an individual. The following table shows the credit scores and ages of 5 randomly selected people. These data have a sample correlation coefficient, rounded to three decimal places, of 0.970. Using this data and a = 0.05, test if the population correlation coefficient between a person's age and credit score is different than zero. What conclusions can you draw? Age 39 27 55 26 34 D Credit score 670 650 755 620 655 What are the correct null and alternative hypotheses? O A. Ho:p0 Hyp=0 O C. Ho: p>0 н: p=0 OB. Ho:p=0 H:p> 0 OD. HO:20 нар О What is the test statistic? ta (Round to two decimal places as needed.) What is the p-value? p-value (Round to three decimal places as needed.)\n\n#### Similar Solved Questions\n\n##### Lancer, Inc. (a U.S.-based company), establishes a subsidiary in a foreign country on January 1, 2016....\nLancer, Inc. (a U.S.-based company), establishes a subsidiary in a foreign country on January 1, 2016. The following account balances for the year ending December 31, 2017, are stated in kanquo (KQ), the local currency: Sales KQ 280,000 Inventory (bought on 3/1/17) 168,000 Equipment (bought on 1/1/1...\n##### Includes yeast and molds; rarely pathogenic. A fungus that can cause thrush. The virus that causes...\nincludes yeast and molds; rarely pathogenic. A fungus that can cause thrush. The virus that causes infectious mononucleosis....\n##### Thirteen people on a softball team show up for a game. (a) How many ways are...\nThirteen people on a softball team show up for a game. (a) How many ways are there to choose 10 players to be in the game? (b) How many ways are there to assign 10 people out of the 13 to the various positions on the field? (c) Of the 13 people, three are college students. How many ways are there to...\n##### Assumptions for the Model 1. The generations are distinct (making this a discrete system). 2. Each...\nAssumptions for the Model 1. The generations are distinct (making this a discrete system). 2. Each rabbit pair will be ready to reproduce at the end of the second generation (born in Generation 1, develop in Generation 2, first offspring in Generation 3). 3. Each pair of mature rabbits has one pair ...\n##### E° (cal.) E° (exp.) 1. Fe(s)| Fe2+ (1 M) || Cu2+ (1 M) | Cu(s) 0.775...\nE° (cal.) E° (exp.) 1. Fe(s)| Fe2+ (1 M) || Cu2+ (1 M) | Cu(s) 0.775 V 2. Pb(s) Pb2+ (1 M) || Cu2+ (1 M) Cu(s) 0.461 V 3. Sn(s) | Sn2+ (1 M) || Cu2+ (1 M) | Cu(s) 0.472 V 4. Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) Cu(s) 1.095 V 5. Zn(s) | Zn2+ (1 M) || Cu2+ (1 M + NH3) | Cu(s) 0.928 V a) The Nernst...\n##### 1. Set up the appropriate form of a particular solution yp, but do not determine the...\n1. Set up the appropriate form of a particular solution yp, but do not determine the values of the coefficients V\" +y = r? + cos2.c 2. Transform the following differential equation into an equivalent system of first order differential equations - 312) - 4x + 2x2 - 2 cost...\n##### How do you solve lnx+ln(x+1)=1?\nHow do you solve lnx+ln(x+1)=1?...\n##### Cost of JUU Cash Inventory 10.UUU Taxes payable 25.000 Land 1.000 Notes payable (due in 6...\nCost of JUU Cash Inventory 10.UUU Taxes payable 25.000 Land 1.000 Notes payable (due in 6 months) Salaries payable 900 4,300 Requirements: A Compute the quick ratio. 2.04 B. Determine the amount of working capital. 18, SOO C. Assume that cash is used to pay the balance due on accounts payable. 1. Co...\n##### A carpeted living room and dining room area measures 32 ft by 10 ft\nA carpeted living room and dining room area measures 32 ft by 10 ft. Mark decides to install wood flooring in the 12 ft by 10 ft dining room. By what percent has he reduced the area that is carpeted?...\n##### How do you solve (x+6)/(x^2-5x-24)>=0?\nHow do you solve (x+6)/(x^2-5x-24)>=0?...\n##### A firm's year-end retained earnings balances are $670,000 and$780,000, for 2018 and 2019 respectively. The...\nA firm's year-end retained earnings balances are $670,000 and$780,000, for 2018 and 2019 respectively. The firm paid $10,000 in dividends in 2019. The firm's net profit after taxes in 2019 was$110,000 -$100,000$100,000 -\\$110,000..."
]
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"https://i.imgur.com/cDOuGt0.png",
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https://answers.everydaycalculation.com/divide-fractions/75-4-divided-by-35-9 | [
"Solutions by everydaycalculation.com\n\n## Divide 75/4 with 35/9\n\n1st number: 18 3/4, 2nd number: 3 8/9\n\n75/4 ÷ 35/9 is 135/28.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 35/9: 9/35\n2. Now, multiply it with the dividend\nSo, 75/4 ÷ 35/9 = 75/4 × 9/35\n3. = 75 × 9/4 × 35 = 675/140\n4. After reducing the fraction, the answer is 135/28\n5. In mixed form: 423/28\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
]
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null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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http://www.summitcontractors.co.uk/woolworths-home-owjq/joule-to-erg-84ae2d | [
"7 joule to ergs = 70000000 ergs. conversion en ligne de Joule (J, Système SI international) en Erg (SGH et unités non-système). Ergs Joules; 40 erg: 0.00 J: 41 erg: 0.00 J: 42 erg: 0.00 J: 43 erg: 0.00 J: 44 erg: 0.00 J: 45 erg: 0.00 J: 46 erg: 0.00 J: 47 erg: 0.00 J: 48 erg: 0.00 J: 49 erg: 0.00 J: 50 erg: 0.00 J: 51 erg: 0.00 J: 52 erg: 0.00 J: 53 erg: 0.00 J: 54 erg: 0.00 J: 55 erg: 0.00 J: 56 erg: 0.00 J: 57 erg: 0.00 J: 58 erg: 0.00 J: 59 erg: 0.00 J Energy Converter. ergs metres squared, grams, moles, feet per second, and many more. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. symbols, abbreviations, or full names for units of length, The first one is from Common Units.The second one is from International System (SI). 7 joule to erg = 70000000 erg. 1 joule is equal to 10000000 ergs. inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, joule to gigawatt-hour. Type in your own numbers in the form to convert the units! as English units, currency, and other data. metres squared, grams, moles, feet per second, and many more. Free online energy conversion. joule to terajoule It is defined as the amount of work done by a force of one dyne exerted for a distance of one centimeter. Joules (J) Ergs (erg) Precision: Reverse conversion? The answer is 1.0E-7. 10 Erg (Cgs Unit) - erg. input convert. Convertissez les unités de énergie. The unit is pronounced to rhyme with \"tool\", and is named in honor of the physicist James Prescott Joule (1818-1889). Type in unit We assume you are converting between joule and erg. 1 joule is equal to 10000000 erg. 1 BTU (BTU) = 10550560000 Erg (Erg) 8.7719298245614E-6 Gallon of Gas (g.G.) joule or ergs to joule, or enter any two units below: joule to dekawatt-hour converter table. Calculate from power into other power unit measures. The first one is from International System (SI).The second one is from Common Units. 6 joule to erg = 60000000 erg. Free Convert joule (J) to erg Converter calculator in energy units, joule to erg conversion table and from joule to other energy units By another definition, the joule is equal to the energy required to pass an electric current of one ampere through a one ohm resistor for one second. Ergs to Joules (or just enter a value in the \"to\" field) Please share if you found this tool useful: Tweet. It has the symbol erg. Energy Conversion. 5 joule to erg = 50000000 erg. Examples include mm, Instant free online tool for joule/second to erg/second conversion or vice versa. 4 joule to erg = 40000000 erg. joule to myriawatt-hour Definition: In relation to the base unit of [power] => (watts), 1 Joules Per Minute (J/min) is equal to 0.016666666666667 watts, while 1 Ergs Per Second (erg/s) = 1.0E-7 watts. Erg : Erg (short for ergon) is a measurement unit of energy and mechanical work used in physics, equal to 10−7 joules. 6 joule to ergs = 60000000 ergs. You can find metric conversion tables for SI units, as well An erg is a unit of energy equal to one 10 millionth of a joule. conversion calculator for all types of measurement units. Joule to erg conversion allow you make a conversion between joule and erg easily. m). 1 joule to erg = 10000000 erg. input convert. Convert joule to erg (J to erg), Fuel Economy metric conversion using Converterin The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. 1 joule to ergs = 10000000 ergs. You can view more details on each measurement unit: 3 joule to ergs = 30000000 ergs. If you need to convert erg to another compatible unit, please pick the one you need on the page below. Unit Descriptions; 1 Joule (SI unit): 1 J = 1 m*N = 1 kg*m 2 /s 2: 1 Erg: Exactly 1 g*cm 2 /s 2. 1 J/min = 166666.66666667 erg/s. The unit name “joule” is in honor of the English physicist James Prescott Joule. Convertir joule en erg. The erg is a unit of energy equal to 10 −7 joules (100 n J). Always check the results; rounding errors may occur. On the surface of the Earth, it takes about 0.98 ergs to lift a 1 milligram object by 1 centimeter. 1 joule is 10000000 ergs 1 erg is 1.0E-7 joules As nouns the difference between joule and erg joule to yottawatthour joule to therm erg to joule, or enter any two units below: joule to kilogram-force meter 1 erg is equal to 1.0E-7 joule. joule to megawatthour The unit is pronounced to rhyme with \"tool\", and is named in honor of the physicist James Prescott Joule (1818-1889). The answer is 1.0E-7. The answer is 10000000. 1 erg = 10 −7 joule = 100 nJ = 100 nanojoules 1 erg = 10 −10 sthène m = 100 psn m = 100 picosthènes-mètres 1 erg = 624,15 GeV = 6,241 5 × 10 11 eV 1 erg = 1 dyne cm = 1 g cm2 s−2 5 joule to ergs = 50000000 ergs. A joule is a unit of energy equal to a newton-meter An erg is a CGS unit of energy equal to a dyne-centimeter, and all named derived units of the CGS are deprecated by the SI Brochure, so I would never do the conversion. You can find the tool in the following. joule to hectowatt-hour Convert joule to erg (J to erg), Energy metric conversion using Converterin The erg is not an SI unit. input; 1 joule = 10 000 000 ergs = 10,000,000.00000000 = 1.0 × 10 7 = 1.0E+7 = 1.0e+7. joule to femtojoule joule to therm Outil gratuit en ligne pour faire vos calculs d'unités. unitsconverters.com helps in the conversion of different units of measurement like J to Erg through multiplicative conversion factors. An erg is a unit of energy equal to one 10 millionth of a joule. Note that rounding errors may occur, so always check the results. joule to teraelectron volt convert from Joule to Erg conversion of Joule to Erg Joule to Erg Joule into Erg Joule in Erg conversion from Joule to Erg J to Erg. An erg is the unit of energy and mechanical work in the centimetre-gram-second (CGS) system of units, symbol \"erg\". joule to megalerg joule to kilopond meter joule to terajoule For example, when a tennis ball is hit by a racket and stops momentarily, the forces acting upon it (like gravity and the resistance of the racket), make it stay still in that position. Diferent power units conversion from joule per second to ergs per second. You can do the reverse unit conversion from Convert 1 Joule (Si Unit) to Erg (Cgs Unit) - 1 J to erg. It originated in the centimetre–gram–second (CGS) system of units. You can do the reverse unit conversion from We assume you are converting between erg and joule. Convert 1 Joule (Si Unit) to Erg (Cgs Unit) how many Joule (Si Unit) to Erg (Cgs Unit) : 1 J = 10.0 erg. It is defined as the amount of work done by a … The SI derived unit for energy is the joule. Energy Conversion. Calculez les ergs en joules, convertir erg vers J . Joule to erg conversion allow you make a conversion between joule and erg easily. 1 x 166666.66666667 erg/s = 166666.66666667 Ergs Per Second. Thermal energy is measured by heat meters. area, mass, pressure, and other types. Please enable Javascript This page features online conversion from erg to joule.These units belong to different measurement systems. Type in unit 3 joule to erg = 30000000 erg. =. m), or in passing an electric current of one ampere through a resistance of one ohm for one second. joule to inch ounce. area, mass, pressure, and other types. converter table. Its name is derived from the Greek word meaning \"work\". 9 joule to erg = 90000000 erg. input; 1 joule = 10 000 000 ergs = 10,000,000.00000000 = 1.0 × 10 7 = 1.0E+7 = 1.0e+7. The SI derived unit for energy is the joule. An erg is the unit of energy and mechanical work in the centimetre-gram-second (CGS) system of units, symbol \"erg\". to use the unit converter. unitsconverters.com helps in the conversion of different units of measurement like Joule to Erg through multiplicative conversion factors. Joules Ergs; 0 J: 0.00 erg: 1 J: 10000000.00 erg: 2 J: 20000000.00 erg: 3 J: 30000000.00 erg: 4 J: 40000000.00 erg: 5 J: 50000000.00 erg: 6 J: 60000000.00 erg: 7 J: 70000000.00 erg: 8 J: 80000000.00 erg: 9 J: 90000000.00 erg: 10 J: 100000000.00 erg: 11 J: 110000000.00 erg: 12 J: 120000000.00 erg: 13 J: 130000000.00 erg: 14 J: 140000000.00 erg: 15 J: 150000000.00 erg: 16 J: 160000000.00 erg: 17 J: 170000000.00 erg: 18 J: 180000000.00 erg: … You can find the tool in the following. inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, Convert joule to erg (J to erg). from to. 9 joule to ergs = 90000000 ergs. Also, explore tools to convert joule/second or erg/second to other power units or learn more about power conversions. Convert From BTU. We assume you are converting between joule and erg. Use this page to learn how to convert between joules and ergs. If you need to convert erg to another compatible unit, please pick the one you need on the page below. It originated in the centimetre–gram–second (CGS) system of units. 8 joule to ergs = 80000000 ergs. You can find metric conversion tables for SI units, as well The joule/second [J/s] to erg/second [erg/s] conversion table and conversion steps are also listed. Please enable Javascript joule to megaelectron volt joule to newton meter erg to joule (—J) measurement units conversion. One joule converted into erg equals = 10,000,000.00 erg 1 J = 10,000,000.00 erg How many joule in 1 ergs? How many erg in 1 joule? This page features online conversion from erg to joule.These units belong to different measurement systems. Examples include mm, The first one is from Common Units.The second one is from International System (SI). 2 joule to ergs = 20000000 ergs. +> with much ♥ by CalculatePlus The erg is a unit of energy and work equal to 10 −7 joules. The other way around, how many ergs per second - erg/s are in one joule per second - J/s unit? It has the symbol erg. Joule is the unit for energy Dimension Formula for Joule is [M¹L²T⁻²] Let n1(joule)=n2(erg) n1(M₁¹L₁²T₁⁻²]=n2[M₂¹L₂²T₂⁻²] 4 joule to ergs = 40000000 ergs. Electrical energy is measured by electricity meters. If you need to convert joule to another compatible unit, please pick the one you need on the page below. Convert joule to ergs (J to erg). 5 Joule (Si Unit) - J. joule to foot poundal How much is joule to erg? Convert 1 J/s into erg per second and joules per second to erg/s. You can view more details on each measurement unit: Note that rounding errors may occur, so always check the results. joule to hectowatt-hour +> with much ♥ by CalculatePlus How many joule in 1 erg? erg to use the unit converter. How much is joule to ergs? Free Convert 2 joule (J) to erg Converter calculator in energy units,2 joule to erg conversion table and from 2 joule to other energy units Exchange reading in joules unit J into ergs unit erg as in an equivalent measurement result (two different units but the same identical physical total value, which is also equal to their proportional parts when divided or multiplied). You can view more details on each measurement unit: erg or joule The SI derived unit for energy is the joule. The erg is not an SI unit. conversion calculator for all types of measurement units. ConvertUnits.com provides an online symbols, abbreviations, or full names for units of length, 1 Ergs to common energy units; 1 ergs = 1.0E-7 joules (J) 1 ergs = 1.0E-10 kilojoules (kJ) 1 ergs = 2.3900573613767E-8 calories (cal) 1 ergs = 2.3900573613767E-11 kilocalories (kcal) 1 ergs = 624149596175.21 electron volt (eV) 1 ergs = 2.7777777777778E-11 watt hour (Wh) 1 ergs = 22937126583.579 atomic unit of energy (au) 1 ergs = 2.3900573613767E-17 tons of TNT (tTNT) 1 ergs Note that rounding errors may occur, so always check the results. Convert joule to erg (J to erg), Fuel Economy metric conversion using Converterin On the surface of the Earth, it takes about 0.98 ergs to lift a 1 milligram object by 1 centimeter. Between J/s and erg/s measurements conversion chart page. as English units, currency, and other data. Erg : Erg (short for ergon) is a measurement unit of energy and mechanical work used in physics, equal to 10−7 joules. Free Convert joule (J) to erg Converter calculator in energy units, joule to erg conversion table and from joule to other energy units Free online energy conversion. from to. The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. 2 joule to erg = 20000000 erg. joule or Convert 5 Joule (Si Unit) to Erg (Cgs Unit) - 5 J to erg. An erg is a unit of energy equal to one 10 millionth of a joule. This page features online conversion from joule to erg.These units belong to different measurement systems. unitsconverters.com helps in the conversion of different units of measurement like J to Erg through multiplicative conversion factors. 8 joule to erg = 80000000 erg. Its name is derived from the Greek word meaning \"work\". Le joule (symbole : J) est une unité dérivée du Système international (SI) pour quantifier l'énergie, le travail et la quantité de chaleur . You are currently converting energy units from erg to joule 1 erg = 1.0⋅10-7 J 1 Joule (Si Unit) - J. ConvertUnits.com provides an online m), or in passing an electric current of one ampere through a resistance of one ohm for one second. On the surface of the Earth, it … Use this page to learn how to convert between joules and ergs. Type in your own numbers in the form to convert the units! Unit, please pick the one you need to convert erg to another compatible unit, please the... About power conversions - erg/s are in one joule converted into erg joule to erg... And conversion steps are also listed to different measurement systems CGS unit ) erg... Pick the one you need on the surface of the Earth, …. One second and work equal to 10 −7 joules each measurement unit: joule or erg the SI derived for... Of length, area, mass, pressure, and other types occur... Of joule to erg units derived from the Greek word meaning `` work '' erg and joule or full names units! One joule per second work '' the Earth, it … Convertissez les unités de.... Are in one joule per second - J/s unit et unités non-système ) joules... ) Precision: Reverse conversion page to learn how to convert the units this... Pressure, and other types or ergs the SI derived unit for energy is the joule the.! De énergie Precision: Reverse conversion 10,000,000.00000000 = 1.0 × 10 7 1.0E+7... Unit for energy is the joule can find metric conversion tables for SI,! Per second to erg/s J to erg word meaning `` work '' conversion table and conversion steps are also.... The Greek word meaning `` work '' how many ergs per second and joules per second and joules second. Or ergs the SI derived unit for energy is the unit converter ] conversion table and conversion steps are listed. X 166666.66666667 erg/s = 166666.66666667 ergs per second in your own numbers in the centimetre–gram–second ( CGS ) system units... En erg ( CGS unit ) - 1 J = 10,000,000.00 erg please enable Javascript to use the of. Milligram object by 1 centimeter can find metric conversion tables for SI units, currency, and types. ), or full names for units of length, area, mass pressure! Other types gratuit en ligne pour faire vos calculs d'unités ) Precision: Reverse conversion conversion en pour. 10 millionth of a joule you can view more details on each measurement unit: joule or ergs the derived... From Common units from erg to joule.These units belong to different measurement systems or erg the SI derived unit energy. And conversion steps are also listed so always check the results ; errors. 10,000,000.00 erg 1 J to erg through multiplicative conversion factors and joule for SI units as... Joule/Second [ J/s ] to erg/second [ erg/s ] conversion table and conversion steps are also listed International! Erg and joule milligram object by 1 centimeter gratuit en ligne de joule ( SI unit ) to ). A force of one ohm for one second for a distance of ampere... Metric conversion tables for SI units, symbol `` erg '' measurement unit: erg or joule SI! Own numbers in the centimetre–gram–second ( CGS ) system of units, currency, and other data,... ), or full names for units of measurement units the conversion of different of..., convertir erg vers J your own numbers in the conversion of different units of length,,... And joules per second, so always check the results for all types of measurement units units or more... 1 centimeter ) ergs ( erg ) 000 ergs = 10,000,000.00000000 = 1.0 × 10 7 = 1.0E+7 =.. Measurement units tables for SI units, currency, and other data systems. ] to erg/second [ erg/s ] conversion table and conversion steps are also.. Defined as the amount of work done by a force of one centimeter unitsconverters.com helps the., it takes about 0.98 ergs to lift a 1 milligram object by centimeter! 1.0E+7 = 1.0E+7 = 1.0E+7 and mechanical work in the centimetre–gram–second ( CGS ) system of units 10,000,000.00 erg J... Mechanical work in the centimetre-gram-second ( CGS ) system of units word ``! To erg/s second to erg/s calculs d'unités different units of length, area mass! A distance of one ampere through a resistance of one ohm for one second lift! Pressure, and other types your own numbers in the centimetre–gram–second ( CGS system! Joule or ergs the SI derived unit for energy is the unit of energy equal to 10 −7.! Per second - erg/s are in one joule per second - erg/s in! 1.0 × 10 7 = 1.0E+7 = 1.0E+7 = 1.0E+7 = 1.0E+7 = 1.0E+7 need on surface... Of work done by a force of one ohm for one second we you! Well as English units, as well as English units, currency, other! Unit converter as the amount of work done by a force of one through... 000 ergs = 10,000,000.00000000 = 1.0 × 10 7 = 1.0E+7 = 1.0E+7 for units of length,,! Free online tool for joule/second to erg/second conversion or vice versa how many per... Much ♥ by CalculatePlus Free online tool for joule/second to erg/second conversion or vice.! = 166666.66666667 ergs per second and joules per second work equal to one 10 of! Measurement units et unités non-système ) learn how to convert the units first one is from Common Units.The second is! To other power units or learn more about power conversions ergs per second joule/second [ J/s ] to conversion. Resistance of one centimeter errors may occur, so always check the results or learn more power... The page below online conversion from erg to another compatible unit, please pick the you! Note that rounding errors may occur, so always check the results ; rounding may... Erg 1 J to erg to different measurement systems ampere through a resistance of one ohm for second... Of different units of length, area, mass, pressure, and types..., abbreviations, or full names for units of length, area, mass, pressure, and data... Done by a force of one ampere through a resistance of one ampere through resistance! Units.The second one is from International system ( SI unit ) to ). Page to learn how to convert the units erg per second - erg/s are one... Erg conversion allow you make a conversion between joule and erg ] to erg/second conversion or vice versa the.... - erg/s are in one joule converted into erg equals = 10,000,000.00 erg 1 J = 10,000,000.00 erg enable... Joule = 10 000 000 ergs = 10,000,000.00000000 = 1.0 × 10 7 = 1.0E+7 and... From erg to joule.These units belong to different measurement systems de joule ( SI unit ) - J... Conversion from erg to joule.These units belong to different measurement systems passing an electric current one... Into erg per second - J/s unit 10 −7 joules converted into per! Also listed = 10550560000 erg ( CGS unit ) to erg ) en erg ( )... - 1 J = 10,000,000.00 erg please enable Javascript to use the unit of energy and work! Check the results SGH et unités non-système ) learn how to convert between joules and ergs joules. … Convertissez les unités de énergie online energy conversion en joules, erg. Or full names for units of length, area, mass, pressure and. Conversion calculator for all types of measurement units one second measurement systems non-système.... Convert 5 joule ( SI unit ) - 5 J to erg ) gratuit! Equals = 10,000,000.00 erg 1 J = 10,000,000.00 erg please enable Javascript to use the of... Units of measurement units or learn more about power conversions power units or learn more power! Ergs the SI derived unit for energy is the joule and other types ( SGH unités. A conversion between joule and erg easily SI ).The second one from... Erg through multiplicative conversion factors other data name is derived from the Greek word meaning `` ''! From Common units it takes about 0.98 ergs to lift a 1 milligram by! To another compatible unit, please pick the one you need on the page below a milligram! ) en erg ( SGH et unités non-système ) the joule/second [ J/s ] to erg/second [ erg/s ] table. Type in unit symbols, abbreviations, or in passing an electric of... From Common Units.The second one is from International system ( SI unit ) - 5 J erg... From International system ( SI ) for one second measurement unit: joule or ergs the derived... One dyne exerted for a distance of one dyne exerted joule to erg a distance one! The SI derived unit for energy is the unit of energy equal to one 10 millionth a. - 5 J to erg ) 8.7719298245614E-6 Gallon of Gas ( g.G. the Earth, it Convertissez. With much ♥ by CalculatePlus Free online tool for joule/second to erg/second [ erg/s ] conversion table and conversion are. Measurement units the surface of the Earth, it … Convertissez les unités énergie... To erg/s to different measurement systems occur, so always check the results second to erg/s of done. Convert 5 joule ( SI ) is from Common Units.The second one is from system!, how many ergs per second - J/s unit for joule/second to erg/second conversion or vice versa, area mass... One second or erg the SI derived unit for energy is the joule to! Centimetre-Gram-Second ( CGS ) system of units, symbol `` erg '' between erg joule. Page features online conversion from joule to erg type in your own numbers in the centimetre-gram-second ( CGS ) of. Around, how many ergs per second - erg/s are in one joule converted into erg second..."
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https://my.vocabularysize.com/example-sentence/delta-distance | [
"# Example sentences for: delta-distance\n\nHow can you use “delta-distance” in a sentence? Here are some example sentences to help you improve your vocabulary:\n\n• Residual accumulated delta-distance (RADD) is defined as window size times the cumulative sum of terms δ*( t ) - mean(δ*), t = 1, 2, ...\n\n• Increasing trend in mean scaled delta-distance\n\n• Weakly consistent patterns of intragenomic variability in delta-distance\n\n• The essentially increasing trend in δ*√ n has an obvious implication for the scalability of standard binning levels used in classifying intra-genomic delta-distance [ 8 ] . These levels cannot be re-scaled by the reciprocal square root of window size without admitting that the profiles seen through smaller windows are statistically closer to the global signature.\n\n• Table 2shows mean scaled delta-distance by sequence ( Abbr ) and window size ( n ). Scaled delta-distance, defined above as δ*√ n , is a statistical invariant for any benchmark sequence generated by a Markov chain exhibiting the same signature as the given sequence.\n\n## Search\n\nSearch for example sentences",
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http://www.numbersaplenty.com/104118 | [
"Search a number\nBaseRepresentation\nbin11001011010110110\n312021211020\n4121122312\n511312433\n62122010\n7612360\noct313266\n9167736\n10104118\n1171253\n1250306\n1338511\n1429d30\n1520cb3\nhex196b6\n\n104118 has 32 divisors (see below), whose sum is σ = 248064. Its totient is φ = 28512.\n\nThe previous prime is 104113. The next prime is 104119. The reversal of 104118 is 811401.\n\nAdding to 104118 its reverse (811401), we get a palindrome (915519).\n\nIt can be divided in two parts, 104 and 118, that added together give a palindrome (222).\n\nIt is a junction number, because it is equal to n+sod(n) for n = 104097 and 104106.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (104113) by changing a digit.\n\nIt is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 1521 + ... + 1587.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (7752).\n\n2104118 is an apocalyptic number.\n\nIt is a practical number, because each smaller number is the sum of distinct divisors of 104118, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (124032).\n\n104118 is an abundant number, since it is smaller than the sum of its proper divisors (143946).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n104118 is a wasteful number, since it uses less digits than its factorization.\n\n104118 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 116.\n\nThe product of its (nonzero) digits is 32, while the sum is 15.\n\nThe square root of 104118 is about 322.6732093001. The cubic root of 104118 is about 47.0444727687.\n\nThe spelling of 104118 in words is \"one hundred four thousand, one hundred eighteen\"."
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http://forums.wolfram.com/mathgroup/archive/2012/Oct/msg00198.html | [
"",
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"Re: Plot in function\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg128435] Re: Plot in function\n• From: Daniel <dosadchy at its.jnj.com>\n• Date: Fri, 19 Oct 2012 02:42:43 -0400 (EDT)\n• Delivered-to: [email protected]\n• Delivered-to: [email protected]\n• Delivered-to: [email protected]\n• Delivered-to: [email protected]\n\n```>\n> I would like to make a function that plots a variable\n> over a parameter range. I tried:\n>\n> test[f_, sol_] :=\n> Table[Plot[\n> f /. DeleteCases[sol, i], {i[], 0.9 i[], 1.1\n> 1.1 i[]}], {i,\n> sol}]\n>\n> But this gives the error:\n>\n> test[x^2 + y^2, {x -> 0, y -> 0}]\n>\n> During evaluation of In:= Plot::write: Tag Part\n> in i[] is Protected. >>\n>\n> I see that i cannot define a part in the variable of\n> a plotfunction, but also if I first assign i[] to\n> a variable, I get a problem with the bounds.\n>\n> test[f_, sol_] :=\n> Table[a = i[];\n> Plot[f /. DeleteCases[sol, i], {a, 0.9 i[], 1.1\n> .1 i[]}], {i,\n> sol}]\n>\n> I don't know of a good solution. Thanks in advance.\n>\n\nFirst, the expression {i[], 0.9 i[], 1.1 i[]} is not good in your case as both limits evaluate to 0.\nOther that this, you're missing Evaluate[]:\n\ntest[f_, sol_] :=\nTable[Plot[f /. DeleteCases[sol, i], {i[], i[] - 1, i[] + 1}//Evaluate], {i, sol}]\n\ntest[x^2 + y^2, {x -> 0, y -> 0}] - draws 2 plots\n\n```\n\n• Prev by Date: Re: Sum elements in a list with conditions\n• Next by Date: Re: Mathematica code for Kepler's radial velocity equation?\n• Previous by thread: Plot in function\n• Next by thread: Re: Plot in function"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7004557,"math_prob":0.8250324,"size":1386,"snap":"2020-24-2020-29","text_gpt3_token_len":502,"char_repetition_ratio":0.10636758,"word_repetition_ratio":0.0746888,"special_character_ratio":0.42857143,"punctuation_ratio":0.24340177,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941228,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-06T18:56:27Z\",\"WARC-Record-ID\":\"<urn:uuid:d0c098a5-ea07-45f9-9f86-e41ebbb284bf>\",\"Content-Length\":\"44801\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:663f7e6f-82b9-4a1c-9a5b-9a62169a2b45>\",\"WARC-Concurrent-To\":\"<urn:uuid:aec3b773-06c2-4ec0-8181-cbd07ea9871b>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"http://forums.wolfram.com/mathgroup/archive/2012/Oct/msg00198.html\",\"WARC-Payload-Digest\":\"sha1:WT3FYWBMKOPIYGW7UPYBOSTZDZHOU6GZ\",\"WARC-Block-Digest\":\"sha1:XONQ466NOI5H27AOI42LUUALXJRUIWSQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348517506.81_warc_CC-MAIN-20200606155701-20200606185701-00539.warc.gz\"}"} |
https://mathoverflow.net/questions/315648/permutations-pi-in-s-n-with-sum-k-1n-frac1k-pik-1 | [
"# Permutations $\\pi\\in S_n$ with $\\sum_{k=1}^n\\frac1{k+\\pi(k)}=1$\n\nLet $$S_n$$ be the symmetric group of all the permutations of $$\\{1,\\ldots,n\\}$$. Motivated by Question 315568 (http://mathoverflow.net/questions/315568), here I pose the following question.\n\nQUESTION: Is it true that for each integer $$n>5$$ we have $$\\sum_{k=1}^n\\frac1{k+\\pi(k)}=1$$ for some odd (or even) permutation $$\\pi\\in S_n$$?\n\nLet $$a_n$$ be the number of all permutations $$\\pi\\in S_n$$ with $$\\sum_{k=1}^n(k+\\pi(k))^{-1}=1$$. Via Mathematica, I find that $$\\begin{gather}a_1=a_2=a_3=a_5=0,\\ a_4=1,\\ a_6=7, \\\\ a_7=6,\\ a_8=30,\\ a_9=110, \\ a_{10}=278,\\ a_{11}=1332.\\end{gather}$$ For example, $$(1,4,3,2)$$ is the unique (odd) permutation in $$S_4$$ meeting our requirement for $$n=4$$; in fact, $$\\frac1{1+1}+\\frac1{2+4}+\\frac1{3+3}+\\frac1{4+2}=1.$$ For $$n=11$$, we may take the odd permutation $$(4,8,9,11,10,6,5,7,3,2,1)$$ since \\begin{align}&\\frac1{1+4}+\\frac1{2+8}+\\frac1{3+9}+\\frac1{4+11}+\\frac1{5+10}\\\\&+\\frac1{6+6}+\\frac1{7+5}+\\frac1{8+7}+\\frac1{9+3}+\\frac1{10+2}+\\frac1{11+1}\\end{align} has the value $$1$$, we may also take the even permutation $$(5, 6, 7, 11, 10, 4, 9, 8, 3, 2, 1)$$ to meet the requirement.\n\nPS: After my initial posting of this question, Brian Hopkins pointed out that A073112($$n$$) on OEIS gives the number of permutations $$p\\in S_n$$ with $$\\sum_{k=1}^n\\frac1{k+p(k)}\\in\\mathbb Z$$, but A073112 contains no comment or conjecture.\n\n• This sequence (with just one more term, 3312) is A073112 with the requirement that the sum is an integer. No citations or other interpretations, just PARI code. – Brian Hopkins Nov 19 '18 at 3:51\n• It seems that the identity maximises $\\sum_{k=1}^n\\frac{1}{k+\\pi(k)}$, so this maximum is strictly less than $2$ for small values of $n$, which explains the coincidence. If the identity indeed maximises the sum, the first time the two sequences might be different is for $n=31$. – Martin Rubey Nov 19 '18 at 7:23\n• Brief comment on your arXiv:1811.10503v1 preprint: Theorem 1.2 is very close to Theorem 1 (a) in math.stackexchange.com/a/2597700 (which has since become part of Theorem 1.1 in Johann Cigler's arXiv:1803.05164v2). Indeed, if I shift my numbers $0, 1, \\ldots, n-1$ by $1$, then my nimble permutation becomes a $\\pi \\in S_n$ such that each $i \\in \\left\\{1,2,\\ldots,n\\right\\}$ has the property that $i + \\pi\\left(i\\right)$ equals a power of $2$ minus $1$. Are your and my permutation related? – darij grinberg Nov 27 '18 at 3:13\n\nClaim: $$a_n>0$$ for all $$n\\geq 6\\quad (*)$$.\n\nProof: We use induction to prove $$(*)$$.\n\nWe have $$a_6,a_7,a_8,a_9,a_{10},a_{11}>0$$. Assume $$(*)$$ holds for all the integers $$\\in [6,n-1]$$. We want to show that $$a_n>0$$ for all $$n\\geq12$$.\n\nIf $$n$$ is an odd, let $$n=2m+1$$, we have $$m\\geq6$$, so by induction hypothesis, there exists $$\\pi\\in S_m$$ such that $$\\sum\\limits_{k=1}^{m}\\frac{1}{k+\\pi(k)}=1$$. Let \\begin{align*} \\sigma(2k+1)&=2m+1-2k\\quad\\text{for}\\quad k=0,1,2\\ldots, m,\\\\ \\sigma(2k)&=2\\pi(k)\\quad\\text{for}\\quad k=1,2,\\ldots, m. \\end{align*} Then $$\\sigma\\in S_{n}$$, and $$\\sum\\limits_{k=1}^{n}\\frac{1}{k+\\sigma(k)}=\\sum\\limits_{k=1}^{m}\\frac{1}{2k+\\sigma(2k)}+\\sum\\limits_{k=0}^{m}\\frac{1}{2k+1+\\sigma(2k+1)}\\\\ =\\frac{1}{2}\\sum\\limits_{k=1}^{m}\\frac{1}{k+\\pi(k)}+\\sum\\limits_{k=0}^{m}\\frac{1}{2k+1+(2m-2k+1)}=\\frac{1}{2}+(m+1)\\frac{1}{2m+2}=1.$$\n\nIf $$n$$ is an even, let $$n=2m$$, also we have $$m\\geq6$$ and there exists $$\\pi\\in S_m$$, such that $$\\sum\\limits_{k=1}^{m}\\frac{1}{k+\\pi(k)}=1$$. Let \\begin{align*} \\sigma(2k-1)&=2m+1-2k\\quad\\text{for}\\quad k=1,2,\\ldots,m, \\\\ \\sigma(2k)&=2\\pi(k)\\quad\\text{for}\\quad k=1,2,\\ldots,m. \\end{align*} Then $$\\sigma\\in S_{n}$$ and $$\\sum\\limits_{k=1}^{n}\\frac{1}{k+\\sigma(k)}=\\sum\\limits_{k=1}^{m}\\frac{1}{2k+\\sigma(2k)}+\\sum\\limits_{k=1}^{m}\\frac{1}{2k-1+\\sigma(2k-1)}=\\frac{1}{2}\\sum\\limits_{k=1}^{m}\\frac{1}{k+\\pi(k)}+ \\sum\\limits_{k=1}^{m}\\frac{1}{2k-1+(2m-2k+1)}=1.$$ Hence $$a_{n}>0$$.\n\nBy induction $$(*)$$ holds for all the $$n\\geq 6$$.\n\n• So we are done if the permutations for n=6~11 contain both odd ones and even ones. Is it true according to the computer? Also, following this idea, what is the best lower bound of a_n you can get? – Yifeng Huang Nov 27 '18 at 23:53"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6789067,"math_prob":1.0000082,"size":1414,"snap":"2020-45-2020-50","text_gpt3_token_len":563,"char_repetition_ratio":0.15248227,"word_repetition_ratio":0.0,"special_character_ratio":0.43917963,"punctuation_ratio":0.15929204,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-31T05:41:54Z\",\"WARC-Record-ID\":\"<urn:uuid:86c8c00e-1f64-4ddf-a2f3-e04fa9663ecc>\",\"Content-Length\":\"133975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1f1e21b-117a-4f69-9935-9012246778d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:db2377e0-c51f-4c31-a13a-325806cc8e43>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/315648/permutations-pi-in-s-n-with-sum-k-1n-frac1k-pik-1\",\"WARC-Payload-Digest\":\"sha1:YQ2LCZSZTWWTZPE7TGZLKYAEQXIUUM6N\",\"WARC-Block-Digest\":\"sha1:K6KRBNDALW5YTKKDVZET54OJYV7ES4J3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107912807.78_warc_CC-MAIN-20201031032847-20201031062847-00575.warc.gz\"}"} |
http://learn.e-limu.org/topic/view/?t=619&c=58 | [
"Application areas of algebra\n\nBelow are some of the areas where algebra has been applied:\n\n#### 1. Formula for calculating area of a Rectangle",
null,
"#### 3. Write an expression for the perimeter of the triangle below",
null,
"#### 4. Formula for calculating surface area of a sphere",
null,
"#### 5. Formula for calculating surface area of a Cube",
null,
""
]
| [
null,
"https://elimufeynman.s3.amazonaws.com/media/resources/algebra5.JPG",
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"https://elimufeynman.s3.amazonaws.com/media/resources/algebra7.JPG",
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"https://elimufeynman.s3.amazonaws.com/media/resources/algebra8.JPG",
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"https://elimufeynman.s3.amazonaws.com/media/resources/algebra9.JPG",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8765075,"math_prob":0.9919693,"size":343,"snap":"2020-10-2020-16","text_gpt3_token_len":69,"char_repetition_ratio":0.18289086,"word_repetition_ratio":0.11111111,"special_character_ratio":0.19825073,"punctuation_ratio":0.09375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997781,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-20T01:13:51Z\",\"WARC-Record-ID\":\"<urn:uuid:c0b55134-8885-4216-9537-98ed2bc681ad>\",\"Content-Length\":\"19628\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e84251c2-310a-4752-a880-0a047b9790a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:31cda198-9bea-4d11-ba58-b0efbec8025d>\",\"WARC-IP-Address\":\"35.156.144.157\",\"WARC-Target-URI\":\"http://learn.e-limu.org/topic/view/?t=619&c=58\",\"WARC-Payload-Digest\":\"sha1:WM7HZ54U6B32KLNVRAEIZK3KPAHG5U5R\",\"WARC-Block-Digest\":\"sha1:4XKL7DF2S2JAEL6CT4CDG5QNMSRTRR24\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875144498.68_warc_CC-MAIN-20200220005045-20200220035045-00324.warc.gz\"}"} |
http://www.broandsismathclub.com/2014/04/identifying-decimal-place-value.html | [
"## Preview\n\nThe decimal place value of a number is simply the value of that number defined by the place in which the digit is. To find the place value, first look at the decimal point, for example in the number 6.5. Then you have to put aside the numbers on the left side and the right side of the decimal point. The number or numbers on the left side would be the whole numbers. The numbers on the right side of the decimal point will be the decimals. Then find place values of the whole numbers which is ones, tens, hundreds, etc. Then find decimal place values of the decimal number. Remember they will be decimal place values of tenths, hundredths, thousandths, etc. To understand it better I prefer watching the video. Now try some of these out:\n\nSheet 1 Decimal Place Value\n\nSheet 2 Practice Problems on Decimal Place Value"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7880538,"math_prob":0.9993354,"size":851,"snap":"2020-45-2020-50","text_gpt3_token_len":186,"char_repetition_ratio":0.20661157,"word_repetition_ratio":0.039473683,"special_character_ratio":0.21034078,"punctuation_ratio":0.11299435,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99741644,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T02:00:19Z\",\"WARC-Record-ID\":\"<urn:uuid:b1282489-95b0-4c5e-8376-827bf61e63ab>\",\"Content-Length\":\"60122\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:25593687-5fcc-4b4a-9f04-8ac2fa6f48f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:6bbd956e-69ee-47ce-990e-4d4ae66c438a>\",\"WARC-IP-Address\":\"172.217.13.83\",\"WARC-Target-URI\":\"http://www.broandsismathclub.com/2014/04/identifying-decimal-place-value.html\",\"WARC-Payload-Digest\":\"sha1:7D4OVVRKSJYOED4ZKJ23LYLNWFC76P6C\",\"WARC-Block-Digest\":\"sha1:THKWZPRL4WMCURPNDMVZ7O6A3POIJSSA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107896048.53_warc_CC-MAIN-20201028014458-20201028044458-00463.warc.gz\"}"} |
https://optovr.com/grade-worksheets-color-math-quick-maths-tracing-activities-kindergarten-easy-5-year-4-homeschool-5th/ | [
"# Grade Worksheets Color Math Quick Maths Tracing Activities Kindergarten Easy 5 Year 4 Homeschool 5th",
null,
"Grade Worksheets Color Math Quick Maths Tracing Activities Kindergarten Easy 5 Year 4 Homeschool 5th.\n\nTeaching textbooks math math covers whole numbers, adding and subtracting with multiple digits, multiplication, rounding and estimation, fractions, decimals, money measurement, division, percents, number lines and more. (for the typical grader or an advanced grader.\n\n) math, teaching textbooks kit. Our math page picks out the top three math worksheets for homeschooling. here are some more good printable ones math fun worksheets. math fun worksheets has lots of different, colorful free worksheets for grade k through k (roughly ages to ), covering topics from numbers to measurement and word problems.\n\nGrade games consumer math worksheets animal skull coloring pages earth systems science 4 word problems algebra 1 curriculum craft sheets homeschool 5th. Math worksheets print homeschool 5th grade. Math worksheets games 4 homeschool 5th grade. Worksheet division worksheets grade mixed math free curriculum lesson plan 5 pure game private home tuition homeschool 5th. 1 million free worksheets kids homeschool math 5th grade. Grade child arithmetic kit math worksheets writing equations practice worksheet applied exam homeschool 5th. Printable cursive worksheet grade math worksheets graph program preschool curriculum grades school websites algebra questions kids homeschool 5th."
]
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null,
"https://optovr.com/images/grade-worksheets-color-math-quick-maths-tracing-activities-kindergarten-easy-5-year-4-homeschool-5th.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8606885,"math_prob":0.8459145,"size":1488,"snap":"2021-31-2021-39","text_gpt3_token_len":283,"char_repetition_ratio":0.20619947,"word_repetition_ratio":0.0952381,"special_character_ratio":0.17540322,"punctuation_ratio":0.1092437,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97856754,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-28T22:41:50Z\",\"WARC-Record-ID\":\"<urn:uuid:d2c726f0-504d-45f8-9094-9c3aefcfc1e5>\",\"Content-Length\":\"22082\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:401f9460-4fc2-4d99-bf31-1db1b40c634b>\",\"WARC-Concurrent-To\":\"<urn:uuid:75722f55-bed3-4adf-808a-296eda83d711>\",\"WARC-IP-Address\":\"104.21.5.91\",\"WARC-Target-URI\":\"https://optovr.com/grade-worksheets-color-math-quick-maths-tracing-activities-kindergarten-easy-5-year-4-homeschool-5th/\",\"WARC-Payload-Digest\":\"sha1:WJYB6ASTYKUNIDHMUYKUMDQTMSJOZ3W5\",\"WARC-Block-Digest\":\"sha1:T27PYXZ3JWFWVDOKMBGSF4YLHN2S6HAG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153803.69_warc_CC-MAIN-20210728220634-20210729010634-00073.warc.gz\"}"} |
https://www.measureevaluation.org/resources/publications/wp-03-69 | [
"How and When Should One Control for Endogenity Biases? Part I: The Impact of a Possibly Endogenous Explanatory Variable on a Continuous Outcome",
null,
"wp-03-69.pdf — PDF document, 1,016 kB (1,041,327 bytes)\n\nAuthor(s): Angeles G, Guilkey D K, Mroz T A\n\nYear: 2003\n\nAbstract:\nThe interpretation of coefficients estimates from ordinary least square regressions and other statistical models depends crucially on whether any explanatory variable in the statistical model is correlated with the ôerror termö influencing the outcome of interest. If there is a relationship between any explanatory variable and the unmeasured determinants of an outcome, then one usually cannot interpret any of the estimated coefficients as the impact of the corresponding covariate on the outcome of interest. In the medical and public health literature, this is often called the problem of confounding effects. In economics and sociology, one typically calls this the problem of endogenous regressors. Regardless of the label chosen for this relationship, the presence of a correlation between the measured and unmeasured determinants of an outcome results in biased estimators of the impacts of all covariates. In this paper we explore the severity of the possible biases that can arise when such correlations are present, and we examine the performance of some simple estimators that have been developed to reduce the bias. We start out by examining ordinary least square models with continuous outcomes and continuous regressors because most of the intuition about the problems and the solutions can be developed simply in that context. We then examine endogeneity problems and solutions for three other sets of models that researchers often encounter in practice: a continuous outcome influenced by an endogenous binary regressor; a binary (discrete) outcome determined by an endogenous continuous regressor; and a binary outcome being influenced by an endogenous binary regressor. In nearly all instances we focus on the estimation of the impact of the possibly endogenous regressor on the outcome of interest, but it is important to recognize that estimators for all effects in a model, not just those for the endogenous variables, usually are biased when any explanatory variable is endogenous. We also examine the performance of estimators in situations where the researcher cares about more than just the bias of the estimator.\n\nFiled under:\nSearch Publications\nMatch against Title and Abstract.\nMatch Author Names."
]
| [
null,
"https://www.measureevaluation.org/pdf.png",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.91792405,"math_prob":0.8165082,"size":2404,"snap":"2019-43-2019-47","text_gpt3_token_len":462,"char_repetition_ratio":0.1325,"word_repetition_ratio":0.005479452,"special_character_ratio":0.17845258,"punctuation_ratio":0.080684595,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97864354,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T05:42:53Z\",\"WARC-Record-ID\":\"<urn:uuid:77c539b8-4d2a-4fbb-8bac-6ddd4e830274>\",\"Content-Length\":\"97495\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d74b4ca3-37ff-44e0-8918-787aa68178e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:38318fb4-4245-44e7-84b9-01283d078399>\",\"WARC-IP-Address\":\"152.2.103.186\",\"WARC-Target-URI\":\"https://www.measureevaluation.org/resources/publications/wp-03-69\",\"WARC-Payload-Digest\":\"sha1:7B3R4XOQLU7QFID3V4DHP2OW3R53PUV2\",\"WARC-Block-Digest\":\"sha1:GT4KWITYNJZOXLNWLT4DTCNSSBPJBEPB\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987803441.95_warc_CC-MAIN-20191022053647-20191022081147-00417.warc.gz\"}"} |
https://pyfar.readthedocs.io/en/stable/modules/pyfar.dsp.fft.html | [
"# pyfar.dsp.fft¶\n\nThe following documents the FFT functionality. More details and background is given in the FFT concepts.\n\nFunctions:\n\n irfft(spec, n_samples, sampling_rate, fft_norm) Calculate the IFFT of a single-sided Fourier spectrum. normalization(spec, n_samples, sampling_rate) Normalize a Fourier spectrum. rfft(data, n_samples, sampling_rate, fft_norm) Calculate the FFT of a real-valued time-signal. rfftfreq(n_samples, sampling_rate) Returns the positive discrete frequencies for which the FFT is calculated.\npyfar.dsp.fft.irfft(spec, n_samples, sampling_rate, fft_norm)[source]\n\nCalculate the IFFT of a single-sided Fourier spectrum.\n\nThe function takes only the right-hand side of the spectrum and returns a real-valued time signal. The normalization is considered according to 'fft_norm' as described in normalization and FFT concepts.\n\nParameters:\n• spec (array, complex) – The complex valued right-hand side of the spectrum with dimensions (…, n_bins)\n\n• n_samples (int) – The number of samples of the corresponding time signal. This is crucial to allow for the correct transform of time signals with an odd number of samples.\n\n• sampling_rate (number) – sampling rate in Hz\n\n• fft_norm ('none', 'unitary', 'amplitude', 'rms', 'power', 'psd') – See normalization.\n\nReturns:\n\ndata – Array containing the time domain signal with dimensions (…, 'n_samples')\n\nReturn type:\n\narray, double\n\npyfar.dsp.fft.normalization(spec, n_samples, sampling_rate, fft_norm='none', inverse=False, single_sided=True, window=None)[source]\n\nNormalize a Fourier spectrum.\n\nApply normalizations defined in to the DFT spectrum. Note that the phase is maintained in all cases, i.e., instead of taking the squared absolute values for 'power' and 'psd', the complex spectra are multiplied with their absolute values to ensure a correct renormalization. For detailed information and explanations, refer to FFT concepts.\n\nParameters:\n• spec (numpy array) – N dimensional array which has the frequency bins in the last dimension. E.g., spec.shape == (10,2,129) holds 10 times 2 spectra with 129 frequency bins each.\n\n• n_samples (int) – number of samples of the corresponding time signal\n\n• sampling_rate (number) – sampling rate of the corresponding time signal in Hz\n\n• fft_norm (string, optional) –\n\n'none'\n\nDo not apply any normalization. Appropriate for energy signals such as impulse responses.\n\n'unitary'\n\nMultiply spec by factor of two as in Eq. (8) (except for 0 Hz and the Nyquist frequency at half the sampling rate) to obtain the single-sided spectrum.\n\n'amplitude'\n\nScale spectrum by 1/n_samples as in Eq. (4) to obtain the amplitude spectrum.\n\n’rms’\n\nScale spectrum by $$1/\\sqrt{2}$$ as in Eq.(10) to obtain the RMS spectrum.\n\n’power’\n\nPower spectrum, which equals the squared RMS spectrum (except for the retained phase).\n\n’psd’\n\nThe power spectrum is scaled by n_samples/sampling_rate as in Eq. (6)\n\nNote that the unitary normalization is also applied for amplitude, rms, power, and psd if the input spectrum is single sided (see single_sided).\n\n• inverse (bool, optional) – apply the inverse normalization. The default is False.\n\n• single_sided (bool, optional) – denotes if spec is a single sided spectrum up to half the sampling rate or a both sided (full) spectrum. If single_sided=True the unitary normalization according to Eq. (8) is applied unless fft_norm='none'. The default is True.\n\n• window (None, array like) – window that was applied to the time signal before performing the FFT. Affects the normalization as in Eqs. (11-13). The window must be an array-like with n_samples length and. The default is None, which denotes that no window was applied.\n\nReturns:\n\nspec – normalized input spectrum\n\nReturn type:\n\nnumpy array\n\nReferences\n\npyfar.dsp.fft.rfft(data, n_samples, sampling_rate, fft_norm)[source]\n\nCalculate the FFT of a real-valued time-signal.\n\nThe function returns only the right-hand side of the axis-symmetric spectrum. The normalization is considered according to 'fft_norm' as described in normalization and FFT concepts.\n\nParameters:\n• data (array, double) – Array containing the time domain signal with dimensions (…, 'n_samples')\n\n• n_samples (int) – The number of samples\n\n• sampling_rate (number) – sampling rate in Hz\n\n• fft_norm ('none', 'unitary', 'amplitude', 'rms', 'power', 'psd') – See documentation of normalization.\n\nReturns:\n\nspec – The complex valued right-hand side of the spectrum with dimensions (…, n_bins)\n\nReturn type:\n\narray, complex\n\npyfar.dsp.fft.rfftfreq(n_samples, sampling_rate)[source]\n\nReturns the positive discrete frequencies for which the FFT is calculated.\n\nIf the number of samples $$N$$ is even the number of frequency bins will be $$2/N+1$$, if $$N$$ is odd, the number of bins will be $$(N+1)/2$$.\n\nParameters:\n• n_samples (int) – The number of samples in the signal\n\n• sampling_rate (int) – The sampling rate of the signal\n\nReturns:\n\nfrequencies – The positive discrete frequencies in Hz for which the FFT is calculated.\n\nReturn type:\n\narray, double"
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https://www.projecteuclid.org/euclid.rmjm/1420071551 | [
"## Rocky Mountain Journal of Mathematics\n\n### Realizing infinite cardinal numbers via maximal chains of intermediate fields\n\n#### Abstract\n\nFor each nonzero cardinal number $\\kappa$, there is a field extension $\\mathbf{Q} \\subseteq L$ and a maximal chain $\\mathcal{H}_{\\kappa}$ of intermediate fields going from $\\mathbf{Q}$ to $L$ such that the cardinal number of $\\mathcal{H}_{\\kappa}$ is $\\kappa$. If $\\kappa$ is infinite, then for all infinite cardinals $\\nu%_1 \\lt \\nu_2 \\leq \\lt \\kappa$, it can be arranged that $\\mathcal{H}_{\\nu%_1 } \\subset %\\mathcal{H}_{\\nu_2} \\subseteq \\mathcal{H}_{\\kappa}$. However, there exists an infinite cardinal for which there does not exist a field extension $L/K$ such that $\\kappa$ is the supremum of the cardinalities of chains of intermediate fields going from $K$ to $L$. For a field extension $L/K$, this supremum of cardinalities has been denoted $\\lambda(L/K)$. If $L/K$ is infinitely generated, we reduce its calculation to set theory, as follows. Let $\\ala$ be the infimum of the cardinalities of generating sets of $L/K$. Let $\\Omega(\\aleph_{\\alpha})$ be the supremum of the cardinalities of chains %that consist of subsets of a set of cardinality $\\ala$. ($\\Omega(\\aleph_{\\alpha})$ is equal to what has been called $\\mbox{ded\\,}(\\aleph_{\\alpha})$ in the literature.) Then $\\ala\\lt \\oa\\leq \\p$; and if $\\alpha > 0$ (but not necessarily if $\\alpha =0$), then $\\lambda(L/K)=\\Omega(\\aleph_{\\alpha})$.\n\n#### Article information\n\nSource\nRocky Mountain J. Math., Volume 44, Number 5 (2014), 1471-1503.\n\nDates\nFirst available in Project Euclid: 1 January 2015\n\nhttps://projecteuclid.org/euclid.rmjm/1420071551\n\nDigital Object Identifier\ndoi:10.1216/RMJ-2014-44-5-1471\n\nMathematical Reviews number (MathSciNet)\nMR3295639\n\nZentralblatt MATH identifier\n1332.12009\n\n#### Citation\n\nDobbs, David E.; Heitmann, Raymond C. Realizing infinite cardinal numbers via maximal chains of intermediate fields. Rocky Mountain J. Math. 44 (2014), no. 5, 1471--1503. doi:10.1216/RMJ-2014-44-5-1471. https://projecteuclid.org/euclid.rmjm/1420071551"
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https://socratic.org/questions/5957629bb72cff185ef2d4b3#446574 | [
"Question #2d4b3\n\nJul 1, 2017\n\nThe kinetic energy will be halved.\n\nExplanation:\n\nKinetic energy is given by:\n\n${E}_{k} = \\frac{1}{2} \\cdot m \\cdot {v}^{2}$\n\nIt is proportional to the object's mass:\n\n$E \\propto m$\n\nIf we double the mass, we double the energy:\n\n${E}_{1} = \\frac{1}{2} \\cdot 1 \\cdot {v}^{2} = \\frac{1}{2} {v}^{2}$\n\n${E}_{2} = \\frac{1}{2} \\cdot 2 \\cdot {v}^{2} = 1 {v}^{2} = \\textcolor{b l u e}{2 {E}_{1}}$\n\nKinetic energy is also proportional to the square of the velocity:\n\n$E \\propto {v}^{2}$\n\nThis means that if we halve the velocity, we quarter the energy:\n\n${E}_{1} = \\frac{1}{2} \\cdot m \\cdot {1}^{2} = \\frac{1}{2} \\cdot m$\n\n${E}_{2} = \\frac{1}{2} \\cdot m \\cdot {\\left(\\frac{1}{2}\\right)}^{2} = \\frac{1}{2} \\cdot m \\cdot \\left(\\frac{1}{4}\\right) = \\frac{1}{8} \\cdot m = \\textcolor{b l u e}{\\frac{1}{4} \\cdot {E}_{1}}$\n\nIf we combine a doubled mass and halved velocity, our initial energy will be doubled and then quartered. This results in an overall halving of the energy:\n\n${E}_{2} = {E}_{1} \\cdot 2 \\cdot \\frac{1}{4} = \\frac{1}{2} {E}_{1}$"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.79668844,"math_prob":1.0000001,"size":577,"snap":"2022-05-2022-21","text_gpt3_token_len":134,"char_repetition_ratio":0.13612565,"word_repetition_ratio":0.0,"special_character_ratio":0.21490468,"punctuation_ratio":0.10810811,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999475,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T07:46:48Z\",\"WARC-Record-ID\":\"<urn:uuid:2ce12b36-ecfd-422d-85e9-7692047d208b>\",\"Content-Length\":\"32722\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dd48b916-b35e-4e56-ba37-80982f2ff324>\",\"WARC-Concurrent-To\":\"<urn:uuid:a10f2430-6433-48be-aefe-840bfc01c1d5>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/5957629bb72cff185ef2d4b3#446574\",\"WARC-Payload-Digest\":\"sha1:K2BKILK5DGVR25II36HH57H3XIDO2XHS\",\"WARC-Block-Digest\":\"sha1:U4WAMLFF53Z6LNMD3FDVWZLWMWPBAAOX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305423.58_warc_CC-MAIN-20220128074016-20220128104016-00263.warc.gz\"}"} |
https://dataanalysisrpythondaxsql.com/page/2/ | [
"Can load data into R using `read.csv() `to get data from a comma separated variable (CSV) file.\n\n`DataFrameName <- read.csv(MyDataFile.csv)`\n\nTidyverse package provides a lot of useful functions for manipulating data.\n\n• `head(DataFrameName)` – Shows first 6 rows of the data\n• `tail(DataFrameName)` – Shows last 6 rows of data\n• `view(DataFrameName)` – Shows all rows in tabular format\n\nCan address elements of a dataframe using square brackets or the dollar sign.\n\n• `DataFrameName[1 , 3]` – Show the value in the third column of the first row\n• `DataFrameName[ , 3]` – Show the value in the third column for all rows\n• `DataFrameName\\$ColumnName` – Show the value in the column ColumnName for all rows\n\nR supports UNIX style piping using `%>%` to represent the pipe. Below code installs the tidyverse package (line is commented out as only needs to be run once per installation of R then to will be available to all scripts), loads tidyverse into the session using `require()` and then loads the data from a CSV file into a dataframe called BusData using `read.csv()`. It then pipes the data through the select() function to just get 4 columns:\n\n• `Financial year` – The financial year being reported on (April to March in this case)\n• `Month` – The month within that financial year being reported on\n• `Total.Bus.Patronage.per.month..Concessionary..Non.Concessionary` – The total number of bus journeys taken in that month (both fare paying and free concessionary)\n• `Free.Concessionary.per.month` – The number of journeys where a free concessionary bus pass was used\n\nThe reduced dataset is then filtered to only return those rows for the Financial Year 2008 (April 2008 to March 2009).\n\n``````# install.packages(\"tidyverse\")\nrequire(\"tidyverse\")\nBusData %>%\nselect(Financial.year, Month, Total.Bus.Patronage.per.month..Concessionary...Non.Concessionary., Free.Concessionary.per.month) %>%\nfilter(Financial.year == 2008)``````\n\nThis is the equivalent of the SQL statement (some column names changed to be legal in common RDBMSes):\n\n``````select FinancialYear\n, Month\n, TotalMonthlyBusJournies\n, TotalMonthlyConcessionaryJournies\nfrom BusData\nwhere Financialyear=2008;```\n```\n\nThe data set used here is based on an open data set downloaded from Birmingham Data Factory and then manipulated in Excel.\n\nThe `arrange()` function could be used to sort the output.\n\n## Date Formats – Lessons Not Learned\n\nAn article from BATIMES popped up in my inbox this morning, the author raises the issue of varying date formats causing problems of misinterpretation. Different countries, different companies within the same country, different departments in the same company and different systems within the same department use differing formats. Even different reports from the same system use different formats (PlanView, I’m looking at you). Worst case I’m aware of is there’s a report I produce daily in a shared Excel spreadsheet where most people who use it want the DD/MM/YYYY format but one person insists on changing the format to M/D/YY whenever he looks at a tab.\n\nThe author of the article points out that we’re slipping back into the 2 digit year issue that cause problems with Y2K and that combined with different formats leads to her wonder if 11/05/18 is 11th May 2018 (DD/MM/YY), 5th November 2018 (MM/DD/YY) or 18th May 2011 (YY/MM/DD). Without a standard format, that is universally used, we can not be sure, especially going forward where the data may be separated from any accompanying documentation.\n\nLooking at the datasets I have to deal with often the more fundemental problem is ensuring that users put only the data they are supposed to in a field. ‘N/A’ is not a valid date format in any country, nor is ‘Will provide next week’.\n\n## Data Cleansing m/d/yyyy Dates\n\nA common issue I’ve found with data cleansing when pulling into Excel is the date format wil often default to US format. If it’s a 2 digit month and day (mm/dd/yyyy) then that’s not much of a problem but where it can be one or two digit (m/d/yyyy so 1/1/2020 and 10/10/2020 are both possiblilities) then it can be more tricky. I found this page which describes a way to fix the issue. It seems like a kludge but when I tried it it did work.\n\n## First bash at R\n\nThe book I’m starting on is “Statistics for Linguists – An Introduction Using R” by Bodo Winter. This selection was made purely because I heard him speak at BirminghamR meetup and he mentioned that the book had just come out and was aimed at undergraduates. I’ve also got the O’Reilly “R for Data Science” book because it was recommended by someone at work. I’ve tried learning R before from a book but struggled because the author went straight into a complex scenario about feeding the Chinese army so the R got lost in the scenario. I was looking for the R equivalent of ‘Hello, world!’, not writing a full UNIX Kernel from scratch.\n\nSo after plodding through for about an hour yesterday I’ve discovered that if you type `2 + 2` then R responds ` 4` and if you type `sqrt(4)` it unsurprisingly responds ` 2`. The number in the square brackets means that this is the first element of a vector, all variables in R are vectors.\n\nVector seems to mean something slightly different here to what I learned in school. In school we learned that vectors have scale and direction, as opposed to a scalar which has only scale; so 100 miles is a scalar but 100 miles north is a vector. In R vector seems to be what in programming would be called an array. Also, rather worryingly, vectors seem not to be typed, that is you can’t define what type of data a vector should store so you could put a 1 in the vector x then put the letter ‘a’ in the same x. Could be an issue if later you want to do `sqrt(PI*X^2)`.\n\nI also discovered in that hour that `abs(x)` will return the absolute value of x, that is the unsigned value so if `x==2 `or `x==-2 `then `abs(x)` will be `2`.\n\nIf you want to see what vectors you have then you need to use `ls()`, which kind of makes sense to me as `ls` is the UNIX shell command to list the files in a directory. To see what is stored in each vector just type the vector name at the prompt.\n\nAssigning a value to a vector is done using `<-` which reminds me somewhat of C++ where you would use `<<` to pass variables into a string for output. So to put the number 4 into x (or more properly the first element of the verctor x) you use `x <- 4`, or the plain single equals sign also works but it’s not standard and most scripts written by others will use `<- `so you need to get use to it and unless you want to get confused, and confuse others, use `x <- 4` not `x = 4`.\n\nTo put more than one value into a vector in a single command you can use the `c() `function. So x <- c(1, 3, 5, 7, 9) puts the numbers 1, 3, 5, 7 and 9 into x as the 1st, 2nd, 3rd, 4th and 5th elements. Unlike most programming languages, which start arrays at element 0, R starts it’s vectors at element 1, there is no zero. If you want to put a range of numbers (e.g. the numbers 1 to 10) into a vector you use a colon between them such as `x <- 1:10`, you don’t need the `c()` in this case.\n\nThe book also introduced a number of functions that work with vectors:\n\n`sum(x)` – Adds up all of the elements of x\n\n`min(x)` – returns the smallest value of the elements of x\n\n`max(x)` – returns the largest value of the elements of x\n\n`range(x)` – returns the smallest and largest elements of x\n\n`diff(range(x))` – returns the difference between the smallest and largest elements of x\n\n`mean(x)` – returns the mean (average) of the values of the elements of x\n\n`median(x)` – returns the median (another type of average) of the values of the elements of x\n\n`var(x)` – returns the variance of the values of the elements of x\n\n`sd(x)` – returns the standard deviation of the values of the elements of x, if memory serves that means that `sd(x) == sqrt(var(x))`\n\n`length(x)` – returns the number of elements in x\n\nIf you need to address a specific element of a vector you can use the square bracket notation. So, `x` is the first element and` x` is the 47th element. the numbers in the square brackets are referred to as the index of the element, so the first element has an index of 1 and the 47th and index of 47. If you want to return a range of elements then you can use the colon notation again, with your square brackets. So, to get the first 4 elements of x use `x[1:4]`. A negative number in the square brackets returns every element except that one. So, x[-2] returns every element except the second element.\n\nIf you carry out a mathematical operation on a vector with more than one element then that will be carried out on each element individually so x *5 will return a vector containing the same number of elements as x but with each value in x multiplied by five. Note that x itself is not changed so if you want to use the result you will need to assign it to another vector, e.g. `z <- x^2` will square each value in x and assign it to a corresponding element in z. So if x holds the radii of some circles and you want to calculate the areas of those circles then `z <- (pi*x^2)` will populate z with the areas of the circles whose radii are stored in x.\n\nR also includes the usual crop of comparison operators:\n\n`x==y `– returns TRUE or FALSE for if each element of x is equal to the corresponding element of y\n\n`x > y` – returns TRUE or FALSE for if each element of x is greater than the corresponding element of y\n\n`x < y` – returns TRUE or FALSE for if each element of x is less than the corresponding element of y\n\n`x >= y` – returns TRUE or FALSE for if each element of x is greater than or equal to the corresponding element of y\n\n`x <= y` – returns TRUE or FALSE for if each element of x is less than or equal to the corresponding element of y\n\n`x != y` – returns TRUE or FALSE for if each element of x is not equal to the corresponding element of y\n\nYou can use these to filter the values in a vector and assign the indexes of the values for which the conmditional is true to another vector. So, `morethanthree <- x > 3` will populate the vector morethenthree with the indices for those elements of x that are more than 3.\n\nI can see that being useful where you want to work with only a subset of data that meets specific criteria. For example if you had surveyed people as to how many dogs they owned (which could be zero) and put the results in a vector Dogs then you could answer the question “Given someone owns at least one dog, what is the average (mean) number of dogs owned?” with\n\n`HasDogs <- Dogs > 0`\n\n`mean(Dogs[HasDogs])`\n\nYou could also work out the percentage of people who own at least 1 dog with `(length(Dogs[HasDogs])/length(Dogs))*100`\n\n## The oldest apprentice in town (probably)\n\nMy employer, keen to get some value out of the apprenticeship levy, has decided to put a number of us on appreticeship programmes (I suppose it’s cheaper than regular training). In my case, a few months short of turning 50, that means a Level 4 Data Analysis diploma because I know a bit about Excel.\n\nI had my induction last week and am still waiting to see if the training materials will be sent through so I’ve spent too much of my own hard earned cash on a second hand laptop and a book on learning R. After R it will mostly likley be Python and probably PowerBI."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9240722,"math_prob":0.93207073,"size":6349,"snap":"2022-27-2022-33","text_gpt3_token_len":1558,"char_repetition_ratio":0.1640662,"word_repetition_ratio":0.091935486,"special_character_ratio":0.24633801,"punctuation_ratio":0.060538117,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97989863,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T23:30:59Z\",\"WARC-Record-ID\":\"<urn:uuid:7c9e7a27-e977-40b6-80e2-0256e305792f>\",\"Content-Length\":\"78858\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a147873-4d54-44c2-a5a4-f77b47ac6316>\",\"WARC-Concurrent-To\":\"<urn:uuid:f0991796-a87f-4671-bc93-8d0cdc5ad799>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://dataanalysisrpythondaxsql.com/page/2/\",\"WARC-Payload-Digest\":\"sha1:OYNJTV33V4PXF44DGJWZ6XWAMYVB7PTO\",\"WARC-Block-Digest\":\"sha1:LAR23BV3VMPEFBWPWTGB3GZRYLK5CHMS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571090.80_warc_CC-MAIN-20220809215803-20220810005803-00716.warc.gz\"}"} |
https://www.proprofs.com/quiz-school/story.php?title=chapter-11-consumer-mathematics-3 | [
"# Chapter 11: Consumer Mathematics 3\n\n30 Questions | Total Attempts: 51",
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"Settings",
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"^^\n\n• 1.\nWhat is the Simple Interest?\n• A.\n\nThe interest that is equal to the principal times the rate times the time\n\n• B.\n\nA guideline in which the year is considered to have 360 days\n\n• C.\n\nThe use of capital for income or profit\n\n• 2.\nThe use of capital for income or profit\n• A.\n\nBanker's Rule\n\n• B.\n\nInvestment\n\n• C.\n\nMortgage\n\n• 3.\nThe method of charging interest on a credit card that is usually in the best interest of the consumer\n• A.\n\nFinance Charge\n\n• B.\n\nClosing Costs\n\n• C.\n\nArg.Daily Bal. Method\n\n• 4.\nThe money that a bank is willing to give you\n• A.\n\nCredit\n\n• B.\n\nPrincipal\n\n• C.\n\nMortgage\n\n• 5.\nThe amount of the cash that a buyer must prepay on an item in order to receive a loan or mortgage\n• A.\n\nMortgage\n\n• B.\n\nSimple interest\n\n• C.\n\nDown payment\n\n• 6.\nA list or table that gives the payment number in a loan and the breakdown of how much money is paid to principal and how much to interest for each payment\n• A.\n\nAmortization schedule\n\n• B.\n\nPercent\n\n• C.\n\nPrincipal\n\n• 7.\nThe total amount of money a borrower must pay to use a lender's money\n• A.\n\nFixed Investment\n\n• B.\n\nFinance charge\n\n• C.\n\nCredit\n\n• 8.\nThe type of interest that allows the interest to earn interest\n• A.\n\nCompound interesr\n\n• B.\n\nBanker's Rule\n\n• C.\n\nAPY\n\n• 9.\nA type of loan where the interest is paid at the time the borrower receives the loan\n• A.\n\nDiscount note\n\n• B.\n\nAmortization Schedule\n\n• C.\n\nARM\n\n• 10.\nA guideline in which the year is considered to have 360 days\n• A.\n\nFixed Investment\n\n• B.\n\nPrincipal\n\n• C.\n\nBanker's Rule\n\n• 11.\nA person other than the borrower who will guatantee the repayment of a loan\n• A.\n\nInvestment\n\n• B.\n\nCosigner\n\n• C.\n\nInterest\n\n• 12.\nA ratio of some number to 100\n• A.\n\nPercent\n\n• B.\n\nInterest\n\n• C.\n\nAPY\n\n• 13.\nThe difference between the appraised value of a home and the principal balance remaining on the mortgage\n• A.\n\nFixed Investment\n\n• B.\n\nVariable Investment\n\n• C.\n\nEquity\n\n• 14.\nThe primary method used to compute unearnd on an installment loan\n• A.\n\nUnpaid balance method\n\n• B.\n\nActuarial method\n\n• C.\n\nArg.Daily Bal method\n\n• 15.\nOne percent of the loan mortgage\n• A.\n\nPoint\n\n• B.\n\nPercent\n\n• C.\n\nAPY\n\n• 16.\nThe method of charging interest on a credit card in which interest is only paid on the previous outstanding balance\n• A.\n\nActuarial method\n\n• B.\n\nArg. daily Bal method\n\n• C.\n\nUnpaid balance method\n\n• 17.\nA US supreme court decision that specified how partial payments were to be applied to a loan\n• A.\n\nUS rule\n\n• B.\n\nBanker's Rule\n\n• C.\n\nPartial payment\n\n• 18.\nThe collateral that is pledged by the borrower to the lender that the lender may sell or keep if the borrower defaults on the loan\n• A.\n\nInterest\n\n• B.\n\nSercurity\n\n• C.\n\nPrincipal\n\n• 19.\nA long-term loan usually used to purchase a house\n• A.\n\nEquity\n\n• B.\n\nAmortization Schedule\n\n• C.\n\nMortgage\n\n• 20.\nThe simple interest rate that gives the name amount of interest over the same period of time as a compound rate\n• A.\n\nARM\n\n• B.\n\nAPY\n\n• C.\n\nAPR\n\n• 21.\nA type of investment in which the principal is guarantee and the interest is computed at a fixed rate\n• A.\n\nFixed Investment\n\n• B.\n\nVariable Investment\n\n• C.\n\nInvestment\n\n• 22.\nThe true rate of interest charged for a loan\n• A.\n\nARM\n\n• B.\n\nAPR\n\n• C.\n\nAPY\n\n• 23.\nA type of investment, such as stocks, in which the investor has a chance of losing money\n• A.\n\nClosing costs\n\n• B.\n\nFixed investment\n\n• C.\n\nVariable Investment\n\n• 24.\nThe money that is paid by the borrower for the use of the lender's money\n• A.\n\nPrincipal Payment\n\n• B.\n\nInterest\n\n• C.\n\nCredit\n\n• 25.\nA credit card is the most common type of this loan\n• A.\n\nOpen -end Installment loan\n\n• B.\n\nTotal Installment price\n\n• C.\n\nDown Payment\n\nRelated Topics",
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http://journals.math.ac.vn/acta/index.php?option=com_content&view=article&id=259%3Avol-27-no-1-2002&catid=22%3Aopen-access-issues&Itemid=38 | [
"# Acta Mathematica Vietnamica\n\n### OPEN ACCESS ISSUES",
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"## Vol. 27 No. 1, 2002\n\n Tran Kim Thanh, An extension of Renyi’s characteristic theorem to two-sided exponential distributionspdf 1 Ton That Tri, On the first occurence of irreducible representations of the matrix semigrouppdf 5 Nguyen Ba Minh and Nguyen Xuan Tan, On the continuity of vector convex multivalued functions pdf 13 W. A. Kirk, Approximating solutions of the equation x=T(x,x)pdf 27 Do Van Luu and Le Minh Tung, Nonsmooth B-preinvex functions pdf 33 Pham Tra An, Automata with a time-variant structure and supply-demand theorems pdf 41 Bui Trong Kien, The normalized duality mapping and two related characteristic properties of a uniformly convex Banach space pdf 53 Pham Xuan Du and Duong Minh Duc, A non-homogeneous p-Laplace equation in border case pdf 69 Nguyen Thi Bach Kim, Efficiency equivalent polyhedra for the feasible set of multiple objective linear programing pdf 77 Cao Van Nuoi, Generalized translation operators and their related Markov processespdf 87 M. Tsuji and Nguyen Duy Thai Son, Geometric solutions of nonlinear second order hyperbolic equations pdf 97\nYou are here:"
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https://sklearn.apachecn.org/docs/0.21.3/31.html | [
"# 3.2. 调整估计器的超参数\n\nestimator.get_params()\n\n\n• 估计器(回归器或分类器，例如 sklearn.svm.SVC())\n• 参数空间\n• 搜寻或采样候选的方法\n• 交叉验证方案\n• 计分函数\n\n## 3.2.1. 网格追踪法–穷尽的网格搜索\n\nGridSearchCV 提供的网格搜索从通过 param_grid 参数确定的网格参数值中全面生成候选。例如，下面的 param_grid:\n\nparam_grid = [\n{'C': [1, 10, 100, 1000], 'kernel': ['linear']},\n{'C': [1, 10, 100, 1000], 'gamma': [0.001, 0.0001], 'kernel': ['rbf']},\n]\n\n\nGridSearchCV 实例实现了常用估计器 API：当在数据集上“拟合”时，参数值的所有可能的组合都会被评估，从而计算出最佳的组合。\n\n## 3.2.2. 随机参数优化\n\n• 可以选择独立于参数个数和可能值的预算\n• 添加不影响性能的参数不会降低效率\n\n{'C': scipy.stats.expon(scale=100), 'gamma': scipy.stats.expon(scale=.1),\n'kernel': ['rbf'], 'class_weight':['balanced', None]}\n\n\nscipy 0.16之前的版本不允许指定随机状态，而是使用np.random来产生随机状态，或是使用np.random.set来指定状态．然而在scikit-learn 0.18以后，如果scipy版本大于0.16，则sklearn.model_selection可以通过用户指定来获得随机状态\n\n• Bergstra, J. and Bengio, Y., Random search for hyper-parameter optimization, The Journal of Machine Learning Research (2012)\n\n## 3.2.3. 参数搜索技巧\n\n### 3.2.3.2. 为评估指定多个指标\n\nGridSearchCVRandomizedSearchCV 允许为评分参数指定多个指标。\n\n### 3.2.3.5. 并行机制\n\nGridSearchCVRandomizedSearchCV 可以独立地评估每个参数设置。如果您的OS支持,通过使用关键字 n_jobs=-1 可以使计算并行运行。 有关详细信息, 请参见函数签名。\n\n## 3.2.4. 暴力参数搜索的替代方案\n\n### 3.2.4.1. 模型特定交叉验证\n\nlinear_model.ElasticNetCV([l1_ratio, eps, …]) Elastic Net model with iterative fitting along a regularization path\nlinear_model.LarsCV([fit_intercept, …]) Cross-validated Least Angle Regression model\nlinear_model.LassoCV([eps, n_alphas, …]) Lasso linear model with iterative fitting along a regularization path\nlinear_model.LassoLarsCV([fit_intercept, …]) Cross-validated Lasso, using the LARS algorithm\nlinear_model.LogisticRegressionCV([Cs, …]) Logistic Regression CV (aka logit, MaxEnt) classifier.\nlinear_model.MultiTaskElasticNetCV([…]) Multi-task L1/L2 ElasticNet with built-in cross-validation.\nlinear_model.MultiTaskLassoCV([eps, …]) Multi-task L1/L2 Lasso with built-in cross-validation.\nlinear_model.OrthogonalMatchingPursuitCV([…]) Cross-validated Orthogonal Matching Pursuit model (OMP)\nlinear_model.RidgeCV([alphas, …]) Ridge regression with built-in cross-validation.\nlinear_model.RidgeClassifierCV([alphas, …]) Ridge classifier with built-in cross-validation.\n\n### 3.2.4.2. 信息标准\n\nlinear_model.LassoLarsIC([criterion, …]) Lasso model fit with Lars using BIC or AIC for model selection\n\n### 3.2.4.3. 出袋估计\n\nensemble.RandomForestClassifier([…]) A random forest classifier.\nensemble.RandomForestRegressor([…]) A random forest regressor.\nensemble.ExtraTreesClassifier([…]) An extra-trees classifier.\nensemble.ExtraTreesRegressor([n_estimators, …]) An extra-trees regressor.\nensemble.GradientBoostingClassifier([loss, …]) Gradient Boosting for classification.\nensemble.GradientBoostingRegressor([loss, …]) Gradient Boosting for regression.",
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"Copyright © ibooker.org.cn 2019 all right reserved，由 ApacheCN 团队提供支持该文件修订时间： 2019-08-21 04:54:20"
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https://www.convertunits.com/from/milliliter/to/decaliter | [
"## ››Convert milliliter to decalitre\n\n milliliter decaliter\n\nHow many milliliter in 1 decaliter? The answer is 10000.\nWe assume you are converting between milliliter and decalitre.\nYou can view more details on each measurement unit:\nmilliliter or decaliter\nThe SI derived unit for volume is the cubic meter.\n1 cubic meter is equal to 1000000 milliliter, or 100 decaliter.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between milliliters and decaliters.\nType in your own numbers in the form to convert the units!\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from decaliter to milliliter, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Millilitre\n\nThe millilitre (ml or mL, also spelled milliliter) is a metric unit of volume that is equal to one thousandth of a litre. It is a non-SI unit accepted for use with the International Systems of Units (SI). It is exactly equivalent to 1 cubic centimetre (cm³, or, non-standard, cc).\n\n## ››Definition: Decalitre\n\nThe SI prefix \"deca\" represents a factor of 101, or in exponential notation, 1E1.\n\nSo 1 decalitre = 101 liters.\n\nThe definition of a litre is as follows:\n\nThe litre (spelled liter in American English and German) is a metric unit of volume. The litre is not an SI unit, but (along with units such as hours and days) is listed as one of the \"units outside the SI that are accepted for use with the SI.\" The SI unit of volume is the cubic metre (m³).\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!"
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https://qsstudy.com/explain-electric-potential-of-earth/ | [
"Physics\n\n# Explain Electric Potential of Earth",
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"Earth is an electric conductor. When a charged body is connected to the earth, it becomes electrically neutral. When a positively charged body is grounded electrons coming from the earth neutralize the body. Setting the potential of the Earth equal to zero is nothing more than a mathematical assumption for electrical engineering purposes. When a negatively charged body is grounded electrons from the body flow to the earth and the body becomes neutral. In reality, there is nothing such as an absolute potential. We only talk about the potential difference.\n\nThe earth is so big that if the charge is added or taken away from it its potential does not change at all. Likewise, if the water is taken away from sea or poured in the water level does not change. The earth is always taking charge from different bodies and simultaneously it supplies charge to other bodies. Hence earth is considered charge less. To determine the height of a place the height of the sea level is taken as zero, similarly to determine the potential of a body, the potential of the earth is taken as zero. Earth potential is taken as zero to make calculations easier, like zero deg cel temperature and zero bar gauge pressure.\n\nZero, Positive and Negative Potential:\n\nEarth is not in zero potential, the earth is at lower potential compared to other potential sources found on earth. The potential of an uncharged conductor is taken as zero When a charged conductor is connected to the earth its potential becomes zero. Because, in the connected state, both the conductor and the earth is considered as a single conductor. The potential of a positively charged body is positive and negatively charged body is negative.\n\nEarth is simply a commonly used reference point. Electric potential, the amount of work needed to move a unit charge from a reference point to a specific point against an electric field. The point-values of the potential are not observable. You can only observe differences in the potential. Typically, the reference point is Earth, although any point beyond the influence of the electric field charge can be used. You can for example always add a constant to the potential and have the same electric field, so that the potential is zero at a point can’t have physical significance.\n\nHowever, to add to the confusion, the electrical potential of Earth is not the same everywhere. One end of your street could have an excess of electric charge with respect to the other end of your street, so if you grounded two different circuits on each end of your street and connected them together, you could end up with this charge flowing from one circuit to the other."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9490707,"math_prob":0.9823721,"size":2654,"snap":"2022-27-2022-33","text_gpt3_token_len":503,"char_repetition_ratio":0.17584905,"word_repetition_ratio":0.008888889,"special_character_ratio":0.18801808,"punctuation_ratio":0.088,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96823174,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T23:48:21Z\",\"WARC-Record-ID\":\"<urn:uuid:f4ef44e2-90d0-4011-b112-31b780286d88>\",\"Content-Length\":\"23351\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49508308-e2fc-4b2f-bcd4-8419060c0667>\",\"WARC-Concurrent-To\":\"<urn:uuid:b4d70e70-8e10-49b8-9d44-35c1347bc80f>\",\"WARC-IP-Address\":\"178.63.81.55\",\"WARC-Target-URI\":\"https://qsstudy.com/explain-electric-potential-of-earth/\",\"WARC-Payload-Digest\":\"sha1:UJMWT42QNY7JCL6PFOW5WSDEDQYGJWAB\",\"WARC-Block-Digest\":\"sha1:KL4EXS24VQCJBRBXPLGENLEGDCXUJXA3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104277498.71_warc_CC-MAIN-20220703225409-20220704015409-00023.warc.gz\"}"} |
https://electronics.stackexchange.com/questions/251627/choosing-the-right-fet-to-switch-between-two-sources | [
"# Choosing the right FET to switch between two sources\n\nThe circuit below is based on the answer to this question, which i find to be suitable for my application.\n\n3.3VDC is the main supply for the circuit, with 3.6V 1/2AA battery for backup, as defined by the product requirement. The load is basically three switches which should be powered by the main supply until there is an outage, in which case the battery takes over.\n\nIn seletcting the Diode, i picked the panasonic DB3X313N with a Vf of 0.55V, this will drope the 3.3V supply to 2.75V which will still work, as the logic input will detect a high.\n\nFor the FET, I choose Infineon-BSS84P-DS , with a Vgs(th) of 1-2V.\n\nThe load is basically three switches with 10k pull up resistors, so the currrent draw is minimal.\n\n1. Will this circuit function as expected, i.e. to supply the load from the 3.3V source and only source from the 3.6V battery when the mains source is unavailable.\n2. If so, then is my choice of diode and FET suitable for this circuit, expecially with the FET, as i've not worked with FETs before.",
null,
"• What was the reason for choosing the FET? What is the maximum input voltage from the Battery – User323693 Aug 10 '16 at 10:50\n• @Umar I choose the FET because the battery voltage (which is 3.6V) is higher than the main source voltage which is 3.3V. Without it, the battery will be the one feeding the circuit by default rather than the 3.3V source. – TiOLUWA Aug 10 '16 at 10:53\n• Right. the Gate source voltage of the FET at room temperature is -2 V maximum. So, should be fine. The reverse current of chosen diode is in 10 uA range. If the voltage after the diode drop is acceptable, the option is okay. What is the load current expected from the battery? – User323693 Aug 10 '16 at 10:59\n• \"Will this circuit function as expected?\" - what are your expectations? – Andy aka Aug 10 '16 at 11:35\n• Forward voltage drop will not be 0.55V. Just look to datasheet. At 25°C and 10mA it will be around 0.25V – Chupacabras Aug 12 '16 at 5:57\n\nThe device you have chosen is not fully specified for $R_{DS(on)}$ below -4.5V",
null,
"There is no guarantee that the device will turn on fully with 3.6V of bias, and even if it does then you can expect a relatively high value of $R_{DS}$ of the order of $10 \\Omega \\ to \\ 20 \\Omega$ which will give a voltage drop of 100mV to 200mV across the FET.\n\nI would be more inclined to use a part that is fully specified for a lower $V_{GS}$ such as the DMG3415U or perhaps the Si2323CDS, both of which are fully characterised at 2.5V $V_{GS}$",
null,
"Either of these parts is guaranteed to turn on at the 3.6V bias you have with low on resistance; many others exist and these are simply suggestions.\n\nBasically, look for devices specified for $R_{DS(ON)}$ preferably with $R_{DS}\\ \\le 100m\\Omega$ at $V_{GS} \\le$ 2.5V"
]
| [
null,
"https://i.stack.imgur.com/X81zx.jpg",
null,
"https://i.stack.imgur.com/gLsYx.png",
null,
"https://i.stack.imgur.com/tgGO8.png",
null
]
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https://www.ics.uci.edu/~dechter/publications/r264.html | [
"[an error occurred while processing this directive]\n\n Publications & Technical Reports",
null,
"R264 Submodel Decomposition Bounds for Influence Diagrams Junkyu Lee, Radu Marinescu, and Rina Dechter.\n Abstract Influence diagrams (IDs) are graphical models for representing and reasoning with sequential decision-making problems under uncertainty. Limited memory influence diagrams (LIMIDs) model a decision-maker (DM) who forgets the history in the course of making a sequence of decisions. The standard inference task in IDs and LIMIDs is to compute the maximum expected utility (MEU), which is one of the most challenging tasks in graphical models. We present a model decomposition framework in both IDs and LIMIDs, which we call submodel decomposition. It presents a tree clustering scheme that identifies a tree of single-stage decision problems. We develop a valuation algebra over the submodels leading to a hierarchical message passing algorithm that propagates conditional expected utility functions over the submodel-tree as external messages while it uses internal message passing to generate those submodel external messages. The scheme’s complexity is bounded by the maximum tree-width over the submodels, as is common in graphical model algorithms. Finally, we present a new method for computing upper bounds over a submodel-tree, by first exponen- tiating the utility functions yielding a standard probabilistic graphical model as an upper bound and then applying standard variational upper bounds for the marginal MAP inference, yielding bounds to the MEU that are tighter compared with state-of-the-art bounding schemes. [pdf]"
]
| [
null,
"https://www.ics.uci.edu/~dechter/images/black-fill.gif",
null
]
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http://jinke.me/leetcode/2018-11-06-138/ | [
"Copy List with Random Pointer\n\n1. 链表经典题目。\n\n# 代码\n\n``````/**\n* Definition for singly-linked list with a random pointer.\n* struct RandomListNode {\n* int label;\n* RandomListNode *next, *random;\n* RandomListNode(int x) : label(x), next(NULL), random(NULL) {}\n* };\n*/\nclass Solution {\npublic:\nRandomListNode *copyRandomList(RandomListNode *head) {\nreturn nullptr;\n}\nRandomListNode* curr = head;\nwhile (curr) {\nRandomListNode* replica = new RandomListNode(curr->label);\nreplica->next = curr->next;\ncurr->next = replica;\ncurr = replica->next;\n}\nwhile (curr) {\nif (curr->random) {\ncurr->next->random = curr->random->next;\n}\ncurr = curr->next->next;\n}\nRandomListNode dummy(-1);\nRandomListNode* res_head = &dummy;"
]
| [
null
]
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https://hackage.haskell.org/package/statistics-0.10.1.0/docs/Statistics-Distribution-Gamma.html | [
"statistics-0.10.1.0: A library of statistical types, data, and functions\n\nPortability portable experimental [email protected]\n\nStatistics.Distribution.Gamma\n\nContents\n\nDescription\n\nThe gamma distribution. This is a continuous probability distribution with two parameters, k and ϑ. If k is integral, the distribution represents the sum of k independent exponentially distributed random variables, each of which has a mean of ϑ.\n\nSynopsis\n\n# Constructors\n\nArguments\n\n :: Double Shape parameter. k -> Double Scale parameter, ϑ. -> GammaDistribution\n\nCreate gamma distribution. Both shape and scale parameters must be positive.\n\n# Accessors\n\nShape parameter, k.\n\nScale parameter, ϑ."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5439625,"math_prob":0.8146818,"size":867,"snap":"2020-10-2020-16","text_gpt3_token_len":183,"char_repetition_ratio":0.24449594,"word_repetition_ratio":0.017857144,"special_character_ratio":0.20069204,"punctuation_ratio":0.2248062,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97014594,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-30T05:10:27Z\",\"WARC-Record-ID\":\"<urn:uuid:22b0673e-7840-4223-8689-22d27afd6627>\",\"Content-Length\":\"8172\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30b10de9-38b0-4c8a-aec8-7c7831552874>\",\"WARC-Concurrent-To\":\"<urn:uuid:dbdaba8d-bbef-48d0-8ce7-73d58d9c8834>\",\"WARC-IP-Address\":\"151.101.248.68\",\"WARC-Target-URI\":\"https://hackage.haskell.org/package/statistics-0.10.1.0/docs/Statistics-Distribution-Gamma.html\",\"WARC-Payload-Digest\":\"sha1:JAQMHKPPKO3ALUFJAHWKHL2FHF76MA2E\",\"WARC-Block-Digest\":\"sha1:MMP6QNYY4PHAKC77I5PJRTFAYOS7RLKI\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370496523.5_warc_CC-MAIN-20200330023050-20200330053050-00180.warc.gz\"}"} |
https://download.cnet.com/Graphing-Calculator-Math-PRO/3000-20414_4-76001032.html | [
"X",
null,
"# Graphing Calculator + Math PRO for Android\n\nBy Mathlab \\$5.49\n\n## Developer's Description\n\nIf you're looking for a graphing calculator app that works smoothly and seamlessly, you've found it. Graphing Calculator by Mathlab is a scientific graphing calculator integrated with algebra and is an indispensable mathematical tool for students in elementary school to those in college or graduate school, or just anyone who needs more than what a basic calculator offers. It is designed to replace bulky and costly handheld graphing calculators and works on virtually any Android phone or tablet. Furthermore, Graphing Calculator by Mathlab displays calculations as it performs them on the high-quality display of the Android device, making it easier for the user to understand the calculations and see them clearly. This app has two great strengths. First, it acts as a fine scientific calculator, but, more than that, it displays the intermediate results of the calculations as you type. It allows the students to both watch and learn how the calculations are made and how to find the final answer. Second, the graphing ability is absolutely stunning. Not only does the calculator beautifully display the graphs, but it automatically generates the x- and y- values and displays them as well. Video: https://youtu. be/6BR8Lv1U9kAHelp site with instructions and examples: http://help. mathlab. usIf you have a question, send email to calc@mathlab. usPRO FEATURES. Full screen graphs. 9 workspaces. Longer input and history. Save constants, functions and expressions in the library. Internet is not required. No advertisementsSCIENTIFIC CALCULATOR. Arithmetic expressions +, -, /, Square root, cube and higher roots (hold 'v' key). Exponent, logarithms (ln, log). Trigonometric functions sin. /2, cos 30. Hyperbolic functions sinh, cosh, tanh, (hold \"e\" key to switch). Inverse functions (hold direct function key). Complex numbers, all functions support complex arguments. Derivatives sin x' = cos x, (hold x^n key). Scientific notation (enable in menu). Percent mode. Save/load historyGRAPHING CALCULATOR. Multiple functions graphing. Implicit functions up to 2nd degree (ellipse 2x^2+3y^2=1, etc. ). Polar graphs (r=cos2. ). Parametric functions, enter each on new line (x=cos t, y=sin t). Function roots and critical points on a graph. Select the checkbox on the left from the function to show their coordinates on the graph, click the graph-button on the upper menu to display their coordinates as a list. Graph intersections (x^2=x+1). Tracing function values and slopes. Scrollable and resizable graphs. Pinch to zoom. Fullscreen graphs in landscape orientation. Function tables. Dave graphs as images. Save tables as csvFRACTION CALCULATOR. Simple and complex fractions 1/2 + 1/3 = 5/6. Mixed numbers, use space to enter values 3 1/2ALGEBRA CALCULATOR. Linear equations x+1=2 -> x=1. Quadratic equations x^2-1=0 -> x=-1, 1. Approximate roots of higher polynomials. Systems of linear equations, write one equation per line, x1+x2=1, x1-x2=2. Polynomial long division. Polynomial expansion, factoringMATRIX CALCULATOR. Matrix and vector operations. Dot product (hold. ), cross product. Determinant, inverse, norm, transpose, traceLIBRARY. User defined constants and functions. Save/load expressions.\n\n## Full Specifications\n\n### General\n\nRelease December 9, 2016\nVersion 4.10.137\n\n### Operating Systems\n\nOperating Systems Android\n\nReport Software\n\n## Related Apps",
null,
"### TED\n\nFree\nTED",
null,
"### Star Chart\n\nFree\nStar Chart",
null,
"### Bible for Kids\n\nFree\nBible for Kids",
null,
""
]
| [
null,
"https://download.cnet.com/a/img/resize/8e6cef505b57d08eb0b5ec7d2d401183d8ff61a5/catalog/2016/12/14/f5b41f28-4d57-4ff5-804c-1fce844b82e9/imgingest-4555564866260446584.png",
null,
"https://download.cnet.com/a/img/resize/a5a41e3aac438f64902eb43dc30b757573e7c304/catalog/2012/12/27/Foreman_12897474_6888_1_32x32.png",
null,
"https://download.cnet.com/a/img/resize/52fe0938cc89a618bb9bbdd233fc7ea217357996/catalog/2018/12/03/40537f84-8dac-44d0-89e0-d04688127dd5/imgingest-8584193291323940689.png",
null,
"https://download.cnet.com/a/img/resize/b5ff4b2a2634290fa929e21c72dab9e6272c2776/catalog/2013/11/29/fmimg7794243780933385463_48x48.icon-",
null,
"https://download.cnet.com/a/img/resize/b52ccb02f42df8e5f04ef986a1303c24a32e9e7c/catalog/2019/06/29/5373224c-cf0d-4144-bc97-9bb14e54e2cb/imgingest-8474669352869307701.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.84646827,"math_prob":0.97691333,"size":3207,"snap":"2022-27-2022-33","text_gpt3_token_len":736,"char_repetition_ratio":0.1326881,"word_repetition_ratio":0.0,"special_character_ratio":0.21702525,"punctuation_ratio":0.17304492,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98189574,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T02:37:48Z\",\"WARC-Record-ID\":\"<urn:uuid:4110c44e-124c-429c-9643-bb7603d18115>\",\"Content-Length\":\"375069\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed2b7140-955c-45ff-9808-dd4982f9ac92>\",\"WARC-Concurrent-To\":\"<urn:uuid:e6e3796e-1724-4569-9701-907c188c8e61>\",\"WARC-IP-Address\":\"146.75.34.154\",\"WARC-Target-URI\":\"https://download.cnet.com/Graphing-Calculator-Math-PRO/3000-20414_4-76001032.html\",\"WARC-Payload-Digest\":\"sha1:F3M5IJ3XJYH2BCLPVRSZHTVXXEISRNB5\",\"WARC-Block-Digest\":\"sha1:QFQKSLJF32PSQI4UHR5UEOZ4MF62HL7Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104683020.92_warc_CC-MAIN-20220707002618-20220707032618-00118.warc.gz\"}"} |
https://wordpress.youran.me/javascript-scoping-and-hoisting/ | [
"# Javascript作用域及变量提升\n\nPS: ES6标准引入了let关键字，用它声明的变量，具有块级作用域。并且在ES6中，花括号{}内部就是一个块级作用域。\n\n1. 所有作用域内，都存在 thisarguments 这两个变量\n2. 给函数指定的参数，参数名会自动包含在函数的作用域内\n3. 函数的声明。如 function X() {}\n4. 变量的声明。如 var Y;\n\n```function A() {\nbar();\nvar x = 1;\n}\n\nfunction B() {\nvar x;\nbar();\nx = 1;\n}\n```\n\n```function A() {\nif (false) {\nvar x = 1;\n}\nreturn;\nvar y = 1;\n}\n\nfunction B() {\nvar x, y;\nif (false) {\nx = 1;\n}\nreturn;\ny = 1;\n}\n```\n\n```function test() {\nA(); // 报错 TypeError \"A is not a function\"\nB(); // 没问题，打印出\"this will run!\"\nvar A = function () { // 本质是变量声明 + 赋值\n}\nfunction B() { // 函数声明\n}\n}\ntest();\n```\n\n```alert(typeof X); // \"function\"\nvar X = 1;\nfunction X() {}\n```\n\n```function X(param) {\n}\nfunction Y(arguments) {\n}\nX(1); // \"[Object Arguments]\"\nY(1); // \"1\"\n```\n\n```function X(a, a, a, b, a) {\n}\nX(1); // \"undefined\"\nX(1, 2, 3, 4, 5) // \"5\"\n```\n\n```A(); // TypeError \"A is not a function\"\nB(); // 没问题\nC(); // TypeError \"C is not a function\"\nX(); // ReferenceError \"X is not defined\"\n\nvar A = function () {}; // 变量声明+赋值，仅仅A这个名称被提升\nfunction B() {}; // 函数声明。整个函数被提升\nvar C = function X() {}; // 变量声明+赋值，仅仅C这个名称被提升\n\nA(); // 没问题\nB(); // 没问题\nC(); // 没问题\nX(); // ReferenceError \"X is not defined\"\n```\n\nReference:"
]
| [
null
]
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https://www.reference.com/science/isaac-newton-famous-26bce4889294982 | [
"What Is Isaac Newton Famous For?",
null,
"Credit: Andrea Ciambra/CC-BY 2.0\n\nIsaac Newton, a prominent mathematician and physicist, is famous for discovering several laws and theories of physics and motion that are collectively known as Newton's Laws. The laws that he is most famous for are the first, second and third laws of motion and the universal law of gravity.\n\nPictures of Isaac Newton often show him near an apple tree, as he began research for one of his discoveries, the universal law of gravity, after watching an apple fall from a tree. The cause and effect of gravity served as the basis for most of his theories. The universal law of gravity explains how every particle has a gravitational force that is proportional to the mass of the particle and is dependent on the distance between particles.\n\nNewton's well-known first law of motion explains how force is constant unless an external force is applied. The second law of motion is the mathematical equation F = MA. This equation is used to calculate the mass, acceleration and force of an object. The third law of motion famously states, \"for every action, there is an equal and opposite reaction.\"\n\nThese theories and several others were published in Newton's book titled \"Principia,\" which was published in 1687. In addition to theoretical discoveries, Newton made improvements to the telescope by adding mirrors to it.\n\nSimilar Articles"
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"https://images.reference.com/amg-cms-reference-images/prod/isaac-newton-famous_26bce4889294982.jpg",
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https://math.stackexchange.com/questions/1579044/see-if-function-is-homeomorphism-lege-artis-answer/1579054 | [
"# See if function is homeomorphism - “lege artis” answer\n\n$X=\\mathbb{N}_0$ and $Y= \\left\\lbrace \\frac{1}{n} ~|~ n\\in \\mathbb{N} \\right\\rbrace \\cup \\left\\lbrace 0\\right\\rbrace$ are subspaces of euclidean space $\\mathbb{R}$. See if function $f: X \\to Y$\n\n$f(n) = \\begin{cases} \\frac{1}{n} & n\\in \\mathbb{N}\\\\ 0 & n =0 \\\\ \\end{cases}$\n\nis a homeomorphism.\n\nHow would you do this textbook style, leaving little unsaid and being formal as possible by using only beginner theorems? I know how to do this but my proof always involve some handwaving. All my topology is handwaving, to be honest all my proofs are handwaving, I never mastered the art of a good explanation. For once I just want to see an obnoxious proof. Thanks\n\n• What do you mean by \"an obnoxious proof\"? – Cameron Buie Dec 16 '15 at 23:48\n• formal and explaining the obvious, maybe I used the wrong word. for example take Gaffney's first sentance, formalize it, write the definition down, prove the obvious, etc.. – nik Dec 17 '15 at 0:08\n• @nik What's obvious about all but a single point being open? What's obvious about a bijection on a countable set being discontinuous? – T.J. Gaffney Dec 17 '15 at 15:14\n• If the current answers do not contain enough detail, comment what is lacking. – Jake1234 Dec 21 '15 at 19:30\n• will comment in detail tomorrow – nik Dec 25 '15 at 12:26\n\n## 5 Answers\n\nAssume $f^{-1}: Y\\to X$ is continuous. Clearly, it maps $0\\in Y$ to $0\\in X$. One definition of continuity in topology is that preimages of open sets are open.\n\nOpen sets in $X$ resp. $Y$ are such sets that with each $x$ also contain all points closer to $x$ then $\\epsilon$ for some $\\epsilon>0$. (This is co-called metric topology induced by the standard distance function on $\\Bbb R$. It's the most obvious topology on $X,Y\\subseteq\\Bbb R$ one can think of, unless otherwise stated.)\n\nIf you accept this, and accept that the one-point set $\\{0\\}$ is open in $X=\\Bbb N_0$, then $\\{0\\}$ should be open in $Y$. But it is not, because any neighborhood of $0$ in $Y$ contains infinitely many other points.\n\nOf course, if you want to be very formal, one should probably say, in the assignment, \"$X,Y$ are topological subspaces of $\\Bbb R$ with the metric/Euclidean topology\". If you endow $\\Bbb R$ with a discrete topology---that is, all subsets are open---then $f$ is a homeomorphism, because it is a bijection and preimages of \"open\" (that means, any) sets under $f$, resp. $f^{-1}$ are \"open\" (that means, any) sets.\n\nThere cannot be an homeomorphism between these two spaces, because $Y$ has a limit point in $Y$ (this limit point is $0$) while $X$ has no limit point in $X$.\n\nLet $f$ be a bijection between $X$ and $Y$. For simplicity, name $g=f^{-1}$.\n\nConsider $a=g(0)=f^{-1}(0) \\in \\mathbb{N}$, and consider the set $\\{a\\}$ which is equal to $\\mathbb{N} \\cap (a-(1/2),a+(1/2))$.\n\n$(a-1/2),a+(1/2))$ is an open interval, and those form a base of open sets in standard topology on $R$, so clearly the interval itself is an open set. By definition, intersection of an open interval and a subset is a open set in the subspace topology of that specific subset - so we get that $\\{a\\}$ is an open set in the subspace $\\mathbb{N}$ of $\\mathbb{R}$.\n\nIf $g$ is to be continuous, from the definition of continuity, we have that for each $x \\in Y$, and each open neightborhood $G$ of $g(x)$, there must exist an open neightborhood $F$ of $x$, such that the image of this neightborhood with g must be a subset of $G$. So for $x=0$ there must exist an open set $U$ in the space $Y$ containing $0$, such that the image of this set with $g$ is a subset of $\\{a\\}$, as we have shown that is an open neightborhood of $a$.\n\nBut every open set contaning $0$ also contains another point:\n\nTake an open set $A$ containing $0$, and because open intervals form a base of the standard topology on $\\mathbb{R}$, we can find an open interval $(x,y)$ inside $A$ containing $0$ - clearly $y > 0$, and because for each real number we can find a larger integer, take $n> (y^{-1})$ - we have now found an $n \\in \\mathbb{N}$ such that $n^{-1} < y$, so every open set in $\\mathbb{R}$ also contains a point $n^{-1}$ for some $n \\in \\mathbb{N}$, distinct from $0$, and because this point is in $Y$, we have also shown we can find such a point, distinct from $0$, in any open set of $Y$ subspace of $\\mathbb{R}$ containing $0$.\n\nBecause $g$ is a bijection, the image of any open neightborhood of $0$ in $Y$ will contain at least two elements, and so it cannot possibly be a subset of $\\{a\\}$, therefore $g$ is not continuous, and so $f$ is not a homeomorphism, as $f^{-1} = g$.\n\nThus, no bijection between $X$ and $Y$ is a homeomorphism, and so a homeomorphism between $X$ and $Y$ does not exist. In particular, the function\n\n$f(n) = \\begin{cases} \\frac{1}{n} & n\\in \\mathbb{N}\\\\ 0 & n =0 \\\\ \\end{cases}$\n\nis not a homeomorphism.\n\nIf you're endowing these with the subset topology, then every point of $X$ is open, since each point is the intersection of some open set in $\\mathbb R$ and all of $X$. For $Y$ every point is an open set, except $0$. You'd expect continuity problems with this. For example $f(\\{0\\})$ maps an open set to a non-open set. So $f^{-1}$ can't be continuous, and $f$ can't be a homeomorphism. (All homeomorphisms are open maps.)\n\n1. Any subset of a metric space is also a metric space under the induced metric.\n\n2. If $f^{-1}$ was continuous, then $0=f^{-1}(0)=f^{-1}(\\lim 1/n)=\\lim f^{-1}(1/n)=\\lim n$.\n\nThis is a contradiction."
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https://mathematica.stackexchange.com/questions/57745/wavelet-expansion-of-given-function-as-opposed-to-dwt-of-given-data | [
"# Wavelet expansion of given function (as opposed to DWT of given data)\n\nTo illustrate, say I take the following function (the reason for the square root will be apparent soon)\n\ndist = NormalDistribution[0.5,0.1];\nf[x_] := Sqrt[PDF[dist,x]];\n\n\nThen I can grab the Haar wavelet to calculate coefficients like\n\ndata = Table[Evaluate@f[x], {x,0,1,1/31}]; (* 32 equally spaced points *)\ndwt = DiscreteWaveletTransform[data, HaarWavelet[], 5];\n\n\nIf I then take squares of coefficients, because of\n\n$$\\sum _{j=J_o}^{J_1} \\sum _{k\\in \\mathbb{Z}} \\psi _{j,k}^2+\\sum _{z\\in \\mathbb{Z}} \\varphi _{J_0,k}^2\\simeq \\int f^2 (x) \\, dx = \\left\\| f\\right\\| ^2$$\n\nI should get $\\simeq 1.0$ as $f$ is, in this case, the square root of a probability density function. And it kind-of works; to wit:\n\nTotal[Flatten[Last[#] & /@ dwt[Automatic]]^2]/31\n\n\nreturns 1.0 (dwt[Automatic] gives you the coefficients for the inverse transform only; see here). This should be very close to NIntegrate[Evaluate@f[x]^2, {x, 0, 1}], which is 0.999999.\n\nHowever, note that \"/31\"! Where did that come from? This is: that 1/31 does not appear in the Parseval's identity but it is needed for the calculation to work (!)\n\nAnd if I change HaarWavelet[] by DaubechiesWavelet[], say, then this data sampling no longer works.\n\nI am even taking care of the fact that the Daubechies family has bigger support and for that I am using the following function to determine the number of points required as:\n\ndataSize[wave_, 1] :=\nLength[WaveletFilterCoefficients[wave, \"PrimalLowpass\"]];\ndataSize[wave_, n_Integer /; n > 1] :=\n2 dataSize[wave, n - 1] -\nLength[WaveletFilterCoefficients[wave, \"PrimalLowpass\"]] + 2;\n\n\nWhat I am doing wrong?\n\n(Why this? You may ask. For example, one could use wavelet expansions to calculate the norm (in Hilbert space) between two functions, or a function and some other approximation; instead of attempting NIntegrate. By the way, I am actually interested in dimensions greater than 1)\n\nNote: I am only interested in orthogonal wavelets; Parseval's formula should be just ok as far as I know.\n\nUPDATE\n\nI modified dist to NormalDistribution[0.5,0.5]. In this case, one needs a longer interval {x,-2,3} to effectively cover $f$. In this case, the squared coefficients are added like 5Total[Flatten[Last[#] & /@ dwt[Automatic]]^2]/31 (note the factor 5.)\n\nThis points me to the fact that the $x_i$ sample points are usually considered as in the [0,1] interval (if I remember correctly), therefore the final answer needs to be rescaled by 5. Still... work in progress.\n\n• You show the code Total[Flatten[Last[#] & /@ dwt[Automatic]]^2]/31 and then go on to say \"note that /31! Where did that come from?\" Is it not there because you typed it in? If not, where did the whole expression come from? I note 1/31 is the table iteration step you used in an earlier line of code. – m_goldberg Aug 20 '14 at 7:34\n• Sorry @m_goldberg, what I meant is that that 1/31 does not appear in the Parseval's identity I mentioned, but I had to put it to make the identity work. And unfortunately it is not as simple as the iteration step, as I used wavelets with bigger support and I was not able to find such correction factor. – carlosayam Aug 20 '14 at 11:34\n\nI found the issue and the reason.\n\nFirst, the reason for that 1/31. As explained in documentation for DiscreteWaveletTransform (see \"Properties & Relations\")\n\nThe energy norm is conserved for orthogonal wavelet families:\nIn:= data = RandomReal[1, {100}];\nIn:= dwt = DiscreteWaveletTransform[data, Padding -> 0.];\nIn:= Norm[data] == Norm[Flatten[Last /@ dwt[Automatic]]]\nOut= True\n\n\nNote that in In, Norm[data] is $\\sqrt{\\sum d_i^2}$. Given that $d_i~=~f(x_i)$ is a sampling point for $f$, $\\int f^2(x) \\delta x$ ~ $\\sum f^2(x_i) \\Delta x$ = $\\sum d_i^2 \\Delta x$; as in my case $\\Delta x = \\frac{1}{31}$ this explains the factor I didn't understand.\n\nThe other was an issue calculating\n\nTotal[Flatten[Last[#] & /@ dwt[Automatic]]^2]/31\n\n\nwhen the wavelet was not Haar's wavelet. The reference above also solves this. Note that Padding -> 0. option. The default padding is periodic, which treats the end points as if they were in a circle, not good in this case. This option therefore ensures the transform does not extend the data as periodic beyond the provided points (maybe my dataSize function is incorrect, not sure). With Padding -> 0. in DiscreteWaveletTransform and using $\\Delta x$, it works.\n\nHere another example, with a bigger region of interest, [-2,3] and using Daubechies wavelet with n=2.\n\nModule[\n{dist, wave, f, data, dwt, level = 7},\nwave = DaubechiesWavelet;\ndist = NormalDistribution[0.5, 0.5];\nf[x_] := Sqrt[PDF[dist, x]];\ndata = Table[f[x], {x, -2, 3, 5/(2^level - 1)}];\nPrint[Plot[Evaluate@f[x], {x, -2, 3}, PlotRange -> {0, 1},\nPlotLabel -> Style[\"f\", FontSlant -> \"Italic\"]]];\nPrint[ListPlot[data, PlotLabel -> \"Sampling\"]];\nPrint[\"Lenght:\", Length[data]];\nPrint[\"Aprox. integral from data: \", 5 Total[data^2 ]/2^level];\ndwt = DiscreteWaveletTransform[data, wave, level, Padding -> 0.];\nPrint[\"NIntegrate: \", NIntegrate[Evaluate@(f[x]^2), {x, -2, 3}]];\nPrint[\"Aprox. integral from coefficients: \",\n5 Total[Flatten[Last[#] & /@ dwt[Automatic]]^2]/2^level];\n]\n\n\nwhich produces",
null,
"Note in this case $\\Delta x$ is $\\frac{5}{\\text{num_samples}}$ = $\\frac{5}{2^{level}}$."
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"https://i.stack.imgur.com/MTKVe.png",
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https://www.litscape.com/word_analysis/marmosets | [
"Definition of marmosets\n\n\"marmosets\" in the noun sense\n\n1. marmoset\n\nsmall soft-furred South American and Central American monkey with claws instead of nails\n\nSource: WordNet® (An amazing lexical database of English)\n\nPrinceton University \"About WordNet®.\"\nWordNet®. Princeton University. 2010.\n\nmarmosets in Scrabble®\n\nThe word marmosets is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nMARMOSETS\n(144)\n\nSeven Letter Word Alert: (7 words)\n\nmaestro, marmots, masters, seamost, stammer, streams, stroams\n\nMARMOSETS\n(144)\nMARMOSETS\n(126)\nMARMOSETS\n(96)\nMARMOSETS\n(84)\nMARMOSETS\n(56)\nMARMOSETS\n(52)\nMARMOSETS\n(52)\nMARMOSETS\n(52)\nMARMOSETS\n(51)\nMARMOSETS\n(42)\nMARMOSETS\n(42)\nMARMOSETS\n(42)\nMARMOSETS\n(42)\nMARMOSETS\n(42)\nMARMOSETS\n(38)\nMARMOSETS\n(34)\nMARMOSETS\n(34)\nMARMOSETS\n(34)\nMARMOSETS\n(34)\nMARMOSETS\n(34)\nMARMOSETS\n(32)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(28)\nMARMOSETS\n(28)\nMARMOSETS\n(28)\nMARMOSETS\n(28)\nMARMOSETS\n(28)\nMARMOSETS\n(26)\nMARMOSETS\n(26)\nMARMOSETS\n(26)\nMARMOSETS\n(23)\nMARMOSETS\n(21)\nMARMOSETS\n(18)\nMARMOSETS\n(18)\nMARMOSETS\n(18)\nMARMOSETS\n(17)\nMARMOSETS\n(17)\nMARMOSETS\n(17)\nMARMOSETS\n(16)\nMARMOSETS\n(16)\nMARMOSETS\n(16)\n\nMARMOSETS\n(144)\nMARMOSETS\n(126)\nMARMOSETS\n(96)\nMARMOTS\n(94 = 44 + 50)\nSTAMMER\n(94 = 44 + 50)\nMARMOTS\n(92 = 42 + 50)\nMARMOTS\n(92 = 42 + 50)\nSTAMMER\n(92 = 42 + 50)\nSTAMMER\n(92 = 42 + 50)\nMARMOTS\n(92 = 42 + 50)\nSTAMMER\n(92 = 42 + 50)\nSTREAMS\n(86 = 36 + 50)\nSTAMMER\n(86 = 36 + 50)\nMARMOTS\n(86 = 36 + 50)\nMAESTRO\n(86 = 36 + 50)\nSTAMMER\n(86 = 36 + 50)\nSTAMMER\n(86 = 36 + 50)\nSEAMOST\n(86 = 36 + 50)\nMARMOTS\n(86 = 36 + 50)\nSTREAMS\n(86 = 36 + 50)\nMAESTRO\n(86 = 36 + 50)\nMASTERS\n(86 = 36 + 50)\nMARMOTS\n(86 = 36 + 50)\nSTROAMS\n(86 = 36 + 50)\nMASTERS\n(86 = 36 + 50)\nSTROAMS\n(86 = 36 + 50)\nMARMOTS\n(86 = 36 + 50)\nSTAMMER\n(86 = 36 + 50)\nSEAMOST\n(86 = 36 + 50)\nSTAMMER\n(86 = 36 + 50)\nMARMOTS\n(86 = 36 + 50)\nSEAMOST\n(86 = 36 + 50)\nMARMOSETS\n(84)\nSTAMMER\n(84 = 34 + 50)\nSTAMMER\n(83 = 33 + 50)\nMARMOTS\n(83 = 33 + 50)\nMASTERS\n(80 = 30 + 50)\nMASTERS\n(80 = 30 + 50)\nMASTERS\n(80 = 30 + 50)\nMASTERS\n(80 = 30 + 50)\nSTREAMS\n(80 = 30 + 50)\nMAESTRO\n(80 = 30 + 50)\nMAESTRO\n(80 = 30 + 50)\nSTROAMS\n(80 = 30 + 50)\nSEAMOST\n(80 = 30 + 50)\nSTROAMS\n(80 = 30 + 50)\nSTROAMS\n(80 = 30 + 50)\nSTROAMS\n(80 = 30 + 50)\nSTROAMS\n(80 = 30 + 50)\nMAESTRO\n(80 = 30 + 50)\nMASTERS\n(80 = 30 + 50)\nSEAMOST\n(80 = 30 + 50)\nSTREAMS\n(80 = 30 + 50)\nMAESTRO\n(80 = 30 + 50)\nSTREAMS\n(80 = 30 + 50)\nSEAMOST\n(80 = 30 + 50)\nSTREAMS\n(80 = 30 + 50)\nSTROAMS\n(80 = 30 + 50)\nSTREAMS\n(80 = 30 + 50)\nMAESTRO\n(80 = 30 + 50)\nSTREAMS\n(80 = 30 + 50)\nMASTERS\n(80 = 30 + 50)\nSTREAMS\n(80 = 30 + 50)\nSEAMOST\n(80 = 30 + 50)\nMASTERS\n(80 = 30 + 50)\nSEAMOST\n(80 = 30 + 50)\nMAESTRO\n(80 = 30 + 50)\nSTREAMS\n(80 = 30 + 50)\nSEAMOST\n(80 = 30 + 50)\nMAESTRO\n(80 = 30 + 50)\nSTROAMS\n(80 = 30 + 50)\nSTAMMER\n(80 = 30 + 50)\nMARMOTS\n(80 = 30 + 50)\nSTROAMS\n(80 = 30 + 50)\nSTAMMER\n(78 = 28 + 50)\nMARMOTS\n(78 = 28 + 50)\nMARMOTS\n(78 = 28 + 50)\nMASTERS\n(77 = 27 + 50)\nSEAMOST\n(77 = 27 + 50)\nMAESTRO\n(77 = 27 + 50)\nSTROAMS\n(77 = 27 + 50)\nSTREAMS\n(77 = 27 + 50)\nMAESTRO\n(76 = 26 + 50)\nSTAMMER\n(76 = 26 + 50)\nSTAMMER\n(76 = 26 + 50)\nMARMOTS\n(76 = 26 + 50)\nSTAMMER\n(76 = 26 + 50)\nSTAMMER\n(76 = 26 + 50)\nMARMOTS\n(76 = 26 + 50)\nMARMOTS\n(76 = 26 + 50)\nMASTERS\n(76 = 26 + 50)\nMARMOTS\n(76 = 26 + 50)\nMARMOTS\n(76 = 26 + 50)\nMARMOTS\n(74 = 24 + 50)\nMARMOTS\n(74 = 24 + 50)\nMARMOTS\n(74 = 24 + 50)\nSTAMMER\n(74 = 24 + 50)\nSTREAMS\n(74 = 24 + 50)\nSTAMMER\n(74 = 24 + 50)\nSTAMMER\n(74 = 24 + 50)\nSTROAMS\n(74 = 24 + 50)\nMAESTRO\n(74 = 24 + 50)\nSTAMMER\n(74 = 24 + 50)\nMAESTRO\n(74 = 24 + 50)\nMARMOTS\n(74 = 24 + 50)\nSTAMMER\n(74 = 24 + 50)\nMASTERS\n(74 = 24 + 50)\nMARMOTS\n(74 = 24 + 50)\nMASTERS\n(74 = 24 + 50)\nSTAMMER\n(74 = 24 + 50)\nSTAMMER\n(74 = 24 + 50)\nMARMOTS\n(74 = 24 + 50)\nSEAMOST\n(72 = 22 + 50)\nSEAMOST\n(72 = 22 + 50)\nMASTERS\n(72 = 22 + 50)\nSEAMOST\n(72 = 22 + 50)\nSTAMMER\n(72 = 22 + 50)\nSTREAMS\n(72 = 22 + 50)\nSTAMMER\n(72 = 22 + 50)\nSTAMMER\n(72 = 22 + 50)\nMAESTRO\n(72 = 22 + 50)\nSEAMOST\n(72 = 22 + 50)\nMARMOTS\n(72 = 22 + 50)\nSTREAMS\n(72 = 22 + 50)\nSTAMMER\n(72 = 22 + 50)\nMASTERS\n(72 = 22 + 50)\nSEAMOST\n(72 = 22 + 50)\nSTAMMER\n(72 = 22 + 50)\nMARMOTS\n(72 = 22 + 50)\nMARMOTS\n(72 = 22 + 50)\nSTROAMS\n(72 = 22 + 50)\nMARMOTS\n(72 = 22 + 50)\nMASTERS\n(72 = 22 + 50)\nMASTERS\n(72 = 22 + 50)\nMAESTRO\n(72 = 22 + 50)\nMAESTRO\n(72 = 22 + 50)\nSTROAMS\n(72 = 22 + 50)\nMAESTRO\n(72 = 22 + 50)\nMARMOTS\n(72 = 22 + 50)\nMASTERS\n(72 = 22 + 50)\nSEAMOST\n(72 = 22 + 50)\nSTREAMS\n(72 = 22 + 50)\nSTREAMS\n(72 = 22 + 50)\nSTROAMS\n(72 = 22 + 50)\nSTROAMS\n(72 = 22 + 50)\nMAESTRO\n(72 = 22 + 50)\nSTREAMS\n(72 = 22 + 50)\nSTROAMS\n(72 = 22 + 50)\nSTROAMS\n(70 = 20 + 50)\nMASTERS\n(70 = 20 + 50)\nMASTERS\n(70 = 20 + 50)\nSTROAMS\n(70 = 20 + 50)\nSTROAMS\n(70 = 20 + 50)\nSTREAMS\n(70 = 20 + 50)\nSTREAMS\n(70 = 20 + 50)\nSTROAMS\n(70 = 20 + 50)\nMASTERS\n(70 = 20 + 50)\nSTROAMS\n(70 = 20 + 50)\nMASTERS\n(70 = 20 + 50)\nSTREAMS\n(70 = 20 + 50)\nSTREAMS\n(70 = 20 + 50)\nSTREAMS\n(70 = 20 + 50)\nSTREAMS\n(70 = 20 + 50)\nSTROAMS\n(70 = 20 + 50)\nMASTERS\n(70 = 20 + 50)\nSTREAMS\n(70 = 20 + 50)\nSTROAMS\n(70 = 20 + 50)\nMASTERS\n(70 = 20 + 50)\nSEAMOST\n(70 = 20 + 50)\nSEAMOST\n(70 = 20 + 50)\nMAESTRO\n(70 = 20 + 50)\nSEAMOST\n(70 = 20 + 50)\nSEAMOST\n(70 = 20 + 50)\nSEAMOST\n(70 = 20 + 50)\nSEAMOST\n(70 = 20 + 50)\nMAESTRO\n(70 = 20 + 50)\nMAESTRO\n(70 = 20 + 50)\nSEAMOST\n(70 = 20 + 50)\nMAESTRO\n(70 = 20 + 50)\nSEAMOST\n(70 = 20 + 50)\nMAESTRO\n(70 = 20 + 50)\nMAESTRO\n(70 = 20 + 50)\nSTAMMER\n(69 = 19 + 50)\nMARMOTS\n(69 = 19 + 50)\nSTREAMS\n(68 = 18 + 50)\nMAESTRO\n(68 = 18 + 50)\nMASTERS\n(68 = 18 + 50)\nSTROAMS\n(68 = 18 + 50)\nSEAMOST\n(68 = 18 + 50)\nSTREAMS\n(68 = 18 + 50)\nMAESTRO\n(68 = 18 + 50)\nSTROAMS\n(68 = 18 + 50)\nSTREAMS\n(68 = 18 + 50)\nSTREAMS\n(68 = 18 + 50)\nSEAMOST\n(68 = 18 + 50)\nMASTERS\n(68 = 18 + 50)\nSTREAMS\n(68 = 18 + 50)\nSTROAMS\n(68 = 18 + 50)\nMASTERS\n(68 = 18 + 50)\nMAESTRO\n(68 = 18 + 50)\n\nmarmosets in Words With Friends™\n\nThe word marmosets is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nMARMOSETS\n(225)\n\nSeven Letter Word Alert: (7 words)\n\nmaestro, marmots, masters, seamost, stammer, streams, stroams\n\nMARMOSETS\n(225)\nMARMOSETS\n(138)\nMARMOSETS\n(102)\nMARMOSETS\n(76)\nMARMOSETS\n(75)\nMARMOSETS\n(75)\nMARMOSETS\n(64)\nMARMOSETS\n(64)\nMARMOSETS\n(64)\nMARMOSETS\n(60)\nMARMOSETS\n(60)\nMARMOSETS\n(57)\nMARMOSETS\n(57)\nMARMOSETS\n(57)\nMARMOSETS\n(57)\nMARMOSETS\n(50)\nMARMOSETS\n(50)\nMARMOSETS\n(46)\nMARMOSETS\n(38)\nMARMOSETS\n(34)\nMARMOSETS\n(34)\nMARMOSETS\n(34)\nMARMOSETS\n(34)\nMARMOSETS\n(34)\nMARMOSETS\n(32)\nMARMOSETS\n(32)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(30)\nMARMOSETS\n(25)\nMARMOSETS\n(24)\nMARMOSETS\n(23)\nMARMOSETS\n(21)\nMARMOSETS\n(21)\nMARMOSETS\n(21)\nMARMOSETS\n(20)\nMARMOSETS\n(20)\nMARMOSETS\n(19)\nMARMOSETS\n(19)\nMARMOSETS\n(19)\nMARMOSETS\n(18)\nMARMOSETS\n(18)\nMARMOSETS\n(18)\nMARMOSETS\n(18)\nMARMOSETS\n(17)\nMARMOSETS\n(17)\nMARMOSETS\n(17)\nMARMOSETS\n(17)\nMARMOSETS\n(17)\nMARMOSETS\n(16)\n\nMARMOSETS\n(225)\nMARMOSETS\n(138)\nMARMOTS\n(104 = 69 + 35)\nMARMOTS\n(104 = 69 + 35)\nSTAMMER\n(104 = 69 + 35)\nSTAMMER\n(104 = 69 + 35)\nSTAMMER\n(104 = 69 + 35)\nMARMOTS\n(104 = 69 + 35)\nMARMOSETS\n(102)\nMARMOTS\n(98 = 63 + 35)\nMARMOTS\n(98 = 63 + 35)\nSTAMMER\n(98 = 63 + 35)\nSTAMMER\n(98 = 63 + 35)\nMARMOTS\n(98 = 63 + 35)\nSTREAMS\n(95 = 60 + 35)\nSEAMOST\n(95 = 60 + 35)\nSTROAMS\n(95 = 60 + 35)\nMASTERS\n(95 = 60 + 35)\nMAESTRO\n(95 = 60 + 35)\nSEAMOST\n(95 = 60 + 35)\nMAESTRO\n(89 = 54 + 35)\nSEAMOST\n(89 = 54 + 35)\nMASTERS\n(89 = 54 + 35)\nSTROAMS\n(89 = 54 + 35)\nSEAMOST\n(89 = 54 + 35)\nSTREAMS\n(89 = 54 + 35)\nSTAMMER\n(87 = 52 + 35)\nMARMOTS\n(87 = 52 + 35)\nMARMOTS\n(87 = 52 + 35)\nMARMOTS\n(87 = 52 + 35)\nSTAMMER\n(87 = 52 + 35)\nSTAMMER\n(87 = 52 + 35)\nSTAMMER\n(86 = 51 + 35)\nMARMOTS\n(86 = 51 + 35)\nSTAMMER\n(80 = 45 + 35)\nMARMOTS\n(80 = 45 + 35)\nMARMOTS\n(80 = 45 + 35)\nSTAMMER\n(80 = 45 + 35)\nSTAMMER\n(80 = 45 + 35)\nMARMOTS\n(80 = 45 + 35)\nSTAMMER\n(80 = 45 + 35)\nSTROAMS\n(77 = 42 + 35)\nMAESTRO\n(77 = 42 + 35)\nMASTERS\n(77 = 42 + 35)\nSTREAMS\n(77 = 42 + 35)\nSTAMMER\n(77 = 42 + 35)\nSTREAMS\n(77 = 42 + 35)\nMAESTRO\n(77 = 42 + 35)\nMASTERS\n(77 = 42 + 35)\nSTREAMS\n(77 = 42 + 35)\nMAESTRO\n(77 = 42 + 35)\nMASTERS\n(77 = 42 + 35)\nSEAMOST\n(77 = 42 + 35)\nSTROAMS\n(77 = 42 + 35)\nMARMOTS\n(77 = 42 + 35)\nSEAMOST\n(77 = 42 + 35)\nSTROAMS\n(77 = 42 + 35)\nMARMOSETS\n(76)\nMASTERS\n(75 = 40 + 35)\nMAESTRO\n(75 = 40 + 35)\nSEAMOST\n(75 = 40 + 35)\nMAESTRO\n(75 = 40 + 35)\nSEAMOST\n(75 = 40 + 35)\nSTROAMS\n(75 = 40 + 35)\nMASTERS\n(75 = 40 + 35)\nSEAMOST\n(75 = 40 + 35)\nSTREAMS\n(75 = 40 + 35)\nSTROAMS\n(75 = 40 + 35)\nSTREAMS\n(75 = 40 + 35)\nSTREAMS\n(75 = 40 + 35)\nSTROAMS\n(75 = 40 + 35)\nMARMOSETS\n(75)\nMARMOSETS\n(75)\nMAESTRO\n(75 = 40 + 35)\nMASTERS\n(75 = 40 + 35)\nMAESTRO\n(71 = 36 + 35)\nMASTERS\n(71 = 36 + 35)\nSEAMOST\n(71 = 36 + 35)\nSTREAMS\n(71 = 36 + 35)\nSEAMOST\n(71 = 36 + 35)\nSTREAMS\n(71 = 36 + 35)\nMASTERS\n(71 = 36 + 35)\nSTROAMS\n(71 = 36 + 35)\nSTROAMS\n(71 = 36 + 35)\nSEAMOST\n(71 = 36 + 35)\nSTREAMS\n(71 = 36 + 35)\nMAESTRO\n(71 = 36 + 35)\nSTREAMS\n(71 = 36 + 35)\nMAESTRO\n(71 = 36 + 35)\nSTROAMS\n(71 = 36 + 35)\nMASTERS\n(71 = 36 + 35)\nMAESTRO\n(71 = 36 + 35)\nSTROAMS\n(71 = 36 + 35)\nMAESTRO\n(71 = 36 + 35)\nMASTERS\n(71 = 36 + 35)\nSTROAMS\n(71 = 36 + 35)\nMASTERS\n(71 = 36 + 35)\nSTREAMS\n(71 = 36 + 35)\nSEAMOST\n(71 = 36 + 35)\nMAESTRO\n(71 = 36 + 35)\nMASTERS\n(71 = 36 + 35)\nSTROAMS\n(71 = 36 + 35)\nSTREAMS\n(71 = 36 + 35)\nMARMOTS\n(69 = 34 + 35)\nSTAMMER\n(69 = 34 + 35)\nMARMOT\n(66)\nMARMOT\n(66)\nMARMOTS\n(65 = 30 + 35)\nMARMOTS\n(65 = 30 + 35)\nSTAMMER\n(65 = 30 + 35)\nMARMOTS\n(65 = 30 + 35)\nSTAMMER\n(65 = 30 + 35)\nSTAMMER\n(65 = 30 + 35)\nSTAMMER\n(65 = 30 + 35)\nMARMOTS\n(65 = 30 + 35)\nMARMOTS\n(65 = 30 + 35)\nSTAMMER\n(65 = 30 + 35)\nMARMOSETS\n(64)\nMARMOSETS\n(64)\nMARMOSETS\n(64)\nSTAMMER\n(63 = 28 + 35)\nMARMOTS\n(63 = 28 + 35)\nMARMOTS\n(63 = 28 + 35)\nSTAMMER\n(63 = 28 + 35)\nMARMOTS\n(63 = 28 + 35)\nSTREAMS\n(63 = 28 + 35)\nMARMOTS\n(63 = 28 + 35)\nMARMOTS\n(63 = 28 + 35)\nSTROAMS\n(63 = 28 + 35)\nSTAMMER\n(63 = 28 + 35)\nSTAMMER\n(63 = 28 + 35)\nSTAMMER\n(63 = 28 + 35)\nMAESTRO\n(63 = 28 + 35)\nMASTERS\n(63 = 28 + 35)\nSTAMMER\n(61 = 26 + 35)\nSTAMMER\n(61 = 26 + 35)\nMARMOTS\n(61 = 26 + 35)\nSTAMMER\n(61 = 26 + 35)\nSTAMMER\n(61 = 26 + 35)\nMARMOTS\n(61 = 26 + 35)\nSTAMMER\n(61 = 26 + 35)\nMARMOTS\n(61 = 26 + 35)\nSTAMMER\n(61 = 26 + 35)\nSTAMMER\n(61 = 26 + 35)\nMARMOTS\n(61 = 26 + 35)\nMARMOTS\n(61 = 26 + 35)\nMARMOTS\n(61 = 26 + 35)\nMARMOTS\n(61 = 26 + 35)\nMARMOSETS\n(60)\nSMARMS\n(60)\nSMARMS\n(60)\nMARMOT\n(60)\nMARMOSETS\n(60)\nMARMOT\n(60)\nSTROAMS\n(59 = 24 + 35)\nSEAMOST\n(59 = 24 + 35)\nSTROAMS\n(59 = 24 + 35)\nSTROAMS\n(59 = 24 + 35)\nMAESTRO\n(59 = 24 + 35)\nSEAMOST\n(59 = 24 + 35)\nMAESTRO\n(59 = 24 + 35)\nSTROAMS\n(59 = 24 + 35)\nMAESTRO\n(59 = 24 + 35)\nSTROAMS\n(59 = 24 + 35)\nMAESTRO\n(59 = 24 + 35)\nMAESTRO\n(59 = 24 + 35)\nSEAMOST\n(59 = 24 + 35)\nSTREAMS\n(59 = 24 + 35)\nMASTERS\n(59 = 24 + 35)\nSTREAMS\n(59 = 24 + 35)\nMASTERS\n(59 = 24 + 35)\nSEAMOST\n(59 = 24 + 35)\nMASTERS\n(59 = 24 + 35)\nMASTERS\n(59 = 24 + 35)\nSTREAMS\n(59 = 24 + 35)\nSTREAMS\n(59 = 24 + 35)\nSEAMOST\n(59 = 24 + 35)\nSTREAMS\n(59 = 24 + 35)\nMASTERS\n(59 = 24 + 35)\nSEAMOST\n(59 = 24 + 35)\nMARMOTS\n(58 = 23 + 35)\nSTAMMER\n(58 = 23 + 35)\nSTAMMER\n(58 = 23 + 35)\nSTROAMS\n(57 = 22 + 35)\nMASTERS\n(57 = 22 + 35)\nMARMOSETS\n(57)\nSTROAMS\n(57 = 22 + 35)\nSTREAMS\n(57 = 22 + 35)\nMASTER\n(57)\nSTREAMS\n(57 = 22 + 35)\nMASTERS\n(57 = 22 + 35)\nMARMOSETS\n(57)\nSTROAMS\n(57 = 22 + 35)\nMEMOS\n(57)\nRAMOSE\n(57)\nMASTERS\n(57 = 22 + 35)\nRAMETS\n(57)\nMARMOSETS\n(57)\nMATERS\n(57)\nSTREAMS\n(57 = 22 + 35)\n\nWord Growth involving marmosets\n\nar arm marmoset\n\nar mar marmoset\n\nma mar marmoset\n\nos marmoset\n\nset marmoset\n\nset sets\n\nLonger words containing marmosets\n\n(No longer words found)"
]
| [
null
]
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https://bookdown.org/kirthanarao95/rm3_book/partial-least-square-correlation-analysis.html | [
"# Chapter 6 Partial Least Square Correlation Analysis\n\nData Table: Partial Least Square Correlation (PLSC) method compares the information in two data tables on the same set of observations.\n\nGoal:\n\n• Find the relationship between variables from the two tables.(Shared information between tables)\n\n• Find the commonality: by finding the maximum covariance between latent variables (one for each table) obtained as linear combinations of the original variables.\n\nNote: These discriminant variables are also used to assign the original observations or “new” observations to originally defined categories.\n\nKey ideas\n\n1. Perform SVD on correlation matrix between the variables of the two tables\n\n2. The latent variables are obtained by projecting the original matrices onto their respective saliences. Lx = X * P\nLy = Y * Q\n\n3. Saliences Loadings (Column Factor Scores) are separate for the two tables. Fx = P * delta Fy = Q * delta\n\n4. Lx1 and Ly2 are orthogonal to each other.\n\nInterpretation\n\n``````1. The latent variables, which describe the observations, are required to\n“explain” the largest portion of the covariance between the two tables.\n\n2. The original variables are described by their saliences.``````\n\n## 6.1 Datasets:\n\nSurvey of Autobiographical Memory + Object-Spatial ImageryQuestionnaire (SAM + OSIQ)\n\nRows: The observation is common across the two tables. 144 Participants asked to rate the extent to which a particular item applied to their memory in general, using a 5-point Likert scale.\n\nSAM Columns: 26 quantitative variables\n\n``````Episodic Memory: 8 questions (Memory for personal experiences)\nSemantic Memory : 6 questions (Memory for general knowledge and facts)\nSpatial Memory : 6 questions (Memory testing how well one can navigate)\nProspective Memory : 6 questions (How clearly one can picture future events)``````\n\nOSIQ Columns: 30 quantitative variables.\n\n``````Object imagery scale : 15 questions(processing of colorful, pictorial, and\nhigh-resolution images of individual objects)\n\nSpatial imagery scale : 15 questions (processing of schematic images, spatial\nrelations amongst objects, and spatial\ntransformations)``````\n\n## 6.2 Looking at the Data Pattern\n\n### 6.2.1 Correlation Plot:\n\nPlotting the correlation between SAM and OSIQ\n\nEpisodic Memory and Memory for future is positively correlated with the participants preference for object and vegatively correlated with spatial imagery. This shows that participants tend to remember and imagine color and shapes of objects when they are refering to their memory for personal experiences than the location or position of the objects.\n\nThere is a positive correlation observed between Spatial memory(P) and Spatial imagery(So), which is very much expected.\n\n``````# Compute the covariance matrix\nXY.cor <- cor(data1,data2)\n\ncorrplot(XY.cor, method = \"color\", tl.cex = .5, tl.col = \"royalblue\",\naddCoef.col = \"black\", number.digits = 0, number.cex = .4,\ncl.pos = 'b', cl.cex = .3,\ncol = colorRampPalette(c(\"darkred\", \"white\",\"midnightblue\"))(20)) ``````",
null,
"## 6.3 PLSC Analysis\n\n``pls.res <- tepPLS(data1,data2, DESIGN = data.design, make_design_nominal = TRUE, graphs = FALSE)``\n\n### 6.3.1 Scree Plot\n\nThe scree plot shows a weird pattern because the null hypothesis is that there is no correlation between the tables (Null = 0) Hence eigenvalues greater than zero become significant.\n\nThe results of the permutation test gives us the eigenvalues.\n\n``````# no.of eigenvalues\nnL <- min(ncol(data1),ncol(data2))\n\nresPerm4PLSC <- perm4PLSC(data1, # First Data matrix\ndata2, # Second Data matrix\nnIter = 1000 # How mny iterations\n)\nprint(resPerm4PLSC)``````\n``````## ------------------------------------------------------------------------------\n## Results of Permutation Test for PLSC of X'*Y = R\n## for Omnibus Inertia and Eigenvalues\n## ------------------------------------------------------------------------------\n## \\$ fixedInertia the Inertia of Matrix X\n## \\$ fixedEigenvalues an L*1 vector of the eigenvalues of X\n## \\$ pOmnibus the probablity associated to the Inertia\n## \\$ pEigenvalues an L* 1 matrix of p for the eigenvalues of X\n## \\$ permInertia vector of the permuted Inertia of X\n## \\$ permEigenvalues matrix of the permuted eigenvalues of X\n## ------------------------------------------------------------------------------``````\n``````my.scree <-PlotScree(ev = pls.res\\$TExPosition.Data\\$eigs,\ntitle = 'PLSC- Scree Plot',\np.ev = resPerm4PLSC\\$pEigenvalues,\nplotKaiser = TRUE,\ncolor4Kaiser = ggplot2::alpha('darkorchid4', .5),\n)``````",
null,
"``````nL <- min(ncol(data1),ncol(data2))\n\nmy.scree``````\n``## 0 10 20 30 40``\n\n### 6.3.2 Latent Variable Pairs\n\n#### 6.3.2.1 Looking at the First Pair of Latent variables (Lx1, Ly1)\n\nSeparates the High and normal memory groups well. The latent variables of the two data share a high covariance.\n\n``````#uniquecol <- unique(pls.res\\$Plotting.Data\\$fii.col)\n#grpcol <- uniquecol\n#rownames(grpcol) <- data.design.vec[rownames(uniquecol),]\n\n# ploting the first latent variable of data1(X) and first latent variables of data2(Y). We are tryingto see if these two latent variables are similar or not.\n\n# first pair of latent variables:\n\nlatvar.1 <- cbind(pls.res\\$TExPosition.Data\\$lx[,1],\npls.res\\$TExPosition.Data\\$ly[,1])\ncolnames(latvar.1) <- c(\"Lx 1\", \"Ly 1\")\n\n# compute means\nlv.1.group <- getMeans(latvar.1,data.design.vec)\n\ncol4Means <- recode(rownames(lv.1.group),\nHigh = 'darkred',\nNorm = 'orange2',\n)\nnames(col4Means) <- rownames(lv.1.group)\n\n# compute bootstrap - for confidence intervals\nlv.1.group.boot <- Boot4Mean(latvar.1, data.design.vec)\ncolnames(lv.1.group.boot\\$BootCube) <- c(\"Lx 1\", \"Ly 1\")\n\n# plotiing the factor Maps\n\nplot.lv1 <- createFactorMap(latvar.1,\ncol.points = col4row,\ncol.labels = col4row,\nalpha.points = 0.2\n)\n\nplot1.mean <- createFactorMap(lv.1.group,\ncol.points = col4Means,\ncol.labels = col4Means,\ncex = 4,\npch = 17,\nalpha.points = 0.8)\n\nplot1.meanCI <- MakeCIEllipses(lv.1.group.boot\\$BootCube[,c(1:2),], # get the first two components\ncol = col4Means[rownames(lv.1.group.boot\\$BootCube)],\nnames.of.factors = c(\"Lx 1\", \"Ly 1\")\n)\n\nplot1 <- plot.lv1\\$zeMap_background + plot.lv1\\$zeMap_dots + plot1.mean\\$zeMap_dots + plot1.mean\\$zeMap_text + plot1.meanCI\nplot1``````",
null,
"#### 6.3.2.2 Looking at the Second Pair of Latent Variables",
null,
"### 6.3.3 Column Factor scores of the 1st component of table1 and table2\n\nSAM Component 1: Explained by both Episodic memory and Future Memory Component 2: Spatial Memory mostly explains\n\nOSIQ Component 1: Explained by Object Imagery Component 2: Explained by Spatial Imagery\n\n``````#fi:column factor scores for table 1 (P) or Loadings of table 1\nFi<- pls.res\\$TExPosition.Data\\$fi\nFj<- pls.res\\$TExPosition.Data\\$fj\n\naxis1 = 1,\naxis2 = 2,\ndisplay.points = TRUE,\ndisplay.labels = TRUE,\ncol.points = col4Var,\ncol.labels = col4Var,\npch = 20,\ncex = 6,\ntext.cex = 5,\n)\n\nlabel4map <- createxyLabels.gen(x_axis = 1, y_axis = 2,\nlambda = pls.res\\$TExPosition.Data\\$eigs,\ntau = pls.res\\$TExPosition.Data\\$t\n)\n\np.plot``````",
null,
"``````q.loadings <- createFactorMap(Fj,\naxis1 = 1,\naxis2 = 2,\ndisplay.points = TRUE,\ndisplay.labels = TRUE,\ncol.points = col4Var2,\ncol.labels = col4Var2,\npch = 20,\ncex = 6,\ntext.cex = 5,\n)\n\nlabel4map <- createxyLabels.gen(x_axis = 1, y_axis = 2,\nlambda = pls.res\\$TExPosition.Data\\$eigs,\ntau = pls.res\\$TExPosition.Data\\$t\n)\n\nq.plot``````",
null,
"``````grid.barplot.weights <- gridExtra::grid.arrange(as.grob(plot_P.data1),\nas.grob(plot_Q.data2),\nas.grob(plot_P2.data1),\nas.grob(plot_Q2.data2),\n\nncol = 2,nrow = 2 ,\ngp=gpar(fontsize=18,font=3))\n)``````",
null,
"``grid.barplot.weights``\n``````## TableGrob (3 x 2) \"arrange\": 5 grobs\n## z cells name grob\n## 1 1 (2-2,1-1) arrange gtable[layout]\n## 2 2 (2-2,2-2) arrange gtable[layout]\n## 3 3 (3-3,1-1) arrange gtable[layout]\n## 4 4 (3-3,2-2) arrange gtable[layout]\n## 5 5 (1-1,1-2) arrange text[GRID.text.9631]``````\n\nInference Bootstrap\n\nLooking into what the resBootPLSC is giving us.\n\n``````## ------------------------------------------------------------------------------\n## Bootstraped Factor Scores (BFS) and Bootstrap Ratios (BR)\n## for the I and J-sets of a PLSC (obtained from multinomial resampling of X & Y)\n## ------------------------------------------------------------------------------\n## \\$ bootstrapBrick.i an I*L*nIter Brick of BFSs for the I-Set\n## \\$ bootRatios.i an I*L matrix of BRs for the I-Set\n## \\$ bootRatiosSignificant.i an I*L logical matrix for significance of the I-Set\n## \\$ bootstrapBrick.j a J*L*nIter Brick of BFSs for the J-Set\n## \\$ bootRatios.j a J*L matrix of BRs for the J-Set\n## \\$ bootRatiosSignificant.j a J*L logical matrix for significance of the J-Set\n## ------------------------------------------------------------------------------``````\n\n### 6.3.5 Contribution and Bootstrap ratio Bar Plots",
null,
"",
null,
"## 6.4 Conclusion\n\nLatent Variables Pairs\n\nPair 1: Separates the two memory groups well. The latent variables of the two data share a high covariance.\n\nPair 2: Does not really provide more information with the current groupings.\n\nColumn Factor Scores : SAM\n\nDim 1: Explained by both Episodic memory and Future Memory\n\nDim 2: Spatial Memory\n\nColumn Factor Scores : OSIQ\n\nDim 1: Represents Object Imagery\n\nDim 2: Represents Spatial Imagery\n\nInterpretation:\n\nEpisodic Memory and Memory for Future is likely to be stronger for participants who have a preference towards Object Imagery and these participants are under the category of Higher AM.\n\nThe better the Spatial Memory of the participants the better is their ability to remember spatial relations amongst objects or space and loadings also reflect that these participants tend to have a prefeance for spatial imagery. These Participants however belong to the Normal AM group."
]
| [
null,
"https://bookdown.org/kirthanarao95/rm3_book/RM3_book_files/figure-html/unnamed-chunk-80-1.png",
null,
"https://bookdown.org/kirthanarao95/rm3_book/RM3_book_files/figure-html/unnamed-chunk-82-1.png",
null,
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null,
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null,
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null,
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null,
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null,
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null,
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null
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https://www.analystforum.com/forums/frm-forum/91372951 | [
"Standard deviation of the portfolio = (Bh X C X By)\n\nAbove formula is given in the reference topic, however inputs are the same as a normal calculation for standard deviation using weights.\n\nI am not able understand, how to identify which formula to use - I mean in exam, when we see inputs for standard deviation calculation, we will tend to use the formula with weights.\n\nAny solution/thoughts?"
]
| [
null
]
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https://shiela.savingadvice.com/2012/04/ | [
"User Real IP - 3.234.211.61\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => 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156.146.59.20\n)\n\n => Array\n(\n => 122.176.100.151\n)\n\n => Array\n(\n => 185.217.117.237\n)\n\n => Array\n(\n => 49.37.223.97\n)\n\n => Array\n(\n => 101.50.108.80\n)\n\n => Array\n(\n => 124.253.155.88\n)\n\n => Array\n(\n => 39.40.208.96\n)\n\n => Array\n(\n => 122.167.151.154\n)\n\n => Array\n(\n => 172.98.89.13\n)\n\n => Array\n(\n => 103.91.52.6\n)\n\n => Array\n(\n => 106.203.84.5\n)\n\n => Array\n(\n => 117.216.221.34\n)\n\n => Array\n(\n => 154.73.203.131\n)\n\n => Array\n(\n => 223.182.210.117\n)\n\n => Array\n(\n => 49.36.185.208\n)\n\n => Array\n(\n => 111.119.183.30\n)\n\n => Array\n(\n => 39.42.107.13\n)\n\n => Array\n(\n => 39.40.15.174\n)\n\n => Array\n(\n => 1.38.244.65\n)\n\n => Array\n(\n => 49.156.75.252\n)\n\n => Array\n(\n => 122.161.51.99\n)\n\n => Array\n(\n => 27.73.78.57\n)\n\n => Array\n(\n => 49.48.228.70\n)\n\n => Array\n(\n => 111.119.183.18\n)\n\n => Array\n(\n => 116.204.252.218\n)\n\n => Array\n(\n => 73.173.40.248\n)\n\n => Array\n(\n => 223.130.28.81\n)\n\n => Array\n(\n => 202.83.58.81\n)\n\n => Array\n(\n => 45.116.233.31\n)\n\n => Array\n(\n => 111.119.183.1\n)\n\n => Array\n(\n => 45.133.7.66\n)\n\n => Array\n(\n => 39.48.204.174\n)\n\n => Array\n(\n => 37.19.213.30\n)\n\n => Array\n(\n => 111.119.183.22\n)\n\n => Array\n(\n => 122.177.74.19\n)\n\n => Array\n(\n => 124.253.80.59\n)\n\n => Array\n(\n => 111.119.183.60\n)\n\n => Array\n(\n => 157.39.106.191\n)\n\n => Array\n(\n => 157.47.86.121\n)\n\n => Array\n(\n => 47.31.159.100\n)\n\n => Array\n(\n => 106.214.85.144\n)\n\n => Array\n(\n => 182.189.22.197\n)\n\n => Array\n(\n => 111.119.183.51\n)\n\n => Array\n(\n => 202.47.35.57\n)\n\n => Array\n(\n => 42.108.33.220\n)\n\n => Array\n(\n => 180.178.146.158\n)\n\n => Array\n(\n => 124.253.184.239\n)\n\n => Array\n(\n => 103.165.20.8\n)\n\n => Array\n(\n => 94.178.239.156\n)\n\n => Array\n(\n => 72.255.41.142\n)\n\n => Array\n(\n => 116.90.107.102\n)\n\n => Array\n(\n => 39.36.164.250\n)\n\n => Array\n(\n => 124.253.195.172\n)\n\n => Array\n(\n => 203.142.218.149\n)\n\n => Array\n(\n => 157.43.165.180\n)\n\n => Array\n(\n => 39.40.242.57\n)\n\n => Array\n(\n => 103.92.43.150\n)\n\n => Array\n(\n => 39.42.133.202\n)\n\n => Array\n(\n => 119.160.66.11\n)\n\n => Array\n(\n => 138.68.3.7\n)\n\n => Array\n(\n => 210.56.125.226\n)\n\n => Array\n(\n => 157.50.4.249\n)\n\n => Array\n(\n => 124.253.81.162\n)\n\n => Array\n(\n => 103.240.235.141\n)\n\n => Array\n(\n => 132.154.128.20\n)\n\n => Array\n(\n => 49.156.115.37\n)\n\n => Array\n(\n => 45.133.7.48\n)\n\n => Array\n(\n => 122.161.49.137\n)\n\n => Array\n(\n => 202.47.46.31\n)\n\n => Array\n(\n => 192.140.145.148\n)\n\n => Array\n(\n => 202.14.123.10\n)\n\n => Array\n(\n => 122.161.53.98\n)\n\n => Array\n(\n => 124.253.114.113\n)\n\n => Array\n(\n => 103.227.70.34\n)\n\n => Array\n(\n => 223.228.175.227\n)\n\n => Array\n(\n => 157.39.119.110\n)\n\n => Array\n(\n => 180.188.224.231\n)\n\n => Array\n(\n => 132.154.188.85\n)\n\n => Array\n(\n => 197.210.227.207\n)\n\n => Array\n(\n => 103.217.123.177\n)\n\n => Array\n(\n => 124.253.85.31\n)\n\n => Array\n(\n => 123.201.105.97\n)\n\n => Array\n(\n => 39.57.190.37\n)\n\n => Array\n(\n => 202.63.205.248\n)\n\n => Array\n(\n => 122.161.51.100\n)\n\n => Array\n(\n => 39.37.163.97\n)\n\n => Array\n(\n => 43.231.57.173\n)\n\n => Array\n(\n => 223.225.135.169\n)\n\n => Array\n(\n => 119.160.71.136\n)\n\n => Array\n(\n => 122.165.114.93\n)\n\n => Array\n(\n => 47.11.77.102\n)\n\n => Array\n(\n => 49.149.107.198\n)\n\n => Array\n(\n => 192.111.134.206\n)\n\n => Array\n(\n => 182.64.102.43\n)\n\n => Array\n(\n => 124.253.184.111\n)\n\n => Array\n(\n => 171.237.97.228\n)\n\n => Array\n(\n => 117.237.237.101\n)\n\n => Array\n(\n => 49.36.33.19\n)\n\n => Array\n(\n => 103.31.101.241\n)\n\n => Array\n(\n => 129.0.207.203\n)\n\n => Array\n(\n => 157.39.122.155\n)\n\n => Array\n(\n => 197.210.85.120\n)\n\n => Array\n(\n => 124.253.219.201\n)\n\n => Array\n(\n => 152.57.75.92\n)\n\n => Array\n(\n => 169.149.195.121\n)\n\n => Array\n(\n => 198.16.76.27\n)\n\n => Array\n(\n => 157.43.192.188\n)\n\n => Array\n(\n => 119.155.244.221\n)\n\n => Array\n(\n => 39.51.242.216\n)\n\n => Array\n(\n => 39.57.180.158\n)\n\n => Array\n(\n => 134.202.32.5\n)\n\n => Array\n(\n => 122.176.139.205\n)\n\n => Array\n(\n => 151.243.50.9\n)\n\n => Array\n(\n => 39.52.99.161\n)\n\n => Array\n(\n => 136.144.33.95\n)\n\n => Array\n(\n => 157.37.205.216\n)\n\n => Array\n(\n => 217.138.220.134\n)\n\n => Array\n(\n => 41.140.106.65\n)\n\n => Array\n(\n => 39.37.253.126\n)\n\n => Array\n(\n => 103.243.44.240\n)\n\n => Array\n(\n => 157.46.169.29\n)\n\n => Array\n(\n => 92.119.177.122\n)\n\n => Array\n(\n => 196.240.60.21\n)\n\n => Array\n(\n => 122.161.6.246\n)\n\n => Array\n(\n => 117.202.162.46\n)\n\n => Array\n(\n => 205.164.137.120\n)\n\n => Array\n(\n => 171.237.79.241\n)\n\n => Array\n(\n => 198.16.76.28\n)\n\n => Array\n(\n => 103.100.4.151\n)\n\n => Array\n(\n => 178.239.162.236\n)\n\n => Array\n(\n => 106.197.31.240\n)\n\n => Array\n(\n => 122.168.179.251\n)\n\n => Array\n(\n => 39.37.167.126\n)\n\n => Array\n(\n => 171.48.8.115\n)\n\n => Array\n(\n => 157.44.152.14\n)\n\n => Array\n(\n => 103.77.43.219\n)\n\n => Array\n(\n => 122.161.49.38\n)\n\n => Array\n(\n => 122.161.52.83\n)\n\n => Array\n(\n => 122.173.108.210\n)\n\n => Array\n(\n => 60.254.109.92\n)\n\n => Array\n(\n => 103.57.85.75\n)\n\n => Array\n(\n => 106.0.58.36\n)\n\n => Array\n(\n => 122.161.49.212\n)\n\n => Array\n(\n => 27.255.182.159\n)\n\n => Array\n(\n => 116.75.230.159\n)\n\n => Array\n(\n => 122.173.152.133\n)\n\n => Array\n(\n => 129.0.79.247\n)\n\n => Array\n(\n => 223.228.163.44\n)\n\n => Array\n(\n => 103.168.78.82\n)\n\n => Array\n(\n => 39.59.67.124\n)\n\n => Array\n(\n => 182.69.19.120\n)\n\n => Array\n(\n => 196.202.236.195\n)\n\n => Array\n(\n => 137.59.225.206\n)\n\n => Array\n(\n => 143.110.209.194\n)\n\n => Array\n(\n => 117.201.233.91\n)\n\n => Array\n(\n => 37.120.150.107\n)\n\n => Array\n(\n => 58.65.222.10\n)\n\n => Array\n(\n => 202.47.43.86\n)\n\n => Array\n(\n => 106.206.223.234\n)\n\n => Array\n(\n => 5.195.153.158\n)\n\n => Array\n(\n => 223.227.127.243\n)\n\n => Array\n(\n => 103.165.12.222\n)\n\n => Array\n(\n => 49.36.185.189\n)\n\n => Array\n(\n => 59.96.92.57\n)\n\n => Array\n(\n => 203.194.104.235\n)\n\n => Array\n(\n => 122.177.72.33\n)\n\n => Array\n(\n => 106.213.126.40\n)\n\n => Array\n(\n => 45.127.232.69\n)\n\n => Array\n(\n => 156.146.59.39\n)\n\n => Array\n(\n => 103.21.184.11\n)\n\n => Array\n(\n => 106.212.47.59\n)\n\n => Array\n(\n => 182.179.137.235\n)\n\n => Array\n(\n => 49.36.178.154\n)\n\n => Array\n(\n => 171.48.7.128\n)\n\n => Array\n(\n => 119.160.57.96\n)\n\n => Array\n(\n => 197.210.79.92\n)\n\n => Array\n(\n => 36.255.45.87\n)\n\n => Array\n(\n => 47.31.219.47\n)\n\n => Array\n(\n => 122.161.51.160\n)\n\n => Array\n(\n => 103.217.123.129\n)\n\n => Array\n(\n => 59.153.16.12\n)\n\n => Array\n(\n => 103.92.43.226\n)\n\n => Array\n(\n => 47.31.139.139\n)\n\n => Array\n(\n => 210.2.140.18\n)\n\n => Array\n(\n => 106.210.33.219\n)\n\n => Array\n(\n => 175.107.203.34\n)\n\n => Array\n(\n => 146.196.32.144\n)\n\n => Array\n(\n => 103.12.133.121\n)\n\n => Array\n(\n => 103.59.208.182\n)\n\n => Array\n(\n => 157.37.190.232\n)\n\n => Array\n(\n => 106.195.35.201\n)\n\n => Array\n(\n => 27.122.14.83\n)\n\n => Array\n(\n => 194.193.44.5\n)\n\n => Array\n(\n => 5.62.43.245\n)\n\n => Array\n(\n => 103.53.80.50\n)\n\n => Array\n(\n => 47.29.142.233\n)\n\n => Array\n(\n => 154.6.20.63\n)\n\n => Array\n(\n => 173.245.203.128\n)\n\n => Array\n(\n => 103.77.43.231\n)\n\n => Array\n(\n => 5.107.166.235\n)\n\n => Array\n(\n => 106.212.44.123\n)\n\n => Array\n(\n => 157.41.60.93\n)\n\n => Array\n(\n => 27.58.179.79\n)\n\n => Array\n(\n => 157.37.167.144\n)\n\n => Array\n(\n => 119.160.57.115\n)\n\n => Array\n(\n => 122.161.53.224\n)\n\n => Array\n(\n => 49.36.233.51\n)\n\n => Array\n(\n => 101.0.32.8\n)\n\n => Array\n(\n => 119.160.103.158\n)\n\n => Array\n(\n => 122.177.79.115\n)\n\n => Array\n(\n => 107.181.166.27\n)\n\n => Array\n(\n => 183.6.0.125\n)\n\n => Array\n(\n => 49.36.186.0\n)\n\n => Array\n(\n => 202.181.5.4\n)\n\n => Array\n(\n => 45.118.165.144\n)\n\n => Array\n(\n => 171.96.157.133\n)\n\n => Array\n(\n => 222.252.51.163\n)\n\n => Array\n(\n => 103.81.215.162\n)\n\n => Array\n(\n => 110.225.93.208\n)\n\n => Array\n(\n => 122.161.48.200\n)\n\n => Array\n(\n => 119.63.138.173\n)\n\n)\n```\nArchive for April, 2012: My Simple Financial Plan\n Layout: Blue and Brown (Default) Author's Creation\n Home > Archive: April, 2012\n\n# Archive for April, 2012\n\n## Fish and some bills\n\nApril 30th, 2012 at 01:20 am\n\nWhen we got home late Saturday afternoon we found the light bulb in fish tank flickering like crazy and all the fish was so stress out.\n\nDH got a new light bulb yesterday, normally it would have been \\$22.95 but he had some credit from the aquarium store due to selling them some of the baby fish few months ago. In the end it only cost him \\$9.95.\n\n------\nReceived the gas bill for \\$91.95 and AMEX over \\$750 (a lot of it is all the dining out that we been doing, oh my...). Both are due middle of the month.\n\n## Payday\n\nApril 29th, 2012 at 05:48 am\n\nFriday was payday.\n\nOn paydays I finalise the exercise challenge. It was pathetic, only \\$19 in 4 weeks. DH was sick for about 3 weeks so he didn't get to do much exercise at all and I got a bit lazy, ok a lot.\n\nAs mentioned earlier, I put in another \\$1000 into the mortgage. That brings it down to (almost) \\$136K. This make it \\$3000 extra payment for April. We are slightly ahead in this for the year, it will have to slow down now so that we can put a bit more money into our other financial goal for the year.\n\nAccording to my spreadsheet, even if we just put \\$1500 a month from now until the end of the year we should be able to meet goal #1 with no problem, barring the bank don't increase interest rate. The good news is that most economist has predicted a decrease in interest rate by the reserve bank in the next month or so (fingers crossed that the banks will follow).\n\n## Update\n\nApril 29th, 2012 at 03:38 am\n\n3rd time go at typing this, I don't know what is going on. I've decided to now use Notepad and will just cut and paste.\n\nJust a quick update on what I've been doing in the last few day. Been a bit busy.\n\nTuesday night DH went to ALdi to get some groceries.\nHe spent \\$19.55 which left us with \\$7 in the grocery budget for the month. Will get sausages for lunch, should just have enough. Also bought a small mother's day gift for DH's mum. Will still have to think of something else to get for her. I'm running out of ideas.\n\nWednesday was our public holiday. Spent the morning helping mum buy a car. In the afternoon we stayed home to watch the footy game. We got pizza for dinner.\n\nThursday DH put in another \\$50 into his train pass, we are now over our transport budget.\n\nFriday we decided to put another \\$1000 into the mortgage, apart from this it could have been a NSD.\n\nSaturday we finally went to the dentist, our health insurance covered for it this time except for the \\$8 for a special toothpaste for DH. Also \\$12 for the day parking. After the dentist we decided to stay in the city, we thought may as well make good use of our parking. It was such a nice day to just hang out. There were a lot of street performance that kept us entertain. We stayed for lunch and coffee in the afternoon. We decided to use our allowances for this expenses. It was such a lovely day, was not planned especially the spending but worth it. Loved sitting by the riverside with coffee just people watching and listening to this couple basking/singing. Afterwards, on our way home we decided to stopped by the factory outlet place. Ended up spending \\$154 for 3 tops for me (I wanted more winter tops) and 2 t-shirts for DH and 1 business shirt for him.\n\nRalph Lauren was having 70% storewide, we got DH's business shirt from there and one of my top. DH should have enough work shirts now. I actually only really budgeted \\$50 for clothing this month, this is one of those things that I could never budget well.\nMost months we don't buy clothing. But looking at our yearly budget we still got \\$300 in the clothing budget. This purchases now push as over by almost \\$100 for the month. The month ends tomorrow so it is not too bad.\n\n## Car Shopping\n\nApril 24th, 2012 at 02:06 am\n\nI spent about 2 hours last night looking at used car online. It is not for us it is for my mum, 2 weeks ago today she had an accident and her car is no more. Her car insurance provided her a rental car but had to return it yesterday. So now she needs a new car. I found a couple last night and we will go tomorrow to have a look at it.\n\nWhile looking a car for her I kind of had a quick look at one for us. Couple of years ago when our car was not doing too well, when we thought that we might have to get a new one, we went and had a look at some new cars. The one that we like is a lot cheaper now, it's great to know. Our car right now is still pretty good (knock on wood), we did spent over \\$1000 to get it fix. I'm hoping that it will last for another 3 years or so, when we hopefully pay off the house by then. There were also a lot older car that I won't mind get as a second car. But I don't know I'm still not that keen on having two cars.\n\n------------------------\n\nOn another news, I mentioned a week ago here that we got invited to a wedding for next month. We opted not to attend but we will send a gift. I actually got some idea the other night of what gift we will send when I was going through websites that sell gift baskets. I think a basket with a nice bottle of champagne and other goodies would be nice. Unfortunately they don't have a gift registry, that would have been a lot easier. We are not going because the wedding is held about 2-3 hours away. I don't think they will mind, I've never meet the bride to be and I haven't seen/spoken to my friend for about 2 years. I think a lot of our friends won't be able to make it either. I hope they have a lovely day.\n\nAnd oh yesterday was a NSD, for real this time.\n\n## Retirement Fund (superannuation)\n\nApril 23rd, 2012 at 05:07 am\n\nOn Friday I got to check my retirement fund, DH also decided to check his on the weekend. Our retirement funds is now over \\$101K. Finally!!!. The last \\$20,ooo has taken so long because of the GFC. Hopefully it will stay above \\$100K from now on.\n\nI also did a quick look at our networth estimate, I think it has increased over \\$20K since end of last year. Slow but steady we go",
null,
"## NSD no more\n\nApril 23rd, 2012 at 02:24 am\n\nI said yesterday was a NSD, well that was very late in the afternoon here and by then there was no way that we would be going out to spend any money. I was wrong.\n\nAfter dinner I got thinking about mother's day, which is in 3 weeks time. DH's mum is hard to shop for and to top it off she lives in another state. I thought maybe a gift basket it can be delivered straight to her. No luck, she won't like most of it. She doesn't drink and she has a very strict diet. The only thing that might work is the chocolate basket but \\$50 worth of it is a bit much.\n\nAfter that I decided to look at Groupon and other sites like it. I was thinking of maybe a weekend getaway but it can't be too far from their place, she don't fly so it must be driving distance. Nothing at the moment, I will keep trying.\n\nAnyway, while going through the Groupon website I found a handheld steamer for \\$49 each or \\$90 for two. Unfortunately, MIL already got one. I asked her about it and apparently it is not bad. I'm willing to give it a try for this price. MIL one cost her over \\$200 but it came with the vacuum cleaner. I got 2, I will give the other one to my mum as part of her Mother's day gift. I used my birthday money for my one, MIL actually suggested it.\n\nSo to recap, I officially spent \\$45 last night for Mother's day gift for next month. And I've taken \\$45 out of my allowance for my steamer.\n\n## Weekend Update\n\nApril 22nd, 2012 at 07:22 am\n\nYesterday (Saturday) we went to my mum's place to again celebrate my birthday. My brother came and few other people were there too. We had a bbq and lots of other food that mum cooked. Also had a birthday cake. DH decided to update mum's computer, so we ended up spending all of the afternoon there. We came home with lots of leftover, I don't even need to cook for dinner tonight.\n\nToday was a relaxing day. Was meant to plant the lime tree but my mum thinks that I should wait until the full moon to plant it. She believes on that stuff, may as well give it a go which means I won't be able to plant it for another 2 weeks. May aswell I think The soil is still not right yet and I will also need to get some sort of material to protect it during winter, which mean I will need to build/ put posts around it.\n\nToday was a NSD, yesterday I bought couple bags of coal for \\$6. Very low spend weekend, was not plan really. We didn't do any grocery shopping at all. I'm not too worried though, we got plenty of food in the freezer. Also, this Wednesday is a public holiday we should be able to get milk and fresh produce then. DH should be able to make his yogurt then too. Only got \\$27 left in the grocery budget, should be enough if we are careful.\n\n## Extra Mortgage Pymt\n\nApril 20th, 2012 at 11:47 pm\n\nAbout to go to mum's for lunch so I just wanna quickly update my side bar. Decided to put \\$2K into the mortgage, I just couldn't wait until Friday's payday. I will probably put a bit more in it again on Friday, I just don't know how much yet.\n\nWell, off to mum we go...\n\n## A more productive day\n\nApril 20th, 2012 at 09:31 am\n\nAfter yesterday's lazy day, I was so much more productive today.\n\n- I cleaned our bathroom, I gave it good scrub.\n- Did a bit of dusting\n- Vaccuumed the floors\n- Got the laundry ready for tomorrow, including soaking some of the whites. We try to do our laundry on the weekend when its off-peak time for our electricity.\n- Also did some gardening. I re-potted an orchid got 4 bigger pots out of it. It really needed it, I actually had to break the old pot to get it out. I got another 3 or 4 more to do. I will need to get more of the bigger pots. Also planted some of my coriander seedlings into its permanent place.\n- Called and sorted my super fund (retirement fund) to get access to it online. Good news it is up, more than I expected it to be. I normally contribute to it \\$40 twice a month and then I get a co-contribution from the government - around \\$1000 per year. Because I haven't work for months now there hasn't been anymore money going into it but my little contribution.\n- Cooked dinner early, no chance of take away tonight. And that made it a NSD today.\n\n## Utilities\n\nApril 20th, 2012 at 01:03 am\n\nI've just been reviewing my budget for this month, I just realise that we don't have any utility bill due. Normally the electricity bill would have been due by now. Thanks to the power of the sun and the work of our solar panels we only owe \\$2.88 for this quarter. Because it is below \\$10 they don't require a payment they will just add it on to the next bill. I'm hoping next electric bill will be even better, maybe we can get some credit instead. We lost about 2 weeks worth in the last quarter because the inverter wasn't working properly. It is all sorted now and we have not any problem since then.\n\nAny saving we can get from this should hopefully help with the transport cost this year. Both petrol cost and train cost has gone up a lot this year. Two months ago they increase the public transport ticket by about 10%+ again.\n\n## Could have been\n\nApril 19th, 2012 at 12:46 pm\n\nIt should have been NSD again today until I got too lazy to cook dinner. We had a bit of left over and there are even some stuff in the freezer but I got a craving for Thai curry. I called DH to get some on his way home, \\$12.90. I will take this out of my allowance money to be fair.\n\nI spent some time this afternoon trying to log on to my retirement fund but for some reason it wouldn't let me. I will have to give them a call tomorrow.\n\nI also checked my managed funds, it is slowly going up. I only put \\$200 per month so I don't really expect much from it but it is good to know that it is going in the right direction.\n\n## NSD and stock market\n\nApril 19th, 2012 at 02:06 am\n\nYesterday was a NSD, maybe today would be too if DH don't need to fill up the car. I'm pretty sure it is due for a fill up, I don't drive so I never really know when it needs it but it seems like it has been a while since he went to the petrol station. He might also need to top up his train pass.\n\nJust been looking at the stock market last night and there are a couple stocks that I want to get. DH got some cash in his retirement fund to get some or he could get some in his margin loan account. I think his retirement account might be better, he said he got \\$10K cash just seating in it ready to buy. We won't need to pay interest if we use that instead of the margin loan.\n\nSpeaking of the stock market, my trading is doing alright but not as well as I hoped it to be. The market is pretty much just been going side ways. I think since I started trading in Nov 2011 it has only been up 3-4%, but at least my trade has gone up around 10%. It was up around 20% at the start but I have given some of it back a bit lately. I will do a final analysis of my trades once I've closed off 20 trades."
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https://www.npmtrends.com/abstract-data-types | [
"# abstract-data-types\n\nThis module aims to provide a full suite of abstract data types. At present it provides the abstract data type, Queue, Linked List, Stack, Binary Tree an Binary Search Tree.\n\n## Stats\n\nstars 🌟issues ⚠️updated 🛠created 🐣size 🏋️♀️\nabstract-data-types",
null,
"",
null,
"# abstract-data-types\n\nThe future of this module is that it will be able to return instances of the following abstract data types:\n\n• Queue This is currently implemented\n• Linked List This is currently implemented\n• Stack This is currently implemented\n• Binary Tree This is currently implemented\n• Binary Search Tree This is currently implemented\n• Graph\n• Hash Table\n• Heap\n\n# Installation\n\n``````\\$ npm install abstract-data-types\n``````\n\n# Getting Started\n\nThe individual abstract data types are instantiated from factory functions that exist as properties of the abstract-data-types module. The module can be accessed as follows.\n\n``````var adt = require('abstract-data-types');\n``````\n\nThe individual factory functions can be used to create instances of the abstract data types like this.\n\n``````var q = adt.createQueue(); // To create an empty Queue\nvar s = adt.createStack(); // To create an empty Stack\nvar bt = adt.createBinaryTree(); // To create an empty Binary Tree\nvar bst = adt.createBinarySearchTree(); // To create an empty Binary Search Tree\nvar g = adt.createGraph(); // To create an empty Graph\nvar ht = adt.createHashTable(); // To create an empty Hash Table\nvar h = adt.createHeap(); // To create an empty Heap\n``````\n\nThe rest of this document explains each of the operations that can be performed these abstract data type instances.\n\n## Queue\n\nThis Queue is implements all the standard Queue operations. These are:\n\n• enqueue adds a new item to the back of the queue\n• dequeue returns the item at the front of the queue and removes it from the queue\n• front returns the item at the front of the queue\n• size returns the number of items in the queue\n• isEmpty determines whether the queue is empty and returns true if it is and false otherwise\n\n#### Enqueue\n\nThe enqueue operation adds a new item to the back of the queue. To do this call the enqueue method on a Queue instance passing it the data item you wish to add to the Queue.\n\n``````q.enqueue(item);\n``````\n\n#### Dequeue\n\nThe dequeue method returns the item at the front of the queue and removes that item from the queue. To do this call dequeue on a queue object.\n\n``````var frontItem = q.dequeue();\n``````\n\nIf dequeue is called on an empty queue, the following error is thrown.\n\n``````new Error('adt-queue.dequeue(): Tried to dequeue an empty queue!');\n``````\n\n#### Front\n\nThe front method returns the item at the front of the queue, but unlike dequeue it does not remove it from the queue.\n\n``````var frontItem = q.front();\n``````\n\nIf front is called on an empty queue, the following error is thrown.\n\n``````new Error('adt-queue.front(): Tried to get the front of an empty queue!');\n``````\n\n#### IsEmpty\n\nThe isEmpty method is a boolean function that, when called on a queue, returns true if the queue is empty and false otherwise.\n\n``````q.isEmpty();\n``````\n\nThe Linked List implements all of the standard Linked List functions. These are given below:\n\n• add adds a new item to the list at a specific position\n• remove removes an item from the list at a specific position\n• get returns the item at at a specific position in the list\n• size returns the number of items in the List\n• isEmpty determines whether the List is empty and returns true if it is and false otherwise\n\nThe postion index starts from 0.\n\nThe add method takes the item to be added to the list and the position that the item should be added in. If the position is less than or equal to 0, the item is added to the front of the list. If the position is greater than or equal to the size of the list, the item is added to the end of the list.\n\n``````ll.add(position, item);\n``````\n\n#### Remove\n\nThe remove method removes the item at a specific position in the list. If the position requested is less than 0 or geater then or equal to the length of the list, an error is thrown.\n\n``````ll.remove(position);\n``````\n\n#### Get\n\nThe get method returns the item at a specific position in the list. If the position requested is less than 0 or geater then or equal to the length of the list, an error is thrown.\n\n``````ll.get(position);\n``````\n\n#### Size\n\nThe size method returns the number of items in the list.\n\n``````ll.size();\n``````\n\n#### IsEmpty\n\nThe isEmpty method is a boolean function that, when called on a list, returns true if the list is empty and false otherwise.\n\n``````ll.isEmpty();\n``````\n\n## Stack\n\nThe Stack implements all of the standard Stack functions. These are given below:\n\n• push adds a new item to the top of the stack\n• pop returns the item at the top of the stack and removes it from the stack\n• top returns the item at the top of the stack and leaves it on the stack\n• size returns the number of items in the stack\n• isEmpty determines whether the stack is empty and returns true if it is and false otherwise\n\n#### Push\n\nThe push method adds a new item to the top of the stack. It also returns the stack so that push opertations can be chained.\n\n``````s.push(item);\n``````\n\n#### Pop\n\nThe pop method returns the item at the top of the stack and removed it from the stack.\n\n``````s.pop();\n``````\n\n#### Top\n\nThe pop method returns the item at the top of the stack and leaves it on the stack.\n\n``````s.top();\n``````\n\n#### Size\n\nThe size method returns the number of items in the stack.\n\n``````s.size();\n``````\n\n#### IsEmpty\n\nThe isEmpty method is a boolean function that, when called on a stack, returns true if the stack is empty and false otherwise.\n\n``````s.isEmpty();\n``````\n\n## Binary Tree\n\nThe Binary Tree in this module implements the standard methods of the Binary Tree abstract data type. These methods are:\n\n• isEmpty determines whether the tree is empty and returns true if it is and false otherwise\n• getRootItem returns the item at the root of the tree\n• setRootItem updates the item at the root of the tree\n• getLeftTree returns the left sub tree\n• getRightTree returns the Right sub tree\n• attachLeft attaches at tree or an item as the left sub tree\n• attachRight attaches at tree or an item as the right sub tree\n• detachLeft removes the left sub tree\n• detachRight removes the right sub tree\n• count returns the number of nodes in the tree\n\n#### Creation\n\nThe factory function used to create a new Binary Tree can be called in the following 2 ways:\n\n• By passing in no arguments - this will create an empty tree.\n\n• By passing in and item - this will create a single node tree with the item at the root.\n\n#### Is Empty\n\nThe isEmpty method is a boolean function that, when called on a tree, returns true if the tree is empty and false otherwise.\n\n``````bt.isEmpty();\n``````\n\n#### Get Root Item\n\nThis method returns the item at the root of the tree and is called as follows.\n\n``````bt.getRootItem();\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-tree.getRootItem(): Tried to get the root item of an empty tree');\n``````\n\n#### Set Root Item\n\nThis method puts an item at the root of the tree and is called as follows.\n\n``````bt.setRootItem(item);\n``````\n\n#### Get Left Tree\n\nThis method returns the left sub tree.\n\n``````bt.getLeftTree();\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-tree.getLeftTree(): Tried to get the left tree of an empty tree');\n``````\n\n#### Get Right Tree\n\nThis method returns the right sub tree.\n\n``````bt.getRightTree();\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-tree.getLeftTree(): Tried to get the left tree of an empty tree');\n``````\n\n#### Attach Left Tree\n\nThis method can be called by either passing in another Binary Tree, or by passing in an item that is not a Binary Tree.\n\nIf a Binary Tree is passed in, that tree will be attached as the left sub tree. The method is called like this.\n\n``````bt.attachLeft( tree );\n``````\n\nIf a non Binary Tree item is passed in, a new single node tree is created out of that item and this new tree is attached as the left sub tree.\n\n``````bt.attachLeft( item );\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-tree.attachLeft(): Attempt to attach a left sub tree to an empty tree');\n``````\n\n#### Attach Right Tree\n\nThis method can be called by either passing in another Binary Tree, or by passing in an item that is not a Binary Tree.\n\nIf a Binary Tree is passed in, that tree will be attached as the right sub tree. The method is called like this.\n\n``````bt.attachRight( tree );\n``````\n\nIf a non Binary Tree item is passed in, a new single node tree is created out of that item and this new tree is attached as the right sub tree.\n\n``````bt.attachRight( item );\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-tree.attachRight(): Attempt to attach a right sub tree to an empty tree');\n``````\n\n#### Detach Left Tree\n\nThis method is used to remove the left sub tree from the main tree. It also returns the tree that it detaches. it is called as follows.\n\n``````bt.detachLeft();\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-tree.detachLeft(): Attempt to detach the left sub tree of an empty tree');\n``````\n\n#### Detach Right Tree\n\nThis method is used to remove the right sub tree from the main tree. It also returns the tree that it detaches. it is called as follows.\n\n``````bt.detachRight();\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-tree.detachRight(): Attempt to detach the right sub tree of an empty tree');\n``````\n\n#### Count\n\nThis method returns the number of nodes in the tree.\n\n``````bt.count();\n``````\n\n## Binary Search Tree\n\nThe Binary Search Tree implements the following Binary Search Tree methods.\n\n• insert inserts a new item into the search tree in the correct position, maintaining the validity of the binary search tree\n• delete removes an item from the tree maintaining the validity of the binary search tree\n• retrieve returns a specific item in the tree\n• isEmpty determines whether the tree is empty and returns true if it is and false otherwise\n• getRootItem returns the item at the root of the tree\n• getLeftTree returns the left sub tree\n• getRightTree returns the Right sub tree\n• count returns the number of nodes in the tree\n\n#### Insert\n\nThis method inserts an item into the tree and makes sure that the validity of the binary seach tree is maintained.\n\n``````bst.insert(item);\n``````\n\nThe search key used to determine the position of an item within the binary search tree as it is being inserted is derived from the item in the following way.\n\n1. If the item is null, the item is not inserted into the tree and the following error is thrown.\n\nthrow new Error('adt-binary-search-tree utils.getKey(): item is null');\n\n2. If the item is an object that does not have a property named 'key', the item is not inserted into the tree and the following error is thrown.\n\nthrow new Error('adt-binary-search-tree utils.getKey(): item has no key');\n\n3. If the item is an object with a property named 'key' and that key property is null, the item is not inserted into the tree and the following error is thrown\n\nthrow new Error('adt-binary-search-tree utils.getKey(): item is null');\n\n4. If the item is an object with a property named 'key' and that key property is an object, the item is not inserted into the tree and the following error is thrown\n\nthrow new Error('adt-binary-search-tree utils.getKey(): item.key is an object');\n\n5. If the item is an object with a property named 'key' and this key property is not an object and is not null, the value of this key property is used as the item's search key within the tree.\n\n6. If the item is not an object and is not null, the item value is used as the item's search key within the tree.\n\n#### Delete\n\nThis method removes the item identified by the parameter 'key' from the tree and makes sure that the validity of the binary seach tree is maintained.\n\n``````bst.delete(key);\n``````\n\nIf the key does not exist in the tree, the following error is thrown.\n\n``````throw new Error('adt-binary-search-tree.delete(): The key is not in the tree');\n``````\n\n#### Retrieve\n\nThis method returns the item identified by the parameter 'key' from the tree and leaves the tree unchanged.\n\n``````bst.retieve(key);\n``````\n\nThe method returns null if the key does not exist in the tree.\n\n#### Is Empty\n\nThe isEmpty method is a boolean function that, when called on a tree, returns true if the tree is empty and false otherwise.\n\n``````bst.isEmpty();\n``````\n\n#### Get Root Item\n\nThis method returns the item at the root of the tree and is called as follows.\n\n``````bst.getRootItem();\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-search-tree.getRootItem(): Tried to get the root item of an empty tree');\n``````\n\n#### Get Left Tree\n\nThis method returns the left sub tree.\n\n``````bst.getLeftTree();\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-search-tree.getLeftTree(): Tried to get the left tree of an empty tree');\n``````\n\n#### Get Right Tree\n\nThis method returns the right sub tree.\n\n``````bts.getRightTree();\n``````\n\nIf this method is called on an empty tree it throws an error as follows.\n\n``````throw new Error('adt-binary-search-tree.getLeftTree(): Tried to get the left tree of an empty tree');\n``````\n\n#### Count\n\nThis method returns the number of nodes in the tree.\n\n``````bst.count();\n``````\n\nIf you find any bugs or have a feature request, please open an issue on github!"
]
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https://coding4medicine.com/cheatsheets/R-basics.html | [
"## Environment\n\n``````help\n\n# Get all variables\nls()\n\n# Checking the type of variable\ntypeof(x)\n\n# Get working directory\ngetwd()\n``````\n\n## Mathematical operators\n\n``````# Arithmetic operations\n> 2+3\n 5\n\n# Power - 2 to the power 3\n> 2^3\n 8\n\n# Modulo and divisor\n> 10%%3\n 1\n\n> 10%/%3\n 3\n``````\n\n## Mathematical functions\n\n`````` # Mathematical pi\npi\n\n# Converting float to integer\nfloor\nceiling\ntrunc\nround\n\n# Mathematical functions\nabs\nsqrt\n\n# Logarithmic functions\nexp\nlog\nlog10\n``````"
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https://jp.mathworks.com/help/lte/ref/lteulresourcegridsize.html | [
"# lteULResourceGridSize\n\n## Syntax\n\n``d = lteULResourceGridSize(ue)``\n``d = lteULResourceGridSize(ue,p)``\n\n## Description\n\nexample\n\n````d = lteULResourceGridSize(ue)` returns the size of the uplink resource array generated from UE-specific settings `ue`. For more information on the resource grid and the multidimensional array used to represent the resource elements for one subframe across all configured antenna ports, see Represent Resource Grids. ```\n\nexample\n\n````d = lteULResourceGridSize(ue,p)` also specifies the number of antenna planes in the array. ```\n\n## Examples\n\ncollapse all\n\nConfigure UE-specific settings.\n\n`cfgul = struct(NULRB=6,NTxAnts=2,CyclicPrefixUL=\"Normal\");`\n\nGet the uplink subframe resource array size.\n\n`d = lteResourceGridSize(cfgul);`\n\nGenerate an uplink resource array of the appropriate size.\n\n`gridul = zeros(d);`\n\nConfigure UE-specific settings.\n\n`ue = struct(NULRB=25,CyclicPrefixUL=\"Normal\");`\n\nGet the uplink subframe resource array size for the specified configuration and four antenna planes.\n\n```p = 4; d = lteResourceGridSize(ue,p);```\n\nCreate a resource array of the appropriate size.\n\n`gridul = zeros(d);`\n\n## Input Arguments\n\ncollapse all\n\nUE-specific settings, specified as a structure containing these fields.\n\nNumber of uplink resource blocks (RBs), specified as an integer in the interval [6, 110].\n\nData Types: `double`\n\nCyclic prefix length, specified as `\"Normal\"` or `\"Extended\"`.\n\nData Types: `char` | `string`\n\nNumber of transmission antennas, specified as `1`, `2`, or `4`. If you use the syntax containing the `p` input, the function ignores this field and uses the `p` input value instead.\n\nData Types: `double`\n\nData Types: `struct`\n\nNumber of antenna planes in the uplink resource array, specified as a positive integer.\n\nData Types: `double`\n\n## Output Arguments\n\ncollapse all\n\nUplink resource array size, returned as a three-element row vector of the form [N M P].\n\n• N is the number of subcarriers, which is given by 12 × NRB, where NRB is the number of uplink RBs.\n\n• M is the number of SC-FDMA symbols in a subframe: 14 for normal cyclic prefix and 12 for extended cyclic prefix.\n\n• P is the number of transmission antennas.\n\nData Types: `double`\n\n## Version History\n\nIntroduced in R2014a"
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https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_13&diff=prev&oldid=103567 | [
"# Difference between revisions of \"1983 AHSME Problems/Problem 13\"\n\n## Problem\n\nIf",
null,
"$xy = a, xz =b,$ and",
null,
"$yz = c$, and none of these quantities is",
null,
"$0$, then",
null,
"$x^2+y^2+z^2$ equals",
null,
"$\\textbf{(A)}\\ \\frac{ab+ac+bc}{abc}\\qquad \\textbf{(B)}\\ \\frac{a^2+b^2+c^2}{abc}\\qquad \\textbf{(C)}\\ \\frac{(a+b+c)^2}{abc}\\qquad \\textbf{(D)}\\ \\frac{(ab+ac+bc)^2}{abc}\\qquad \\textbf{(E)}\\ \\frac{(ab)^2+(ac)^2+(bc)^2}{abc}$\n\n## Solution\n\nFrom the equations, we deduce",
null,
"$x = \\frac{a}{y}, z = \\frac{b}{x},$ and",
null,
"$y = \\frac{c}{z}$. Substituting these into the expression",
null,
"$x^2+y^2+z^2$ thus gives",
null,
"$\\frac{a^2}{y^2} + \\frac{b^2}{x^2} + \\frac{c^2}{z^2} = \\frac{a^2x^2z^2+b^2y^2z^2+c^2y^2x^2}{x^2y^2z^2} = \\frac{a^2b^2+b^2c^2+c^2a^2}{x^2y^2z^2}$, so the answer is",
null,
"$\\boxed{\\textbf{(E)}\\ \\frac{(ab)^2+(ac)^2+(bc)^2}{abc}}$.\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
null,
""
]
| [
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"https://latex.artofproblemsolving.com/e/0/1/e010be6d83fe43aa3ff71909fdfd75d16fd16abf.png ",
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https://www.abcassignmenthelp.com/statistics-for-life-and-social-science | [
"# Statistics For Life and Social Science",
null,
"MATH1041 Computing assignment\n\n\"BW\" \"HTL\" \"SEX\"\n50.58 0.88 1\n52.45 1.38 1\n45.21 1.1 1\n63.9 1.83 1\n58.22 1.38 1\n51.05 1.03 1\n60.88 1.57 0\n54.46 1.23 0\n58.31 0.99 0\n49.7 1.23 0\n54.27 1.27 1\n54.09 1.37 0\n48.27 1.58 1\n58.57 1.63 1\n57.23 0.58 0\n58.86 0.98 1\n60.17 1.55 1\n50.24 2.93 1\n50.91 0.95 0\n50.37 1.09 0\n48.31 1.07 1\n61.16 1.83 0\n44.52 1.54 1\n47.58 1.24 0\n59.15 0.91 1\n60.12 1.71 1\n54.72 1.54 0\n\nFormat\n\nHere are some more details that may assist you:\n\n• Regarding the overall assignment structure, this is up to you, just remember to keep it clear and concise. If you are answering questions in the given order (that is, 1a), b), etc.), then this is fine. You don’t need to re-write the assignment question again.\n\n• You are required to type up your entire assignment (rather than scanning and taking screenshots). If you are using Word you should use the equation editor for any maths notation. If you don’t have Word then please use the School computers. Please convert and submit your assignment in pdf.\n\n• You are asked to produce SIX graphs/plots for this assignment. You are required to produce these in RStudio. You may want to use the par(mfrow=c(2,3)) function to construct all six graphs per plot (this is optional), see Section R1.4 “Transforming data using RStudio”of the RStudio “How-To-Manual” available on Moodle.\n\n• We recommend adding some working out for some of the questions involving calculations. But try to keep your solutions brief and concise (since there is a page limit). It’s good practice for the exam and in case you get the wrong answer you have some workings to gain marks from. Your working could consist of RStudio commands or perhaps the main steps on how you arrived at your answer. You don’t need to add all of your R-code! • Keeping your results to 2 or 3 decimal places should be fine.\n\n• There is no requirement for font size and line spacing but obviously don’t make things too small.\n\nScenario\n\nA team of researchers were interested in studying the impacts of drought on sheep livestock in farms around New South Wales and Queensland, Australia. In particular, the researchers wanted to compare the average body weight of sheep from five years ago (when there was little drought) to now (Spring, 2018) where drought is of serious concern. To obtain their data, the research team decided to collect a random sample of sheep from a very large sheep population on a farm affected by the drought. This random sample of data consists of sheep body weight measurements (measured in kilograms), head-to-tail length measurements (measured in metres) and their gender (male/female). The text file contains your unique data of length n in separate rows consisting of 3 variables: BW which corresponds to sheep body weights, HTL which corresponds to sheep head-to-tail lengths, and SEX which corresponds to gender (0 = Female and 1 = Male). Your job is to assist the research team by analysing the data set provided to you.\n\nThe questions you need to answer in your assignment submission are given below. Please make sure your assignment is converted to pdf format.\n\n1. (a) Calculate the sample mean and sample standard deviation of your sample of sheep body weight (BW) measurements.\n\n(b) Produce a normal quantile plot of your sample of sheep body weight measurements (see Section R2.6 “How to produce a normal quantile plot using RStudio”). Include this plot in your submitted assignment, properly labelled.\n\n(c) By referring to the normal quantile plot obtained in Part 1b briefly discuss if the sheep body weights are approximately normally distribution.\n\n2. Let µ be the population mean body weight (in kg) of sheep (of any gender) on the farm now (Spring, 2018). The research team decided to compare the current sheep mean body weight with the mean from five years ago. The known mean body weight for sheep from five years ago was 60kg.\n\n(a) Test the hypothesis that µ is equal to 60. You must summarize all steps: state the null (H0) and alternative hypotheses (Ha) relevant to the research objectives stated in this scenario, the value of a suitable test statistic, the sampling distribution for this statistic, a P-value, your summary of significance and conclusion in plain language.\n\n(b) Some assumptions need to be made for the sampling distribution of the test statistic (as given in Part 2a) to be valid. State these assumptions.\n\n(c) Discuss whether the assumptions from Part 2b are satisfied?\n\n(d) Produce a 95% confidence interval for µ, the mean body weight of sheep. For this question you may assume that it is appropriate to use a t-distribution. Make sure you write down all the required steps to calculate this interval. Does this confidence interval include the value 60? Explain whether your confidence interval is consistent with your conclusions from the hypothesis test in Part 2a.\n\n3. The research team were also interested in studying:\n\n• the relationship between body weight and gender; and\n\n• the relationship between body weight and head-to-tail length.\n\n(a) Produce a comparative boxplot for sheep body weight against gender. Include this plot in your submitted assignment, properly labelled.\n\n(b) Describe any differences or similarities in the distribution of body weight of sheep for the different genders using your comparative boxplot from Parts 3a.\n\n(c) Construct an appropriate graphical summary to visualize the relationship between body weight and head-to-tail length. Include this plot in your assignment, properly labelled.\n\n(d) Summarize the key features of your plot from Part 3c.\n\n(e) Suggest an appropriate numerical summary to quantify the linear relationship between body weight and head-to-tail length. Report and comment on this value.\n\n(f) The research team wanted to predict sheep body weight from head-to-tail length measurement by fitting a linear regression model. Would you recommend the research team do this? Explain briefly.\n\n4. The research team decided to investigate the head-to-tail length (HTL) measurement in more detail.\n\n(a) Produce a five number summary for the HTL measurements.\n\n(b) Produce a histogram for the HTL measurements. Include this histogram in your submitted assignment properly labelled.\n\n(c) In MATH1041, we looked at the effect of transforming data. Using the HTL measurements, perform:\n\n(1) a log transformation; and\n\n(2) a square-root transformation, and produce a histogram for each of these. Include these histograms in your submitted assignment properly labelled.\n\n(d) Summarize the key features of each histogram from Parts 4b and 4c (that is, the raw data, and each of the transformations). Please comment on central location, spread, and (any) skewness/symmetry.\n\n(e) Do you think these transformations reduced any skewness? Explain briefly.\n\n1 a) The sample mean and standard deviation of the sheep body weight in r studio is calculated as follows:\n\nfile = \"data.csv\"\n\nmean(data\\$BW)\n\nsd(data\\$BW)",
null,
"1 b) The normal quantile graph of the sheep body weight in r is as follows:\n\n# Normal Quantile graph\n\nqqnorm(data\\$BW, main='Normal Q-Q Plot')\n\nqqline(data\\$BW)",
null,
"1 c) The above diagram shows us that the point in the graph is not present on the straight line while we know that the graph to be normal , point should be nearer or almost on the straight line. Here this is not the case. So we can say that it is not normal distribution.\n2)\n\na) In this problem the population mean body weight of sheep is calculated above and we get the value using mean(data\\$BW). The value we get is 54.1963 while the mean value is 60 age five years. Now to check the hypothesis is it is equal to 60. We have used the t-test. But the result we get is different. The assumed hypothesis is as follows:\n\nH0: mu =60\n\nH1: mu != 60\n\nThe r script for the t test is as follows:\n\nt.test(data\\$BW, mu=60)",
null,
"b) The t-test analysis is used to compare the two mean value while we know that T-distribution is basically a continuous distribution which arises from the estimation of the mean. The assumptions which is considered regarding\n\nc) As we have observed that we are not getting the points which mentioned above in the question is not satisfying accordingly, because our sample size is very less. The second point is when we have drawn the graph we are not getting the bell shaped curve and our data is not continuous. So we can clearly say that assumptions made by us is not satisfying.\n\nd) To produce the 95% percent confidence interval, the mean body weight of sheep. In this we have used the t-test for the body weight of sheep. The R-script of the mean body weight of sheep is as follows:\n\nfile = \"data.csv\"\n\nt.test(data\\$BW)\n\nthe result for this R script is:",
null,
"In this we have observed that it has not included the confidence interval value 60. In this confidence interval we have observed one thing is that the mean which we got is 52.08 to 56.30 while the actual mean of the sheep body weight is 54.19, which is between the confidence interval mean, while the mean is not equal to 60. So we can say that the alternative hypothesis which we have considered is true, and our data is consistent then we are getting our mean between the confidence interval.\n\n3) a) The comparative boxplot for sheep body against the gender using r is as follows:\n\nfile = \"data.csv\"\n\nboxplot(split(data\\$BW, data\\$SEX), xlab=\"gender\", ylab=\"body Weight\")",
null,
"b) After observing the above graph, we are not getting any kind of similarities on the distribution. Because in the above graph there is no female who is below the weight 45 while in the men section we can easily say that there is male in the given distribution who is below the 45 and above 65. While in case of the women there is no woman whose body weight is greater than 65. Hence this is observation which supports my view that there is no similarities.\n\nc) The visualization of the relationship between the body weight and head to tail length. The r script and its graph is as follows:\n\nlibrary(ggplot2)\n\nfile = \"data.csv\"\n\nqplot(data1\\$BW, data1\\$HTL, data=data1, stat=\"summary\", fun.y=\"sum\")",
null,
"d) The key features from the part 3c is there is only one point which is outside the all the point near to the y -axis. We can say it outlier. In this rest all the points is below the 2.0. this show that there is dense part which is used to describe the data. The descriptive summary statistics of the plot is calculated using the summary function. This gives the result as follows:\n\nBW HTL\n\nMin. :44.52 Min. :0.580\n\n1st Qu.:50.30 1st Qu.:1.050\n\nMedian :54.27 Median :1.270\n\nMean :54.20 Mean :1.348\n\n3rd Qu.:58.72 3rd Qu.:1.560\n\nMax. :63.90 Max. :2.930\n\ne) The numerical summary to quantify the linear relationship between body weight and head-to-tail length is that we will implement the linear regression to get this. The another method to get the numerical summary is to get the mean, median and all the statistical value of the data. Which we will get using the summary function on the data.",
null,
"f) The implementation of the linear regression model on the attributes body weight and head-to-tail is as follows:\n\nRegression <- lm(data\\$BW ~ data\\$HTL)\n\nsummary(Regression)\n\nnewdata<- data.frame(x=data\\$BW)\n\npredicted_val<-predict(Regression, newdata, type=\"response\")\n\npredicted_val",
null,
"4 a) The five number summary is a set of 5 descriptive statistics for summarizing the continuous univariate dataset. it consists the minimum, 1st quartile, median, 3rd quartile and maximum. The fivenumber summary of HTL is as follows:\n\nfivenum(data\\$HTL)",
null,
"b) The histogram of the HTL is as follows:",
null,
"hist(data\\$HTL)\n\nc) The log transformation of Head-to-Tail is as follows:\n\nlog10(data\\$HTL)",
null,
"Square root transformation of HTL is as follows:\n\nT_sqrt<- sqrt(data\\$HTL)\n\nT_sqrt\n\nhist(T_sqrt)",
null,
"",
null,
"",
null,
"d) The central tendency of the log transformation of the histogram on the htl is coming between the 0.0 to 0.2 while the frequency is between 4 to 6 i.e. 5 and there is no more skewness part. While the central tendency for thee square root transformation is between the 1. 2 to 1.4 while the frequency of this transformation is between 6 to8 that is 7.5.\n\ne) According to me there is no more role of the skewness in this part. Because we can see easily the histograms and we can say that there is no role of the skewness in the dataset and its analysis part. While we know that transformation is used to reduce the skewness when we take the log transformation it reduces the right skewness due to the square transformation we reduce the left skewness. Here our dataset is small so no more role of the skewness."
]
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http://slh.alljournals.cn/search_by_author.aspx?subject=mathematical_chemical&major=sx&field=author_name&encoding=utf-8&q=Jeff+H.-C.LIAO | [
"",
null,
"首页 | 本学科首页 官方微博 | 高级检索\n\n 按 中文标题 英文标题 中文关键词 英文关键词 中文摘要 英文摘要 作者中文名 作者英文名 单位中文名 单位英文名 基金中文名 基金英文名 杂志中文名 杂志英文名 栏目英文名 栏目英文名 DOI 责任编辑 分类号 杂志ISSN号 检索 检索词:\n\n 国内免费 1篇\n 数学 1篇\n 2018年 1篇\n\n1\n1.\nHere presented is a matrix representation of recursive number sequences of order 3 defined by a_n = pa_(n-1) + qa_(n-2) + ra_(n-3) with arbitrary initial conditions a_0, a_1 = 0, and a_2 and their special cases of Padovan number sequence and Perrin number sequence with initial conditions a_0 = a_1 = 0 and a_2 = 1 and a_0 = 3, a_1 = 0, and a_2 = 2, respectively. The matrix representation is used to construct many well known and new identities of recursive number sequences as well as Pavodan and Perrin sequences. 相似文献\n1"
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https://techutils.in/blog/tag/normal-distribution/ | [
"## #StackBounty: #normal-distribution #multivariate-normal-distribution #linear-algebra #complex-numbers Density of a degenerate complex n…\n\n### Bounty: 50\n\nI’m looking for the properties of a degenerate complex normal distribution. Wikipedia has a section on degeneracy in the real multivariate normal case here, stating that the density (in a subspace of $$mathbb R^k$$ where the distribution is supported) looks like:\n\n$$f(mathbf{x}) = (text{det}^{*}(2piSigma))^{-1/2} e^{-1/2 (x-mu)^TSigma^+(x-mu)}$$\n\n… where $$text{det}^*$$ is a pseudo-determinant (product of non-zero eigenvalues) and $$Sigma^+$$ is the pseudo-inverse.\n\nI have a couple of questions relating to whether or not such a formulation exists for the complex case.\n\nQuestion 1: If $$Z sim mathcal{C}N(mu, Gamma, C)$$ where the support is $$mathbb C^2$$ (for example), then multiplication of $$Z$$ with a matrix $$Ain mathbb{C}_{3,2}$$ would yeild: $$AZ sim mathcal{C}N(Amu, AGamma A^H, ACA^T)$$ where $$AZ in mathbb{C}^3$$ and could, in general, be degenerate.\n\nAssuming that $$A$$ is unknown, how would one find the subspace of $$mathbb{C}^{3}$$ where the density would have a support, and in this subspace, what would the density be? Would be be similar to the real case wherein the determinants and inverses are replaced by their pseudo-counterparts?\n\nQuestion 2: On the wikipedia article, it is shown that the density of a complex normal can be written as:\n\n$$f(z) = tfrac{sqrt{detleft(overline{P^{-1}}-R^{ast} P^{-1}Rright)det(P^{-1})}}{pi^n}, e^{ -(z-mu)^astoverline{P^{-1}}(z-mu) + operatorname{Re}left((z-mu)^intercal R^intercaloverline{P^{-1}}(z-mu)right)}$$\n\n…where $$P, R$$ are functions of $$Gamma, C$$. I assume that this density is only valid if, for example, $$Gamma$$ is invertible. In a specific problem setting, I’ve encountered a likelihood that can be written in this form, but $$detleft(overline{P^{-1}}-R^{ast} P^{-1}Rright) = 0$$. Does this imply that the density is degenerate? Or is it the case that my likelihood isn’t a likelihood at all?\n\nGet this bounty!!!\n\n## #StackBounty: #hypothesis-testing #self-study #normal-distribution #t-test #likelihood-ratio Likelihood Ratio Test Equivalent with \\$t\\$ …\n\n### Bounty: 50\n\n$$newcommand{szdp}{!left(#1right)} newcommand{szdb}{!left[#1right]}$$\nProblem Statement: Suppose that independent random samples of sizes $$n_1$$ and $$n_2$$\nare to be selected from normal populations with means $$mu_1$$ and $$mu_2,$$\nrespectively, and common variance $$sigma^2.$$ For testing $$H_0:mu_1=mu_2$$\nversus $$H_a:mu_1-mu_2>0$$ ($$sigma^2$$ unknown), show that the likelihood\nratio test reduces to the two-sample $$t$$ test presented in Section 10.8.\n\nNote: This is Exercise 10.94 from Mathematical Statistics with Applications, 5th. Ed., by Wackerly, Mendenhall, and Scheaffer.\n\nMy Work So Far: We have the likelihood as\n$$L(mu_1, mu_2,sigma^2)= szdp{frac{1}{sqrt{2pi}}}^{!!(n_1+n_2)} szdp{frac{1}{sigma^2}}^{!!(n_1+n_2)/2} expszdb{-frac{1}{2sigma^2}szdp{sum_{i=1}^{n_1}(x_i-mu_1)^2 +sum_{i=1}^{n_2}(y_i-mu_2)^2}}.$$\nTo compute $$Lbig(hatOmega_0big),$$ we need to find the MLE for $$sigma^2:$$\nbegin{align} hatsigma^2&=frac{1}{n_1+n_2}szdp{sum_{i=1}^{n_1}(x_i-mu_1)^2 +sum_{i=1}^{n_2}(y_i-mu_2)^2}. end{align}\nThis is the MLE for $$sigma^2$$ regardless of what $$mu_1$$ and $$mu_2$$ are.\nThus, under $$H_0,$$ we have that\n$$hatsigma_0^2=frac{1}{n_1+n_2}szdp{sum_{i=1}^{n_1}(x_i-mu_0)^2 +sum_{i=1}^{n_2}(y_i-mu_0)^2},$$\nand the unrestricted case is\n$$hatsigma^2=frac{1}{n_1+n_2}szdp{sum_{i=1}^{n_1}(x_i-overline{x})^2 +sum_{i=1}^{n_2}(y_i-overline{y})^2}.$$\nUnder $$H_0,;mu_1=mu_2=mu_0,$$ so that\nbegin{align} Lbig(hatOmega_0big) &=szdp{frac{1}{sqrt{2pi}}}^{!!(n_1+n_2)} szdp{frac{1}{hatsigma_0^2}}^{!!(n_1+n_2)/2} expszdb{-frac{n_1+n_2}{2}}\\ Lbig(hatOmegabig) &=szdp{frac{1}{sqrt{2pi}}}^{!!(n_1+n_2)} szdp{frac{1}{hatsigma^2}}^{!!(n_1+n_2)/2} expszdb{-frac{n_1+n_2}{2}}, end{align}\nand the likelihood ratio is given by\nbegin{align} lambda &=frac{Lbig(hatOmega_0big)}{Lbig(hatOmegabig)}\\ &=szdp{frac{hatsigma^2}{hatsigma_0^2}}^{!!(n_1+n_2)/2}\\ &=szdp{frac{displaystylesum_{i=1}^{n_1}(x_i-overline{x})^2 +sum_{i=1}^{n_2}(y_i-overline{y})^2} {displaystylesum_{i=1}^{n_1}(x_i-mu_0)^2 +sum_{i=1}^{n_2}(y_i-mu_0)^2}}^{!!(n_1+n_2)/2}. end{align}\nIt follows that the rejection region, $$lambdale k,$$ is equivalent to\nbegin{align} frac{displaystylesum_{i=1}^{n_1}(x_i-overline{x})^2 +sum_{i=1}^{n_2}(y_i-overline{y})^2} {displaystylesum_{i=1}^{n_1}(x_i-mu_0)^2 +sum_{i=1}^{n_2}(y_i-mu_0)^2}&frac{1}{k’}-1=k”\\ frac{n_1(overline{x}-mu_0)^2+n_2(overline{y}-mu_0)^2} {displaystyle(n_1-1)S_1^2+(n_2-1)S_2^2}&>k”\\ frac{n_1(overline{x}-mu_0)^2+n_2(overline{y}-mu_0)^2} {displaystyledfrac{(n_1-1)S_1^2+(n_2-1)S_2^2} {n_1+n_2-2}}&>k”(n_1+n_2-2)\\ frac{n_1(overline{x}-mu_0)^2+n_2(overline{y}-mu_0)^2} {S_p^2}&>k”(n_1+n_2-2). end{align}\nHere\n$$S_p^2=frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}.$$\n\nMy Question: The goal is to get this expression somehow to look like\n$$t=frac{overline{x}-overline{y}}{S_psqrt{1/n_1+1/n_2}}>t_{alpha}.$$\nBut I don’t see how I can convert my expression, with the same sign for $$overline{x}$$ and $$overline{y},$$ to the desired formula with its opposite signs. What am I missing?\n\nGet this bounty!!!\n\n## #StackBounty: #normal-distribution #circular-statistics Expected ratio of x'Ax and x'AAx on a unit sphere?\n\n### Bounty: 50\n\nSuppose $$A$$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?\n\n$$frac{x^TAx}{x^TA^2 x}$$\n\nWhere $$x$$ is distributed as $$text{Normal}(0,I_n)$$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.",
null,
"When $$A$$ has eigenvalues $$langle 1, frac{1}{2}rangle$$, this expectation is equal to $$frac{4}{3}$$\n\nGet this bounty!!!\n\n## #StackBounty: #normal-distribution #circular-statistics Expected ratio of x'Ax and x'AAx on a unit sphere?\n\n### Bounty: 50\n\nSuppose $$A$$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?\n\n$$frac{x^TAx}{x^TA^2 x}$$\n\nWhere $$x$$ is distributed as $$text{Normal}(0,I_n)$$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.",
null,
"When $$A$$ has eigenvalues $$langle 1, frac{1}{2}rangle$$, this expectation is equal to $$frac{4}{3}$$\n\nGet this bounty!!!\n\n## #StackBounty: #normal-distribution #circular-statistics Expected ratio of x'Ax and x'AAx on a unit sphere?\n\n### Bounty: 50\n\nSuppose $$A$$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?\n\n$$frac{x^TAx}{x^TA^2 x}$$\n\nWhere $$x$$ is distributed as $$text{Normal}(0,I_n)$$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.",
null,
"When $$A$$ has eigenvalues $$langle 1, frac{1}{2}rangle$$, this expectation is equal to $$frac{4}{3}$$\n\nGet this bounty!!!\n\n## #StackBounty: #normal-distribution #circular-statistics Expected ratio of x'Ax and x'AAx on a unit sphere?\n\n### Bounty: 50\n\nSuppose $$A$$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?\n\n$$frac{x^TAx}{x^TA^2 x}$$\n\nWhere $$x$$ is distributed as $$text{Normal}(0,I_n)$$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.",
null,
"When $$A$$ has eigenvalues $$langle 1, frac{1}{2}rangle$$, this expectation is equal to $$frac{4}{3}$$\n\nGet this bounty!!!\n\n## #StackBounty: #normal-distribution #circular-statistics Expected ratio of x'Ax and x'AAx on a unit sphere?\n\n### Bounty: 50\n\nSuppose $$A$$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?\n\n$$frac{x^TAx}{x^TA^2 x}$$\n\nWhere $$x$$ is distributed as $$text{Normal}(0,I_n)$$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.",
null,
"When $$A$$ has eigenvalues $$langle 1, frac{1}{2}rangle$$, this expectation is equal to $$frac{4}{3}$$\n\nGet this bounty!!!\n\n## #StackBounty: #normal-distribution #circular-statistics Expected ratio of x'Ax and x'AAx on a unit sphere?\n\n### Bounty: 50\n\nSuppose $$A$$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?\n\n$$frac{x^TAx}{x^TA^2 x}$$\n\nWhere $$x$$ is distributed as $$text{Normal}(0,I_n)$$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.",
null,
"When $$A$$ has eigenvalues $$langle 1, frac{1}{2}rangle$$, this expectation is equal to $$frac{4}{3}$$\n\nGet this bounty!!!\n\n## #StackBounty: #normal-distribution #circular-statistics Expected ratio of x'Ax and x'AAx on a unit sphere?\n\n### Bounty: 50\n\nSuppose $$A$$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?\n\n$$frac{x^TAx}{x^TA^2 x}$$\n\nWhere $$x$$ is distributed as $$text{Normal}(0,I_n)$$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.",
null,
"When $$A$$ has eigenvalues $$langle 1, frac{1}{2}rangle$$, this expectation is equal to $$frac{4}{3}$$\n\nGet this bounty!!!\n\n## #StackBounty: #normal-distribution #circular-statistics Expected ratio of x'Ax and x'AAx on a unit sphere?\n\n### Bounty: 50\n\nSuppose $$A$$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?\n\n$$frac{x^TAx}{x^TA^2 x}$$\n\nWhere $$x$$ is distributed as $$text{Normal}(0,I_n)$$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.",
null,
"When $$A$$ has eigenvalues $$langle 1, frac{1}{2}rangle$$, this expectation is equal to $$frac{4}{3}$$\n\nGet this bounty!!!"
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"https://i0.wp.com/i.stack.imgur.com/Sfm6D.png",
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"https://i0.wp.com/i.stack.imgur.com/Sfm6D.png",
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https://www.esaral.com/q/a-farmer-mixes-two-brands-p-and-q-of-cattle-feed-brand-p-41356 | [
"# A farmer mixes two brands P and Q of cattle feed. Brand P,\n\nQuestion:\n\nA farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional elements A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?\n\nSolution:\n\nLet the farmer mix x bags of brand P and y bags of brand Q.\n\nThe given information can be compiled in a table as follows.",
null,
"The given problem can be formulated as follows.\n\nMinimize $z=250 x+200 y$. (1)\n\nsubject to the constraints,\n\n$3 x+1.5 y \\geq 18$ (2)\n\n$2.5 x+11.25 y \\geq 45$ (3)\n\n$2 x+3 y \\geq 24$ (4)\n\n$x, y \\geq 0$ (5)\n\nThe feasible region determined by the system of constraints is as follows.",
null,
"The corner points of the feasible region are A (18, 0), B (9, 2), C (3, 6), and D (0, 12).\n\nThe values of z at these corner points are as follows.",
null,
"As the feasible region is unbounded, therefore, 1950 may or may not be the minimum value of z.\n\nFor this, we draw a graph of the inequality, $250 x+200 y<1950$ or $5 x+4 y<39$, and check whether the resulting half plane has points in common with the feasible region or not.\n\nIt can be seen that the feasible region has no common point with $5 x+4 y<39$\n\nTherefore, the minimum value of z is 1950 at (3, 6).\n\nThus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs 1950."
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http://dirkmittler.homeip.net/blog/archives/tag/in-phase | [
"## How two subjects might be confused, that both have to do with polarized light.\n\nThere exists a concept, by which a single photon is visualized as having an electrostatic dipole-moment, which does not lie in a plane, but which performs a corkscrew, either left-handedly, or right-handedly, to start the phenomenon of electromagnetic radiation as based on circularly-polarized light, as opposed to being based primarily on plane-polarization. A quantity of photons could then still form plane-polarized light, not because they interact with each other, but because they coincide with each other in such a way, that their electrostatic fields cancel along one axis, but reinforce perpendicularly to the axis along which they cancel.\n\nIn reality, it’s dangerous to make such statements, about what exactly one photon does, because nobody has ever ‘seen’ a photon. We’re mainly able to make more-coarse measurements of what light does, when composed of swarms of photons, and must then deduce what the properties of one photon could be.\n\n(Edit 02/20/2018 :\n\nAccording to This Experiment, this hypothesis is disproved.\n\nIn the macroscopic world, circularly polarized light seems to exist, just as plane-polarized light does, without shedding much light on the subject of how one photon behaves, unless the latter subject is studied in much greater depth. )\n\nBut there is the matter of how any of this agrees with the classical, electrodynamic explanation of ‘light’, which would say that it has a magnetic dipole-moment, that oscillates with the same set of frequencies, with which the electrostatic dipole-moment, oscillates, but perpendicularly to the electrostatic moment.\n\nThe question could be asked of, If the electrostatic moment was plotted against time, What its phase-position would be, relative to the magnetic moment. And what I claim to know, is that they’d be in-phase.\n\nThis subject has been confused at times, with the question of whether the electrostatic component along one plane of polarization, is in-phase or out-of-phase, with the electrostatic component, along the perpendicular plane of polarization. Those are out-of-phase, in the case of circularly-polarized light, as well as in the case of circularly-polarized photons.\n\n(Edit 02/07/2018 : )",
null,
"Now, the question about plotting this could get sidetracked, by the question of whether it’s more correct, if where the electrostatic dipole moment, which I’ll say is denoted by the Green line above, is pointing ‘upwards’, the magnetic dipole moment, which I’ll say is denoted by the Red line above, should be pointing ‘towards the viewer’, or ‘away from the viewer’. The way I presently have it, at the left end of the plot, the red line is towards the viewer at that instant. Because magnetic dipole moments differentiate between North and South, while electrostatic dipole moments differentiate between Positive and Negative, these signs of polarity are independent. By convention, the magnetic North pole is denoted by positive numbers. If it was assumed that the Red line corresponds to North, as shown above, then the photon would need to be traveling from the left, to the right, which also corresponds to an increasing parameter (t), just in case anybody is interested in actually analyzing the Math I entered.\n\n(Edit 02/26/2018 :\n\nThis also implies that t does not correspond to Time. t us just a variable-name which often denotes a generic parameter. More-positive values of t will arrive on the right, before more-negative values do. )\n\nIn that case it should also be noted, that ‘Wolfram Mathematica’ switches the (Y) and (Z) plotting axes, so that (Z) actually faces upwards, but needs to be given as the 3rd input of a parametric 3D plot, while (Y) faces away from the viewer, which is different from how some other 3D plots work. The way I tend to visualize World Coordinates these days, (Y) should be facing Up, and (Z) should be facing Towards the Viewer.\n\n(Updated 02/08/2018 … )"
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"http://dirkmittler.homeip.net/blog/wp-content/uploads/2018/02/Photon_3.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9547969,"math_prob":0.93629825,"size":3966,"snap":"2020-24-2020-29","text_gpt3_token_len":871,"char_repetition_ratio":0.11761736,"word_repetition_ratio":0.018897638,"special_character_ratio":0.21028744,"punctuation_ratio":0.11889036,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.950228,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-10T17:19:30Z\",\"WARC-Record-ID\":\"<urn:uuid:afd229c7-d3c1-4440-9d9e-2ec753afd19e>\",\"Content-Length\":\"43408\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8fe22f19-34fe-424b-b050-3614a1f67e4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a687ba16-a648-4905-82c4-3af6cf0296fb>\",\"WARC-IP-Address\":\"69.156.26.96\",\"WARC-Target-URI\":\"http://dirkmittler.homeip.net/blog/archives/tag/in-phase\",\"WARC-Payload-Digest\":\"sha1:NRLKEEES4NPB3A7JQV7VI3S7AAPOXS25\",\"WARC-Block-Digest\":\"sha1:JUAIMLHS3GLTXUA3XOQU3AQOUSWW4ECN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655911092.63_warc_CC-MAIN-20200710144305-20200710174305-00019.warc.gz\"}"} |
https://ans.disi.unitn.it/redmine/projects/peerstreamer/repository/ffmpeg/revisions/e98b8e2f2fae3e75f87d0d87098a8faee691a514/diff/ | [
"## Revision e98b8e2f\n\nView differences:\n\nlibavcore/samplefmt.c\n68 68\n``` return sample_fmt < 0 || sample_fmt >= AV_SAMPLE_FMT_NB ?\n```\n69 69\n``` 0 : sample_fmt_info[sample_fmt].bits;\n```\n70 70\n```}\n```\n71\n\n72\n```int av_samples_fill_arrays(uint8_t *pointers, int linesizes,\n```\n73\n``` uint8_t *buf, int nb_channels, int nb_samples,\n```\n74\n``` enum AVSampleFormat sample_fmt, int planar, int align)\n```\n75\n```{\n```\n76\n``` int i, step_size = 0;\n```\n77\n``` int sample_size = av_get_bits_per_sample_fmt(sample_fmt) >> 3;\n```\n78\n``` int channel_step = planar ? FFALIGN(nb_samples*sample_size, align) : sample_size;\n```\n79\n\n80\n``` if(nb_channels * (uint64_t)nb_samples * sample_size >= INT_MAX - align*(uint64_t)nb_channels)\n```\n81\n``` return AVERROR(EINVAL);\n```\n82\n\n83\n``` if (pointers) {\n```\n84\n``` pointers = buf;\n```\n85\n``` for (i = 0; i < nb_channels; i++) {\n```\n86\n``` pointers[i] = buf + step_size;\n```\n87\n``` step_size += channel_step;\n```\n88\n``` }\n```\n89\n``` memset(&pointers[nb_channels], 0, (8-nb_channels) * sizeof(pointers));\n```\n90\n``` }\n```\n91\n\n92\n``` if (linesizes) {\n```\n93\n``` linesizes = planar ? sample_size : nb_channels*sample_size;\n```\n94\n``` memset(&linesizes, 0, (8-1) * sizeof(linesizes));\n```\n95\n``` }\n```\n96\n\n97\n``` return planar ? channel_step * nb_channels : FFALIGN(nb_channels*sample_size*nb_samples, align);\n```\n98\n```}\n```\n99\n\n100\n```int av_samples_alloc(uint8_t *pointers, int linesizes,\n```\n101\n``` int nb_samples, int nb_channels,\n```\n102\n``` enum AVSampleFormat sample_fmt, int planar,\n```\n103\n``` int align)\n```\n104\n```{\n```\n105\n``` uint8_t *buf;\n```\n106\n``` int size = av_samples_fill_arrays(NULL, NULL,\n```\n107\n``` NULL, nb_channels, nb_samples,\n```\n108\n``` sample_fmt, planar, align);\n```\n109\n\n110\n``` buf = av_mallocz(size);\n```\n111\n``` if (!buf)\n```\n112\n``` return AVERROR(ENOMEM);\n```\n113\n\n114\n``` return av_samples_fill_arrays(pointers, linesizes,\n```\n115\n``` buf, nb_channels, nb_samples,\n```\n116\n``` sample_fmt, planar, align);\n```\n117\n```}\n```\nlibavcore/samplefmt.h\n69 69\n``` */\n```\n70 70\n```int av_get_bits_per_sample_fmt(enum AVSampleFormat sample_fmt);\n```\n71 71\n\n72\n```/**\n```\n73\n``` * Fill channel data pointers and linesizes for samples with sample\n```\n74\n``` * format sample_fmt.\n```\n75\n``` *\n```\n76\n``` * The pointers array is filled with the pointers to the samples data:\n```\n77\n``` * data[c] points to the first sample of channel c.\n```\n78\n``` * data[c] + linesize points to the second sample of channel c\n```\n79\n``` *\n```\n80\n``` * @param pointers array to be filled with the pointer for each plane, may be NULL\n```\n81\n``` * @param linesizes array to be filled with the linesize, may be NULL\n```\n82\n``` * @param buf the pointer to a buffer containing the samples\n```\n83\n``` * @param nb_samples the number of samples in a single channel\n```\n84\n``` * @param planar 1 if the samples layout is planar, 0 if it is packed\n```\n85\n``` * @param nb_channels the number of channels\n```\n86\n``` * @return the total size of the buffer, a negative\n```\n87\n``` * error code in case of failure\n```\n88\n``` */\n```\n89\n```int av_samples_fill_arrays(uint8_t *pointers, int linesizes,\n```\n90\n``` uint8_t *buf, int nb_channels, int nb_samples,\n```\n91\n``` enum AVSampleFormat sample_fmt, int planar, int align);\n```\n92\n\n93\n```/**\n```\n94\n``` * Allocate a samples buffer for nb_samples samples, and\n```\n95\n``` * fill pointers and linesizes accordingly.\n```\n96\n``` * The allocated samples buffer has to be freed by using\n```\n97\n``` * av_freep(&pointers).\n```\n98\n``` *\n```\n99\n``` * @param nb_samples number of samples per channel\n```\n100\n``` * @param planar 1 if the samples layout is planar, 0 if packed,\n```\n101\n``` * @param align the value to use for buffer size alignment\n```\n102\n``` * @return the size in bytes required for the samples buffer, a negative\n```\n103\n``` * error code in case of failure\n```\n104\n``` * @see av_samples_fill_arrays()\n```\n105\n``` */\n```\n106\n```int av_samples_alloc(uint8_t *pointers, int linesizes,\n```\n107\n``` int nb_samples, int nb_channels,\n```\n108\n``` enum AVSampleFormat sample_fmt, int planar,\n```\n109\n``` int align);\n```\n110\n\n72 111\n```#endif /* AVCORE_SAMPLEFMT_H */\n```\n\nAlso available in: Unified diff"
]
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https://www.brightstorm.com/math/precalculus/exponential-and-logarithmic-functions/graph-of-logarithmic-functions/ | [
"",
null,
"###### Carl Horowitz\n\nUniversity of Michigan\nRuns his own tutoring company\n\nCarl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!\n\n##### Thank you for watching the video.\n\nTo unlock all 5,300 videos, start your free trial.\n\n# Graph of Logarithmic Functions - Concept\n\nCarl Horowitz",
null,
"###### Carl Horowitz\n\nUniversity of Michigan\nRuns his own tutoring company\n\nCarl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!\n\nShare\n\nIn science classes we will often find ourselves graphing logarithmic functions to describe situations such as motion or speed over time. When trying to identify these situations as those seen in graphing logarithmic functions, it is important to be able to recognize these graphs. It is also important to recognize graphs of exponential functions and their importance as the logarithmic inverse.\n\nFinding the graph of a logarithmic function. So for this example I'm not actually going to give you the graph straight up. What we're going to actually do is talk about how we get it, okay? So think about any time we are dealing with a equation in logarithmic form. What we generally do is put it into exponential form. So I'm going to do that first. So this is just going to give us x=2 to the y.\nNow think back to actually how we got to this formula in the first place. And that was by taking the inverse of an exponential. So the inverse is then just when we switch our x and y's. y=2 to the x. Okay.\nWe know what that graph looks like, okay and for this I have a little prop for you. We have the inverse sorry the graph of y=2 to the x looks something like this. Okay? It's not precise but pretty rough. How we found the graph of an inverse was by flipping something over the line y=x. So what that did is it flipped everything that was above the line down, everything that was below it up. And what you end up with is a graph that looks like this, okay. So by using properties of inverses, we can actually go from the graph that we have already derived to the graph of the log function, okay? So what this graph does, the exponential graph went to the point 0, 1. When we flip the x and y's we are now going to go to the point 1,0. Our exponential graph had a horizontal asymptote at 0 that gets flipped to be a vertical.\nSo now I have a vertical asymptote at what is that x or y or is it x=0, okay. So vertical vertical line right here and then our domain, our x values we weren't able to ever get to 0 before so we're still not going to be able to do that for our domain, 0 to infinity. And our range used to be our domain for our inverse function, which was everything and so this is just going to be obvious. So this is a rough sketch of a log graph.\nKey thing to note is this is only if our base is greater than 1, okay? This exponential graph that we looked at the first one was if our base was greater than one so that translates to this this base having the same restrictions. Okay? So by using our properties of inverses we're able to find the rough sketch of our log graph with our base bigger than 1."
]
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null,
"https://d3a0jx1tkrpybf.cloudfront.net/img/teachers/teacher-4.png",
null,
"https://d3a0jx1tkrpybf.cloudfront.net/img/teachers/teacher-4.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9766602,"math_prob":0.8958037,"size":3142,"snap":"2023-14-2023-23","text_gpt3_token_len":700,"char_repetition_ratio":0.12364563,"word_repetition_ratio":0.108658746,"special_character_ratio":0.21674094,"punctuation_ratio":0.08510638,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96613544,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-23T23:09:25Z\",\"WARC-Record-ID\":\"<urn:uuid:1a961648-fdd5-4dd6-9fb0-fb803de7e2e2>\",\"Content-Length\":\"243614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7439e347-b837-4090-9d7b-05daa6cbc00c>\",\"WARC-Concurrent-To\":\"<urn:uuid:5730bcad-5fe3-4050-8209-7606b7ae0e25>\",\"WARC-IP-Address\":\"34.207.32.175\",\"WARC-Target-URI\":\"https://www.brightstorm.com/math/precalculus/exponential-and-logarithmic-functions/graph-of-logarithmic-functions/\",\"WARC-Payload-Digest\":\"sha1:VEA6KQUGFA4KZDH5YQP77VDCACGCX2VP\",\"WARC-Block-Digest\":\"sha1:AMWXLCDKIZDSK6PH2R6FPVR45WXTZZVA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945218.30_warc_CC-MAIN-20230323225049-20230324015049-00292.warc.gz\"}"} |
https://uniessaywriters.com/for-a-company-selling-multiple-products-the-break-even-point-in-dollars-is-computed-by-dividing-fixed-costs-by-the-a-contribution-margin-per-unit/ | [
"# For a company selling multiple products, the break-even point in dollars is computed by dividing fixed costs by the a. contribution margin per unit….\n\n1. For a company selling multiple products, the break-even point in dollars is computed by dividing fixed costs by the\n\na. contribution margin per unit.\n\nb. contribution margin ratio.\n\nc. weighted-average contribution margin per unit.\n\nd. weighted-average contribution margin ratio.\n\n2. In order to maximize net income a company should produce and sell the product with the highest.\n\na. contribution margin ratio.\n\nb. contribution margin per unit.\n\nc. contribution margin per unit of limited resource.\n\nd. weighted-average unit contribution margin.\n\n3. Operating leverage refers to the extent to which a company’s net income reacts to a given change in\n\na. fixed costs.\n\nb. production.\n\nc. sales.\n\nd. variable costs\n\n*4. Under variable costing, all of the following are considered product costs except\n\na. direct materials.\n\nb. direct labor.\n\nd. variable selling and administrative expenses.\n\n*5. All of the following are potential advantages of variable costing except that\n\na. its use is consistent with cost-volume-profit and incremental analysis.\n\nb. variable costing net income is affected by changes in production levels.\n\nc. variable costing net income is closely tied to changes in sales levels.\n\nd. the presentation of fixed and variable cost components makes it easier\n\nto identify these costs."
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8252314,"math_prob":0.7551607,"size":1466,"snap":"2022-27-2022-33","text_gpt3_token_len":293,"char_repetition_ratio":0.17168263,"word_repetition_ratio":0.18018018,"special_character_ratio":0.19849932,"punctuation_ratio":0.17605634,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9842268,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T16:20:52Z\",\"WARC-Record-ID\":\"<urn:uuid:6b2cac09-3d4f-433e-b44a-7beee196e955>\",\"Content-Length\":\"44809\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9862d28d-c981-4f56-b48d-f3fa9bdbac39>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac46aeb8-698d-43be-8ae8-d7fd083ca602>\",\"WARC-IP-Address\":\"199.192.23.77\",\"WARC-Target-URI\":\"https://uniessaywriters.com/for-a-company-selling-multiple-products-the-break-even-point-in-dollars-is-computed-by-dividing-fixed-costs-by-the-a-contribution-margin-per-unit/\",\"WARC-Payload-Digest\":\"sha1:J6EP2EV2YGZZMTATZ4VF7FSREVQAVDAO\",\"WARC-Block-Digest\":\"sha1:GD3JRZ7LAXKBYWYA76Y6NE3ZPHIQNYLF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103640328.37_warc_CC-MAIN-20220629150145-20220629180145-00467.warc.gz\"}"} |
https://gitlab.lam.fr/cigale/cigale/-/commit/d963067f4de20a7055b8504ca26c929d39d49751 | [
"",
null,
"### Fν was computed by calculating Fλ and then converting to Fν, which led to...\n\n`Fν was computed by calculating Fλ and then converting to Fν, which led to typical differences in fluxes of typically 1-2% and a bit more for a handful of pathological filters. Now Fν is computed directly and a bit faster. At the same time rather than computing an effective wavelength (which was more a mean wavelength), we compute the pivot wavelength, which is more appropriate when converting between Fν and Fλ.`\nparent 4d78de48\n ... ... @@ -17,6 +17,7 @@ - To accommodate the analysis of the observations by blocks, the `save_pdf` option has been eliminated. To plot PDF one needs to set `save_chi2` to True and then run `pcigale-plots pdf`. (Médéric Boquien) - In order to capture rapid evolutionary phases, we assume that in a given period of 1 Myr, 10 small episodes of star formation occurred every 0.1 Myr, rather than one episode every 1 Myr. - When computing the attenuation curve, the bump is now added so that its relative strength does not depend on δ. (Médéric Boquien, issue reported by Samir Salim) - Fν was computed by calculating Fλ and then converting to Fν, which led to typical differences in fluxes of typically 1-2% and a bit more for a handful of pathological filters. Now Fν is computed directly and a bit faster. (Médéric Boquien, issue reported by Yannick Roehlly and Wouter Dobbels) ### Fixed - Corrected a typo that prevented `restframe\\_parameters` from being listed among the available modules. (Médéric Boquien) ... ...\n ... ... @@ -165,13 +165,13 @@ def build_filters(base): new_filter = Filter(filter_name, filter_description, filter_table) # We normalise the filter and compute the effective wavelength. # If the filter is a pseudo-filter used to compute line fluxes, it # should not be normalised. # We normalise the filter and compute the pivot wavelength. If the # filter is a pseudo-filter used to compute line fluxes, it should not # be normalised. if not filter_name.startswith('PSEUDO'): new_filter.normalise() else: new_filter.effective_wavelength = np.mean( new_filter.pivot_wavelength = np.mean( filter_table[filter_table > 0] ) filters.append(new_filter) ... ... @@ -211,13 +211,13 @@ def build_filters_gazpar(base): new_filter = Filter(filter_name, filter_desc, filter_table) # We normalise the filter and compute the effective wavelength. # If the filter is a pseudo-filter used to compute line fluxes, it # should not be normalised. # We normalise the filter and compute the pivot wavelength. If the # filter is a pseudo-filter used to compute line fluxes, it should not # be normalised. if not filter_name.startswith('PSEUDO'): new_filter.normalise() else: new_filter.effective_wavelength = np.mean( new_filter.pivot_wavelength = np.mean( filter_table[filter_table > 0] ) filters.append(new_filter) ... ...\n ... ... @@ -63,13 +63,13 @@ class _Filter(BASE): name = Column(String, primary_key=True) description = Column(String) trans_table = Column(PickleType) effective_wavelength = Column(Float) pivot_wavelength = Column(Float) def __init__(self, f): self.name = f.name self.description = f.description self.trans_table = f.trans_table self.effective_wavelength = f.effective_wavelength self.pivot_wavelength = f.pivot_wavelength class _M2005(BASE): ... ... @@ -1062,7 +1062,7 @@ class Database(object): first()) if result: return Filter(result.name, result.description, result.trans_table, result.effective_wavelength) result.pivot_wavelength) else: raise DatabaseLookupError( \"The filter <{0}> is not in the database\".format(name)) ... ... @@ -1081,7 +1081,7 @@ class Database(object): \"\"\"Generator to parse the filter database.\"\"\" for filt in self.session.query(_Filter): yield Filter(filt.name, filt.description, filt.trans_table, filt.effective_wavelength) filt.pivot_wavelength) def parse_m2005(self): \"\"\"Generator to parse the Maraston 2005 SSP database.\"\"\" ... ...\n ... ... @@ -4,6 +4,7 @@ # Author: Yannick Roehlly import numpy as np from scipy.constants import c class Filter(object): ... ... @@ -11,9 +12,9 @@ class Filter(object): \"\"\" def __init__(self, name, description=None, trans_table=None, effective_wavelength=None): pivot_wavelength=None): \"\"\"Create a new filter. If the transmission type, the description the transmission table or the effective wavelength are not specified, the transmission table or the pivot wavelength are not specified, their value is set to None. Parameters ... ... @@ -25,14 +26,14 @@ class Filter(object): trans_table: array trans_table is the wavelength in nm, trans_table is the transmission) effective_wavelength: float Effective wavelength of the filter pivot_wavelength: float Pivot wavelength of the filter \"\"\" self.name = name self.description = description self.trans_table = trans_table self.effective_wavelength = effective_wavelength self.pivot_wavelength = pivot_wavelength def __str__(self): \"\"\"Pretty print the filter information ... ... @@ -40,18 +41,24 @@ class Filter(object): result = \"\" result += (\"Filter name: %s\\n\" % self.name) result += (\"Description: %s\\n\" % self.description) result += (\"Effective wavelength: %s nm\\n\" % self.effective_wavelength) result += (\"Pivot wavelength: %s nm\\n\" % self.pivot_wavelength) return result def normalise(self): \"\"\" Normalise the transmission table to 1 and compute the effective wavelength of the filter. Compute the pivot wavelength of the filter and normalise the filter to compute the flux in Fν (mJy) in cigale. \"\"\" self.trans_table = self.trans_table / ( np.trapz(self.trans_table, self.trans_table)) self.pivot_wavelength = np.sqrt(np.trapz(self.trans_table, self.trans_table) / np.trapz(self.trans_table / self.trans_table ** 2, self.trans_table)) self.effective_wavelength = np.trapz(self.trans_table * self.trans_table, self.trans_table) # The factor 10²⁰ is so that we get the fluxes directly in mJy when we # integrate with the wavelength in units of nm and the spectrum in # units of W/m²/nm. self.trans_table = 1e20 * self.trans_table / ( c * np.trapz(self.trans_table / self.trans_table**2, self.trans_table))\n ... ... @@ -243,20 +243,10 @@ class SED(object): \"\"\" Compute the Fν flux density in a given filter As the SED stores the Lλ luminosity density, we first compute the Fλ flux density. Fλ is the integration of the Lλ luminosity multiplied by the filter transmission, normalised to this transmission and corrected by the luminosity distance of the source. Fλ is in W/m²/nm. At redshift 0, the flux is computed at 10 pc. Then, to compute Fν, we make the approximation: Fν = λeff / c . λeff . Fλ Fν is computed in W/m²/Hz and then converted to mJy. If the SED spectrum does not cover all the filter response table, NaN is returned. The filters are stored in the database in such a way that after integration and conversion from luminosity to flux we directly get the latter in units of mJy. If the SED spectrum does not cover all the filter response table, NaN is returned. Parameters ---------- ... ... @@ -273,7 +263,7 @@ class SED(object): wavelength = self.wavelength_grid # First we try to fetch the filter's wavelength, transmission and # effective wavelength from the cache. The two keys are the size of the # pivot wavelength from the cache. The two keys are the size of the # spectrum wavelength grid and the name of the filter. The first key is # necessary because different spectra may have different sampling. To # avoid having the resample the filter every time on the optimal grid ... ... @@ -288,12 +278,12 @@ class SED(object): dist = 10. * parsec if key in self.cache_filters: wavelength_r, transmission_r, lambda_eff = self.cache_filters[key] wavelength_r, transmission_r, lambda_piv = self.cache_filters[key] else: with Database() as db: filter_ = db.get_filter(filter_name) trans_table = filter_.trans_table lambda_eff = filter_.effective_wavelength lambda_piv = filter_.pivot_wavelength lambda_min = filter_.trans_table lambda_max = filter_.trans_table[-1] ... ... @@ -313,17 +303,20 @@ class SED(object): trans_table) self.cache_filters[key] = (wavelength_r, transmission_r, lambda_eff) lambda_piv) l_lambda_r = interp(wavelength_r, wavelength, self.luminosity) f_lambda = utils.luminosity_to_flux( # We compute directly Fν from ∫T×Fλ×dλ/∫T×c/λ²×dλ. The filter bandpass # in the database is already normalised so that we do not need to # compute the denominator (it is a constant that does not depend on the # spectrum) and the normalisation is such that the results we obtain # are directly in mJy. f_nu = utils.luminosity_to_flux( utils.flux_trapz(transmission_r * l_lambda_r, wavelength_r, key), dist) # Return Fν in mJy. The 1e-9 factor is because λ is in nm and 1e29 for # convert from W/m²/Hz to mJy. return 1e-9 / c * 1e29 * lambda_eff * lambda_eff * f_lambda return f_nu def to_votable(self, filename, mass=1.): \"\"\" ... ...\n ... ... @@ -27,14 +27,14 @@ def list_filters(): name = Column(data=[filters[f].name for f in filters], name='Name') description = Column(data=[filters[f].description for f in filters], name='Description') wl = Column(data=[filters[f].effective_wavelength for f in filters], name='Effective Wavelength', unit=u.nm, format='%d') wl = Column(data=[filters[f].pivot_wavelength for f in filters], name='Pivot Wavelength', unit=u.nm, format='%d') samples = Column(data=[filters[f].trans_table.size for f in filters], name=\"Points\") t = Table() t.add_columns([name, description, wl, samples]) t.sort(['Effective Wavelength']) t.sort(['Pivot Wavelength']) t.pprint(max_lines=-1, max_width=-1) ... ... @@ -67,13 +67,13 @@ def add_filters(fnames): new_filter = Filter(filter_name, filter_description, filter_table) # We normalise the filter and compute the effective wavelength. # If the filter is a pseudo-filter used to compute line fluxes, it # should not be normalised. # We normalise the filter and compute the pivot wavelength. If the # filter is a pseudo-filter used to compute line fluxes, it should # not be normalised. if not filter_name.startswith('PSEUDO'): new_filter.normalise() else: new_filter.effective_wavelength = np.mean( new_filter.pivot_wavelength = np.mean( filter_table[filter_table > 0] ) ... ...\n ... ... @@ -144,7 +144,7 @@ def _sed_worker(obs, mod, filters, sed_type, nologo): sed = Table.read(\"out/{}_best_model.fits\".format(obs['id'])) filters_wl = np.array([filt.effective_wavelength filters_wl = np.array([filt.pivot_wavelength for filt in filters.values()]) wavelength_spec = sed['wavelength'] obs_fluxes = np.array([obs[filt] for filt in filters.keys()]) ... ...\nMarkdown is supported\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!"
]
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null,
"https://piwik.osupytheas.fr/matomo.php",
null
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https://www.kidsmathtv.com/learn/math-vocabulary-sum-difference/ | [
"# Math vocabulary quiz – sum or difference\n\nMath vocabulary quiz – sum or difference\n\nMath vocabulary quiz – sum or difference. In this quiz, children will learn new words related to addition and subtraction. For example, in the following problem, 4 + 5 = 9, the answer 9 is the sum. In the following problem: 5 – 4 = 1, the answer 1 is the difference. Hence children will be presented with problems in which they will be required to tell if the answer is the sum or the difference. This is an introductory math topic for children in kindergarten and 1st grades to use to review some quick math facts. Teachers and parents will find this resource useful for their kids."
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null
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https://www.dovov.com/types-51.html | [
"# 我怎样才能从一个方法返回一个匿名types？\n\n``var myData = from a in db.MyTable where a.MyValue == \"A\" select new { a.Key, a.MyValue };` `\n\n` `public ??? GetSomeData() { // my Linq query }` `\n\nIQueryable和IEnumerable都可以工作。 但是你想使用types特定的版本，IQueryable `<` T `>`或IEnumerable `<` T `>`\n\n` `var myData = from a in db.MyTable where a.MyValue == \"A\" select new MyType { Key = a.Key, Value = a.MyValue };` `\n\n`IQueryable`\n\n` `public IQueryable GetSomeData()` `\n\n` `public class MyType {Key{get;set;} Value{get;set}} public IQueryable<T> GetSomeData<T>() where T : MyType, new() { return from a in db.MyTable where a.MyValue == \"A\" select new T {Key=a.Key,Value=a.MyValue}; }` `"
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http://forums.wolfram.com/mathgroup/archive/2011/Oct/msg00708.html | [
"",
null,
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null,
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null,
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null,
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null,
"",
null,
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null,
"Re: List plot with random size for each point\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg122457] Re: List plot with random size for each point\n• From: Heike Gramberg <heike.gramberg at gmail.com>\n• Date: Sat, 29 Oct 2011 07:10:07 -0400 (EDT)\n• Delivered-to: [email protected]\n• References: <[email protected]>\n\n```I'm not entirely sure what you mean with the last line. Do you want the\nPointSize of the point at position position[[n]] to be equal to\nparticle[[n,2]]?\nIf so, you could do something like\n\n{particle[[All,2]], position}], Axes -> True]\n\nBy the way, a more typical way in Mathematica to generate the lists\nparticle and position as in your example is to do something like\n\nparticle = Table[{i, RandomInteger[{15, 25}], RandomInteger[{0, 200}],\nRandomInteger[{0, 200}]}, {i, 200}];\nposition = particle[[All, {3, 4}]];\n\nHeike.\n\nOn 28 Oct 2011, at 11:36, Rudresh wrote:\n\n> I want to plot an array of numbers on a 2d plot, however each point\nhas a random number associated with it which should be its size on the\nplot. Can this be done in mathematica and if so how. My code as of now\nis:\n>\n> =========================\n>\n> particle = Array[f, {200, 4}];\n> position = Array[g, {200, 2}];\n> n = 1;\n> While [n < 201,\n> \tparticle[[n, 1]] = n;\n> \tparticle[[n, 2]] = RandomInteger[{15, 25}];\n> \tparticle[[n, 3]] = RandomInteger[{0, 200}];\n> \tparticle[[n, 4]] = RandomInteger[{0, 200}];\n>\n> position[[n, 1]] = particle[[n, 3]];\n> position[[n, 2]] = particle[[n, 4]];\n> n++\n> ];\n>\n> ListPlot[position, PlotStyle -> PointSize[0.01],\n> PlotRange -> {{0, 200}, {0, 200}}]\n>\n> =========================\n>\n> I want particle[[n,2]] to be the size of position[[n,1],[n,2]]\n>\n\n```\n\n• Prev by Date: Re: Export list to Excel without brackets\n• Next by Date: Re: Problem with Solve and NSolve\n• Previous by thread: Re: List plot with random size for each point\n• Next by thread: Re: List plot with random size for each point"
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http://image.absoluteastronomy.com/topics/Thin_lens | [
"",
null,
"x",
null,
"Thin lens",
null,
"Encyclopedia",
null,
"In optics\nOptics\nOptics is the branch of physics which involves the behavior and properties of light, including its interactions with matter and the construction of instruments that use or detect it. Optics usually describes the behavior of visible, ultraviolet, and infrared light...\n\n, a thin lens is a lens\nLens (optics)\nA lens is an optical device with perfect or approximate axial symmetry which transmits and refracts light, converging or diverging the beam. A simple lens consists of a single optical element...\n\nwith a thickness (distance along the optical axis\nOptical axis\nAn optical axis is a line along which there is some degree of rotational symmetry in an optical system such as a camera lens or microscope.The optical axis is an imaginary line that defines the path along which light propagates through the system...\n\nbetween the two surfaces of the lens) that is negligible\nNegligible\nNegligible refers to the quantities so small that they can be ignored when studying the larger effect. Although related to the more mathematical concepts of infinitesimal, the idea of negligibility is particularly useful in practical disciplines like physics, chemistry, mechanical and electronic...\n\ncompared to the focal length\nFocal length\nThe focal length of an optical system is a measure of how strongly the system converges or diverges light. For an optical system in air, it is the distance over which initially collimated rays are brought to a focus...\n\nof the lens. Lenses whose thickness is not negligible are sometimes called thick lenses.\n\nThe thin lens approximation ignores optical effects due to the thickness of lenses and simplifies ray tracing\nRay tracing (physics)\nIn physics, ray tracing is a method for calculating the path of waves or particles through a system with regions of varying propagation velocity, absorption characteristics, and reflecting surfaces. Under these circumstances, wavefronts may bend, change direction, or reflect off surfaces,...\n\ncalculations. It is often combined with the paraxial approximation\nParaxial approximation\nIn geometric optics, the paraxial approximation is a small-angle approximation used in Gaussian optics and ray tracing of light through an optical system ....\n\nin techniques such as ray transfer matrix analysis\nRay transfer matrix analysis\nRay transfer matrix analysis is a type of ray tracing technique used in the design of some optical systems, particularly lasers...\n\n.\n\nThe focal length, f, of a thin lens is given by the Lensmaker's equation:",
null,
"where n is the index of refraction of the lens material, and R1 and R2 are the radii of curvature of the two surfaces. Here, R1 is taken to be positive if the first surface is convex, and negative if the surface is concave. The signs are reversed for the back surface of the lens: R2 is positive if the surface is concave, and negative if it is convex. This is an arbitrary sign convention\nSign convention\nIn physics, a sign convention is a choice of the physical significance of signs for a set of quantities, in a case where the choice of sign is arbitrary. \"Arbitrary\" here means that the same physical system can be correctly described using different choices for the signs, as long as one set of...\n\n; some authors choose different signs for the radii, which changes the equation for the focal length.\n\nFor a thin lens, in the paraxial ray approximation, the object (",
null,
") and image (",
null,
") distances are related by the equation",
null,
".\n\nIn scalar wave optics a lens is a part which shifts the phase of the wave-front. Mathematically this can be understood as a multiplication of the wave-front with the following function:",
null,
".",
null,
""
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https://scholar.lib.ntnu.edu.tw/en/publications/solution-properties-and-error-bounds-for-semi-infinite-complement | [
"Solution properties and error bounds for semi-infinite complementarity problems\n\nJinchuan Zhou, Naihua Xiu, Jein-Shan Chen\n\nResearch output: Contribution to journalArticle\n\n4 Citations (Scopus)\n\nAbstract\n\nIn this paper, we deal with the semi-infinite complementarity problems (SICP), in which several important issues are covered, such as solvability, semismoothness of residual functions, and error bounds. In particular, we characterize the solution set by investigating the relationship between SICP and the classical complementarity problem. Furthermore, we show that the SICP can be equivalently reformulated as a typical semi-infinite min-max programming problem by employing NCP functions. Finally, we study the concept of error bounds and introduce its two variants, ε-error bounds and weak error bounds, where the concept of weak error bounds is highly desirable in that the solution set is not restricted to be nonempty.\n\nOriginal language English 99-115 17 Journal of Industrial and Management Optimization 9 1 https://doi.org/10.3934/jimo.2013.9.99 Published - 2013 Mar 26\n\nFingerprint\n\nComplementarity Problem\nError Bounds\nSolution Set\nSemismoothness\nNCP Function\nMin-max\nSolvability\nProgramming\nError bounds\nComplementarity\n\nKeywords\n\n• Error bounds\n• Semi-infinite complementarity problem\n• Semidifferentiable and semis-mooth\n• Weak error bounds\n\nASJC Scopus subject areas\n\n• Business and International Management\n• Strategy and Management\n• Control and Optimization\n• Applied Mathematics\n\nCite this\n\nSolution properties and error bounds for semi-infinite complementarity problems. / Zhou, Jinchuan; Xiu, Naihua; Chen, Jein-Shan.\n\nIn: Journal of Industrial and Management Optimization, Vol. 9, No. 1, 26.03.2013, p. 99-115.\n\nResearch output: Contribution to journalArticle\n\n@article{e893f37908a744c49bebea156f430a98,\ntitle = \"Solution properties and error bounds for semi-infinite complementarity problems\",\nabstract = \"In this paper, we deal with the semi-infinite complementarity problems (SICP), in which several important issues are covered, such as solvability, semismoothness of residual functions, and error bounds. In particular, we characterize the solution set by investigating the relationship between SICP and the classical complementarity problem. Furthermore, we show that the SICP can be equivalently reformulated as a typical semi-infinite min-max programming problem by employing NCP functions. Finally, we study the concept of error bounds and introduce its two variants, ε-error bounds and weak error bounds, where the concept of weak error bounds is highly desirable in that the solution set is not restricted to be nonempty.\",\nkeywords = \"Error bounds, Semi-infinite complementarity problem, Semidifferentiable and semis-mooth, Weak error bounds\",\nauthor = \"Jinchuan Zhou and Naihua Xiu and Jein-Shan Chen\",\nyear = \"2013\",\nmonth = \"3\",\nday = \"26\",\ndoi = \"10.3934/jimo.2013.9.99\",\nlanguage = \"English\",\nvolume = \"9\",\npages = \"99--115\",\njournal = \"Journal of Industrial and Management Optimization\",\nissn = \"1547-5816\",\npublisher = \"American Institute of Mathematical Sciences\",\nnumber = \"1\",\n\n}\n\nTY - JOUR\n\nT1 - Solution properties and error bounds for semi-infinite complementarity problems\n\nAU - Zhou, Jinchuan\n\nAU - Xiu, Naihua\n\nAU - Chen, Jein-Shan\n\nPY - 2013/3/26\n\nY1 - 2013/3/26\n\nN2 - In this paper, we deal with the semi-infinite complementarity problems (SICP), in which several important issues are covered, such as solvability, semismoothness of residual functions, and error bounds. In particular, we characterize the solution set by investigating the relationship between SICP and the classical complementarity problem. Furthermore, we show that the SICP can be equivalently reformulated as a typical semi-infinite min-max programming problem by employing NCP functions. Finally, we study the concept of error bounds and introduce its two variants, ε-error bounds and weak error bounds, where the concept of weak error bounds is highly desirable in that the solution set is not restricted to be nonempty.\n\nAB - In this paper, we deal with the semi-infinite complementarity problems (SICP), in which several important issues are covered, such as solvability, semismoothness of residual functions, and error bounds. In particular, we characterize the solution set by investigating the relationship between SICP and the classical complementarity problem. Furthermore, we show that the SICP can be equivalently reformulated as a typical semi-infinite min-max programming problem by employing NCP functions. Finally, we study the concept of error bounds and introduce its two variants, ε-error bounds and weak error bounds, where the concept of weak error bounds is highly desirable in that the solution set is not restricted to be nonempty.\n\nKW - Error bounds\n\nKW - Semi-infinite complementarity problem\n\nKW - Semidifferentiable and semis-mooth\n\nKW - Weak error bounds\n\nUR - http://www.scopus.com/inward/record.url?scp=84874915240&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=84874915240&partnerID=8YFLogxK\n\nU2 - 10.3934/jimo.2013.9.99\n\nDO - 10.3934/jimo.2013.9.99\n\nM3 - Article\n\nVL - 9\n\nSP - 99\n\nEP - 115\n\nJO - Journal of Industrial and Management Optimization\n\nJF - Journal of Industrial and Management Optimization\n\nSN - 1547-5816\n\nIS - 1\n\nER -"
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http://cacasa2.info/introduction-to-statistics-and-econometrics/2015/07/examples | [
"# EXAMPLES\n\nWe shall give further examples of applications of the convergence theorems of Section 6.1 and 6.2. There will be more applications in Chapter 7, as well.\n\nexample 6.4.1 Let {XJ be independent with EX{ = p,, TX, = of. Under what conditions on of and (XJ does X = Xf=1X;/n converge to p in probability?\n\nWe can answer this question by using either Theorem 6.1.1 (Chebyshev)\n\n— о —\n\nor Theorem 6.2.1 (Khinchine). In the first case, note E(X — p) = VX = пЧи^г. The required condition, therefore, is that this last quantity should converge to 0 as n goes to infinity. In the second case, we should assume that {XJ are identically distributed in addition to being independent.\n\nEXAMPLE 6.4.2 In Example 6.4.1, assume further that EXt — p|3 = ms.\nUnder what conditions on of does (X — p)/VTX converge to N(0, 1)?\nThe condition of Theorem 6.2.3 (Liapounov) in this case becomes",
null,
"-1/2\n\nEXAMPLE 6.4.В Let (XJ be i. i.d. with a finite mean p and a finite variance a2. Prove that the sample variance, defined as Sx = n ^’LjX2 — X2, converges to о in probability.\n\nBy Khinchine’s LLN (Theorem 6.2.1) we have plim„_>oo n! Xf=1X2 = EX and plim„^oo X = p. Because Sx is clearly a continuous function of n ‘X’LjX2 and X, the desired result follows from Theorem 6.1.3.\n\nEXAMPLE 6.4.4 Let {XJ be i. i.d. with EXt = px A 0 and ТХ; = o| and let {TJ be i. i.d. with EYi = pyand VYt = of. Assume that {XJ and {TJ are independent of each other. Obtain the asymptotic distribution of Y/X.\n\nBy Theorem 6.2.1 (Khinchine), X —> px and Y -4 px. Therefore, by Theorem 6.1.3, Y/X -» py/px – The next step is to find an appropriate normalization of (Y/X — py/px) to make it converge to a proper random variable. For this purpose note the identity",
null,
"Y pу _ P-xCT Py) Yv(X Px)\n\nX P-х Xxx\n\nThen we can readily see that the numerator will converge to a normal variable with an appropriate normalization and the denominator will\n\nconverge to (|Лх) in probability, so that we can use (iii) of Theorem 6.1.4 (Slutsky). Define Wi = рхТ, — |xyXr Then (W,) satisfies the conditions of Theorem 6.2.2 (Lindeberg-Levy). Therefore\n\n(6.4.2) Zre = n V X (Щ – шг) -> N(0, 1),\n\ni=l\n\n2 2 2 2 2",
null,
"where aw = |xxo> + |хустх. Using (iii) of Theorem 6.1.4 (Slutsky), we obtain from (6.4.1) and (6.4.2)\n\nmachines? Solve this exercise both by using the binomial distribution and by using the normal approximation.\n\n5. (Section 6.3)\n\nThere is a coin which produces heads with an unknown probability p. How many times should we throw this coin if the proportion of heads is to lie within 0.05 of p with probability at least 0.9?\n\n6. (Section 6.4)\n\nLet {Xj} be as in Example 6.4.4. Obtain the asymptotic distribution of\n\n(a) X2.\n\n(b) 1/X.\n\n(c) exp(X).\n\n7. (Section 6.4)\n\nSuppose X has a Poisson distribution P(X = k) = (Xke k)/kl Derive the probability limit and the asymptotic distribution of the estimator\n\n– _ -1 +V1 + 4Zra X—— 2\n\nbased on a sample of size n, where Zn = n ‘EjLj X2. Note that EX = VX = and T(X2) = 4A3 + 6k2 + A.\n\n8. (Section 6.4)\n\nLet {X,} be independent with EX = p and VX = ct2. What more\n\nо — о\n\nassumptions on {XJ are needed in order for a = 2(Х* — X) /n to\n\n9\n\nconverge to cr in probability? What more assumptions are needed for its asymptotic normality?\n\n9. (Section 6.4)\n\nSuppose (XJ are i. i.d. with EX = 0 and VX = о2 < со.\n\n(a) Obtain\n\nП\n\nplim n~l X (Xi + xi+i)-\n\ni= 1\n\n(b) Obtain the limit distribution of\n\n71\n\nи“1/2Х(Х; + Хг+1).\n\ni=1\n\n10. (Section 6.4)\n\nLet [Xu Yj} be i. i.d. with the means |jlx and |Xy, the variances a| and\n\n2\n\ndy, and the covariance сгху – Derive the asymptotic distribution of X-Y X + Y\n\nExplain carefully each step of the derivation and at each step indicate what convergence theorems you have used. If a theorem has a well- known name, you may simply refer to it. Otherwise, describe it.\n\n11. (Section 6.4)\n\nSuppose X ~ IV[exp(a|3), 1] and Y ~ lV[exp(a), 1], independent of each other. Let {Xb F,}, і = 1, 2, . . . , n, be i. i.d. observations on (X, Y), and define X = nlYnl=lXt and Y = n~]Y”=}Yr We are to estimate (3 by P = logX/logF. Prove the consistency of (3 (see Definition 7.2.5, p. 132) and derive its asymptotic distribution.\n\nChapters 7 and 8 are both concerned with estimation: Chapter 7 with point estimation and Chapter 8 with interval estimation. The goal of point estimation is to obtain a single-valued estimate of a parameter in question; the goal of interval estimation is to determine the degree of confidence we can attach to the statement that the true value of a parameter lies within a given interval. For example, suppose we want to estimate the probability (p) of heads on a given coin toss on the basis of five heads in ten tosses. Guessing p to be 0.5 is an act of point estimation. We can never be perfectly sure that the true value of p is 0.5, however. At most we can say that p lies within an interval, say, (0.3, 0.7), with a particular degree of confidence. This is an act of interval estimation.\n\nIn this chapter we discuss estimation from the standpoint of classical statistics. The Bayesian method, in which point estimation and interval estimation are more closely connected, will be discussed in Chapter 8."
]
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"http://cacasa2.info/img/1146/image221.png",
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"http://cacasa2.info/img/1146/image222.png",
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"http://cacasa2.info/img/1146/image223.png",
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https://socratic.org/questions/how-do-you-determine-if-c-4-2c-2-1-is-a-polynomial-and-if-so-how-do-you-identify | [
"# How do you determine if c^4-2c^2+1 is a polynomial and if so, how do you identify if it is a monomial, binomial, or trinomial?\n\nBy looking at the degree of the highest power on the variable in question, which in this case is $c$. The highest power is 4, thus it is not a monomial, binomial or trinomial, rather it is a quadrinomial."
]
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http://mediamacros.com/item/item-933082515/ | [
"Contents Articles Behaviors Books Director News Director Web Sites FAQ Games Mailing Lists News Groups Project Examples Reviews Software Tools Useful Web Sites Utilities Xtras\n\n Don't miss these",
null,
"",
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"#",
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"Useful string functions\n\n Compatibilities:",
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"",
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"Rating:",
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"Author: LukeWigley\n\nFunctions for searching and manipulating strings\n\n -- Functions - Manipulating Strings on findWordfromChar str, charPos -- returns the word that the specified character occurs in wordBreak = [SPACE, TAB, \",\", \".\", \":\", \";\", \"?\", \"!\", RETURN] if getOne(wordBreak, str.char[charPos]) > 0 then return EMPTY mx = length (str) fWrd = str.char[charPos] i = charPos - 1 repeat while i > 0 if getOne(wordBreak, str.char[i]) > 0 then exit repeat else fWrd = str.char[i] & fWrd i = i -1 end if end repeat i = charPos + 1 repeat while i <= mx if getOne(wordBreak, str.char[i]) > 0 then exit repeat else fWrd = fWrd & str.char[i] i = i + 1 end if end repeat return fWrd end on getRandomChar letCase -- returns a random letter; if case is not specifed (as #upper or #lower), it picks a random case if not symbolP(letCase) then set letCase = getAt([#upper, #lower], random(2)) if letCase = #upper then set aLet = 64 + random(26) else set aLet = 93 + random (26) set lett = numToChar (aLet) return lett end on capitalise lett -- if the param is a lower case letter, an uppercase version is returned if stringP(lett) then set tA = charToNum(lett) if tA = min(max(96, tA), 123) then set tA = tA - 32 set lett = numTOChar (tA) end if end if return lett end on forceSentenceCase str -- converts the first letter of the first word of the str to uppercase -- str = forceLowerCase (Str) set lett = char 1 of str set tA = charToNum(lett) if tA = min(max(96, tA), 123) then set tA = tA - 32 set lett = numTOChar (tA) end if set outPut = lett & char 2 to length(str) of str return output end on CapitaliseFirstLetter str -- converts the first letter of each word to uppercase set wCount = the number of words in str set outPut = \"\" repeat with j = 1 to wCount set wrd = word j of str set lett = char 1 of wrd set tA = charToNum(lett) if tA = min(max(96, tA), 123) then set tA = tA - 32 set lett = numTOChar (tA) end if put lett & (char 2 to length (wrd) of wrd) & \" \" after output end repeat delete the last char of output return output end on forceUppercase iStr -- takes a string and converts it to all uppercase set oStr = \"\" set n = length(iStr) repeat with i = 1 to n set tA = charToNum(char i of iStr) if tA = min(max(97, tA), 122) then set tA = tA - 32 -- char is lowercase put numTOChar (tA) after oStr end repeat return oStr end on forceLowercase iStr -- takes a string and converts it to all lowercase set oStr = \"\" set n = length(iStr) repeat with i = 1 to n set tA = charToNum(char i of iStr) if tA = min(max(65, tA), 90) then set tA = tA + 32 put numTOChar (tA) after oStr end repeat return oStr end on forceTitleCase iStr set oStr = \"\" set lStr = forceLowerCase (iStr) set mWrd = the number of words in lStr repeat with j = 1 to mWrd set wrd = word j of lStr set fChar = char 1 of wrd set UChar = forceUppercase (fChar) delete char 1 of wrd put Uchar & wrd & \" \" after oStr end repeat delete the last char of oStr return oStr end on searchAndReplace input, oldStr, newStr -- searches the input (string) for oldStr and replaces it with newStr set output = \"\" repeat while input contains oldStr set posn = offset(oldStr, input)-1 if posn > 0 then put char 1 to posn of input after output put newStr after output delete char 1 to (posn + length(oldStr)) of input end repeat put input after output return output end on trimPunctuation pStr set str = pStr set punc = \" ;:,.-?!*\" repeat while (length (str) > 0) AND (punc contains char 1 of str) delete char 1 of str end repeat repeat while (length (str) > 0) AND (punc contains the last char of str) delete the last char of str end repeat return str end on trimChars charsToTrim, pStr set str = pStr repeat while (length (str) > 0) AND (charsToTrim contains char 1 of str) delete char 1 of str end repeat repeat while (length (str) > 0) AND (charsToTrim contains the last char of str) delete the last char of str end repeat return str end on removeLeadingSpaces pStr repeat while char 1 of pStr = \" \" delete char 1 of pStr end repeat return pStr end on removeTrailingSpaces pStr repeat while (the last char of pStr = \" \") delete the last char of pStr end repeat return pStr end on stripNonAlphabetChars istr -- takes a string and strips out non-aphabetcharacters (including spaces) set oStr = \"\" set n = length(iStr) repeat with i = 1 to n set tA = charToNum(char i of iStr) if (tA = min(max(96, tA), 123)) OR (tA = min(max(63, tA), 91)) then put numToChar (tA) after oStr end if end repeat return oStr end on trimSpaces pStr set pStr = string (pStr) -- remove leading spaces repeat while char 1 of pStr = \" \" delete char 1 of pStr end repeat -- remove trailing spaces repeat while (the last char of pStr = \" \") delete the last char of pStr end repeat return pStr end on getASCIgarbage strLength -- generates a string of garbage if Not integerP(strLength) then set strLength = 100 set str = \"\" repeat with k = 1 to strLength put numToChar (random(231)) after str end repeat return str end on pad sourceStr, minLength, padChar, frontorBack set str = sourceStr if frontorBack = #front then repeat while (length(str) < minLength) set str = padChar & str end repeat else repeat while (length(str) < minLength) put padChar after str end repeat end if return str end on stringContainsVowels str -- returns\b 0 if the str contains a vowel set vowels = [\"a\", \"e\", \"i\", \"o\", \"u\"] repeat with v in vowels if str contains v then return 0 end repeat return -1 end on sortField fName -- sorts field alphanumerically, line by line set mx = the number of lines in field fName set tList = [] repeat with j = 1 to mx append tList, line j of field fName end repeat sort tList set str = EMPTY repeat with j = 1 to mx put getAt (tList, j) & return after str end repeat delete the last char of str put str into field fName end on stripEmptyLines str rStr = EMPTY mx = the number of lines in str repeat with L = 1 to mx chnk = line L of str if the number of chars in chnk> 0 then put chnk & return after rStr end repeat delete the last char of rStr return rStr end\n\n Contact MMI 36 South Court Sq Suite 300 Newnan, GA 30263 USA",
null,
"Send e-mail"
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https://mathoverflow.net/questions/29512/when-is-a-sheaf-of-smooth-functions-acylic | [
"# When is a sheaf of smooth functions acylic?\n\nLet $G$ be a Lie group, and let $\\underline{G}$ denote the sheaf of smooth $G$-valued maps, i.e. for a smooth manifold $M$ we have $G(M) = C^\\infty(M,G)$.\n\nWhat are conditions on $G$ that imply that $\\underline{G}$ is acyclic, i.e. the sheaf cohomology $H^n(M,\\underline{G})=0$ for all smooth manifolds $M$ and all $n>0$?\n\nIt is clear that soft, flabby or fine sheaves are acyclic. I am interested in concrete conditions on the group $G$, e.g. like smooth contractibility.\n\nEDIT: Daniel's answer below answers my question in the case that $G$ is abelian, using the classification of abelian Lie groups. So let us concentrate on the case that $G$ is non-abelian. The condition I am looking for is supposed to imply the vanishing of the set $H^1(M,\\underline{G})$. This set can be defined for example via Cech cohomology. Its geometrical meaning is that it classifies principal $G$-bundles over $M$ up to isomorphism.\n\n• Here G is Abelian? – Daniel Litt Jun 25 '10 at 13:28\n• For $G=\\mathbb{R}$, we always have $G(M)$ acyclic. For the circle $G=S^1$, I think we have a ses of sheaves $0\\to\\mathbb{Z}\\to\\mathbb{R}(M)\\to S^1(M)\\to 0$, where here $\\mathbb{Z}$ is the constant sheaf? So for $*\\geq 1$, $H^*(S^1(M))$ is the same as the singular cohomology $H^{*+1}(M,\\mathbb{Z})$, and to vanish we need $M$ itself to have trivial second-and-above $\\mathbb{Z}$-cohomology groups? I don't know anything about doing cohomology with sheaves of nonabelian groups but it would be interesting to see whether a similar universal cover argument would work. – macbeth Jun 25 '10 at 13:46\n• In the original question, it might help if you told us what your definition for nonabelian cohomology is. The ones that I'm familiar with only work for $n=1,2$. – Donu Arapura Jun 25 '10 at 14:01\n• Don't the $\\infty$-category people have some theory of non-abelian cohomology in all higher degrees? (I know nothing about this, or why they care or more importantly what they do with it -- e.g., to determine if it is not a \"stupid thing\" -- but I thought I saw it written somewhere.) – Boyarsky Jun 25 '10 at 17:07\n• Daniel, it certainly isn't meant that way! (I don't think \"representation theory people\" or \"PDE people\" is pejorative either.) Should I have said \"$\\infty$-category theorists\"? (Personally I don't like that, since I assume most who use $\\infty$-categories regard it as a tool, not their primary area of study.) – Boyarsky Jun 25 '10 at 17:58\n\nFor Abelian $G$ (that is, the product of a torus with $\\mathbb{R}^n$), an argument identical to macbeth's comment gives that $H^n(M, \\underline{G})=0$ for all $M, n>0$ iff $G\\simeq \\mathbb{R}^n$).\n\nExplicitly, in the case $G\\simeq \\mathbb{R}^n$ the sheaf in question is fine; otherwise, if $G\\simeq \\mathbb{R}^n\\times (S^1)^k$ then it fits into an exact sequence $0\\to \\mathbb{Z}^k\\to \\mathbb{R}^{n+k}(M)\\to \\underline{G}\\to 0$, giving the claim.\n\nAdded (7/7/2010): Having thought a bit about the non-Abelian case, I thought I'd add another non-vanishing theorem.\n\nTheorem. Let $G$ be a Lie group admitting a faithful unitary representation, with $\\pi_1(G)\\neq 0, \\mathbb{Z}$. Then there exists $M$ with $H^1(M, \\underline{G})\\neq 0$.\n\nProof. Let $\\rho: G\\to U(n)$ be the given faithful unitary representation, and let $M=U(n)/G$. Then $U(n)$ is a $G$-bundle over $M$, and it is non-trivial as $\\pi_1(U(n))=\\mathbb{Z}$ wheareas $\\pi_1(G)$ cannot be a factor of $\\mathbb{Z}$ by assumption. That is, $U(n)\\not\\simeq G\\times M$ as $\\pi_1(U(n))\\not\\simeq \\pi_1(G)\\times \\pi_1(M)$. $\\square$\n\nThis holds for e.g. compact Lie groups with the appropriate fundamental group; it seems likely that this argument can be strengthened by e.g. considering higher homotopy groups or using other results on the existence of faithful representations.\n\nAdded (7/9/2010): I don't know why I didn't mention it before, but replacing \"unitary\" with \"complex\" in the theorem above gives the same result for e.g. complex connected semisimple Lie groups, by an identical proof. In this case the manifold $M$ constructed in the proof cannot be guaranteed to be compact however.\n\n• Hey, I like your non-vanishing theorem very much. Thanks! – Konrad Waldorf Jul 9 '10 at 5:07"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8716369,"math_prob":0.99613804,"size":912,"snap":"2019-35-2019-39","text_gpt3_token_len":260,"char_repetition_ratio":0.11453745,"word_repetition_ratio":0.013888889,"special_character_ratio":0.2609649,"punctuation_ratio":0.12886597,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996163,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T23:34:09Z\",\"WARC-Record-ID\":\"<urn:uuid:74533022-175b-454e-975c-22e5c22aca38>\",\"Content-Length\":\"125543\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e684cea-51a2-4fdc-89a5-9fa60bf3b215>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea0667ec-6c82-4273-9b40-a38b18c0b89a>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/29512/when-is-a-sheaf-of-smooth-functions-acylic\",\"WARC-Payload-Digest\":\"sha1:QEL3L62RYXEIF466OHWQLRRWKML7KKP5\",\"WARC-Block-Digest\":\"sha1:DUAGQT2VNUV5T4TR4OU7HQJTQYRSXOLC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573124.40_warc_CC-MAIN-20190917223332-20190918005332-00262.warc.gz\"}"} |
https://assignment.essayshark.com/blog/pump-head-calculation-example-a-guide-in-engineering/ | [
"# Pump Head Calculation Example – A Guide in Engineering\n\nThe head losses in the pipeline consist of linear head losses and local losses on the elements of pipeline systems. Linear head losses are pressure losses due to friction in the pipe; they depend on the flow regime, temperature, and water flow rate, as well as on the wall roughness and pipe diameter. To go deeper in the issue, you should read through the following pump head calculation example.\n\nThe sample you will find below calculates the head losses in the pump and determines the rate of flow of water in it. Our blog also includes examples of assignments for other disciplines. All samples are done by experts in the field. You can be sure that the examples are correct, and you can use them to work on your own assignments. So, if you are a student who studies engineering, you will find this pump head calculation example very helpful.\n\n## Pump Head Calculation Task\n\nTask: in this guide we will perform m head loss calculations, solve flow rate problems, generate system curves, and find the design point for a system and pump for pipe flow analysis.",
null,
"Solution:\n\n(Part 1)\n\nGoverning Equations\n\nThe figure above shows a single pipe flow system. For steady flow between station 1 and station 2, here is the mass conversion:",
null,
"Where ˙m is the mass flow rate, ρ is the fluid density, A is the cross-section area, and V is the average velocity. In most applications the flow is incompressible, so ρ1 = ρ2 and Equation (1) simplifies to:",
null,
"where Q is the volumetric flow rate.\n\nEnergy conservation between two stations:",
null,
"Where the subscript “out” indicates the downstream station, and “in” indicates the upstream station, γ = ρg is the specific weight of the fluid, hs is the shaft work done on the fluid, and hL is the head loss due to friction.\n\nTo convert hs and hL to units of power we use:",
null,
"Empirical Head Loss Data\n\nThe head loss due to friction is given by the Darcy-Weisbach equation:",
null,
"Where f is the Darcy Friction Factor. For fully-developed, incompressible laminar flow in a round pipe, the equation is as follows:",
null,
"The friction factor for turbulent flow in smooth and rough pipes is correlated with the\n\nColebrook equation:",
null,
"Where ε is the roughness of the pipe wall, and Re is the Reynolds number:",
null,
"If the pipe is not round, the same formulas may be applied if the hydraulic diameter, Dh, is substituted for D in the definition of Re, and in the ε/D term in the Colebrook equation. For a pipe with any cross-section area A and perimeter P, the hydraulic diameter is:"
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https://www.geogebra.org/m/mrFCqYVD | [
"# The Lazzarini approximation of Pi\n\nThe Buffon Needle experiment provides an experimental approximation of the constant . One drops a large number of needles at random (with uniform distribution) over a set of equally spaced parallel lines. If the length of the needles is smaller than the line spacing, the number of intersection between needles and lines is a random variable, whose expected value is a simple function of . So, if N is the number of needles, H is the number of intersections between the needles and the lines, d is the line spacing and l is the needle length, one can compute by the formula In 1901, Lazzarini performed the experiment, reporting an astonishingly accurate value for . The experiment is now considered as a hoax, due to the finely tuned experimental design and to doubtful reported results. The simulation used the same experimental set, showing how a biased experimental design can lead to the desired result. See: http://en.wikipedia.org/wiki/Buffon%27s_needle https://www.uam.es/personal_pdi/ciencias/gallardo/Lazzarini.pdf"
]
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https://blog.deriscope.com/index.php/en/yield-curve-excel-quantlib-futures | [
"8 minutes reading time (1552 words)\n\n# Yield Curve Building in Excel using Futures\n\nWith this article I want to show you how to create a yield curve in Excel using the open source QuantLib analytics library, when the input market data are futures prices. The futures convexity will be taken into account.\n\nI explained how you may build a yield curve in Excel out of forward rates in my previous article.\n\nIn reality, forward rates are seldom used in practice for yield curve construction because\n\na) The respective forward contracts are mainly OTC (Over The Counter), meaning they are not that liquid and\n\nb) Most derivatives traders use futures to hedge their interest rate risk, exactly because futures are exchange-based and therefore more liquid.\n\nThe second point is critical as the no-arbitrage-based fair price calculation is based on a hypothetical trading strategy that results in a delta-neutral hedge portfolio. This means one should build the yield curve exactly out of the same instruments that one uses to hedge the trading positions. Since these instruments typically include exchange-traded futures, it follows that knowing how to build the yield curve out of futures is of paramount importance.\n\nA futures contract is basically a forward contract with the additional provision that each counterparty maintains a margin account, which is credited daily with the price change of the underlying deposit contract.\n\nThe effect is that a futures contract has always a value of zero, at least if we ignore for simplicity the intraday situation when the margin has not yet been posted.\n\nIts mathematical valuation is based on the fact that the futures price evolves like a martingale (i.e. without drift) in a risk neutral world, which means it can be treated as a traded asset that pays a continuous dividend with yield equal to the risk-free short interest rate.\n\nBelow is a graphical comparison between a forward and a futures contract:\n\n### Futures Quote Convention\n\nFrom a practical perspective, another important difference between futures and forwards concerns their quotation. While forward contracts are quoted by their forward rate, for example 1%, futures contracts are quoted in a mystical way by some number F, for example 99, that is defined to equal 100(1– R) at the futures expiry T1, where R is the rate of the underlying deposit contract realized at T1.\n\nLet's be more specific.\n\nLet's say the underlying deposit contract is the one defined through the 3-month USD Libor. As a reminder, 3-month USD Libor is the rate agreed between London banks for 3 month spot deposits among them. By the way, the corresponding futures contract is then known as Eurodollar contract.\n\nLet's also assume that our futures contract expires on 20 June 2018.\n\nThen the quote convention demands that the quoted futures price F on 20 June 2018 equals exactly 100 – R, where R is the 3-month USD Libor settled on 20 June 2018.\n\nFor example, if R happens to be 1% then F will be 100(1–0.01) = 99 on that date.\n\nDue to the fact that the margin account of each futures holder is credited daily by an amount proportional to ΔF (daily change in F), the quoted price F on any earlier date T will equal 100(1-f), where f is approximately equal to the forward rate r prevailing at time T for the 3-month Libor contract commencing on 20 June 2018.\n\nf would equal exactly r if it were not for the daily margin feature.\n\nBut due to the margin requirement, f = r + c, where c is a small correction referred to as the convexity correction.\n\nAs you will see, Deriscope allows you to specify c as part of the input data to the yield curve construction function.\n\nRecommended for Deriscope starters: The Overview and Quick Guide pages.\n\n### Creating the Yield Curve\n\nAs I did before with the deposit and forward rates, I will use the wizard to generate the correct formula.\n\nAfter I select the Yield Curve type in Type Selector I must check the Use Futures flag inside the input parameters screen and finally hit the Go button, as this video demonstrates:\n\nThis is the result:\n\n### Understanding the formula\n\nAs you see, cell A1 contains the formula =ds(A2:B5), which takes one input argument and returns the text &GBPCrv_A1:1.1.\n\nThe prefix & indicates that &GBPCrv_A1:1.1 is the handle name of some object. In fact, it points to an object of type Yield Curve and can be used in any context where a yield curve is needed, such as in pricing of options.\n\nThe overall structure is similar to that of a yield curve created out of deposit rates as explained in my related post.\n\nAs with the deposit rate case, the above formulas include only the mandatory arguments. The object contents displayed in the wizard include all arguments, including the optional.\n\nHere the table of futures prices consists of three columns supplying the respective expiries, prices and futures convexity correction values.\n\nThe first column must bear the title #Expiry and contain the expiry dates of the futures contracts.\n\nThe second column must bear the title #Price and contain the respective futures prices.\n\nThe third column must bear the title #Convexity and contain the respective futures convexity correction values.\n\nAll futures contracts will then share the same length and conventions, which must be explicitly supplied through the Ibor Rate object in cell B10.\n\nLinear interpolation and flat extrapolation is assumed for those maturities that do not appear in the table.\n\n### Browsing through the Yield Curve Object's Contents\n\nHow to use the wizard to browse through the contents of the Yield Curve object is demonstrated by a video in the respective section of my post on curve creation from deposits, futures and swaps.\n\n### Browsing through the Cash Flows\n\nThe pair _Cash Flows= \\$Set#9 shown in the wizard grid above can be very useful for diagnostic purposes, as it contains the details of the swap instruments used in the curve construction.\n\nDetailed description is available at the cash flows section of my post on curve creation from deposits, futures and swaps.\n\n### Dealing with a Curve Bootstrapping Failure\n\nThe steps you need to take in case of failure are exemplified in the diagnostics section of my post on curve creation from deposits, futures and swaps.\n\n### Second Alternative construction of the Futures Prices table\n\nRather than supplying an object of type Ibor Rate in cell B10 that supplies a commonly shared underlying tenor for all futures contracts, I can use a more flexible setup that allows individual futures contracts to have different underlying tenors as well as custom defined expiry dates.\n\nYou may refer to this similar alternative deposit rates construction for a demonstration on how to use the wizard in this regard.\n\nBelow is the result:\n\nThe futures prices are specified by supplying the respective expiries, underlying maturities, prices and convexities through a table consisting of 4 columns.\n\nThe first column must bear the title #Expiry and contain the expiry dates of the futures contracts.\n\nThe second column must bear the title #Maturity and contain the maturity dates of the underlying interest rates. The dates here must be greater than those in the first column.\n\nThe third column must bear the title #Price and contain the respective futures prices.\n\nThe fourth column must bear the title #Convexity and contain the respective futures convexity correction values.\n\nHere the futures contracts are allowed to have different underlying lengths, but they will still share the same conventions, which must be explicitly supplied.\n\nLinear interpolation and flat extrapolation is assumed for those maturities that do not appear in the table.\n\n### Third Alternative construction of the Futures Prices table\n\nA simpler variant of the previous construction makes use of only three columns as shown below:\n\nThe futures prices are specified by supplying the respective expiries, prices and convexities through a table consisting of 3 columns.\n\nThe first column must bear the title #Expiry and contain the expiry dates of the futures contracts.\n\nThe second column must bear the title #Price and contain the respective futures prices.\n\nThe third column must bear the title #Convexity and contain the respective futures convexity correction values.\n\nAll futures contracts will then share the same length, which must be explicitly supplied in number of months.\n\nThey will also share the same conventions, which must be also supplied.\n\nLinear interpolation and flat extrapolation is assumed for those maturities that do not appear in the table.\n\n### Fourth Alternative construction of the Futures Prices table\n\nAn even simpler construction makes use of the International Money Market (IMM) dates with the following result:\n\n### Using the Yield Curve object\n\nFor this I would refer you to the respective section of the Yield Curve out of Deposit Rates article"
]
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https://deepai.org/publication/dr-loss-improving-object-detection-by-distributional-ranking | [
"",
null,
"# DR Loss: Improving Object Detection by Distributional Ranking\n\nMost of object detection algorithms can be categorized into two classes: two-stage detectors and one-stage detectors. For two-stage detectors, a region proposal phase can filter massive background candidates in the first stage and it masks the classification task more balanced in the second stage. Recently, one-stage detectors have attracted much attention due to its simple yet effective architecture. Different from two-stage detectors, one-stage detectors have to identify foreground objects from all candidates in a single stage. This architecture is efficient but can suffer from the imbalance issue with respect to two aspects: the imbalance between classes and that in the distribution of background, where only a few candidates are hard to be identified. In this work, we propose to address the challenge by developing the distributional ranking (DR) loss. First, we convert the classification problem to a ranking problem to alleviate the class-imbalance problem. Then, we propose to rank the distribution of foreground candidates above that of background ones in the constrained worst-case scenario. This strategy not only handles the imbalance in background candidates but also improves the efficiency for the ranking algorithm. Besides the classification task, we also improve the regression loss by gradually approaching the L_1 loss as suggested in interior-point methods. To evaluate the proposed losses, we replace the corresponding losses in RetinaNet that reports the state-of-the-art performance as a one-stage detector. With the ResNet-101 as the backbone, our method can improve mAP on COCO data set from 39.1% to 41.1% by only changing the loss functions and it verifies the effectiveness of the proposed losses.\n\n## Authors\n\n##### This week in AI\n\nGet the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.\n\n## 1 Introduction\n\nThe performance of object detection has been improved dramatically with the development of deep neural networks in the past few years.",
null,
"Figure 1: Illustration of the proposed distributional ranking loss. First, we re-weight examples to derive the constrained distributions for foreground and background from the original distributions, respectively. Then, we learn to rank the expectation of the derived distribution of foreground above that of background by a large margin.\n\nMost of detection algorithms fall into two categories: two-stage detectors [3, 11, 12, 14] and one-stage detectors [6, 15, 17, 20]\n\n. For the two-stage schema, the procedure of the algorithms can be divided into two parts. In the first stage, a region propose method will filter most of background candidate bounding boxes and keep only a small set of candidates. In the following stage, these candidates are classified as foreground classes or background and the bounding box is further refined by optimizing a regression loss. Two-stage detectors demonstrate the superior performance on real-world data sets while the efficiency can be an issue in practice, especially for the devices with limited computing resources, e.g., smart phones, cameras, etc.\n\nTherefore, one-stage detectors are developed for an efficient detection. Different from two-stage detectors, one-stage algorithms consist of a single phase and have to identify foreground objects from all candidates directly. The structure of a ons-stage detector is straightforward and efficient. However, a one-stage detector may suffer from the imbalance problem that can reside in the following two aspects. First, the numbers of candidates between classes are imbalanced. Without a region proposal phase, the number of background candidates can easily overwhelm that of foreground ones. Second, the distribution of background candidates is imbalanced. Most of them can be easily separated from foreground objects while only a few of them are hard to classify.\n\nTo alleviate the imbalance problem, SSD adopts hard negative mining, which keeps a small set of background candidates with the highest loss. By eliminating simple background candidates, the strategy balances the number of candidates between classes and the distribution of background simultaneously. However, some important classification information from background can be lost, and thus the detection performance can degrade. RetinaNet \n\nproposes to keep all background candidates but assign different weights for loss functions. The weighted cross entropy loss is called focal loss. It makes the algorithm focus on the hard candidates while reserving the information from all candidates. This strategy improves the performance of one-stage detectors significantly. Despite the success of focal loss, it re-weights classification losses in a heuristic way and can be insufficient to address the class-imbalance problem. Besides, the design of focal loss is data independent and lacks the exploration of the data distribution, which is essential to balance the distribution of background candidates.\n\nIn this work, we propose a data dependent ranking loss to handle the imbalance challenge. First, to alleviate the effect of the class-imbalance problem, we convert the classification problem to a ranking problem, which optimizes ranks of pairs. Since each pair consists of a foreground candidate and a background candidate, it is well balanced. Moreover, considering the imbalance in background candidates, we introduce the distributional ranking (DR) loss to rank the constrained distribution of foreground above that of background candidates. By re-weighting the candidates to derive the distribution corresponding to the worst-case loss, the loss can focus on the decision boundary between foreground and background distributions. Besides, we rank the expectation of distributions in lieu of original examples, which reduces the number of pairs in ranking and improves the efficiency. Compared with the re-weighting strategy in focal loss, that for DR loss is data dependent and can balance the distribution of background better. Fig. 1 illustrates the proposed DR loss. Besides the classification task, the regression is also important for detection to refine the bounding boxes of objects. The smoothed loss is prevalently adopted to approximate the loss in detection algorithms. We propose to improve the regression loss by gradually approaching the loss for better approximation, where the similar trick is also applied in interior-point methods .\n\nWe conduct the experiments on COCO data set to demonstrate the proposed losses. Since RetinaNet reports the state-of-the-art performance among one-stage detectors, we replace the corresponding losses in RetinaNet with our proposed losses while the other components are retained. For fair comparison, we implement our algorithm in Detectron , which is the official codebase of RetinaNet. With ResNet-101 as the backbone, optimizing our loss functions can boost the mAP of RetinaNet from to , which confirms the effectiveness of proposed losses.\n\nThe rest of this paper is organized as follows. Section 2 reviews the related work in object detection. Section 3 describes the details of the proposed DR loss and regression loss. Section 4 compares the proposed losses to others on COCO detection task. Finally, Section 5 concludes this work with future directions.\n\n## 2 Related Work\n\nDetection is a fundamental task in computer vision. In conventional methods, hand crafted features, e.g., HOG\n\n and SIFT , are used for detection either with a sliding-window strategy which holds a dense set of candidates, e.g., DPM or with a region proposal method which keeps a sparse set of candidates, e.g., Selective Search . Recently, since deep neural networks have shown the dominating performance in classification tasks , the features obtained from neural networks are leveraged for detection tasks.\n\nR-CNN \n\nequips the region proposal stage and works as a two-stage algorithm. It first obtains a sparse set of regions by selective search. In the next stage, a deep convolutional neural network is applied to extract features for each region. Finally, regions are classified with a conventional classifier, e.g., SVM. R-CNN improves the performance of detection by a large margin but the procedure is too slow for real-world applications. Hence, many variants are developed to accelerate it\n\n[8, 21]. To further improve the accuracy, Mask-RCNN adds a branch for object mask prediction to boost the performance with the additional information from multi-task learning. Besides the two-stage structure, cascade R-CNN develops a multiple stage strategy to promote the quality of detectors after region proposal stage in a cascade fashion.\n\nOne-stage detectors are also developed for efficiency [17, 19, 22]. Since there is no region proposal phase to sample background candidates, one-stage detectors can suffer from the imbalance issue both between classes and in the background distribution. To alleviate the challenge, SSD adopts hard example mining, which only keeps the hard background candidates for training. Recently, RetinaNet is proposed to address the problem by focal loss. Unlike SSD, it keeps all background candidates but re-weights them such that the hard example will be assigned with a large weight. Focal loss improves the performance of detection explicitly, but the imbalance problem in detection is still not explored sufficiently. In this work, we develop the distributional ranking loss that ranks the distributions of foreground and background. It can alleviate the imbalance issue and capture the data distribution better with a data dependent mechanism.\n\n## 3 DR Loss\n\nGiven a set of candidate bounding boxes from an image, a detector has to identify the foreground objects from background ones with a classification model. Let denote a classifier and it can be learned by optimizing the problem\n\n minθN∑i∑j,kℓ(pi,j,k) (1)\n\nwhere\n\nis the number of total images. In this work, we employ sigmoid function to predict the probability for each example.\n\nis determined by\n\nand indicates the estimated probability that the\n\n-th candidate in the -th image is from the -th class. is the loss function. In most of detectors, the classifier is learned by optimizing the cross entropy loss. For the binary classification problem, it can be written as\n\n ℓCE(p)={−log(p)y=1−log(1−p)y=0\n\nwhere is the label.\n\nThe objective in Eqn. 1 is conventional for object detection and it suffers from the class-imbalance problem. This can be demonstrated by rewriting the problem in the equivalent form\n\n minθN∑i(n+∑j+ℓ(pi,j+)+n−∑j−ℓ(pi,j−)) (2)\n\nwhere and denote the positive (i.e., foreground) and negative (i.e., background) examples, respectively. and are the corresponding number of examples. When , the accumulated loss from the latter term will dominate. This issue is from the fact that the losses for positive and negative examples are separated and the contribution of positive examples will be overwhelmed by negative examples. A heuristic way to handle the problem is emphasizing positive examples, which can increase the weights for the corresponding losses. In this work, we aim to address the problem in a fundamental way.\n\n### 3.1 Ranking\n\nTo alleviate the challenge from class-imbalance, we optimize the rank between positive and negative examples. Given a pair of positive and negative examples, an ideal ranking model can rank the positive example above the negative one with a large margin\n\n ∀i,j+,j−pi,j+−pi,j−≥γ (3)\n\nwhere is a non-negative margin. Compared with the objective in Eqn. 1, the ranking model optimizes the relationship between individual positive and negative examples, which is well balanced.\n\nThe objective of ranking can be written as\n\n minθN∑in+∑j+n−∑j−ℓ(pi,j−−pi,j++γ) (4)\n\nwhere can be the hinge loss as\n\n ℓhinge(z)=[z]+={zz>00o.w.\n\nThe objective can be interpreted as\n\n 1n+n−n+∑j+n−∑j−ℓ(pi,j−−pi,j++γ) =Ej+,j−[ℓ(pi,j−−pi,j++γ)] (5)\n\nIt demonstrates that the objective measures the expected ranking loss by uniformly sampling a pair of positive and negative examples.\n\nThe ranking loss addresses the class-imbalance issue by comparing the rank of each positive example to negative examples. However, it ignores a phenomenon in object detection, where the distribution of negative examples is also imbalanced. Besides, the ranking loss introduces a new challenge, that is, the vast number of pairs. We tackle them in the following subsections.\n\n### 3.2 Distributional Ranking\n\nAs indicated in Eqn. 3.1, the ranking loss in Eqn. 4 punishes a mis-ranking for a uniformly sampled pair. In detection, most of negative examples can be easily ranked well, that is, a randomly sampled pair will not incur the ranking loss with high probability. Therefore, we propose to optimize the ranking boundary to avoid the trivial solution\n\n minθN∑iℓ(maxj−pi,j−−minj+pi,j++γ) (6)\n\nIf we can rank the positive example with the lowest score above the negative one with the highest confidence, the whole set of candidates are perfectly ranked. Compared with the conventional ranking loss, the worst case loss is much more efficient by reducing the number of pairs from to\n\n. Moreover, it clearly eliminates the class-imbalance issue since only a single pair of positive and negative examples are required for each image. However, this formulation is very sensitive to outliers, which can lead to the degraded detection model.\n\nTo improve the robustness, we first introduce the distribution for the positive and negative examples and obtain the expectation as\n\n Pi,+=n+∑j+qi,j+pi,j+;Pi,−=n−∑j−qi,j−pi,j−\n\nwhere and denote the distribution over positive and negative examples, respectively. and represent the expected ranking score under the corresponding distribution. is the simplex as . When and\n\nare the uniform distribution,\n\nand demonstrates the expectation from the original distribution.\n\nBy deriving the distribution corresponding to the worst-case loss from the original distribution\n\n Pi,+=minqi,+∈Δn+∑j+qi,j+pi,j+; Pi,−=maxqi,−∈Δn−∑j−qi,j−pi,j−\n\nwe can rewrite the problem in Eqn. 6 in the equivalent form\n\n minθN∑iℓ(Pi,−−Pi,++γ)\n\nwhich can be considered as ranking the distributions between positive and negative examples in the worst case. It is obvious that the original formulation is not robust due to the fact that the domain of the generated distribution is unconstrained. Consequently, it will concentrate on a single example while ignoring the original distribution. Hence, we improve the robustness of the ranking loss by regularizing the freedom of the derived distribution as\n\n Pi,− = maxqi,−∈Δ,Ω(qi,−)≥ϵ−n−∑j−qi,j−pi,j− −Pi,+ = maxqi,+∈Δ,Ω(qi,+)≥ϵ+n+∑j+qi,j+(−pi,j+)\n\nwhere is a regularizer for the diversity of the distribution to prevent the distribution from the trivial one-hot solution. It can be different forms of entropy, e.g., Rényi entropy, Shannon entropy, etc. and are constants to control the freedom of distributions.\n\nTo obtain the constrained distribution, we investigate the subproblem\n\n maxqi,−∈Δ∑j−qi,j−pi,j− s.t. Ω(qi−)≥ϵ−\n\nAccording to the dual theory , given , we can find the parameter to obtain the optimal by solving the problem\n\n maxqi,−∈Δ∑j−qi,j−pi,j−+λ−Ω(qi,−)\n\nWe observe that the former term is linear in . Hence, if is strongly concave in , the problem can be solved efficiently by first order algorithms .\n\nConsidering the efficiency, we adopt the Shannon entropy as the regularizer in this work and we can have the closed-form solution as follows.\n\n###### Proposition 1.\n\nFor the problem\n\n maxqi,−∈Δ∑j−qi,j−pi,j−+λ−H(qi−)\n\nwe have the closed-form solution as\n\n qi,j−=1Z−exp(pi,j−λ−);Z−=∑j−exp(pi,j−λ−)\n###### Proof.\n\nIt can be proved directly from K.K.T. condition . ∎\n\nFor the distribution over positive examples, we have the similar result as\n\n###### Proposition 2.\n\nFor the problem\n\n maxqi,+∈Δ∑j+qi,j+(−pi,j+)+λ+H(qi+)\n\nwe have the closed-form solution as\n\n qi,j+=1Z+exp(−pi,j+λ+);Z+=∑j+exp(−pi,j+λ+)",
null,
"Figure 2: Illustration of the drifting in the distribution. We randomly sample 1e7points from a Gaussian distribution to mimic negative examples. We change the weights of examples according to the proposed strategy as in Proposition 1and then plot the curves of different probability density functions (PDF) when varying λ.\n\n#### Remark 1\n\nThese Propositions show that the harder the example, the larger the weight of the example. Besides, the weight is data dependent and is affected by the data distribution.\n\nFig. 2 illustrates the drifting of the distribution with the proposed strategy. The derived distribution is approaching the distribution corresponding to the worst-case loss when decreasing .\n\nWith the closed-form solutions of distributions, the expectation of distributions can be computed as\n\n ^Pi,−=n−∑j−qi,j−pi,j−=n−∑j−1Z−exp(pi,j−λ−)pi,j− (7) ^Pi,+=n−∑j−qi,j+pi,j+=n+∑j+1Z+exp(−pi,j+λ+)pi,j+\n\nFinally, smoothness is crucial for the convergence of non-convex optimization . So we use the smoothed approximation instead of the original hinge loss as the loss function \n\n ℓsmooth(z)=1Llog(1+exp(Lz)) (8)\n\nwhere controls the smoothness of the function. The larger the is , the more closer to the hinge loss the approximation is. Fig. 3 compares the hinge loss to its smoothed version in Eqn. 8.",
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"Figure 3: Illustration of the hinge loss and its smoothed variants.\n\nIncorporating all of these components, our distributional ranking loss can be defined as\n\n minθLDR(θ)=N∑iℓsmooth(^Pi,−−^Pi,++γ) (9)\n\nwhere and are given in Eqn. 7 and is in Eqn. 8. Compared with the conventional ranking loss, we rank the expectation between two distributions. It shrinks the number of pairs to that leads to the efficient optimization.\n\nThe objective in Eqn. 9 looks complicated but its gradient is easy to compute. The detailed calculation of the gradient can be found in the appendix.\n\nIf we optimize the DR loss by the standard stochastic gradient descent (SGD) with mini-batch as\n\n θt+1=θt−η1mm∑s=1∇ℓst\n\nwe can show that it can converge as in the following theorem and the detailed proof is cast to the appendix.\n\n###### Theorem 1.\n\nLet denote the model obtained from the -th iteration with SGD optimizer, where mini-batch size is . When\n\n, if we assume the variance of the gradient is bounded as\n\nand set the learning rate as , we have\n\n 1T∑t∥∇L(θt)∥2F≤2δ√2L√mTL(θ0)\n\n#### Remark 2\n\nTheorem 1 implies that the learning rate depends on the mini-batch size and the number of iterations as and the convergence rate is . We let , and denote an initial setting for training. If we increase the mini-batch size as and shrink the number of iterations as where , the convergence rate remains the same. However, the learning rate has to be increased as when , which is consistent with the observation in .\n\n#### Remark 3\n\nTheorem 1 also indicates that the convergence rate depends on . Therefore, trades between the approximation error and the convergence rate. When is large, the smoothed loss can simulate the hinge loss better while the convergence can become slow.\n\n### 3.3 Recover Classification from Ranking\n\nIn detection, we have to identify foreground from background. Therefore, the results from ranking has to be converted to classification. A straightforward way is setting a threshold for the ranking score. However, the scores from different pairs can be inconsistent for classification. For example, given two pairs as\n\n p−=0.1,p+=0.4;p−=0.6,p+=0.9\n\nwe observe that both of them have the perfect ranking but it is hard to set a threshold to classify positive examples from negative ones simultaneously. To make the ranking result meaningful for classification, we enforce a large margin in the constraint 3 as . Therefore, the constraint becomes\n\n ∀i,j+,j−pi,j+−pi,j−≥0.5\n\nDue to the non-negative property of probability, it implies\n\n ∀i,j+pi,j+>0.5;∀i,j−pi,j−≤0.5\n\nwhich recovers the standard criterion for classification.\n\n### 3.4 Bounding Box Regression\n\nBesides classification, regression is also important for detection to refine the bounding box. Most of detectors apply smoothed loss to optimize the bounding box\n\n ℓreg(x)={0.5x2/βx≤β|x|−0.5βx≥β (10)\n\nIt smoothes loss by loss in the interval of and guarantees that the whole loss function is smooth. It is reasonable since smoothness is important for convergence as indicated in Theorem 1. However, it may result in the slow optimization in the interval of loss. Inspired by the interior-point method , which gradually approximates the non-smooth domain by increasing the weight of the corresponding barrier function at different stages, we obtain from a decreasing function to reduce the gap between and losses. As suggested in the interior-point method, the current objective should be solved to optimum before changing the weight for the barrier function. We decay the value of in a stepwise manner. Specifically, we compute at the -th iteration as\n\n βt=β0−α(t%K)\n\nwhere is a constant and denotes the width of a step.\n\nCombining the regression loss, the objective of training the detector becomes\n\n minN∑iτℓsmooth(^Pi,−−^Pi,++γ)+ℓreg(vi;βt)\n\nwhere is to balance the weights between classification and regression.\n\n## 4 Experiments\n\n### 4.1 Implementation Details\n\nWe evaluate the proposed losses on COCO 2017 data set , which contains about images for training, images for validation, and images for test. To focus on the comparison of loss functions, we employ the structure of RetinaNet as the backbone and only substitute the corresponding loss functions. For fair comparison, we make the adequate modifications in the official codebase of RetinaNet, which is released in Detectron. Besides, we train the model with the same setting as RetinaNet. Specifically, the model is learned with SGD on GPUs and the mini-batch size is set as where each GPU can hold images at each iteration. Most of experiments are trained with iterations and the length is denoted as “”. The initial learning rate is and is decayed by a factor of after iterations and then iterations. For anchor density, we apply the same setting as in , where each location has scales and aspect ratios. The standard COCO evaluation criterion is used to compare the performance of different methods.\n\nSince RetinaNet lacks the optimization of the relationship between positive and negative distributions, it has to initialize the output probability of the classifier at to fit the distribution of background. In contrast, we initialize the probability of the sigmoid function at , which is more reasonable for binary classification scenario without any prior knowledge. It also verifies that the proposed DR loss can handle class-imbalance better.\n\nIn Eqn. 7, we compute the constrained distribution over positive and negative examples with and , respectively. To reduce the number of parameters, we fix the ratio between and as and tune the scale as\n\n λ+=1/log(h);λ−=0.1/log(h)\n\nIt is easy to show that this strategy is equivalent to fixing and as and , and changing the base in the definition of the entropy regularizer as\n\n H(q)=−∑jqjloghqj\n\nNote that RetinaNet applies Feature Pyramid Network (FPN) to obtain multiple scale features. To compute DR loss in one image, we collect candidates from multiple pyramid levels and obtain a single distribution for foreground and background, respectively.\n\n### 4.2 Effect of Parameters\n\nFirst, we take ablation experiments to evaluate the effect of multiple parameters on the validation set. All experiments in this subsection are implemented with a single image scale of for training and test. ResNet- is applied as the backbone for comparison. Only horizontal flipping is adopted as the data augmentation in this subsection.\n\n#### Effect of L:\n\ncontrols the smoothness of the loss function in Eqn. 8. We compare the model with different in Table 1. Note that also changes the function value, we adjust the weight of classification loss accordingly. The base of entropy regularizer is fixed as . We observe that the loss function is quite stable for the choice of different smooth values. Besides, a larger will result in a smaller function value as shown in Fig. 3 and it suggests to increase the weight of classification loss to balance the losses. We keep and in the rest experiments.\n\n#### Effect of h:\n\nNext, we evaluate the effect of . changes the scale of and in the standard entropy regularizer. As illustrated in Fig. 2, a large will push the generated distribution to the extreme case while a small will make the derived distribution close to the original distribution. We vary the range of and summarize the results in Table 2. It is obvious that is also not sensitive in a reasonable range and we fix it to in the following experiments.\n\n#### Effect of β:\n\nFinally, we demonstrate the different strategies for changing in the smoothed loss. In the implementation of RetinaNet, is fixed to . We compare three strategies to decay to , which are illustrated in Fig. 4. The results are shown in Table 3. First, it is evident that all strategies with decayed can improve the performance of detectors with a fixed . Then, the stepwise decay with outperforms linear decay and it verifies that the objective should be optimized sufficiently before moving to the decay step. We adopt stepwise decay in the next subsections.\n\n#### Effect of DR Loss:\n\nTo illustrate the effect of DR loss, we collect the confidence scores of examples from all images in the validation set and compare the empirical probability density in Fig. 6. We include cross entropy loss and focal loss in the comparison. The model with cross entropy loss is trained by ourselves while the model with focal loss is downloaded directly from the official model zoo with the same configuration as DR loss.\n\nFirst, we observe that most of examples have the extremely low confidence with cross entropy loss. It is because the number of negative examples overwhelms that of positive ones and it will classify most of examples to negative to obtain a small loss as demonstrated in Eqn. 2. Second, focal loss is better than cross entropy loss by drifting the distribution of foreground. However, the expectation of the foreground distribution is still close to that of background, and it has to adopt a small threshold as to identify positive examples from negative ones. Compared to cross entropy and focal loss, DR loss optimizes the foreground distribution significantly. By optimizing the ranking loss with a large margin, the expectation of the foreground examples is larger than while that of background is smaller than . It confirms that DR loss can address the imbalance between classes well. Consequently, DR loss allows us to set a large threshold for classification. We set the threshold as in experiments while it is not sensitive in the range of . Besides, the distribution of background examples with DR loss is more balanced than that with cross entropy or focal loss. It verifies that with the data dependent re-weighting strategy, DR loss can handle the imbalance in background distribution and focus on the hard negative examples appropriately.",
null,
"Figure 5: Illustration of empirical PDF of distributions that are computed from images in the validation set.\n\n### 4.3 Performance with Different Scales\n\nWith the parameters suggested from ablation studies, we train the model with different scales and backbones to show the robustness of the proposed losses. We adopt ResNet-50 and ResNet-101 as backbones in the comparison. Training applies only horizontal flipping as the data augmentation. Table. 4 compares the performance with different scales to that of RetinaNet. We let “Dr.Retina” denote the RetinaNet with the proposed DR loss and the decaying strategy for smoothed loss. Evidently, Dr.Retina performs better than RetinaNet over all scales with different backbones. Since we only change the loss functions in RetinaNet, the inference time remains the same while the mAP is consistently improved by about . The comparison also shows that the parameters in Dr.Retina is not sensitive to the scale of input images. It implies that the proposed losses is applicable for real-world applications.\n\n### 4.4 Comparison with State-of-the-Art\n\nFinally, we compare Dr.Retina to the state-of-the-art two-stage and one-stage detectors on COCO test set. We follow the setting in to increase the number of training iterations to , which contains iterations, and applies scale jitter in as the additional data augmentation for training. Note that we still use a single image scale and a single crop for test as above. Table 5 summarizes the comparison for Dr.Retina. To emphasize the effectiveness of DR loss, we first train a model with the original regression loss, which is denoted as “Dr.Retina”. With ResNet-101 as the backbone, we can observe that Dr.Retina improves AP from to and it confirms that DR loss can handle the imbalance issue in detection better than focal loss. With gradually approaching regression loss, Dr.Retina gains another improvement and surpasses RetinaNet by . Equipped with ResNeXt-32x8d-101 and training, the performance of Dr.Retina can achieve as a one-stage detector on COCO detection task.\n\n## 5 Conclusion\n\nIn this work, we propose the distributional ranking loss to address the imbalance challenge in one-stage object detection. It first converts the original classification problem to a ranking problem, which balances the classes of foreground and background. Furthermore, we propose to rank the expectation of derived distributions in lieu of original examples to focus on the hard examples, which balances the distribution of background. Besides, we improve the regression loss by developing the strategy to optimize loss better. Experiments on COCO verifies the effectiveness of the proposed losses. Since RPN also has the imbalance issue in two-stage detectors, applying DR loss for that can be our future work.\n\n## References\n\n• S. Boyd and L. Vandenberghe. Convex optimization. Cambridge university press, 2004.\n• Z. Cai and N. 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In CVPR, pages 580–587, 2014.\n• P. Goyal, P. Dollár, R. B. Girshick, P. Noordhuis, L. Wesolowski, A. Kyrola, A. Tulloch, Y. Jia, and K. He. Accurate, large minibatch SGD: training imagenet in 1 hour. CoRR, abs/1706.02677, 2017.\n• K. He, G. Gkioxari, P. Dollár, and R. B. Girshick. Mask R-CNN. In ICCV, pages 2980–2988, 2017.\n• K. He, X. Zhang, S. Ren, and J. Sun. Deep residual learning for image recognition. In CVPR, pages 770–778, 2016.\n• A. Krizhevsky, I. Sutskever, and G. E. Hinton. Imagenet classification with deep convolutional neural networks. In NIPS, pages 1106–1114, 2012.\n• T. Lin, P. Dollár, R. B. Girshick, K. He, B. Hariharan, and S. J. Belongie. Feature pyramid networks for object detection. In CVPR, pages 936–944, 2017.\n• T. Lin, P. Goyal, R. B. Girshick, K. He, and P. Dollár. Focal loss for dense object detection. In ICCV, pages 2999–3007, 2017.\n• T. Lin, M. Maire, S. J. Belongie, J. Hays, P. Perona, D. Ramanan, P. Dollár, and C. L. Zitnick. Microsoft COCO: common objects in context. In ECCV, pages 740–755, 2014.\n• W. Liu, D. Anguelov, D. Erhan, C. Szegedy, S. E. Reed, C. Fu, and A. C. Berg. SSD: single shot multibox detector. In ECCV, pages 21–37, 2016.\n• D. G. Lowe. Distinctive image features from scale-invariant keypoints. International Journal of Computer Vision, 60(2):91–110, 2004.\n• J. Redmon, S. K. Divvala, R. B. Girshick, and A. Farhadi. You only look once: Unified, real-time object detection. In CVPR, pages 779–788, 2016.\n• J. Redmon and A. Farhadi. YOLO9000: better, faster, stronger. In CVPR, pages 6517–6525, 2017.\n• S. Ren, K. He, R. B. Girshick, and J. Sun. Faster R-CNN: towards real-time object detection with region proposal networks. In NIPS, pages 91–99, 2015.\n• P. Sermanet, D. Eigen, X. Zhang, M. Mathieu, R. Fergus, and Y. LeCun. Overfeat: Integrated recognition, localization and detection using convolutional networks. In ICLR, 2014.\n• J. R. R. Uijlings, K. E. A. van de Sande, T. Gevers, and A. W. M. Smeulders. Selective search for object recognition. International Journal of Computer Vision, 104(2):154–171, 2013.\n• S. Xie, R. B. Girshick, P. Dollár, Z. Tu, and K. He. Aggregated residual transformations for deep neural networks. In CVPR, pages 5987–5995, 2017.\n• T. Zhang and F. J. Oles. Text categorization based on regularized linear classification methods. Inf. Retr., 4(1):5–31, 2001.\n\n## Appendix A Gradient of DR Loss\n\nWe define the DR loss as\n\n minθLDR(θ)=N∑iℓsmooth(^Pi,−−^Pi,++γ)\n\nwhere\n\n ℓsmooth(z)=1Llog(1+exp(Lz)) (11)\n\nand\n\n ^Pi,− = n−∑j−1Z−exp(pi,j−λ−)pi,j−=n−∑j−qi,j−pi,j− ^Pi,+ = n+∑j+1Z+exp(−pi,j+λ+)pi,j+=n+∑j+qi,j+pi,j+\n\nIt looks complicated but its gradient is easy to compute. Here we give the detailed gradient form. For , we have\n\n ∂ℓ∂pi,j−=11+exp(−Lz)∂z∂pi,j− =qi,j−1+exp(−Lz)(1+pi,j−λ−−1λ−(∑j−qi,j−pi,j−))\n\nwhere .\n\nFor , we have\n\n ∂ℓ∂pi,j+=11+exp(−Lz)∂z∂pi,j+ =qi,j+1+exp(−Lz)(−1+pi,j+λ+−1λ+(∑j+qi,j+pi,j+))\n\n## Appendix B Proof of Theorem 1\n\n###### Proof.\n\nWe assume that the loss in Eqn. 9 is -smoothness, so we have\n\n E[L(θt+1)]≤E[L(θt)+⟨∇L(θt),θt+1−θt⟩ +L2∥θt+1−θt∥2F] =E[L(θt)+⟨∇L(θt),−ηmm∑s=1∇ℓst⟩ +Lη22∥1mm∑s=1∇ℓst∥2F]\n\nAccording to the definition, we have\n\n ∀s,E[∇ℓst]=∇L(θt)\n\nIf we assume that the variance is bounded as\n\n ∀s,∥∇ℓst−∇Lt∥F≤δ\n\nthen we have\n\n E[L(θt+1)]≤E[L(θt)−η∥∇Lt∥2F +Lη22∥1mm∑s=1∇ℓst−∇Lt+∇Lt∥2F] ≤E[L(θt)−η∥∇Lt∥2F+Lη22(δ2m+∥∇Lt∥2F)\n\nTherefore, we have\n\n (η−Lη22)∥∇L(θt)∥2F≤E[L(θt)]−E[L(θt+1)]+Lη2δ22m\n\nBy assuming and adding from to , we have\n\n ∑t∥∇L(θt)∥2F≤2L(θ0)η+LηTδ2m\n\nWe finish the proof by letting\n\n η=√2mL(θ0)δ√LT\n\n## Appendix C Experiments\n\n#### Effect of DR Loss:\n\nWe illustrate the empirical PDF of foreground and background from DR loss in Fig. 6. Fig. 6 (a) show the original density of foreground and background. To make the results more explicit, we decay the density of background by a factor of and demonstrate the result in Fig. 6 (b). It is obvious that DR loss can separate the foreground and background with a large margin in the imbalance scenario.",
null,
"Figure 6: Illustration of empirical PDF of distributions from DR loss."
]
| [
null,
"https://deepai.org/static/images/logo.png",
null,
"https://deepai.org/publication/None",
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"https://deepai.org/publication/None",
null,
"https://deepai.org/publication/None",
null,
"https://deepai.org/publication/None",
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"https://deepai.org/publication/None",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.88120407,"math_prob":0.953638,"size":32775,"snap":"2021-43-2021-49","text_gpt3_token_len":7996,"char_repetition_ratio":0.1500412,"word_repetition_ratio":0.028195847,"special_character_ratio":0.25891685,"punctuation_ratio":0.16646902,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.986787,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T02:42:31Z\",\"WARC-Record-ID\":\"<urn:uuid:69163a7f-a7af-46d7-9aad-cf2c724dccde>\",\"Content-Length\":\"819450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:45be0119-0826-41f7-b07a-4ae05383d28b>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca8d5489-1e55-4f02-99ac-e030297cb95d>\",\"WARC-IP-Address\":\"52.12.119.3\",\"WARC-Target-URI\":\"https://deepai.org/publication/dr-loss-improving-object-detection-by-distributional-ranking\",\"WARC-Payload-Digest\":\"sha1:BBEZ4OZ7UPGL47V3RJAJFP3FO3ODS5CR\",\"WARC-Block-Digest\":\"sha1:PXELPUKSUE2CKULMPFTNZTIEULPHAQUO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587608.86_warc_CC-MAIN-20211024235512-20211025025512-00531.warc.gz\"}"} |
https://www.expertsmind.com/questions/slope-intercept-form-301138063.aspx | [
"## slope-intercept form, Algebra\n\nAssignment Help:\n(-5,-2) y=5/2x+3\n\n#### Unitary method, What are the pre conditions to applying unitary method to a...\n\nWhat are the pre conditions to applying unitary method to a given problem? e.g. We know that 37 degrees celsius is equal to 98.6 degrees fahrenheit, but 1 degrees celsius is not eq\n\n#### Percentages, in a cloths shop reduces it prices by 20% how much is it on sa...\n\nin a cloths shop reduces it prices by 20% how much is it on sale\n\n#### Inequalities, How do i know where to shade on an inequalitie graph\n\nHow do i know where to shade on an inequalitie graph\n\n#### Alligation Method, You need to prepare a 30 mL solution of a 1:6 syrup solu...\n\nYou need to prepare a 30 mL solution of a 1:6 syrup solution. You have on hand a 50% syrup solution and a 1:200 soda solution. How many mL of each solution will you need?\n\nb^4* b^6\n\nhow do you do it\n\n#### Equations, how to to a equations ?\n\nhow to to a equations ?\n\n#### Which quantity is bigger?, Sam is twice as old as John was two years ago. I...\n\nSam is twice as old as John was two years ago. If the difference between their ages be 2 years, how old is Sam today?",
null,
"",
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""
]
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null,
"https://www.expertsmind.com/questions/CaptchaImage.axd",
null,
"https://www.expertsmind.com/prostyles/images/3.png",
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https://socratic.org/questions/how-do-you-factor-by-grouping-3p-2-2p-5 | [
"# How do you factor by grouping 3p^2 - 2p - 5?\n\nMay 29, 2015\n\n$3 {p}^{2} - 2 p - 5 = 3 {p}^{2} + 3 p - 5 p - 5$\n\n$= \\left(3 {p}^{2} + 3 p\\right) - \\left(5 p + 5\\right)$\n\n$= 3 p \\left(p + 1\\right) - 5 \\left(p + 1\\right)$\n\n$= \\left(3 p - 5\\right) \\left(p + 1\\right)$\n\nThe trick here is how to split the middle term into two so that the ratio of the resulting 1st and 2nd term is the same as the ratio of the 3rd and 4th term. In this particular example it was easy to spot, but in general you may like to use a variant of the AC Method to find a suitable pair to use.\n\nTo see how this works, identify the coefficients of the terms ignoring the signs: $A = 3$, $B = 2$, $C = 5$. Noticing that the sign of the last, constant, term is $-$, we then look for pairs of factors of $A C = 15$ whose difference is $B$. The pair $5 , 3$ satisfies:\n$5 \\times 3 = 15 = A C$ and $5 - 3 = 2 = B$. So the pair of coefficients to split the middle term into is $5$ and $3$, as we have done above."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9473175,"math_prob":0.9999919,"size":486,"snap":"2019-43-2019-47","text_gpt3_token_len":123,"char_repetition_ratio":0.12655602,"word_repetition_ratio":0.0,"special_character_ratio":0.24279836,"punctuation_ratio":0.057692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998735,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T18:11:48Z\",\"WARC-Record-ID\":\"<urn:uuid:475bc073-dc69-4ce9-bddf-5c50eff38eac>\",\"Content-Length\":\"34082\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:de9ef207-a4db-4eff-a123-01eddd35b36c>\",\"WARC-Concurrent-To\":\"<urn:uuid:5311c7e5-1178-41dd-a7eb-30536629d899>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-factor-by-grouping-3p-2-2p-5\",\"WARC-Payload-Digest\":\"sha1:IQOQT4RP5OXBZ3ANCL74ESLC32W5NLN3\",\"WARC-Block-Digest\":\"sha1:SSMYS6QCBFJQG3MUESKQTSBPZPP2XGLG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671411.14_warc_CC-MAIN-20191122171140-20191122200140-00191.warc.gz\"}"} |
https://isabelle.in.tum.de/repos/isabelle/rev/a3035d56171d | [
"author hoelzl Thu, 03 Mar 2011 10:55:41 +0100 changeset 41874 a3035d56171d parent 41873 250468a1bd7a child 41875 e3cd0dce9b1a child 41879 7f9c48c17d2a\nfinally remove upper_bound_finite_set\n```--- a/src/HOL/Multivariate_Analysis/Integration.thy\tThu Mar 03 15:59:44 2011 +1100\n+++ b/src/HOL/Multivariate_Analysis/Integration.thy\tThu Mar 03 10:55:41 2011 +0100\n@@ -4470,24 +4470,6 @@\n\nsubsection {* monotone convergence (bounded interval first). *}\n\n-lemma upper_bound_finite_set:\n- assumes fS: \"finite S\"\n- shows \"\\<exists>(a::'a::linorder). \\<forall>x \\<in> S. f x \\<le> a\"\n-proof(induct rule: finite_induct[OF fS])\n- case 1 thus ?case by simp\n-next\n- case (2 x F)\n- from \"2.hyps\" obtain a where a:\"\\<forall>x \\<in>F. f x \\<le> a\" by blast\n- let ?a = \"max a (f x)\"\n- have m: \"a \\<le> ?a\" \"f x \\<le> ?a\" by simp_all\n- {fix y assume y: \"y \\<in> insert x F\"\n- {assume \"y = x\" hence \"f y \\<le> ?a\" using m by simp}\n- moreover\n- {assume yF: \"y\\<in> F\" from a[rule_format, OF yF] m have \"f y \\<le> ?a\" by (simp add: max_def)}\n- ultimately have \"f y \\<le> ?a\" using y by blast}\n- then show ?case by blast\n-qed\n-\nlemma monotone_convergence_interval: fixes f::\"nat \\<Rightarrow> 'n::ordered_euclidean_space \\<Rightarrow> real\"\nassumes \"\\<forall>k. (f k) integrable_on {a..b}\"\n\"\\<forall>k. \\<forall>x\\<in>{a..b}.(f k x) \\<le> (f (Suc k) x)\"```"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5324946,"math_prob":0.907467,"size":1135,"snap":"2021-31-2021-39","text_gpt3_token_len":418,"char_repetition_ratio":0.11229001,"word_repetition_ratio":0.022346368,"special_character_ratio":0.40969163,"punctuation_ratio":0.18775511,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99540824,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T11:25:25Z\",\"WARC-Record-ID\":\"<urn:uuid:3685b458-f391-4178-810e-3db48b13b24b>\",\"Content-Length\":\"7723\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5dec89d8-98ce-4cbf-acb3-acc94a050e0b>\",\"WARC-Concurrent-To\":\"<urn:uuid:5de39f2e-b28b-47ac-a23c-4fd2666fd31c>\",\"WARC-IP-Address\":\"131.159.46.82\",\"WARC-Target-URI\":\"https://isabelle.in.tum.de/repos/isabelle/rev/a3035d56171d\",\"WARC-Payload-Digest\":\"sha1:7TI5AIKXKLKBQVGK6TZ5YRJLR7A6SOAK\",\"WARC-Block-Digest\":\"sha1:ZR2S7XZJMBJ7GUOV635L6NE3JBY4WA3G\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056856.4_warc_CC-MAIN-20210919095911-20210919125911-00442.warc.gz\"}"} |
https://r-pkg.org/pkg/modmarg | [
"# Calculating Marginal Effects and Levels with Errors\n\nCalculate predicted levels and marginal effects, using the delta method to calculate standard errors. This is an R-based version of the 'margins' command from Stata.\n\nCalculate predicted levels and marginal effects using the delta method to calculate standard errors. This is an R-based version of Stata's 'margins' command.\n\nFeatures:\n\n• Calculate predictive levels and margins for `glm` and `ivreg` objects (more models to be added - PRs welcome) using closed-form derivatives\n\n• Add custom variance-covariance matrices to all calculations to add, e.g., clustered or robust standard errors (for more information on replicating Stata analyses, see here)\n\n• Frequency weights are incorporated into margins and effects\n\n# Usage\n\nTo install this package from CRAN, please run\n\n``````install.packages('modmarg')\n``````\n\nTo install the development version of this package, please run\n\n``````devtools::install_github('anniejw6/modmarg', build_vignettes = TRUE)\n``````\n\nHere is an example of estimating predicted levels and effects using the `iris` dataset:\n\n``````data(iris)\n\nmod <- glm(Sepal.Length ~ Sepal.Width + Species,\ndata = iris, family = 'gaussian')\n\n# Predicted Levels\nmodmarg::marg(mod, var_interest = 'Species', type = 'levels')\n\n# Predicted Effects\nmodmarg::marg(mod, var_interest = 'Species', type = 'effects')\n``````\n\nThere are two vignettes included:\n\n``````vignette('usage', package = 'modmarg')\nvignette('delta-method', package = 'modmarg')\n``````\n\n# Version 0.9.2\n\n• Change variance covariance function for compatibility with R-devel\n\n# Version 0.9.0\n\n• Add generic functions for `glm`\n• Add functionality for `ivreg`\n\n# Version 0.7.0\n\n• Refactors to use generic functions\n\n# Version 0.6.0\n\n• Incorporates frequency weights into predictive margins and levels\n\n# Version 0.5.0\n\n• This is the initial CRAN release of `modmarg`\n\n# Reference manual\n\ninstall.packages(\"modmarg\")\n\n0.9.6 by Annie Wang, 5 months ago\n\nhttps://github.com/anniejw6/modmarg\n\nReport a bug at https://github.com/anniejw6/modmarg/issues\n\nBrowse source code at https://github.com/cran/modmarg\n\nAuthors: Alex Gold [aut] , Nat Olin [aut] , Annie Wang [aut, cre]\n\nDocumentation: PDF Manual"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.70055,"math_prob":0.6450116,"size":2046,"snap":"2021-21-2021-25","text_gpt3_token_len":453,"char_repetition_ratio":0.11018609,"word_repetition_ratio":0.09219858,"special_character_ratio":0.19745846,"punctuation_ratio":0.12238806,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96028906,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-06T04:35:18Z\",\"WARC-Record-ID\":\"<urn:uuid:520785fa-7715-4148-b580-8c2ff5451662>\",\"Content-Length\":\"21812\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28d3a3c3-c820-4912-b6eb-de9d9be6111d>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b4ee485-2233-4fb4-aeaa-8e9b3b7047d4>\",\"WARC-IP-Address\":\"104.21.46.21\",\"WARC-Target-URI\":\"https://r-pkg.org/pkg/modmarg\",\"WARC-Payload-Digest\":\"sha1:V2G5T7OVPY2LWDEHKCUY2RTNR4NQZEHF\",\"WARC-Block-Digest\":\"sha1:FCRUMZUQXQ25VVR5YZSLR2WAQWHT5FJQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988725.79_warc_CC-MAIN-20210506023918-20210506053918-00016.warc.gz\"}"} |
https://www.archimedes-lab.com/tag/visual-proof/ | [
"## Visual Proof (sum of cubes)\n\nThe sum of the sequence of the first n cubes equals [n(n+1)/2]² as shown below:\n1³+2³+3³+…+n³ = (1+2+3+…+n)² = [n(n+1)/2]²",
null,
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"https://www.archimedes-lab.com//wp-content/uploads/2020/02/cubes_squares.jpg",
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https://lists.boost.org/Archives/boost/2001/08/16372.php | [
"",
null,
"# Boost :\n\nFrom: Peter Dimov (pdimov_at_[hidden])\nDate: 2001-08-23 06:10:42\n\nWhen compiling another_tuple_test_bench.cpp, I get internal compiler errors\nin type_traits:\n\ntemplate <typename T>\nstruct is_const\n{\nprivate:\nstatic T t;\npublic:\nBOOST_STATIC_CONSTANT(bool, value = (sizeof(detail::yes_type) ==\nsizeof(detail::is_const_helper(&t)))); // error here\n};\n\nLeaving aside whether the above is legal (I don't see is_const<>::t defined\nanywhere, this modification fixes the problem:\n\ntemplate <typename T>\nstruct is_const\n{\nprivate:\nstatic T * t();\npublic:\nBOOST_STATIC_CONSTANT(bool, value = (sizeof(detail::yes_type) ==\nsizeof(detail::is_const_helper(t()))));\n};\n\nis_volatile suffers from the same problem.\n\nis_array also fails to compile due to the same 'static T t' problem, but I\ndon't see a simple way to fix it. For now I've made it return false. ;-)\n\n```--\nPeter Dimov\nMulti Media Ltd.\n```"
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"https://lists.boost.org/boost/images/boost.png",
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https://forsaljningavaktierixtp.web.app/58967/9496.html | [
"Brun matta\n\nProblem 1.5. Let (Xn) be a Now we turn to quantities of fundamental interest in discussions of the prop- erties of random walkers and random walks. The most widely used, and generally random-walk technique random-walk technique A method of random sampling in which the direction taken and the distance moved between sample points is Aug 27, 2020 In addition, random walks are models of diffusion. ProblemMonte Carlo simulation of a one-dimensional random walk Program RandomWalk1d A one-dimensional random walk. Let us reformulate the previous problem in terms of diffusion of a molecule in a dilute gas.\n\nThis indicator was originally developed by Michael Poulos. As you can see, the result is very similar to the Vortex Indicator Branched Random Walk. GitHub Gist: instantly share code, notes, and snippets. Bolagsöversikt. Telefonnummer. Inga telefonnummer registrerade.\n\nRandom Walk Index indicator script. This indicator was originally developed by Michael Poulos.\n\n## Random Walk Imaging AB - 556683-3280 - Lund - Se - Proff\n\nThe random walk hypothesis states that stock market prices change in a random manner, and therefore, you can't predict what price movements will occur in advance. A one-dimensional random walk is a Markov chain whose state space is a finite or infinite subset a, a + 1, …, b of the integers, in which the particle, if it is in state i, can in a single transition either stay in i or move to one of the neighboring states i − 1, i + 1.\n\n### Matematisk statistik Stockholms universitet Random\n\n Another test that Weber ran that contradicts the random walk hypothesis, was finding stocks that have had an upward revision for earnings outperform other stocks in the following six months. 20 Random Walks Random Walks are used to model situations in which an object moves in a sequence of steps in randomly chosen directions. Many phenomena can be modeled as a random walk and we will see several examples in this chapter.\n\nThe random walk index (RWI) is a technical indicator that compares a security's price movements to random movements in an effort to determine if it's in a statistically significant trend.\nValuta danska kronan",
null,
"Recurrence of simple random walk in two dimensions. First, we recall two well- known proofs of recurrence of two-dimensional simple ran- dom walk: the Random walks and electric networks.\n\nWe now know that simple random walk on the integers Random Walk. A random process consisting of a sequence of discrete steps of fixed length. The random thermal perturbations in a liquid are responsible for a random walk phenomenon known as Brownian motion, and the collisions of molecules in a gas are a random walk responsible for diffusion.\nSms tecken hjärta",
null,
"trafikprov järfälla\nklappträ bastu\nspecifikke varmekapacitet luft\nvägledningscentrum i malmö\nvad kostar en hemsida per år\n\n### Application of continuous-time random walk to statistical\n\n8 years ago It`s not violent at all!Rather funny 8 years ago U mean t The simple random walk process is a minor modification of the Bernoulli trials process. Nonetheless, the process has a number of very interesting properties, Random walk definition is - a process (such as Brownian motion or genetic drift) consisting of a sequence of steps (such as movements or changes in gene Feb 9, 2018 Introduction A random walk is a mathematical object, known as a stochastic or random process, that describes a path that consists of a Random walk definition, the path taken by a point or quantity that moves in steps, where the direction of each step is determined randomly. See more.\n\nStaples kalender\nmusik stockholm november 2021\n\n### Random walk på engelska EN,SV lexikon Tyda\n\nSee more. We study the evolution of a random walker on a conservative dynamic random environment composed of independent particles performing simple symmetric Random walk. Theory that stock price changes from day to day are accidental or haphazard; changes are independent of each other and have the same Think of the random walk as a game, where the player starts at the origin (i.e. all coordinates equal 0 0 0) and at each move, he is required to make one step on The nervous systems of foraging and predatory animals may prompt them to move along a special kind of random path called a Lévy walk to find food efficiently Nov 22, 2017 Random walks are ubiquitous in the sciences, and they are The term “random walk” was coined by Karl Pearson , and the study of RWs A “random walk” is a statistical phenomenon where a variable follows no discernible trend and moves seemingly at random."
]
| [
null,
"https://picsum.photos/800/639",
null,
"https://picsum.photos/800/615",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.87663555,"math_prob":0.94586265,"size":4606,"snap":"2023-14-2023-23","text_gpt3_token_len":1003,"char_repetition_ratio":0.14124294,"word_repetition_ratio":0.0131233595,"special_character_ratio":0.20017369,"punctuation_ratio":0.0823389,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9573324,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T19:05:06Z\",\"WARC-Record-ID\":\"<urn:uuid:12786540-7883-41de-895d-0a2226da4ce9>\",\"Content-Length\":\"9492\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ff6b883-233b-4534-a7fe-db8ed3f35d1f>\",\"WARC-Concurrent-To\":\"<urn:uuid:d229e085-9964-49b7-88fd-8c48a5ebd96c>\",\"WARC-IP-Address\":\"199.36.158.100\",\"WARC-Target-URI\":\"https://forsaljningavaktierixtp.web.app/58967/9496.html\",\"WARC-Payload-Digest\":\"sha1:B25OJXZFCF4IUX7PAMFYLQELADT5NVYG\",\"WARC-Block-Digest\":\"sha1:UATSRMXH3E7PIN3H3LK2GTK6URGH4ZZF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656788.77_warc_CC-MAIN-20230609164851-20230609194851-00568.warc.gz\"}"} |
https://whatisconvert.com/478-long-tons-in-short-tons | [
"# What is 478 Long Tons in Short Tons?\n\n## Convert 478 Long Tons to Short Tons\n\nTo calculate 478 Long Tons to the corresponding value in Short Tons, multiply the quantity in Long Tons by 1.12 (conversion factor). In this case we should multiply 478 Long Tons by 1.12 to get the equivalent result in Short Tons:\n\n478 Long Tons x 1.12 = 535.36 Short Tons\n\n478 Long Tons is equivalent to 535.36 Short Tons.\n\n## How to convert from Long Tons to Short Tons\n\nThe conversion factor from Long Tons to Short Tons is 1.12. To find out how many Long Tons in Short Tons, multiply by the conversion factor or use the Mass converter above. Four hundred seventy-eight Long Tons is equivalent to five hundred thirty-five point three six Short Tons.",
null,
"## Definition of Long Ton\n\nA long ton is defined as exactly 2,240 pounds. The long ton arises from the traditional British measurement system: A long ton is 20 cwt, each of which is 8 stone (1 stone = 14 pounds). Thus a long ton is 20 × 8 × 14 lb = 2,240 lb. Long ton, also known as the imperial ton or displacement ton is the name for the unit called the \"ton\" in the avoirdupois or Imperial system of measurements standardised in the thirteenth century that is used in the United Kingdom\n\n## Definition of Short ton\n\nThe short ton is a unit of weight equal to 2,000 pounds (907.18474 kg), that is most commonly used in the United States where it is known simply as the ton.\n\n### Using the Long Tons to Short Tons converter you can get answers to questions like the following:\n\n• How many Short Tons are in 478 Long Tons?\n• 478 Long Tons is equal to how many Short Tons?\n• How to convert 478 Long Tons to Short Tons?\n• How many is 478 Long Tons in Short Tons?\n• What is 478 Long Tons in Short Tons?\n• How much is 478 Long Tons in Short Tons?\n• How many ton are in 478 uk ton?\n• 478 uk ton is equal to how many ton?\n• How to convert 478 uk ton to ton?\n• How many is 478 uk ton in ton?\n• What is 478 uk ton in ton?\n• How much is 478 uk ton in ton?"
]
| [
null,
"https://whatisconvert.com/images/478-long-tons-in-short-tons",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.88380086,"math_prob":0.9558976,"size":1928,"snap":"2019-26-2019-30","text_gpt3_token_len":518,"char_repetition_ratio":0.21829522,"word_repetition_ratio":0.112860896,"special_character_ratio":0.2873444,"punctuation_ratio":0.09307876,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99086666,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-24T03:00:53Z\",\"WARC-Record-ID\":\"<urn:uuid:a4998189-4258-4aad-b646-78689cf899a2>\",\"Content-Length\":\"27456\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b150c671-fbf2-417b-901b-ee861828f45e>\",\"WARC-Concurrent-To\":\"<urn:uuid:d01dbc82-81e1-4f6c-9605-cd50456dc4a8>\",\"WARC-IP-Address\":\"104.31.71.53\",\"WARC-Target-URI\":\"https://whatisconvert.com/478-long-tons-in-short-tons\",\"WARC-Payload-Digest\":\"sha1:KWY2K2RUSSOAE357V34YAGIRY3FULP4P\",\"WARC-Block-Digest\":\"sha1:FFD56XK57J5L7HGKZPNXLBILGCGTDQGD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195530250.98_warc_CC-MAIN-20190724020454-20190724042454-00110.warc.gz\"}"} |
https://math.stackexchange.com/questions/2204429/greatest-common-divisor-algorithm | [
"# Greatest common divisor algorithm\n\nI am having some issues understanding the gcd. I am using \"Art of Proof,\" and in this book, the gcd(m,n) is defined as being:\n\n\"the smallest element of the set S = {k $\\in$ N : k = mx+ny for some x, y $\\in$ Z}\n\nI guess my issue is that I don't really understand what k=mx+ny is in the context of gcd.\n\nTo provide some context for my confusion, we are asked to use the gcd definition when we prove various things about prime numbers and specifically Euclid's lemma. I have been struggling with this and I think part of it is because I don't really understand the algorithmic definition of gcd.\n\n• Well, the point is that this definition coincides with the traditional one...though that isn't at all obvious. Surely your text proves that?\n– lulu\nMar 26 '17 at 21:53\n• Hmm...I'm not really sure what you are saying? Mar 26 '17 at 21:55\n• See this answer for a more conceptual view. Mar 28 '17 at 1:48\n\nIf we take as definition of the g.c.d. of $m$ and $n$: ‘the greatest natural number which is a divisor of both $m$ and $n$, the smallest element of $S$ satisfies this definition.\n\nIndeed, let's denote $d$ this smallest element non-zero of $S$. Thus, we have $d=xm+yn$ for some $x,y\\in\\mathbf Z$. Perform the Euclidean division of $m$ by $d$: $$m=qd+r,\\quad 0\\le r<d$$ We deduce that $$r=m-qd=(1-qx)m-(qy)n\\in S.$$ As $d$ is the smallest positive element of S, this relation implies $r=0$. In other words, $d$ divides $m$. For similar reasons, $d$ divides $n$.\n\nOn ther other hand, $d$ is the greatest common divisor of $m$ and $n$, since if $d'\\mid m$ and $d'\\mid n$, $d'$ divides any linear combination of $m$ and $n$ – in particular, it divides $d$.\n\nA theorem known as Bezut's Lemma says that that definition is equivalent to the following definition:\n\nDefinition (gcd): A GCD of two integers, $a,b$ is a number $c$ such that $c$ is a common divisor of $a$ and $b$, and for any other common divisor, $d$, we have that $d|c$. We denote this as $gcd(a,b)=c$\n\nThe intuition for this theorem comes from lattices. Think about the number line, and put a penny on $0$. If you are only allowed to move up and down the number line in steps of size $a$, then it should be clear that you can only reach multiples of $a$.\n\nNow imagine that we are allowed to use two different step sizes, $a$ and $b$. By moving up $a$ and then down $b$ we wind up closer to $0$ than either $a$ or $b$ alone. The set of all points we can reach with these two step sizes is guaranteed to have a smallest positive value, since every non-empty set of positive integers does. Let's call this number $d$. We got to $d$ using steps of size $a$ and $b$, so $ax+by=d$ holds for some numbers $x$ and $y$. Thus thus number $d$ is the smallest element of the set in your definition.\n\nThat's the intuition for what the set and what it's smallest positive element means. Just like in the case of steps of size $a$, the set in question turns out to be the set of all multiples of $d$. Proving that this is equal to the other sense of GCD isn't particularly hard, but does require a little work and the Euclidean Algorithm. You can easily find such proofs by searching this website or google for Bezut's Lemma."
]
| [
null
]
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http://es.wikidoc.org/index.php/Moment-generating_function | [
"# Moment-generating function\n\nIn probability theory and statistics, the moment-generating function of a random variable X is\n\n$M_{X}(t)=E\\left(e^{tX}\\right),\\quad t\\in \\mathbb {R} ,$",
null,
"wherever this expectation exists. The moment-generating function generates the moments of the probability distribution.\n\nFor vector-valued random variables X with real components, the moment-generating function is given by\n\n$M_{X}(\\mathbf {t} )=E\\left(e^{\\langle \\mathbf {t} ,\\mathbf {X} \\rangle }\\right)$",
null,
"where t is a vector and $\\langle \\mathbf {t} ,\\mathbf {X} \\rangle$",
null,
"is the dot product.\n\nProvided the moment-generating function exists in an interval around t = 0, the nth moment is given by\n\n$E\\left(X^{n}\\right)=M_{X}^{(n)}(0)=\\left.{\\frac {\\mathrm {d} ^{n}M_{X}(t)}{\\mathrm {d} t^{n}}}\\right|_{t=0}.$",
null,
"If X has a continuous probability density function f(x) then the moment generating function is given by\n\n$M_{X}(t)=\\int _{-\\infty }^{\\infty }e^{tx}f(x)\\,\\mathrm {d} x$",
null,
"$=\\int _{-\\infty }^{\\infty }\\left(1+tx+{\\frac {t^{2}x^{2}}{2!}}+\\cdots \\right)f(x)\\,\\mathrm {d} x$",
null,
"$=1+tm_{1}+{\\frac {t^{2}m_{2}}{2!}}+\\cdots ,$",
null,
"where $m_{i}$",
null,
"is the ith moment. $M_{X}(-t)$",
null,
"is just the two-sided Laplace transform of f(x).\n\nRegardless of whether the probability distribution is continuous or not, the moment-generating function is given by the Riemann-Stieltjes integral\n\n$M_{X}(t)=\\int _{-\\infty }^{\\infty }e^{tx}\\,dF(x)$",
null,
"where F is the cumulative distribution function.\n\nIf X1, X2, ..., Xn is a sequence of independent (and not necessarily identically distributed) random variables, and\n\n$S_{n}=\\sum _{i=1}^{n}a_{i}X_{i},$",
null,
"where the ai are constants, then the probability density function for Sn is the convolution of the probability density functions of each of the Xi and the moment-generating function for Sn is given by\n\n$M_{S_{n}}(t)=M_{X_{1}}(a_{1}t)M_{X_{2}}(a_{2}t)\\cdots M_{X_{n}}(a_{n}t).$",
null,
"Related to the moment-generating function are a number of other transforms that are common in probability theory, including the characteristic function and the probability-generating function.\n\nThe cumulant-generating function is the logarithm of the moment-generating function.",
null,
""
]
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null,
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null,
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null,
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null,
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null,
"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/95ec8e804f69706d3f5ad235f4f983220c8df7c2",
null,
"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e5374386463a31864730a75f330e8148bcc41485",
null,
"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a1c01f220becd870d525aad5dd549b0ed9a33796",
null,
"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/55bf1ba75357e53861251060d7841d9a94909266",
null,
"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/0fc8febaa69f32405de30ea3e7991f7d32f79d52",
null,
"http://es.wikidoc.org/images/6/67/Linked-in.jpg",
null
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https://infiniteadz.com/qa/quick-answer-where-is-maximum-power-transfer-used.html | [
"",
null,
"# Quick Answer: Where Is Maximum Power Transfer Used?\n\n## Under what situation maximum power transfer theorem is not applicable?\n\nNow, if the source impedance is zero or near zero then the load voltage is nearly the same as the source voltage and the maximum power transfer theorem need not be used.\n\nNotice that modern power amplifiers have near-zero output impedances and the power transfer theorem is not used in their case..\n\n## How do you calculate power consumption?\n\nHow to calculate my energy consumptionDevice Wattage (watts) x Hours Used Per Day = Watt-hours (Wh) per Day.Device Usage (Wh) / 1000 (Wh/kWh) = Device Usage in kWh.Daily Usage (kWh) x 30 (Days) = Approximate Monthly Usage (kWh/Month)\n\n## How do you solve maximum power transfer theorem problems?\n\nSteps To Solve Maximum Power Transfer TheoremStep 1: Remove the load resistance of the circuit.Step 2: Find the Thevenin’s resistance (RTH) of the source network looking through the open-circuited load terminals.More items…\n\n## What is the current in the circuit?\n\nCurrent is the rate at which charge flows. Charge will not flow in a circuit unless there is an energy source capable of creating an electric potential difference and unless there is a closed conducting loop through which the charge can move. 2. Current has a direction.\n\n## Where and why maximum power transfer theorem is applied?\n\nThe Maximum Power Transfer Theorem is another useful circuit analysis method to ensure that the maximum amount of power will be dissipated in the load resistance when the value of the load resistance is exactly equal to the resistance of the power source.\n\n## How do you find maximum power transfer?\n\nCondition for Maximum Power Transfer Therefore, the condition for maximum power dissipation across the load is RL=RTh. That means, if the value of load resistance is equal to the value of source resistance i.e., Thevenin’s resistance, then the power dissipated across the load will be of maximum value.\n\n## What is maximum output power?\n\nThe maximum output power = the maximum output current × the rated output voltage so there is no problem if it is confirmed that one of them is not exceeded. (2) When setting the output voltage higher than the rated output voltage.\n\n## What are the limitations of maximum power transfer theorem?\n\nOne of the limitation of maximum power theorem is the efficiency is only 50% and therefore it can’t be used in power systems where efficiency is the main concern. It is applicable to all circuits and whenever we build circuit according to this principle efficiency will drop by 50%.\n\n## What is power transfer efficiency?\n\nThe efficiency of power transfer (ratio of output power to input power) from the source to the load increases as the load resistance is increased.\n\n## How is rated power calculated?\n\nCalculate power rating in KVA when you know voltage and output resistance. Use the formula: P(KVA) = (V^2/R)/1000 where R is resistance in ohms. For example, if V is 120 volts and R is 50 ohms, P(KVA) = V^2/R/1000 = (14400/50)/1000 = 288/1000 = 0.288 KVA.\n\n## Can maximum power transfer theorem be applied to AC sources?\n\nMaximum power transfer theorem can be applied to both DC and AC circuits, but the only difference is that the resistance is replaced with impedance in AC circuit. … Therefore in order to have maximum power transfer the load must possess same value of reactance but it should be of opposite type.\n\n## What power factor means?\n\nPower factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA). … PF expresses the ratio of true power used in a circuit to the apparent power delivered to the circuit.\n\n## Why maximum power transfer is not always possible?\n\nThe Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. … A load impedance that is too low will not only result in low power output but possibly overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance.\n\n## What are the advantages and disadvantages of using superposition theorem?\n\nAdvantages – It is applicable to the elements of the network as well as to the sources. It is very useful for circuit analysis. It is utilized to convert any circuit into its Thevenin equivalent or Norton equivalent. Disadvantages – Superposition is applicable to current and voltage but not to power.\n\n## Where is maximum power transfer theorem used?\n\nMPTT is applied in Radio communications, where the power amplifier transmits the maximum amount of signal to the antenna if and only if load impedance in the circuit is equal to the source impedance. It is also applied in audio systems, where the voice is to be transmitted to the speaker.\n\n## Is it always possible to operate at maximum power transfer condition?\n\nIt is always possible to operate a machine under maximum power condition. You just have to figure out the amount of load it shares when operating in parallel with other machines. It is always possible to operate a machine under maximum power condition.\n\n## How do you calculate current?\n\nOhms Law and PowerTo find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)\n\n## How much power is dissipated in the resistor?\n\nTo find out, we need to be able to calculate the amount of power that the resistor will dissipate. If a current I flows through through a given element in your circuit, losing voltage V in the process, then the power dissipated by that circuit element is the product of that current and voltage: P = I × V."
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"https://mc.yandex.ru/watch/68557528",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90744483,"math_prob":0.98103946,"size":6540,"snap":"2021-04-2021-17","text_gpt3_token_len":1429,"char_repetition_ratio":0.19124848,"word_repetition_ratio":0.13024476,"special_character_ratio":0.22079511,"punctuation_ratio":0.083059214,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998411,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-13T23:27:49Z\",\"WARC-Record-ID\":\"<urn:uuid:5c33f4ec-37ba-4bf8-80b7-7b0176436252>\",\"Content-Length\":\"38565\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:408c0c90-4196-4a09-8b35-c91acc7c56fe>\",\"WARC-Concurrent-To\":\"<urn:uuid:1fa4bec3-8da7-4065-8693-9a12eee8970f>\",\"WARC-IP-Address\":\"45.130.40.27\",\"WARC-Target-URI\":\"https://infiniteadz.com/qa/quick-answer-where-is-maximum-power-transfer-used.html\",\"WARC-Payload-Digest\":\"sha1:35BJQF6TL264STYSGQOCCGZKRYMFHPH2\",\"WARC-Block-Digest\":\"sha1:O2JCE2LSGQYNZTJHWTIE5UQGJ4FNGGVH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038075074.29_warc_CC-MAIN-20210413213655-20210414003655-00147.warc.gz\"}"} |
https://it.mathworks.com/help/econ/monte-carlo-forecasting.html | [
"## Monte Carlo Forecasting of Conditional Mean Models\n\n### Monte Carlo Forecasts\n\nYou can use Monte Carlo simulation to forecast a process over a future time horizon. This is an alternative to minimum mean square error (MMSE) forecasting, which provides an analytical forecast solution. You can calculate MMSE forecasts using `forecast`.\n\nTo forecast a process using Monte Carlo simulations:\n\n• Fit a model to your observed series using `estimate`.\n\n• Use the observed series and any inferred residuals and conditional variances (calculated using `infer`) for presample data.\n\n• Generate many sample paths over the desired forecast horizon using `simulate`.\n\n### Advantage of Monte Carlo Forecasting\n\nAn advantage of Monte Carlo forecasting is that you obtain a complete distribution for future events, not just a point estimate and standard error. The simulation mean approximates the MMSE forecast. Use the 2.5th and 97.5th percentiles of the simulation realizations as endpoints for approximate 95% forecast intervals."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8118268,"math_prob":0.9517031,"size":921,"snap":"2020-45-2020-50","text_gpt3_token_len":187,"char_repetition_ratio":0.16030534,"word_repetition_ratio":0.0,"special_character_ratio":0.18566775,"punctuation_ratio":0.09210526,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9799981,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-20T04:23:20Z\",\"WARC-Record-ID\":\"<urn:uuid:a5e31be4-e3e2-4e93-87bc-53e2200d495c>\",\"Content-Length\":\"64788\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81ea445e-fd65-4b7b-81ae-75e68adaf92a>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b32ec6b-068b-4be4-ae7b-6bbb55379df3>\",\"WARC-IP-Address\":\"184.24.72.83\",\"WARC-Target-URI\":\"https://it.mathworks.com/help/econ/monte-carlo-forecasting.html\",\"WARC-Payload-Digest\":\"sha1:TXQRNAEC5GXWA3CZNSNTQDUJEYX7NUXF\",\"WARC-Block-Digest\":\"sha1:GX6M3QPFBCWYX2SFUCL2WJ2ATAZFXOWY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107869785.9_warc_CC-MAIN-20201020021700-20201020051700-00529.warc.gz\"}"} |
https://edoc.unibas.ch/view/contributors_basel_id/110360.html | [
"Browse by Basel Contributors ID",
null,
"Up a level\n Export as ASCII CitationBibTeXDublin CoreEP3 XMLEndNoteHTML CitationJSONMETSObject IDsOpenURL ContextObjectRDF+N-TriplesRDF+N3RDF+XMLReferReference Manager\nGroup by: Date | Item Type | Refereed | No Grouping\nJump to: 2009 | 2007 | 2006 | 2005 | 2004 | 2001\n\n2009\n\nBonnet, Philippe and Vénéreau, Stéphane. (2009) Relations between the leading terms of a polynomial automorphism. Journal of algebra, Vol. 322, H. 2. pp. 579-599.\n\n2007\n\nvan den Essen, Arno and Maubach, Stefan and Vénéreau, Stéphane. (2007) The special automorphism group of R[t]/(t(m))[x₁,…,x(n)] and coordinates of a subring of R[t][x₁,…,x(n)]. Journal of pure and applied algebra, Vol. 210, H. 1. pp. 141-146.\n\nVénéreau, Stéphane. (2007) A parachute for the degree of a polynomial in algebraically independent ones.\n\n2006\n\nVénéreau, Stéphane. (2006) New bad lines in R[x, y] and optimization of the Epimorphism Theorem. Journal of Algebra, Vol. 302, H. 2. pp. 729-749.\n\n2005\n\nVenereau, S.. (2005) Hyperplanes of the form f(1) (x, y)z(1)+center dot center dot center dot+f(k)(x, y)z(k)+g(x, y) are variables. Canadian mathematical bulletin = Bulletin canadien de mathématiques, Vol. 48, H. 4. pp. 622-635.\n\n2004\n\nKaliman, Shulim and Vénéreau, Stéphane and Zaidenberg, Mihail. (2004) Simple birational extensions of the polynomial algebra ℂ³. Transactions of the American Mathematical Society, Vol. 356, H. 2. pp. 509-555.\n\n2001\n\nEdo, Eric and Vénéreau, Stéphane. (2001) Length 2 variables of A[x,y] and transfer. Annales Polonici Mathematici, Vol. 76, H. 1-2. pp. 67-76.\n\nKaliman, Shulim and Vénéreau, Stéphane and Zaidenberg, Mihail. (2001) Extensions birationnelles simples de l'anneau de polynômes ℂ³ = Simple birational extensions of the polynomial ring ℂ³. Comptes rendus de l'Académie des sciences. Série 1, Mathématique. Series 1, Mathematics, Vol. 333, H. 4. pp. 319-322."
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https://plainmath.net/2473/determine-statement-displaystyle-quadtext-displaystyle-hyperbolais | [
"# Question",
null,
"# Determine whether the statement If displaystyle{D}ne{0}{quadtext{or}quad}{E}ne{0}, then the graph of displaystyle{y}^{2}-{x}^{2}+{D}{x}+{E}{y}={0} is a hyperbolais true or false. If it is false, explain why or give an example that shows it is false.\n\nConic sections\nANSWERED",
null,
"Determine whether the statement If $$\\displaystyle{D}\\ne{0}{\\quad\\text{or}\\quad}{E}\\ne{0}$$,\nthen the graph of $$\\displaystyle{y}^{2}-{x}^{2}+{D}{x}+{E}{y}={0}$$ is a hyperbolais true or false. If it is false, explain why or give an example that shows it is false.",
null,
"2021-02-16\nStep 1\nWe have given a statement:\nIf $$\\displaystyle{D}\\ne{0}{\\quad\\text{or}\\quad}{E}\\ne{0}$$,\nthen graph of $$\\displaystyle{y}^{2}-{x}^{2}+{D}{x}+{E}{y}={0}$$ is a hyperbolais.\nStep 2\nWe know the general form of conic section:\n$$\\displaystyle{A}{x}^{2}+{B}{x}{y}+{C}{y}^{2}+{D}{x}+{E}{y}+{F}={0}$$\nTo find the type of conic section we solve for $$\\displaystyle{B}^{2}-{4}{A}{C}:$$\n(i) If $$\\displaystyle{B}^{2}-{4}{A}{C}<{0}$$</span> then the conic section is ellipse.\n(ii) If $$\\displaystyle{B}^{2}-{4}{A}{C}<{0}{\\quad\\text{and}\\quad}{A}={C},{B}={0}$$</span> then we have a perfect circle.\n(iii) If $$\\displaystyle{B}^{2}-{4}{A}{C}={0}$$, then we have a parabola.\n$$\\displaystyle{B}^{2}-{4}{A}{C}>{0}$$, then we have a hyperbola\nHyperbola defined as:\n$$\\displaystyle\\frac{{{\\left({x}-{h}\\right)}^{2}}}{{a}^{2}}+\\frac{{{\\left({y}-{k}\\right)}^{2}}}{{b}^{2}}={1}$$\nWhere (h, k) are center.\nWhen $$\\displaystyle{E},{D}={0},\\text{then}{\\left({h},{k}\\right)}={\\left({0},{0}\\right)}$$\nThen the center will be at (0,0)\nStep 3\nHence, the given statement is incorrect since the given condition is not mandatory for a conic section to be a hyperbola."
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"https://plainmath.net/",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5545802,"math_prob":0.99997413,"size":2276,"snap":"2021-31-2021-39","text_gpt3_token_len":914,"char_repetition_ratio":0.17957747,"word_repetition_ratio":0.9722222,"special_character_ratio":0.42794377,"punctuation_ratio":0.089361705,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99985576,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-25T00:20:57Z\",\"WARC-Record-ID\":\"<urn:uuid:4ee7e14e-21a6-45ae-9176-f14c4664fe1e>\",\"Content-Length\":\"42368\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9601df7-65e2-4ab5-b5e2-72259d36f418>\",\"WARC-Concurrent-To\":\"<urn:uuid:1277fc0a-8771-47ec-9bdb-a2b536f1c968>\",\"WARC-IP-Address\":\"172.67.217.47\",\"WARC-Target-URI\":\"https://plainmath.net/2473/determine-statement-displaystyle-quadtext-displaystyle-hyperbolais\",\"WARC-Payload-Digest\":\"sha1:NIHFX3ZQY6V7BW7W74B5P5ZPFQBKFWAL\",\"WARC-Block-Digest\":\"sha1:PKYK3KK7FPF6HSYMEOIPCCTDG6JDUGTQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046151531.67_warc_CC-MAIN-20210724223025-20210725013025-00080.warc.gz\"}"} |
https://de.mathworks.com/matlabcentral/answers/1923965-sinusoidal-curve-from-discrete-dataset?s_tid=prof_contriblnk | [
"# Sinusoidal curve from discrete dataset\n\n6 Ansichten (letzte 30 Tage)\nAshfaq Ahmed am 6 Mär. 2023\nKommentiert: William Rose am 17 Mai 2023\nHi!\nI have a few discrete data-set and when I plot them it looks like this -",
null,
"Can anyone please tell me how can I fit these discrete points into a sinusoidal curve like this?",
null,
"The data is a small 23x4000 matrix and I added the .mat file to the question. Any feedback will be really helpful. Thank you!\n##### 0 Kommentare-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden\n\nMelden Sie sich an, um zu kommentieren.\n\n### Akzeptierte Antwort\n\nWilliam Rose am 6 Mär. 2023\nBearbeitet: William Rose am 7 Mär. 2023\n[edit: I changed the comments in the code for t1 and t. I added a line to display the fitted values.]\n[rows,cols]=size(Sinusoidal); % read data form file\nt1=(1:rows)'; % t1=vector\nt=repmat(t1,1,cols); % t=array\nSinusoidal=reshape(Sinusoidal,rows*cols,1); % convert array to vector\nT=Sinusoidal(~isnan(Sinusoidal)); % remove NaNs\nt=t(~isnan(Sinusoidal)); % remove corresponding times\n% next: define the model equation\nmyfittype=fittype('a*sin(w*t+p)','dependent',{'T'},'independent',{'t'},'coefficients',{'a','w','p'});\nx0=[30,.1,0]; % initial guess for [a,b,c]\nmyfit=fit(t,T,myfittype,'StartPoint',x0); % fit the data\nplot(myfit,'-b',t,T,'r.'); % plot results\nxlabel('Time'); ylabel('Temperature')",
null,
"% Display best-fit parameters on console.\nfprintf('y=a*sin(wt+p): a=%.4f, w=%.4f, p=%.4f\\n',myfit.a,myfit.w,myfit.p)\ny=a*sin(wt+p): a=22.4003, w=0.1006, p=-0.1809\nTry it.\n##### 6 Kommentare5 ältere Kommentare anzeigen5 ältere Kommentare ausblenden\nWilliam Rose am 17 Mai 2023\n@Ashfaq Ahmed, This is rather complicated to do properly, because the data in T.mat has values in datetime format, for the sample times. It also has the numeric values for time. Matlab is telling users to stop using numeric times, and start using datetime values. Therefore I am trying to adapt. The data in T.mat spans approximately 10 years, with uneven time intervals between samples (long data gaps every winter). Please contact me through Matab Central by clicking on my name above, then click on the envelope icon in the pop-up box. Please provide your email in the message, so we can discuss offline.\n\nMelden Sie sich an, um zu kommentieren.\n\n### Kategorien\n\nMehr zu Lengths and Angles finden Sie in Help Center und File Exchange\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5489179,"math_prob":0.7718518,"size":2291,"snap":"2023-40-2023-50","text_gpt3_token_len":630,"char_repetition_ratio":0.10013118,"word_repetition_ratio":0.012048192,"special_character_ratio":0.26625928,"punctuation_ratio":0.17718941,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9875114,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T21:37:07Z\",\"WARC-Record-ID\":\"<urn:uuid:9bb5465b-977d-49f6-9722-0ade01f1cb04>\",\"Content-Length\":\"165041\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b34d05ad-49d0-4f7d-a268-d43510f68069>\",\"WARC-Concurrent-To\":\"<urn:uuid:d8d2fd6c-0667-4a79-a9e5-a63073fef098>\",\"WARC-IP-Address\":\"23.48.20.88\",\"WARC-Target-URI\":\"https://de.mathworks.com/matlabcentral/answers/1923965-sinusoidal-curve-from-discrete-dataset?s_tid=prof_contriblnk\",\"WARC-Payload-Digest\":\"sha1:Y4LK52N476S6M26PYWO25XS64LTCAPP4\",\"WARC-Block-Digest\":\"sha1:6GBUJMFCLNDXS2RTUXZAY53CJ3VEUH3N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510454.60_warc_CC-MAIN-20230928194838-20230928224838-00802.warc.gz\"}"} |
https://eps.leeds.ac.uk/maths-masters-module-information/doc/statistics-msc | [
"# Statistics MSc\n\nYou will study 180 or 185 credits in total during your Statistics MSc. A standard module is typically worth 15 credits and the research project is worth 60 credits. These are the modules studied in 2020. If you are starting in September 2021, these will give you a flavour of the modules you are likely to study. All modules are subject to change.\n\n### Compulsory modules\n\nIndependent Learning and Skills Project - 15 credits\nStudents will be able to develop a systematic search strategy to find material on a given topic, using Mathematical word processing and evaluation of material, referencing conventions.\n\nStatistical Computing - 15 credits\nThe use of computers in mathematics and statistics has opened up a wide range of tech- niques for studying otherwise intractable problems and for analysing very large data sets.\"Statistical computing\" is the branch of mathematics which concerns these techniques for situations which either directly involve randomness, or where randomness is used as part of a mathematical model.\n\nDissertation in Statistics - 60 credits\n\nEach student will discuss with an individual supervisor a suitable research project. The title and objectives of the project will be approved by the Programme Manager.\n\n### Optional modules include\n\nLinear Regression and Robustness - 15 credits\n\nThis module will examine ways of predicting one particular variable from the remaining measurements using the linear regression model. The general theory of linear regression models will be covered, including variable selection, tests and diagnostics. Robust methods will be introduced to deal with the presence of outliers.\n\nTime Series and Spectral Analysis - 15 credits\n\nThe module will concentrate on techniques for model identification, parameter estimation, diagnostic checking and forecasting within the autoregressive moving average family of models and their extensions.\n\nMultivariate and Cluster Analysis - 15 credits\n\nMultivariate datasets are common to all research areas: it is typical that experimental units are measured for (or questioned about) more than one variable at a time. This module covers the extension of univariate statistical techniques for continuous data to a multivariate setting and introduces methods designed specifically for multivariate data analysis (cluster analysis, principal component analysis, multidimensional scaling and factor analysis).\n\nStochastic Calculus for Finance - 15 credits\n\nThis module provides a mathematical introduction to stochastic calculus in continuous time with applications to finance. Students will learn material in areas of mathematical analysis and probability theory. This knowledge will be used to derive expressions for prices of derivatives in financial markets under uncertainty.\n\nGeneralised Linear and Additive Models - 15 credits\n\nLinear regression is a tremendously useful statistical technique but is very limited. Generalised linear models extend linear regression in many ways - allowing us to analyse more complex data sets. In this module we will see how to combine continuous and categorical predictors, analyse binomial response data and model count data.\n\nRisk Management - 15 credits\n\nThis module covers the different sorts of risk to which financial investments are exposed, basic and sophisticated derivates commonly used for hedging, expected utility theory, models of incomplete markets, Value-at-Risk and other risk measures, credit risks and credit derivatives, methods to determine the effectiveness of a hedge, stress-testing of risky investment portfolios.\n\nThe full list of optional modules can be read in the course catalogue."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8717651,"math_prob":0.73681295,"size":3628,"snap":"2020-45-2020-50","text_gpt3_token_len":651,"char_repetition_ratio":0.1227925,"word_repetition_ratio":0.005628518,"special_character_ratio":0.17502756,"punctuation_ratio":0.07491289,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96739304,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T06:40:36Z\",\"WARC-Record-ID\":\"<urn:uuid:74ff5a1c-ea9d-40ec-8ce6-39c3760ac6a7>\",\"Content-Length\":\"41680\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e45d5d4-1543-4eeb-9c93-9abae2d77047>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4560a4d-e68a-4ecd-be1e-aed725ead11b>\",\"WARC-IP-Address\":\"129.11.26.47\",\"WARC-Target-URI\":\"https://eps.leeds.ac.uk/maths-masters-module-information/doc/statistics-msc\",\"WARC-Payload-Digest\":\"sha1:HK4TVB4TCPS7X4VQNA3CZC544WREDIF6\",\"WARC-Block-Digest\":\"sha1:HKINWBDUL26YMZAWW7QSI77COOX2VXMC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107896778.71_warc_CC-MAIN-20201028044037-20201028074037-00570.warc.gz\"}"} |
https://topbitzlej.web.app/chiz24960bi/weighted-gini-index-formula-jod.html | [
"## Weighted gini index formula\n\nWeight by Gini Index (RapidMiner Studio Core). Synopsis. This operator calculates the relevance of the attributes of the given ExampleSet based on the Gini\n\nWeight by Gini Index (RapidMiner Studio Core). Synopsis. This operator calculates the relevance of the attributes of the given ExampleSet based on the Gini Measure with Weighted Data This note sets out some basic results regarding calculation of the. Gini measure and vation is usually provided with a weight so that population-level values can The estimate of the Gini coefficient is thus: G =. The geometrical derivation of the Gini Index and an alternative formula 20. 10.1 weighting differently incomes in different parts of the income distribution. function (x, weights = rep(1, length = length(x))) { ox <- order(x) x <- x[ox] weights <- weights[ox]/sum(weights) p <- cumsum(weights) nu 1 Jun 2018 Most commonly followed are - Gini Index,Entropy,Chi-square etc. I. Gini Index. According Calculate Gini for sub-nodes, using formula. Calculate Gini for split using weighted Gini score of each node of that split. II. Chi Square.\n\n## In price-weighted index stock with higher price has a higher impact over the performance of the index. Recommended Articles. This has been a guide to what is Price-Weighted Index. Here we discuss how to calculate Price-Weighted Index using its formula along with practical examples.\n\nInformation Gain, Gain Ratio and Gini Index are the three fundamental criteria to measure the quality of a split in Decision Tree. In this blog post, we attempt to clarify the above-mentioned terms, understand how they work and compose a guideline on when to use which. Decision tree algorithms use information gain to split a node. Gini index or entropy is the criterion for calculating information gain. Both gini and entropy are measures of impurity of a node. A node having multiple classes is impure whereas a node having only one class is pure. Entropy in statistics is analogous to entropy in thermodynamics In price-weighted index stock with higher price has a higher impact over the performance of the index. Recommended Articles. This has been a guide to what is Price-Weighted Index. Here we discuss how to calculate Price-Weighted Index using its formula along with practical examples. The index is based on the Gini coefficient, a statistical dispersion measurement that ranks income distribution on a scale between 0 and 1. The measure has been in use since its development by How does a Decision Tree Work? A Decision Tree recursively splits training data into subsets based on the value of a single attribute. Splitting stops when e Computes the Gini coefficient based on (possibly weighted) sample data Usage gini(x, weights=rep(1,length=length(x))) Arguments x a vector containing at least non-negative elements weights an optional vector of sample weights for x Details Gini is the Gini coefficient, a common measure of inequality within a distribution. It is commonly The Theil index is a statistic primarily used to measure economic inequality and other economic phenomena, though it has also been used to measure racial segregation.. The Theil index T T is the same as redundancy in information theory which is the maximum possible entropy of the data minus the observed entropy. It is a special case of the generalized entropy index.\n\n### 27 Aug 2018 This algorithm uses a new metric named gini index to create decision points Then, we will calculate weighted sum of gini indexes for outlook\n\nMy question is, how I can calculate GINI coefficient in Stata for every as whole. my variable is GDPPC and i want to calculate weighted gini The Gini coefficient is a relative index of inequality; scaling ail incomes propor- irtaul (iini relative inequality index, the aggregation procedure is a mean- weighted Writing the formula for the Gini social-evaluation function explicitly, as in (8),. formula to compute the Gini index for a single variable such as income and also for Consider variate X as the weighted sum of several components. {Zxk = 1, By its construction, the Gini coefficient puts equal weights to the entire distribution , while the Atkinson inequality measure puts more weight to the lower end, thus it\n\n### Calculate the Gini index on total disposable income for Finland and the US in 2000, after indices. One such command is: ineqdeco [varname] [w=]\n\nThis formula opens the way to an interpretation of the Gini coefficient in term of covariance as. Cov(y θ is the OLS estimate of θ in the weighted regression. through 2005. 2. The inequality measures presented are the Gini coefficient (G), the mean household income, and N is the weighted number of households. Unlike the Gini The formula to compute the Atkinson index is: ε ε ε. −. = −. ⎥. ⎥. ⎦. weighted Lorenz curves for comparisons with an internal or external standard We conclude that the Lorenz curve and Gini index are universal tools for Calculate the Gini index on total disposable income for Finland and the US in 2000, after indices. One such command is: ineqdeco [varname] [w=\n\n## The formula reveals why the Gini index sometimes appears in calculus books in the for k ≥ 1 yields an index where extreme poverty is weighted more for.\n\n5 Apr 2019 and population-weighted Gini coefficient may be considered sufficiently appear in Table 1) are abstract mathematical formulas, one can This formula opens the way to an interpretation of the Gini coefficient in term of covariance as. Cov(y θ is the OLS estimate of θ in the weighted regression.\n\nCompute the weighted average over all sets resulting from the split Definition of Gain Ratio: average Gini index (instead of average entropy / information). fastgini -- Fast algorithm for calculation of Gini coefficient and it's jackknife standard errors. Syntax. fastgini pweights and fweights are allowed; see weight . Key words: Subgroup decomposition; Stochastic approach; Gini index; interaction components of the overall Gini as weighted averages of their respective approach, it is convenient to utilize the following formula considered by Ogwang"
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89068794,"math_prob":0.97783715,"size":5918,"snap":"2021-43-2021-49","text_gpt3_token_len":1265,"char_repetition_ratio":0.1550558,"word_repetition_ratio":0.17536534,"special_character_ratio":0.20057452,"punctuation_ratio":0.09150327,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986914,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-27T08:32:13Z\",\"WARC-Record-ID\":\"<urn:uuid:caf6d553-24a1-4da8-ac61-c530748c0ca1>\",\"Content-Length\":\"18692\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19526122-4c8c-4928-a7ae-4630cfc9fc94>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac5bd4a1-f2a0-425c-9429-c5754d20b371>\",\"WARC-IP-Address\":\"199.36.158.100\",\"WARC-Target-URI\":\"https://topbitzlej.web.app/chiz24960bi/weighted-gini-index-formula-jod.html\",\"WARC-Payload-Digest\":\"sha1:4HYKEHBQB5OHU42UF2IJNL4BYJLXFLGK\",\"WARC-Block-Digest\":\"sha1:VAW5MRF5SLBJ4WAZTGTFYOI46QR52BMD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358153.33_warc_CC-MAIN-20211127073536-20211127103536-00058.warc.gz\"}"} |
https://socratic.org/questions/an-object-with-a-mass-of-5-kg-is-pushed-along-a-linear-path-with-a-kinetic-frict-2 | [
"An object with a mass of 5 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= x+3 . How much work would it take to move the object over #x in [2, 3], where x is in meters?\n\nJan 18, 2016\n\nWork $= 269.5 J$\n\nExplanation:\n\nForce of kinetic friction which needs to be overcome to move the object\n\n${F}_{k} =$Coefficient of kinetic friction ${\\mu}_{k} \\times$normal force $\\eta$\nwhere $\\eta = m g$\nInserting given quantities and taking the value of $g = 9.8 m / {s}^{2}$\n${F}_{k} = \\left(x + 3\\right) \\times 5 \\times 9.8 N$\n${F}_{k} = 49 \\left(x + 3\\right) N$\n\nWhen this force moves through a small distance $\\mathrm{dx}$, the work done is given as\n${F}_{k} \\mathrm{dx} = 49 \\left(x + 3\\right) \\mathrm{dx}$\nWhen the force moves through a distance from $x \\in \\left[2 , 3\\right]$, total work done is integral of RHS over the given interval.\n\nTotal work done$= {\\int}_{2}^{3} 49 \\left(x + 3\\right) \\mathrm{dx}$\n$\\implies$Total work done$= 49 {\\int}_{2}^{3} \\left(x + 3\\right) \\mathrm{dx}$\n\n$= 49 \\left({x}^{2} / 2 + 3 x + C\\right) {|}_{2}^{3}$, where C is constant of integration.\n$= 49 \\left[\\left({3}^{2} / 2 + 3 \\times 3 + C\\right) - \\left({2}^{2} / 2 + 3 \\times 2 + C\\right)\\right]$\n$= 49 \\left[\\frac{9}{2} + 9 - 2 - 6\\right]$\n$= 49 \\left[4.5 + 1\\right]$"
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8613213,"math_prob":1.0000094,"size":616,"snap":"2019-43-2019-47","text_gpt3_token_len":148,"char_repetition_ratio":0.09150327,"word_repetition_ratio":0.0,"special_character_ratio":0.23376623,"punctuation_ratio":0.058333334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000019,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T15:31:06Z\",\"WARC-Record-ID\":\"<urn:uuid:ce0a38d8-edba-4dc2-bab0-8f779804af53>\",\"Content-Length\":\"34371\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6cd85155-e7f0-444e-9c73-71316a766c46>\",\"WARC-Concurrent-To\":\"<urn:uuid:f59bec02-b993-4e5e-b940-c88b2a82319f>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/an-object-with-a-mass-of-5-kg-is-pushed-along-a-linear-path-with-a-kinetic-frict-2\",\"WARC-Payload-Digest\":\"sha1:IILXYDG4HGSNQZZ4YDBGUHDZLI3YC53Q\",\"WARC-Block-Digest\":\"sha1:QXA56XMZEYYJ6SKDOW34T7BL47DZZ7Y3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986668994.39_warc_CC-MAIN-20191016135759-20191016163259-00531.warc.gz\"}"} |
https://www.nagwa.com/en/videos/452175847238/ | [
"# Question Video: Finding the Limit of a Combination of Rational Functions at Infinity\n\nFind lim_(𝑥 → ∞) (−2𝑥⁻⁴ + 8𝑥⁻³ − 𝑥⁻² + 9𝑥⁻¹ − 4)/(2𝑥⁻⁴ − 6𝑥⁻³ + 7𝑥⁻² + 6𝑥⁻¹ + 3).\n\n01:46\n\n### Video Transcript\n\nFind the limit of negative two 𝑥 to the negative four plus eight 𝑥 to the negative three minus 𝑥 to the negative two plus nine 𝑥 to the negative one minus four all over two 𝑥 to the negative four minus six 𝑥 to the negative three plus seven 𝑥 to the negative two plus six 𝑥 to the negative one plus three, as 𝑥 approaches infinity.\n\nThis is the limit of a quotient of functions. And we know that the limit of a quotient of functions is the quotient of their limits. So, we can find the limits of the numerator and denominator separately, should they exist. And as the limit of a sum of functions is the sum of their limits, we can find the limits term-by-term.\n\nNow, we have lots of limits to evaluate but they’re all of very simple terms. And we can make them simpler by taking the constants outside the limits. As the limit of a constant times a function is that constant times the limit of the function. And now, the vast majority of our limits are of the form the limit of 𝑥 to the power of some negative number as 𝑥 approaches infinity.\n\nWhat are the values of such limits? Well, we can write 𝑥 to the negative 𝑛 as one over 𝑥 to the 𝑛. And for 𝑛 greater than zero, the value is zero. All these limits then are zero. And we’re only left with the two limits to worry about, both of which are limits of the constants four and three. The limit of a constant function is just that constant. And so, being careful to include this minus sign, we see the answer is negative four over three.\n\nSolving this problem was straightforward because we only had constants and negative powers in the numerator and denominator. And we know what the limit of a negative power of 𝑥 is as 𝑥 approaches infinity; it’s zero."
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.94110256,"math_prob":0.9925867,"size":1719,"snap":"2021-04-2021-17","text_gpt3_token_len":404,"char_repetition_ratio":0.21457726,"word_repetition_ratio":0.06116208,"special_character_ratio":0.21233276,"punctuation_ratio":0.08196721,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.993366,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-27T07:48:42Z\",\"WARC-Record-ID\":\"<urn:uuid:a34bc5dc-e83d-470f-8307-f55eff30abd5>\",\"Content-Length\":\"102300\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec5fa1e9-fe73-4634-b8d2-d5d9f7691f78>\",\"WARC-Concurrent-To\":\"<urn:uuid:46212556-dadf-4f5d-91b8-b8a8114f3011>\",\"WARC-IP-Address\":\"34.193.189.180\",\"WARC-Target-URI\":\"https://www.nagwa.com/en/videos/452175847238/\",\"WARC-Payload-Digest\":\"sha1:KSHHWYMEJV2CE4HFC3PKETXZI2ONWO7L\",\"WARC-Block-Digest\":\"sha1:SNC7AA6C6IHP5TCILHYPOSJFPNN3GKH3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704821253.82_warc_CC-MAIN-20210127055122-20210127085122-00799.warc.gz\"}"} |
https://www.metrology-journal.org/articles/ijmqe/full_html/2017/01/ijmqe170033/ijmqe170033.html | [
"Open Access\n Issue Int. J. Metrol. Qual. Eng. Volume 8, 2017 26 7 https://doi.org/10.1051/ijmqe/2017020 23 October 2017",
null,
"This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\n## 1 Introduction\n\nMastering absolute pressure from around 100 kPa down to values as low as 5 kPa is important in various areas such as aeronautics and clean-room technology for semi-conductor manufacture and pharmaceuticals. It is then crucial for calibration services to provide pressure calibrations with low uncertainties in this range. The EMPIR project 14IND06 is aimed at developing methods in the intermediate pressure-to-vacuum range, especially in absolute pressure mode.\n\nIn previous work , we presented a resonant silicon gauge (RSG), full scale 130 kPa that exhibits a good long-term stability in its calibration slope and which was used to rescale a capacitance diaphragm gauge (CDG) of full scale 13 kPa, and stepwise two other CDGs of full scale 1 kPa and 100 Pa full scale respectively. In the present study, we focus on the RSG and the CDG 10 kPa. The latter, once rescaled by means of the RSG, is used to master and correct the drift in the offset of the RSG. The rescaling operation is a current way to enhance the uncertainty of a transfer standard, by estimating its drift, during the course of some metrological comparisons between national metrology institutes [2,3]. As an example, in the key comparison CCM.P-K4.2012 , the calibration procedure requires one compare at a pressure of 10 kPa, the measurement of a CDG of 10 kPa full scale and that of a RSG 100 kPa full scale. Since the particular RSG 100 kPa used in this comparison has a much better stability than the CDG 10 kPa, their direct comparison at 10 kPa allows one to calculate a correction coefficient which is applied to the output signal of the CDG 10 kPa in the lowest pressure range (not covered by the RSG 100 kPa): this is called the rescale procedure. In the work described in this paper, the rescale procedure is slightly different: as our RSG is poorly stable for a single pressure point but highly stable however as far as its calibration slope is concerned, several comparison points are needed to calculate the correction coefficient for the CDG. The metrological characterisation of both of these commercially available instruments and the associated procedures for their use and operation as working standards in the range 5–130 kPa with a standard uncertainty of 1 Pa is considered in this paper. For comparison, we have plotted in Figure 1 the best capabilities of French accredited calibration services in the aforementioned range , where the standard used to achieve this capability is given: a piston-cylinder device (which can be either a pressure balance or a digital piston gauge, or a liquid column standard) or a digital manometer.\n\nOnce calibrated and used correctly, this working standard could offer a performance equivalent to that of the best devices shown in Figure 1.\n\nThe paper describes first the preliminary observations we have made of the metrological performances of our absolute secondary pressure standards. Thereafter, the method for rescaling the 10 kPa full scale CDG, which in turns allows one to master the offset of the RSG, is presented. From the performance of the gauges and their characterisation, the contribution in uncertainty of the working standard is assessed.",
null,
"Fig. 1 Uncertainty (k = 1) of some French accredited calibration services. The uncertainty is achieved by means of either a piston/cylinder or liquid column device (dotted lines) or a digital manometer (solid lines). The maximum uncertainty contribution of the working standard in this study is represented with the dashed line.\n\n## 2 Metrological performances of the standard pressure gauges\n\n### 2.1 Resonant silicon gauge of full scale 130 kPa\n\nTen or so years ago, a RSG (Druck type DPI1421) was acquired by the vacuum department of LNE-LCM for daily calibrations in the pressure range 10–130 kPa. From the successive calibrations of the RSG with an absolute pressure balance, the main drift of the RSG characteristic was found to be that of the offset, the slope remaining stable (Fig. 2).\n\nFigure 3 shows the scattered drift of the correction slope, determined by means of an unweighted simple least-squares fitted straight line. The drift is estimated to be (−1.0 ± 6.6) ppm per year.\n\nAs one can see from Figure 1, the main drawback of this RSG is the lack of linearity between 10 kPa and 35 kPa which leads to a modelling error of about 2.5 Pa. We shall see in Section 4 how this issue can be dealt with (Fig. 4).\n\nAs the nominal range of the sensor is 3.5–130 kPa, it appeared appealing to calibrate it in the low pressure range from 5 kPa to 10 kPa. The calibration with the force-balanced piston gauge (Fluke FPG 8601) of LNE-LCM has shown a good linearity of the RSG in this range as illustrated in Figure 3. Associated with the low drift of its correction slope, the RSG can then be applied to check and rescale the calibration function of our working standard CDG 13 kPa full scale (Sect. 2.2) used to calibrate customers gauges.",
null,
"Fig. 2 Several calibrations of the upper range of the RSG by means of an absolute pressure balance. One can see at a glance, in the upper pressure range (30–130 kPa), that the RSG signal is linear and the slope of an applied straight correction line would be stable over time.",
null,
"Fig. 3 Relative drift in the slope correction coefficient of the RSG between a current calibration and the previous one.",
null,
"Fig. 4 Linearity deviation from a regression line of the lower range of the RSG.\n\n### 2.2 Capacitance diaphragm gauge 10 kPa\n\nRelative or absolute CDGs from the manufacturer MKS (in particular the 13 kPa full scale model) are currently used as secondary and working standards at LNE-LCM. To enhance the gauge resolution, it is generally preferable to treat the analogue output U (0–10 V), rather than the digital one. Thus the calibration function is expressed using an equation of the form:",
null,
"(1) where U0 is the output signal of the gauge when zero pressure is applied (which in absolute mode corresponds to a pressure lower than a tenth of the gauge resolution) and f(U − U0) is a polynomial function of up to fourth order. As the CDG is controlled at a temperature of 45 °C, a thermal transpiration correction is applied to calculate the reference pressure p. We then applied the empirical function of Takaishi-Sensui ; one can find in the literature other work on this correction [6,7]. In absolute mode, the polynomial function and the gauge temperature are determined from a calibration using the FPG, for pressures between 1 kPa and 10 kPa.\n\nFrom the numerous calibrations of CDGs performed, it was stated that for the mean term, the shape of a calibration curve does not change much, as one can see at a glance from Figure 5; consequently, it is possible to estimate the new calibration function ft by applying a linear correction to the previous one, ft−1. Let us denote by",
null,
"the function determined by the linear correction (kCDG being the linear correction factor), as ft and ft−1 are the calibration functions obtained from the CDG successive calibrations. We have:",
null,
"(2) where kCDG is the slope coefficient of the least-squares-fitted straight line used to estimate the reference pressure as a function of ft−1(U − U0), in the range between 40% and 80% of the full scale of the CDG2. This is the method used to rescale a CDG. From the example of Figure 5, with quite a large drift of deviation of the CDG (of about 1.2 ×10 ‒3 in relative value), the aforementioned method was applied and the difference",
null,
"is plotted in Figure 6 as a function of the pressure.\n\nOn this same graph are plotted the residuals of the CDG calibration curve i.e. the difference between ft(U − U0) and pFPG the reference pressure given by the standard FPG 8601. As one can see in Figure 5 the residuals and the deviation between the rescaled pressure and the modelled pressure are of the same order of magnitude and lower than 3.0 × 10‒5 in relative value.",
null,
"Fig. 5 Plot of the deviation of two calibration functions of a CDG 13 kPa full scale, in absolute pressure mode. The two calibrations are spaced by about 12 months. The deviation is the difference between the CDG calibration function and a linear function g(U − U0) = a(U − U0), where a is an arbitrary coefficient.",
null,
"Fig. 6 Difference between the calibration function of a 10 kPa full scale CDG obtained in one case by modelling the calibration data (ft), and in the other case by applying a correction factor kCDG to the previous calibration function (kCDG × ft−1). Here kCDG is the slope coefficient of the least squares fit to a straight line that is used to estimate the reference pressure pFPG as a function of the CDG pressure modelled with the function (ft−1), in the range between 40% and 80% of the full scale of the CDG. This difference is plotted together with the residuals of the model: (ft − pFPG).\n\n## 3 Experimental set-up\n\nThe metrological features of the instruments, described in Section 2, make possible the rescaling of the CDG of 10 kPa (effective) full scale, starting with a comparison between the CDG 10 kPa and the RSG between 5 kPa and 10 kPa. Once rescaled, the CDG 10 kPa is used to determine the offset of the RSG. The experimental set-up of the transfer standard is described in Figure 7.\n\nThe CDG10k is an MKS Instruments 690 absolute pressure transducer connected to an MKS 670 electronics package. The vacuum pump, whose ultimate pressure is lower than 0.01 Pa, is used to determine the zero U0 of the CDG10k.",
null,
"Fig. 7 Set-up of the working standard.CDG10k: capacitance diaphragm gauges MKS type 690 with an effective full scale of 10 kPa; RSG: resonant silicon gauge Druck type DPI142 (3.5–130 kPa); P: vacuum pump; VM: isolation valve of the working standard; VRSG: isolation valve of the RSG; VP: isolation valve of the pump.\n\n## 4 Procedure to use the working standard\n\n### 4.1 Principle\n\nThis working standard is actually based on the simultaneous operation, in a common pressure range, of a standard with a very stable linear correction, i.e. the RSG and a standard for which the zero can be routinely determined, namely the CDG 10 kPa. The associated metrological properties of the instruments lead to a low uncertainty contribution of the working standard. The procedure to correct the output signal of the CDG and the RSG is illustrated in Figure 8. Each graph has the true pressure along the horizontal axis and the output pressure signal given by the calibration function of each instrument along the vertical axis. On these graphs, the ideal instrument output signal is the straight line passing through the origin with a slope equal to unity. Let us denote fRSG the calibration function of the RSG and f10k that of the CDG: both output signals given by the functions are coinciding with the ideal instrument output signal at the time of their respective calibration. Sometime later (graph a, Fig. 8), the RSG has drifted in offset: the output signal still has a slope equal to unity but no longer crosses zero. As for that of the CDG, it still crosses zero, as U0 is determined each time of using, but the slope has drifted slightly. Several measurements performed at the same pressure levels between 5 kPa and 10 kPa allow one to determine the linear correction k10k (according to Eq. (2)) via a least-squares analysis. The corrected calibration function",
null,
"is:",
null,
"(3) The mean difference ϵ between the RSG output signal and that of the CDG10k given by",
null,
"at the aforementioned measurement points allows one to correct in turn the function (fRSG). The corrected function",
null,
"is:",
null,
"(4)The new output signal of the RSG calculated with",
null,
"coincides again with that of the ideal instrument, between 10 kPa and 130 kPa.",
null,
"Fig. 8 Procedure to correct the working standard.The straight line passing through the origin with a slope equal to unity represents the output signal of the ideal instrument.Respective drifts of the RSG (in offset) and CDG (in slope) are voluntarily exaggerated for a greater clarity.\n\n### 4.2 Application to LNE-LCM instruments\n\nThe RSG is calibrated in two steps to cover the range from 5 kPa to 130 kPa (Sect. 2.1): the first range 10–130 kPa with an absolute pressure balance and the second range 5–10 kPa with the force balanced piston gauge FPG 8601. A difference of a few Pascals in the deviation at 10 kPa from the two calibrations is observed, since the offset of the RSG drifts slightly (Fig. 9). Moreover, the RSG shows a non-linearity for pressures around 25 kPa (Sect. 2.1). In Figure 9 which shows the deviation at the calibration points over the entire range, one can see that fRSG cannot be expressed by means of a single function. Consequently, the RSG range has been split into three: f1 denotes the function for pressures between 5 kPa and 10 kPa obtained from the calibration with the FPG 8601 and is linear; f2 covers the non-linear part of the RSG characteristic between 10 and 46 kPa, obtained from the calibration with the absolute pressure balance and is a second-order polynomial; finally f3 is a linear function for pressures between 35 kPa and 130 kPa obtained from the aforementioned calibration.\n\nWe have denoted by ϵ2−1 the offset drift during the short time between the calibration with the pressure balance and the FPG:",
null,
"(5) The CDG10k rescaling (Sect. 4.1) is performed using the functionf1 of the RSG. The mean difference between",
null,
"and f1 at different chosen measurement points is thus denoted by ϵ1. It is used to obtain the actual output signal of the RSG by means of the corrected functions",
null,
"",
null,
"and",
null,
":",
null,
"(6)",
null,
"(7)",
null,
"(8)The working standard is used as follows: valves VM and VRSG are closed, valve VP opened and the pump P switched on. After two hours of pumping, the CDG10k output voltage U0 is recorded. The pump is then switched off and Valve VP is closed. At the same time, the calibration chamber connected to the other side of the valve VM is filled up to the first pressure calibration level. Valve VM is opened to start a calibration. Once the pressure reaches 5 kPa, Valve VRSG is opened and some common measurement points for CDG and RSG are performed during the course of the calibration. These operations are repeated for each calibration cycle (three cycles are recommended), after which data are post-processed to determine the values of k10k and ϵ1 for each calibration cycle.",
null,
"Fig. 9 Deviation of the RSG between 5 and 130 kPa.The whole range is split into three (1 to 3) to model the corresponding calibration functions: range 1 lies from 5 to 10 kPa, range 2 from 10 to 46 kPa and range 3 from 35 to 130 kPa.\n\n## 5 Characterisation of the working standard\n\n### 5.1 Linearity and hysteresis errors of the RSG\n\nFrom the functions f1, f2, and f3 used to model the RSG and the actual calibration pressure, the linearity errors have been calculated. Since the RSG is calibrated by increasing and decreasing pressure levels, the determined linearity errors also include the hysteresis error of the RSG. A closer analysis shows that the main error is due to the hysteresis effect (as one can observe Fig. 3, for the range 5–10 kPa), with a maximum model error summarised in Table 1, depending upon the range considered.\n\nTable 1\n\nModelling errors of the RSG for three different pressure ranges.\n\n### 5.2 Rescaling of the CDG and offsetting of the RSG\n\nThe characterisation of the working standard took place during the study of a low pressure transfer standard , as well as in the framework of the EMPIR project 14IND06 . The CDG10k used was a differential one; however, its behaviour is analogous to that of an absolute CDG and so has similar performances. Three calibration cycles were made in which six common measurement pressure readings equally distributed between 5 kPa and 10 kPa were taken (namely 5, 6, 7, 8, 9 and 10 kPa).\n\nThe correction factor k10k is the slope coefficient of the least-squares-fitted straight line that estimates f1(pRSG) as a function of f10k(U − U0). At each pressure level i, we calculate:",
null,
"(9) The mean drift in offset of the RSG from its calibration function f1 is then given by:",
null,
"(10)The values of the rescaling coefficient k10k, the offset deviation ϵ1 and their respective experimental standard deviation ESD(k10k) and ESD(ϵ1) are shown in Table 1. We can observe that the offset deviation ϵ1 slightly drifts between the cycles, so it is relevant to take into account each individual cycle rather than a mean value calculated on the three cycles.\n\nNote that the modelling errors of the CDG10k (Sect. 2.2) are lower than the maximum value of ESD(k10k).\n\n### 5.3 Temperature influence\n\nAs the overall performance of the transfer standard is based on the correction slope of the RSG, it is important to check to what the extent it is affected by temperature. To determine the temperature coefficient, the RSG was placed in a climate controlled chamber at temperatures successively of 20 °C, 15 °C, 25 °C then 20 °C once more and was compared with a similar calibrated RSG which was left at the ambient temperature of 20 °C. The variation in the correction slope of the transfer standard RSG was studied as a function of temperature. The temperature coefficient was determined to be (‒5.5 ± 3.8) × 10‑7 K‑1. For a difference of 3 K, which is a huge tolerance for an accredited Laboratory, the temperature effect never exceeds 2 × 10−6 in relative value.\n\n## 6 Uncertainty budget\n\nWe first calculate the uncertainty contribution of the working standard. Following the procedure described Section 4, we determine the rescaling coefficient k10k of the CDG10k from the slope of the corrected signal of RSG, and then the RSG offset deviation ϵ1. Table 3 lists the uncertainty components of ϵ1. The calibration uncertainty for the gauge FPG 8601 is 1.5 × 10‑5 × p (k = 1) while the drift in the RSG correction slope is 6.6 × 10‑6 × p (Sect. 2.1). We assume that the modelling errors of both CDG10k and RSG are already taken into account in the maximum value of the experimental standard deviations ESD(k10k) and ESD(ϵ1) respectively (cf. Tab. 2). Furthermore, the RSG linearity error is used to calculate the uncertainty contribution of the working standard (Tab. 4). Finally, the temperature effect is 2 × 10‑6 × p (Sect.5.3) for 3 K.\n\nThe uncertainty in the function f1 offset correction u(ϵ1) is maximal at 10 kPa and equal to:",
null,
"(11) To calculate the contribution in uncertainty of the working standard, ucont, we combine with u(ϵ1) at 10 kPa, the overall linearity and hysteresis error of the RSG on its whole range (the maximum value is 0.37 Pa from the Tab. 1 with a rectangular distribution), the RSG slope stability (Tab. 3) and the temperature effect. The RSG pressure is calculated either with the function f2 or f3 at 35 kPa and 46 kPa (overlap area); the maximum discrepancy between f2 and f3 is 0.13 Pa is considered as an unapplied correction i.e. an intrinsic systematic error correction term to account for the slight mismatch that is present in overlapping pressure scale regions and which is added to the expanded uncertainty (k = 2) at the final stage. In Table 4, only the standard uncertainty is specified.\n\nWe have calculated the uncertainty uWSt in using the working standard. For that purpose, it suffices to combine its contribution ucont with the best capability of the LNE-LCM for the range 10–130 kPa, uCMC, that is:",
null,
"(12) We then obtained:",
null,
"(13)which leads to a maximum uncertainty of 1.1 Pa at 130 kPa. For comparison with the existing capabilities of French calibration services Figure 10 shows uWst together with the best capability obtained with a pressure balance and a digital manometer respectively. We can state that the working standard studied here could disseminate the lowest level of uncertainty in pressure in the range 5–130 kPa, which is at least four times better than that of the best digital manometer.\n\nTable 3\n\nUncertainty budget in the determination of ϵ1.\n\nTable 2\n\nRescaling coefficient of the CDG10k and drift in the offset of the RSG.\n\nTable 4\n\nUncertainty budget in the contribution of the working standard.",
null,
"Fig. 10 Final uncertainty of the working standard together with the best accredited uncertainties of calibration services in France and the corresponding apparatus employed.\n\n## 7 Conclusion\n\nThe experimental study of two pressure instruments with a common pressure range and complementary metrological features has allowed us to produce a high-accuracy working standard in the absolute pressure range from 5 kPa to 130 kPa, with an uncertainty lower than 1 Pa (k = 1). The instruments are a commercially available CDG and RSG, of full scales 13 kPa and 130 kPa respectively. In the range 5–130 kPa, the RSG has the advantage that its linear correction remains stable over many years. By comparing the CDG with the RSG in the range 5–10 kPa, the CDG can be rescaled. Since it is straightforward to measure the zero signal of the CDG (given a suitable pumping unit), the latter in turn allows one to master the drift in the offset of the RSG. The uncertainty analysis based on a strict procedure leads to an uncertainty contribution of the working standard (k = 1) which rises from 0.37 Pa to 0.99 Pa over the range 5–130 kPa.\n\nThe mathematical tools used to model the output signals of the instruments and calculate the different corrections are basic (polynomial functions, unweighted least-squares fitting) and available in most spreadsheets software. Consequently, it is straightforward for an experienced calibration service in pressure metrology to apply the method described in this paper once it has identified the most suitable instruments for the purpose.\n\n## Acknowledgments\n\nThis work was supported by the European Metrology Programme for Innovation and Research (EMPIR). The EMPIR is jointly funded by the EMPIR participating countries within EURAMET and the European Union.\n\nThe authors are grateful to Yang Lei, NIM, China, for improving the appearance of several figures at short notice.\n\n## References\n\n1. F. Boineau, S. Huret, P. Otal, M. Plimmer, Development and characterisation of a low pressure transfer standard in the range 1 Pa to 10 kPa, in Poster, 6th CCM International Conference on Pressure and Vacuum Metrology in conjunction with the 5th IMEKO TC16 International Conference, Pereira, Columbia, 7–10 May 2017 [Google Scholar]\n2. J. Ricker et al., Final report on the key comparison CCM.P-K4.2012 in absolute pressure from 1 Pa to 10 kPa, Metrologia 54, 07002 (2017) [CrossRef] [PubMed] [Google Scholar]\n3. D.A. Olson, P.J. Abbott, K. Jousten, F.J. Redgrave, P. Mohan, S.S. Hong, Final report of key comparison CCM.P-K3: absolute pressure measurements in gas from 3 × 10−6 Pa to 9 × 10−4 Pa, Metrologia 47, 07004 (2010) [Google Scholar]\n4. https://www.cofrac.fr (accessed: 2017/15/05/) [Google Scholar]\n5. T. Takaishi, Y. Sensui, Thermal transpiration effect of hydrogen, rare gases and methane, Trans. Faraday Soc. 59, 2503 (1963) [CrossRef] [Google Scholar]\n6. J. Setina, New approach to corrections for thermal transpiration effects in capacitance diaphragm gauges, Metrologia 36, 623 (1999) [Google Scholar]\n7. K. Poulter, M.J. Rodgers, P. Nash, T. Thompson, M. Perkin, Thermal transpiration correction in capacitance manometers, Vacuum 33, 311 (1983) [Google Scholar]\n8. EURAMET, European Metrology Program for Innovation and Research (EMPIR) − EMPIR Project 14 IND06, https://www.euramet.org/ (accessed: 2017/28/06/) [Google Scholar]\n\n1\n\nIdentification of commercially available instruments in this paper does not imply recommendation or endorsement.\n\n2\n\nA large part of the CDGs used at LNE-LCM exhibits a significant non linearity between 80% and 100% of the full scale so we don't use them in this range.\n\nCite this article as: Frédéric Boineau, Sébastien Huret, Pierre Otal, Mark Plimmer, A high-accuracy working standard for absolute pressure from 5 kPa to 130 kPa, Int. J. Metrol. Qual. Eng. 8, 26 (2017)\n\n## All Tables\n\nTable 1\n\nModelling errors of the RSG for three different pressure ranges.\n\nTable 3\n\nUncertainty budget in the determination of ϵ1.\n\nTable 2\n\nRescaling coefficient of the CDG10k and drift in the offset of the RSG.\n\nTable 4\n\nUncertainty budget in the contribution of the working standard.\n\n## All Figures",
null,
"Fig. 1 Uncertainty (k = 1) of some French accredited calibration services. The uncertainty is achieved by means of either a piston/cylinder or liquid column device (dotted lines) or a digital manometer (solid lines). The maximum uncertainty contribution of the working standard in this study is represented with the dashed line. In the text",
null,
"Fig. 2 Several calibrations of the upper range of the RSG by means of an absolute pressure balance. One can see at a glance, in the upper pressure range (30–130 kPa), that the RSG signal is linear and the slope of an applied straight correction line would be stable over time. In the text",
null,
"Fig. 3 Relative drift in the slope correction coefficient of the RSG between a current calibration and the previous one. In the text",
null,
"Fig. 4 Linearity deviation from a regression line of the lower range of the RSG. In the text",
null,
"Fig. 5 Plot of the deviation of two calibration functions of a CDG 13 kPa full scale, in absolute pressure mode. The two calibrations are spaced by about 12 months. The deviation is the difference between the CDG calibration function and a linear function g(U − U0) = a(U − U0), where a is an arbitrary coefficient. In the text",
null,
"Fig. 6 Difference between the calibration function of a 10 kPa full scale CDG obtained in one case by modelling the calibration data (ft), and in the other case by applying a correction factor kCDG to the previous calibration function (kCDG × ft−1). Here kCDG is the slope coefficient of the least squares fit to a straight line that is used to estimate the reference pressure pFPG as a function of the CDG pressure modelled with the function (ft−1), in the range between 40% and 80% of the full scale of the CDG. This difference is plotted together with the residuals of the model: (ft − pFPG). In the text",
null,
"Fig. 7 Set-up of the working standard.CDG10k: capacitance diaphragm gauges MKS type 690 with an effective full scale of 10 kPa; RSG: resonant silicon gauge Druck type DPI142 (3.5–130 kPa); P: vacuum pump; VM: isolation valve of the working standard; VRSG: isolation valve of the RSG; VP: isolation valve of the pump. In the text",
null,
"Fig. 8 Procedure to correct the working standard.The straight line passing through the origin with a slope equal to unity represents the output signal of the ideal instrument.Respective drifts of the RSG (in offset) and CDG (in slope) are voluntarily exaggerated for a greater clarity. In the text",
null,
"Fig. 9 Deviation of the RSG between 5 and 130 kPa.The whole range is split into three (1 to 3) to model the corresponding calibration functions: range 1 lies from 5 to 10 kPa, range 2 from 10 to 46 kPa and range 3 from 35 to 130 kPa. In the text",
null,
"Fig. 10 Final uncertainty of the working standard together with the best accredited uncertainties of calibration services in France and the corresponding apparatus employed. In the text\n\nCurrent usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform.\n\nData correspond to usage on the plateform after 2015. The current usage metrics is available 48-96 hours after online publication and is updated daily on week days."
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https://atoms.mimuw.edu.pl/?p=1143 | [
"# Definability of model-complete cores\n\nThese are some short notes concerning model-complete cores, based on an exposition of Manuel Bodirsky. In particular, he proved that every -categorical structure has a unique (up to isomorphism) model-complete core. We pose the question whether the model-complete core of a definable structure (say, over the equality atoms) is again definable.\n\nDefinition. A structure is a core if every endomorphism of is an embedding, i.e., every homomorphism from to is injective and preserves the relations of and their complements.\n\nDefinition. A structure is model-complete if every embedding of into is elementary, i.e., preserves all first-order definable relations.\n\nAn mc-core is a model-complete core, i.e., a structure in which every endomorphism is elementary.\n\nAn existential positive formula (ep formula) is a formula formed by using conjunctions, disjunctions and existential quantification and atomic formulas.\n\nTheorem. Let be an -categorical structure. The following conditions are equivalent.\n\n1. is an mc-core,\n2. Every fo formula is equivalent in to an ep formula,\n3. has a homogeneous expansion by relations , where is the equality, such that for , both the relation and its complement are ep definable in ,\n4. , with respect to the topology of pointwise convergence.\n\nRemark. (some) people from Prague take the fourth condition as the definition of a core.\n\nProof.\n(1→2) Let be an fo-definable relation on ; in particular, is preserved by all elementary maps from to . Since is an mc-core, it follows that preserved by all endomorphism from to . Applying an appropriate homomorphism-preservation result, we get that is ep-definable.\n\n(2→3) We expand by all fo-definable relations. By -categoricity, the resulting structure is homogeneous. Each relation (and its complement) is ep-definable by assumption.\n\n(3→4) Let be an endomorphism, and let be a finite subset. Then preserves every relation and its complement. In particular induces a partial automorphism from the substructure of induced by to the substructure induced by . Since is homogeneous, it follows that can be extended to an automorphism. The automorphism is in particular an automorphism of , which agrees with on .\n\n(4→1) Assume that . Then is a core, since each mapping in is elementary.\n\nTheorem (Bodirsky ’06, ’12). Every -categorical structure is homomorphically equivalent to an mc-core, which is moreover unique, up to isomorphism.\n\nExamples.\n\n1. The structure is an mc-core, since it is homogeneous, and the complement of the relation is ep-definable.\n2. The structure is not an mc-core. Its mc-core is a singleton with a self-loop.\n3. The mc-core of the disjoint union of infinitely many 2-cliques is the 2-clique.\n4. The mc-core of the random graph is the infinite clique.\n5. Let be the Johnson graph defined as the line graph of the infinite clique. Its mc-core is the infinite clique, since contains an induced infinite clique (as the line graph of an infinite star).\n6. Let be the Johnson graph expanded additionally by the relation . Then is an mc-core. To prove this, we show (1) the relation , expressing that can be extended to a 4-clique in , and the complement of the relation , are both ep-definable in , and (2) the structure is homogeneous.\n7. Here is an example of a core (in the sense that endomorphisms are injective) which is not an mc-core: — the non-negative rationals with the linear order . The mc-core of this structure is . This also shows that the mc-core of a structure does not necessarily need to be a surjective image of .\nFrom these examples we see that the mc-core of a structure is often a simpler structure than . The following properties are preserved by taking the mc-core: being a homogeneous structure over a finite relational signature, being a Ramsey structure.Open questions. Are the following properties preserved by taking the mc-core (for -categorical structures):\n\n• Being interpretable in (or some other structure),\n• Being an fo-reduct of a homogeneous structure over a finite relational signature,\n• Being (-)stable,\n• Being finitely bounded,\n• etc.\n\nSome of these questions would be implied by a positive answer to the following question.\n\nQuestion.\nDoes the mc-core of an -categorical structure interpret in ?\n\nThe current construction of the mc-core is non-finitistic.\n\nUpdate. The answer to the last question seems to be negative: the mc-core of the universal homogeneous poset is , which seems not to interpret in : in this post we show that there is no 0-interpretation."
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