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UltraData-Math-L1 · 86M rows

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train · 86M rows

content stringlengths 86 994k	meta stringlengths 288 619
Compute digital filter coefficients from frequency response values. [b,a] = invfreqz(h,f,nb,na) [b,a] = invfreqz(h,f,nb,na,w) The complex frequency response values. Type: double Dimension: vector The frequencies corresponding to h. The values in Hz must be normalized relative to fs/(2pi), where fs is the sampling frequency, so that the Nyquist frequency corresponds to a value of pi. Type: double Dimension: vector The filter numerator polynomial order. Type: integer Dimension: scalar The filter denominator polynomial order. Type: integer Dimension: scalar Optional weights applied to achieve a weighted fitting of the response values. Type: double Dimension: vector The estimated numerator polynomial coefficients of the filter. Type: double Dimension: vector The estimated denominator polynomial coefficients of the filter. Type: double Dimension: vector Recover coefficients from the output of a digital Chebyshev I filter. order = 3; fc = 200; fs = 1000; [b1,a1] = cheby1(order, 1, fc/(fs/2), 'z') f = [0:0.2:2] fc; h = freqz(b1,a1,f,fs); [b2,a2] = invfreqz(h,pi*f/(fs/2),order,order) b1 = [Matrix] 1 x 4 0.07360 0.22079 0.22079 0.07360 a1 = [Matrix] 1 x 4 1.00000 -0.97613 0.85676 -0.29186 b2 = [Matrix] 1 x 4 0.07360 0.22079 0.22079 0.07360 a2 = [Matrix] 1 x 4 1.00000 -0.97613 0.85676 -0.29186 It is recommended to use freqz to assess the quality of the fitted filter coefficients.	{"url":"https://www.openmatrix.org/help/topics/reference/oml_language/SignalProcessing/invfreqz.htm","timestamp":"2024-11-09T16:10:18Z","content_type":"application/xhtml+xml","content_length":"10317","record_id":"<urn:uuid:3b179b6c-698a-4c3e-ba5f-b9f839d29edb>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00222.warc.gz"}
A Partial Depth-Search Heuristic for Packing Spheres Hakim Akeb A Partial Depth-Search Heuristic for Packing Spheres This paper proposes a new heuristic for packing non-identical spheres into a three-dimensional container of fixed dimensions. Given a set that contains n spheres, the objective is to place a subset of spheres so as to maximize the volume occupied by these ones. The proposed heuristic is based on an idea that applies a two-level look-forward search. The computational investigation indicates that the heuristic is effective since it improves most of the best known results in the literature on the used instances. Packing problems, Packing spheres, Heuristic, Look-Forward, Knapsack [1] R. Alvarez-Valdes, A. Martinez and J. M. Tamarit, A branch & bound algorithm for cutting and packing irregularly shaped pieces, Int. J. Prod. Econ. 145, 2013, pp. 463–477. [1] R. Alvarez-Valdes, A. Martinez and J. M. Tamarit, A branch & bound algorithm for cutting and packing irregularly shaped pieces, Int. J. Prod. Econ. 145, 2013, pp. 463–477. [2] J. A. George, J. M. George and B. W. Lamar, Packing different-sized circles into a rectangular container, European J. Oper. Res. 84, 1995, pp. 693–712. [3] D. He, N. N. Ekere and L. Cai, Computer simulation of random packing of unequal particles. Phys. Rev. E. 60, 1999, 7098. [4] K. Hitti and M. Bernacki, Optimized droping and rolling (ODR) method for packing of polydisperse spheres, Appl. Math. Model. 37, 2013, pp. 5715–5722. [5] W. Q. Huang, Y. Li, H. Akeb and C. M. Li, Greedy algorithms for packing unequal circles into a rectangular container. J. Oper. Res. Soc. 56, 2005, pp. 539–548. [6] Y.–K. Joung and S. D. Noh, Intelligent 3D packing using a grouping algorithm for automative container engineering, J. Comput. Des. Eng. 1, 2014, pp. 140–151. [7] T. Kubach, A. Bortfeldt, T. Tilli and H. Gehring, Greedy algorithms for packing unequal sphere into a cuboidal strip or a cuboid, Asia Pac. J. Oper. Res. 28, 2011, pp. 739–753. [8] Y. Li and W. Ji, Stability and convergence analysis of a dynamics–based collective method for random sphere packing, J. Comput. Phy. 250, 2013, pp. 373–387. [9] C. O. Lopes and J. E. Beasley, A formulation ´ space search heuristic for packing unequal circles in a fixed size circular container, European J. Oper. Res. 251, 2016, pp. 64–73. [10] S. Martello and M. Monaci, Models and algorithm for packing rectangles into the smallest square, Comput. Oper. Res. 63, 2015, pp. 161– 171. [11] A. Martinez-Sykora, R. Alvarez-Valdes, J. Bennell and J. M. Tamarit, Constructive procedures to solve 2-dimensional bin packing problems with irregular pieces and guillotine cuts, Omega 52, 2015, pp. 15–32. [12] R. M’Hallah, A. Alkandari and N. Mladenovic,´ Packing unit spheres into the smallest sphere using VNS and NLP. Comput. Oper. Res. 40, 2013, pp. 603–615. [13] K. Soontrapa and Y. Chen, Mono-sized sphere packing algorithm development using optimized Monte Carlo technique. Ad. Powder Technol. 24, 2013, pp. 955–961. [14] W. Visscher and M. Bolsterli, Random packing of equal and unequal spheres in two and three dimensions, Nature. 239, 1972, pp. 504–507. [15] T. Zauner, Application of a force field algorithm for creating stringly correlated multiscale sphere packings. J. Comput. Phys. In Press, 2016. [16] J. Wang, Packing of unequal spheres and automated radiosurgical treatment planning. J. Comb. Optim. 3, 1999, pp. 453–463. [17] Y. Wu, W. Li, M. Goh and R. Souza, Threedimensional bin packing problem with variable bin height, European J. Oper. Res. 202, 2010, pp. 347–355. Cite this paper Hakim Akeb. (2016) A Partial Depth-Search Heuristic for Packing Spheres. International Journal of Mathematical and Computational Methods, 1, 120-127 Copyright © 2016 Author(s) retain the copyright of this article. This article is published under the terms of the Creative Commons Attribution License 4.0	{"url":"https://www.iaras.org/journals/caijmcm/a-partial-depth-search-heuristic-for-packing-spheres","timestamp":"2024-11-14T04:25:44Z","content_type":"text/html","content_length":"41525","record_id":"<urn:uuid:e1ffa9b6-f5b8-45e2-a034-0b7995468861>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00358.warc.gz"}
Eliminating Blank Cells In A Range This page describes formulas and VBA functions to remove blank cells from a range. It is not uncommon that you have a range of data containing both values and blank cells and you want to eliminate the blank cells. This page describes worksheet array formulas and VBA code to create a new range that contains only the non-blank elements of the original range. Due to the way array formulas work, it is necessary that the original range and the new range be referenced by a defined name rather than directly by cell references. We will use BlanksRange to refer to the original range that contains both values and blanks, and we will use NoBlanksRange to reference the new range that contains only the non-blank values of the original range. BlanksRange that contains a combination of values and blank cells. Although the values are in alphabetical order, this is by no means necessary. It is for illustration only. The values will be extracted and will appear in the no-blanks range in the order in which the appear in the original data. To use the formula, paste it into the first cell of NoBlanksRange and then copy it down to fill that range. The NoBlanksRange should have as many rows as BlanksRange. Any unused cells in BlanksRange will contain empty values. This is an array formula, so you must press CTRL SHIFT ENTER rather than just ENTER when you first enter the formula and whenever you edit it later, but you do not array enter it into the entire range at once. Array enter the formula into the first cell of NoBlanksRange and then fill down to the last cell of NoBlanksRange. The formula is: The formula above is split into several lines for readability. In practice, it should be entered as a single line. A simpler method is available in Excel 2007 and later versions, using the IFERROR Enter this formula into the first cell of NoBlanksRange and copy it down through the last cell of NoBlanksRange. Like the other formulas, this is an array formula, so enter it with CTRL SHIFT ENTER rather than just ENTER. This formula is for extracting the non-blank elements to a vertical range -- a range in a single column that spans several rows. If you want the results in a single row spanning several columns, use the following array formula, where the result range is named NoBlanksRow. Array enter this formula into the first cell of NoBlanksRow and fill to the right through the last cell of NoBlanksRow. If the formulas above seem overly complex, you might want to opt for a much simpler VBA function. The NoBlanks function is shown below. Function NoBlanks(RR As Range) As Variant Dim Arr() As Variant Dim R As Range Dim N As Long Dim L As Long If RR.Rows.Count > 1 And RR.Columns.Count > 1 Then NoBlanks = CVErr(xlErrRef) Exit Function End If If Application.Caller.Cells.Count > RR.Cells.Count Then N = Application.Caller.Cells.Count N = RR.Cells.Count End If ReDim Arr(1 To N) N = 0 For Each R In RR.Cells If Len(R.Value) > 0 Then N = N + 1 Arr(N) = R.Value End If Next R For L = N + 1 To UBound(Arr) Arr(L) = vbNullString Next L ReDim Preserve Arr(1 To L) If Application.Caller.Rows.Count > 1 Then NoBlanks = Application.Transpose(Arr) NoBlanks = Arr End If End Function This code does not require the use of any defined names. Simply array enter the formula NoBlanks into the entire range that is to get the results, passing to the function the range from which the blank elements are to be extracted. To array enter the formula into all the cells, first select the entire range that is to receive the results, enter =NoBlanks(A1:A10) in the first cell and press CTRL SHIFT ENTER. The code is written to be entered into a single column spanning several rows or a single row spanning several columns. If the number of rows and the number of columns of the input range are both greater than 1, the function will return a #REF error. The function will orient itself so that there is no difference between entering it into a row or into a column. Moreover, if the function is called from a range with more elements than there are non-blank elements, the results at the end of the result list will filled out with empty string to the full length of cells into which it was entered. You can download the workbook file with all the example formulas and code on this page. This page last updated: 5-February-2011.	{"url":"https://www.cpearson.com/Excel/noblanks.aspx","timestamp":"2024-11-14T04:31:55Z","content_type":"text/html","content_length":"35740","record_id":"<urn:uuid:790ab0ff-6304-4cfb-a7b2-b211ea046f02>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00192.warc.gz"}
Now showing items 1-10 of 51 I-cesàro summability of sequences of sets (Electronic Journal of Mathematical Analysis and Applications, 2017-01) In this paper, we de ned concept of Wijsman I-Cesàro summability for sequences of sets and investigate the relationships between the concepts of Wijsman strongly I-Cesàro summability, Wijsman strongly I-lacunary summability, ... The second laplace-beltrami operator on rotational hypersurfaces in the euclidean 4-space We consider rotational hypersurface in the four dimensional Euclidean space. We calculate the mean curvature and the Gaussian curvature, and some relations of the rotational hypersurface. Moreover, we define the second ... On I_{σ}-convergence of folner sequence on amenable semigroups (New Trends in Mathematical Sciences, 2018-04) In this paper, the concepts of-uniform density of subsetsAof the setof positive integers and corresponding-convergence of functions defined on discrete countable amenable semigroups were introduced. Furthermore, for any ... I-cesaro summability of a sequence of order α of random variables in probability (Fundamental Journal of Mathematics and Applications, 2018-12) In this paper, we define four types of convergence of a sequence of random variables, namely, I-statistical convergence of order a, I-lacunary statistical convergence of order a, strongly I-lacunary convergence of order ... Weierstrass representation degree and classes of the surfaces in the four dimensional Euclidean space (Celal Bayar University, 2017) We study two parameters families of Bour-type and Enneper-type minimal surfaces using the Weierstrass representation in the four dimensional Euclidean space. We obtain implicit algebraic equations, degree and classes of ...	{"url":"https://acikerisim.bartin.edu.tr/handle/11772/2/discover?filtertype_0=dateIssued&filter_relational_operator_0=equals&filter_0=%5B2010+TO+2019%5D&filtertype=author&filter_relational_operator=equals&filter=Ki%C5%9Fi%2C+%C3%96mer","timestamp":"2024-11-13T13:08:47Z","content_type":"text/html","content_length":"56448","record_id":"<urn:uuid:b520ba73-4327-4b5e-8780-006020cf7c1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00782.warc.gz"}
rmthm – Use a roman font for theorem statements Modifies the theorem environment of LaTeX 2.09 to cause the statement of a theorem to appear in \rm. LaTeX 2e users should use a more general package (such as ntheorem) to support this requirement. Sources /macros/latex209/contrib/misc/rmthm.sty Version 1991-06-14 Maintainer Nico Verwer Topics Maths theorem Community Comments Maybe you are interested in the following packages as well.	{"url":"https://ctan.org/pkg/rmthm","timestamp":"2024-11-05T19:42:31Z","content_type":"text/html","content_length":"14645","record_id":"<urn:uuid:eaa613c2-5d9f-47b9-b0b3-da6871ac9dff>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00196.warc.gz"}
WhoMadeWhat – Learn Something New Every Day and Stay Smart How do you make a column calculate in Excel? – Create a table. … – Insert a new column into the table. … – Type the formula that you want to use, and press Enter. … – When you press Enter, the formula is automatically filled into all cells of the column â€” above as well as below the cell where you entered the formula. – Create a table. … – Insert a new column into the table. … – Type the formula that you want to use, and press Enter. … – When you press Enter, the formula is automatically filled into all cells of the column â€” above as well as below the cell where you entered the formula. Subsequently, How do you apply formula to entire column in Excel without dragging? Instead, you can accomplish the same copy with a double-click instead of a drag. Set up your formula in the top cell, position the mouse in the lower right-hand corner of the cell until you see the plus, and double-click. Note that this option can copy the formula down as far as Excel finds data to the left. Also, How do I create a formula for multiple cells in Excel? – Select all the cells where you want to enter the formula. To select non-contiguous cells, press and hold the Ctrl key. – Press F2 to enter the edit mode. – Input your formula in one cell, and press Ctrl + Enter instead of Enter. That’s it! How do I apply a formula to an entire column in Excel? Select the cell with the formula and the adjacent cells you want to fill. Click Home > Fill, and choose either Down, Right, Up, or Left. Keyboard shortcut: You can also press Ctrl+D to fill the formula down in a column, or Ctrl+R to fill the formula to the right in a row. Last Review : 8 days ago. How do you apply a formula to an entire column quickly? Select the cell with the formula and the adjacent cells you want to fill. Click Home > Fill, and choose either Down, Right, Up, or Left. Keyboard shortcut: You can also press Ctrl+D to fill the formula down in a column, or Ctrl+R to fill the formula to the right in a row. How do you apply a formula to an entire column in Excel? Just select the cell F2, place the cursor on the bottom right corner, hold and drag the Fill handle to apply the formula to the entire column in all adjacent cells. How do I apply a formula to an entire column in numbers? Select that cell. You will see a small circle in the bottom-right corner of the cell. Click and drag that down and all cells below will auto-fill with the number 50 (or a formula if you have that in a cell). How do I apply a formula to all cells in Excel? Select the cell with the formula and the adjacent cells you want to fill. Click Home > Fill, and choose either Down, Right, Up, or Left. Keyboard shortcut: You can also press Ctrl+D to fill the formula down in a column, or Ctrl+R to fill the formula to the right in a row. How do I apply a formula to an entire column automatically? Dragging the AutoFill handle is the most common way to apply the same formula to an entire column or row in Excel. Firstly type the formula of =(A13+8)/5 in Cell C1, and then drag the AutoFill Handle down to the bottom in Column C, then the formula of =(A13+8)/5 is applied in the whole Column C. How do I apply a formula to an entire column except the first row? Select the header or the first row of your list and press Shift + Ctrl + â†“(the drop down button), then the list has been selected except the first row. How do I keep the formula from dragging in Excel? Select the cell with the formula and the adjacent cells you want to fill. Click Home > Fill, and choose either Down, Right, Up, or Left. Keyboard shortcut: You can also press Ctrl+D to fill the formula down in a column, or Ctrl+R to fill the formula to the right in a row. How do I automatically insert rows in Excel and keep formulas? – Auto fill formula when inserting blank rows with creating a table. – Auto fill formula when inserting blank rows with VBA code. – Select the data range that you want to auto fill formula, and then click Insert > Table, see screenshot: How do I lock and drag formulas in Excel? Drag or copy formula and lock the cell value with the F4 key For locking the cell reference of a single formula cell, the F4 key can help you easily. Select the formula cell, click on one of the cell reference in the Formula Bar, and press the F4 key. Then the selected cell reference is locked. How do I autofill a new row in Excel with formulas? – Step 1: In excel ribbon, click Insert->Table. – Step 2: In pops up ‘Create Table’ dialog, select the table range ($A$1:$C$6 in this case) as your table. … – Step 3: Click OK. … – Step 4: Insert a new row for test. How do I apply a formula to an entire column on a Mac? – You can also press Ctrl+D to fill the formula down in a column. First select the cell that has the formula you want to fill, then select the cells underneath it, and then press Ctrl+D. – You can also press Ctrl+R to fill the formula to the right in a row. How do I keep a formula from moving in Excel? Keep formula cell reference constant with the F4 key 1. Select the cell with the formula you want to make it constant. 2. In the Formula Bar, put the cursor in the cell which you want to make it constant, then press the F4 key. How do I apply a formula to multiple cells in numbers? – Select all the cells where you want to enter the formula. To select non-contiguous cells, press and hold the Ctrl key. – Press F2 to enter the edit mode. – Input your formula in one cell, and press Ctrl + Enter instead of Enter. That’s it! Why do my formulas keep changing in Excel? The behaviour you are seeing is as designed. It is related to editing options. Excel sees that you are entering data in cells that are adjacent to, but not included in the range of the formula, and expands the range in the formula as an aid for you. How do I keep a formula in Excel when adding a row? – Insert the new row. – Copy the source row. – Select the newly created target row, right click and paste special. – Paste as formulas. [advanced_iframe use_shortcode_attributes_only=”true” src=”about:blank” height=”800″ width=”800″ change_parent_links_target=”a#link1″ show_iframe_as_layer=”external” enable_ios_mobile_scolling= Spread the word ! Don’t forget to share.	{"url":"https://whomadewhat.org/how-do-you-make-a-column-calculate-in-excel/","timestamp":"2024-11-14T07:11:31Z","content_type":"text/html","content_length":"51562","record_id":"<urn:uuid:11b74d9a-0297-4473-902a-2b84ad817781>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00684.warc.gz"}
What’s Cooler Than Being Cool? You may have seen some news reports over the last week or two saying that scientists had made a substance with the hottest temperature ever recorded — but that temperature was somehow below absolute zero, a negative temperature on the Kelvin scale. Weirdly enough, this is absolutely true. A lot of the other stuff that showed up in those stories was completely false — my favorite was a statement that this would let us build a 100% efficient engine, breaking the laws of thermodynamics — but there is such a thing as a negative absolute temperature, and those negative temperatures are hotter than any positive temperature. In fact, these scientists pushed a substance a few billionths of a Kelvin below absolute zero, which is far and away the hottest temperature ever recorded. But the surprise is that they managed to do it with this substance in this particular way, not that negative temperatures are so hot. The idea of negative absolute temperature has been around for decades, and this isn’t the first substance to be prodded into a negative temperature. So what is “negative absolute temperature”? High school physics and chemistry teach us that temperature is a measure of molecular motion, or molecular kinetic energy: the faster the molecules in a substance are moving, the hotter that substance is. Contrariwise, as you make something cooler, its constituent molecules move more and more slowly, until they stop moving altogether, at which point you’ve reached absolute zero, 0 This definition of temperature worked pretty well until quantum mechanics showed up in the early 20th century. (Quantum mechanics: screwing things up for a hundred years.) The problem is that, according to quantum mechanics, nothing ever really stops moving. This falls directly out of the Heisenberg Uncertainty Principle, which says that you can’t measure both a particle’s position and its momentum to arbitrary precision.1 If you got something to stop moving altogether, you’d know its momentum with perfect precision and its position with some finite precision, and that’s Not Allowed. So quantum mechanics says that, in general, molecules still have kinetic energy when they hit absolute zero — this is the “zero-point energy” that you may have heard about elsewhere (and that’s another subject that’s prone to silly & overblown reporting with breathless statements about breaking the laws of thermodynamics).2 This is a problem, though, because now we need a new definition of “absolute zero” — if it’s not the temperature where things stop moving, then what is it? The best way to preserve the idea of “absolute zero” while still remaining consistent with quantum mechanics is a pretty intuitive re-definition: absolute zero is just the temperature where all the heat energy that you can possibly get out of the system has been taken from it; i.e. the molecules have as little kinetic energy as they can possibly have. But this minimum energy, the zero-point energy, varies from substance to substance — a molecule in a brick will have a different amount of zero-point energy than a molecule in a block of ice, and so on. So you can’t just point to the kinetic energy of the molecules in a substance and “read” the temperature off of that, because different substances are working off of a different baseline of minimum kinetic energy per molecule. This means temperature can’t be a measure of the kinetic energy of molecules anymore. So quantum mechanics forces us to re-define “temperature” too. What’s the new definition of temperature? Quantum mechanics famously states that the amount of energy a molecule (or atom, or electron, or whatever) can have is quantized — it comes in tiny packets of a particular size. The new definition for temperature relies on this. This is where things get a little tricky, and we need an analogy. Instead of talking about the kinetic energy of a collection of molecules, let’s talk about a bunch of rock climbers climbing up a cliff face. Better yet, there are many cliffs, each with lots of climbers on them, and some are climbing up and and some are climbing down. The cliffs are also infinitely high — climbers don’t actually reach the top, they just stop at some point and go back down. The climbers are our molecules, their height on their cliff is their kinetic energy, and each cliff corresponds to a diffWerner came along and screwed everything up, it did work. When Werner showed up, he made your life (and the lives of the climbers) totally miserable. First, he made every cliff face totally sheer — no purchase anywhere at all. Then he attached a ladder to each cliff — but the ladders were all different. Some of them had huge spacings between their rungs, while some of them had tiny spacings; some had their first rung way off the ground, while others had their first rung nearly touching the ground (i.e. the zero-point energy varies from substance to substance). And while most of the ladders were infinitely long, just like the cliff faces, others weren’t — they just stopped partway up the cliff, so you couldn’t go any higher. But that wasn’t even the worst of it. Werner put some kind of spell on the climbers, forcing them to remain on the ladders for all eternity. They could never go lower than the lowest rung, and they could never leave the cliff faces. So now you have a serious problem. You could still try to measure the average height of climbers on each cliff, but you’re not going to get the same information out of it. Some climbers are on ladders with the first rung way off the ground, and other climbers are on ladders that have their first rung very close to the ground, so it’s not fair to just measure their heights off the ground and be done with it. And since the distance between rungs also varies, you can’t just measure the average distance with respect to the first rung, especially if all the climbers are only on the first few rungs. So what do you do? You decide that what really matters is how the climbers are arranged on the rungs. It stands to reason that the more climbers there are on lower rungs, the harder the ladder is to climb. Thinking a little more carefully, you decide that, in the simple case of a finite ladder that’s really easy to climb, every climber would reach the top, then turn around and climb back down, meaning that you’d have about the same number of climbers on every rung. So if a ladder that’s really hard to climb has everyone on the bottom rung, and a really easy ladder would have an equal number of people on every rung, all you have to do is figure out a way to “measure” between these two extremes, and you’ll have a new way of determining which ladders are easy to climb (hot) and which ones are hard to climb (cold). As it turns out that there’s a (relatively) easy way to do this with math, even though most of the ladders are infinitely long. So you try this out, and lo, it works! You can use the same formula for the finite and the infinite ladders, and surprisingly, when you use it on the special cliffs that Werner hardly affected, you get the same answer that you did before he came along. (In other words, the new quantum-mechanical version of temperature gives the same answer as the old-fashioned version of temperature when you’re not in the quantum regime.) So the new definition for temperature (difficulty of climbing a cliff/ladder) relies on the arrangement of molecules (climbers) in their various possible discrete energy states (rungs on the ladder) that are allowed them by quantum mechanics (Werner, the jerk). If the molecules in a substance are all clustered down near their minimum energy (bottom of the ladder), then the substance is cold — but if they’re spread out more evenly among the allowed states (rungs), then the substance is hot. Now we have a definition of temperature that plays nicely with quantum mechanics. Back to the original question: what is negative absolute temperature? Back to the cliffs. Some of the ladders are finite. On those ladders, there really can be an equal number of climbers on each rung, the theoretical ideal of the easiest (hottest) ladder. If that happened, those ladders would be marked as “infinitely easy” (∞ Kelvin). But one day, when you look at one of the finite ladders, you notice something weird: the climbers have organized a conference at the top of the ladder. So there’s nobody at the bottom, and nearly everyone on the few rungs closest to the top. Therefore, your algorithm should say that while the conference is going on, the ladder is easier than the easiest possible ladder (i.e. hotter than ∞ K)! Curious, you punch this arrangement of climbers into your algorithm, half-expecting your calculator to go up in smoke. Instead, you find something truly strange: the ease of this ladder comes out to be negative. Going back to the real world: when there are more molecules near the top of the “ladder” of allowed energies than there are at the bottom, then a substance is hotter than ∞ K. For mathematical reasons, this means the temperature of the substance is said to be negative. What does this really mean? It means that if you artificially pump all the molecules in a substance up to their most energetic state, to the point where you have more molecules near the top than the bottom of the allowed range, then you’ve heated that substance to a negative temperature. Is there a reason this is called a negative temperature that goes beyond the mathematical? Sort of. The true definition of temperature, which comes from statistical mechanics, has to do with the relationship between energy and entropy (i.e. disorder). This sounds arcane, but in 99% of cases, this ends up reducing to exactly the same familiar definition of temperature as the average energy of molecular motion. As the molecules spread out across the ladder, occupying more rungs, the entropy of the system goes up, because there’s more disorder — stuff is spread out across the range of allowed energies, rather than clustered in one place. As you add more energy to the system, the entropy goes up. Temperature is inversely proportional to the increase in entropy as you add energy to the system. When your system doesn’t have much energy, entropy is low because the molecules are all at roughly the same (low) energy. A little bit of extra energy goes a long way when there’s not much to begin with, so the entropy of the system goes up a lot as you add a little energy, and therefore the temperature is low (because it’s inversely proportional to the change in entropy). On the other hand, when there’s lots of energy in the system, the molecules have a wide range of energies, meaning entropy is high, and adding a little more energy only causes the entropy to increase a little bit. Therefore, the temperature is high. But if there’s a maximum energy — if the ladder has a top — and if all the molecules are clustered way up there near the maximum, then entropy is low because the energies of the molecules are clustered, and adding energy actually clusters the molecules’ energies more, decreasing the entropy. Since temperature is inversely proportional to the increase in entropy as you add energy, decreasing entropy means that you get a negative temperature. Finally: how can there be a top to the “ladder” of allowed energies for molecules? The short answer is “quantum mechanics,” and the long answer is long. Very long. Well beyond the scope of this post. The important thing to remember is this: absolute zero is definitely the coldest possible temperature, but it’s not the lowest possible temperature — there are temperatures lower than absolute zero, all of which are intensely hot, due to the true nature of temperature. 1. Put the philosophical issues with that off to the side: I have some qualms with that way of stating the HUP, but that doesn’t really matter for now. [↩] 2. Zero-point energy is also a real favorite among crackpots and charlatans, as you can see from the comments in the previous link. In fact, I’ll go out on a limb and predict right now that I’m likely to get crackpot comments on this post simply because I’m using the phrase “zero-point energy” three times. [↩] One thought on “What’s Cooler Than Being Cool?” 1. Best teaching analogy ever!	{"url":"http://freelanceastrophysicist.com/2013/01/whats-cooler-than-being-cool/","timestamp":"2024-11-09T10:39:39Z","content_type":"text/html","content_length":"57497","record_id":"<urn:uuid:d990a57d-cf48-4026-becc-9c8d90dd1609>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00071.warc.gz"}
In mathematics, a topological space is an ordered pair ${\displaystyle (X,{\mathcal {T}})}$ where ${\displaystyle X}$ is a set and ${\displaystyle {\mathcal {T}}}$ is a certain collection of subsets of ${\displaystyle X}$ called the open sets or the topology of ${\displaystyle X}$. The topology of ${\displaystyle X}$ introduces an abstract structure of space in the set ${\displaystyle X}$, which allows to define general notions such as of a point being surrounded by a set (by a neighborhood) or belonging to its boundary, of convergence of sequences of elements of ${\displaystyle X}$, of connectedness, of a space or set being contractible, etc. A topological space is an ordered pair ${\displaystyle (X,{\mathcal {T}})}$ where ${\displaystyle X}$ is a set and ${\displaystyle {\mathcal {T}}}$ is a collection of subsets of ${\displaystyle X}$ (i.e., any element ${\displaystyle A\in {\mathcal {T}}}$ is a subset of X) with the following three properties: 1. ${\displaystyle X}$ and ${\displaystyle \varnothing }$ (the empty set) are in ${\displaystyle {\mathcal {T}}}$ 2. The union of any family (infinite or otherwise) of elements of ${\displaystyle {\mathcal {T}}}$ is again in ${\displaystyle {\mathcal {T}}}$ 3. The intersection of two elements of ${\displaystyle {\mathcal {T}}}$ is again in ${\displaystyle {\mathcal {T}}}$ Elements of the set ${\displaystyle {\mathcal {T}}}$ are called open sets of ${\displaystyle X}$. We often simply write ${\displaystyle X}$ instead of ${\displaystyle (X,{\mathcal {T}})}$ once the topology ${\displaystyle {\mathcal {T}}}$ is established. Once we have a topology in ${\displaystyle X}$, we define the closed sets of ${\displaystyle X}$ to be the complements (in ${\displaystyle X}$) of the open sets; the closed sets of ${\displaystyle X} $ have the following characteristic properties: 1. ${\displaystyle X}$ and ${\displaystyle \varnothing }$ (the empty set) are closed 2. The intersection of any family of closed sets is closed 3. The union of two closed sets is closed Alternatively, notice that we could have defined a structure of closed sets (having the properties above as axioms) and defined the open sets relative to that structure as complements of closed sets. Then such a family of open sets obeys the axioms for a topology; we obtain a one to one correspondence between topologies and structures of closed sets. Similarly, the axioms for systems of neighborhoods (described below) give rise to a collection of "open sets" verifying the axioms for a topology, and conversely --- every topology defines the systems of neighborhoods; for every set ${\ displaystyle X}$ we obtain a one to one correspondence between topologies in ${\displaystyle X}$ and systems of neighborhoods in ${\displaystyle X}$. These correspondences allow one to study the topological structure from different viewpoints. The category of topological spaces Given that topological spaces capture notions of geometry, a good notion of isomorphism in the category of topological spaces should require that equivalent spaces have equivalent topologies. The correct definition of morphisms in the category of topological spaces is the continuous homomorphism. A function ${\displaystyle f:X\to Y}$ is continuous if ${\displaystyle f^{-1}(V)}$ is open in ${\displaystyle X}$ for every open in ${\displaystyle Y}$. Continuity can be shown to be invariant with respect to the representation of the underlying topology; e.g., if ${\displaystyle f^{-1}(Z)}$ is closed in ${\displaystyle X}$ for each closed subset ${\displaystyle Z}$ of Y, then ${\displaystyle f}$ is continuous in the sense just defined, and conversely. Isomorphisms in the category of topological spaces (often referred to as a homeomorphism) are bijective and continuous with continuous inverses. The category of topological spaces has a number of nice properties; there is an initial object (the empty set), subobjects (the subspace topology) and quotient objects (the quotient topology), and products and coproducts exist as well. The necessary topologies to define on the latter two objects become clear immediately; if they're going to be universal in the category of topological spaces, then the topologies should be the coarsest making the canonical maps commute. 1. Let ${\displaystyle X=\mathbb {R} }$ where ${\displaystyle \mathbb {R} }$ denotes the set of real numbers. The open interval ]a, b[ (where a < b) is the set of all numbers between a and b: ${\displaystyle {\mathopen {]}}a,b{\mathclose {[}}=\{y\in \mathbb {R} \mid a<y<b\}.}$ Then a topology ${\displaystyle {\mathcal {T}}}$ can be defined on ${\displaystyle X=\mathbb {R} }$ to consist of ${\displaystyle \emptyset }$ and all sets of the form: ${\displaystyle \bigcup _{\gamma \in \Gamma }{\mathopen {]}}a_{\gamma },b_{\gamma }{\mathclose {[}},}$ where ${\displaystyle \Gamma }$ is any arbitrary index set, and ${\displaystyle a_{\gamma }}$ and ${\displaystyle b_{\gamma }}$ are real numbers satisfying ${\displaystyle a_{\gamma }<b_{\gamma }}$ for all ${\displaystyle \gamma \in \Gamma }$. This is the familiar topology on ${\displaystyle \mathbb {R} }$ and probably the most widely used in the applied sciences. However, in general one may define different inequivalent topologies on a particular set ${\displaystyle X}$ and in the next example another topology on ${\displaystyle \mathbb {R} }$, albeit a relatively obscure one, will be 2. Let ${\displaystyle X=\mathbb {R} }$ as before. Let ${\displaystyle {\mathcal {T}}}$ be a collection of subsets of ${\displaystyle \mathbb {R} }$ defined by the requirement that ${\displaystyle A\ in {\mathcal {T}}}$ if and only if ${\displaystyle A=\emptyset }$ or ${\displaystyle A}$ contains all except at most a finite number of real numbers. Then it is straightforward to verify that ${\ displaystyle {\mathcal {T}}}$ defined in this way has the three properties required to be a topology on ${\displaystyle \mathbb {R} }$. This topology is known as the cofinite topology or Zariski 3. Every metric ${\displaystyle d}$ on ${\displaystyle X}$ gives rise to a topology on ${\displaystyle X}$. The open ball with centre ${\displaystyle x\in X}$ and radius ${\displaystyle r>0}$ is defined to be the set ${\displaystyle B_{r}(x)=\{y\in X\mid d(x,y)<r\}.}$ A set ${\displaystyle A\subset X}$ is open if and only if for every ${\displaystyle x\in A}$, there is an open ball with centre ${\displaystyle x}$ contained in ${\displaystyle A}$. The resulting topology is called the topology induced by the metric ${\displaystyle d}$. The standard topology on ${\displaystyle \mathbb {R} }$, discussed in Example 1, is induced by the metric ${\displaystyle d 4. For a given set ${\displaystyle X}$, the family ${\displaystyle {\mathcal {T}}=\{\emptyset ,X\}}$ is a topology: the indiscrete or weakest topology. 5. For a given set ${\displaystyle X}$, the family ${\displaystyle {\mathcal {T}}={\mathcal {P}}X}$ of all subsets of ${\displaystyle X}$ is a topology: the discrete topology. Given a topological space ${\displaystyle (X,{\mathcal {T}})}$ of opens, we say that a subset ${\displaystyle N}$ of ${\displaystyle X}$ is a neighborhood of a point ${\displaystyle x\in X}$ if ${\ displaystyle N}$ contains an open set ${\displaystyle U\in {\mathcal {T}}}$ containing the point ${\displaystyle x}$ ^[1] If ${\displaystyle N_{x}}$ denotes the system of neighborhoods of ${\displaystyle x}$ relative to the topology ${\displaystyle {\mathcal {T}}}$, then the following properties hold: 1. ${\displaystyle N_{x}}$ is not empty for any ${\displaystyle x\in X}$ 2. If ${\displaystyle U}$ is in ${\displaystyle N_{x}}$ then ${\displaystyle x\in U}$ 3. The intersection of two elements of ${\displaystyle N_{x}}$ is again in ${\displaystyle N_{x}}$ 4. If ${\displaystyle U}$ is in ${\displaystyle N_{x}}$ and ${\displaystyle V\subset X}$ contains ${\displaystyle U}$, then ${\displaystyle V}$ is again in ${\displaystyle N_{x}}$ 5. If ${\displaystyle U}$ is in ${\displaystyle N_{x}}$ then there exists a ${\displaystyle V\in N_{x}}$ such that ${\displaystyle V}$ is a subset of ${\displaystyle U}$ and ${\displaystyle U\in N_ {y}}$ for all ${\displaystyle y\in V}$ Conversely, if we define a topology of neighborhoods on ${\displaystyle X}$ via the above properties, then we can recover a topology of opens whose neighborhoods relative to that topology give rise to the neighborhood topology we started from: ${\displaystyle U}$ is open if it is in ${\displaystyle N_{x}}$ for all ${\displaystyle x\in U}$. Moreover, the opens relative to a topology of neighborhoods form a topology of opens whose neighborhoods are the same as those we started from. All this just means that a given topological space is the same, regardless of which axioms we choose to start from. The neighborhood axioms lend themselves especially well to the study of topological abelian groups and topological rings because knowing the neighborhoods of any point is equivalent to knowing the neighborhoods of 0 (since the operations are presumed continuous). For example, the ${\displaystyle I}$-adic topology on a ring ${\displaystyle A}$ is Hausdorff if and only if ${\displaystyle \bigcap I^{n}=0}$, thus a topological property is equivalent to an algebraic property which becomes clear when thinking in terms of neighborhoods. Bases and sub-bases A basis for the topology ${\displaystyle {\mathcal {T}}}$ on X is a collection ${\displaystyle {\mathcal {B}}}$ of open sets such that every open set is a union of elements of ${\displaystyle {\ mathcal {B}}}$. For example, in a metric space the open balls form a basis for the metric topology. A sub-basis ${\displaystyle {\mathcal {S}}}$ is a collection of open sets such that the finite intersections of elements of ${\displaystyle {\mathcal {S}}}$ form a basis for ${\displaystyle {\mathcal {T}}}$. Some topological notions This section introduces some important topological notions. Throughout, ${\displaystyle X}$ will denote a topological space with the topology ${\displaystyle {\mathcal {T}}}$. Partial list of topological notions The closure in ${\displaystyle X}$ of a subset ${\displaystyle E}$ is the intersection of all closed sets of ${\displaystyle X}$ which contain ${\displaystyle E}$ as a subset. The interior in ${\displaystyle X}$ of a subset ${\displaystyle E}$ is the union of all open sets of ${\displaystyle X}$ which are contained in ${\displaystyle E}$ as a subset. Limit point A point ${\displaystyle x\in X}$ is a limit point of a subset ${\displaystyle A}$ of ${\displaystyle X}$ if any open set in ${\displaystyle {\mathcal {T}}}$ containing ${\displaystyle x}$ also contains a point ${\displaystyle y\in A}$ with ${\displaystyle yeq x}$. An equivalent definition is that ${\displaystyle x\in X}$ is a limit point of ${\displaystyle A}$ if every neighbourhood of ${\ displaystyle x}$ contains a point ${\displaystyle y\in A}$ different from ${\displaystyle x}$. Open cover A collection ${\displaystyle {\mathcal {U}}}$ of open sets of ${\displaystyle X}$ is said to be an open cover for ${\displaystyle X}$ if each point ${\displaystyle x\in X}$ belongs to at least one of the open sets in ${\displaystyle {\mathcal {U}}}$. A path ${\displaystyle \gamma }$ is a continuous function ${\displaystyle \gamma :[0,1]\rightarrow X}$. The point ${\displaystyle \gamma (0)}$ is said to be the starting point of ${\displaystyle \gamma }$ and ${\displaystyle \gamma (1)}$ is said to be the end point. A path joins its starting point to its end point. Hausdorff/separability property ${\displaystyle X}$ has the Hausdorff (or separability, or T2) property if for any pair ${\displaystyle x,y\in X}$ there exist disjoint sets ${\displaystyle U}$ and ${\displaystyle V}$ with ${\ displaystyle x\in U}$ and ${\displaystyle y\in V}$. ${\displaystyle X}$ is noetherian if it satisfies the descending chain condition for closed set: any descending chain of closed subsets ${\displaystyle Y_{0}\supseteq Y_{1}\supseteq \ldots }$ is eventually stationary; i.e., if there is an index ${\displaystyle i\geq 0}$ such that ${\displaystyle Y_{i+r}=Y_{i}}$ for all ${\displaystyle r\geq 0}$. ${\displaystyle X}$ is connected if given any two disjoint open sets ${\displaystyle U}$ and ${\displaystyle V}$ such that ${\displaystyle X=U\cup V}$, then either ${\displaystyle X=U}$ or ${\ displaystyle X=V}$. ${\displaystyle X}$ is path-connected if for any pair ${\displaystyle x,y\in X}$ there exists a path joining ${\displaystyle x}$ to ${\displaystyle y}$. A path connected topological space is also connected, but the converse need not be true. ${\displaystyle X}$ is said to be compact if any open cover of ${\displaystyle X}$ has a finite sub-cover. That is, any open cover has a finite number of elements which again constitute an open cover for ${\displaystyle X}$. A topological space with the Hausdorff, connectedness, path-connectedness property is called, respectively, a Hausdorff (or separable), connected, path-connected topological space. Induced topologies A topological space can be used to define a topology on any particular subset or on another set. These "derived" topologies are referred to as induced topologies. Descriptions of some induced topologies are given below. Throughout, ${\displaystyle (X,{\mathcal {T}}_{X})}$ will denote a topological space. Some induced topologies Relative topology If ${\displaystyle A}$ is a subset of ${\displaystyle X}$ then open sets may be defined on ${\displaystyle A}$ as sets of the form ${\displaystyle U\cap A}$ where ${\displaystyle U}$ is any open set in ${\displaystyle {\mathcal {T}}_{X}}$. The collection of all such open sets defines a topology on ${\displaystyle A}$ called the relative topology of ${\displaystyle A}$ as a subset of ${\ displaystyle X}$ Quotient topology If ${\displaystyle Y}$ is another set and ${\displaystyle q}$ is a surjective function from ${\displaystyle X}$ to ${\displaystyle Y}$ then open sets may be defined on ${\displaystyle Y}$ as subsets ${\displaystyle U}$ of ${\displaystyle Y}$ such that ${\displaystyle q^{-1}(U)=\{x\in X\mid q(x)\in U\}\in {\mathcal {T}}_{X}}$. The collection of all such open sets defines a topology on ${\ displaystyle Y}$ called the quotient topology induced by ${\displaystyle q}$. Product topology If ${\displaystyle (X_{\lambda },{\mathcal {T}}_{\lambda })_{\lambda \in \Lambda }}$ is a family of topological spaces, then the product topology on the Cartesian product ${\displaystyle \prod _{\ lambda \in \Lambda }X_{\lambda }}$ has as sub-basis the sets of the form ${\displaystyle \prod _{\lambda \in \Lambda }U_{\lambda }}$ where each ${\displaystyle U_{\lambda }\in {\mathcal {T}}_{\lambda }}$ and ${\displaystyle U_{\lambda }=X_{\lambda }}$ for all but finitely many ${\displaystyle \lambda \in \Lambda }$. See also 1. ↑ Some authors use a different definition, in which a neighborhood N of x is an open set containing x. 1. K. Yosida, Functional Analysis (6 ed.), ser. Classics in Mathematics, Berlin, Heidelberg, New York: Springer-Verlag, 1980 2. D. Wilkins, Lecture notes for Course 212 - Topology, Trinity College Dublin, URL: [1]	{"url":"https://en.citizendium.org/wiki/Topological_space","timestamp":"2024-11-07T13:07:27Z","content_type":"text/html","content_length":"218242","record_id":"<urn:uuid:41310e75-bf2f-4f16-8052-c01b3765f841>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00640.warc.gz"}
Jump to navigation Jump to search Interval information Factorization 2^47 × 3^4 × 7^-19 Monzo [47 4 0 -19⟩ Size in cents 0.12778055¢ Name revopentisma Color name ssr^19-8, sasa-neru negative octave FJS name [math]\text{6d}{-8}_{7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7}[/math] Special properties reduced Tenney height (log[2] nd) 106.68 Weil height (log[2] max(n, d)) 106.68 Wilson height (sopfr (nd)) 239 Harmonic entropy ~1.19837 bits (Shannon, [math]\sqrt{nd}[/math]) Comma size unnoticeable open this interval in xen-calc Revopentisma is a 7-limit (and also 2.3.7 subgroup) unnoticeable comma associated with very high accuracy temperaments. Revopentisma is the comma shared by 31edo and 270edo in the 2.3.7 subgroup and is notable as supported by many mega-edos. The associated temperament is named revopent, and it's remarkable in its short generator chains for edos in the thousands, with 7/4 being just four generators down and 3/2 being just nineteen, with octave reduction. Tempering it out in the full 7-limit produces rank-3 revopentismic temperament, as well as the 2.3.7 rank-2 revopentic temperament. The rank-2 revopent temperament adds mapping for 5 where it is reached in 287 generators. Comma's name is derived from two edos in the temperament-merging, number 1848 being associated with a year with a large number of revolutions in Europe, and 3125 being equal to 5 to the power of 5, hence pent- meaning five.	{"url":"https://en.xen.wiki/w/Revopentisma","timestamp":"2024-11-12T15:52:07Z","content_type":"text/html","content_length":"25320","record_id":"<urn:uuid:4c6acef5-223a-4972-8a9d-73722c92295b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00306.warc.gz"}
Increasing Kinetic Energy: Acceleration and Deceleration in context of kinetic energy to work 31 Aug 2024 Increasing Kinetic Energy: Acceleration and Deceleration Kinetic energy is the energy of motion, which is a fundamental concept in physics. As an object moves from rest to motion or vice versa, its kinetic energy changes accordingly. This article explores the relationship between acceleration, deceleration, and kinetic energy, highlighting the formulae that govern these interactions. Acceleration: Increasing Kinetic Energy When an object accelerates, its velocity increases, resulting in a corresponding increase in kinetic energy. The formula for kinetic energy is: K = (1/2)mv^2 where K is the kinetic energy, m is the mass of the object, and v is its velocity. As an object accelerates, its velocity increases, causing its kinetic energy to increase according to the above formula. The rate at which kinetic energy increases depends on the acceleration, which is defined as: a = Δv / Δt where a is the acceleration, Δv is the change in velocity, and Δt is the time over which the acceleration occurs. Deceleration: Decreasing Kinetic Energy Conversely, when an object decelerates, its velocity decreases, resulting in a decrease in kinetic energy. The formula for kinetic energy remains the same: K = (1/2)mv^2 However, as the object decelerates, its velocity decreases, causing its kinetic energy to decrease accordingly. The rate at which kinetic energy decreases depends on the deceleration, which is defined as: a = -Δv / Δt where a is the deceleration, Δv is the change in velocity, and Δt is the time over which the deceleration occurs. Work: Energy Transfer When an object accelerates or decelerates, it performs work on its surroundings. The formula for work is: W = F \* s where W is the work done, F is the force applied, and s is the distance over which the force is applied. As an object accelerates, it performs positive work, transferring energy from its surroundings to itself. Conversely, as it decelerates, it performs negative work, transferring energy from itself to its surroundings. In conclusion, this article has explored the relationship between acceleration, deceleration, and kinetic energy. The formulae for kinetic energy, acceleration, and deceleration have been presented, highlighting the interplay between these concepts. Understanding these relationships is crucial in various fields, including physics, engineering, and sports. • Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons. • Serway, R. A., & Jewett, J. W. (2014). Physics for Scientists and Engineers. Cengage Learning. Related articles for ‘kinetic energy to work ‘ : Calculators for ‘kinetic energy to work ‘	{"url":"https://blog.truegeometry.com/tutorials/education/df2077d57f3e3fbe7041a1b0f674cb8c/JSON_TO_ARTCL_Increasing_Kinetic_Energy_Acceleration_and_Deceleration_in_contex.html","timestamp":"2024-11-02T10:56:53Z","content_type":"text/html","content_length":"17836","record_id":"<urn:uuid:69ca34fe-d193-4b95-b439-3f01ac6f1fb2>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00803.warc.gz"}
Grade 6 Chapter 3 Rational Numbers Exercise 3.1 Solutions - Math Book Answers Grade 6 Chapter 3 Rational Numbers Exercise 3.1 Solutions Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers Page 11 Problem 1 Answer We have given the number as 0.3 Asked to write each rational number in the form a/b To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as 0.3 So here we get the fractional form as 0.3/1=3/10 For the given number 0.3 we get the fractional form as3/10 Page 11 Problem 2 Answer We have given the number as 2 x 7/8 Asked to write each rational number in the form a/b To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as ⇒ $2 \frac{7}{8}$ So here we get the fractional form as ⇒ $2 \frac{7}{8}$ ⇒ $=\frac{23}{8}$ For the given number 2×7/8 we get the fractional form as23/8 Page 11 Problem 3 Answer We have given the number as −5 and Asked you to write each rational number in the form To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as −5 So here we get the fractional form as −5/1 For the given number −5 we get the fractional form as−5/1 Page 11 Problem 4 Answer We have given the number as 16 and Asked to write each rational number in the forma/b To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as 16 So here we get the fractional form as 16/1 For the given number 16 we get the fractional form as16/1 Page 11 Problem 5 Answer We have given the number as −1×3/4 Asked to write each rational number in the forma/b To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as ⇒ $-1 \frac{3}{4}$ So here we get the fractional form as ⇒ $ -1 \frac{3}{4}$ ⇒ $\frac{-7}{4}$ For the given number −1×3/4 we get the fractional form as−7/4 Page 11 Problem 6 Answer We have given the number as −4.5 Asked to write each rational number in the form a/b To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as So here we get the fractional form as ⇒ $\frac{-4.5}{1}$ ⇒ $ \frac{-45}{10}$ ⇒ $ \frac{-9}{2}$ For the given number −4.5 we get the fractional form as−9/2 Page 11 Problem 7 Answer We have given the number as 3 Asked to write each rational number in the form a/b To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as 3 So here we get the fractional form as 3/1 For the given number 3 we get the fractional form as 3/1 Page 11 Problem 8 Answer We have given the number as 0.11 Asked to write each rational number in the form a/b To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number a So here we get the fractional form as ⇒ $\frac{0.11}{1} $ ⇒ $\frac{11}{100}$ For the given number 0.11 we get the fractional form as11/100 Page 11 Problem 9 Answer We have given the number as −13 Asked to place each number in the correct place on the Venn diagram.To get the right place for the number in the diagram we have to analyze the number first. To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as −13 The nature of the given number is an integer The given number −13 is an integer and placed in the diagram is given as Page 11 Problem 10 Answer We have given the number as 1/6 Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first. To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as 1/6 The nature of the given number is a rational number The given number1/6 is a rational number and placed in the diagram is given Page 11 Problem 11 Answer We have given the number as 0. Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first. To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as 0 The nature of the given number is a whole number The given number 0 is a whole number and the place in the diagram is given as Page 11 Problem 12 Answer We have given the number as 0.99 Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first. To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as 0.99 The nature of the given number is rational number The given number 0.99 is a rational number and the place in the diagram is given as Page 11 Problem 13 Answer We have given the number as −6.7 Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first. To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as −6.7 The nature of the given number is a rational number The given number 6.7 is a rational number and its place in the diagram is given as Page 11 Problem 14 Answer We have given the number as 34 Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first. To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as 34 The nature of the given number is a whole number The given number 34 is a whole number and the place in the diagram is given as Page 11 Problem 15 Answer We have given the number as −14x/2 and Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first. To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as −14×1/2 The nature of the given number is a rational number The given number −14×1/2 is a rational number and placed in the diagram is given Page 12 Exercise 1 Answer We have given the number as−12 So here we get the fractional form as−12/1 For the given number −12 we get the fractional form as −12/1 and the nature of the number as Page 12 Exercise 2 Answer We have given the number as 7.3 Asked to write each rational number in the form a/b, then circle the name of each set to which the number belongs. o, get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as So here we get the fractional form as ⇒ $\frac{7.3}{1}$ ⇒ $\frac{73}{10}$ For the given number 7.3 we get the fractional form as73/10 and the nature of the n Page 12 Exercise 3 Answer We have given the number as 0.41 Asked to write each rational number in the form a/b, then circle the name of each set to which the number belongs. To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as So here we get the fractional form as ⇒ $\frac{0.41}{1}$ ⇒ $\frac{41}{100}$ For the given number 0.41 we get the fractional form as 41/100 and the nature of the number as Page 12 Exercise 4 Answer We have given the number 6 Asked to write each rational number in the form a/b, then circle the name of each set to which the number belongs. To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as 6 So here we get the fractional form as 6/1 For the given number 6 we get the fractional form as 6/1 and the nature of the number Page 12 Exercise 5 Answer We have given the number as 3×1/2 Asked to write each rational number in the form a/b, then circle the name of each set to which the number belongs. To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one. We have given the number as ⇒ $3 \frac{1}{2}$ So here we get the fractional form as ⇒ $3 \frac{1}{2}$ ⇒ $\frac{7}{2}$ For the given number 3×1/2 we get the fractional form as 7/2 and the nature of the number as Go Math Answer Key Leave a Comment	{"url":"https://mathbookanswers.com/go-math-practice-fluency-workbook-grade-6-chapter-3-ex-3-1-answer-key/","timestamp":"2024-11-04T01:05:16Z","content_type":"text/html","content_length":"110974","record_id":"<urn:uuid:86da3f89-9246-445a-a78a-de6b83a2c98c>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00716.warc.gz"}
Homoscedasticity - (Intro to Business Analytics) - Vocab, Definition, Explanations \| Fiveable from class: Intro to Business Analytics Homoscedasticity refers to the condition in regression analysis where the variance of the residuals or errors is constant across all levels of the independent variable(s). This concept is crucial for ensuring that the results of regression analyses are reliable and valid, as violations of this assumption can lead to biased estimates and incorrect conclusions. In both simple and multiple linear regression, recognizing and addressing homoscedasticity helps in making sound business decisions based on statistical outputs. congrats on reading the definition of Homoscedasticity. now let's actually learn it. 5 Must Know Facts For Your Next Test 1. Homoscedasticity is essential for ordinary least squares (OLS) regression because it ensures that the estimates are efficient and unbiased. 2. Visual methods, like scatter plots of residuals against predicted values, can help detect homoscedasticity issues. 3. When homoscedasticity is violated, techniques such as weighted least squares or transforming variables may be applied to correct it. 4. In business decision-making, ignoring homoscedasticity can lead to incorrect interpretations of how factors influence outcomes, impacting strategic decisions. 5. Statistical tests like Breusch-Pagan or White's test can be used to formally check for homoscedasticity in regression models. Review Questions • How does homoscedasticity impact the reliability of regression results? □ Homoscedasticity ensures that the variance of errors remains constant across all levels of the independent variables. When this condition is met, it leads to efficient and unbiased parameter estimates in regression analysis. If homoscedasticity is violated, it can result in distorted standard errors and confidence intervals, ultimately affecting the reliability of conclusions drawn from the data. • What are some visual methods used to identify potential issues with homoscedasticity in a regression model? □ To identify potential issues with homoscedasticity, analysts often use scatter plots that display residuals against predicted values or independent variables. In a properly functioning model with homoscedasticity, these residuals should appear randomly scattered around zero without forming any discernible pattern. If a pattern emerges, such as a funnel shape or systematic distribution, it indicates heteroscedasticity, prompting further investigation and potential model adjustments. • Evaluate the implications of ignoring homoscedasticity when interpreting regression results in a business context. □ Ignoring homoscedasticity can lead to significant misinterpretations in a business context, as it may result in invalid conclusions regarding relationships between variables. For instance, if a company relies on flawed regression outputs due to heteroscedastic errors, it might misestimate costs or revenues linked to certain factors. This could lead to poor strategic decisions, inefficient resource allocation, and ultimately affect the companyโ s performance and competitiveness in the market. ยฉ 2024 Fiveable Inc. All rights reserved. APยฎ and SATยฎ are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.	{"url":"https://library.fiveable.me/key-terms/introduction-to-business-analytics/homoscedasticity","timestamp":"2024-11-07T21:57:06Z","content_type":"text/html","content_length":"178623","record_id":"<urn:uuid:c782639e-7bd0-40e4-906b-4f186a29ceec>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00527.warc.gz"}
Demographic and Health Survey 2017-2018 - IPUMS Subset Reference ID National Institute of Population Studies (NIPS) [Pakistan], and ICF., Minnesota Population Center Last modified May 14, 2020 Sample weight for persons (C_PERWEIGHT) Type: Continuous Decimal: 6 Start: 173 End: 180 Width: 8 Range: - Format: Numeric PERWEIGHT (V005) is an 8-digit variable with 6 implied decimal places, which should be used as a weighting factor to produce representative numbers accurately describing the surveyed population. While the DHS Recode Manuals direct the researcher to divide the original weight variable by 1,000,000 before applying the weighting factor to the original DHS data files, it is not necessary to modify the value of PERWEIGHT before applying this weight to cases in IPUMS-DHS. PERWEIGHT should be used to weight nearly all tabulations made using IPUMS-DHS data. Occasionally, as with the domestic violence variables, a subset of respondents are randomly selected to answer questions from a survey module, and a specialized weight such as DVWEIGHT should be used instead. Note: The 6 implied decimal places in PERWEIGHT mean that the last six digits of the eight-digit variable are decimal digits, but there is no actual decimal in the data. var_concept.title Vocabulary Weights and subsample selection Variables -- TOPICS IPUMS Imputation and derivation PERWEIGHT is an 8-digit numeric variable with 6 implied decimal places. See the variable description for directions on the use of PERWEIGHT.	{"url":"https://microdata.worldbank.org/index.php/catalog/3683/variable/C/C_PERWEIGHT?name=C_PERWEIGHT","timestamp":"2024-11-10T20:33:31Z","content_type":"text/html","content_length":"40868","record_id":"<urn:uuid:ac8c89ec-a00b-44e0-a4e5-f85d15ad26a7>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00195.warc.gz"}
t-Test, Chi-Square, ANOVA, Regression, Correlation... Cohen's Kappa Cohen's Kappa is a measure of agreement between two dependent categorical samples, and you use it whenever you want to know if two raters' measurements are in agreement. In the case of Cohen's Kappa, the variable to be measured by the two rates is a nominal variable. So if you have a nominal variable and you want to know how much agreement there is between two raters, you would use Cohen's Kappa. If you have an ordinal variable and two raters, you would use Kendall's tau or the weighted Cohen's Kappa, and if you have a metric variable, you would use Pearson's correlation. If you have more than two nominal dependent samples, the Fleiss Kappa is used. Cohen's Kappa Example Let's say you have developed a measurement tool, for example a questionnaire, that doctors can use to determine whether a person is depressed or not. Now you give this tool to a doctor and ask her to assess 50 people with it. For example, your method shows that the first person is depressed, the second person is depressed and the third person is not depressed. The big question now is: Will a second doctor come to the same So, with a second doctor, the result could now look like this: For the first person, both doctors come to the same result, but for the second person, the result differs. You're interested in how big the agreement of the doctors are, and this is where Cohen's Kappa comes in. Inter-rater reliability If the assessments of the two doctors agree very well, the inter-rater reliability is high. And it is this inter-rater reliability that is measured by Cohen's Kappa. Cohen's Kappa (κ) is a statistical measure used to quantify the level of agreement between two raters (or judges, observers, etc.) who each classify items into categories. It's especially useful in situations where decisions are subjective and the categories are nominal (i.e., they do not have a natural order). Cohen's Kappa is therefore a measure of how reliably two raters measure the same thing. Use cases for Cohen's Kappa So far we have considered the case where two people measure the same thing. However, Cohen's Kappa can also be used when the same rater makes the measurement at two different times. In this case, the Cohen's Kappa score indicates how well the two measurements from the same person agree. Measuring the agreement Cohen's Kappa measures the agreement between two dependent categorical samples. Cohen's Kappa reliability and validity It is important to note that the Cohen's Kappa coefficient can only tell you how reliably both raters are measuring the same thing. It does not tell you whether what the two raters are measuring is the right thing! In the first case we speak of reliability (whether both are measuring the same thing) and in the second case we speak of validity (whether both are measuring the right thing). Cohen's Kappa can only be used to measure reliability. Calculate Cohen's Kappa Now the question arises, how is Cohen's Kappa calculated? This is not difficult! We create a table with the frequencies of the corresponding answers. For this we take our two raters, each of whom has rated whether a person is depressed or not. Now we count how often both have measured the same and how often not. So we make a table with Rater 1 with "not depressed" and "depressed" and Rater 2 with "not depressed" and "depressed". Now we simply keep a tally sheet and count how often each combination occurs. Let's say our final result is as follows: 17 people rated both raters as "not depressed." For 19 people, both chose the rating "depressed." So if both raters measured the same thing, that person is on the diagonal, if they measured something different, that person is on the edge. Now we want to know how often both raters agree and how often they don't. Rater 1 and Rater 2 agree that 17 patients are not depressed and 19 are depressed. So both raters agree in 36 cases. In total, 50 people were assessed. With these numbers, we can now calculate the probability that both raters are measuring the same thing in a person. We do this by dividing 36 by 50. This gives us the following result: In 72% of the cases, both raters assess the same, in 28% of the cases they rate it differently. This gives us the first part we need to calculate Cohen's Kappa. Cohen's Kappa is given by this formula: So we just calculated p[o], what is p[e]? If both doctors were to answer the question of whether a person is depressed or not purely by chance, by simply tossing a coin, they would probably come to the same conclusion in some cases, purely by chance. And that is exactly what p[e] indicates: The hypothetical probability of a random match. But how do you calculate p[e]? To calculate p[e], we first need the sums of the rows and columns. Then we can calculate p[e]. In the first step, we calculate the probability that both raters would randomly arrive at the rating "not depressed." • Rater 1 rated 25 out of 50 people as "not depressed", i.e. 50%. • Rater 2 rated 23 out of 50 people as "not depressed", i.e. 46%. The overall probability that both raters would say "not depressed" by chance is: 0.5 * 0.46 = 0.23 In the second step, we calculate the probability that the raters would both say "depressed" by chance. • Rater 1 says "depressed" in 25 out of 50 persons, i.e. 50%. • Rater 2 says "depressed" in 27 out of 50 people, i.e. 54%. The total probability that both raters say "depressed" by chance is: 0.5 * 0.54 = 0.27. Now we can calculate p[e]. If both values are now added, we get the probability that the two raters coincidentally agree. p[e] is therefore 0.23 + 0.27 which is equal to 0.50. Therefore, if the doctors had no guidance and simply rolled the dice, the probability of such a match is 50%. Now we can calculate the Cohen's Kappa coefficient. We simply substitute p[o] and p[e] and we get a Kappa value of 0.4 in our example. By the way, in p[o] the o stands for "observed". And in p[e], the e stands for "expected". Therefore, p[o] is what we actually observed and p[e] is what we would expect if it were purely random. Cohen's Kappa interpretation Now, of course, we would like to interpret the calculated Cohens Kappa coefficient. The table of Landis & Koch (1977) can be used as a guide. >0.8 Almost Perfect >0.6 Substantial >0.4 Moderate >0.2 Fair 0-0,2 Slight <0 Poor Therefore, the calculated Cohen's Kappa coefficient of 0.44 indicates moderate reliability or agreement. Cohen's Kappa Standard Error (SE) The Standard Error (SE) of a statistic, like Cohen's Kappa, is a measure of the precision of the estimated value. It indicates the extent to which the calculated value would vary if the study were repeated multiple times on different samples from the same population. Therefore it is a measure of the variability or uncertainty around the Kappa statistic estimate. Calculating Standard Error of Cohen's Kappa: The calculation of the SE for Cohen's Kappa involves somewhat complex formulas that account for the overall proportions of each category being rated and the distribution of ratings between the raters. The general formula for the SE of Cohen's Kappa is: Where n is the total number of items being rated. Interpreting Standard Error Small Standard Error: A small SE suggests that the sample estimate is likely to be close to the true population value. The smaller the SE, the more precise the estimate is considered to be. Large Standard Error: A large SE indicates that there is more variability in the estimates from sample to sample and, therefore, less precision. It suggests that if the study were repeated, the resulting estimates could vary widely. Weighted Cohen's Kappa Cohen's Kappa takes into account the agreement between two raters, but it is only relevant whether both raters measure the same or not. In the case of an ordinal variable, i.e. a variable with a ranking, such as school grades, it is of course desirable that the gradations are also considered. A difference between "very good" and "satisfactory" is greater than between "very good" and "good". To take this into account, the weighted Kappa can be calculated. Here, the deviation is included in the calculation. The differences can be taken into account linearly or quadratically. Calculate Cohen's Kappa with DATAtab Now we will discuss how you can easily calculate Cohen's Kappa for your data online using DATAtab. Simply go to the Cohen's Kappa calculator and copy your own data into the table. Now click on the tab "Reliability". All you have to do is click on the variables you want to analyse and Cohen's Kappa will be displayed automatically. First you will see the crosstab and then you can read the calculated Cohen's Kappa coefficient. If you don't know how to interpret the result, just click on interpretations in words. An inter-rater reliability analysis was performed between the dependent samples Rater1 and Rater2. For this, Cohen's Kappa was calculated, which is a measure of the agreement between two related categorical samples. The Cohen's Kappa showed that there was moderate agreement between the samples Rater1 and Rater2 with κ= 0.23. Statistics made easy • many illustrative examples • ideal for exams and theses • statistics made easy on 412 pages • 5rd revised edition (April 2024) • Only 8.99 € Free sample "It could not be simpler" "So many helpful examples"	{"url":"https://datatab.net/tutorial/cohens-kappa","timestamp":"2024-11-04T23:54:29Z","content_type":"text/html","content_length":"70055","record_id":"<urn:uuid:fb0d87ed-5e7c-4d45-bf1b-8b49e8404d38>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00548.warc.gz"}
Erik D. Demaine Paper by Erik D. Demaine Erik D. Demaine, Francisco Gomez-Martin, Henk Meijer, David Rappaport, Perouz Taslakian, Godfried T. Toussaint, Terry Winograd, and David R. Wood, “The Distance Geometry of Deep Rhythms and Scales”, in Proceedings of the 17th Canadian Conference on Computational Geometry (CCCG 2005), Windsor, Ontario, Canada, August 10–12, 2005, pages 163–166. We characterize which sets of k points chosen from n points spaced evenly around a circle have the property that, for each i = 1, 2, …, k − 1, there is a nonzero distance along the circle that occurs as the distance between exactly i pairs from the set of k points. Such a set can be interpreted as the set of onsets in a rhythm of period n, or as the set of pitches in a scale of n tones, in which case the property states that, for each i = 1, 2, …, k − 1, there is a nonzero time [tone] interval that appears as the temporal [pitch] distance between exactly i pairs of onsets [pitches]. Rhythms with this property are called Erdős-deep. The problem is a discrete, one-dimensional (circular) analog to an unsolved problem posed by Erdős in the plane. The paper is 4 pages. The paper is available in PostScript (324k), gzipped PostScript (148k), and PDF (184k). Related papers: DeepRhythms_CGTA (The Distance Geometry of Music) See also other papers by Erik Demaine. These pages are generated automagically from a BibTeX file. Last updated July 23, 2024 by Erik Demaine.	{"url":"https://erikdemaine.org/papers/DeepRhythms_CCCG2005/","timestamp":"2024-11-04T13:47:38Z","content_type":"text/html","content_length":"5926","record_id":"<urn:uuid:d26be1c4-d9e6-4312-bd3d-480e1bd47d82>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00251.warc.gz"}
Differentiating Inverse Trigonometric Functions Q&As - Calculus \| HIX Tutor Differentiating Inverse Trigonometric Functions Understanding inverse trigonometric functions is essential in mathematics, particularly in calculus and trigonometry. These functions provide a means to find angles associated with specific trigonometric ratios. Differentiating inverse trigonometric functions involves applying differentiation rules to these functions, which allows for the calculation of rates of change and slopes in various contexts, such as physics, engineering, and optimization problems. Mastery of this topic is crucial for solving advanced mathematical problems and gaining insight into the behavior of trigonometric functions and their inverses.	{"url":"https://tutor.hix.ai/subject/calculus/differentiating-inverse-trigonometric-functions","timestamp":"2024-11-14T05:34:45Z","content_type":"text/html","content_length":"561758","record_id":"<urn:uuid:504063c6-367e-4d87-8007-afcba35e8fd5>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00621.warc.gz"}
Angles In A Triangle Worksheet Answers With Work - TraingleWorksheets.com Angles In A Triangle Worksheet With Answers – Triangles are one of the most fundamental forms in geometry. Understanding triangles is important for learning more advanced geometric concepts. In this blog it will explain the different types of triangles including triangle angles and the methods to determine the size and perimeter of a triangle, and present specific examples on each. Types of Triangles There are three kinds of triangulars: Equilateral, isosceles, and scalene. Equilateral triangles are … Read more	{"url":"https://www.traingleworksheets.com/tag/angles-in-a-triangle-worksheet-answers-with-work/","timestamp":"2024-11-10T19:08:31Z","content_type":"text/html","content_length":"48481","record_id":"<urn:uuid:d3b9278d-4dd0-4784-b77a-677396d6840a>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00427.warc.gz"}
RE: [tlaplus] Checking that all expected values of a variable are seen Hi Tom, Unfortunately, there’s no easy way to check this property in TLA+. It’s a consequence of "Sometimes" is Sometimes "Not Never"; there are certain properties that branching time logics can express that linear time logics cannot. That said, it’s possible to work around this. We can show a state is reachable via counterexample. We can also instantiate the spec in a larger “controller” spec. The controller has an auxiliary variable that tracks which states the machine reached. We also add an additional Reset action, which returns the machine to its initial state without changing the aux variable. This is then equivalent to simulating multiple different behaviors of the machine, to see if any combination of them reach all of the states. VARIABLES MachineState, seen_states vars == <<MachineState, seen_states>> Machine == INSTANCE CycleExample WITH MachineState <- MachineState Init == /\ Machine!Init /\ seen_states = {MachineState} Reset == /\ MachineState' = "a" Next == /\ seen_states' = seen_states \union {MachineState} /\ \/ Reset \/ Machine!Next Spec == Init /\ [][Next]_vars AllStatesNotVisited == seen_states /= Machine!MachineStates If AllStatesNotVisited is violated then we know that some combination of executions reach all possible states. From: tlaplus@xxxxxxxxxxxxxxxx <tlaplus@xxxxxxxxxxxxxxxx> On Behalf Of tlabonte@xxxxxxxxx Sent: Friday, November 15, 2019 3:11 PM To: tlaplus <tlaplus@xxxxxxxxxxxxxxxx> Subject: [tlaplus] Checking that all expected values of a variable are seen I am new to TLA+ and am trying to figure out if there's a way to write a check that guarantees that the spec is defined such that a variable is able to take on all expected values. (In other words, in the set of all states defined by the spec, the variable of interest is seen to take on all expected values.) This would be used to determine if all states of a hardware state machine are The complication is that the (simplified) hardware state machine is defined to have an outer cycle, but also a smaller inner cycle, and can run indefinitely (no terminal state). The hardware state machine states and arcs look like: a <-> b -> c -> (back to a) Here is the TLA code demonstrating this: ---------------------------- MODULE CycleExample ---------------------------- MachineStates == { "a", "b", "c" } VARIABLES MachineState TypeOK == MachineState \in MachineStates Init == MachineState = "a" Next == MachineState = "a" -> MachineState' = "b" [] MachineState = "b" -> MachineState' \in {"a", "c"} [] MachineState = "c" -> MachineState' = "a" [] OTHER -> MachineState' = "error" Spec == Init /\ [][Next]_MachineState /\ WF_MachineState(Next) AllSMStatesVisited == <>(\A x \in MachineStates : (MachineState = x)) I've attempted to write the temporal property, AllSMStatesVisited, to perform this check, but it fails because a valid behavior is for the state machine to cycle between a and b infinitely and never visit c. From what I understand, invariants must be true in every state of the specification and temporal properties must be true for all steps in the specification. Given that, I see no way to write a check that will do what I want. I understand that I can use TLC to show that a state is reachable by counterexample, but in a large specification, this is not practical. I'm really looking for a way to write a self-contained module that contains the specification and all checks to guarantee that it functions as intended. Is it possible to achieve this with TLA+ and TLC? Thank you in advance. You received this message because you are subscribed to the Google Groups "tlaplus" group. To unsubscribe from this group and stop receiving emails from it, send an email to tlaplus+unsubscribe@xxxxxxxxxxxxxxxx. To view this discussion on the web visit https://groups.google.com/d/msgid/tlaplus/4b4e005e-79cc-4770-8fbb-0eaf4ba9950f%40googlegroups.com.	{"url":"https://discuss.tlapl.us/msg03231.html","timestamp":"2024-11-11T05:20:10Z","content_type":"text/html","content_length":"22233","record_id":"<urn:uuid:10f6e9e8-ac3d-4230-8430-c0f6e07d1aaf>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00087.warc.gz"}
Background and Motivation Learn why elementary mathematical logic is essential for everyone. We'll cover the following Why mathematical logic? Mathematical logic is the study of how we reason. Reasoning allows us to infer new conclusions from the current body of knowledge that we already have. We want this reasoning to be sound and precise. By using tools from logic, we can precisely formulate statements, state our assumptions, ask questions, and express our existing knowledge. We can make valid conclusions about the phenomenon that we are studying. Logic allows us to examine any particular line of reasoning and its soundness. In other words, it will enable us to identify flawed reasoning. Real-world applications Logic is indispensable in every avenue of human life. In a court of law, judges must make sure that the arguments presented in a court case are sound and that the conclusions suggested by the lawyers are logical and valid. Logic is a question of life and death. Lawmakers use logic to develop constitutions and promulgate laws. Scientists, engineers, and doctors all strive to adhere to the rules of logic. We need logic in “weaving, mining, farming, building ships, or baking cakes.Words taken from the poem “On Death, without Exaggeration” by Wislawa Szymborska— Polish poet and Nobel Laureate.” It takes a lot of imagination to think of a human pursuit that does not require logic. As a simple example, look at the following two sentences and appreciate that it is logic that enables us to conclude the equivalence of the two. • If John works hard, then he will succeed. • Either John does not work hard, or he will succeed. Every learner needs at least a rudimentary knowledge of elementary logic. We can avoid logical fallacies by being aware of the rules of logic and how to conclude from given facts. It makes us better professionals and helps us lead a more mature and wholesome life. Simply put— every learner would benefit from the contents of this course. By the end of this course, you will acquire the skills to differentiate between a sound, well-formed argument versus verbal bamboozlement. You will have the ability to decide if the views presented to you are valid. More importantly, you will be able to construct a logical series of statements that support your ideas. In today’s world, where we are constantly bombarded with information, theories, and explanations, logic will serve as a compass and a guiding light. It will allow you to construct a coherent This course, without exaggeration, is very important. We can constantly apply its principles during our entire lifetime.	{"url":"https://www.educative.io/courses/introduction-to-logic-basics-of-mathematical-reasoning/background-and-motivation","timestamp":"2024-11-14T04:31:32Z","content_type":"text/html","content_length":"752566","record_id":"<urn:uuid:8bde0a7a-55ae-4903-b636-c801fde6182c>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00104.warc.gz"}
How could I amplify the translation of a transform var? Hi everyone, I wonder how to amplify the translation of a transform var? Seems theres the compose transform function only, but I don’t know how to make it. Your help is to be greatly appreciated! What do you mean? What exactly is it that you are trying to do? Could you provide an example? Well, for example, I have a transform var now, whose translation value x=1.0, y=1.0, z=1.0, and I wanna multiply all the three by 10, how could I make it? Thanks! Break (drag a node and type break) or split (right click on the transform pin) the transform node, take the location vector and multiply it with a float value and then create the transform again and plug all the values back in. I’ll try it, you have my thanks:D	{"url":"https://forums.unrealengine.com/t/how-could-i-amplify-the-translation-of-a-transform-var/44902","timestamp":"2024-11-11T17:44:23Z","content_type":"text/html","content_length":"24392","record_id":"<urn:uuid:b4c0e69a-92f2-4686-98a0-d395ed9c7a0c>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00224.warc.gz"}
Lesson 3 Creating Cross Sections by Dilating • Let’s create cross sections by doing dilations. 3.1: Dilating, Again Dilate triangle $BCD$ using center $P$ and a scale factor of 2. Look at your drawing. What do you notice? What do you wonder? 3.2: Pyramid Mobile Your teacher will give you sheets of paper. Each student in the group should take one sheet of paper and complete these steps: 1. Locate and mark the center of your sheet of paper by drawing diagonals or another method. 2. Each student should choose one scale factor from the table. On your paper, draw a dilation of the entire sheet of paper, using the center you marked as the center of dilation. 3. Measure the length and width of your dilated rectangle and calculate its area. Record the data in the table. 4. Cut out your dilated rectangle and make a small hole in the center. │scale factor, $k$│length of scaled rectangle │width of scaled rectangle│area of scaled rectangle│ │$k=0.25$ │ │ │ │ │$k=0.5$ │ │ │ │ │$k=0.75$ │ │ │ │ │$k=1$ │ │ │ │ Now the group as a whole should complete the remaining steps: 1. Cut 1 long piece of string (more than 30 centimeters) and 4 shorter pieces of string. Make 4 marks on the long piece of string an equal distance apart. 2. Thread the long piece of string through the hole in the largest rectangle. Tie a shorter piece of string beneath it where you made the first mark on the string. This will hold up the rectangle. 3. Thread the remaining pieces of paper onto the string from largest to smallest, tying a short piece of string beneath each one at the marks you made. 4. Hold up the end of the string to make your cross sections resemble a pyramid. As a group, you may have to steady the cross sections for the pyramid to clearly appear. Is dilating a square using a factor of 0.9, then dilating the image using scale factor 0.9 the same as dilating the original square using a factor of 0.8? Explain or show your reasoning. Imagine a triangle lying flat on your desk, and a point $P$ directly above the triangle. If we dilate the triangle using center $P$ and scale factor $k=\frac12$ or 0.5, together the triangles resemble cross sections of a pyramid. We can add in more cross sections. This image includes two more cross sections, one with scale factor $k=0.25$ and one with scale factor $k=0.75$. The triangle with scale factor $k=1$ is the base of the pyramid, and if we dilate with scale factor $k=0$ we get a single point at the very top of the pyramid. Each triangle’s side lengths are a factor of $k$ times the corresponding side length in the base. For example, for the cross section with $k=\frac12$, each side length is half the length of the base’s side lengths. • axis of rotation A line about which a two-dimensional figure is rotated to produce a three-dimensional figure, called a solid of rotation. The dashed line is the axis of rotation for the solid of rotation formed by rotating the green triangle. • cone A cone is a three-dimensional figure with a circular base and a point not in the plane of the base called the apex. Each point on the base is connected to the apex by a line segment. • cross section The figure formed by intersecting a solid with a plane. • cylinder A cylinder is a three-dimensional figure with two parallel, congruent, circular bases, formed by translating one base to the other. Each pair of corresponding points on the bases is connected by a line segment. • face Any flat surface on a three-dimensional figure is a face. A cube has 6 faces. • prism A prism is a solid figure composed of two parallel, congruent faces (called bases) connected by parallelograms. A prism is named for the shape of its bases. For example, if a prism’s bases are pentagons, it is called a “pentagonal prism.” • pyramid A pyramid is a solid figure that has one special face called the base. All of the other faces are triangles that meet at a single vertex called the apex. A pyramid is named for the shape of its base. For example, if a pyramid’s base is a hexagon, it is called a “hexagonal pyramid.” • solid of rotation A three-dimensional figure formed by rotating a two-dimensional figure using a line called the axis of rotation. The axis of rotation is the dashed line. The green triangle is rotated about the axis of rotation line to form a solid of rotation. • sphere A sphere is a three-dimensional figure in which all cross-sections in every direction are circles.	{"url":"https://im.kendallhunt.com/HS/students/2/5/3/index.html","timestamp":"2024-11-01T22:51:14Z","content_type":"text/html","content_length":"115670","record_id":"<urn:uuid:0ab871e2-feff-458d-8f3c-c3a5ebc0c0fa>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00519.warc.gz"}
Section: Application Domains Self calibration problem & Gear fault diagnosis $-$ collaboration with Safran Tech Self calibration problem Due to numerous applications (e.g. sensor network, mobile robots), sources and sensors localization has intensively been studied in the literature of signal processing. The anchor position self calibration problem, a well-known problem in signal processing, consists in estimating the positions of both the moving sources and a set of fixed sensors (anchors) when only the distance information between the points from the two different sets is available. The position self-calibration problem is a particular case of the Multidimensional Unfolding (MDU) problem for the Euclidean space of dimension 3. Based on computer algebra methods for polynomial systems, we have recently proposed a new approach for the MDU problem which yields closed-form solutions and an efficient algorithm for the estimation of the positions [56] only based on linear algebra techniques. This first result, obtained in collaboration with Dagher (Research Engineer, Inria Chile) and Zheng (Defrost , Inria Lille - Nord Europe), yields a recent patent [55]. Real tests are now carried out. Our first results will be further developed, improved, tested, and demonstrated. The MDU problem is just one instance of localization problems: more problems can be addressed for which a computer algebra expertise can brought new interesting results, especially in finding closed-form solutions, yielding new estimation techniques which avoid the use of optimization algorithms as commonly done in the signal processing literature. The main differences between these localization problems can essentially be read on a certain matrix of distance called the Euclidean distance matrix [56]. Gear fault diagnosis We have a collaboration with Barau (Safran Tech) and Hubert (Safran Tech), and Dagher (Research Engineer, Inria Chile) on the symbolic-numeric study of the new multi-carrier demodulation method developed in [71]. Gear fault diagnosis is an important issue in aeronautics industry since a damage in a gearbox, which is not detected in time, can have dramatic effects on the safety of a plane. Since the vibrations of a spur gear can be modeled as a product of two periodic functions related to the gearbox kinematic, [71] has proposed to recover each function from the global signal by means of an optimal reconstruction problem which, by means of Fourier analysis, can be rewritten as ${\mathrm{argmin}}_{u\in {ℂ}^{n},{v}_{1},{v}_{2}\in {ℂ}^{m}}\parallel M-u\phantom{\rule{0.166667em}{0ex}}{v}_{1}^{☆}-D\phantom{\rule{0.166667em}{0ex}}u\phantom{\rule{0.166667em}{0ex}}{v}_{2}^{☆}{\ parallel }_{F},$ where $M\in {ℂ}^{n×m}$ (resp. $D\in {ℂ}^{n×n}$) is a given (resp. diagonal) matrix with a special shape, $\parallel ·{\parallel }_{F}$ denotes the Frobenius norm, and ${v}^{☆}$ the Hermitian transpose of $v$. Based on closed-form solutions of the exact problem $-$ which are defined by a system of polynomial equations in the unknowns $-$ we have recently proposed efficient numerical algorithms to numerically solve the problem. The first results are interesting and they will be further developed and tested on different data sets. Finally, we shall continue to study the extremal solutions of the corresponding polynomial problem by means of symbolic and numeric methods, etc.	{"url":"https://radar.inria.fr/report/2018/gaia/uid39.html","timestamp":"2024-11-07T16:08:55Z","content_type":"text/html","content_length":"39645","record_id":"<urn:uuid:4d02f0dc-a999-4835-a766-da7897dcff3b>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00351.warc.gz"}
Meitner Audio PRE Preamplifier Measurements Link: reviewed by Phil Gold on SoundStage! Hi-Fi on January 15, 2024 General information All measurements taken using an Audio Precision APx555 B Series analyzer. The PRE was conditioned for 30 minutes at 2Vrms at the output before any measurements were taken. All measurements were taken with both channels driven. The PRE offers two sets of line-level unbalanced (RCA) inputs, one set of line-level balanced (XLR) inputs, one set each of unbalanced (RCA) and balanced (XLR) outputs (both always on). The PRE offers a maximum of 6dB of gain from input to output for the same input type. That is to say, if the volume is set to unity gain, an input of 2Vrms will yield 2Vrms at the output for the unbalanced input/output scenario, and the balanced input/output scenario. For the unbalanced in/balanced out scenario, 12dB gain is available. For the balanced in/unbalanced out scenario, 0dB of gain is Based on the accuracy and non-repeatable nature of the channel deviation (table below), the volume control is in the analog domain, but digitally controlled. It offers between 2 and 3dB step increments for the first 12 volume steps. From steps 12 to 22, 1dB steps were measured. Beyond level 22 up to 100, the volume control offers 0.5 dB steps. Overall gain was measured at -68.7dB for volume step one, up to +6dB at the maximum position (100). Volume channel tracking proved exquisite, ranging from 0.000dB to 0.008dB. There is a difference in terms of THD and noise between unbalanced and balanced signals in the PRE (see both the main table and FFTs below). The balanced outputs have about 6dB more uncorrelated thermal noise, whereas using the balanced inputs yields about 10dB less THD compared to the unbalanced inputs. Unfortunately, the lower distortion is only apparent in the FFTs, because they allow averages over multiple data runs, which averages out and lowers the noise floor, making the very low distortion peaks visible. During normal real-time THD measurements, the analyzer is set to measure for 2-3 seconds (maximum) and cannot assign a THD value below the measured noise floor. This explains why in the primary table below, THD appears lower for the unbalanced input/output compared to the balanced input/output. The true THD ratio figure for the balanced configuration, based on the balanced input/output FFT, is an astounding 0.00002% (about -135dB), compared to the 0.00007% (about -123dB) or so for the unbalanced input. Unless otherwise stated, balanced input and output was evaluated, with an input and output of 2Vrms into a 200k ohm-load, with the analyzer’s input bandwidth filter set to 10Hz to 22.4kHz (exceptions include FFTs and THD vs frequency sweeps where the bandwidth is extended to 90kHz, and frequency and squarewave response where the bandwidth is extended from DC to 1MHz). Volume-control accuracy (measured at preamp outputs): left-right channel tracking Volume position Channel deviation 1 0.003dB 10 0.000dB 20 0.008dB 30 0.001dB 40 0.001dB 50 0.003dB 60 0.005dB 70 0.005dB 80 0.004dB 90 0.002dB 100 0.001dB Published specifications vs. our primary measurements The table below summarizes the measurements published by Meitner for the PRE compared directly against our own. The published specifications are sourced from Meitner’s website, either directly or from the manual available for download, or a combination thereof. With the exception of frequency response, where the Audio Precision bandwidth was set at its maximum (DC to 1MHz), assume, unless otherwise stated, a 1kHz sinewave, 2Vrms input and output into 200k ohms load, 10Hz to 22.4kHz bandwidth, and the worst-case measured result between the left and right channels. Parameter Manufacturer SoundStage! Lab SNR (4Vrms output, 20Hz-20kHz BW) >116dB 108.1dB Gain control range 74dB 74.6dB THD (1kHz) 0.004% <0.0001% Frequency range 0Hz-200kHz 0Hz-200kHz (0/-0.14dB) System gain 6dB 6dB Maximum input level 6.2Vrms 13.5Vrms Input impedance (XLR) 20k ohms 47.9k ohms Input impedance (RCA) 10k ohms 11.6k ohms Output impedance (XLR) 150 ohms 149.4 ohms Output impedance (RCA) 75 ohms 150.7 ohms SNR measured with unbalanced in/out = 115.3dB *SNR calculated with residual noise (volume at 0) and unbalanced in/out = 118.6dB Our primary measurements revealed the following using the balanced line-level inputs (unless otherwise specified, assume a 1kHz sinewave, 2Vrms input and output into 200k ohms load, 10Hz to 22.4kHz Parameter Left channel Right channel Crosstalk, once channel driven (10kHz) -122.2dB -89.1dB DC offset <-1.7mV <0.6mV Gain (default) 6dB 6dB IMD ratio (CCIF, 18kHz + 19kHz stimulus tones, 1:1) <-113dB <-113dB IMD ratio (SMPTE, 60Hz + 7kHz stimulus tones, 4:1 ) <-100dB <-100dB Input impedance (balanced) 47.9k ohms 47.9k ohms Input impedance (unbalanced) 11.5k ohms 11.6k ohms Maximum output voltage (at clipping 1% THD+N) 20.2Vrms 20.2Vrms Maximum output voltage (at clipping 1% THD+N into 600 ohms) 16Vrms 16Vmrs Noise level (with signal, A-weighted) <12uVrms <12uVrms Noise level (with signal, 20Hz to 20kHz) <15uVrms <15uVrms Noise level (no signal, volume min, A-weighted) <7.9uVrms <7.9uVrms Noise level (no signal, volume min, 20Hz to 20kHz) <10uVrms <10uVrms Output impedance (balanced) 149.4 ohms 149.9 ohms Output impedance (unbalanced) 150.7 ohms 150.7 ohms Signal-to-noise ratio (2Vrms out, A-weighted, 2Vrms in) 104.2dB 104.1dB Signal-to-noise ratio (2Vrms out, 20Hz to 20kHz, 2Vrms in) 102.1dB 102.3dB Signal-to-noise ratio (2Vrms out, A-weighted, max volume) 100.7dB 100.7dB THD (unweighted, balanced) <0.0001% <0.0001% THD (unweighted, unbalanced) <0.00009% <0.00009% THD+N (A-weighted) <0.0006% <0.0006% THD+N (unweighted) <0.00082% <0.00082% Frequency response In our measured frequency-response plot above, the PRE is perfectly flat within the audioband (0dB at 20Hz and 20kHz). The PRE appears to be DC-coupled, as it yielded 0dB of deviation at 5Hz. The PRE can certainly be considered an extended-bandwidth audio device, as it is only 0.14dB down at 200kHz. In the graph above and most of the graphs below, only a single trace may be visible. This is because the left channel (blue or purple trace) is performing identically to the right channel (red or green trace), and so they perfectly overlap, indicating that the two channels are ideally Phase response Above is the phase-response plot from 20Hz to 20kHz. The PRE does not invert polarity, and exhibited zero phase shift within the audioband. THD ratio (unweighted) vs. frequency The plot above shows THD ratios at the output as a function of frequency (20Hz to 20kHz) for a sinewave input stimulus. The blue and red plots are for the left and right channels into 200k ohms, while purple/green (L/R) are into 600 ohms. THD values were flat across most of the audioband at 0.0001% into 600 ohms and 200k ohms, with a small rise to 0.0002% at 20kHz. This shows that the PRE’s outputs are robust and would yield identical THD performance feeding an amplifier with either a high or low input impedance. THD ratio (unweighted) vs. output voltage The plot above shows THD ratios measured at the output of the PRE as a function of output voltage into 200k ohms with a 1kHz input sinewave. At the 10mVrms level, THD values measured around 0.03%, dipping down to around 0.00006% at 6-8Vrms, followed by a rise to 0.0003% at around 18Vrms. The 1% THD point is reached at 20.2Vrms. It’s also important to mention that anything above 2-4Vrms is not typically required to drive most power amps to full power. THD+N ratio (unweighted) vs. output voltage The plot above shows THD+N ratios measured at the output of the PRE as a function of output voltage into 200k ohms with a 1kHz input sinewave. At the 10mVrms level, THD+N values measured around 0.2%, dipping down to around 0.0003% at 10-18Vrms. FFT spectrum – 1kHz (balanced in, balanced out) Shown above is the fast Fourier transform (FFT) for a 1kHz input sinewave stimulus, measured at the output into a 200k-ohm load, for the balanced inputs and outputs. We see that the signal’s second harmonic, at 2kHz, is extremely low at around -135dBrA, or 0.00002%, and subsequent signal harmonics are not visible above the -145dBrA noise floor. Below 1kHz, we can see very small peaks at 60, 120, 148, 180, and 300Hz. These peaks are all below the -130dBrA, or 0.00003%, level. This is a very clean FFT. FFT spectrum – 1kHz (unbalanced in, balanced out) Shown above is the fast Fourier transform (FFT) for a 1kHz input sinewave stimulus, measured at the output into a 200k-ohm load for the unbalanced inputs and balanced outputs. The main difference here compared to the FFT above is the higher second signal harmonic, at -125dBRa, or 0.00006%, versus the -135dBrA 2kHz peak seen when the balanced inputs are used. Noise peaks left of the signal peak are at similar levels as the FFT above. FFT spectrum – 1kHz (unbalanced in, unbalanced out) Shown above is the fast Fourier transform (FFT) for a 1kHz input sinewave stimulus, measured at the output into a 200k-ohm load for the unbalanced inputs and outputs. The same distortion profile with the higher 2kHz peaks can be seen here as with the FFT above. The common denominator is the use of the unbalanced inputs. The overall noise floor is at its lowest here, at -150dBrA. FFT spectrum – 1kHz (balanced in, unbalanced out) Shown above is the fast Fourier transform (FFT) for a 1kHz input sinewave stimulus, measured at the output into a 200k-ohm load for the balanced inputs and unbalanced outputs. The same distortion profile with the lower 2kHz peaks can be seen here as with the first FFT above. The common denominator is the use of the balanced inputs. FFT spectrum – 50Hz Shown above is the FFT for a 50Hz input sinewave stimulus measured at the output into a 200k-ohm load. The X axis is zoomed in from 40Hz to 1kHz, so that peaks from noise artifacts can be directly compared against peaks from the harmonics of the signal. Signal-related peaks can be seen at the second (100Hz) and third (150Hz) harmonics, at an extremely low -140dBrA, or 0.00001%. Noise-related peaks are all below -135dBrA, or 0.00002%. Intermodulation distortion FFT (18kHz + 19kHz summed stimulus) Shown above is an FFT of the intermodulation distortion (IMD) products for an 18kHz + 19kHz summed sinewave stimulus tone measured at the output into a 200k-ohm load. The input RMS values are set at -6.02dBrA so that, if summed for a mean frequency of 18.5kHz, would yield 2Vrms (0dBrA) at the output. We find that the second-order modulation product (i.e., the difference signal of 1kHz) is at -125dBrA, or 0.0006%, while the third-order modulation products, at 17kHz and 20kHz, are at roughly the same level. This, like the 1kHz FFTs, is an extremely clean result. Intermodulation distortion FFT (line-level input, APx 32 tone) Shown above is the FFT of the output of the PRE into 200k ohms with the APx 32-tone signal applied to the input. The combined amplitude of the 32 tones is the 0dBrA reference, and corresponds to 2Vrms into 200k ohms. The intermodulation products—i.e., the “grass” between the test tones—are distortion products from the amplifier. Distortion products are at a vanishingly low -140dBrA, or 0.00001%. Thus, even with a complex input signal, the PRE does not add any audible coloration to the input signal. Square-wave response (10kHz) Above is the 10kHz squarewave response at the output into 200k ohms. Due to limitations inherent to the Audio Precision APx555 B Series analyzer, this graph should not be used to infer or extrapolate the PRE’s slew-rate performance. Rather, it should be seen as a qualitative representation of its relatively high bandwidth. An ideal squarewave can be represented as the sum of a sinewave and an infinite series of its odd-order harmonics (e.g., 10kHz + 30kHz + 50kHz + 70kHz . . .). A limited bandwidth will show only the sum of the lower-order harmonics, which may result in noticeable undershoot and/or overshoot, and softening of the edges. The PRE’s reproduction of the 10kHz squarewave is extremely clean with sharp corners. Diego Estan Electronics Measurement Specialist	{"url":"https://soundstagenetwork.com/index.php?option=com_content&view=article&id=2967:meitner-audio-pre-preamplifier&catid=433:preamplifier-measurements&Itemid=153","timestamp":"2024-11-02T07:43:00Z","content_type":"text/html","content_length":"107242","record_id":"<urn:uuid:4c95f988-686e-4faf-81ed-f9b049b4b34e>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00551.warc.gz"}
A class of equations with peakon and pulson solutions (with an Appendix by Harry Braden and John Byatt-Smith) Holm, Darryl D., Hone, Andrew N.W. (2005) A class of equations with peakon and pulson solutions (with an Appendix by Harry Braden and John Byatt-Smith). Journal of Nonlinear Mathematical Physics, 12 (Sup.1). pp. 380-394. ISSN 1402-9251. (doi:10.2991/jnmp.2005.12.s1.31) (KAR id:41497) We consider a family of integro-differential equations depending upon a parameter b as well as a symmetric integral kernel g(x). When b=2 and g is the peakon kernel (i.e. g(x)=exp(?\|x\|) up to rescaling) the dispersionless Camassa-Holm equation results, while the Degasperis-Procesi equation is obtained from the peakon kernel with b=3. Although these two cases are integrable, generically the corresponding integro-PDE is non-integrable. However,for b=2 the family restricts to the pulson family of Fringer & Holm, which is Hamiltonian and numerically displays elastic scattering of pulses. On the other hand, for arbitrary b it is still possible to construct a nonlocal Hamiltonian structure provided that g is the peakon kernel or one of its degenerations: we present a proof of this fact using an associated functional equation for the skew-symmetric antiderivative of g. The nonlocal bracket reduces to a non-canonical Poisson bracket for the peakon dynamical system, for any value of b?1. • Depositors only (login required):	{"url":"https://kar.kent.ac.uk/41497/","timestamp":"2024-11-03T12:33:13Z","content_type":"application/xhtml+xml","content_length":"29485","record_id":"<urn:uuid:11c2f7be-300a-4a8d-9477-361b653bd786>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00771.warc.gz"}
Mastering Formulas In Excel: What Must All Formulas Begin With Excel formulas are powerful tools that allow users to perform complex calculations and automate tasks within spreadsheets. Mastering formulas in Excel is a crucial skill for anyone who works with data, as it can greatly increase efficiency and accuracy in data analysis and reporting. In this blog post, we will explore the essential components of Excel formulas and what all formulas must begin Key Takeaways • Starting all Excel formulas with the equal sign is crucial for their functionality and accuracy • Mastering Excel formulas is essential for anyone working with data for increased efficiency and accuracy • Understanding different types of cell references is important for correct formula calculations • Knowledge of basic mathematical operators and built-in functions is key for creating effective formulas • Exploring nested functions can lead to advanced and complex calculations in Excel The equal sign When it comes to mastering formulas in Excel, one of the most important things to remember is that all formulas must begin with the equal sign (=). This simple symbol is the starting point for every formula you create in Excel, and it is essential for ensuring that your formulas are recognized and processed correctly by the software. A. Explanation of the equal sign as the starting point for all formulas The equal sign serves as the signal to Excel that what follows is a formula, rather than a piece of text or a number. Without the equal sign, Excel will interpret the contents of the cell as a simple piece of data, rather than a calculation to be performed. This is why it is crucial to always begin your formulas with the equal sign, to avoid any confusion or errors in your spreadsheet. B. Importance of remembering to start all formulas with the equal sign It can be easy to forget to start a formula with the equal sign, especially if you are working with a large and complex spreadsheet. However, it is important to develop the habit of always including the equal sign at the beginning of your formulas, as this will help to avoid any potential issues with your calculations. Additionally, starting all formulas with the equal sign also makes it easier for others to understand and interpret your spreadsheet, as it clearly indicates which cells contain formulas and which contain raw data. Understanding cell references When mastering formulas in Excel, it is crucial to have a comprehensive understanding of cell references. Excel formulas always begin with an equal sign (=) followed by the cell references and mathematical operators. The cell references indicate the location of the data that the formula is using to perform calculations. Explanation of different types of cell references Relative: A relative cell reference in a formula changes based on the position where it is copied or filled. For example, if a formula in cell C2 references cell A1 as =A1, when the formula is copied to cell C3, it will automatically change to =A2. Absolute: An absolute cell reference is fixed and does not change when copied to other cells. It is denoted by adding a dollar sign ($) in front of the column and row references (e.g., $A$1). Mixed: A mixed cell reference contains an absolute or relative reference. For example, $A1 is an absolute column reference and a relative row reference. How to use cell references in formulas Cell references are used in formulas to perform calculations. For example, to add the values in cell A1 and A2, the formula would be =A1+A2. By using cell references, the formula can dynamically update based on the values in the referenced cells. Importance of accurate cell referencing for correct formula calculations Accurate cell referencing is essential for ensuring that formulas produce the correct calculations. Using the wrong cell references or neglecting to use references can lead to errors in the results. By understanding the different types of cell references and how to use them effectively, users can improve the accuracy and reliability of their Excel formulas. Mastering Formulas in Excel: What All Formulas Must Begin With When it comes to mastering formulas in Excel, understanding the basic mathematical operators is essential. These operators form the foundation of all formulas and play a crucial role in performing calculations within the spreadsheet. Explanation of Basic Operators The basic mathematical operators in Excel include addition (+), subtraction (-), multiplication (), and division (/). These operators are used to perform mathematical calculations within cells and are essential for creating effective formulas. Demonstrating How to Use These Operators in Formulas For example, to add the values in cell A1 and cell B1, you would use the formula =A1+B1. Similarly, for subtraction, multiplication, and division, you would use the operators - , , and / respectively. These operators can also be used in combination within a single formula to perform complex calculations. Importance of Understanding Mathematical Operators for Creating Effective Formulas Mastering the basic mathematical operators is crucial for creating effective formulas in Excel. By understanding how to use these operators, users can perform a wide range of calculations and analyze data more efficiently. It also lays the groundwork for more advanced functions and formulas within Excel. Built-in functions When it comes to mastering formulas in Excel, understanding and effectively using built-in functions is crucial to performing various calculations and analyses. Let’s take a closer look at the importance of built-in functions and how to start formulas with them. Overview of common Excel functions Excel offers a wide range of built-in functions that cater to different mathematical, statistical, and logical operations. Some of the most commonly used functions include SUM, AVERAGE, MAX, MIN, and many more. Each function serves a specific purpose and can be utilized to streamline the process of performing calculations within a spreadsheet. How to start formulas with built-in functions Starting a formula with a built-in function involves typing an equal sign (=) followed by the function name and its respective arguments. For example, to calculate the sum of a range of cells, you would begin by typing =SUM(, followed by the range of cells, and then close the parentheses. This initiates the formula and prompts Excel to perform the specified function on the designated data. Importance of knowing various Excel functions for complex calculations Having a comprehensive understanding of various Excel functions is essential for tackling complex calculations and data analysis tasks. By familiarizing yourself with a wide array of functions, you can efficiently manipulate data, extract valuable insights, and make informed business decisions. Furthermore, knowing the right function to use can significantly expedite the process of performing calculations, thereby increasing productivity and accuracy. Nested functions in Excel When it comes to mastering formulas in Excel, understanding how to use nested functions is crucial. Nested functions allow you to combine multiple functions within a single formula, enabling you to perform more complex calculations and manipulate data in various ways. Explanation of nested functions and their uses Nested functions in Excel are simply functions that are used within another function. This allows you to perform different operations within a single formula. For example, you can use the IF function within the SUM function to add specific values based on certain conditions. How to start formulas with nested functions Starting formulas with nested functions is relatively straightforward. You simply begin by typing the equal sign (=), followed by the function you want to use. Then, within that function, you can add additional functions to create the nested structure. It’s important to pay attention to the order of operations and ensure that the functions are properly nested to avoid errors. Importance of nested functions for advanced calculations Nested functions are essential for performing advanced calculations in Excel. They allow you to build complex formulas that can manipulate and analyze large sets of data. Whether you’re working with financial models, statistical analysis, or any other type of data manipulation, understanding how to use nested functions will greatly enhance your ability to perform sophisticated calculations in Mastering Formulas in Excel is crucial for efficient data analysis and manipulation. As we have discussed, all formulas must begin with the equal sign to ensure accurate calculation in Excel. Understanding and practicing formulas is essential for anyone working with spreadsheets regularly. It's important to emphasize the significance of mastering formulas for both personal and professional use. The ability to quickly and accurately perform calculations and manipulate data can greatly enhance productivity and decision-making. We encourage you to continue practicing and exploring Excel formulas to expand your skill set and become more proficient in using this powerful tool. With dedication and practice, you can further enhance your capabilities in Excel and streamline your spreadsheet tasks. ONLY $99 Immediate Download MAC & PC Compatible Free Email Support	{"url":"https://dashboardsexcel.com/blogs/blog/mastering-formulas-in-excel-what-must-all-formulas-begin-with","timestamp":"2024-11-14T17:54:45Z","content_type":"text/html","content_length":"211916","record_id":"<urn:uuid:dc16c10f-4cf3-40c9-8aa8-0d510c9a4535>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00045.warc.gz"}
How do you calculate per capita growth rate? The complete formula for annual per capita growth rate is: ((G / N) * 100) / t, where t is the number of years. Finding the annual per capita growth rate, as opposed to only the rate for the entire time period, makes it easier to predict future population changes because it relates to both time and overall population. What is per capita growth rate in biology? First, population size is influenced by the per capita population growth rate, which is the rate at which the population size changes per individual in the population. This growth rate is determined by the birth, death, emigration, and migration rates in the population. How do you calculate growth rate biology? (Science: biology, cell culture, Ecology) The rate, or speed, at which the number of organisms in a population increases. this can be calculated by dividing the change in the number of organisms from one point in time to another by the amount of time in the interval between the points of time. What is per capita in biology? โ ข The per capita birth rate is number of offspring produced per unit time. โ ข The per capita death rate is the number of individuals that die per unit time (mortality. What is the per capita growth rate r for the population? The Population Growth Rate (r ) The population growth rate (sometimes called the rate of increase or per capita growth rate, r) equals the birth rate (b) minus the death rate (d) divided by the initial population size (N0). What is the per capita growth rate of the global human population? Global human population growth amounts to around 83 million annually, or 1.1% per year. The global population has grown from 1 billion in 1800 to 7.9 billion in 2020. How do you calculate per capita output? The GDP per capita formula calculates the average of the nation’s economic output when divided by the total population. How do you calculate population in biology? What is measured by per capita GDP quizlet? Per capita GDP is a measure of the total output of a country that takes gross domestic product (GDP) and divides it by the number of people in the country. This is helpful when comparing one country to another. You just studied 6 terms! How do you calculate real output growth rate? It can be calculated by (1) finding real GDP for two consecutive periods, (2) calculating the change in GDP between the two periods, (3) dividing the change in GDP by the initial GDP, and (4) multiplying the result by 100 to get a percentage. How do you calculate per capita in Excel? 1. = (100 * 4,50,000) + (5,000 * 35,000) 2. = $4,50,00,000 + $17,50,00,000. 3. Total Income = $220,000,000. What are the 4 methods used to estimate population size? Here we compare estimates produced by four different methods for estimating population size, i.e. aerial counts, hunter observations, pellet group counts and cohort analysis. Which factors are used to calculate population growth? Population growth rate depends on birth rates and death rates, as well as migration. First, we will consider the effects of birth and death rates. You can predict the growth rate by using this simple equation: growth rate = birth rate โ death rate. How do you calculate the population of a quadrat? The average number of individual organisms within the quadrat area is called the population density. The quadrat equation uses the population density to calculate the estimated total population or N: N = (A/a) x n, where A is the total study area, a is the area of the quadrat, and n is the population density. What does per capita mean quizlet? per capita. definition: The term per capita is used in the field of statistics to indicate the average per person for any given concern, e.g. income, crime. usage:The per capita income of public school teachers in America is approximately $45,000. aka per person things. What is the difference between GDP and GDP per capita? 1. GDP is a measure of a nationร s economic health while GDP per capita takes into account the reflection of such economic health into an individual citizenร s perspective. 2. GDP measures the nationร s wealth while GDP per capita roughly determines the standard of living in a particular country. What is growth rate of population? The annual average rate of change of population size, for a given country, territory, or geographic area, during a specified period. It expresses the ratio between the annual increase in the population size and the total population for that year, usually multiplied by 100. How do you calculate per capita nominal GDP? 1. Nominal GDP = $6 trillion + $10 trillion + $4 trillion + ($2.5 trillion โ $1 trillion) 2. Nominal GDP = $21.5 trillion. What is the most accurate method of determining population size? Counting all individuals in a population is the most accurate way to determine its size. How do biologists estimate population size answers? A technique called sampling can be used to estimate population size. In this procedure, the organisms in a few small areas are counted and projected to the entire area. How do you calculate sample size for a population? 1. Determine the population size (if known). 2. Determine the confidence interval. 3. Determine the confidence level. 4. Determine the standard deviation (a standard deviation of 0.5 is a safe choice where the figure is unknown) 5. Convert the confidence level into a Z-Score. How do you calculate average growth rate over time? The formula used for the average growth rate over time method is to divide the present value by the past value, multiply to the 1/N power and then subtract one. “N” in this formula represents the number of years. How are quadrats used to estimate the population size of an organism? Quadrats are small plots, of uniform shape and size, placed in randomly selected sites for sampling purposes. By counting the number of individuals within each sampling plot, we can see how the density of individuals changes from one part of the habitat to another. How do you calculate abundance of plant species why it should be calculated in quadrat study? Answer: Relative species abundance is calculated by dividing the number of species from one group by the total number of species from all groups. Quadrats are useful for studying both the distribution of ant hills within a larger area and ant behavior within the sample area. … How do you calculate quadrats in biology? For example, if the meadow measured 10 m by 10 m, then its total area is 10 m ร 10 m = 100 m 2. Step 4 – Divide the total area of the habitat by the area of one quadrat. = 400. This gives you the total number of quadrats that could fit into the habitat.	{"url":"https://scienceoxygen.com/how-do-you-calculate-per-capita-growth-rate/?query-1-page=2","timestamp":"2024-11-05T06:10:05Z","content_type":"text/html","content_length":"308341","record_id":"<urn:uuid:058565ad-6d65-4260-9d8e-33a0a4eb8199>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00636.warc.gz"}
Main function for ACTCD package npar.CDM {ACTCD} R Documentation Main function for ACTCD package This function is used to classify examinees into labeled classes given responses and the Q-matrix. npar.CDM(Y, Q, cluster.method = c("HACA", "Kmeans"), Kmeans.centers = NULL, Kmeans.itermax = 10, Kmeans.nstart = 1, HACA.link = c("complete", "ward", "single", "average", "mcquitty", "median", "centroid"), HACA.cut = NULL, label.method = c("2b", "2a", "1", "3"),perm=NULL) Y A required N \times J response matrix with binary elements (1=correct, 0=incorrect), where N is the number of examinees and J is the number of items. A required J \times K binary item-by-attribute association matrix (Q-matrix), where K is the number of attributes. The j^{th} row of the matrix is an indicator vector, 1 indicating Q attributes are required and 0 indicating attributes are not required to master item j. cluster.method The cluster algorithm used to classify data. Two options are available, including "Kmeans" and "HACA", where "HACA" is the default method. See cd.cluster for details. Kmeans.centers The number of clusters when "Kmeans" argument is selected. It must be not less than 2 and not greater than 2^K where K is the number of attributes. The default is 2^K. Kmeans.itermax The maximum number of iterations allowed when "Kmeans" argument is selected. Kmeans.nstart The number of random sets to be chosen when "Kmeans" argument is selected. HACA.link The link to be used with HACA. It must be one of "ward", "single", "complete", "average", "mcquitty", "median" or "centroid". The default is "complete". HACA.cut The number of clusters when "HACA" argument is selected. It must be not less than 2 and not greater than 2^K, where K is the number of attributes. The default is 2^K. The algorithm used for labeling. It should be one of "1","2a", "2b" and "3" corresponding to different labeling methods in Chiu and Ma (2013). The default is "2b". See labeling for label.method details. perm The data matrix of the partial orders of the attribute patterns. att.pattern A N \times K binary attribute patterns, where N is the number of examinees and K is the number of attributes. att.dist A 2^K \times 2 data frame, where the first column is the attribute pattern, the second column is its frequency. cluster.size A set of integers, indicating the sizes of latent clusters. cluster.class A vector of estimated memberships for examinees. See Also print.npar.CDM, cd.cluster,labeling # Classification based on the simulated data and Q matrix # Information about the dataset N <- nrow(sim.dat) #number of examinees J <- nrow(sim.Q) #number of items K <- ncol(sim.Q) #number of attributes # Compare the difference in results among different labeling methods # Note that the default cluster method is HACA labeled.obj.2a <- npar.CDM(sim.dat, sim.Q, label.method="2a") labeled.obj.2b <- npar.CDM(sim.dat, sim.Q, label.method="2b") labeled.obj.3 <- npar.CDM(sim.dat, sim.Q, label.method="3") labeled.obj.1 <- npar.CDM(sim.dat, sim.Q, label.method="1",perm=perm3) #User-specified number of latent clusters M <- 5 labeled.obj.2b <- npar.CDM(sim.dat, sim.Q, cluster.method="HACA", HACA.cut=M, label.method="2b") labeled.obj.2a <- npar.CDM(sim.dat, sim.Q, cluster.method="HACA", HACA.cut=M, label.method="2a") #The attribute pattern for each examinee attpatt <- labeled.obj.2b$att.pattern version 1.3-0	{"url":"https://search.r-project.org/CRAN/refmans/ACTCD/html/npar.CDM.html","timestamp":"2024-11-05T06:04:53Z","content_type":"text/html","content_length":"6988","record_id":"<urn:uuid:eb05a630-2dfb-4cea-ae51-69f55883317a>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00017.warc.gz"}
Work, Energy Mechanics: Work, Energy and Power Work, Energy and Power: Problem Set Problem 1: Renatta Gass is out with her friends. Misfortune occurs and Renatta and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel station. Determine the work done on the car. Problem 2: Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35 degrees above the horizontal to drag his backpack a horizontal distance of 129 meters to the right. Determine the work (in Joules) done upon the backpack. Problem 3: Lamar Gant, U.S. powerlifting star, became the first man to deadlift five times his own body weight in 1985. Deadlifting involves raising a loaded barbell from the floor to a position above the head with outstretched arms. Determine the work done by Lamar in deadlifting 300 kg to a height of 0.90 m above the ground. Problem 4: Sheila has just arrived at the airport and is dragging her suitcase to the luggage check-in desk. She pulls on the strap with a force of 190 N at an angle of 35° to the horizontal to displace it 45 m to the desk. Determine the work done by Sheila on the suitcase. Problem 5: While training for breeding season, a 380 gram male squirrel does 32 pushups in a minute, displacing its center of mass by a distance of 8.5 cm for each pushup. Determine the total work done on the squirrel while moving upward (32 times). Problem 6: During the Powerhouse lab, Jerome runs up the stairs, elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed. a. Determine the work done by Jerome in climbing the stair case. b. Determine the power generated by Jerome. Problem 7: A new conveyor system at the local packaging plan will utilize a motor-powered mechanical arm to exert an average force of 890 N to push large crates a distance of 12 meters in 22 seconds. Determine the power output required of such a motor. Problem 8: The Taipei 101 in Taiwan is a 1667-foot tall, 101-story skyscraper. The skyscraper is the home of the world’s fastest elevator. The elevators transport visitors from the ground floor to the Observation Deck on the 89th floor at speeds up to 16.8 m/s. Determine the power delivered by the motor to lift the 10 passengers at this speed. The combined mass of the passengers and cabin is 1250 Problem 9: The ski slopes at Bluebird Mountain make use of tow ropes to transport snowboarders and skiers to the summit of the hill. One of the tow ropes is powered by a 22-kW motor which pulls skiers along an icy incline of 14° at a constant speed. Suppose that 18 skiers with an average mass of 48 kg hold onto the rope and suppose that the motor operates at full power. a. Determine the cumulative weight of all these skiers. b. Determine the force required to pull this amount of weight up a 14° incline at a constant speed. c. Determine the speed at which the skiers will ascend the hill. Problem 10: The first asteroid to be discovered is Ceres. It is the largest and most massive asteroid is our solar system’s asteroid belt, having an estimated mass of 3.0 x 10^21 kg and an orbital speed of 17900 m/s. Determine the amount of kinetic energy possessed by Ceres. Problem 11: A bicycle has a kinetic energy of 124 J. What kinetic energy would the bicycle have if it had … a. … twice the mass and was moving at the same speed? b. … the same mass and was moving with twice the speed? c. … one-half the mass and was moving with twice the speed? d. … the same mass and was moving with one-half the speed? e. … three times the mass and was moving with one-half the speed? Problem 12: A 78-kg skydiver has a speed of 62 m/s at an altitude of 870 m above the ground. a. Determine the kinetic energy possessed by the skydiver. b. Determine the potential energy possessed by the skydiver. c. Determine the total mechanical energy possessed by the skydiver. Problem 13: Lee Ben Fardest (esteemed American ski jumper), has a mass of 59.6 kg. He is moving with a speed of 23.4 m/s at a height of 44.6 meters above the ground. Determine the total mechanical energy of Lee Ben Fardest. Problem 14: Chloe leads South’s varsity softball team in hitting. In a game against New Greer Academy this past weekend, Chloe slugged the 181-gram softball so hard that it cleared the outfield fence and landed on Lake Avenue. At one point in its trajectory, the ball was 28.8 m above the ground and moving with a speed of 19.7 m/s. Determine the total mechanical energy of the softball. Problem 15: Olive Udadi is at the park with her father. The 26-kg Olive is on a swing following the path as shown. Olive has a speed of 0 m/s at position A and is a height of 3.0-m above the ground. At position B, Olive is 1.2 m above the ground. At position C (2.2 m above the ground), Olive projects from the seat and travels as a projectile along the path shown. At point F, Olive is a mere picometer above the ground. Assume negligible air resistance throughout the motion. Use this information to fill in the table. Position Height (m) PE (J) KE (J) TME (J) Speed (m/s) A 3.0 0.0 B 1.2 C 2.2 F 0 Problem 16: Suzie Lavtaski (m=56 kg) is skiing at Bluebird Mountain. She is moving at 16 m/s across the crest of a ski hill located 34 m above ground level at the end of the run. a. Determine Suzie's kinetic energy. b. Determine Suzie's potential energy relative to the height of the ground at the end of the run. c. Determine Suzie's total mechanical energy at the crest of the hill. d. If no energy is lost or gained between the top of the hill and her initial arrival at the end of the run, then what will be Suzie's total mechanical energy at the end of the run? e. Determine Suzie's speed as she arrives at the end of the run and prior to braking to a stop. Problem 17: Nicholas is at The Noah's Ark Amusement Park and preparing to ride on The Point of No Return racing slide. At the top of the slide, Nicholas (m=72.6 kg) is 28.5 m above the ground. a. Determine Nicholas' potential energy at the top of the slide. b. Determine Nicholas's kinetic energy at the top of the slide. c. Assuming negligible losses of energy between the top of the slide and his approach to the bottom of the slide (h=0 m), determine Nicholas's total mechanical energy as he arrives at the bottom of the slide. d. Determine Nicholas' potential energy as he arrives at the bottom of the slide. e. Determine Nicholas' kinetic energy as he arrives at the bottom of the slide. f. Determine Nicholas' speed as he arrives at the bottom of the slide. Problem 18: Ima Scaarred (m=56.2 kg) is traveling at a speed of 12.8 m/s at the top of a 19.5-m high roller coaster loop. a. Determine Ima's kinetic energy at the top of the loop. b. Determine Ima's potential energy at the top of the loop. c. Assuming negligible losses of energy due to friction and air resistance, determine Ima's total mechanical energy at the bottom of the loop (h=0 m). d. Determine Ima's speed at the bottom of the loop. Problem 19: Justin Thyme is traveling down Lake Avenue at 32.8 m/s in his 1510-kg 1992 Camaro. He spots a police car with a radar gun and quickly slows down to a legal speed of 20.1 m/s. a. Determine the initial kinetic energy of the Camaro. b. Determine the kinetic energy of the Camaro after slowing down. c. Determine the amount of work done on the Camaro during the deceleration. Problem 20: Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fills a pitcher full of Cola, places it on the counter top and gives the 2.6-kg pitcher a 8.8 N forward push over a distance of 48 cm to send it to a customer at the end of the counter. The coefficient of friction between the pitcher and the counter top is 0.28. a. Determine the work done by Pete on the pitcher during the 48 cm push. b. Determine the work done by friction upon the pitcher . c. Determine the total work done upon the pitcher . d. Determine the kinetic energy of the pitcher when Pete is done pushing it. e. Determine the speed of the pitcher when Pete is done pushing it. Problem 21: The Top Thrill Dragster stratacoaster at Cedar Point Amusement Park in Ohio uses a hydraulic launching system to accelerate riders from 0 to 53.6 m/s (120 mi/hr) in 3.8 seconds before climbing a completely vertical 420-foot hill. a. Jerome (m=102 kg) visits the park with his church youth group. He boards his car, straps himself in and prepares for the thrill of the day. What is Jerome's kinetic energy before the acceleration b. The 3.8-second acceleration period begins to accelerate Jerome along the level track. What is Jerome's kinetic energy at the end of this acceleration period? c. Once the launch is over, Jerome begins screaming up the 420-foot, completely vertical section of the track. Determine Jerome's potential energy at the top of the vertical section. (GIVEN: 1.00 m = 3.28 ft) d. Determine Jerome's kinetic energy at the top of the vertical section. e. Determine Jerome's speed at the top of the vertical section. Problem 22: Paige is the tallest player on South's Varsity volleyball team. She is in spiking position when Julia gives her the perfect set. The 0.226-kg volleyball is 2.29 m above the ground and has a speed of 1.06 m/s. Paige spikes the ball, doing 9.89 J of work on it. a. Determine the potential energy of the ball before Paige spikes it. b. Determine the kinetic energy of the ball before Paige spikes it. c. Determine the total mechanical energy of the ball before Paige spikes it. d. Determine the total mechanical energy of the ball upon hitting the floor on the opponent's side of the net. e. Determine the speed of the ball upon hitting the floor on the opponent's side of the net. Problem 23: According to ABC's Wide World of Sports show, there is the thrill of victory and the agony of defeat. On March 21 of 1970, Vinko Bogataj was the Yugoslavian entrant into the World Championships held in former West Germany. By his third and final jump of the day, heavy and persistent snow produced dangerous conditions along the slope. Midway through the run, Bogataj recognized the danger and attempted to make adjustments in order to terminate his jump. Instead, he lost his balanced and tumbled and flipped off the slope into the dense crowd. For nearly 30 years thereafter, footage of the event was included in the introduction of ABC's infamous sports show and Vinco has become known as the agony of defeat icon. a. Determine the speed of 72-kg Vinco after skiing down the hill to a height which is 49 m below the starting location. b. After descending the 49 m, Vinko tumbled off the track and descended another 15 m down the ski hill before finally stopping. Determine the change in potential energy of Vinko from the top of the hill to the point at which he stops. c. Determine the amount of cumulative work done upon Vinko's body as he crashes to a halt. Problem 24: Nolan Ryan reportedly had the fastest pitch in baseball, clocked at 100.9 mi/hr (45.0 m/s) If such a pitch had been directed vertically upwards at this same speed, then to what height would it have Problem 25: In the Incline Energy lab, partners Anna Litical and Noah Formula give a 1.00-kg cart an initial speed of 2.35 m/s from a height of 0.125 m above the lab table. Determine the speed of the cart when it is located 0.340 m above the lab table. Problem 26: In April of 1976, Chicago Cub slugger Dave Kingman hit a home run which cleared the Wrigley Field fence and hit a house located 530 feet (162 m) from home plate. Suppose that the 0.145-kg baseball left Kingman's bat at 92.7 m/s and that it lost 10% of its original energy on its flight through the air. Determine the speed of the ball when it cleared the stadium wall at a height of 25.6 m. Problem 27: Dizzy is speeding along at 22.8 m/s as she approaches the level section of track near the loading dock of the Whizzer roller coaster ride. A braking system abruptly brings the 328-kg car (rider mass included) to a speed of 2.9 m/s over a distance of 5.55 meters. Determine the braking force applied to Dizzy's car. Problem 28: A 6.8-kg toboggan is kicked on a frozen pond, such that it acquires a speed of 1.9 m/s. The coefficient of friction between the pond and the toboggan is 0.13. Determine the distance which the toboggan slides before coming to rest. Problem 29: Connor (m=76.0 kg) is competing in the state diving championship. He leaves the springboard from a height of 3.00 m above the water surface with a speed of 5.94 m/s in the upward direction. a. Determine Connor's speed when he strikes the water. b. Connor's body plunges to a depth of 2.15 m below the water surface before stopping. Determine the average force of water resistance experienced by his body. Problem 30: Gwen is baby-sitting for the Parker family. She takes 3-year old Allison to the neighborhood park and places her in the seat of the children's swing. Gwen pulls the 1.8-m long chain back to make a 26° angle with the vertical and lets 14-kg Allison (swing mass included) go. Assuming negligible friction and air resistance, determine Allison's speed at the lowest point in the trajectory. Problem 31: Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approaches a long embankment inclined upward at 16° above the horizontal. As she slides up the embankment, she encounters a coefficient of friction of 0.128. Determine the height to which she will travel before coming to rest. Problem 32: Matthew starts from rest on top of 8.45 m high sledding hill. He slides down the 32-degree incline and across the plateau at its base. The coefficient of friction between the sled and snow is 0.128 for both the hill and the plateau. Matthew and the sled have a combined mass of 27.5 kg. Determine the distance which Matthew will slide along the level surface before coming to a complete stop. Return to Overview View Audio Guided Solution for Problem:	{"url":"https://staging.physicsclassroom.com/calcpad/energy/problems","timestamp":"2024-11-11T05:00:54Z","content_type":"application/xhtml+xml","content_length":"236280","record_id":"<urn:uuid:e1e2b20c-2052-4e07-807d-b228d96bca18>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00261.warc.gz"}
Binary Search Archives - TO THE INNOVATION Here, We see Find K Closest Elements LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches. List of all LeetCode Solution Find K Closest Elements LeetCode Solution Problem Statement Given a sorted integer array arr, two integers k and x, return the k closest integers to x in the array. The result should also […] Find K Closest Elements LeetCode Solution Read More » Leetcode Solution Find Peak Element LeetCode Solution Here, We see Find Peak Element LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches. List of all LeetCode Solution Find Peak Element LeetCode Solution Problem Statement A peak element is an element that is strictly greater than its neighbors. 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This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches. List of all LeetCode Solution Kth Smallest Element in a Sorted Matrix LeetCode Solution Problem Statement Given an n x n matrix where each of the rows and Kth Smallest Element in a Sorted Matrix LeetCode Solution Read More » Leetcode Solution Split Array Largest Sum LeetCode Solution Here, We see Split Array Largest Sum LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches. List of all LeetCode Solution Split Array Largest Sum LeetCode Solution Problem Statement Given an integer array nums and an integer k, split nums into k non-empty subarrays such that the largest sum of any Split Array Largest Sum LeetCode Solution Read More » Leetcode Solution Russian Doll Envelopes LeetCode Solution Here, We see Russian Doll Envelopes LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches. List of all LeetCode Solution Russian Doll Envelopes LeetCode Solution Problem Statement You are given a 2D array of integers envelopes where envelopes[i] = [wi, hi] represents the width and the height of Russian Doll Envelopes LeetCode Solution Read More » Leetcode Solution	{"url":"https://totheinnovation.com/tag/binary-search/","timestamp":"2024-11-02T11:35:57Z","content_type":"text/html","content_length":"206719","record_id":"<urn:uuid:c0e52e57-3ba2-4681-890c-372c9fc8abcb>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00421.warc.gz"}
Assumptions and Insights of k-epsilon Low Re Yang-Shih Turbulence Model Key Takeaways • A new time scale-based k-epsilon (k-ɛ) model for near-wall turbulence is discussed. In this model, • The Kolmogorov time scale bounds the time scale. When this time scale is used to reformulate the dissipation equation, there is no singularity at the wall, and introducing a pseudo-dissipation rate is avoided. • A damping function, proposed as a function of The k-ɛ model is one of the most widely used turbulence models. Although not very efficient in cases involving large adverse pressure gradients, it is a two-equation model, incorporating two additional transport equations to describe the turbulent properties of the flow (Wilcox et al. 1998). The number of equations denotes the number of additional partial differential equations (PDEs) being solved. This enables the model to account for historical effects such as the convection and diffusion of turbulent energy. The first transported variable is the turbulent kinetic energy, k, which determines the energy in turbulence. The second transported variable is the turbulent dissipation, ε, which determines the scale of turbulence. There are two significant formulations of k-ɛ models (Jones et al. 1972 and Launder et al. 1974). The formulation by Launder et al. 1974 is typically called the standard k-ɛ model. The original motivation for developing the k-ɛ model was to improve the mixing length model and provide an alternative to algebraically describing turbulent length scales in moderate to highly complex flows. The k-ɛ model is useful for free shear layer flows with relatively small pressure gradients (Bardina et al. 1997). Similarly, the model produced good results for wall-bounded and internal flows only when the mean pressure gradients were small. Experimental evidence indicates that its accuracy diminishes for flows containing large adverse pressure gradients. The standard k-ɛ model, designed for high Reynolds number turbulent flows, is traditionally used with a wall function for wall-bounded turbulent flows. However, no universal wall function exists for complex flows. Consequently, a low Reynolds number (LRN) Yang-Shih k-ɛ model was introduced to extend the model's applicability down to the wall. Approaches for the Efficient Implementation of the k-ɛ Turbulence Model The transport equations for turbulent kinetic energy, k, and turbulent dissipation rate, ɛ, are modeled as follows: where P is the turbulence production, T is the turbulent time scale, E is a term specific to the Yang-Shih, 1993, and -ɛ model. Away from the wall, the present model reduces to the standard k-ɛ model. Thus, it is only necessary to assess the performance of the model for near-wall turbulence. Denoting the trace of a matrix product by NOTE: In the Cadence Fidelity Open Solver, the incompressible formulation of the production is considered, even when compressible flows are simulated. This assumption guarantees the production term never becomes negative since A singularity would arise if the standard k-ɛ model were applied all the way to the wall due to the vanishing of turbulent kinetic energy, k, at the wall, which causes the time scale As the wall approaches, the turbulent length and velocity scales approach zero. However, the turbulent timescale, defined as the ratio of the length scale of energy-containing eddies to the turbulent velocity scale, remains non-zero and matches the Kolmogorov timescale due to dominant viscous dissipation near the wall. Therefore, the turbulent time scale is given by Now the time scale given by the modified turbulent time scale equation is bounded by the Kolmogorov time scale that is always positive. When this time scale is used in the dissipation equation, there will be no singularity at the wall. A Kolmogorov-Prandtl type formula defines the turbulent eddy viscosity The damping function in the present model is taken to be a function of These constants are devised by comparing the performance of the model prediction and the direct numerical simulation (DNS) data for turbulent channel flow at The damping function's use of In near-wall turbulence, the mean field's inhomogeneity creates a secondary source term in the dissipation equation, aside from the turbulence itself. The secondary source term E is given by the following equation: Yap correction (Yap. C. J. 1987) modifies the epsilon equation by adding a source term, The Yap correction, active in non-equilibrium flows, stabilizes the turbulence length scale, often enhancing predictions. It notably improves results with the k-ɛ in separated flows and stagnation regions. When implementing the Yap correction, it is only common to activate the correction when the source term is positive. Hence: • Wilcox, David C (1998). "Turbulence Modeling for CFD". Second edition. Anaheim: DCW Industries, 1998. pp. 174. • Jones, W. P., and Launder, B. E. (1972), "The Prediction of Laminarization with a Two-Equation Model of Turbulence", International Journal of Heat and Mass Transfer, vol. 15, 1972, pp. 301-314. • Launder, B. E., and Sharma, B. I. (1974), "Application of the Energy Dissipation Model of Turbulence to the Calculation of Flow Near a Spinning Disc", Letters in Heat and Mass Transfer, vol. 1, no. 2, pp. 131-138. • Bardina, J.E., Huang, P.G., Coakley, T.J. (1997), "Turbulence Modeling Validation, Testing, and Development," NASA Technical Memorandum 110446. • Lam, C. K. G., and Bremhorst, K., (1981)"A Modified Form of the k-e Model for Predicting Wall Turbulences," Journal of Fluids Engineering, Vol. 103, No. 3, pp. 456-460. • Yang Z. and Shih T. H., (1993), "A k-epsilon model for turbulence and transitional boundary layer", Near-Wall Turbulent Flows, R.M.C. So., C.G. Speziale and B.E. Launder(Editors), Elsevier-Science Publishers B. V., pp. 165-175. • Hanjalic, K., and Launder, B. E., (1976) "Contribution Towards a Reynolds-Stress Closure for Low-Reynolds-Number Turbulence," Journal of Fluid Mechanics, Vol. 74, Pt. 4, pp. 593-610. • Yap, C. J. (1987), Turbulent Heat and Momentum Transfer in Recirculating and Impinging Flows, PhD Thesis, Faculty of Technology, University of Manchester, United Kingdom. Theoretical Foundations of k-ω Menter-Shear Stress Transport Turbulence Model - Computational Fluid Dynamics - Cadence Blogs - Cadence Community Introducing Fidelity CharLES - Accelerated, Accurate Large Eddy Simulation - Computational Fluid Dynamics - Cadence Blogs - Cadence Community Cadence CFD Technology Update – Turbomachinery Workflow - Computational Fluid Dynamics - Cadence Blogs - Cadence Community	{"url":"https://community.cadence.com/cadence_blogs_8/b/cfd/posts/physical-interpretations-and-assumptions-behind-the-k-epsilon-k--low-reynolds-number-lrn-yang-shih-turbulence-model","timestamp":"2024-11-15T04:42:54Z","content_type":"text/html","content_length":"88362","record_id":"<urn:uuid:370285f5-6ae1-4622-8687-e678301b1af1>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00881.warc.gz"}
Spring Module but on CFrames IS there any spring module that works with Vector3 and CFrame ? There’s a spring constraint which Roblox provides. The attachments can be positioned with Vector3. No, not what I am talking about, I mean … you know that’s a lua module… and roblox runs on a modified version of lua…? that should work. that works only for vector3 or numbers, not for cframes Then you would have to implement orientation into the mix. I believe Roblox springs are good enough, and you can always rotate the Attachments to rotate the springs. As of now I don’t think there is something like this that you are looking for. 1 Like A CFrame is a constructor made out of vector3 values. If you really need, you can get relative position using CFrame.Position. It should work for you with positions anyway. Alright, so I went and did some research and simply you can and cannot. While you cannot spring CFrames you can spring their Position and Angles. Simply create a function that will linearize the CFrame like this; local function getPositionAndAnglesFromCFrame(cf) return cf.Position, Vector3.new(cf:ToEulerAnglesXYZ()) Assign these to variables like so; local camera_part_position, camera_part_angles = getPositionAndAnglesFromCFrame(camera_part.CFrame) Now, this function will return two things one being the CFrame’s position and two being the CFrames angles. These two variables are both vector3’s. Now simply have two springs and plug these two vector3’s into the springs like this; spring_pos.Target = camera_part_position spring_rot.Target = camera_part_angles now finally combine them into one and BOOM; camera.CFrame = CFrame.new(spring_pos.Position) * CFrame.Angles(spring_rot.Position.X, spring_rot.Position.Y, spring_rot.Position.Z) Notice I did spring_rot.Position.X this is because since its a Vector3 it has X,Y,Z variables in it, and CFrame.Angles won’t take a vector3, but we can just grab the linear X,Y,Z values out of said vector 3 and CFrame.Angles will take it just fine. If this helped please mark solution.(Also I just realised you posted this a year ago so I hope this still finds you and helps.) 2 Likes You could turn the Vector3 to CFrame like this CFrame.new(spring data here)	{"url":"https://devforum.roblox.com/t/spring-module-but-on-cframes/834438","timestamp":"2024-11-08T02:00:56Z","content_type":"text/html","content_length":"38721","record_id":"<urn:uuid:4e45e512-041e-40c9-8a8d-a4f4b78c0e36>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00482.warc.gz"}
Set Theory First published Wed Oct 8, 2014; substantive revision Tue Jan 31, 2023 Set theory is the mathematical theory of well-determined collections, called sets, of objects that are called members, or elements, of the set. Pure set theory deals exclusively with sets, so the only sets under consideration are those whose members are also sets. The theory of the hereditarily-finite sets, namely those finite sets whose elements are also finite sets, the elements of which are also finite, and so on, is formally equivalent to arithmetic. So, the essence of set theory is the study of infinite sets, and therefore it can be defined as the mathematical theory of the actual—as opposed to potential—infinite. The notion of set is so simple that it is usually introduced informally, and regarded as self-evident. In set theory, however, as is usual in mathematics, sets are given axiomatically, so their existence and basic properties are postulated by the appropriate formal axioms. The axioms of set theory imply the existence of a set-theoretic universe so rich that all mathematical objects can be construed as sets. Also, the formal language of pure set theory allows one to formalize all mathematical notions and arguments. Thus, set theory has become the standard foundation for mathematics, as every mathematical object can be viewed as a set, and every theorem of mathematics can be logically deduced in the Predicate Calculus from the axioms of set theory. Both aspects of set theory, namely, as the mathematical science of the infinite, and as the foundation of mathematics, are of philosophical importance. Set theory, as a separate mathematical discipline, begins in the work of Georg Cantor. One might say that set theory was born in late 1873, when he made the amazing discovery that the linear continuum, that is, the real line, is not countable, meaning that its points cannot be counted using the natural numbers. So, even though the set of natural numbers and the set of real numbers are both infinite, there are more real numbers than there are natural numbers, which opened the door to the investigation of the different sizes of infinity. See the entry on the early development of set theory for a discussion of the origin of set-theoretic ideas and their use by different mathematicians and philosophers before and around Cantor’s time. According to Cantor, two sets $A$ and $B$ have the same size, or cardinality, if they are bijectable, i.e., the elements of $A$ can be put in a one-to-one correspondence with the elements of \ (B\). Thus, the set $\mathbb{N}$ of natural numbers and the set $\mathbb{R}$ of real numbers have different cardinalities. In 1878 Cantor formulated the famous Continuum Hypothesis (CH), which asserts that every infinite set of real numbers is either countable, i.e., it has the same cardinality as $\mathbb{N}$, or has the same cardinality as $\mathbb{R}$. In other words, there are only two possible sizes of infinite sets of real numbers. The CH is the most famous problem of set theory. Cantor himself devoted much effort to it, and so did many other leading mathematicians of the first half of the twentieth century, such as Hilbert, who listed the CH as the first problem in his celebrated list of 10 (later expanded to 23 in the published version) unsolved mathematical problems presented in 1900 at the Second International Congress of Mathematicians, in Paris. The attempts to prove the CH led to major discoveries in set theory, such as the theory of constructible sets, and the forcing technique, which showed that the CH can neither be proved nor disproved from the usual axioms of set theory. To this day, the CH remains open. Early on, some inconsistencies, or paradoxes, arose from a naive use of the notion of set; in particular, from the deceivingly natural assumption that every property determines a set, namely the set of objects that have the property. One example is Russell’s Paradox, also known to Zermelo: consider the property of sets of not being members of themselves. If the property determines a set, call it $A$, then $A$ is a member of itself if and only if $A$ is not a member of itself. Thus, some collections, like the collection of all sets, the collection of all ordinals numbers, or the collection of all cardinal numbers, are not sets. Such collections are called proper classes. In order to avoid the paradoxes and put it on a firm footing, set theory had to be axiomatized. The first axiomatization was due to Zermelo (1908) and it came as a result of the need to spell out the basic set-theoretic principles underlying his proof of Cantor’s Well-Ordering Principle. Zermelo’s axiomatization avoids Russell’s Paradox by means of the Separation axiom, which is formulated as quantifying over properties of sets, and thus it is a second-order statement. Further work by Skolem and Fraenkel led to the formalization of the Separation axiom in terms of formulas of first-order, instead of the informal notion of property, as well as to the introduction of the axiom of Replacement, which is also formulated as an axiom schema for first-order formulas (see next section). The axiom of Replacement is needed for a proper development of the theory of transfinite ordinals and cardinals, using transfinite recursion (see Section 3). It is also needed to prove the existence of such simple sets as the set of hereditarily finite sets, i.e., those finite sets whose elements are finite, the elements of which are also finite, and so on; or to prove basic set-theoretic facts such as that every set is contained in a transitive set, i.e., a set that contains all elements of its elements (see Mathias 2001 for the weaknesses of Zermelo set theory). A further addition, by von Neumann, of the axiom of Foundation, led to the standard axiom system of set theory, known as the Zermelo-Fraenkel axioms plus the Axiom of Choice, or ZFC. Other axiomatizations of set theory, such as those of von Neumann-Bernays-Gödel (NBG), or Morse-Kelley (MK), allow also for a formal treatment of proper classes. ZFC is an axiom system formulated in first-order logic with equality and with only one binary relation symbol $\in$ for membership. Thus, we write $A\in B$ to express that $A$ is a member of the set $B$. See the Supplement on Basic Set Theory for further details. See also the Supplement on Zermelo-Fraenkel Set Theory for a formalized version of the axioms and further comments. We state below the axioms of ZFC informally. • Extensionality: If two sets $A$ and $B$ have the same elements, then they are equal. • Null Set: There exists a set, denoted by ${\varnothing}$ and called the empty set, which has no elements. • Pair: Given any sets $A$ and $B$, there exists a set, denoted by $\{ A,B\}$, which contains $A$ and $B$ as its only elements. In particular, there exists the set $\{ A\}$ which has \ (A\) as its only element. • Power Set: For every set $A$ there exists a set, denoted by $\mathcal{P}(A)$ and called the power set of $A$, whose elements are all the subsets of $A$. • Union: For every set $A$, there exists a set, denoted by $\bigcup A$ and called the union of $A$, whose elements are all the elements of the elements of $A$. • Infinity: There exists an infinite set. In particular, there exists a set $Z$ that contains ${\varnothing}$ and such that if $A\in Z$, then $\bigcup\{ A, \{ A\}\}\in Z$. • Separation: For every set $A$ and every given property, there is a set containing exactly the elements of $A$ that have that property. A property is given by a formula $\varphi$ of the first-order language of set theory. Thus, Separation is not a single axiom but an axiom schema, that is, an infinite list of axioms, one for each formula $\varphi$. • Replacement: For every given definable function with domain a set $A$, there is a set whose elements are all the values of the function. Replacement is also an axiom schema, as definable functions are given by formulas. • Foundation: Every non-empty set $A$ contains an $\in$-minimal element, that is, an element such that no element of $A$ belongs to it. These are the axioms of Zermelo-Fraenkel set theory, or ZF. The axioms of Null Set and Pair follow from the other ZF axioms, so they may be omitted. Also, Replacement implies Separation. Finally, there is the Axiom of Choice (AC): • Choice: For every set $A$ of pairwise-disjoint non-empty sets, there exists a set that contains exactly one element from each set in $A$. The AC was, for a long time, a controversial axiom. On the one hand, it is very useful and of wide use in mathematics. On the other hand, it has rather unintuitive consequences, such as the Banach-Tarski Paradox, which says that the unit ball can be partitioned into finitely-many pieces, which can then be rearranged to form two unit balls. The objections to the axiom arise from the fact that it asserts the existence of sets that cannot be explicitly defined. But Gödel’s 1938 proof of its consistency, relative to the consistency of ZF, dispelled any suspicions left about it. The Axiom of Choice is equivalent, modulo ZF, to the Well-ordering Principle, which asserts that every set can be well-ordered, i.e., it can be linearly ordered so that every non-empty subset has a minimal element. Although not formally necessary, besides the symbol $\in$ one normally uses for convenience other auxiliary defined symbols. For example, $A\subseteq B$ expresses that $A$ is a subset of $B$, i.e., every member of $A$ is a member of $B$. Other symbols are used to denote sets obtained by performing basic operations, such as $A\cup B$, which denotes the union of $A$ and $B$, i.e., the set whose elements are those of $A$ and $B$; or $A\cap B$, which denotes the intersection of $A$ and $B$, i.e., the set whose elements are those common to $A$ and $B$. The ordered pair $(A,B)$ is defined as the set $\{ \{ A\},\{ A,B\}\}$. Thus, two ordered pairs $(A,B)$ and $(C,D)$ are equal if and only if $A=C$ and $B=D$. And the Cartesian product $A\times B$ is defined as the set of all ordered pairs $(C,D)$ such that $C\in A$ and $D\in B$. Given any formula $\varphi(x,y_1,\ldots ,y_n)$, and sets $A,B_1,\ldots ,B_n$, by the axiom of Separation one can form the set of all those elements of $A$ that satisfy the formula $\varphi(x,B_1,\ldots ,B_n)$. This set is denoted by $\{ a\in A: \varphi(a,B_1,\ldots ,B_n)\}$. In ZF one can easily prove that all these sets exist. See the Supplement on Basic Set Theory for further discussion. In ZFC one can develop the Cantorian theory of transfinite (i.e., infinite) ordinal and cardinal numbers. Following the definition given by Von Neumann in the early 1920s, the ordinal numbers, or ordinals, for short, are obtained by starting with the empty set and performing two operations: taking the immediate successor, and passing to the limit. Thus, the first ordinal number is ${\ varnothing}$. Given an ordinal $\alpha$, its immediate successor, denoted by $\alpha +1$, is the set $\alpha \cup \{ \alpha \}$. And given a non-empty set $X$ of ordinals such that for every $\alpha \in X$ the successor $\alpha +1$ is also in $X$, one obtains the limit ordinal $\bigcup X$. One shows that every ordinal is (strictly) well-ordered by $\in$, i.e., it is linearly ordered by $\in$ and there is no infinite $\in$-descending sequence. Also, every well-ordered set is isomorphic to a unique ordinal, called its order-type. Note that every ordinal is the set of its predecessors. However, the class $ON$ of all ordinals is not a set. Otherwise, $ON$ would be an ordinal greater than all the ordinals, which is impossible. The first infinite ordinal, which is the set of all finite ordinals, is denoted by the Greek letter omega ($\omega$). In ZFC, one identifies the finite ordinals with the natural numbers. Thus, ${\varnothing}=0$, $\{ {\varnothing}\}=1$, $\{ {\varnothing}, \{ {\varnothing}\}\}=2$, etc., hence $\omega$ is just the set $\mathbb{N}$ of natural numbers. One can extend the operations of addition and multiplication of natural numbers to all the ordinals. For example, the ordinal $\alpha +\beta$ is the order-type of the well-ordering obtained by concatenating a well-ordered set of order-type $\alpha$ and a well-ordered set of order-type $\beta$. The sequence of ordinals, well-ordered by $\in$, starts as follows 0, 1, 2,…, $n$,…, $\omega$, $\omega+1$, $\omega+2$,…, $\omega +\omega$,…, $\omega \cdot n$, …, $\omega \cdot \omega$,…, $\omega^n$, …, $\omega^\omega$, … The ordinals satisfy the principle of transfinite induction: suppose that $C$ is a class of ordinals such that whenever $C$ contains all ordinals $\beta$ smaller than some ordinal $\alpha$, then $\alpha$ is also in $C$. Then the class $C$ contains all ordinals. Using transfinite induction one can prove in ZFC (and one needs the axiom of Replacement) the important principle of transfinite recursion, which says that, given any definable class-function $G$ (namely a definable operation that takes any set $x$ to a set $G(x)$), one can define a class-function $F$ with domain the class $ON$ of ordinals, such that $F(\alpha)$ is the value of the function $G$ applied to the function $F$ restricted to $\alpha$. One uses transfinite recursion, for example, in order to define properly the arithmetical operations of addition, product, and exponentiation on the ordinals. Recall that an infinite set is countable if it is bijectable, i.e., it can be put into a one-to-one correspondence, with $\omega$. All the ordinals displayed above are either finite or countable. But the set of all finite and countable ordinals is also an ordinal, called $\omega_1$, and is not countable. Similarly, the set of all ordinals that are bijectable with some ordinal less than or equal to $\omega_1$ is also an ordinal, called $\omega_2$, and is not bijectable with $\omega_1$, and so on. A cardinal is an ordinal that is not bijectable with any smaller ordinal. Thus, every finite ordinal is a cardinal, and $\omega$, $\omega_1$, $\omega_2$, etc. are also cardinals. The infinite cardinals are represented by the letter aleph ($\aleph$) of the Hebrew alphabet, and their sequence is indexed by the ordinals. It starts like this $\aleph_0$, $\aleph_1$, $\aleph_2$, …, $\aleph_\omega$, $\aleph_{\omega +1}$, …, $\aleph_{\omega +\omega}$, …, $\aleph_{\omega^2}$, …, $\aleph_{\omega^\omega}$, …, $\aleph_{\ omega_1}$, …, $\aleph_{\omega_2}$, … Thus, $\omega=\aleph_0$, $\omega_1=\aleph_1$, $\omega_2=\aleph_2$, etc. For every cardinal there is a bigger one, and the limit of an increasing sequence of cardinals is also a cardinal. Thus, the class of all cardinals is not a set, but a proper class. An infinite cardinal $\kappa$ is called regular if it is not the union of less than $\kappa$ smaller cardinals. Thus, $\aleph_0$ is regular, and so are all infinite successor cardinals, such as $\aleph_1$. Non-regular infinite cardinals are called singular. The first singular cardinal is $\aleph_\omega$, as it is the union of countably-many smaller cardinals, namely $\aleph_\omega =\ bigcup \{\aleph_n : n<\omega\} $. The cofinality of a cardinal $\kappa$, denoted by $cf(\kappa)$ is the smallest cardinal $\lambda$ such that $\kappa$ is the union of $\lambda$-many smaller ordinals. Thus, $cf(\aleph_\ By the AC (in the form of the Well-Ordering Principle), every set \(A$ can be well-ordered, hence it is bijectable with a unique cardinal, called the cardinality of $A$. Given two cardinals $\ kappa$ and $\lambda$, the sum $\kappa +\lambda$ is defined as the cardinality of the set consisting of the union of any two disjoint sets, one of cardinality $\kappa$ and one of cardinality \ (\lambda\). And the product $\kappa \cdot \lambda$ is defined as the cardinality of the Cartesian product $\kappa \times \lambda$. The operations of sum and product of infinite cardinals are trivial, for if $\kappa$ and $\lambda$ are infinite cardinals, then $\kappa +\lambda =\kappa \cdot \lambda = maximum \{ \kappa ,\lambda\}$. In contrast, cardinal exponentiation is highly non-trivial, for even the value of the simplest non-trivial infinite exponential, namely $2^{\aleph_0}$, is not known and cannot be determined in ZFC (see below). The cardinal $\kappa^\lambda$ is defined as the cardinality of the Cartesian product of $\lambda$ copies of $\kappa$; equivalently, as the cardinality of the set of all functions from $\lambda$ into $\kappa$. König’s theorem asserts that $\kappa^{cf(\kappa)}>\kappa$, which implies that the cofinality of the cardinal $2^{\aleph_0}$, whatever that cardinal is, must be uncountable. But this is essentially all that ZFC can prove about the value of the exponential $2^{\aleph_0}$. In the case of exponentiation of singular cardinals, ZFC has a lot more to say. In 1989, Shelah proved the remarkable result that if $\aleph_\omega$ is a strong limit, that is, $2^{\aleph_n}<\ aleph_\omega$, for every $n<\omega$, then $2^{\aleph_\omega}<\aleph_{\omega_4}$ (see Shelah (1994)). The technique developed by Shelah to prove this and similar theorems, in ZFC, is called pcf theory (for possible cofinalities), and has found many applications in other areas of mathematics. A posteriori, the ZF axioms other than Extensionality—which needs no justification because it just states a defining property of sets—may be justified by their use in building the cumulative hierarchy of sets. Namely, in ZF we define using transfinite recursion the class-function that assigns to each ordinal $\alpha$ the set $V_\alpha$, given as follows: • $V_0={\varnothing}$ • $V_{\alpha +1}=\mathcal{P}(V_\alpha)$ • $V_\alpha =\bigcup \{ V_\beta : \beta <\alpha \}$, whenever $\alpha$ is a limit ordinal. The Power Set axiom is used to obtain $V_{\alpha +1}$ from $V_\alpha$. Replacement and Union allow one to form $V_\alpha$ for $\alpha$ a limit ordinal. Indeed, consider the function that assigns to each $\beta <\alpha$ the set $V_\beta$. By Replacement, the collection of all the $V_\beta$, for $\beta <\alpha$, is a set, hence the Union axiom applied to that set yields $V_\ alpha$. The axiom of Infinity is needed to prove the existence of $\omega$ and hence of the transfinite sequence of ordinals. Finally, the axiom of Foundation is equivalent, assuming the other axioms, to the statement that every set belongs to some $V_\alpha$, for some ordinal $\alpha$. Thus, ZF proves that the set theoretic universe, denoted by $V$, is the union of all the $V_\ alpha$, $\alpha$ an ordinal. The proper class $V$, together with the $\in$ relation, satisfies all the ZFC axioms, and is thus a model of ZFC. It is the intended model of ZFC, and one may think of ZFC as providing a description of $V$, a description however that is highly incomplete, as we shall see below. One important property of $V$ is the so-called Reflection Principle. Namely, for each formula $\varphi(x_1,\ldots ,x_n)$, ZFC proves that there exists some $V_\alpha$ that reflects it, that is, for every $a_1,\ldots,a_n\in V_\alpha$, $\varphi(a_1,\ldots ,a_n)$ holds in $V$ if and only if $\varphi(a_1,\ldots,a_n)$ holds in $V_\alpha$. Thus, $V$ cannot be characterized by any sentence, as any sentence that is true in $V$ must be also true in some initial segment $V_\alpha$. In particular, ZFC is not finitely axiomatizable, for otherwise ZFC would prove that, for unboundedly many ordinals $\alpha$, $V_\alpha$ is a model of ZFC, contradicting Gödel’s second incompleteness theorem (see Section 5.2). The Reflection Principle encapsulates the essence of ZF set theory, for as shown by Levy (1960), the axioms of Extensionality, Separation, and Foundation, together with the Reflection Principle, formulated as the axiom schema asserting that each formula is reflected by some set that contains all elements and all subsets of its elements (note that the $V_\alpha$ are like this), is equivalent to ZF. Every mathematical object may be viewed as a set. For example, the natural numbers are identified with the finite ordinals, so $\mathbb{N}=\omega$. The set of integers $\mathbb{Z}$ may be defined as the set of equivalence classes of pairs of natural numbers under the equivalence relation $(n, m)\equiv (n',m')$ if and only if $n+m'=m+n'$. By identifying every natural number $n$ with the equivalence class of the pair $(n,0)$, one may extend naturally the operations of sum and product of natural numbers to $\mathbb{Z}$ (see Enderton (1977) for details, and Levy (1979) for a different construction). Further, one may define the rationals $\mathbb{Q}$ as the set of equivalence classes of pairs $(n,m)$ of integers, where $m\ne 0$, under the equivalence relation \ ((n,m)\equiv (n',m')\) if and only if $n\cdot m'=m\cdot n'$. Again, the operations $+$ and $\cdot$ on $\mathbb{Z}$ may be extended naturally to $\mathbb{Q}$. Moreover, the ordering $\leq_ {\mathbb{Q}}$ on the rationals is given by: $r \leq_{\mathbb{Q}} s$ if and only if there exists $t\in \mathbb{Q}$ such that $s=r+t$. The real numbers may be defined as Dedekind cuts of $\ mathbb{Q}$, namely, a real number is given by a pair $(A,B)$ of non-empty disjoint sets such that $A\cup B=\mathbb{Q}$, $A$ has no greatest element, and $a \lt_{\mathbb{Q}} b$ for every $a\ in A$ and $b\in B$. One can then extend again the operations of $+$ and $\cdot$ on $\mathbb{Q}$, as well as the ordering $\leq_{\mathbb{Q}}$, to the set of real numbers $\mathbb{R}$. Let us emphasize that it is not claimed that, e.g., real numbers are Dedekind cuts of rationals, as they could also be defined using Cauchy sequences, or in other different ways. What is important, from a foundational point of view, is that the set-theoretic version of $\mathbb{R}$, together with the usual algebraic operations, satisfies the categorical axioms that the real numbers satisfy, namely those of a complete ordered field. The metaphysical question of what the real numbers really are is irrelevant here. Algebraic structures can also be viewed as sets, as any $n$-ary relation on the elements of a set $A$ can be viewed as a set of ordered $n$-tuples $(a_1,\ldots ,a_n)$ of elements of $A$. And any $n$-ary function $f$ on $A$, with values on some set $B$, can be seen as the set of ordered $n+1$-tuples $((a_1,\ldots ,a_n),b)$ such that $b$ is the value of $f$ on $(a_1,\ ldots ,a_m)$. Thus, for example, a group is just a triple $(A, +, 0)$, where $A$ is a non-empty set, $+$ is a binary function on $A$ that is associative, $0$ is an element of $A$ such that $a+0=0+a=a$, for all $a\in A$, and for every $a\in A$ there is an element of $A$, denoted by $-a$, such that $a+(-a)=(-a)+a=0$. Also, a topological space is just a set $X$ together with a topology $\tau$ on it, i.e., $\tau$ is a subset of $\mathcal{P}(X)$ containing $X$ and ${\varnothing}$, and closed under arbitrary unions and finite intersections. Any mathematical object whatsoever can always be viewed as a set, or a proper class. The properties of the object can then be expressed in the language of set theory. Any mathematical statement can be formalized into the language of set theory, and any mathematical theorem can be derived, using the calculus of first-order logic, from the axioms of ZFC, or from some extension of ZFC. It is in this sense that set theory provides a foundation for mathematics. The foundational role of set theory for mathematics, while significant, is by no means the only justification for its study. The ideas and techniques developed within set theory, such as infinite combinatorics, forcing, or the theory of large cardinals, have turned it into a deep and fascinating mathematical theory, worthy of study by itself, and with important applications to practically all areas of mathematics. The remarkable fact that virtually all of mathematics can be formalized within ZFC, makes possible a mathematical study of mathematics itself. Thus, any questions about the existence of some mathematical object, or the provability of a conjecture or hypothesis can be given a mathematically precise formulation. This makes metamathematics possible, namely the mathematical study of mathematics itself. So, the question about the provability or unprovability of any given mathematical statement becomes a sensible mathematical question. When faced with an open mathematical problem or conjecture, it makes sense to ask for its provability or unprovability in the ZFC formal system. Unfortunately, the answer may be neither, because ZFC, if consistent, is incomplete. Gödel’s completeness theorem for first-order logic implies that ZFC is consistent—i.e., no contradiction can be derived from it—if and only if it has a model. A model of ZFC is a pair $(M,E)$, where $M$ is a non-empty set and $E$ is a binary relation on $M$ such that all the axioms of ZFC are true when interpreted in $(M,E)$, i.e., when the variables that appear in the axioms range over elements of $M$, and $\in$ is interpreted as $E$. Thus, if $\varphi$ is a sentence of the language of set theory and one can find a model of ZFC in which $\varphi$ holds, then its negation $\neg \varphi$ cannot be proved in ZFC. Hence, if one can find a model of $\varphi$ and also a model of $\neg \varphi$, then $\varphi$ is neither provable nor disprovable in ZFC, in which case we say that $\varphi$ is undecidable in, or independent of, ZFC. In 1931, Gödel announced his striking incompleteness theorems, which assert that any reasonable formal system for mathematics is necessarily incomplete. In particular, if ZFC is consistent, then there are propositions in the language of set theory that are undecidable in ZFC. Moreover, Gödel’s second incompleteness theorem implies that the formal (arithmetical) statement $CON(ZFC)$, which asserts that ZFC is consistent, while true, cannot be proved in ZFC. And neither can its negation. Thus, $CON(ZFC)$ is undecidable in ZFC. If ZFC is consistent, then it cannot prove the existence of a model of ZFC, for otherwise ZFC would prove its own consistency. Thus, a proof of consistency or of undecidability of a given sentence \ (\varphi\) is always a relative consistency proof. That is, one assumes that ZFC is consistent, hence it has a model, and then one constructs another model of ZFC where the sentence $\varphi$ is true. We shall see several examples in the next sections. From Cantor and until about 1940, set theory developed mostly around the study of the continuum, that is, the real line $\mathbb{R}$. The main topic was the study of the so-called regularity properties, as well as other structural properties, of simply-definable sets of real numbers, an area of mathematics that is known as Descriptive Set Theory. Descriptive Set Theory is the study of the properties and structure of definable sets of real numbers and, more generally, of definable subsets of $\mathbb{R}^n$ and other Polish spaces (i.e., topological spaces that are homeomorphic to a separable complete metric space), such as the Baire space $\mathcal{N}$ of all functions $f:\mathbb{N} \to \mathbb{N}$, the space of complex numbers, Hilbert space, and separable Banach spaces. The simplest sets of real numbers are the basic open sets (i.e., the open intervals with rational endpoints), and their complements. The sets that are obtained in a countable number of steps by starting from the basic open sets and applying the operations of taking the complement and forming a countable union of previously obtained sets are the Borel sets. All Borel sets are regular, that is, they enjoy all the classical regularity properties. One example of a regularity property is the Lebesgue measurability: a set of reals is Lebesgue measurable if it differs from a Borel set by a null set, namely, a set that can be covered by sets of basic open intervals of arbitrarily-small total length. Thus, trivially, every Borel set is Lebesgue measurable, but sets more complicated than the Borel ones may not be. Other classical regularity properties are the Baire property (a set of reals has the Baire property if it differs from an open set by a meager set, namely, a set that is a countable union of sets that are not dense in any interval), and the perfect set property (a set of reals has the perfect set property if it is either countable or contains a perfect set, namely, a nonempty closed set with no isolated points). In ZFC one can prove that there exist non-regular sets of reals, but the AC is necessary for this (Solovay 1970). The analytic sets, also called $\mathbf{\Sigma}^1_1$, are the continuous images of Borel sets. And the co-analytic, or $\mathbf{\Pi}^1_1$, sets are the complements of analytic sets. Starting from the analytic (or the co-analytic) sets and applying the operations of projection (from the product space $\mathbb{R}\times \mathcal{N}$ to $\mathbb{R}$) and complementation, one obtains the projective sets. The projective sets form a hierarchy of increasing complexity. For example, if $A\subseteq \mathbb{R}\times \mathcal{N}$ is co-analytic, then the projection $\{ x\in \ mathbb{R}: \exists y\in \mathcal{N}((x,y)\in A)\}$ is a projective set in the next level of complexity above the co-analytic sets. Those sets are called $\mathbf{\Sigma}^1_2$, and their complements are called $\mathbf{\Pi}^1_2$. The projective sets come up very naturally in mathematical practice, for it turns out that a set $A$ of reals is projective if and only if it is definable in the structure \[\mathcal{R}=(\mathbb{R}, +,\cdot ,\mathbb{Z}).\] That is, there is a first-order formula $\varphi ( x, y_1,\ldots, y_n)$ in the language for the structure such that for some $r_1,\ldots ,r_n\in \mathbb{R}$, \[A=\{ x\in \mathbb{R}: \mathcal{R}\models \varphi(x,r_1,\ldots ,r_n)\}.\] ZFC proves that every analytic set, and therefore every co-analytic set, is Lebesgue measurable and has the Baire property. It also proves that every analytic set has the perfect set property. But the perfect set property for co-analytic sets implies that the first uncountable cardinal, $\aleph_1$, is a large cardinal in the constructible universe $L$ (see Section 7), namely a so-called inaccessible cardinal (see Section 10), which implies that one cannot prove in ZFC that every co-analytic set has the perfect set property. The theory of projective sets of complexity greater than co-analytic is completely undetermined by ZFC. For example, in $L$ there is a $\mathbf{\Sigma}^1_2$ set that is not Lebesgue measurable and does not have the Baire property, whereas if Martin’s axiom holds (see Section 11), every such set has those regularity properties. There is, however, an axiom, called the axiom of Projective Determinacy, or PD, that is consistent with ZFC, modulo the consistency of some large cardinals (in fact, it follows from the existence of some large cardinals), and implies that all projective sets are regular. Moreover, PD settles essentially all questions about the projective sets. See the entry on large cardinals and determinacy for further details. A regularity property of sets that subsumes all other classical regularity properties is that of being determined. For simplicity, we shall work with the Baire space $\mathcal{N}$. Recall that the elements of $\mathcal{N}$ are functions $f:\mathbb{N} \to \mathbb{N}$, that is, sequences of natural numbers of length $\omega$. The space $\mathcal{N}$ is topologically equivalent (i.e., homeomorphic) to the set of irrational points of $\mathbb{R}$. So, since we are interested in the regularity properties of subsets of $\mathbb{R}$, and since countable sets, such as the set of rationals, are negligible in terms of those properties, we may as well work with $\mathcal{N}$, instead of $\mathbb{R}$. Given $A\subseteq \mathcal{N}$, the game associated to $A$, denoted by ${\mathcal G}_A$, has two players, I and II, who play alternatively $n_i\in \mathbb{N}$: I plays $n_0$, then II plays $n_1$, then I plays $n_2$, and so on. So, at stage $2k$, player I plays $n_{2k}$ and at stage $2k+1$, player II plays $n_{2k+1} $. We may visualize a run of the game as follows: $\mathbf{I}$ $n_0$ $n_2$ $n_4$ $\cdots$ $n_{2k}$ $\cdots$ $\mathbf{II}$ $n_1$ $n_3$ $\cdots$ $\cdots$ $n_{2k+1}$ $\cdots$ After infinitely many moves, the two players produce an infinite sequence $n_0, n_1,n_2,\ldots$ of natural numbers. Player I wins the game if the sequence belongs to $A$. Otherwise, player II The game ${\mathcal{G}}_A$ is determined if there is a winning strategy for one of the players. A winning strategy for one of the players, say for player II, is a function $\sigma$ from the set of finite sequences of natural numbers into $\mathbb{N}$, such that if the player plays according to this function, i.e., she plays $\sigma (n_0,\ldots ,n_{2k})$ at the $k$-th turn, she will always win the game, no matter what the other player does. We say that a subset $A$ of $\mathcal{N}$ is determined if and only if the game ${\mathcal{G}}_A$ is determined. One can prove in ZFC—and the use of the AC is necessary—that there are non-determined sets. Thus, the Axiom of Determinacy (AD), which asserts that all subsets of $\mathcal{N}$ are determined, is incompatible with the AC. But Donald Martin proved, in ZFC, that every Borel set is determined. Further, he showed that if there exists a large cardinal called measurable (see Section 10), then even the analytic sets are determined. The axiom of Projective Determinacy (PD) asserts that every projective set is determined. It turns out that PD implies that all projective sets of reals are regular, and Woodin has shown that, in a certain sense, PD settles essentially all questions about the projective sets. Moreover, PD seems to be necessary for this. Another axiom, $AD^{L(\Bbb R )}$, asserts that the AD holds in $L(\Bbb R)$, which is the least transitive class that contains all the ordinals and all the real numbers, and satisfies the ZF axioms (see Section 7). So, $AD^{L(\Bbb R )}$ implies that every set of reals that belongs to $L(\Bbb R)$ is regular. Also, since $L(\Bbb R)$ contains all projective sets, $AD^{L(\Bbb R )}$ implies PD. The Continuum Hypothesis (CH), formulated by Cantor in 1878, asserts that every infinite set of real numbers has cardinality either $\aleph_0$ or the same cardinality as $\mathbb{R}$. Thus, the CH is equivalent to $2^{\aleph_0}=\aleph_1$. Cantor proved in 1883 that closed sets of real numbers have the perfect set property, from which it follows that every uncountable closed set of real numbers has the same cardinality as $\mathbb{R} $. Thus, the CH holds for closed sets. More than thirty years later, Pavel Aleksandrov extended the result to all Borel sets, and then Mikhail Suslin to all analytic sets. Thus, all analytic sets satisfy the CH. However, the efforts to prove that co-analytic sets satisfy the CH would not succeed, as this is not provable in ZFC. In 1938 Gödel proved the consistency of the CH with ZFC. Assuming that ZF is consistent, he built a model of ZFC, known as the constructible universe, in which the CH holds. Thus, the proof shows that if ZF is consistent, then so is ZF together with the AC and the CH. Hence, assuming ZF is consistent, the AC cannot be disproved in ZF and the CH cannot be disproved in ZFC. See the entry on the continuum hypothesis for the current status of the problem. Gödel’s constructible universe, denoted by $L$, is defined by transfinite recursion on the ordinals, similarly as $V$, but at successor steps, instead of taking the power set of $V_\alpha$ to obtain $V_{\alpha +1}$, one only takes the set of those subsets of $L_\alpha$ that are definable in $L_\alpha$, using elements of $L_\alpha$ as parameters. Thus, letting $\mathcal{P}^{Def} (X)$ to denote the set of all the subsets of $X$ that are definable in the structure $(X,\in )$ by a formula of the language of set theory, using elements of $X$ as parameters of the definition, we let • $L_0={\varnothing}$ • $L_{\alpha +1}=\mathcal{P}^{Def}(L_\alpha)$ • $L_\lambda =\bigcup \{L_\alpha :\alpha <\lambda\}$, whenever $\lambda$ is a limit ordinal. Then $L$ is the union of all the $L_\alpha$, for $\alpha$ an ordinal. Gödel showed that $L$ satisfies all the ZFC axioms, and also the CH. In fact, it satisfies the Generalized Continuum Hypothesis (GCH), namely $2^{\aleph_\alpha}=\aleph_{\alpha +1}$, for every ordinal $\alpha$. The statement $V=L$, called the axiom of constructibility, asserts that every set belongs to $L$. It holds in $L$, hence it is consistent with ZFC, and implies both the AC and the GCH. The proper class $L$, together with the $\in$ relation restricted to $L$, is an inner model of ZFC, that is, a transitive (i.e., it contains all elements of its elements) class that contains all ordinals and satisfies all the ZFC axioms. It is in fact the smallest inner model of ZFC, as any other inner model contains it. More generally, given any set $A$, one can build the smallest transitive model of ZF that contains $A$ and all the ordinals in a similar manner as $L$, but now starting with the transitive closure of $\{A \}$, i.e., the smallest transitive set that contains $A$, instead of ${\varnothing}$. The resulting model, $L(A)$, need not be however a model of the AC. One very important such model is $L(\mathbb{R})$, the smallest transitive model of ZF that contains all the ordinals and all the real numbers. The theory of constructible sets owes much to the work of Ronald Jensen. He developed the so-called fine structure theory of $L$ and isolated some combinatorial principles, such as the diamond ($\ diamondsuit$) and square ($\Box$), which can be used to carry out complicated constructions of uncountable mathematical objects. Fine structure theory plays also an important role in the analysis of bigger $L$-like models, such as $L(\mathbb{R})$ or the inner models for large cardinals (see Section 10.1). In 1963, twenty-five years after Gödel’s proof of the consistency of the CH and the AC, relative to the consistency of ZF, Paul Cohen (1966) proved the consistency of the negation of the CH, and also of the negation of the AC, relative to the consistency of ZF. Thus, if ZF is consistent, then the CH is undecidable in ZFC, and the AC is undecidable in ZF. To achieve this, Cohen devised a new and extremely powerful technique, called forcing, for expanding countable transitive models of ZF, or of ZFC. Since the axiom $V=L$ implies the AC and the CH, any model of the negation of the AC or the CH must violate $V=L$. So, let’s illustrate the idea of forcing in the case of building a model for the negation of $V=L$. We start with a transitive model $M$ of ZFC, which we may assume, without loss of generality, to be a model of $V=L$. To violate $V=L$ we need to expand $M$ by adding a new set $r$ so that, in the expanded model, $r$ will be non-constructible. Since all hereditarily-finite sets are constructible, we aim to add an infinite set of natural numbers. The first problem we face is that $M$ may contain already all subsets of $\omega$. Fortunately, by the Löwenheim-Skolem theorem for first-order logic, $M$ has an elementary submodel which is isomorphic to a countable transitive model $N$. So, since we are only interested in the statements that hold in $M$, and not in $M$ itself, we may as well work with $N$ instead of $M$, and so we may assume that $M$ itself is countable. Then, since $\mathcal{P}(\omega)$ is uncountable, there are plenty of subsets of $\omega$ that do not belong to $M$. But, unfortunately, we cannot just pick any infinite subset $r$ of $\omega$ that does not belong to $M$ and add it to $M$. The reason is that $r$ may encode a lot of information, so that when added to $M$, $M$ is no longer a model of ZF, or it is still a model of $V=L$. To avoid this, one needs to pick $r$ with great care. The idea is to pick $r$ generic over $M$, meaning that $r$ is built from its finite approximations in such a way that it does not have any property that is definable in $M$ and can be avoided. For example, by viewing $r$ as an infinite sequence of natural numbers in the increasing order, the property of $r$ containing only finitely-many even numbers can be avoided, because given any finite approximation to $r$—i.e., any finite increasing sequence of natural numbers—one can always extend it by adding more even numbers, so that at the end of the construction $r$ will contain infinitely-many even numbers; while the property of containing the number 7 cannot be avoided, because when a finite approximation to $r$ contains the number 7, then it stays there no matter how the construction of $r$ proceeds. Since $M$ is countable, there are such generic $r$. Then the expanded model $M[r]$, which includes $M$ and contains the new set $r$, is called a generic extension of $M$. Since we assumed $M$ is a transitive model of $V=L$, the model $M[r]$ is just $L_\alpha (r)$, where $\alpha$ is the supremum of the ordinals of $M$. Then one can show, using the forcing relation between finite approximations to $r$ and formulas in the language of set theory expanded with so-called names for sets in the generic extension, that $M[r]$ is a model of ZFC and $r$ is not constructible in $M[r]$, hence the axiom of constructibility $V=L$ fails. In general, a forcing extension of a model $M$ is obtained by adding to $M$ a generic subset $G$ of some partially ordered set $\mathbb{P}$ that belongs to $M$. In the above example, $\ mathbb{P}$ would be the set of all finite increasing sequences of natural numbers, seen as finite approximations to the infinite sequence $r$, ordered by $\subseteq$; and $G$ would be the set of all finite initial segments of $r$. In the case of the consistency proof of the negation of the CH, one starts from a model $M$ of ZFC, as before, and adds $\aleph_2$ new subsets of $\omega$, so that in the generic extension the CH fails. In this case one needs to use an appropriate partial ordering $\mathbb{P}$ so that the $\aleph_2$ of $M$ is not collapsed, i.e., it is the same as the $\aleph_2$ of the generic extension, and thus the generic extension $M[G]$ will satisfy the sentence that says that there are $\aleph_2$ real numbers. Besides the CH, many other mathematical conjectures and problems about the continuum, and other infinite mathematical objects, have been shown undecidable in ZFC using the forcing technique. One important example is Suslin’s Hypothesis (SH). Cantor had shown that every linearly ordered set $S$ without endpoints that is dense (i.e., between any two different elements of $S$ there is another one), complete (i.e., every subset of $S$ that is bounded above has a supremum), and with a countable dense subset is isomorphic to the real line. Suslin conjectured that this is still true if one relaxes the requirement of containing a countable dense subset to being ccc, i.e., every collection of pairwise-disjoint intervals is countable. In the early 1970s, Thomas Jech produced a consistent counterexample using forcing, and Ronald Jensen showed that a counterexample exists in $L$. About the same time, Robert Solovay and Stanley Tennenbaum (1971) developed and used for the first time the iterated forcing technique to produce a model where the SH holds, thus showing its independence from ZFC. In order to make sure that the SH holds in the generic extension, one needs to destroy all counterexamples, but by destroying one particular counterexample one may inadvertently create new ones, and so one needs to force again and again; in fact one needs to go on for at least $\omega_2$-many steps. This is why a forcing iteration is needed. Among other famous mathematical problems that have been shown undecidable in ZFC thanks to the forcing technique, especially using iterated forcing and sometimes combined with large cardinals, we may mention the Measure Problem and the Borel Conjecture in measure theory, Kaplansky’s Conjecture on Banach algebras, and Whitehead’s Problem in group theory. As a result of 60 years of development of the forcing technique, and its applications to many open problems in mathematics, there are now literally hundreds of problems and questions, in practically all areas of mathematics, that have been shown independent of ZFC. These include almost all important questions about the structure of uncountable sets. One might say that the undecidability phenomenon is pervasive, to the point that the investigation of the uncountable has been rendered nearly impossible in ZFC alone (see however Shelah (1994) for remarkable exceptions). This prompts the question about the truth-value of the statements that are undecided by ZFC. Should one be content with them being undecidable? Does it make sense at all to ask for their truth-value? There are several possible reactions to this. One is the skeptic’s position: the statements that are undecidable in ZFC have no definite answer; and they may even be inherently vague. Another, the common one among mathematicians, is Gödel’s position: the undecidability only shows that the ZFC system is too weak to answer those questions, and therefore one should search for new axioms that once added to ZFC would answer them. The search for new axioms has been known as Gödel’s Program. See Hauser (2006) for a thorough philosophical discussion of the Program, and also the entry on large cardinals and determinacy for philosophical considerations on the justification of new axioms for set theory. A central theme of set theory is thus the search and classification of new axioms. These fall currently into two main types: the axioms of large cardinals and the forcing axioms. One cannot prove in ZFC that there exists a regular limit cardinal $\kappa$, for if $\kappa$ is such a cardinal, then $L_\kappa$ is a model of ZFC, and so ZFC would prove its own consistency, contradicting Gödel’s second incompleteness theorem. Thus, the existence of a regular limit cardinal must be postulated as a new axiom. Such a cardinal is called weakly inaccessible. If, in addition $\kappa$ is a strong limit, i.e., $2^\lambda <\kappa$, for every cardinal $\lambda <\kappa$, then $\kappa$ is called strongly inaccessible. A cardinal $\kappa$ is strongly inaccessible if and only if it is regular and $V_\kappa $ is a model of ZFC. If the GCH holds, then every weakly inaccessible cardinal is strongly inaccessible. Large cardinals are uncountable cardinals satisfying some properties that make them very large, and whose existence cannot be proved in ZFC. The first weakly inaccessible cardinal is just the smallest of all large cardinals. Beyond inaccessible cardinals there is a rich and complex variety of large cardinals, which form a linear hierarchy in terms of consistency strength, and in many cases also in terms of outright implication. See the entry on independence and large cardinals for more details. To formulate the next stronger large-cardinal notion, let us say that a subset $C$ of an infinite cardinal $\kappa$ is closed if every limit of elements of $C$ which is less than $\kappa$ is also in $C$; and is unbounded if for every $\alpha <\kappa$ there exists $\beta\in C$ greater than $\alpha$. For example, the set of limit ordinals less than $\kappa$ is closed and unbounded. Also, a subset $S$ of $\kappa$ is called stationary if it intersects every closed unbounded subset of $\kappa$. If $\kappa$ is regular and uncountable, then the set of all ordinals less than $\kappa$ of cofinality $\omega$ is an example of a stationary set. A regular cardinal $\kappa$ is called Mahlo if the set of strongly inaccessible cardinals smaller than $\kappa$ is stationary. Thus, the first Mahlo cardinal is much larger than the first strongly inaccessible cardinal, as there are $\kappa$-many strongly inaccessible cardinals smaller than $\kappa$. Much stronger large cardinal notions arise from considering strong reflection properties. Recall that the Reflection Principle (Section 4), which is provable in ZFC, asserts that every true sentence (i.e., every sentence that holds in $V$) is true in some $V_\alpha$. A strengthening of this principle to second-order sentences yields some large cardinals. For example, $\kappa$ is strongly inaccessible if and only if every $\Sigma^1_1$ sentence (i.e., existential second-order sentence in the language of set theory, with one additional predicate symbol) true in any structure of the form $(V_\kappa ,\in, A)$, where $A\subseteq V_\kappa$, is true in some $(V_\alpha ,\in ,A\cap V_\alpha)$, with $\alpha <\kappa$. The same type of reflection, but now for $\Pi^1_1$ sentences (i.e., universal second-order sentences), yields a much stronger large cardinal property of $\kappa$, called weak compactness. Every weakly compact cardinal $\kappa$ is Mahlo, and the set of Mahlo cardinals smaller than $\kappa$ is stationary. By allowing reflection for more complex second-order, or even higher-order, sentences one obtains large cardinal notions stronger than weak compactness. The most famous large cardinals, called measurable, were discovered by Stanisław Ulam in 1930 as a result of his solution to the Measure Problem. A (two-valued) measure, or ultrafilter, on a cardinal $\kappa$ is a subset $U$ of $\mathcal{P}(\kappa)$ that has the following properties: (i) the intersection of any two elements of $U$ is in $U$; (ii) if $X\in U$ and $Y$ is a subset of \ (\kappa\) such that $X\subseteq Y$, then $Y\in U$; and (iii) for every $X\subseteq\kappa$, either $X\in U$ or $\kappa -X\in U$, but not both. A measure $U$ is called $\kappa$-complete if every intersection of less than $\kappa$ elements of $U$ is also in $U$. And a measure is called non-principal if there is no $\alpha <\kappa$ that belongs to all elements of $U$. A cardinal $\kappa$ is called measurable if there exists a measure on $\kappa$ that is $\kappa$-complete and non-principal. Measurable cardinals can be characterized by elementary embeddings of the universe $V$ into some transitive class $M$. That such an embedding $j:V\to M$ is elementary means that $j$ preserves truth, i.e., for every formula $\varphi (x_1,\ldots ,x_n)$ of the language of set theory, and every $a_1,\ldots ,a_n$, the sentence $\varphi (a_1,\ldots ,a_n)$ holds in $V$ if and only if $\ varphi (j(a_1),\ldots ,j(a_n))$ holds in $M$. It turns out that a cardinal $\kappa$ is measurable if and only if there exists an elementary embedding $j:V\to M$, with $M$ transitive, so that $\kappa$ is the first ordinal moved by $j$, i.e., the first ordinal such that $j(\kappa)\ne \kappa$. We say that $\kappa$ is the critical point of $j$, and write $crit(j)=\kappa$. The embedding $j$ is definable from a $\kappa$-complete non-principal measure on $\kappa$, using the so-called ultrapower construction. Conversely, if $j:V\to M$ is an elementary embedding, with $M$ transitive and $\kappa=crit(j)$, then the set $U=\{ X\subseteq \kappa:\kappa\in j(X)\}$ is a $\kappa$-complete non-principal ultrafilter on $\kappa$. A measure $U$ obtained in this way from $j$ is called normal. Every measurable cardinal $\kappa$ is weakly compact, and there are many weakly compact cardinals smaller than $\kappa$. In fact, below $\kappa$ there are many cardinals that are totally indescribable, i.e., they reflect all sentences, of any complexity, and in any higher-order language. If $\kappa$ is measurable and $j:V\to M$ is given by the ultrapower construction, then $V_\kappa \subseteq M$, and every sequence of length less than or equal to $\kappa$ of elements of $M$ belongs to $M$. Thus, $M$ is quite similar to $V$, but it cannot be $V$ itself. Indeed, a famous theorem of Kenneth Kunen shows that there cannot be any elementary embedding $j:V\to V$, other than the trivial one, i.e., the identity. All known proofs of this result use the Axiom of Choice, and it is an outstanding important question if the axiom is necessary. Kunen’s Theorem opens the door to formulating large cardinal notions stronger than measurability by requiring that $M$ is closer to $V$. For example, $\kappa$ is called strong if for every ordinal $\alpha$ there exists an elementary embedding $j:V\to M$, for some $M$ transitive, such that $\kappa =crit(j)$ and $V_\alpha \ subseteq M$. Another important, and much stronger large cardinal notion is supercompactness. A cardinal $\kappa$ is supercompact if for every $\alpha$ there exists an elementary embedding $j:V\to M$, with \ (M\) transitive and critical point $\kappa$, so that $j(\kappa)>\alpha$ and every sequence of elements of $M$ of length $\alpha$ belongs to $M$. Woodin cardinals fall between strong and supercompact. Every supercompact cardinal is Woodin, and if $\delta$ is Woodin, then $V_\delta$ is a model of ZFC in which there is a proper class of strong cardinals. Thus, while a Woodin cardinal $\delta$ need not be itself very strong—the first one is not even weakly compact—it implies the existence of many large cardinals in $V_\delta$. Beyond supercompact cardinals we find the extendible cardinals, the huge, the super huge, etc. Kunen’s theorem about the non-existence of a non-trivial elementary embedding $j:V\to V$ actually shows that there cannot be an elementary embedding $j:V_{\lambda +2}\to V_{\lambda +2}$ different from the identity, for any $\lambda$. The strongest large cardinal notions not known to be inconsistent, modulo ZFC, are the following: • There exists an elementary embedding $j:V_{\lambda +1}\to V_{\lambda +1}$ different from the identity. • There exists an elementary embedding $j:L(V_{\lambda +1})\to L(V_{\lambda +1})$ different from the identity. Large cardinals form a linear hierarchy of increasing consistency strength. In fact they are the stepping stones of the interpretability hierarchy of mathematical theories. See the entry on independence and large cardinals for more details. Typically, given any sentence $\varphi$ of the language of set theory, exactly one the following three possibilities holds about the theory ZFC plus $\varphi$: • ZFC plus $\varphi$ is inconsistent. • ZFC plus $\varphi$ is equiconsistent with ZFC (i.e., ZFC is consistent if and only if so is ZFC plus $\varphi$). • ZFC plus $\varphi$ is equiconsistent with ZFC plus the existence of some large cardinal. Thus, large cardinals can be used to prove that a given sentence $\varphi$ does not imply another sentence $\psi$, modulo ZFC, by showing that ZFC plus $\psi$ implies the consistency of some large cardinal, whereas ZFC plus $\varphi$ is consistent assuming the existence of a smaller large cardinal, or just assuming the consistency of ZFC. In other words, $\psi$ has higher consistency strength than $\varphi$, modulo ZFC. Then, by Gödel’s second incompleteness theorem, ZFC plus $\varphi$ cannot prove $\psi$, assuming ZFC plus $\varphi$ is consistent. As we already pointed out, one cannot prove in ZFC that large cardinals exist. But everything indicates that their existence not only cannot be disproved, but in fact the assumption of their existence is a very reasonable axiom of set theory. For one thing, there is a lot of evidence for their consistency, especially for those large cardinals for which it is possible to construct a canonical inner model. An inner model of ZFC is a transitive proper class that contains all the ordinals and satisfies all ZFC axioms. Thus, $L$ is the smallest inner model, while $V$ is the largest. Some large cardinals, such as inaccessible, Mahlo, or weakly-compact, may exist in $L$. That is, if $\kappa$ has one of these large cardinal properties, then it also has the property in $L$. But some large cardinals cannot exist in $L$. Indeed, Scott (1961) showed that if there exists a measurable cardinal $\kappa$, then $V\ne L$. It is important to notice that $\kappa$ does belong to $L $, since $L$ contains all ordinals, but it is not measurable in $L$ because a $\kappa$-complete non-principal measure on $\kappa$ cannot exist there. If $\kappa$ is a measurable cardinal, then one can construct an $L$-like model in which $\kappa$ is measurable by taking a $\kappa$-complete non-principal and normal measure $U$ on $\kappa $, and proceeding as in the definition of $L$, but now using $U$ as an additional predicate. The resulting model, called $L[U]$, is an inner model of ZFC in which $\kappa$ is measurable, and in fact $\kappa$ is the only measurable cardinal. The model is canonical, in the sense that any other normal measure witnessing the measurability of $\kappa$ would yield the same model, and has many of the properties of $L$. For instance, it has a projective well ordering of the reals, and it satisfies the GCH. Building similar $L$-like models for stronger large cardinals, such as strong, or Woodin, is much harder. Those models are of the form $L[E]$, where $E$ is a sequence of extenders, each extender being a coherent system of measures, that encode the relevant elementary embeddings. The largest $L$-like inner models for large cardinals that have been obtained so far can contain Woodin cardinals that are limits of Woodin cardinals (Neeman 2002). However, building an $L$-like model for a supercompact cardinal is still a challenge. The supercompact barrier seems to be the crucial one, for Woodin has shown that for a natural kind of inner model for a supercompact cardinal \ (\kappa\), which he calls a weak extender model for the supercompactness of $ \kappa $, all stronger large cardinals greater than $\kappa$ that may exist in $V$, such as extendible, huge, etc. would also exist in the model. The existence of large cardinals has dramatic consequences, even for simply-definable small sets, like the projective sets of real numbers. For example, Solovay (1970) proved, assuming that there exists a measurable cardinal, that all $\mathbf{\Sigma}^1_2$ sets of reals are Lebesgue measurable and have the Baire property, which cannot be proved in ZFC alone. And Shelah and Woodin (1990) showed that the existence of a proper class of Woodin cardinals implies that the theory of $L(\mathbb{R})$, even with real numbers as parameters, cannot be changed by forcing, which implies that all sets of real numbers that belong to $L(\mathbb{R})$ are regular. Further, under a weaker large-cardinal hypothesis, namely the existence of infinitely many Woodin cardinals, Martin and Steel (1989) proved that every projective set of real numbers is determined, i.e., the axiom of PD holds, hence all projective sets are regular. Moreover, Woodin showed that the existence of infinitely many Woodin cardinals, plus a measurable cardinal above all of them, implies that every set of reals in $L(\mathbb{R})$ is determined, i.e., the axiom $AD^{L(\mathbb{R})}$ holds, hence all sets of real numbers that belong to $L(\mathbb{R})$, and therefore all projective sets, are regular. He also showed that Woodin cardinals provide the optimal large cardinal assumptions by proving that the following two statements: 1. There are infinitely many Woodin cardinals. 2. $AD^{L({\Bbb R})}$. are equiconsistent, i.e., ZFC plus 1 is consistent if and only if ZFC plus 2 is consistent. See the entry on large cardinals and determinacy for more details and related results. Another area in which large cardinals play an important role is the exponentiation of singular cardinals. The so-called Singular Cardinal Hypothesis (SCH) completely determines the behavior of the exponentiation for singular cardinals, modulo the exponentiation for regular cardinals. The SCH follows from the GCH, and so it holds in $L$. A consequence of the SCH is that if $2^{\aleph_n}<\ aleph_\omega$, for all finite $n$, then $2^{\aleph_{\omega}}=\aleph_{\omega +1}$. Thus, if the GCH holds for cardinals smaller than $\aleph_\omega$, then it also holds at $\aleph_\omega$. The SCH holds above the first supercompact cardinal (Solovay). But Magidor (1977) showed that, remarkably, assuming the existence of large cardinals it is possible to build a model of ZFC where the GCH first fails at $\aleph_\omega$, hence the SCH fails. Large cardinals stronger than measurable are actually needed for this. In contrast, however, ZFC alone suffices to prove that if the SCH holds for all cardinals smaller than $\aleph_{\omega_1}$, then it also holds for $\aleph_{\omega_1}$. Moreover, if the SCH holds for all singular cardinals of countable cofinality, then it holds for all singular cardinals (Silver). Forcing axioms are axioms of set theory that assert that certain existential statements are absolute between the universe $V$ of all sets and its (ideal) forcing extensions, i.e., some existential statements that hold in some forcing extensions of $V$ are already true in $V$. The first forcing axiom was formulated by Donald Martin in the wake of the Solovay-Tennenbaum proof of the consistency of Suslin’s Hypothesis, and is now known as Martin’s Axiom (MA). Before we state it, let us say that a partial ordering is a non-empty set $P$ together with a binary relation $\leq$ on $P$ that is reflexive and transitive. Two elements, $p$ and $q$, of $P$ are called compatible if there exists $r\in P$ such that $r\leq p$ and $r\leq q$. An antichain of $P$ is a subset of $P$ whose elements are pairwise-incompatible. A partial ordering $P$ is called ccc if every antichain of $P$ is countable. A non-empty subset $G$ of $P$ is called a filter if (i) every two elements of $G$ are compatible, and (ii) if $p\in G$ and $p\leq q$, then also $q\in G$. Finally, a subset $D$ of $P$ is called dense if for every $p\in P$ there is $q\in D$ such that $q\leq p$. MA asserts the following: For every ccc partial ordering $P$ and every set $\{ D_\alpha :\alpha <\kappa \}$, where $\kappa \lt 2^{\aleph_0}$, of dense subsets of $P$, there exists a filter $G\subseteq P$ that is generic for the set, i.e., $G\cap D_\alpha \ne {\varnothing}$, for all $\alpha <\kappa$. Since MA follows easily from the CH, MA is only of interest if the CH fails. Martin and Solovay (1970) proved that MA plus the negation of the CH is consistent with ZFC, using iterated forcing with the ccc property. At first sight, MA may not look like an axiom, namely an obvious, or at least reasonable, assertion about sets, but rather like a technical statement about ccc partial orderings. It does look more natural, however, when expressed in topological terms, for it is simply a generalization of the well-known Baire Category Theorem, which asserts that in every compact Hausdorff topological space the intersection of countably-many dense open sets is non-empty. Indeed, MA is equivalent to: In every compact Hausdorff ccc topological space, the intersection of less than $2^{\aleph_0}$-many dense open sets is non-empty. MA has many different equivalent formulations and has been used very successfully to settle a large number of open problems in other areas of mathematics. For example, MA plus the negation of the CH implies Suslin’s Hypothesis and that every $\mathbf{\Sigma}^1_2$ set of reals is Lebesgue measurable and has the Baire property. It also implies that $2^{\aleph_0}$ is a regular cardinal, but it does not decide what cardinal it is. See Fremlin (1984) for many more consequences of MA and other equivalent formulations. In spite of this, the status of MA as an axiom of set theory is still unclear. Perhaps the most natural formulation of MA, from a foundational point of view, is in terms of generic absoluteness. Namely, MA is equivalent to the following: For every ccc partial ordering $P$, if an existential statement, containing subsets of some cardinal less than $2^{\aleph_0}$ as parameters, holds in an (ideal) generic extension of $V$ obtained by forcing with $P$, then the statement is true, i.e., it holds in $V$. In other words, if a set having a property that depends only on bounded subsets of $ 2^{\aleph_0}$ exists in some (ideal) generic extension of $V$ obtained by forcing with a ccc partial ordering, then a set with that property already exists in $V$. The notion of ideal generic extension of $V$ can be made precise in terms of so-called Boolean-valued models, which provide an alternative version of forcing. Much stronger forcing axioms than MA (for $\omega_1$) were introduced in the 1980s, such as J. Baumgartner’s Proper Forcing Axiom (PFA), and the stronger Martin’s Maximum (MM) of Foreman, Magidor, and Shelah (1988). Both the PFA and MM are consistent relative to the existence of a supercompact cardinal. The PFA asserts the same as MA, but for partial orderings that have a property weaker than the ccc, called properness, introduced by Shelah. And MM asserts the same for the wider class of partial orderings that, when forcing with them, do not destroy stationary subsets of $\omega_1$. Strong forcing axioms, such as the PFA and MM imply that all projective sets of reals are determined (PD), and have many other strong consequences in infinite combinatorics. Notably, they imply that the cardinality of the continuum is $\aleph_2$. A different forcing axiom is Woodin’s Axiom $(\ast)$, which resulted from his analysis of $\mathbb{P}_{{\rm max}}$ forcing extensions over models of AD. Even though a stronger form of MM known as ${\rm MM}^{++}$ and Axiom $(\ast)$ have similar and very rich consequences, both their motivation and their formulation are very different, to the point that they were seen as competing axioms. Their connection was far from being clear, until Asperó and Schindler (2021) showed that, in the presence of large cardinals, Axiom $(\ast)$ is equivalent to a bounded form of ${\rm MM}^{++}$, and therefore to a natural generic absoluteness principle. • Asperó, D. and Schindler, R. D., 2021, “Martin’s Maximum$^{++}$ implies Woodin’s Axiom $(\ast)$”, Annals of Mathematics, 193(3): 793–835. • Bagaria, J., 2008, “Set Theory”, in The Princeton Companion to Mathematics, edited by Timothy Gowers; June Barrow-Green and Imre Leader, associate editors. Princeton: Princeton University Press. • Cohen, P.J., 1966, Set Theory and the Continuum Hypothesis, New York: W. A. Benjamin, Inc. • Enderton, H.B., 1977, Elements of Set Theory, New York: Academic Press. • Ferreirós, J., 2007, Labyrinth of Thought: A History of Set Theory and its Role in Modern Mathematics, Second revised edition, Basel: Birkhäuser. • Foreman, M., Magidor, M., and Shelah, S., 1988, “Martin’s maximum, saturated ideals and non-regular ultrafilters”, Part I, Annals of Mathematics, 127: 1–47. • Fremlin, D.H., 1984, “Consequences of Martin’s Axiom”, Cambridge tracts in Mathematics #84. Cambridge: Cambridge University Press. • Gödel, K., 1931, “Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme I,” Monatshefte für Mathematik Physik, 38: 173–198. English translation in Gödel 1986, • –––, 1938, “The consistency of the axiom of choice and of the generalized continuum hypothesis”, Proceedings of the National Academy of Sciences, U.S.A. 24: 556–557. • –––, 1986, Collected Works I. Publications 1929–1936, S. Feferman et al. (eds.), Oxford: Oxford University Press. • Hauser, K., 2006, “Gödel’s program revisited, Part I: The turn to phenomenology”, Bulletin of Symbolic Logic, 12(4): 529–590. • Jech, T., 2003, Set theory, 3d Edition, New York: Springer. • Jensen, R.B., 1972, “The fine structure of the constructible hierarchy”, Annals of Mathematical Logic, 4(3): 229–308. • Kanamori, A., 2003, The Higher Infinite, Second Edition. Springer Monographs in Mathematics, New York: Springer. • Kechris, A.S., 1995, Classical Descriptive Set Theory, Graduate Texts in Mathematics, New York: Springer Verlag. • Kunen, K., 1980, Set Theory, An Introduction to Independence Proofs, Amsterdam: North-Holland. • Levy, A., 1960, “Axiom schemata of strong infinity in axiomatic set theory”, Pacific Journal of Mathematics, 10: 223–238. • –––, 1979, Basic Set Theory, New York: Springer. • Magidor, M., 1977, “On the singular cardinals problem, II”, Annals of Mathematics, 106: 514–547. • Martin, D.A. and R. Solovay, 1970, “Internal Cohen Extensions”, Annals of Mathematical Logic, 2: 143–178. • Martin, D.A. and J.R. Steel, 1989, “A proof of projective determinacy”, Journal of the American Mathematical Society, 2(1): 71–125. • Mathias, A.R.D., 2001, “Slim models of Zermelo Set Theory”, Journal of Symbolic Logic, 66: 487–496. • Neeman, I., 2002, “Inner models in the region of a Woodin limit of Woodin cardinals”, Annals of Pure and Applied Logic, 116: 67–155. • Scott, D., 1961, “Measurable cardinals and constructible sets”, Bulletin de l’Académie Polonaise des Sciences. Série des Sciences Mathématiques, Astronomiques et Physiques, 9: 521–524. • Shelah, S., 1994, “Cardinal Arithmetic”, Oxford Logic Guides, 29, New York: The Clarendon Press, Oxford University Press. • –––, 1998, Proper and improper forcing, 2nd Edition, New York: Springer-Verlag. • Shelah, S. and W.H. Woodin, 1990, “Large cardinals imply that every reasonably definable set of reals is Lebesgue measurable”, Israel Journal of Mathematics, 70(3): 381–394. • Solovay, R., 1970, “A model of set theory in which every set of reals is Lebesgue measurable”, Annals of Mathematics, 92: 1–56. • Solovay, R. and S. Tennenbaum, 1971, “Iterated Cohen extensions and Souslin’s problem”, Annals of Mathematics (2), 94: 201–245. • Todorcevic, S., 1989, “Partition Problems in Topology”, Contemporary Mathematics, Volume 84. American Mathematical Society. • Ulam, S., 1930, ‘Zur Masstheorie in der allgemeinen Mengenlehre’, Fundamenta Mathematicae, 16: 140–150. • Woodin, W.H., 1999, The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal, De Gruyter Series in Logic and Its Applications 1, Berlin-New York: Walter de Gruyter. • –––, 2001, “The Continuum Hypothesis, Part I”, Notices of the AMS, 48(6): 567–576, and “The Continuum Hypothesis, Part II”, Notices of the AMS 48(7): 681–690. • Zeman, M., 2001, Inner Models and Large Cardinals, De Gruyter Series in Logic and Its Applications 5, Berlin-New York: Walter de Gruyter. • Zermelo, E., 1908, “Untersuchungen über die Grundlagen der Mengenlehre, I”, Mathematische Annalen 65: 261–281. Reprinted in Zermelo 2010: 189–228, with a facing-page English translation, and an Introduction by Ulrich Felgner (2010). English translation also in van Heijenoort 1967: 201–215. Academic Tools How to cite this entry. Preview the PDF version of this entry at the Friends of the SEP Society. Look up topics and thinkers related to this entry at the Internet Philosophy Ontology Project (InPhO). Enhanced bibliography for this entry at PhilPapers, with links to its database. • Jech, Thomas, “Set Theory”, The Stanford Encyclopedia of Philosophy (Fall 2014 Edition), Edward N. Zalta (ed.), URL = <https://plato.stanford.edu/archives/fall2014/entries/set-theory/>. [This was the previous entry on set theory in the Stanford Encyclopedia of Philosophy — see the version history.]	{"url":"https://plato.stanford.edu/ENTRIES/set-theory/index.html","timestamp":"2024-11-04T04:13:38Z","content_type":"text/html","content_length":"95139","record_id":"<urn:uuid:97837177-3285-41cf-a3f3-9fa11a759d76>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00691.warc.gz"}
Free vector categories \| Mathematical patterns In this category we will be publishing a variety of interesting mathematical patterns, diagrams, curves etc. Particular attention will be paid to the naturally occurring mathematical patterns in the world around us (fractals, Fibonacci numbers, Voronoi diagrams, tessellations,…), i.e. natural forms that can be explained by mathematical laws. If you have an idea for some interesting mathematical pattern or mathematical natural form, please send us your proposal and we will try to draw and publish it within this category. All Mathematical 2D patterns are under Creative commons attribution licenses; depending on the license type, some patterns are only for personal use and some for commercial use as well. Before you make use of these patterns in your project, please check and observe the license that we have prescribed. If you need Mathematical 3D surfaces, you can find them here: Mathematical 3D surfaces Black and white vector portrait of the famous Greek mathematician, physicist, inventor, and astronomer Archimedes of Syracuse. On this page you can download free Cairo pentagonal tiling pattern. Pentagons are not regular pentagons, their sides are different lengths. Here you can download mathematical patterns that are created when drawing cardioids (6 examples resemble a chessboard and 6 are radial patterns). This is a large collection of Voronoi diagrams that were parametrically generated using an algorithm in 2D CAD software. This collection contains parametrically generated examples of all 17 wallpaper groups, and to make the differences between them more obvious, they were all generated using the same design element. Here you can download Fibonacci number 2D patterns. All three of them are examples of integration of mathematics and nature. Here you can download a collection of various versions of the pattern known in applied mathematics as Gilbert's Tessellation. This collection is called Infinite Twisted Tunnel Patterns, and this type of pattern can also be found under the name ‘tunnel optical illusion’ or ‘abstract tunnel’. Here lovers of parametric design can download patterns generated from random lines using an algorithm and 2D CAD software. One of the frequently seen mathematical patterns in nature is the Voronoi pattern whose 4 variations can be downloaded on this page.	{"url":"http://craftsmanspace.com/free-vectors/mathematical-patterns","timestamp":"2024-11-03T03:35:57Z","content_type":"text/html","content_length":"54723","record_id":"<urn:uuid:c402ea85-78a4-4bb6-bf53-9eec73c18bb0>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00304.warc.gz"}
lroundf: round to nearest integer value - Linux Manuals (3p) lroundf (3p) - Linux Manuals lroundf: round to nearest integer value This manual page is part of the POSIX Programmer's Manual. The Linux implementation of this interface may differ (consult the corresponding Linux manual page for details of Linux behavior), or the interface may not be implemented on Linux. lround, lroundf, lroundl - round to nearest integer value #include <math.h> long lround(double x); long lroundf(float x); long lroundl(long double x); These functions shall round their argument to the nearest integer value, rounding halfway cases away from zero, regardless of the current rounding direction. An application wishing to check for error situations should set errno to zero and call feclearexcept(FE_ALL_EXCEPT) before calling these functions. On return, if errno is non-zero or fetestexcept (FE_INVALID \| FE_DIVBYZERO \| FE_OVERFLOW \| FE_UNDERFLOW) is non-zero, an error has occurred. Upon successful completion, these functions shall return the rounded integer value. If x is NaN, a domain error shall occur and an unspecified value is returned. If x is +Inf, a domain error shall occur and an unspecified value is returned. If x is -Inf, a domain error shall occur and an unspecified value is returned. If the correct value is positive and too large to represent as a long, a domain error shall occur and an unspecified value is returned. If the correct value is negative and too large to represent as a long, a domain error shall occur and an unspecified value is returned. These functions shall fail if: Domain Error The x argument is NaN or ±Inf, or the correct value is not representable as an integer. If the integer expression (math_errhandling & MATH_ERRNO) is non-zero, then errno shall be set to [EDOM]. If the integer expression (math_errhandling & MATH_ERREXCEPT) is non-zero, then the invalid floating-point exception shall be raised. The following sections are informative. On error, the expressions (math_errhandling & MATH_ERRNO) and (math_errhandling & MATH_ERREXCEPT) are independent of each other, but at least one of them must be non-zero. These functions differ from the lrint() functions in the default rounding direction, with the lround() functions rounding halfway cases away from zero and needing not to raise the inexact floating-point exception for non-integer arguments that round to within the range of the return type. Portions of this text are reprinted and reproduced in electronic form from IEEE Std 1003.1, 2003 Edition, Standard for Information Technology -- Portable Operating System Interface (POSIX), The Open Group Base Specifications Issue 6, Copyright (C) 2001-2003 by the Institute of Electrical and Electronics Engineers, Inc and The Open Group. In the event of any discrepancy between this version and the original IEEE and The Open Group Standard, the original IEEE and The Open Group Standard is the referee document. The original Standard can be obtained online at feclearexcept(), fetestexcept(), llround(), the Base Definitions volume of IEEE Std 1003.1-2001, Section 4.18, Treatment of Error Conditions for Mathematical Functions, <math.h>	{"url":"https://www.systutorials.com/docs/linux/man/3p-lroundf/","timestamp":"2024-11-04T21:46:38Z","content_type":"text/html","content_length":"10555","record_id":"<urn:uuid:6b129531-b9a4-484b-a89d-ea7bdcabd7bb>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00876.warc.gz"}
Programming Assignment 2: Decomposition of Graphs Welcome to your second programming assignment of the Algorithms on Graphs class! In this assignment, we focus on directed graphs and their parts. In this programming assignment, the grader will show you the input and output data if your solution fails on any of the tests. This is done to help you to get used to the algorithmic problems in general and get some experience debugging your programs while knowing exactly on which tests they fail. However, for all the following programming assignments, the grader will show the input data only in case your solution fails on one of the first few tests (please review the questions 6.4 and 6.5 in the FAQ section for a more detailed explanation of this behavior of the grader). Learning Outcomes Upon completing this programming assignment you will be able to: 1. check consistency of Computer Science curriculum; 2. find an order of courses that is consistent with prerequisite dependencies; 3. check whether any intersection of a city is reachable from any other intersection. Passing Criteria: 2 out of 3 Passing this programming assignment requires passing at least 2 out of 3 code problems from this assignment. In turn, passing a code problem requires implementing a solution that passes all the tests for this problem in the grader and does so under the time and memory limits specified in the problem statement. 1 Graph Representation in Programming Assignments 3 2 Problem: Checking Consistency of CS Curriculum 5 3 Problem: Determining an Order of Courses 7 4 Advanced Problem: Checking Whether Any Intersection in a City is Reachable from Any Other 9 5 General Instructions and Recommendations on Solving Algorithmic Problems 11 5.1 Reading the Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 5.2 Designing an Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 5.3 Implementing Your Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 5.4 Compiling Your Program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 5.5 Testing Your Program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 5.6 Submitting Your Program to the Grading System . . . . . . . . . . . . . . . . . . . . . . . . . 13 5.7 Debugging and Stress Testing Your Program . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 6 Frequently Asked Questions 14 6.1 I submit the program, but nothing happens. Why? . . . . . . . . . . . . . . . . . . . . . . . . 14 6.2 I submit the solution only for one problem, but all the problems in the assignment are graded. Why? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 6.3 What are the possible grading outcomes, and how to read them? . . . . . . . . . . . . . . . . 14 6.4 How to understand why my program fails and to fix it? . . . . . . . . . . . . . . . . . . . . . 15 6.5 Why do you hide the test on which my program fails? . . . . . . . . . . . . . . . . . . . . . . 15 6.6 My solution does not pass the tests? May I post it in the forum and ask for a help? . . . . . . 16 6.7 My implementation always fails in the grader, though I already tested and stress tested it a lot. Would not it be better if you give me a solution to this problem or at least the test cases that you use? I will then be able to fix my code and will learn how to avoid making mistakes. Otherwise, I do not feel that I learn anything from solving this problem. I am just stuck. . . . 16 1 Graph Representation in Programming Assignments In programming assignments, graphs are given as follows. The first line contains non-negative integers 𝑛 and 𝑚 — the number of vertices and the number of edges respectively. The vertices are always numbered from 1 to 𝑛. Each of the following 𝑚 lines defines an edge in the format u v where 1 ≤ 𝑢, 𝑣 ≤ 𝑛 are endpoints of the edge. If the problem deals with an undirected graph this defines an undirected edge between 𝑢 and 𝑣. In case of a directed graph this defines a directed edge from 𝑢 to 𝑣. If the problem deals with a weighted graph then each edge is given as u v w where 𝑢 and 𝑣 are vertices and 𝑤 is a weight. It is guaranteed that a given graph is simple. That is, it does not contain self-loops (edges going from a vertex to itself) and parallel edges. ∙ An undirected graph with four vertices and five edges: ∙ A directed graph with five vertices and eight edges. ∙ A directed graph with five vertices and one edge. Note that the vertices 1, 2, and 5 are isolated (have no adjacent edges), but they are still present in the graph. ∙ A weighted directed graph with three vertices and three edges. 1 2 -2 2 Problem: Checking Consistency of CS Curriculum Problem Introduction A Computer Science curriculum specifies the prerequisites for each course as a list of courses that should be taken before taking this course. You would like to perform a consistency check of the curriculum, that is, to check that there are no cyclic dependencies. For this, you construct the following directed graph: vertices correspond to courses, there is a directed edge (𝑢, 𝑣) is the course 𝑢 should be taken before the course 𝑣. Then, it is enough to check whether the resulting graph contains a cycle. Problem Description Task. Check whether a given directed graph with 𝑛 vertices and 𝑚 edges contains a cycle. Input Format. A graph is given in the standard format. Constraints. 1 ≤ 𝑛 ≤ 103 , 0 ≤ 𝑚 ≤ 103 Output Format. Output 1 if the graph contains a cycle and 0 otherwise. Time Limits. language C C++ Java Python C# Haskell JavaScript Ruby Scala time (sec) 1 1 1.5 5 1.5 2 5 5 3 Memory Limit. 512MB. Sample 1. This graph contains a cycle: 3 → 1 → 2 → 3. Sample 2. There is no cycle in this graph. This can be seen, for example, by noting that all edges in this graph go from a vertex with a smaller number to a vertex with a larger number. Starter Files The starter solutions for this problem read the input data from the standard input, pass it to a blank procedure, and then write the result to the standard output. You are supposed to implement your algorithm in this blank procedure if you are using C++, Java, or Python3. For other programming languages, you need to implement a solution from scratch. Filename: acyclicity What To Do To solve this problem, it is enough to implement carefully the corresponding algorithm covered in the lectures. 3 Problem: Determining an Order of Courses Problem Introduction Now, when you are sure that there are no cyclic dependencies in the given CS curriculum, you would like to find an order of all courses that is consistent with all dependencies. For this, you find a topological ordering of the corresponding directed graph. Problem Description Task. Compute a topological ordering of a given directed acyclic graph (DAG) with 𝑛 vertices and 𝑚 edges. Input Format. A graph is given in the standard format. Constraints. 1 ≤ 𝑛 ≤ 105 , 0 ≤ 𝑚 ≤ 105 . The given graph is guaranteed to be acyclic. Output Format. Output any topological ordering of its vertices. (Many DAGs have more than just one topological ordering. You may output any of them.) Time Limits. language C C++ Java Python C# Haskell JavaScript Ruby Scala time (sec) 2 2 3 10 3 4 10 10 6 Memory Limit. 512MB. Sample 1. Sample 2. Sample 3. Starter Files The starter solutions for this problem read the input data from the standard input, pass it to a blank procedure, and then write the result to the standard output. You are supposed to implement your algorithm in this blank procedure if you are using C++, Java, or Python3. For other programming languages, you need to implement a solution from scratch. Filename: toposort What To Do To solve this problem, it is enough to implement carefully the corresponding algorithm covered in the lectures. 4 Advanced Problem: Checking Whether Any Intersection in a City is Reachable from Any Other We strongly recommend you start solving advanced problems only when you are done with the basic problems (for some advanced problems, algorithms are not covered in the video lectures and require additional ideas to be solved; for some other advanced problems, algorithms are covered in the lectures, but implementing them is a more challenging task than for other problems). Problem Introduction The police department of a city has made all streets one-way. You would like to check whether it is still possible to drive legally from any intersection to any other intersection. For this, you construct a directed graph: vertices are intersections, there is an edge (𝑢, 𝑣) whenever there is a (one-way) street from 𝑢 to 𝑣 in the city. Then, it suffices to check whether all the vertices in the graph lie in the same strongly connected component. Problem Description Task. Compute the number of strongly connected components of a given directed graph with 𝑛 vertices and 𝑚 edges. Input Format. A graph is given in the standard format. Constraints. 1 ≤ 𝑛 ≤ 104 , 0 ≤ 𝑚 ≤ 104 Output Format. Output the number of strongly connected components. Time Limits. language C C++ Java Python C# Haskell JavaScript Ruby Scala time (sec) 1 1 1.5 5 1.5 2 5 5 3 Memory Limit. 512MB. Sample 1. This graph has two strongly connected components: {1, 3, 2}, {4}. Sample 2. This graph has five strongly connected components: {1}, {2}, {3}, {4}, {5}. Starter Files The starter solutions for this problem read the input data from the standard input, pass it to a blank procedure, and then write the result to the standard output. You are supposed to implement your algorithm in this blank procedure if you are using C++, Java, or Python3. For other programming languages, you need to implement a solution from scratch. What To Do To solve this problem, it is enough to implement carefully the corresponding algorithm covered in the lectures. 5 General Instructions and Recommendations on Solving Algorithmic Problems Your main goal in an algorithmic problem is to implement a program that solves a given computational problem in just few seconds even on massive datasets. Your program should read a dataset from the standard input and write an answer to the standard output. Below we provide general instructions and recommendations on solving such problems. Before reading them, go through readings and screencasts in the first module that show a step by step process of solving two algorithmic problems: link. 5.1 Reading the Problem Statement You start by reading the problem statement that contains the description of a particular computational task as well as time and memory limits your solution should fit in, and one or two sample tests. In some problems your goal is just to implement carefully an algorithm covered in the lectures, while in some other problems you first need to come up with an algorithm yourself. 5.2 Designing an Algorithm If your goal is to design an algorithm yourself, one of the things it is important to realize is the expected running time of your algorithm. Usually, you can guess it from the problem statement (specifically, from the subsection called constraints) as follows. Modern computers perform roughly 108–109 operations per second. So, if the maximum size of a dataset in the problem description is 𝑛 = 105 , then most probably an algorithm with quadratic running time is not going to fit into time limit (since for 𝑛 = 105 , 𝑛 2 = 1010) while a solution with running time 𝑂(𝑛 log 𝑛) will fit. However, an 𝑂(𝑛 ) solution will fit if 𝑛 is up to 103 = 1000, and if 𝑛 is at most 100, even 𝑂(𝑛 ) solutions will fit. In some cases, the problem is so hard that we do not know a polynomial solution. But for 𝑛 up to 18, a solution with 𝑂(2𝑛𝑛 ) running time will probably fit into the time limit. To design an algorithm with the expected running time, you will of course need to use the ideas covered in the lectures. Also, make sure to carefully go through sample tests in the problem description. 5.3 Implementing Your Algorithm When you have an algorithm in mind, you start implementing it. Currently, you can use the following programming languages to implement a solution to a problem: C, C++, C#, Haskell, Java, JavaScript, Python2, Python3, Ruby, Scala. For all problems, we will be providing starter solutions for C++, Java, and Python3. If you are going to use one of these programming languages, use these starter files. For other programming languages, you need to implement a solution from scratch. 5.4 Compiling Your Program For solving programming assignments, you can use any of the following programming languages: C, C++, C#, Haskell, Java, JavaScript, Python2, Python3, Ruby, and Scala. However, we will only be providing starter solution files for C++, Java, and Python3. The programming language of your submission is detected automatically, based on the extension of your submission. We have reference solutions in C++, Java and Python3 which solve the problem correctly under the given restrictions, and in most cases spend at most 1/3 of the time limit and at most 1/2 of the memory limit. You can also use other languages, and we’ve estimated the time limit multipliers for them, however, we have no guarantee that a correct solution for a particular problem running under the given time and memory constraints exists in any of those other languages. Your solution will be compiled as follows. We recommend that when testing your solution locally, you use the same compiler flags for compiling. This will increase the chances that your program behaves in the same way on your machine and on the testing machine (note that a buggy program may behave differently when compiled by different compilers, or even by the same compiler with different flags). ∙ C (gcc 5.2.1). File extensions: .c. Flags: gcc – pipe – O2 – std = c11 < filename > – lm ∙ C++ (g++ 5.2.1). File extensions: .cc, .cpp. Flags: g ++ – pipe – O2 – std = c ++14 < filename > – lm If your C/C++ compiler does not recognize -std=c++14 flag, try replacing it with -std=c++0x flag or compiling without this flag at all (all starter solutions can be compiled without it). On Linux and MacOS, you most probably have the required compiler. On Windows, you may use your favorite compiler or install, e.g., cygwin. ∙ C# (mono 3.2.8). File extensions: .cs. Flags: ∙ Haskell (ghc 7.8.4). File extensions: .hs. Flags: ghc – O2 ∙ Java (Open JDK 8). File extensions: .java. Flags: javac – encoding UTF -8 java – Xmx1024m ∙ JavaScript (Node v6.3.0). File extensions: .js. Flags: ∙ Python 2 (CPython 2.7). File extensions: .py2 or .py (a file ending in .py needs to have a first line which is a comment containing “python2”). No flags: ∙ Python 3 (CPython 3.4). File extensions: .py3 or .py (a file ending in .py needs to have a first line which is a comment containing “python3”). No flags: ∙ Ruby (Ruby 2.1.5). File extensions: .rb. ∙ Scala (Scala 2.11.6). File extensions: .scala. 5.5 Testing Your Program When your program is ready, you start testing it. It makes sense to start with small datasets (for example, sample tests provided in the problem description). Ensure that your program produces a correct result. You then proceed to checking how long does it take your program to process a massive dataset. For this, it makes sense to implement your algorithm as a function like solve(dataset) and then implement an additional procedure generate() that produces a large dataset. For example, if an input to a problem is a sequence of integers of length 1 ≤ 𝑛 ≤ 105 , then generate a sequence of length exactly 105 , pass it to your solve() function, and ensure that the program outputs the result quickly. Also, check the boundary values. Ensure that your program processes correctly sequences of size 𝑛 = 1, 2, 105 . If a sequence of integers from 0 to, say, 106 is given as an input, check how your program behaves when it is given a sequence 0, 0, . . . , 0 or a sequence 106 , 106 , . . . , 106 . Check also on randomly generated data. For each such test check that you program produces a correct result (or at least a reasonably looking In the end, we encourage you to stress test your program to make sure it passes in the system at the first attempt. See the readings and screencasts from the first week to learn about testing and stress testing: link. 5.6 Submitting Your Program to the Grading System When you are done with testing, you submit your program to the grading system. For this, you go the submission page, create a new submission, and upload a file with your program. The grading system then compiles your program (detecting the programming language based on your file extension, see Subsection 5.4) and runs it on a set of carefully constructed tests to check that your program always outputs a correct result and that it always fits into the given time and memory limits. The grading usually takes no more than a minute, but in rare cases when the servers are overloaded it might take longer. Please be patient. You can safely leave the page when your solution is uploaded. As a result, you get a feedback message from the grading system. The feedback message that you will love to see is: Good job! This means that your program has passed all the tests. On the other hand, the three messages Wrong answer, Time limit exceeded, Memory limit exceeded notify you that your program failed due to one these three reasons. Note that the grader will not show you the actual test you program have failed on (though it does show you the test if your program have failed on one of the first few tests; this is done to help you to get the input/output format right). 5.7 Debugging and Stress Testing Your Program If your program failed, you will need to debug it. Most probably, you didn’t follow some of our suggestions from the section 5.5. See the readings and screencasts from the first week to learn about debugging your program: link. You are almost guaranteed to find a bug in your program using stress testing, because the way these programming assignments and tests for them are prepared follows the same process: small manual tests, tests for edge cases, tests for large numbers and integer overflow, big tests for time limit and memory limit checking, random test generation. Also, implementation of wrong solutions which we expect to see and stress testing against them to add tests specifically against those wrong solutions. Go ahead, and we hope you pass the assignment soon! 6 Frequently Asked Questions 6.1 I submit the program, but nothing happens. Why? You need to create submission and upload the file with your solution in one of the programming languages C, C++, Java, or Python (see Subsections 5.3 and 5.4). Make sure that after uploading the file with your solution you press on the blue “Submit” button in the bottom. After that, the grading starts, and the submission being graded is enclosed in an orange rectangle. After the testing is finished, the rectangle disappears, and the results of the testing of all problems is shown to you. 6.2 I submit the solution only for one problem, but all the problems in the assignment are graded. Why? Each time you submit any solution, the last uploaded solution for each problem is tested. Don’t worry: this doesn’t affect your score even if the submissions for the other problems are wrong. As soon as you pass the sufficient number of problems in the assignment (see in the pdf with instructions), you pass the assignment. After that, you can improve your result if you successfully pass more problems from the assignment. We recommend working on one problem at a time, checking whether your solution for any given problem passes in the system as soon as you are confident in it. However, it is better to test it first, please refer to the reading about stress testing: link. 6.3 What are the possible grading outcomes, and how to read them? Your solution may either pass or not. To pass, it must work without crashing and return the correct answers on all the test cases we prepared for you, and do so under the time limit and memory limit constraints specified in the problem statement. If your solution passes, you get the corresponding feedback “Good job!” and get a point for the problem. If your solution fails, it can be because it crashes, returns wrong answer, works for too long or uses too much memory for some test case. The feedback will contain the number of the test case on which your solution fails and the total number of test cases in the system. The tests for the problem are numbered from 1 to the total number of test cases for the problem, and the program is always tested on all the tests in the order from the test number 1 to the test with the biggest number. Here are the possible outcomes: Good job! Hurrah! Your solution passed, and you get a point! Wrong answer. Your solution has output incorrect answer for some test case. If it is a sample test case from the problem statement, or if you are solving Programming Assignment 1, you will also see the input data, the output of your program and the correct answer. Otherwise, you won’t know the input, the output, and the correct answer. Check that you consider all the cases correctly, avoid integer overflow, output the required white space, output the floating point numbers with the required precision, don’t output anything in addition to what you are asked to output in the output specification of the problem statement. See this reading on testing: link. Time limit exceeded. Your solution worked longer than the allowed time limit for some test case. If it is a sample test case from the problem statement, or if you are solving Programming Assignment 1, you will also see the input data and the correct answer. Otherwise, you won’t know the input and the correct answer. Check again that your algorithm has good enough running time estimate. Test your program locally on the test of maximum size allowed by the problem statement and see how long it works. Check that your program doesn’t wait for some input from the user which makes it to wait forever. See this reading on testing: link. Memory limit exceeded. Your solution used more than the allowed memory limit for some test case. If it is a sample test case from the problem statement, or if you are solving Programming Assignment 1, you will also see the input data and the correct answer. Otherwise, you won’t know the input and the correct answer. Estimate the amount of memory that your program is going to use in the worst case and check that it is less than the memory limit. Check that you don’t create too large arrays or data structures. Check that you don’t create large arrays or lists or vectors consisting of empty arrays or empty strings, since those in some cases still eat up memory. Test your program locally on the test of maximum size allowed by the problem statement and look at its memory consumption in the system. Cannot check answer. Perhaps output format is wrong. This happens when you output something completely different than expected. For example, you are required to output word “Yes” or “No”, but you output number 1 or 0, or vice versa. Or your program has empty output. Or your program outputs not only the correct answer, but also some additional information (this is not allowed, so please follow exactly the output format specified in the problem statement). Maybe your program doesn’t output anything, because it crashes. Unknown signal 6 (or 7, or 8, or 11, or some other). This happens when your program crashes. It can be because of division by zero, accessing memory outside of the array bounds, using uninitialized variables, too deep recursion that triggers stack overflow, sorting with contradictory comparator, removing elements from an empty data structure, trying to allocate too much memory, and many other reasons. Look at your code and think about all those possibilities. Make sure that you use the same compilers and the same compiler options as we do. Try different testing techniques from this reading: Internal error: exception… Most probably, you submitted a compiled program instead of a source Grading failed. Something very wrong happened with the system. Contact Coursera for help or write in the forums to let us know. 6.4 How to understand why my program fails and to fix it? If your program works incorrectly, it gets a feedback from the grader. For the Programming Assignment 1, when your solution fails, you will see the input data, the correct answer and the output of your program in case it didn’t crash, finished under the time limit and memory limit constraints. If the program crashed, worked too long or used too much memory, the system stops it, so you won’t see the output of your program or will see just part of the whole output. We show you all this information so that you get used to the algorithmic problems in general and get some experience debugging your programs while knowing exactly on which tests they fail. However, in the following Programming Assignments throughout the Specialization you will only get so much information for the test cases from the problem statement. For the next tests you will only get the result: passed, time limit exceeded, memory limit exceeded, wrong answer, wrong output format or some form of crash. We hide the test cases, because it is crucial for you to learn to test and fix your program even without knowing exactly the test on which it fails. In the real life, often there will be no or only partial information about the failure of your program or service. You will need to find the failing test case yourself. Stress testing is one powerful technique that allows you to do that. You should apply it after using the other testing techniques covered in this reading. 6.5 Why do you hide the test on which my program fails? Often beginner programmers think by default that their programs work. Experienced programmers know, however, that their programs almost never work initially. Everyone who wants to become a better programmer needs to go through this realization. When you are sure that your program works by default, you just throw a few random test cases against it, and if the answers look reasonable, you consider your work done. However, mostly this is not enough. To make one’s programs work, one must test them really well. Sometimes, the programs still don’t work although you tried really hard to test them, and you need to be both skilled and creative to fix your bugs. Solutions to algorithmic problems are one of the hardest to implement correctly. That’s why in this Specialization you will gain this important experience which will be invaluable in the future when you write programs which you really need to get right. It is crucial for you to learn to test and fix your programs yourself. In the real life, often there will be no or only partial information about the failure of your program or service. Still, you will have to reproduce the failure to fix it (or just guess what it is, but that’s rare, and you will still need to reproduce the failure to make sure you have really fixed it). When you solve algorithmic problems, it is very frequent to make subtle mistakes. That’s why you should apply the testing techniques described in this reading to find the failing test case and fix your program. 6.6 My solution does not pass the tests? May I post it in the forum and ask for a help? No, please do not post any solutions in the forum or anywhere on the web, even if a solution does not pass the tests (as in this case you are still revealing parts of a correct solution). Recall the third item of the Coursera Honor Code: “I will not make solutions to homework, quizzes, exams, projects, and other assignments available to anyone else (except to the extent an assignment explicitly permits sharing solutions). This includes both solutions written by me, as well as any solutions provided by the course staff or others” 6.7 My implementation always fails in the grader, though I already tested and stress tested it a lot. Would not it be better if you give me a solution to this problem or at least the test cases that you use? I will then be able to fix my code and will learn how to avoid making mistakes. Otherwise, I do not feel that I learn anything from solving this problem. I am just stuck. First of all, you always learn from your mistakes. The process of trying to invent new test cases that might fail your program and proving them wrong is often enlightening. This thinking about the invariants which you expect your loops, ifs, etc. to keep and proving them wrong (or right) makes you understand what happens inside your program and in the general algorithm you’re studying much more. Also, it is important to be able to find a bug in your implementation without knowing a test case and without having a reference solution. Assume that you designed an application and an annoyed user reports that it crashed. Most probably, the user will not tell you the exact sequence of operations that led to a crash. Moreover, there will be no reference application. Hence, once again, it is important to be able to locate a bug in your implementation yourself, without a magic oracle giving you either a test case that your program fails or a reference solution. We encourage you to use programming assignments in this class as a way of practicing this important skill. If you have already tested a lot (considered all corner cases that you can imagine, constructed a set of manual test cases, applied stress testing), but your program still fails and you are stuck, try to ask for help on the forum. We encourage you to do this by first explaining what kind of corner cases you have already considered (it may happen that when writing such a post you will realize that you missed some corner cases!) and only then asking other learners to give you more ideas for tests cases.	{"url":"https://codingprolab.com/answer/programming-assignment-2-decomposition-of-graphs/","timestamp":"2024-11-02T14:40:01Z","content_type":"text/html","content_length":"169996","record_id":"<urn:uuid:ea4d6134-6201-47ae-ba19-c2c67dd86c4e>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00312.warc.gz"}
Statistics Archives - Page 3 Of 3 - Insightoriel There are four types of analytics segment broadly. Those are Descriptive analytics, Diagnostic analytics, Predictive Analytics & Prescriptive Analytics. So, we’ll discuss what is descriptive analytics. Among all four types of analytics, Descriptive Analytics is the starting point for any kind of analytics to find out what has happened in the past. Also, descriptive analytics … Read more What is Bell Curve \| Explained Bell Curve with Standard Deviation \| 4 Steps to Create Bell Curve in Excel Bell Curve is a methodology to understand normal distribution of data. In other words, Bell Curve is the graphical form of normal distribution. In Bell Curve, X axis represents the values of distribution and Y axis represents the frequency of each value. What is Boxplot \| Box and Whisker Plot \| 5 Advantages of Boxplot \| Create Boxplot in Excel & R In 21st century, probably Box and Whisker plot or Boxplot is the mostly used graph in descriptive analytics. Box and Whisker plot is showing five-point summary of a data.	{"url":"https://insightoriel.com/data-science/statistics/page/3/","timestamp":"2024-11-13T11:56:47Z","content_type":"text/html","content_length":"179359","record_id":"<urn:uuid:1c40a76a-be26-4222-ac0a-a2b5c78babc9>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00107.warc.gz"}
National Curriculum (Vocational) Mathematics Level 3 Space, shape and measurement: Solve problems by constructing and interpreting trigonometric models Unit 5: Apply the area rule Dylan Busa Unit outcomes By the end of this unit you will be able to: • Apply the area rule in 2-D triangles. What you should know Before you start this unit, make sure you can: • Calculate the area of a triangle using the formula [latex]\scriptsize \text{area}=\displaystyle \frac{1}{2}\times b\times h[/latex]. • Use a calculator to calculate the sine of a given angle. Refer to level 2 subject outcome 3.6 unit 2 if you need help with this. • Use a calculator to calculate the angle from a given ratio for sine. Refer to level 2 subject outcome 3.6 unit 2 if you need help with this. Trigonometry is not just useful for finding the lengths of unknown sides or the sizes of unknown angles in right-angled triangles. It can also be put to work in non-right-angled triangles and it can be used to find measures other than the lengths of sides and the sizes of angles. The area rule, for example, is a useful trigonometric identity (remember those from unit 3?) that can be used to find the area of any triangle, even triangles that have no [latex]\scriptsize {{90}^\ circ}[/latex] angles. But why do we need another way to calculate the area of a triangle if we already have the well-known formula [latex]\scriptsize \text{Area}=\displaystyle \frac{1}{2}\times b\ times h[/latex]? Well, sometimes, we might not know the length of the perpendicular height or the base. Instead, we might know the size of one of the angles inside the triangle. In this unit, we are going to derive the area rule and see how to use it to find the areas of all sorts of different triangles in different contexts. Derive the area rule Deriving the area rule is not too difficult. You will be led through this process in Activity 5.1. Activity 5.1: Derive the area rule Time required: 15 minutes What you need: • a piece of paper • a pen or pencil • a ruler What to do: 1. On your piece of paper, draw any triangle [latex]\scriptsize \text{ABC}[/latex] and label the sides in relation to the opposite vertices (see Figure 1). The triangle can be any size and shape. 2. Now drop a perpendicular from [latex]\scriptsize \text{B}[/latex] to the opposite side and call this perpendicular [latex]\scriptsize h[/latex] (see Figure 2). 3. Write the expression for the area of triangle [latex]\scriptsize \text{ABC}[/latex] based on its base and perpendicular height. 4. Write an expression for [latex]\scriptsize \sin \text{A}[/latex]. 5. Write a new expression for the area of triangle [latex]\scriptsize \text{ABC}[/latex] that includes the term [latex]\scriptsize \sin A[/latex]. Hint: look for a substitution. 6. Now, write an expression for [latex]\scriptsize \sin \text{C}[/latex] and then write a similar expression for the area of the triangle in terms of[latex]\scriptsize \sin \text{C}[/latex]. 7. What do you think the area of triangle [latex]\scriptsize \text{ABC}[/latex] would be in terms of [latex]\scriptsize \sin \text{B}[/latex]? What did you find? 1. Remember that your triangle [latex]\scriptsize \text{ABC}[/latex] can be any size and shape. Figure 1 shows just one possible triangle. Figure 1: Triangle [latex]\scriptsize \text{ABC}[/latex] 2. Remember that dropping a perpendicular means that the new line meets the other line at [latex]\scriptsize {{90}^\circ}[/latex]. Figure 2 shows what your perpendicular line may look like. Figure 2: Triangle [latex]\scriptsize \text{ABC}[/latex] with perpendicular [latex]\scriptsize h[/latex] 3. The area of triangle [latex]\scriptsize \text{ABC}[/latex] is given by [latex]\scriptsize A=\displaystyle \frac{1}{2}\times b\times h[/latex] where the base is side [latex]\scriptsize \text{b}[/ latex] and the perpendicular height is [latex]\scriptsize \text{h}[/latex]. Therefore, area of [latex]\scriptsize \Delta \text{ABC}[/latex] is [latex]\scriptsize \displaystyle \frac{1}{2}\times \ text{b}\times \text{h}[/latex]. 4. [latex]\scriptsize \sin \text{A}=\displaystyle \frac{{\text{opposite}}}{{\text{hypotenuse}}}=\displaystyle \frac{\text{h}}{\text{c}}[/latex]. 5. [latex]\scriptsize \sin \text{A}=\displaystyle \frac{\text{h}}{\text{c}}\therefore \text{h}=\text{c}\times \sin \text{A}[/latex]. But area [latex]\scriptsize \Delta \text{ABC}[/latex] is [latex]\ scriptsize \displaystyle \frac{1}{2}\times \text{b}\times \text{h}[/latex]. Therefore, area [latex]\scriptsize \Delta \text{ABC}=\displaystyle \frac{1}{2}\times \text{b}\times \text{c}\times \sin text{A}=\displaystyle \frac{1}{2}\text{bc}\sin \text{A}[/latex]. 6. [latex]\scriptsize \sin \text{C}=\displaystyle \frac{{\text{opposite}}}{{\text{hypotenuse}}}=\displaystyle \frac{\text{h}}{\text{a}}[/latex]. Therefore [latex]\scriptsize \text{h}=\text{a}\times \sin \text{C}[/latex] and area [latex]\scriptsize \Delta \text{ABC}=\displaystyle \frac{1}{2}\times \text{b}\times \text{a}\times \sin\text{C}=\displaystyle \frac{1}{2}\text{ab}\sin \text{C}[/ 7. In each case above, the expression for the area of the triangle is the product of the length of two sides and the sine of the angle between these two sides. Therefore, area [latex]\scriptsize \ Delta \text{ABC}=\displaystyle \frac{1}{2}\text{ac}\sin \text{B}[/latex]. The area rule states that in any [latex]\scriptsize \Delta ABC[/latex], the area is given by: Take note! The area rule works for any triangle for which you know the length of any two sides and the size of the included angle (the angle between the two known sides). Use the area rule Let’s look at some examples of how to apply the area rule. Example 5.2 Find the area of [latex]\scriptsize \Delta \text{ABC}[/latex] correct to two decimal places. We are not given the lengths of two sides and the included angle, so we cannot use the area rule with the information we currently have. We need to find the size of [latex]\scriptsize \hat{B}[/ latex]. To do this, we can use the fact that [latex]\scriptsize AB=BC[/latex] (given) and therefore, [latex]\scriptsize \hat{A}=\hat{C}={{43}^\circ}[/latex] (angles opposite equal sides). Then we can use that fact that angles in a triangle are supplementary to find [latex]\scriptsize \hat{B}[/latex]. [latex]\scriptsize AB=BC[/latex] (given) [latex]\scriptsize \therefore \hat{A}=\hat{C}={{43}^\circ}[/latex] (angles opposite equal sides) [latex]\scriptsize \begin{align}\therefore \hat{B} & ={{180}^\circ}-\hat{A}-\hat{C}\\ & ={{180}^\circ}-{{43}^\circ}-{{43}^\circ}\\ & ={{94}^\circ}\end{align}[/latex] (angles in a triangle are Now we can use the area rule. [latex]\scriptsize \begin{align}\text{Area }\Delta ABC&=\displaystyle \frac{1}{2}ac\sin B\\&=\displaystyle \frac{1}{2}\times 8\times 8\times \sin {{94}^\circ}\\&=31.92\ {{\text{m}}^{2}}\end{align} Remember to round off your final answer and include the correct area units. Example 5.3 A parallelogram has adjacent sides of [latex]\scriptsize 16\ \text{cm}[/latex] and [latex]\scriptsize 23\ \text{cm}[/latex]. The angle between them is [latex]\scriptsize {{43}^\circ}[/latex]. Calculate the area of the parallelogram correct to two decimal places. If no diagram is given, it is always best to draw your own using the given information. This does not have to be accurate. We know that [latex]\scriptsize ABCD[/latex] is a parallelogram. Therefore, if we can find the area of [latex]\scriptsize \Delta ACD[/latex] we can multiply this by [latex]\scriptsize 2[/latex] to find the area of [latex]\scriptsize ABCD[/latex]. In [latex]\scriptsize \Delta ACD[/latex], we know two sides and the included angle so we can use the area rule. [latex]\scriptsize \begin{align}\text{Area }\Delta ADC&=\displaystyle \frac{1}{2}ac\sin D\\&=\displaystyle \frac{1}{2}\times 23\times 16\times \sin {{43}^\circ}\\&=125.49\ \text{c}{{\text{m}}^{2}}\ The area of parallelogram [latex]\scriptsize ABCD=2\times 125.49\ \text{c}{{\text{m}}^{2}}=250.98\ \text{c}{{\text{m}}^{2}}[/latex]. Example 5.4 If [latex]\scriptsize \Delta EFG[/latex] has an area of [latex]\scriptsize 167.43\ \text{c}{{\text{m}}^{2}}[/latex] and [latex]\scriptsize e=17\ \text{cm}[/latex] and [latex]\scriptsize f=24.35\ \ text{cm}[/latex], what are the two possible sizes of [latex]\scriptsize \hat{G}[/latex], correct to two decimal places. This question gives us the area and asks us to calculate the size of [latex]\scriptsize \hat{G}[/latex]. [latex]\scriptsize \begin{align}\text{Area }\Delta EFG & =\displaystyle \frac{1}{2}ef\sin G\\\therefore 167.43 & =\displaystyle \frac{1}{2}\times 17\times 24.35\times \sin G\\\therefore \sin G & = We have found that [latex]\scriptsize \hat{G}={{53.99}^\circ}[/latex] but the question mentions two possible solutions. Remember that sine is also positive in the second quadrant, in other words, for angles [latex]\scriptsize 90{}^\circ \le \theta \le {{180}^\circ}[/latex]. Therefore, [latex]\scriptsize \hat{G}={{180}^\circ}-{{53.99}^\circ}={{126.01}^\circ}[/latex] as well. But does this make physical sense? Does such a triangle exist? Figure 3 shows both possible triangles with these dimensions. Figure 3: Both possible [latex]\scriptsize \Delta EFG[/latex] Exercise 5.1 1. Calculate the area of [latex]\scriptsize \Delta ABC[/latex] to two decimal places. 2. Calculate the area of [latex]\scriptsize \Delta ABC[/latex] given [latex]\scriptsize a=10\ \text{cm}[/latex], [latex]\scriptsize c=8\ \text{cm}[/latex] and [latex]\scriptsize \hat{B}={{35}^\circ} [/latex], to three decimal places. 3. Determine the area of equilateral [latex]\scriptsize \Delta PQR[/latex] with [latex]\scriptsize p=14\ \text{cm}[/latex] to two decimal places. 4. Determine, to two decimal places, the area of a parallelogram in which two adjacent sides are [latex]\scriptsize 10\ \text{cm}[/latex] and [latex]\scriptsize 13\ \text{cm}[/latex] and the angle between them is [latex]\scriptsize {{55}^\circ}[/latex]. 5. Determine the length of [latex]\scriptsize AC[/latex] (to one decimal place) if the area of [latex]\scriptsize \Delta ABC=16.18\ {{\text{m}}^{2}}[/latex]. The full solutions are at the end of the unit. In this unit you have learnt the following: • What the area rule is. • How to use the area rule to find the area of any triangle where two sides and the included angle are known. Unit 5: Assessment Suggested time to complete: 40 minutes 1. Calculate the area of [latex]\scriptsize \Delta KLM[/latex] to two decimal places. 2. Calculate, to two decimal places, the area of [latex]\scriptsize \Delta ABC[/latex] given [latex]\scriptsize b=19\ \text{cm}[/latex], [latex]\scriptsize c=18\ \text{cm}[/latex] and [latex]\ scriptsize \hat{A}={{49}^\circ}[/latex]. 3. Determine, to two decimal places, the area of [latex]\scriptsize \Delta PQR[/latex] with [latex]\scriptsize p=r=9\ \text{m}[/latex]. 4. Determine the area of a parallelogram [latex]\scriptsize PQRS[/latex] to two decimal places. 5. Determine two possible values for [latex]\scriptsize \hat{X}[/latex] (to one decimal place) if the area of [latex]\scriptsize \Delta XYZ=26.72\ {{\text{m}}^{2}}[/latex]. The full solutions are at the end of the unit. Unit 5: Solutions Exercise 5.1 1. . [latex]\scriptsize \begin{align}\text{Area }\Delta ABC&=\displaystyle \frac{1}{2}ab\sin C\\&=\displaystyle \frac{1}{2}\times 12\times 9\times \sin {{25}^\circ}\\&=22.82\ {{\text{u}}^{2}}\end Note: No units were given so we include [latex]\scriptsize {{\text{u}}^{2}}[/latex] for units squared. 2. . [latex]\scriptsize \begin{align}\text{Area }\Delta ABC&=\displaystyle \frac{1}{2}ac\sin B\\&=\displaystyle \frac{1}{2}\times 10\times 8\times \sin {{35}^\circ}\\&=22.943\ \text{c}{{\text{m}}^ 3. [latex]\scriptsize \Delta PQR[/latex] is an equilateral triangle. Therefore, all three sides have length [latex]\scriptsize 14\ \text{cm}[/latex] and all three angles are [latex]\scriptsize {{60} ^\circ}[/latex]. We can choose any combination of sides and angle. [latex]\scriptsize \begin{align}\text{Area }\Delta ABC&=\displaystyle \frac{1}{2}pq\sin R\\&=\displaystyle \frac{1}{2}\times 14\times 14\times \sin {{60}^\circ}\\&=84.87\ \text{c}{{\text{m}}^ 4. . [latex]\scriptsize \begin{align}\text{Area }\Delta ACD&=\displaystyle \frac{1}{2}ac\sin D\\&=\displaystyle \frac{1}{2}\times 10\times 13\times \sin {{55}^\circ}\\&=53.24\ \text{c}{{\text{m}}^ Area of parallelogram [latex]\scriptsize ABCD=2\times 53.24\ \text{c}{{\text{m}}^{2}}=106.49\ \text{c}{{\text{m}}^{2}}[/latex] 5. . [latex]\scriptsize \begin{align}\text{Area }\Delta ABC & =16.18\ {{\text{m}}^{2}}\\\therefore 16.18 & =\displaystyle \frac{1}{2}\times AC\times 8\times \sin {{54}^\circ}\\\therefore AC & =4.045\ sin {{54}^\circ}\\\therefore AC & =3.3\ \text{m}\end{align}[/latex] Unit 5: Assessment 1. . [latex]\scriptsize \begin{align}\text{Area }\Delta KLM&=\displaystyle \frac{1}{2}kl\sin M\\&=\displaystyle \frac{1}{2}\times 12\times 4\times \sin {{103}^\circ}\\&=23.38\ {{\text{u}}^{2}}\end 2. . [latex]\scriptsize \begin{align}\text{Area }\Delta ABC&=\displaystyle \frac{1}{2}bc\sin A\\&=\displaystyle \frac{1}{2}\times 19\times 18\times \sin {{49}^\circ}\\&=129.06\ \text{c}{{\text{m}}^ 3. [latex]\scriptsize PQ=RQ[/latex] (given) [latex]\scriptsize \therefore \hat{R}=\hat{P}={{34}^\circ}[/latex] (angles opposite equal sides) [latex]\scriptsize \begin{align}\therefore \hat{Q} & ={{180}^\circ}-R-P\\ & ={{180}^\circ}-{{34}^\circ}-{{34}^\circ}\\ & ={{112}^\circ}\end{align}[/latex] (angles in a triangle are [latex]\scriptsize \begin{align}\text{Area }\Delta PQR&=\displaystyle \frac{1}{2}pr\sin Q\\&=\displaystyle \frac{1}{2}\times 9\times 9\times \sin {{112}^\circ}\\&=37.55\ {{\text{m}}^{2}}\end 4. [latex]\scriptsize PQRS[/latex] is a parallelogram (given) [latex]\scriptsize \therefore QR=PS=22[/latex] (opposite sides of parallelogram) [latex]\scriptsize \begin{align}\text{Area }\Delta QRS&=\displaystyle \frac{1}{2}qs\sin R\\&=\displaystyle \frac{1}{2}\times 15\times 22\times \sin {{135}^\circ}\\&=116.67\ {{\text{u}}^{2}}\end [latex]\scriptsize \text{Area }PQRS=2\times 116.67=233.35\ {{\text{u}}^{2}}[/latex](rounding off in the final step) 5. [latex]\scriptsize \text{Area }\Delta XYZ=26.72\ {{\text{m}}^{2}}[/latex] [latex]\scriptsize \begin{align}\therefore 26.72 & =\displaystyle \frac{1}{2}yz\sin X\\ & =\displaystyle \frac{1}{2}\times 8\times 8\times \sin X\\\therefore \sin X & =0.835\\\therefore \hat{X} & ={{56.6}^\circ}\text{ or }\hat{X} & ={{180}^\circ}-{{56.62}^\circ}={{123.4}^\circ}\end{align}[/latex]	{"url":"http://ncvm3.books.nba.co.za/chapter/unit-5-apply-the-area-rule/","timestamp":"2024-11-12T02:00:28Z","content_type":"text/html","content_length":"117314","record_id":"<urn:uuid:905657b0-ec19-47b0-a509-6ed705774fa3>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00666.warc.gz"}
Authors, Title and Abstract Paper Talk ABSTRACT. We consider basic symbolic heaps, i.e. symbolic heaps without any inductive predicates, but within a memory model featuring permissions. We propose a complete proof system for Jul this logic that is entirely independent of the permission model. This is an ongoing work towards a complete proof system for symbolic heaps with lists, and more generally towards a proof 13 theory of permissions in separation logics with recursive predicates and Boolean BI with abstract predicates. 12:00 ABSTRACT. We present Sloth, a solver for separation logic modulo theory constraints specified in the separation logic SL∗, a propositional separation logic that we recently introduced in our IJCAR'18 paper "A Separation Logic with Data: Small Models and Automation." SL∗ admits NP decision procedures despite its high expressiveness; features of the logic include support Jul for list and tree fragments, universal data constraints about both lists and trees, and Boolean closure of spatial formulas. Sloth solves SL* constraints via an SMT encoding of SL* that 13 is based on the logic's small-model property. We argue that, while clearly still a work in progress, Sloth already demonstrates that SL* lends itself to the automation of nontrivial 16:10 examples. These results complement the theoretical work presented in the IJCAR'18 paper. ABSTRACT. We first show that infinite satisfiability can be reduced to finite satisfiability for all prenex formulas of Separation Logic with $k\geq1$ selector fields ($\seplogk{k}$). Second, we show that this entails the decidability of the finite and infinite satisfiability problem for the class of prenex formulas of $\seplogk{1}$, by reduction to the first-order Jul theory of one unary function symbol and unary predicate symbols. We also prove that the complexity is not elementary, by reduction from the first-order theory of one unary function 13 symbol. Finally, we prove that the Bernays-Sch\"onfinkel-Ramsey fragment of prenex $\seplogk{1}$ formulae with quantifier prefix in the language $\exists^\forall^$ is \pspace-complete. 15:20 The definition of a complete (hierarchical) classification of the complexity of prenex $\seplogk{1}$, according to the quantifier alternation depth is left as an open problem. ABSTRACT. Separation Logic (SL) is a logical formalism for reasoning about programs that use pointers to mutate data structures. SL has proven itself successful in the field of program verification over the past fifteen years as an assertion language to state properties about memory heaps using Hoare triples. Since the full logic is not recursively enumerable, most of Jul the proof-systems and verification tools for SL focus on the decidable but rather restricted symbolic heaps fragment. Moreover, recent proof-systems that go beyond symbolic heaps allow 13 either the full set of connectives, or the definition of arbitrary predicates, but not both. In this work, we present a labelled proof-system called GM SL that allows both the definition 11:30 of arbitrary inductive predicates and the full set of SL connectives. ABSTRACT. We propose the strong wand and the Factor rule in a cyclic-proof system for the separation-logic entailment checking problem with general inductive predicates. The strong wand is similar to but stronger than the magic wand and is defined syntactically. The Factor rule, which uses the strong wand, is an inference rule for spatial factorization to expose an Jul implicitly allocated cell in inductive predicates. By this rule, we can void getting stuck in the Unfold-Match-Remove strategy. We show a semi-decision algorithm of proof search in the 13 cyclic-proof system with the Factor rule and the experimental results of its prototype implementation. 11:00 ABSTRACT. We investigate the complexity consequences of adding pointer arithmetic to separation logic. Specifically, we study an extension of the points-to fragment of symbolic-heap separation logic with sets of simple "difference constraints" of the form x <= y + k, where x and y are pointer variables and k is an integer offset. This extension can be considered a practically minimal language for separation logic with pointer arithmetic. Most significantly, we find that, even for this minimal language, polynomial-time decidability is already impossible: satisfiability becomes NP-complete, while quantifier-free entailment 13 becomes CoNP-complete and quantified entailment becomes \Pi^P_2-complete (where \Pi^P_2 is the second class in the polynomial-time hierarchy). 14:50 However, the language does satisfy the small model property, meaning that any satisfiable formula A has a model of size polynomial in A, whereas this property fails when richer forms of arithmetical constraints are permitted.	{"url":"http://t-news.cn/Floc2018/FLoC2018-pages/volume26-abstracts.html","timestamp":"2024-11-13T13:14:04Z","content_type":"application/xhtml+xml","content_length":"17237","record_id":"<urn:uuid:4ddc0e77-d7e9-4bd5-8a3d-644212b916ab>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00215.warc.gz"}
If $$(x + y)^{m+n} = x^{m}.y^{n}$$, then (a) $\frac{dy}{dx}$ is independent of $m$ but dependent of $n$ (b) $\frac{dy}{dx} = \frac{m}{n}$ (c) $\frac{dy}{dx}$ is independent of $n$ but dependent of $m$ (d) $\frac{dy}{dx}$ does not dependent on $m$ or $n$. $$\lim_{x \to 2} \frac{f(4) - f(x^{2}}{x - 2}$$ is (given $$f(x)$$ is differentiable and $f'(4) = 5$) equal to	{"url":"https://ganitcharcha.com/quiz-page-name-view-Higher-Secondary-Level-Quiz-January-2015.php","timestamp":"2024-11-14T21:15:28Z","content_type":"text/html","content_length":"31618","record_id":"<urn:uuid:e2659c6e-e91f-4765-b1df-129e5b1f12dc>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00252.warc.gz"}
How to Schedule Panel Locations on a Flat Rectangular Surface in Revit This post demonstrates how it is possible to schedule repeater component cell numbers on an orthogonal pattern within the Revit Conceptual Massing or Adaptive Component environment: To set this up, first you need a rectangular surface, which has been divided; then you need to place an adaptive component onto the node(s) of the surface; once arrayed in two directions using the repeater function you can schedule the cell (column and row) numbers on the divided surface. Of course it isn't that simple - the adaptive component has to be aware of its location relative to a fixed point, by doing some calculations: • The example here is perhaps a little more complicated than it needs be to demonstrate the principle, but it does a few other fun things too. • The adaptive component to be repeated must be set to “Shared” so it can be scheduled; it needs to use shared parameters so they can be scheduled too. • In this case it is a 5 point adaptive – one for each corner of the base of a rectangle placed on each cell of the divided surface; the fifth point works as a “Reactor” – it tracks the distance of the component from a base point. • This rectangle is made more interesting by having a pyramid on top that has an apex that moves depending where it is in the repeater pattern. To create the adaptive component: 1. Start a new generic adaptive family. • Place four points in a rectangular shape; make them adaptive • Join the four points with reference lines (make sure 3D snapping is on) 2. Set up the geometry for the pyramid (optional): • Place a point on each line (if correctly hosted it displays as a small point) • For each point assign a (Shared) parameter to its “Normalised Curve Parameter” – for the notional Y axis point, make it “Y_Ratio”; its point on the opposite side will have its “Measure From” value set to End, rather than beginning, so that they line up. • Assign a parameter “X_Ratio” to the notional X axis points (one from beginning, one from end, depending on which direction you drew the reference lines) • Join the two opposing points with reference lines • Place another point on one of the linking reference lines • Select the point and the option bar should show: • Click on “Host by Intersection”, then select the opposing reference line; it should move to the intersection of the reference lines • Set the “Show Reference Planes” property of the point to Always. • Set the work plane to the horizontal plane of the hosted intersection point (it only shows as a single line) • Place another point on top of the one on the intersection (ensure 3D snapping is on); ignore the error message about duplicate points; • Select the new point and drag it up in the zed axis – its ”Offset” property should change. If you are lucky, it should be a positive value (if it is zero, then the workplane or hosting went • Assign a “Height” parameter to it – if it was a negative value you’ll need to assign an interim parameter then convert it to positive with a formula, for the end user to understand. • This point becomes the apex of the pyramid; • Join the point to the four corners of the base with four reference lines; • Flex the X_Ratio and Y_Ratio and Height parameters • Create a Form (surface) on each of the four sides of the pyramid 3. Setting up the “Reactor” controls (Important): • Place a fifth adaptive point to the left of point 1 (bottom left corner of rectangle) • Use a reference line to join it to point 1 • Join points 5 and 4 (Bottom right corner of rectangle) with another reference line – to create a triangle of lines. • Place 3 dimensions between the adaptive points 5 & 1, 1 &4, 4 & 5 – for each one make sure to set the relevant reference line as the work plane for the dimension; it is vital to snap the dimensions to the adaptive points, rather than to line ends, surface corners etc (if not then you can’t use the dimensions later on in formulas) • Make each dimension as a reporting instance parameter, for use in the “cosine law” in trigonometry. We need to calculate the angle in the triangle • The Cosine Law for calculating an angle when 3 sides are known, is: Angle Î³ = Arcos( (A² + B² - C²) / 2AB) • Revit version of Cosines formula: Angle Î³ = Acos( (A^2 + B^2 - C^2) / 2AB) • This calculation assumes that the angle of the line between P1 and P4 is orthogonal – so it only works with a rectangular repeater pattern. 4. Calculation of column and row numbers: • This requires knowledge of how many rows/columns there will be in the repeater, and the overall size of the repeater • Parameters for these need to be built in to the component as shared parameters (for scheduling), as shown below; these can subsequently be linked to the parent family parameters for these values. X Number and Y Number should be integers (count of repeats in each direction) Column Number and Row Number should also be integers. X Ratio and Y Ratio should be number parameters • You may need a couple of checks to take care of when it has zero values for the distances. 5. Creation of the repeater: • The pyramid family needs to be loaded into another family that can support a divided surface – this could be an adaptive component, a mass family or an in-place mass family in a project. In this example it will be an in-place mass family. • Load the pyramid into the project • Start an in-place mass family • Draw a rectangle of reference lines, • Give the rectangle dimension parameters of Length and Width • Generate a form (surface) from the lines • Select the surface and Divide Surface • Make the nodes visible on the surface (Surface representation) • Assign parameters to U Number and V Number on the surface • Place a point just to the left of the bottom left corner of the surface – this will become the control point for measuring distance. It is important for getting the “Reactor” effect working. • Place an instance of the pyramid component by snapping the first four placement points onto four adjacent nodes in the same order as you originally created the adaptive points (say clockwise); place the fifth point on the external point – it is vital that it does not go onto a node of the surface • Link 4 parameters to equivalent parent parameters: Y_Number to V Number X_Number to U Number TotalX to Length TotalY to Width • Hide the nodes on the surface (Using Surface Representation - they cannot be controlled by any view settings) • Select the free control point and move it very close to the bottom left corner of the surface. • Select the Pyramid and turn it into a repeater • All being well, the pyramid will array itself over the whole surface, but each instance will look slightly different as the apex point is changed depending on its distance from the control point in the bottom left corner. If not, it could be caused by a problem with the component itself, or else the fifth adaptive point might be hosted on the same point as adaptive point one, in which case it would move with it. • Select (tab) any one of the pyramids; it should display properties including its correct column and row number • Finish the in-place mass family 6. Create a generic Schedule: • Add the column and row numbers, and any other parameters you require • You will be able to edit the values for Comments and Mark, but no other instance values • If you edit the mass family, it allows you to manually select any of the components in the repeater. Then you can set it to “No Component” or to any other 5 point adaptive component. When you edit the mass family, the schedule temporarily goes blank. Sadly this means that it is not possible to drive the geometry from the schedule. However, it does make it easier to identify and label components in a schedule – useful for adding and editing Mark values to match column and row numbers. This technique will not work with a curved surface because it all works by calculating the actual distance from the control point to the BL corner of the pyramid and relating that to the overall length of the surface. It will only work on an orthogonal divided surface, unless you are a mathematical genius and can write formulas to handle more complex geometry!	{"url":"http://revitcat.blogspot.com/2013/10/how-to-schedule-panel-locations-on-flat.html","timestamp":"2024-11-09T04:46:33Z","content_type":"text/html","content_length":"409504","record_id":"<urn:uuid:7dab4dfa-99c2-42ef-982f-f94b051f0bb8>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00811.warc.gz"}
You need 10+ years if you pay just the minimum due on credit cards! India is adapting to the culture of credit cards at a staggering pace. While credit cards may seem like a means to financial freedom and accessibility, in many cases, they can be dangerous for your financial health. My cousin, for instance, is an IT professional living in Bangalore who enjoys a comfortable lifestyle. He has a stable job and a good salary. He often travels and dines out. Last year, to support his lifestyle, he applied for a credit card. Initially, he used it for small purchases but gradually began charging larger expenses to it, including vacations, electronics, and home improvements—common habits for many credit card holders. As his expenses grew, like many others, he tried the minimum payment option. He loved it because it didn’t add immediate financial strain. He kept paying the minimum amount due, believing this strategy was practical since it allowed him to allocate his income to other areas. A year into using his credit card, however, he noticed that despite regular payments, his debt hadn’t decreased. His outstanding balance ballooned due to high interest rates. He then realized he was only paying off the interest component to a large extent and barely touching the principal amount. So, what really happens when you pay only the minimum amount due? When you apply for a credit card, you are generally given a 30-day window to utilize its benefits. This 30-day window is also called the billing cycle. Let’s say you applied for the credit card and received it on the 1st of April. In this case, your billing cycle would be from the 1st of April to the 30th of April. In every billing cycle, all transactions made during that 30-day period are recorded, and a statement is generated. This statement indicates the amount you owe and the due date by which you need to make the payment. The due date is generally 20-25 days from the issue date of the statement, and the period between the statement generated and the due date is also called the ‘grace period’. Considering the above scenario, the due date here would be the 25th of May. If you make the full payment within that grace period, there are no consequences, and your credit score remains healthy. This is the ideal practice. However, if you pay just the minimum amount due, which in most instances is 5% of your total outstanding amount, you may avoid late payment charges. Still, the remaining unpaid amount will start attracting interest, which can go up to 42% per annum. To remind you, if you have a good credit score, a personal loan will be available to you at 11% – 18% per annum. Not just the unpaid balance, all future spending on the credit card also starts attracting interest. Let me explain this to you with an example: Assume you have just received a new credit card with a credit limit of ₹1 lakh. The card has an annual interest rate of 36%, which translates to a monthly interest rate of 3%. You start using the card for various expenses, and by the end of the first billing cycle, you have spent ₹50,000 on your credit card. 1. Billing Cycle: 30 days 2. Statement Date: 30th of the month 3. Due Date: 20 days from the statement date (20th of the next month) On the First Billing Cycle: 1. The Outstanding Balance is ₹50,000 2. Let’s assume the minimum payment due is 5% of the outstanding balance, i.e., ₹2,500 Here’s how the accrued interest and balance would look at the end of each month if you only paid the minimum amount due – Do note that Interest calculation starts from the day an expense was made. For simplicity, all spends are assumed to be in the middle of the month. Therefore, the accrued interest of 1750 in the first row is for 35 days (from 15 Jan to 20 Feb). As you can see, after 12 months, your principal amount has come down by just ₹10,000 when you had paid about ₹27,000 in the form of minimum dues. If you continue like this, you might need over a decade to fully pay off your dues, and you will have spent huge amounts as interest in the meantime. In fact, if you made any more spends before paying off the dues, the new spends will attract 3% monthly interest from the very day when they were made. Paying only the minimum amount due each month keeps you in a prolonged cycle of debt and accrues high interest costs. Moreover, because of the higher balance outstanding and future credit card transactions, you will be utilizing a larger portion of your total available credit limit. This can impact your credit score. I have explained how credit score works in an earlier post. To sum it up, avoid making just the minimum due payments to avoid financial stress. Instead, budget well so you never fall into such debt traps. Borrow mindfully! Post a comment	{"url":"https://zerodha.com/z-connect/varsity/you-need-10-years-if-you-pay-just-the-minimum-due-on-credit-cards","timestamp":"2024-11-11T04:08:58Z","content_type":"text/html","content_length":"47453","record_id":"<urn:uuid:9addff4c-4782-4f0e-acf2-09a4c12c662d>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00855.warc.gz"}
28th December Mathematicians Of The Day 28th December On this day in 1612, Galileo observed Neptune, but did not recognize it as a planet. The "discovery" of Neptune happened more than 200 years later. See THIS LINK. The postage stamp of one of today's mathematicians at THIS LINK was issued in 1992. Proof is the idol before whom the pure mathematician tortures himself.	{"url":"https://mathshistory.st-andrews.ac.uk/OfTheDay/oftheday-12-28/","timestamp":"2024-11-11T05:18:59Z","content_type":"text/html","content_length":"19298","record_id":"<urn:uuid:3511f565-23de-4600-a216-26f8142bf597>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00575.warc.gz"}
PICUP Exercise Sets: Efficiency of a Water Turbine (3D Printing Lab) Efficiency of a Water Turbine (3D Printing Lab) Developed by Deva O'Neil, Benjamin Hancock, and Benjamin Hanks - Published June 7, 2021 DOI: 10.1119/PICUP.Exercise.waterturbine This Exercise Set describes one way to incorporate 3D printing into lab sessions in Physics I: Students design and print a water-wheel, and measure its efficiency in lifting a load. An optional exercise at the end uses video analysis to verify that the system is approximately in equilibrium for almost all of the lift process. Concepts applied include power, energy, and efficiency. Subject Mechanics, Experimental / Labs, and Fluids Level First Year Students who complete this set of exercises will be able to: * Apply the concept of density to calculate the potential energy of a water reservoir (Exercise 1) * Calculate work done in Learning lifting a load vertically and relate it to power (Exercise 2) * Calculate efficiency and identify sources of energy loss (Exercise 3) * Practice design thinking and develop CAD Objectives skills in designing and printing a water turbine (Exercise 4) * Use video analysis to verify that the system is in equilibrium (Exercise 5) Time to 200 min These exercises are not tied to a specific programming language. Example implementations are provided under the Code tab, but the Exercises can be implemented in whatever platform you wish to use (e.g., Excel, Python, MATLAB, etc.). Exercise 1: Finding Potential Energy of Water Flowing Through a Turbine - Not shown to scale In this lab, we will be placing a water wheel under a water reservoir. The amount of energy available to the wheel depends on the potential energy of the water in the reservoir. The equation for potential energy of the water must incorporate the measurable dimensions $V$ and $h$, and will also depend on $g$. Using the numbers in the figure above, calculate the potential energy of the water; you'll have to look up the density that water would have in your lab room environment. ** Exercise 2: Finding the Power Exhibited by a Water Turbine Your 3D-printed water wheel will be lifting a small weight. In order to find the power of the wheel, it must perform work on the object by moving it through a distance $d$. The time the wheel is exerting a force on the mass is also needed, since $P_{out} = W/t$, where $W$ is the work being done on the block by the string. To get started, we will assume that in the course of 2 minutes, the falling water induced the wheel to lift a 10 gram block 70 cm into the air. Assume the entirety of the water was used for this action. We need to calculate $W$, the work being done on the block by the string, so it is necessary to find the force on the block. Use a free-body diagram to determine the force on the block due to the string. -- What condition must be true to make the tension on the string be equal to $mg$? Using this information, calculate the power of the water turbine, listing all assumptions that must be made. ___ Exercise 3: Finding the Efficiency of the Water Turbine As with all machines, efficiency ($\eta$) is an important factor in determining a water wheel's viability in a real world scenario. It compares the amount of useful energy (work) generated compared to the amount of energy that went into the machine. $$\eta=W_{block} / E_{water}. $$ - Efficiency can also be expressed as a power ratio: $$\eta = P_{out} / P_{in} .$$ Using the potential energy of the water and the wheel's work output, calculate the efficiency of the water wheel. If not all of the water contributes to the lifting process, then one should not count all of it as an energy input. In this scenario, we will assume that all of the water contributes. According to the Law of Energy Conservation, final and initial energy in the universe is the same. Since the efficiency of the wheel is not 1, energy must have left the system. In what ways could it have left the system? Exercise 4: Efficiency of a 3D-Printed Water Wheel** In this exercise, you'll need to design and 3D print a water wheel that will lift a load. Before starting your drawing in a CAD program, discuss with your instructor options for mounting the wheel so that it can rotate freely. In designing your wheel, your goal is to optimize the efficiency. As you create the wheel, size it appropriately in the CAD program so you can predict what its dimensions will be when printed. If there are holes in your shape, keep in mind that the filament may expand during printing so that holes generally end up 1 mm smaller in diameter than the design dimension predicts. Keep in mind also the following constraints when designing the axle: * a string will be attached to the axle; it must wind around the axle as the wheel rotates. * the wheel must be able to be mounted in a stable way Maximize the lifted mass so that it will maintain a constant velocity of the wheel after motion has been established. Using the process indicated by exercises 1-3, calculate the efficiency of your 3D-printed water wheel. You'll need to decide which efficiency equation (power ratio or energy ratio) best applies to your experiment, based on whether you can assume that all the potential energy of the water contributed to the lift. If the lift occurred over 30 seconds but it took 50 seconds for the water to drain, using power would be a better choice. Thinking back to the calculation in Exercise 2, answer the following: - Why is it desirable to maintain a constant velocity? Exercise 5: Analysis of Acceleration by Tracker One of the primary assumptions made to justify the calculations used was that the angular acceleration of the wheel is approximately zero once motion is established. Using video analysis, measure the acceleration as a function of time and determine to what degree the system is in equilibrium. The instructions in the Experiment Tab guide you through this process in detail, if you have not done video analysis before. - What average acceleration do you obtain? Does the answer vary depending on what part of the motion you capture? - Relate your observations to the assumptions made earlier in this lab. - Discuss what factors could have resulted in error. The software can be downloaded at [Tracker](https://physlets.org/tracker/). ___ Download Options Share a Variation Did you have to edit this material to fit your needs? Share your changes by Creating a Variation Credits and Licensing Deva O'Neil, Benjamin Hancock, and Benjamin Hanks, "Efficiency of a Water Turbine (3D Printing Lab)," Published in the PICUP Collection, June 2021, https://doi.org/10.1119/ DOI: 10.1119/PICUP.Exercise.waterturbine The instructor materials are ©2021 Deva O'Neil, Benjamin Hancock, and Benjamin Hanks. 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Summing Over All Paths for Multiple Openings You now know that phase is important when there are two openings because the phases at a screen point determine how amplitudes associated with paths through each opening add to give a total probability. You can see this again in another interactive figure by clicking at two different locations in the region between the barrier and the screen so as to create two openings. The resulting black curve shows the probability for observing an electron at various locations on the screen. Note the minima and maxima due to the constructive and destructive addition of the amplitudes to give the total probability. You can also see here the effects of varying the distance between the openings. We have made a simplifying assumption in this simulation that we have not made before. To find an amplitude for a given opening and final point on the screen, we previously summed over an explicit set of 100 paths. Here we have used the mathematical result from when one does a full sum over all possible paths. Hence, we have not drawn any specific paths as we did earlier. By doing this, we obtain accurate results and significantly reduce the computation time. This also makes it possible to ask the question of what happens if the amplitude and phase of the electron at each opening are varied. By clicking in the circle coresponding to a particular opening you can set the amplitude (length) and phase (orientation) at each opening. Notice that varying the phase at one opening shifts the probability pattern on the screen right and left, while changing the amplitude at one opening alters the minimum probability you can obtain. See how this is also displayed in the vector addition of the arrows on the right. By clicking multiple times in the region between the barrier and the screen you can create up to five openings spaced any way you want. For each, you can vary the phase and amplitude of the electron. As you'll see, quite complicated and intriguing patterns can result from the phase interactions.	{"url":"https://serendipstudio.org/sci_edu/interactphysics/feynman/feynmand.html","timestamp":"2024-11-04T02:32:51Z","content_type":"text/html","content_length":"3169","record_id":"<urn:uuid:15984d67-7b63-4eda-8169-3a59e68247c6>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00236.warc.gz"}
10.7 Quadratic Word Problems: Age and Numbers Quadratic-based word problems are the third type of word problems covered in MATQ 1099, with the first being linear equations of one variable and the second linear equations of two or more variables. Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you will end up constructing a quadratic equation. To find the solution, you will be required to either factor the quadratic equation or use substitution. The sum of two numbers is 18, and the product of these two numbers is 56. What are the numbers? First, we know two things: [latex]\begin{array}{l} \text{smaller }(S)+\text{larger }(L)=18\Rightarrow L=18-S \\ \\ S\times L=56 \end{array}[/latex] Substituting [latex]18-S[/latex] for [latex]L[/latex] in the second equation gives: Multiplying this out gives: Which rearranges to: Second, factor this quadratic to get our solution: [latex]\begin{array}{rrrrrrl} S^2&-&18S&+&56&=&0 \\ (S&-&4)(S&-&14)&=&0 \\ \\ &&&&S&=&4, 14 \end{array}[/latex] [latex]\begin{array}{l} S=4, L=18-4=14 \\ \\ S=14, L=18-14=4 \text{ (this solution is rejected)} \end{array}[/latex] The difference of the squares of two consecutive even integers is 68. What are these numbers? The variables used for two consecutive integers (either odd or even) is [latex]x[/latex] and [latex]x + 2[/latex]. The equation to use for this problem is [latex](x + 2)^2 - (x)^2 = 68[/latex]. Simplifying this yields: [latex]\begin{array}{rrrrrrrrr} &&(x&+&2)^2&-&(x)^2&=&68 \\ x^2&+&4x&+&4&-&x^2&=&68 \\ &&&&4x&+&4&=&68 \\ &&&&&-&4&&-4 \\ \hline &&&&&&\dfrac{4x}{4}&=&\dfrac{64}{4} \\ \\ &&&&&&x&=&16 \end{array}[/ This means that the two integers are 16 and 18. The product of the ages of Sally and Joey now is 175 more than the product of their ages 5 years prior. If Sally is 20 years older than Joey, what are their current ages? The equations are: [latex]\begin{array}{rrl} (S)(J)&=&175+(S-5)(J-5) \\ S&=&J+20 \end{array}[/latex] Substituting for S gives us: [latex]\begin{array}{rrrrrrrrcrr} (J&+&20)(J)&=&175&+&(J&+&20-5)(J&-&5) \\ J^2&+&20J&=&175&+&(J&+&15)(J&-&5) \\ J^2&+&20J&=&175&+&J^2&+&10J&-&75 \\ -J^2&-&10J&&&-&J^2&-&10J&& \\ \hline &&\dfrac{10J} {10}&=&\dfrac{100}{10} &&&&&& \\ \\ &&J&=&10 &&&&&& \end{array}[/latex] This means that Joey is 10 years old and Sally is 30 years old. For Questions 1 to 12, write and solve the equation describing the relationship. 1. The sum of two numbers is 22, and the product of these two numbers is 120. What are the numbers? 2. The difference of two numbers is 4, and the product of these two numbers is 140. What are the numbers? 3. The difference of two numbers is 8, and the sum of the squares of these two numbers are 320. What are the numbers? 4. The sum of the squares of two consecutive even integers is 244. What are these numbers? 5. The difference of the squares of two consecutive even integers is 60. What are these numbers? 6. The sum of the squares of two consecutive even integers is 452. What are these numbers? 7. Find three consecutive even integers such that the product of the first two is 38 more than the third integer. 8. Find three consecutive odd integers such that the product of the first two is 52 more than the third integer. 9. The product of the ages of Alan and Terry is 80 more than the product of their ages 4 years prior. If Alan is 4 years older than Terry, what are their current ages? 10. The product of the ages of Cally and Katy is 130 less than the product of their ages in 5 years. If Cally is 3 years older than Katy, what are their current ages? 11. The product of the ages of James and Susan in 5 years is 230 more than the product of their ages today. What are their ages if James is one year older than Susan? 12. The product of the ages (in days) of two newborn babies Simran and Jessie in two days will be 48 more than the product of their ages today. How old are the babies if Jessie is 2 days older than Doug went to a conference in a city 120 km away. On the way back, due to road construction, he had to drive 10 km/h slower, which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference? The first equation is [latex]r(t) = 120[/latex], which means that [latex]r = \dfrac{120}{t}[/latex] or [latex]t = \dfrac{120}{r}[/latex]. For the second equation, [latex]r[/latex] is 10 km/h slower and [latex]t[/latex] is 2 hours longer. This means the second equation is [latex](r - 10)(t + 2) = 120[/latex]. We will eliminate the variable [latex]t[/latex] in the second equation by substitution: Multiply both sides by [latex]r[/latex] to eliminate the fraction, which leaves us with: Multiplying everything out gives us: [latex]\begin{array}{rrrrrrrrr} 120r&+&2r^2&-&1200&-&20r&=&120r \\ &&2r^2&+&100r&-&1200&=&120r \\ &&&-&120r&&&&-120r \\ \hline &&2r^2&-&20r&-&1200&=&0 \end{array}[/latex] This equation can be reduced by a common factor of 2, which leaves us with: [latex]\begin{array}{rrl} r^2-10r-600&=&0 \\ (r-30)(r+20)&=&0 \\ r&=&30\text{ km/h or }-20\text{ km/h (reject)} \end{array}[/latex] Mark rows downstream for 30 km, then turns around and returns to his original location. The total trip took 8 hr. If the current flows at 2 km/h, how fast would Mark row in still water? If we let [latex]t =[/latex] the time to row downstream, then the time to return is [latex]8\text{ h}- t[/latex]. The first equation is [latex](r + 2)t = 30[/latex]. The stream speeds up the boat, which means [latex]t = \dfrac{30}{(r + 2)}[/latex], and the second equation is [latex](r - 2)(8 - t) = 30[/latex] when the stream slows down the boat. We will eliminate the variable [latex]t[/latex] in the second equation by substituting [latex]t=\dfrac{30}{(r+2)}[/latex]: Multiply both sides by [latex](r + 2)[/latex] to eliminate the fraction, which leaves us with: Multiplying everything out gives us: [latex]\begin{array}{rrrrrrrrrrr} (r&-&2)(8r&+&16&-&30)&=&30r&+&60 \\ &&(r&-&2)(8r&+&(-14))&=&30r&+&60 \\ 8r^2&-&14r&-&16r&+&28&=&30r&+&60 \\ &&8r^2&-&30r&+&28&=&30r&+&60 \\ &&&-&30r&-&60&&-30r&-&60 \\ \hline &&8r^2&-&60r&-&32&=&0&& \end{array}[/latex] This equation can be reduced by a common factor of 4, which will leave us: [latex]\begin{array}{rll} 2r^2-15r-8&=&0 \\ (2r+1)(r-8)&=&0 \\ r&=&-\dfrac{1}{2}\text{ km/h (reject) or }r=8\text{ km/h} \end{array}[/latex] For Questions 13 to 20, write and solve the equation describing the relationship. 13. A train travelled 240 km at a certain speed. When the engine was replaced by an improved model, the speed was increased by 20 km/hr and the travel time for the trip was decreased by 1 hr. What was the rate of each engine? 14. Mr. Jones visits his grandmother, who lives 100 km away, on a regular basis. Recently, a new freeway has opened up, and although the freeway route is 120 km, he can drive 20 km/h faster on average and takes 30 minutes less time to make the trip. What is Mr. Jones’s rate on both the old route and on the freeway? 15. If a cyclist had travelled 5 km/h faster, she would have needed 1.5 hr less time to travel 150 km. Find the speed of the cyclist. 16. By going 15 km per hr faster, a transit bus would have required 1 hr less to travel 180 km. What was the average speed of this bus? 17. A cyclist rides to a cabin 72 km away up the valley and then returns in 9 hr. His speed returning is 12 km/h faster than his speed in going. Find his speed both going and returning. 18. A cyclist made a trip of 120 km and then returned in 7 hr. Returning, the rate increased 10 km/h. Find the speed of this cyclist travelling each way. 19. The distance between two bus stations is 240 km. If the speed of a bus increases by 36 km/h, the trip would take 1.5 hour less. What is the usual speed of the bus? 20. A pilot flew at a constant speed for 600 km. Returning the next day, the pilot flew against a headwind of 50 km/h to return to his starting point. If the plane was in the air for a total of 7 hours, what was the average speed of this plane? Find the length and width of a rectangle whose length is 5 cm longer than its width and whose area is 50 cm^2. First, the area of this rectangle is given by [latex]L\times W[/latex], meaning that, for this rectangle, [latex]L\times W=50[/latex], or [latex](W+5)W=50[/latex]. Multiplying this out gives us: Which rearranges to: Second, we factor this quadratic to get our solution: [latex]\begin{array}{rrrrrrl} W^2&+&5W&-&50&=&0 \\ (W&-&5)(W&+&10)&=&0 \\ &&&&W&=&5, -10 \\ \end{array}[/latex] We reject the solution [latex]W = -10[/latex]. This means that [latex]L = W + 5 = 5+5= 10[/latex]. If the length of each side of a square is increased by 6, the area is multiplied by 16. Find the length of one side of the original square. The relationship between these two is: [latex]\begin{array}{rrl} \text{larger area}&=&16\text{ times the smaller area} \\ (x+12)^2&=&16(x)^2 \end{array}[/latex] Simplifying this yields: [latex]\begin{array}{rrrrrrr} x^2&+&24x&+&144&=&16x^2 \\ -16x^2&&&&&&-16x^2 \\ \hline -15x^2&+&24x&+&144&=&0 \end{array}[/latex] Since this is a problem that requires factoring, it is easiest to use the quadratic equation: [latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=-15, b=24\text{ and }c=144[/latex] Substituting these values in yields [latex]x = 4[/latex] or [latex]x=-2.4[/latex] (reject). Nick and Chloe want to surround their 60 by 80 cm wedding photo with matting of equal width. The resulting photo and matting is to be covered by a 1 m^2 sheet of expensive archival glass. Find the width of the matting. [latex](L+2x)(W+2x)=1\text{ m}^2[/latex] Or, in cm: [latex](80\text{ cm }+2x)(60\text{ cm }+2x)=10,000\text{ cm}^2[/latex] Multiplying this out gives us: Which rearranges to: Which reduces to: [latex]x^2 + 70x - 1300 = 0[/latex] Second, we factor this quadratic to get our solution. It is easiest to use the quadratic equation to find our solutions. [latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=1, b=70\text{ and }c=-1300[/latex] Substituting the values in yields: [latex]x=\dfrac{-70\pm \sqrt{70^2-4(1)(-1300)}}{2(1)}\hspace{0.5in}x=\dfrac{-70\pm 10\sqrt{101}}{2}[/latex] [latex]x=-35+5\sqrt{101}\hspace{0.75in} x=-35-5\sqrt{101}\text{ (rejected)}[/latex] For Questions 21 to 28, write and solve the equation describing the relationship. 21. Find the length and width of a rectangle whose length is 4 cm longer than its width and whose area is 60 cm^2. 22. Find the length and width of a rectangle whose width is 10 cm shorter than its length and whose area is 200 cm^2. 23. A large rectangular garden in a park is 120 m wide and 150 m long. A contractor is called in to add a brick walkway to surround this garden. If the area of the walkway is 2800 m^2, how wide is the walkway? 24. A park swimming pool is 10 m wide and 25 m long. A pool cover is purchased to cover the pool, overlapping all 4 sides by the same width. If the covered area outside the pool is 74 m^2, how wide is the overlap area? 25. In a landscape plan, a rectangular flowerbed is designed to be 4 m longer than it is wide. If 60 m^2 are needed for the plants in the bed, what should the dimensions of the rectangular bed be? 26. If the side of a square is increased by 5 units, the area is increased by 4 square units. Find the length of the sides of the original square. 27. A rectangular lot is 20 m longer than it is wide and its area is 2400 m^2. Find the dimensions of the lot. 28. The length of a room is 8 m greater than its width. If both the length and the width are increased by 2 m, the area increases by 60 m^2. Find the dimensions of the room.	{"url":"https://opentextbc.ca/intermediatealgebraberg/chapter/quadratic-word-problems-age-and-numbers/","timestamp":"2024-11-13T03:12:17Z","content_type":"text/html","content_length":"137026","record_id":"<urn:uuid:178afec7-bc2d-41d7-b093-eb041e0f8213>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00805.warc.gz"}
Translating Word Phrases into Expressions With Integers Using Multiplication and Division Learning Outcomes • Translate word phrases involving multiplication or division to algebraic expressions and simplify Once again, all our prior work translating words to algebra transfers to phrases that include both multiplying and dividing integers. Remember that the key word for multiplication is product and for division is quotient. Translate to an algebraic expression and simplify if possible: the product of [latex]-2[/latex] and [latex]14[/latex]. The word product tells us to multiply. the product of [latex]-2[/latex] and [latex]14[/latex] Translate. [latex]\left(-2\right)\left(14\right)[/latex] Simplify. [latex]-28[/latex] Translate to an algebraic expression and simplify if possible: the quotient of [latex]-56[/latex] and [latex]-7[/latex]. Show Solution The following video shows more examples of how to translate expressions that contain integer multiplication and division.	{"url":"https://courses.lumenlearning.com/wm-developmentalemporium/chapter/translating-word-phrases-into-expressions-with-integers-using-multiplication-and-division/","timestamp":"2024-11-07T22:33:03Z","content_type":"text/html","content_length":"49952","record_id":"<urn:uuid:57f52d24-8f82-4ab9-94a2-174b84efc49e>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00303.warc.gz"}
Jonathan Borwein - APM Institute We can only grieve for the loss of the great Jonathan Borwein. He was APM member since November 2015. We relay the words of his close friend and collaborator, David Bailey. “Jonathan Borwein dies at 65 It is my sad duty to report that our colleague Jonathan Borwein, Laureate Professor of Mathematics at the University of Newcastle, Australia, has passed away at the age of 65. He is survived by his wife Judith and three daughters. What can one say about Jon’s professional accomplishments? Adjectives such as “profound,” “vast” and “far-ranging” don’t really do justice to his work, the sheer volume of which is astounding: 388 published journal articles, plus another 103 articles in refereed or invited conference proceedings (according to his CV, dated one day before his death). The ISI Web of Knowledge lists 6,593 citations from 351 items; one paper has been cited 666 times. The Google Citation Tracker finds over 22,048 citations. But volume is not the only remarkable feature of Jon’s work. Another is the amazing span of his work. In an era when academic researchers in general, and mathematicians in particular, focus ever more tightly on a single specialty, Jon ranged far and wide, with significant work in pure mathematics, applied mathematics, optimization theory, computer science, mathematical finance, and, of course, experimental mathematics, in which he has been arguably the world’s premier authority. Unlike many in the field, Jon tried at every turn to do research that is accessible, and to highlight aspects of his and others’ work that a broad audience (including both researchers and the lay public) could appreciate. This was, in part, behind his long-running interest in Pi, and in the computation and analysis of Pi — this topic, like numerous others he has studied, is one whose wonder and delight can be shared with millions. This desire to share mathematics and science with the outside world led to his writing numerous articles on mathematics, science and society for the Math Drudge blog, the Conversation and the Huffington Post. He was not required to do this, nor, frankly, is such writing counted for professional prestige; instead he did it to share the facts, discoveries and wonder of modern science with the rest of the world. Jon was a mentor par excellence, having guided 30 graduate students and 42 post-doctoral scholars. Working with Jon is not easy — he is a demanding colleague (as the present author will attest), but for those willing to apply themselves, the rewards have been great, as they become first-hand partners in ground-breaking work. There is much, much more that could be mentioned, including his tireless and often thankless service on numerous committees and organizational boards, including Governor at large of the Mathematical Association of America (2004–07), President of the Canadian Mathematical Society (2000–02), Chair of the Canadian National Science Library Advisory Board (2000–2003) and Chair of the Scientific Advisory Committee of the Australian Mathematical Sciences Institute (AMSI). But Jon was more than a scholar. He was a devoted husband and father. He and Judi have been married for nearly 40 years, and they have three lovely and accomplished daughters. They have endured some incredible hardships, but Jon has made some equally incredible sacrifices on their behalf. Jon has also been devoted to his own father and mother, often collaborating on research work with his father David Borwein (also a well-known mathematician), and following the work of his mother, a scholar in her own right. I myself am at a loss of what to say at Jon’s passing. What can I say? I have collaborated with Jon for over 31 years, with over 80 papers and five books with Jon as a co-author. Thus my personal debt to Jon is truly enormous. My work will forever be connected with (and certainly subservient to) that of Jon’s. I am humbled beyond measure and grieve deeply at his passing. Jon’s passing is an incalculable loss to the field of mathematics in general, and to experimental mathematics in particular. Jon is arguably the world’s leading researcher in the field of experimental mathematics, and his loss will be very deeply felt. We will be reading his papers and following his example for decades to come.” by David Bailey (Math Drudge)	{"url":"https://apminstitute.org/jonathan-borwein/","timestamp":"2024-11-14T13:40:53Z","content_type":"text/html","content_length":"57490","record_id":"<urn:uuid:1972b65b-64e8-474f-b198-49b8fd59f478>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00873.warc.gz"}
Probabilistic vs Deterministic Thinking If the weather forecast says there is a 70% chance of rain and it rains: was the forecast “right” or “wrong” or neither? Many people think and talk about the forecast as being “right” when it does rain, but if it doesn’t rain people think and talk about the forecast as being “wrong” (Tetlock & Gardner, 2015). But the forecast cannot be right or wrong because it is a probability of a future event happening based on the data that we had at the moment it was made. This example of how we think and talk about common daily experiences with data-based information I think is illuminating about some of the struggles students have working with data. Let’s explore. There are two common ways of thinking about information: • Deterministic Thinking – For a situation, question, scenario, etc. there is a ”right” and a “wrong” answer. The forecast must be “right” if it rained and “wrong” if it didn’t rain. • Probabilistic Thinking – For a situation, question, scenario, etc. we make conclusions based on what is supported or not with existing evidence. The evidence indicates that there was a higher likelihood that it would rain then it not rain, but what actually happened was dependent upon the evidence that went into the forecast when it was made and many other components after the forecast was made. I think this is relevant to data-based work in a few ways that can be helpful to think about for our students: 1) humans like to think deterministically (and are positively reinforced for that thinking both personally and in school a lot), and 2) we can never make deterministic claims from data. And here in lies the rub. Let’s use an example to explore. Which plant is taller? —> There is “an” answer to this question because 1) we can measure the height of each plant (here given in centimeters), and 2) we can use arithmetic to subtract 52 from 80 to get 28. So we can use deterministic thinking to answer this number fact question. <— Which group of plants are taller? There is no one answer to this question. Even though we can measure each plant (like before) we then need to make a decision on how to think about what the “height of the group” (e.g., maximum? minimum? range? average?) for each is and then how we want to compare those “group heights”. In the second example, we have data (multiple measured values of the variable) and thus we have to use probabilistic thinking to answer the question. Which also means that someone could report a different answer to the same question with the same data (aka they made a different decision on how to calculate the “group height”). These two kinds of thinking also have ramifications on how we think about scientific conclusions... explain more and connect to science literacy • Deterministic Thinking – “more” research/work and/or a lack of “proof/proving” something suggests that there is something wrong with what is already known. • Probabilistic Thinking – “more” research/work is always needed to have more evidence to use and we of course we never prove anything because we cannot possibly have all of the information. So, the question is how can we help our students think more probabilistically rather than deterministically about data? By embracing statistical thinking in all of our work with data! What does that even mean? Great question! Join us at the next Data Literacy Series: Embrace Statistical Thinking session to find out :)	{"url":"https://www.dataspire.org/blog/probabilistic-vs-deterministic-thinking","timestamp":"2024-11-09T05:59:59Z","content_type":"text/html","content_length":"50456","record_id":"<urn:uuid:00a870d4-6e72-43c0-9a3c-4854c34c7137>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00883.warc.gz"}
Mastering Formulas In Excel: What The Formula To Get A Percentage Understanding formulas in Excel is crucial for anyone who wants to efficiently analyze and manipulate data. Being able to master formulas not only saves time but also helps in making accurate calculations and generating valuable insights. There are different types of formulas in Excel, each serving a specific purpose to meet the needs of various data analysis tasks. Key Takeaways • Understanding formulas in Excel is crucial for efficient data analysis. • Mastering formulas saves time and ensures accurate calculations. • Different types of formulas in Excel serve specific purposes for data analysis tasks. • Avoid common errors when using percentage formulas, such as forgetting to format the cell as a percentage. • Practice, utilize tutorials, and keep track of common mistakes to master percentage formulas in Excel. Understanding the basic formula structure When it comes to mastering formulas in Excel, understanding the basic formula structure is essential. This foundational knowledge sets the stage for more complex formulas, including those used to calculate percentages. Here are the components of a basic formula in Excel and how to input cell references and operators into a formula. A. The components of a basic formula in Excel Excel formulas are built using cell references, operators, and functions. Cell references are used to identify the location of the data you want to use in a formula. Operators are used to perform mathematical operations such as addition, subtraction, multiplication, and division. Functions are predefined formulas that perform calculations using specific values. Understanding how these components work together is crucial for formulating the correct percentage formula in Excel. B. How to input cell references and operators into a formula When inputting cell references into a formula, you must start with the equal sign (=) followed by the cell reference or range of cell references you want to use. For example, to add the values in cell A1 and A2, you would input =A1+A2. Operators such as +, -, , and / are used to perform calculations within a formula. Knowing how to correctly input these components is essential for creating a percentage formula that yields accurate results. Different ways to calculate percentages in Excel When working with Excel, it is essential to know how to calculate percentages efficiently. There are several methods to do this, and each has its own benefits and drawbacks. Below are the different ways to calculate percentages in Excel: A. Using the percentage symbol • The percentage symbol (%) can be used to calculate percentages in Excel. • This method involves dividing the part by the whole and then multiplying by 100 to get the percentage. • For example, if cell A1 contains the total and cell B1 contains the part, the formula in cell C1 would be =B1/A1100. B. Multiplying by the percentage as a decimal • This method involves converting the percentage to a decimal and then multiplying it by the whole. • To convert a percentage to a decimal, divide by 100. • For example, if the percentage is 20% and the whole is in cell A1, the formula in cell B1 would be =A120%/100. C. Utilizing the Excel percentage formula • Excel has a built-in percentage formula that can be used to calculate percentages. • The formula is =A1(B1/100), where A1 is the whole and B1 is the percentage. • This method can be particularly useful for large sets of data where using the percentage symbol or multiplying by the percentage as a decimal may be too time-consuming. Common Errors to Avoid When Using Percentage Formulas When working with percentage formulas in Excel, it's important to be aware of common errors that can occur. By understanding these pitfalls, you can ensure that your calculations are accurate and • Forgetting to format the cell as a percentage One common mistake when using percentage formulas in Excel is forgetting to format the cell as a percentage. When you enter a percentage formula, it's essential to ensure that the cell displaying the result is formatted correctly. If this step is overlooked, the result may not be displayed as a percentage, leading to misleading data. • Incorrectly inputting the formula syntax Another error to avoid is incorrectly inputting the formula syntax. When using percentage formulas, it's crucial to input the formula correctly to obtain accurate results. Common mistakes include missing parenthesis, using incorrect operators, or referencing the wrong cells. Taking the time to double-check the formula syntax can save you from potential errors. • Not understanding the order of operations in Excel Understanding the order of operations in Excel is essential for accurate percentage calculations. Failing to grasp the sequence in which Excel performs calculations can lead to errors in your percentage formulas. By familiarizing yourself with the order of operations, you can ensure that your percentage formulas produce the correct results. Tips for mastering percentage formulas When it comes to working with percentages in Excel, it's important to have a solid understanding of how to use formulas effectively. Here are some tips to help you master percentage formulas: A. Practice using different methods to calculate percentages • 1. Understand the basic formula • 2. Try using the percentage format • 3. Experiment with the different functions available B. Utilize Excel tutorials and online resources • 1. Take advantage of Excel tutorials • 2. Explore online resources and forums • 3. Consider enrolling in online courses C. Keep track of common mistakes and how to troubleshoot them • 1. Look out for errors in cell references • 2. Check for divided by zero errors • 3. Use the trace error feature to identify mistakes Advanced percentage formula applications Excel offers a wide range of formula applications, including advanced percentage formulas that can be used for various purposes. Here are some ways in which you can utilize these formulas to enhance your data analysis and visualization. A. Using percentage formulas in conditional formatting Conditional formatting is a powerful feature in Excel that allows you to visually highlight cells based on certain conditions. You can use percentage formulas within conditional formatting to emphasize specific data points or trends within your spreadsheet. 1. Highlighting above or below average percentages You can use conditional formatting with percentage formulas to automatically highlight cells that are above or below the average percentage in a given range. This can be useful for quickly identifying outliers or trends within your data. 2. Color-coding percentage ranges Another way to use percentage formulas in conditional formatting is to color-code different percentage ranges within a dataset. This can help to visually segment and organize your data, making it easier to interpret at a glance. B. Analyzing data with percentage change formulas Percentage change formulas are essential for analyzing trends and fluctuations in data over time. In Excel, you can use these formulas to calculate the percentage change between two values, which can provide valuable insights into the performance of a particular metric. 1. Calculating month-over-month or year-over-year percentage changes By using percentage change formulas, you can easily compare the performance of a specific metric from one period to another. This can be particularly useful for tracking business performance, financial trends, or any other time-based data. 2. Visualizing percentage change with charts and graphs Once you have calculated the percentage changes in your data, you can visualize these trends using Excel's chart and graph tools. This can help to illustrate the patterns and fluctuations in your data, making it easier to communicate your findings to others. In conclusion, mastering percentage formulas in Excel is crucial for accurate data analysis and reporting. Understanding how to use formulas such as the percentage formula can greatly enhance your proficiency in Excel and make you more efficient in your work. It is important to continue practicing and learning about Excel formulas to become proficient in using them. The more you practice, the more comfortable you will become with using these formulas, and the more efficient you will be in your data analysis and reporting. ONLY $99 Immediate Download MAC & PC Compatible Free Email Support	{"url":"https://dashboardsexcel.com/blogs/blog/mastering-formulas-in-excel-what-the-formula-to-get-a-percentage","timestamp":"2024-11-09T03:43:38Z","content_type":"text/html","content_length":"210655","record_id":"<urn:uuid:ebdf4460-e9aa-49d8-af05-946722bc0adf>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00873.warc.gz"}
[Short-CourseSRM] Advances in parametric models of interest in reliability - NOVA Math Inmaculada Barranco-Chamorro (Department of Statistics and Operations Research, Institute of Mathematics, University of Sevilha, Sevilha, Spain) Title: Advances in parametric models of interest in reliability Location: Room 2.1., building VII General Abstract: This course deals with recent advances in parametric models of interest in reliability. It covers two topics. The first one is devoted to the application of LINEX loss function in reliability models such as the generalized half-logistic and Burr type-XII models. The second one deals with advances in slash distributions and the use of these models to obtain generalisations of Birnbaum-Saunders distribution. The course is structured in two sessions of 3 hours each, whose details are next listed. Material (slides) will be provided to the audience in advance. Topics and Abstract for session: Session 1: (June 17, 2019, 9.00-12.00) 1 st part: Models and censoring schemes of interest in reliability. (a) Models for reliability data. (b) Censoring schemes. (c) Progressive type-II right censoring. (d) Introduction to Bayesian methods. Summary: We introduce parametric models of interest in reliability along with common censoring mechanisms. We highlight progressive type-II right censoring scheme. We recall classical methods of estimation and Bayesian methods of inference. 2 nd part: On the use of LINEX loss function. Posterior and Bayes risk under LINEX loss function. (a) Asymmetric loss functions in Bayesian Statistics. LINEX loss function versus quadratic loss function. (b) Main results. Posterior risks and Bayes risks in previous setting. (c) Applications and simulations by using R software. Summary: To carry out estimation in Bayesian Statistics, a loss function must be specified. The most widely used loss is squared error. This function has the drawback that it is symmetric, that is, positive and negative errors of the same magnitude have the same penalty. Quite often this assumption is not realistic enough, since the loss function may not just be a measure of inaccuracy but a real loss, for example, financial. In these cases, an asymmetric loss function may be more appropriate, and following Zellner (1986), the Linex loss may be a good choice. This loss function allows us to penalize, in a different way, positive and negative errors, and it is still mathematically manageable. So, appealing results can be obtained from its use. In spite of this fact, there exist hardly papers dealing with this function. In this course, results will be presented for certain parametric models of interest in reliability, such as the generalized half-logistic distribution and the Burr Type-II distribution, under progressive type-II censoring. Specifically the posterior and Bayes risks of Bayes estimators are obtained in these models, since, following Lehmann and Casella (1998), risks are crucial to assess the performance of an estimator and compare competing estimators. We obtain Bayes and posterior risks of Bayes estimators under quadratic and Linex loss function when the prior distribution is a conjugate prior, along with other intermediate results of interest (the marginal distribution that we need to obtain the risks and the relationship between the Bayes estimators). Simulations are carried out by using R, which illustrate the behaviour of proposed estimators and their risks, along with the importance of different features involved in the progressive censoring scheme. An application to a real data set is included. Session 2: (June 17, 2019, 14.00-17.00) 1 st part: Advances in slash distributions: Generalized Modified Slash (GMS) Distribution. (a) Slash methodology. (b) Generalized Modified Slash (GMS) Distribution. (c) Methods of estimation in GMS model: EM-algorithm. (d) Simulations and applications. Summary: In real-world data, it is quite common to find symmetrical and unimodal histograms with heavy tails that do not _t well to a normal distribution. Slash models are a good option to deal with this kind of situations, in which departures of Gaussianity are a serious problem for the data analyst. This is one the main reason why slash distributions have received a great deal of attention during the last decades. In this context, we face the problem of improving slash models by introducing a generalisation able to model more kurtosis than other slash's previously proposed in literature. In slash models, the emphasis is on kurtosis, because as Moors (1988) pointed out the presence of heavy tails produces high kurtosis. In this session, we will focus on univariate symmetrical slash models. Specifically on the Generalised Modified Slash (GMS) model introduced in Reyes et al. (2019). The GMS model is defined. A closed expression for its pdf is given in terms of the confluent hypergeometric function; GMS model is expressed as a scale mixture; the convergence in law to a normal distribution is proven; moments are obtained, with emphasis on the kurtosis coefficient; and comparisons with other slash models previously introduced in the literature are presented. As for inference, we focus on iterative and EM maximum likelihood estimation methods. A simulation study will be showed which allow to assess the performance of our results. Applications to two real datasets are included. 2 nd part: Birnbaum-Saunders model based on GMS distribution. (a) Birnbaum-Saunders model. (b) Birnbaum-Saunders model based on GMS distribution. (c) Maximum likelihood based on EM-algorithm in these models. (d) Applications and simulations by using R software. Summary: The Birnbaum-Saunders (BS) distribution was introduced by Birnbaum and Saunders (1969). The aim of this distribution is to model the fatigue in lifetime of certain materials. Nowadays its use is being spread out to other contexts such as economic and environmental data. In these new applications, it is quite common to find real datasets in which a BS model with heavier tails would be suitable. Slash models are a good option to deal with this kind of situations. In this context, we briefly describe the BS-model and the generalisation proposed. Maximum likelihood based on EM-algorithm is carried out. Simulations and applications of interest in reliability are given. Barranco Chamorro, Inmaculada, Luque Calvo, Pedro Luis, Jimenez Gamero, Maria Dolores, Alba Fernández, M. Virtudes. A study of risks of Bayes estimators in the generalized half-logistic distribution for progressively type-II censored samples. In: Mathematics and Computers in Simulation. 2017. Vol. 137. Pag. 130-147.10.1016/j.matcom.2016.09.003 Reyes, Jimmy, Barranco Chamorro, Inmaculada, Gallardo, Diego I., Gómez, Héctor W.: Generalized Modified Slash Birnbaum-Saunders Distribution. In: Symmetry. 2018. Vol. 10. Num. 12.10.3390/ Reyes, Jimmy, Barranco Chamorro, Inmaculada, G_omez, H_ector W.: Generalized modified slash distribution with applications. 2019. In: Communications in Statistics – Theory and Methods. https:// and references therein.	{"url":"https://novamath.fct.unl.pt/short-coursesrm-advances-in-parametric-models-of-interest-in-reliability/","timestamp":"2024-11-10T16:22:02Z","content_type":"text/html","content_length":"141462","record_id":"<urn:uuid:eca88da5-604b-47ec-bd1f-39e0a85a395d>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00748.warc.gz"}
What is the derivative of f(x)=(log_6(x))^2 ? \| Socratic What is the derivative of #f(x)=(log_6(x))^2# ? 1 Answer Method 1: We will begin by using the change-of-base rule to rewrite $f \left(x\right)$ equivalently as: $f \left(x\right) = {\left(\ln \frac{x}{\ln} 6\right)}^{2}$ We know that $\frac{d}{\mathrm{dx}} \left[\ln x\right] = \frac{1}{x}$. (if this identity looks unfamiliar, check some of the videos on this page for further explanation) So, we will apply the chain rule: $f ' \left(x\right) = 2 \cdot {\left(\ln \frac{x}{\ln} 6\right)}^{1} \cdot \frac{d}{\mathrm{dx}} \left[\ln \frac{x}{\ln} 6\right]$ The derivative of $\ln \frac{x}{6}$ will be $\frac{1}{x \ln 6}$: $f ' \left(x\right) = 2 \cdot {\left(\ln \frac{x}{\ln} 6\right)}^{1} \cdot \frac{1}{x \ln 6}$ Simplifying gives us: $f ' \left(x\right) = \frac{2 \ln x}{x {\left(\ln 6\right)}^{2}}$ Method 2: The first thing to note is that only $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$ where $\ln = {\log}_{e}$. In other words, only if the base is $e$. We must therefore convert the ${\log}_{6}$ to an expression having only ${\log}_{e} = \ln$. This we do using the fact ${\log}_{a} b = \frac{{\log}_{n} b}{{\log}_{n} a} = \frac{\ln b}{\ln} a$ when $n = e$ Now, let $z = \left(\ln \frac{x}{\ln} 6\right)$ so that $f \left(x\right) = {z}^{2}$ Therefore, $f ' \left(x\right) = \frac{d}{\mathrm{dx}} {z}^{2} = \left(\frac{d}{\mathrm{dz}} {z}^{2}\right) \left(\frac{\mathrm{dz}}{\mathrm{dx}}\right) = 2 z \frac{d}{\mathrm{dx}} \left(\ln \frac{x} {\ln} 6\right)$ $= \frac{2 z}{\ln 6} \frac{d}{\mathrm{dx}} \ln x = \frac{2 z}{\ln 6} \frac{1}{x}$ $= \left(\frac{2}{\ln} 6\right) \left(\ln \frac{x}{\ln} 6\right) \left(\frac{1}{x}\right) = \frac{2 \ln x}{x \cdot {\left(\ln 6\right)}^{2}}$ Impact of this question 5962 views around the world	{"url":"https://api-project-1022638073839.appspot.com/questions/what-is-the-derivative-of-f-x-ln-x-2#108173","timestamp":"2024-11-13T09:36:21Z","content_type":"text/html","content_length":"35564","record_id":"<urn:uuid:cfab97b3-d483-43cc-bfce-88091a6becbd>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00053.warc.gz"}
quant doubts from CL Please give me the solution of following questions? 2. The smallest positive integer X with 24 divisors is a. 480 b. 420 c. 864 d. None of these 3. 5 coins are tossed. If two of them show heads, then the probability that all 5 coins show head is a. 1/32 b. 1/10 c. 1/26 d. 1/13 i have good reason to i have good reason to believe the answer is 1/26 lets see how: 1/32 is the probability of getting all five heads when there are nor restrictions on the sample space, meaning its the worst probability!!! getting 2 heads for given can only improve this probability. but how much ? that is the question. The answer is pretty simple as ican see. the number of ways of getting all 5 heads is always 1 no matter how many heads are given, which in this case is two. the number of ways of getting at least 2 heads = total number of ways - ways to get exactly 0 heads - ways to get exactly 1 head = 32 - 1 - 5 = 26 hence the answer 1/26	{"url":"http://cat4mba.com/comments?id=1587&title=quant%20doubts%20from%20CL","timestamp":"2024-11-10T06:37:53Z","content_type":"text/html","content_length":"11573","record_id":"<urn:uuid:22170f8f-ff10-4327-b6c2-dc5d7dcf2096>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00580.warc.gz"}
Example 9.25: It’s been a mighty warm winter? (Plot on a circular axis) \| R-bloggersExample 9.25: It’s been a mighty warm winter? (Plot on a circular axis) Example 9.25: It’s been a mighty warm winter? (Plot on a circular axis) [This article was first published on SAS and R , and kindly contributed to ]. (You can report issue about the content on this page ) Want to share your content on R-bloggers? if you have a blog, or if you don't. Updated (see below) People here in the northeast US consider this to have been an unusually warm winter. Was it? The University of Dayton and the US Environmental Protection Agency maintain an archive of daily average temperatures that’s reasonably current. In the case of Albany, NY (the most similar of their records to our homes in the Massachusetts’ Pioneer Valley), the data set as of this writing includes daily records from 1995 through March 12, 2012. In this entry, we show how to use R to plot these temperatures on a circular axis, that is, where January first follows December 31st. We’ll color the current winter differently to see how it compares. We’re not aware of a tool to enable this in SAS. It would most likely require a bit of algebra and manual plotting to make it work. The work of plotting is done by the function in the plotrix package. But there are a number of data management tasks to be employed first. Most notably, we need to calculate the relative portion of the year that’s elapsed through each day. This is trickier than it might be, because of leap years. We’ll read the data directly via URL, which we demonstrate in Example 8.31 . That way, when the unseasonably warm weather of last week is posted, we can update the plot with trivial ease. temp1 = read.table("http://academic.udayton.edu/kissock/http/ leap = c(0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1) days = rep(365, 18) + leap monthdays = c(31,28,31,30,31,30,31,31,30,31,30,31) temp1$V3 = temp1$V3 - 1994 , and vectors identify leap years, count the corrrect number of days in each year, and have the number of days in the month in non-leap years, respectively. We need each of these to get the elapsed time in the year for each day. The columns in the data set are the month, day, year, and average temperature (in Fahrenheit). The years are renumbered, since we’ll use them as indexes later. function, below, counts the proportion of days elapsed. yearpart = function(daytvec,yeardays,mdays=monthdays){ part = (sum(mdays[1:(daytvec[1]-1)], (daytvec[1] > 2) * (yeardays[daytvec[3]]==366)) + daytvec[2] - ((daytvec[1] == 1)31)) / yeardays[daytvec[3]] argument to the function will be a row from the data set. The function works by first summing the days in the months that have passed ( ) adding one if it’s February and a leap year ( (daytvec[1] > 2) (yeardays[daytvec[3]]==366)) ). Then the days passed so far in the current month are added. Finally, we subtract the length of January, if it’s January. This is needed, because sum(1:0) = 1 , the result of which is that that January is counted as a month that has “passed” when the function quoted above is calculated for January days. Finally, we just divide by the number of days in the current year. The rest is fairy simple. We calculate the radians as the portion of the year passed * 2 * pi, using the function to repeat across the rows of the data set. Then we make matrices with time before and time since this winter started, admittedly with some ugly logical expressions (section 1.14.11), and use function to make the plots. The options to the function are fairly self-explanatory. temp2 = as.matrix(temp1) radians = 2* pi * apply(temp2,1,yearpart,days,monthdays) t3old = matrix(c(temp1$V4[temp1$V4 != -99 & ((temp1$V3 < 18) \| (temp1$V2 < 12))], radians[temp1$V4 != -99 & ((temp1$V3 < 18) \| (temp1$V2 < 2))]),ncol=2) t3now= matrix(c(temp1$V4[temp1$V4 != -99 & ((temp1$V3 == 18) \| (temp1$V3 == 17 & temp1$V1 == 12))], radians[temp1$V4 != -99 & ((temp1$V3 == 18) \| (temp1$V3 == 17 & temp1$V1 == 12))]),ncol=2) # from plottrix library radial.plot(t3old[,1],t3old[,2],rp.type="s", point.col = 2, point.symbols=46, clockwise=TRUE, start = pi/2, label.pos = (1:12)/6 * (pi), labels=c("February 1","March 1","April 1","May 1","June 1", "July 1","August 1","September 1","October 1","November 1", "December 1","January 1"), radial.lim=c(-20,10,40,70,100)) radial.plot(t3now[,1],t3now[,2],rp.type="s", point.col = 1, point.symbols='*', clockwise=TRUE, start = pi/2, add=TRUE, radial.lim=c(-20,10,40,70,100)) The result is shown at the top. The dots ( is like so 20 is a point (section 5.2.2) show the older data, while the asterisks are the current winter. An alternate plot can be created with the option, which makes a line plot. The result is shown below, but the lines connecting the dots get most of the ink and are not what we care about today. Either plot demonstrates clearly that a typical average temperature in Albany is about 60 to 80 in August and about 10 to 35 in January, the coldest monthttp://www.blogger.com/img/blank.gifh. The top figure shows that it has in fact been quite a warm winter-- most of the black asterisks are near the outside of the range of red dots. Updating with more recent weeks will likely increase this impression. In the first edition of this post, the option was omitted, which resulted in different axes in the original and "add" calls to . This made the winter look much cooler. Many thanks to Robert Allison for noticing the problem in the main plot. Robert has made many hundreds of beautiful graphics in SAS, which can be found . He also has a . Robert also created a version of the plot above in SAS, which you can find , with code . Both SAS and R (not to mention a host of other environments) are sufficiently general and flexible that you can do whatever you want to do-- but varying amounts of expertise might be required. An unrelated note about aggregators We love aggregators! Aggregators collect blogs that have similar coverage for the convenience of readers, and for blog authors they offer a way to reach new audiences. SAS and R is aggregated by with our permission, and by at least 2 other aggregating services which have never contacted us. If you read this on an aggregator that does not credit the blogs it incorporates, please come visit us SAS and R . We answer comments there and offer direct subscriptions if you like our content. In addition, no one is allowed to profit by this work under our ; if you see advertisements on this page, the aggregator is violating the terms by which we publish our work.	{"url":"https://www.r-bloggers.com/2012/04/example-9-25-its-been-a-mighty-warm-winter-plot-on-a-circular-axis/","timestamp":"2024-11-09T10:12:33Z","content_type":"text/html","content_length":"154465","record_id":"<urn:uuid:cd02c88c-6f02-4949-bf41-1c242847225b>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00160.warc.gz"}
motion tracking request Ian Ross has just released a book on Anim8or. It's perect for a beginner and a good reference for experienced users. It contains detailed chapters on every aspect, with many examples. Get your own copy here: "Anim8or Tutorial Book" if you want something like they do for real movies anim8or is not capable yet, and i cannot find anything like voodoo for anim8or There isn't anything for that that I know of. However if voodoo or blender can output a motion file I might be able to read it into Anim8or. Are there any specs on what voodoo does? Now that I look at it closer, I doubt it will help. Oh well. I don't actually know anything about it, but it looks like you simply need to delete the "qkey" text. Then, download the Ic2An8 program here, this converts camera track data into Anim8or camera data, which you can then paste into the scene in Notepad.http://hanes.250free.com/Ic2An8.zipA small tutorial on how to add the data into Anim8or is included.	{"url":"https://www.anim8or.com/smf/index.php?topic=108.msg2209","timestamp":"2024-11-09T19:37:33Z","content_type":"application/xhtml+xml","content_length":"53721","record_id":"<urn:uuid:6becce15-28d1-4a0b-9eda-a61669c281c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00268.warc.gz"}
What is the value of 89 what is the value of 89 Related topics: formula for turning decimals to fractions accounting equation worksheet download quadratic equation examples Answers For Algebra Problems how are equations with one variable like eqations with two variables? abstract algebra john fraleigh ppt intermediate combination math samples simplifying specific radicals online polynomial function calculator rational expressions,2 t common multiple caculater Author Message Hetorax64 Posted: Saturday 02nd of Jun 09:34 Can anybody help me? I have a math test coming up next week and I am completely confused. I need help specially with some problems in what is the value of 89 that are quite confusing . I don’t wish to go to any tutorial and I would really appreciate any help in this area. Thanks! Back to top ameich Posted: Monday 04th of Jun 08:42 Hi friend , what is the value of 89 can be really difficult if your concepts are not clear. I know this software, Algebrator which has helped a lot of newbies clarify their concepts. I have used this software a couple of times when I was in college and I recommend it to every novice . From: Prague, Czech Back to top TihBoasten Posted: Wednesday 06th of Jun 08:07 Hello , Algebrator is one awesome tool ! I started using it when I was in my high school. It’s been years since then, but I still use it frequently . Mark my word for it, it will really help you. Back to top 3Di Posted: Wednesday 06th of Jun 20:41 Algebrator is a very easy to use product and is surely worth a try. You will also find lot of interesting stuff there. I use it as reference software for my math problems and can say that it has made learning math more enjoyable. From: 45°26' N, 09°10' E Back to top	{"url":"https://www.softmath.com/algebra-software/multiplying-fractions/what-is-the-value--of-89.html","timestamp":"2024-11-08T20:09:19Z","content_type":"text/html","content_length":"38909","record_id":"<urn:uuid:f495cb69-5830-4263-9703-243aa892f7e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00892.warc.gz"}
Excel Formula: MAX, INDIRECT, and MATCH Functions In this article, we will explain how to use the MAX, INDIRECT, and MATCH functions in Excel to find the maximum value in a range of cells in a different sheet. This formula is particularly useful when you need to analyze data from multiple sheets and find the highest value. By combining these functions, you can dynamically create a range reference based on the values in specific cells and retrieve the maximum value within that range. To understand this formula, let's break it down step by step. First, the INDIRECT function is used to create a cell reference by combining the sheet name, row number, and column range. The sheet name is obtained from cell I5. Next, the MATCH function is used to find the position of a value in cell M3 of the 'Complaints' sheet within column B of the 'Daily data' sheet. The 0 as the last argument indicates an exact match. Once the necessary references are obtained, the range for the MAX function is constructed by combining the sheet name, starting row number (7), and ending row number (determined by the MATCH function). The TRUE argument in the INDIRECT function is used to indicate that the range reference is in A1 notation. Finally, the MAX function is used to find the maximum value within the constructed range. It's important to note that the actual result of the formula will depend on the specific data in the referenced range. Therefore, the maximum value could be any number depending on the values in the range. To see this formula in action, let's consider a scenario where the value in cell I5 is 'Daily data' and the value in cell M3 of the 'Complaints' sheet is 10. In this case, the INDIRECT function will create the range reference 'Daily data'!D7:D10', and the MATCH function will return the value 4, indicating that the value in cell M3 is found in the 4th row of column B in the 'Daily data' sheet. The constructed range for the MAX function will be 'Daily data'!D7:D10, and the MAX function will find the maximum value within that range. In conclusion, the MAX, INDIRECT, and MATCH functions in Excel provide a powerful way to find the maximum value in a range of cells in a different sheet. By understanding how these functions work together, you can perform complex data analysis tasks and retrieve valuable insights from your Excel spreadsheets. Formula Explanation The given formula is an array formula that uses the MAX function in combination with other functions to find the maximum value in a range of cells in a different sheet. Here is a step-by-step explanation of the formula: 1. The INDIRECT function is used to create a cell reference by combining the sheet name, row number, and column range. The sheet name is obtained from cell I5. 2. The MATCH function is used to find the position of the value in cell M3 of the "Complaints" sheet within column B of the "Daily data" sheet. The 0 as the last argument indicates an exact match. 3. The range for the MAX function is constructed by combining the sheet name, starting row number (7), and ending row number (determined by the MATCH function). 4. The TRUE argument in the INDIRECT function is used to indicate that the range reference is in A1 notation. 5. The MAX function is used to find the maximum value within the constructed range. Let's assume the following scenario: • The value in cell I5 is "Daily data". • The value in cell M3 of the "Complaints" sheet is 10. Based on this scenario, the formula will be evaluated as follows: 1. The INDIRECT function will create the range reference "'Daily data'!D7:D10". 2. The MATCH function will return the value 4, indicating that the value in cell M3 is found in the 4th row of column B in the "Daily data" sheet. 3. The constructed range for the MAX function will be "'Daily data'!D7:D10". 4. The MAX function will find the maximum value within the range, which could be any number depending on the actual values in the range. Please note that the actual result of the formula will depend on the specific data in the referenced range.	{"url":"https://codepal.ai/excel-formula-explainer/query/Kt6t2y3y/excel-formula-max-indirect","timestamp":"2024-11-08T02:32:57Z","content_type":"text/html","content_length":"94549","record_id":"<urn:uuid:772f5a92-a565-4cc4-b827-d4cf99622a1c>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00009.warc.gz"}
Course Info for Math 459 Course: Math 459 Instructor: Prof. Joe Borzellino Time: TR 4:10 6:00pm, Mathematics & Science 38-201 Office Hours: MF 10:00am-11:00am TR 3:00pm 4:00pm, (or by appt) in FOE 25-302 Phone: 756 5192 E Mail: jborzell Web Page: http://www.calpoly.edu/~jborzell/index.html Course Text: Keith Devlin, Mathematics: the new golden age Alternative Sources: • Keith Devlin, The Millennium Problems: The Seven Greatest Unsolved Mathematical Puzzles of Our Time • E.T. Bell, Men of Mathematics • The American Mathematical Monthly (QA 1 A515 - old issues available online through PolyCat) • College Mathematics Journal (QA11 .A1 T9) • Your senior project topic • Anything else that is appropriate (consult with me first) Course Description: Simply speaking, this is a seminar course in which you will be an active participant. Each of you will be required to make two 2-hour presentations to the class. Presentation One will be a lecture on (a) typical topic(s) that you find interesting from any 300/400 level undergraduate mathematics course. Your goal is to pretend that you are presenting this topic to an undergraduate audience for the first time as an instructor. Since this is a 2-hour presentation, you may need to present two separate topics instead of one. Since a rule of thumb is that "you don't know it, 'til you teach it", you may want to consider something that was confusing to you the first time you learned it. More ambitious/difficult topics will reflect positively on your presentation grade, and you should clear your topic with me first. Presentation Two will be based on an investigation into a mathematical topic of your choice. You must get approval for the topic from me before you begin your investigation. This presentation should contain all relevant background information, important theorems and their proofs (when appropriate). You are giving this presentation to you classmates and thus it should be understandable to them. Ideas for topics can come from any of the resources listed above. If you use an article from the Monthly or CMJ as the basis for your presentation, please make copies to hand out to the class before your presentation. Important Before giving a presentation, you must schedule an appointment with me to go over the details of what you are going to present. This should be scheduled enough in advance so that if our discussion warrants changes to your presentation, those changes can be incorporated. If you do not meet with me before your presentation you will receive a failing grade for the course. Grading: The quality of your presentations will be the primary factor in determining your grade on the presentations. Presentations should be well-prepared both in content and delivery. Since this is a seminar course, attendance will also be considered. A good seminar is one that is well attended. Thus, if you miss three or more classes you will receive no credit for attendance; otherwise, you will receive full credit. Your grade for the course will be determined as follows: Presentation One 40% Presentation Two 40% Attendance 20% Other Dates to Remember: January 17 MLK B-Day (Monday) February 21 GW B-Day (Monday) March 11 Last Day of Classes (Friday)	{"url":"http://orbifolds.com/Courses/Year%2004-05/Winter%202005/Math459.html","timestamp":"2024-11-07T07:45:22Z","content_type":"text/html","content_length":"11755","record_id":"<urn:uuid:6c105ac0-5a02-4628-a42f-eb751ada24ea>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00822.warc.gz"}
Sampling Method - Dr Loo (Principal Math Tutor) When attending statistics class, students are often need to learn to design and conduct a survey on a specified group of respondents. When the respondents census consist of too many people, it is often wiser to conduct the survey on a sample. Below are some of the sampling method one can use. Different types of sampling method Simple random sampling In this method, all possible sample size of n are equally likely to be selected. Example of simple random sampling method. Systematic sampling To use this method, one has to first arrange the population in certain order (eg. alphabetically), then choose every kth member from the list after each sampling point. Example of how systematic sampling method is performed If systematic sampling method is to be apply to choose a sample of 100 samples out of the population size of 5000. One has first to determine an appropriate interval by taking 5000 divide by 100 which is equal to 50. Using a random starting point (eg. 15), selected the 15th entry of the population as the first sample, then 65, 115, etc. More about systematic random sampling method. Stratified sampling This method will first create sub-groups population based on certain characteristics (eg. occupation), then one can perform random sampling to select approriate size of sample from each stratum. Cluster sampling Population are allocated into sub-groups called cluster. About the Statistics Tutor - Dr Loo I am a PhD holder with 8 years of teaching experience at secondary school and university. My expertizes are mathematics, statistics, econometrics, finance and machine learning. Currently I do provide consultation and private tuition in Singapore. For those who are looking for mathematics tuition teacher in Singapore or statistics tuition teacher in Singapore, please feel free to contact me at +65-85483705 (SMS/Whatsapp/Telegram). Some of the math tuition in Singapore I provide includes Singapore secondary school math tuition and also mathematics tuition class for Singapore A-level (JC H1 Math Tuition Class & JC H2 Math Tuition). On the other hand, my statistics tuition in Singapore mostly focus on pre-university level until postgraduate level.	{"url":"https://math-tutor-singapore.com/sampling-method","timestamp":"2024-11-02T14:26:12Z","content_type":"text/html","content_length":"133743","record_id":"<urn:uuid:4e241890-c375-49e4-a0ba-8474c3709cfd>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00711.warc.gz"}
cgehrd: reduces a complex general matrix A to upper Hessenberg form H by an unitary similarity transformation - Linux Manuals (l) cgehrd (l) - Linux Manuals cgehrd: reduces a complex general matrix A to upper Hessenberg form H by an unitary similarity transformation CGEHRD - reduces a complex general matrix A to upper Hessenberg form H by an unitary similarity transformation N, ILO, IHI, A, LDA, TAU, WORK, LWORK, INFO ) INTEGER IHI, ILO, INFO, LDA, LWORK, N COMPLEX A( LDA, * ), TAU( * ), WORK( * ) CGEHRD reduces a complex general matrix A to upper Hessenberg form H by an unitary similarity transformation: Qaq * A * Q = H . N (input) INTEGER The order of the matrix A. N >= 0. ILO (input) INTEGER IHI (input) INTEGER It is assumed that A is already upper triangular in rows and columns 1:ILO-1 and IHI+1:N. ILO and IHI are normally set by a previous call to CGEBAL; otherwise they should be set to 1 and N respectively. See Further Details. A (input/output) COMPLEX array, dimension (LDA,N) On entry, the N-by-N general matrix to be reduced. On exit, the upper triangle and the first subdiagonal of A are overwritten with the upper Hessenberg matrix H, and the elements below the first subdiagonal, with the array TAU, represent the unitary matrix Q as a product of elementary reflectors. See Further Details. LDA (input) INTEGER The leading dimension of the array A. LDA >= max TAU (output) COMPLEX array, dimension (N-1) The scalar factors of the elementary reflectors (see Further Details). Elements 1:ILO-1 and IHI:N-1 of TAU are set to zero. WORK (workspace/output) COMPLEX array, dimension (LWORK) On exit, if INFO = 0, WORK(1) returns the optimal LWORK. LWORK (input) INTEGER The length of the array WORK. LWORK >= max(1,N). For optimum performance LWORK >= NNB, where NB is the optimal blocksize. If LWORK = -1, then a workspace query is assumed; the routine only calculates the optimal size of the WORK array, returns this value as the first entry of the WORK array, and no error message related to LWORK is issued by XERBLA. INFO (output) INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value. The matrix Q is represented as a product of (ihi-ilo) elementary reflectors = H(ilo) H(ilo+1) . . . H(ihi-1). Each H(i) has the form H(i) = I - tau v * vaq where tau is a complex scalar, and v is a complex vector with v(1:i) = 0, v(i+1) = 1 and v(ihi+1:n) = 0; v(i+2:ihi) is stored on exit in A(i+2:ihi,i), and tau in TAU(i). The contents of A are illustrated by the following example, with n = 7, ilo = 2 and ihi = 6: on entry, on exit, ( a a a a a a a ) ( a a h h h h a ) ( a a a a a a ) ( a h h h h a ) ( a a a a a a ) ( h h h h h h ) ( a a a a a a ) ( v2 h h h h h ) ( a a a a a a ) ( v2 v3 h h h h ) ( a a a a a a ) ( v2 v3 v4 h h h ) ( a ) ( a ) where a denotes an element of the original matrix A, h denotes a modified element of the upper Hessenberg matrix H, and vi denotes an element of the vector defining H(i). This file is a slight modification of LAPACK-3.0aqs CGEHRD subroutine incorporating improvements proposed by Quintana-Orti and Van de Geijn (2005).	{"url":"https://www.systutorials.com/docs/linux/man/l-cgehrd/","timestamp":"2024-11-07T08:49:35Z","content_type":"text/html","content_length":"10919","record_id":"<urn:uuid:9a3d9ca2-4d33-4e73-98cc-a503ecb8b559>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00766.warc.gz"}
Inverse fast Fourier transform X = ifft(Y) computes the inverse discrete Fourier transform of Y using a fast Fourier transform algorithm. X is the same size as Y. • If Y is a vector, then ifft(Y) returns the inverse transform of the vector. • If Y is a matrix, then ifft(Y) returns the inverse transform of each column of the matrix. • If Y is a multidimensional array, then ifft(Y) treats the values along the first dimension whose size does not equal 1 as vectors and returns the inverse transform of each vector. X = ifft(Y,n) returns the n-point inverse Fourier transform of Y by padding Y with trailing zeros to length n. X = ifft(Y,n,dim) returns the inverse Fourier transform along the dimension dim. For example, if Y is a matrix, then ifft(Y,n,2) returns the n-point inverse transform of each row. X = ifft(___,symflag) specifies the symmetry of Y in addition to any of the input argument combinations in previous syntaxes. For example, ifft(Y,'symmetric') treats Y as conjugate symmetric. Inverse Transform of Vector The Fourier transform and its inverse convert between data sampled in time and space and data sampled in frequency. Create a vector and compute its Fourier transform. X = [1 2 3 4 5]; Y = fft(X) Y = 1×5 complex 15.0000 + 0.0000i -2.5000 + 3.4410i -2.5000 + 0.8123i -2.5000 - 0.8123i -2.5000 - 3.4410i Compute the inverse transform of Y, which is the same as the original vector X. Inverse Transform of Single-Sided Spectrum Find the inverse Fourier transform of the single-sided spectrum that is the Fourier transform of a real signal. Load the single-sided spectrum in the frequency domain. Show the sampling frequency and the sampling period of this spectrum. Plot the complex magnitude of the single-sided spectrum. L1 = length(Y1); f = Fs/(2L1-1)(0:L1-1); xlabel("f (Hz)") The discrete Fourier transform of a time-domain signal has a periodic nature, where the first half of its spectrum is in positive frequencies and the second half is in negative frequencies, with the first element reserved for the zero frequency. For real signals, the discrete Fourier transform in the frequency domain is a two-sided spectrum, where the spectrum in the positive frequencies is the complex conjugate of the spectrum in the negative frequencies with half the peak amplitudes of the real signal in the time domain. To find the inverse Fourier transform of a single-sided spectrum, convert the single-sided spectrum to a two-sided spectrum. Y2 = [Y1(1) Y1(2:end)/2 fliplr(conj(Y1(2:end)))/2]; Find the inverse Fourier transform of the two-sided spectrum to recover the real signal in the time domain. Plot the signal. t = (0:length(X)-1)T; xlabel("t (seconds)") Padded Inverse Transform of Matrix The ifft function allows you to control the size of the transform. Create a random 3-by-5 matrix and compute the 8-point inverse Fourier transform of each row. Each row of the result has length 8. Y = rand(3,5); n = 8; X = ifft(Y,n,2); Conjugate Symmetric Vector For nearly conjugate symmetric vectors, you can compute the inverse Fourier transform faster by specifying the 'symmetric' option, which also ensures that the output is real. Nearly conjugate symmetric data can arise when computations introduce round-off error. Create a vector Y that is nearly conjugate symmetric and compute its inverse Fourier transform. Then, compute the inverse transform specifying the 'symmetric' option, which eliminates the nearly 0 imaginary parts. Y = [1 2:4+eps(4) 4:-1:2] X = 1×7 2.7143 -0.7213 -0.0440 -0.0919 -0.0919 -0.0440 -0.7213 Xsym = ifft(Y,'symmetric') Xsym = 1×7 2.7143 -0.7213 -0.0440 -0.0919 -0.0919 -0.0440 -0.7213 Input Arguments Y — Input array vector \| matrix \| multidimensional array Input array, specified as a vector, a matrix, or a multidimensional array. If Y is of type single, then ifft natively computes in single precision, and X is also of type single. Otherwise, X is returned as type double. Data Types: double \| single \| int8 \| int16 \| int32 \| uint8 \| uint16 \| uint32 \| logical Complex Number Support: Yes n — Inverse transform length [] (default) \| nonnegative integer scalar Inverse transform length, specified as [] or a nonnegative integer scalar. Padding Y with zeros by specifying a transform length larger than the length of Y can improve the performance of ifft. The length is typically specified as a power of 2 or a product of small prime numbers. If n is less than the length of the signal, then ifft ignores the remaining signal values past the nth entry and returns the truncated result. If n is 0, then ifft returns an empty matrix. Data Types: double \| single \| int8 \| int16 \| int32 \| uint8 \| uint16 \| uint32 \| logical dim — Dimension to operate along positive integer scalar Dimension to operate along, specified as a positive integer scalar. By default, dim is the first array dimension whose size does not equal 1. For example, consider a matrix Y. • ifft(Y,[],1) returns the inverse Fourier transform of each column. • ifft(Y,[],2) returns the inverse Fourier transform of each row. Data Types: double \| single \| int8 \| int16 \| int32 \| uint8 \| uint16 \| uint32 \| logical symflag — Symmetry type 'nonsymmetric' (default) \| 'symmetric' Symmetry type, specified as 'nonsymmetric' or 'symmetric'. When Y is not exactly conjugate symmetric due to round-off error, ifft(Y,'symmetric') treats Y as if it were conjugate symmetric by ignoring the second half of its elements (that are in the negative frequency spectrum). For more information on conjugate symmetry, see Algorithms. More About Discrete Fourier Transform of Vector Y = fft(X) and X = ifft(Y) implement the Fourier transform and inverse Fourier transform, respectively. For X and Y of length n, these transforms are defined as follows: $\begin{array}{l}Y\left(k\right)=\sum _{j=1}^{n}X\left(j\right)\text{\hspace{0.17em}}{W}_{n}^{\left(j-1\right)\text{}\left(k-1\right)}\\ X\left(j\right)=\frac{1}{n}\sum _{k=1}^{n}Y\left(k\right)\ ${W}_{n}={e}^{\left(-2\pi i\right)/n}$ is one of n roots of unity. • The ifft function tests whether the vectors in Y are conjugate symmetric. If the vectors in Y are conjugate symmetric, then the inverse transform computation is faster and the output is real. A function $g\left(a\right)$ is conjugate symmetric if $g\left(a\right)={g}^{}\left(-a\right)$. However, the fast Fourier transform of a time-domain signal has one half of its spectrum in positive frequencies and the other half in negative frequencies, with the first element reserved for the zero frequency. For this reason, a vector v is conjugate symmetric when v(2:end) is equal to conj(v(end:-1:2)). Extended Capabilities C/C++ Code Generation Generate C and C++ code using MATLAB® Coder™. Usage notes and limitations: • Unless symflag is 'symmetric', the output is always complex even if all imaginary parts are zero. • Code generation uses the symmetric algorithm only when the option symmetric is specified. • For limitations related to variable-size data, see Variable-Sizing Restrictions for Code Generation of Toolbox Functions (MATLAB Coder). • For MEX output, MATLAB^® Coder™ uses the library that MATLAB uses for FFT algorithms. For standalone C/C++ code, by default, the code generator produces code for FFT algorithms instead of producing FFT library calls. To generate calls to a specific installed FFTW library, provide an FFT library callback class. For more information about an FFT library callback class, see coder.fftw.StandaloneFFTW3Interface (MATLAB Coder). • For simulation of a MATLAB Function block, the simulation software uses the library that MATLAB uses for FFT algorithms. For C/C++ code generation, by default, the code generator produces code for FFT algorithms instead of producing FFT library calls. To generate calls to a specific installed FFTW library, provide an FFT library callback class. For more information about an FFT library callback class, see coder.fftw.StandaloneFFTW3Interface (MATLAB Coder). • Using the Code Replacement Library (CRL), you can generate optimized code that runs on ARM^® Cortex^®-A processors with Neon extension. To generate this optimized code, you must install the Embedded Coder^® Support Package for ARM Cortex-A Processors (Embedded Coder). The generated code for ARM Cortex-A uses the Ne10 library. For more information, see Ne10 Conditions for MATLAB Functions to Support ARM Cortex-A Processors (Embedded Coder). • Using the Code Replacement Library (CRL), you can generate optimized code that runs on ARM Cortex-M processors. To generate this optimized code, you must install the Embedded Coder Support Package for ARM Cortex-M Processors (Embedded Coder). The generated code for ARM Cortex-M uses the CMSIS library. For more information, see CMSIS Conditions for MATLAB Functions to Support ARM Cortex-M Processors (Embedded Coder). GPU Code Generation Generate CUDA® code for NVIDIA® GPUs using GPU Coder™. Usage notes and limitations: Thread-Based Environment Run code in the background using MATLAB® backgroundPool or accelerate code with Parallel Computing Toolbox™ ThreadPool. This function fully supports thread-based environments. For more information, see Run MATLAB Functions in Thread-Based Environment. GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. The ifft function supports GPU array input with these usage notes and limitations: • Unless symflag is 'symmetric', the output is always complex even if all imaginary parts are zero. For more information, see Run MATLAB Functions on a GPU (Parallel Computing Toolbox). Distributed Arrays Partition large arrays across the combined memory of your cluster using Parallel Computing Toolbox™. This function fully supports distributed arrays. For more information, see Run MATLAB Functions with Distributed Arrays (Parallel Computing Toolbox). Version History Introduced before R2006a	{"url":"https://it.mathworks.com/help/matlab/ref/ifft.html","timestamp":"2024-11-13T23:16:36Z","content_type":"text/html","content_length":"116377","record_id":"<urn:uuid:74ea4b14-253f-4908-9404-90a9c90df7dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00405.warc.gz"}
Rosenberg lab - about the image ABOUT THIS IMAGE. This image illustrates the strict upper bound that exists on the value of F[ST] at a locus given the frequency of the most frequent allele. The y-axis represents F[ST], and the x-axis is the frequency M of the most frequent allele. Both axes range from 0 to 1. The gray region represents the set of allowable values of F[ST] given M. The colored rectangles represent counts of the number of microsatellite loci that have values of (M, F[ST]) in particular 0.01 x 0.01 regions, for data on 101 Africans and 63 Native Americans. Darker points represent larger numbers of loci. The figure is based on the work of Jakobsson, Edge &amp Rosenberg (The relationship between F[ST] and the frequency of the most frequent allele; Genetics 193: 515-528, 2013). [Image gallery 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 ]	{"url":"https://rosenberglab.stanford.edu/FstImage.html","timestamp":"2024-11-06T21:27:18Z","content_type":"text/html","content_length":"7361","record_id":"<urn:uuid:1025341c-df3d-4061-b9dd-0e9f85704a6b>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00855.warc.gz"}
Hunting for Turing Machines at the Wolfram Science Summer School—Wolfram Blog Hunting for Turing Machines at the Wolfram Science Summer School This year is the 100th birthday of Alan Turing, so at the 2012 Wolfram Science Summer School we decided to turn a group of 40 unassuming nerds into ferocious hunters. No, we didn’t teach our geeks to take down big game. These are laptop warriors. And their prey? Turing machines! In this blog post, I’m going to teach you to be a fellow hunter-gatherer in the computational universe. Your mission, should you choose to accept it, is to FIND YOUR FAVORITE TURING MACHINE. First, I’ll show you how a Turing machine works, using pretty pictures that even my grandmother could understand. Then I’ll show you some of the awesome Turing machines that our summer school students found using Mathematica. And I’ll describe how I did an über-search through 373 million Turing machines using my Linux server back home, and had it send me email whenever it found an interesting one, for two weeks straight. I’ll keep the code to a minimum here, but you can find it all in the attached Mathematica notebook. Excited? Primed for the hunt? Let me break it down for you. The rules of Turing machines are actually super simple. There’s a row of cells called the tape: Then there’s a head that sits on the tape like so: The head can point in different directions to show what state it’s in. Let’s say there are just three possible states, so the head can point in three different directions: In addition, each cell on the tape has a color. In the simplest case there are just two possible colors, black or white (binary): So for each of the three head states, there are two possible cell colors, giving six different situations: The rule table tells the Turing machine what to do in each situation: You can see how it works: the cell under the head changes color, then the head moves left or right and updates its state. So let’s say the head is in state 1 on a blank tape: We look up the case in the rule table where the head is on a white cell in state 1: The rule says the cell under the head changes to black; the head moves left, and goes into state 3: Notice we are showing the updated tape right below the old one. That way we can see where the head was on the previous step and where it is currently. Now the head is in state 3 on a white cell, so we look that case up: The rule says the cell changes to black; the head moves right, and goes into state 2: Now the head is in state 2 and this time it’s on a black cell: The rule says the cell stays black; the head moves right, and goes into state 3: Getting the hang of it? Pretty simple, right? Here’s what the Turing machine evolution looks like after 20 steps: You can use this handy little CDF to follow along. It shows the evolution step by step, indicating which case in the rule table is used each time: Now let’s use the built-in TuringMachine function to run this Turing machine rule. Here are the rules, in the form {state, cell color} -> {new state, new color, head movement}: We just plug in the rules with the initial condition—a blank tape with the head in state 1 (note that states 1, 2, and 3 are denoted 0, 1, and 2)—and the number of time steps to run for: Here’s what the tape looks like after 200 steps: See the pattern? The head zigzags back and forth, and as soon as it reaches further left than ever before, it zooms to the right and repeats the process. Another way to view the evolution is to look at the position of the head over time: The pattern is clearer if we run it for more steps: The overall growth goes in sections of zigzags, where each section takes longer than the last. When the head reaches the left side, it zooms to the right and starts a new section. Now let’s try compressing it to show just the steps where the head goes further to the right than ever before: The compressed tape is easier to visualize. The head in the compressed evolution is just steadily moving to the right: Let’s try changing the rule slightly to see if we can make it do something more interesting. Here’s the rule table again: Now we are going to change the rule so when the head is in state 1 on a white cell, it goes to the right rather than the left: Here’s what this new rule does: Boring! The head always moves right and turns the cell black, so it just creates a growing black triangle (at least for this initial condition). So now you might be wondering, what happens when the head goes all the way to the right edge of the tape? Does it stop or keep going? Well, it kind of depends on who you ask. At the summer school, we use the convention that the head just wraps around and keeps going from the left: We call this a “periodic boundary”. The other convention you could choose is called the “halting” condition, where the head just stops at the right edge of the tape and the game is over. Side note: this seemingly trivial detail has actually gotten a LOT of attention. It turns out that there are Turing machines (like the one Turing constructed in his famous 1936 paper) where it is impossible to predict (“undecidable”) whether or not the head will reach the right edge and halt. This is called the “halting problem”. And there are even so-called “busy beaver” competitions where people try to find Turing machines that will run the longest before eventually halting. So is it even possible to get something more interesting by changing one of the cases in the rule table? How about a pattern that grows more slowly? Or one that never repeats? Well, it’s kind of hard to tell just by thinking about it, isn’t it? I mean, you have to basically run the rule in your head to figure it out. This happens so often with Turing machines that they gave it a name: computational irreducibility. But that doesn’t mean we’re stuck, because we have Mathematica! Let’s just enumerate all the possible point changes to the rule and show them all in a table like so (see the attached notebook for the Each one of these is a different Turing machine with a rule table that differs by just a single change from the original rule (there are 24 similar rules in all, in a 6×4 grid). While there is quite a variety, they are all pretty predictable. Nothing earth-shattering. But these 24 rules are just a small sample of all the possible Turing machine rules with three states and two colors. Just think about it: for each of the six cases in the rule table, there is a choice of what the new color should be, which state the head will be in, and which direction it will go. If you do the math, it turns out that there are over 2 million of them: And if you think that’s bad, the faculty at the Wolfram Science Summer School decided to be really diabolical and make the students choose their favorite 4-state, 4-color Turing machine. And just how many rules is that to choose from?: Needless to say, an exhaustive search was pretty much out of the question. :) So the students had to figure out some tricks for searching for Turing machines with the kinds of features they were interested in. And with so many rules to sort through, it was definitely a needle-in-a-haystack situation. But the students were up to the challenge, and they got really creative. They ran searches using Mathematica looking for things like: the slowest possible growth of the tape: patterns on the tape with high entropy: patterns too complex to be digitally compressed: and rules that sound interesting when their features are sonified (check out WolframTones): These are just a few of the 39 student submissions. And here are some of the compressed outputs from the student submissions: I decided I wanted to do a search too. I wrote a Mathematica script to do an automated search for Turing machines that met certain criteria: The script chooses random rule numbers to run and test, and then writes the rule number and some statistics to a file called tm-4-4-search.txt for each Turing machine rule that passes through the various filters. The filters are run in stages, with faster filters running before the slower ones, to minimize the time spent running the slower filters on rules that don’t have a chance of making it through anyway. The final filter measures the complexity of the compressed tape using this block-entropy measure (which was thought up by one our fine high school students this year): A Turing machine with very random-looking compressed output would have an entropy closer to 1, and one that has a lot of structure or organization to it would have an entropy closer to 0. If a rule makes it through this final entropy filter, the script outputs a little summary image of the rule as a .png file. I made sure the summary image was small enough for me to easily glance at in Mail.app on my iPhone 4. Here’s an example: The compressed tape is shown at the top, followed by a few statistics (more on those in a second), and then a ListPlot of the head movement for the first 5,000 time steps. I ran the Mathematica script on my Linux server back at the Wolfram headquarters in Champaign, Illinois. To alert me when interesting rules were found, I wrote a simple daemon shell script to poll the log file for new results and email me whenever an interesting new rule was found: The search ran for 16 days, searching a total of 373 million Turing machines, at a rate of about 270 Turing machines per second. It ended up finding about 13,000 interesting rules. On average, the search script emailed me an interesting rule about once every two minutes or so. That was way too many emails for me to look through exhaustively. So after the summer school was over, I took all the winning rule numbers and jammed their compressed outputs into little 5×10 grids like so: This way I could look at grids of 50 rules at a time, and the total result set of 13,188 rules only occupied 264 grids. I could get a pretty good feeling for the rules in one grid in about 10 seconds, so the entire process of looking through all 13,000 rules only took about 45 minutes. After all that, I chose my favorite 4-state 4-color Turing machine, rule number 348,371,888,147,308,385,687,045. Here’s what the compressed outputs look like, compressing by keeping just the steps where the head had moved further left than ever before: And here’s what the tape output looks like when compressing by keeping just the steps where the head has moved further right than ever before: This rule popped up in my search because it had irregular head movement: I quantified this by measuring the number of distinct distances the head traveled continuously in one direction before turning around and going the other direction (which I call a “run length”): But in this case there was also a rather large number of run length frequencies, meaning the number of run lengths that appeared once, twice, three times, etc: A strictly nested pattern will often have a lot of different run lengths but a small number of run length frequencies. For example, the head might just be bouncing from left to right on the tape, going a little farther each time. In that case, there would be a slightly larger run length every time it changed direction. But there would only be one run length frequency, equal to 1, because every run length would occur exactly once. This rule, however, has four different run length frequencies, which means the head probably wasn’t just moving in a steadily growing pattern, but was zigzagging around in an irregular way. But despite having fairly complex head movement, there’s obviously quite a lot of structure to the compressed tape output—it certainly doesn’t look random. This is the distribution of entropies in the 13,000 Turing machines returned by my über-search: My favorite rule has a relatively low entropy of 0.12 (on my normalized scale), well below the average (which is about 0.24), indicating that it is highly organized: And you can see that structure clearly in the compressed output, with the three distinct domains of behavior you see scanning the tape from left to right: It would be interesting to study these three domains of behavior in detail, and find out what this Turing machine is ultimately computing. Could we program it to behave like a fluctuating molecule, or a flock of birds, or a human brain? Stephen Wolfram’s Principle of Computational Equivalence says that a Turing machine with such complex-looking output is probably capable of universal computation, meaning we could use it to compute anything we want to with a suitable encoding of the initial conditions. I’ve shown you how awesome Turing machines are, and how much fun it is to search for them in Mathematica. I think Turing would have been proud to see all the cool Turing machines unearthed by our students this year. If you want to take your hunting skills to the next level, you can apply to Wolfram Science Summer School to try your hand at exploring your favorite part of the computational universe. Here’s the hunting party from last summer: See you at the Wolfram Science Summer School in 2013! (No Turing machines were harmed in the making of this blog post.) Download this ZIP file that includes the blog post as a CDF and all code used in the post. Join the discussion 3 comments 1. That’s very cool; I’ll have to play around with this myself sometime. I’d love to see an extension into two dimensions (“turmites”) – I know the maths is pretty much the same but it might be more obvious to the human eye that something “interesting” is happening when the turing machine output is an image rather than a line. 2. Hi Ruben The colors in each row not affected by the currently applicable rule are inherited, i.e., copied down, from the previous row (set of states). There is only one tape; the display is meant to show how the tape evolves, step by step on each successive row. Cheers.	{"url":"https://blog.wolfram.com/2012/12/20/hunting-for-turing-machines-at-the-wolfram-science-summer-school/","timestamp":"2024-11-07T06:18:45Z","content_type":"text/html","content_length":"213962","record_id":"<urn:uuid:8fa8a3d0-bd0a-44c1-bb9a-63faf2d661f2>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00275.warc.gz"}
seminars - Five lectures on Geometric Invariant Theory (4) Speaker: Victoria Hoskins (Freie Universität Berlin) Time: 12/7~11, 1600~1700 1. (Location: 129-301) Instability stratifications in GIT: quick overview of GIT and Hilbert-Mumford criterion, Kempf's theorem on adapted 1-PSs, the construction of the stratification. 2. (Location: 129-104) Symplectic quotients and their relation to GIT: moment maps and symplectic reduction, the Kempf-Ness Theorem. 3. (Location: 129-301) The Morse stratification is the GIT instability stratification: the norm square of the moment map as a Morse function, a description of the associated Morse srata, and comparison with GIT stratification. 4. (Location: 129-301) Stratifications for quiver representations: King's GIT construction of moduli spaces of quivers and the Harder-Narasimhan stratification, and the result that all three stratifications agree. 5. (Location: 27-325) Stratifications for sheaves: Simpson's GIT construction of moduli spaces of sheaves and the Harder-Narasimhan stratification, then comparison results.	{"url":"https://www.math.snu.ac.kr/board/index.php?mid=seminars&page=54&l=en&sort_index=Time&order_type=asc&document_srl=761793","timestamp":"2024-11-09T20:21:52Z","content_type":"text/html","content_length":"48415","record_id":"<urn:uuid:07a6a80f-c864-4883-91b0-d7faabafa9b6>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00832.warc.gz"}
CAOSP abstracts, Volume: 41, No.: 1, year: 2011 • Author(s): Neslušan, L. • Journal: Contributions of the Astronomical Observatory Skalnaté Pleso, vol. 41, no. 1, p. 35-38. • Date: 5/2011 • Title: A note on the impulse addition of two colliding spherical objects • Keyword(s): planet formation, celestial mechanics • Pages: 35 -- 38 Abstract: In collisions between macroscopic spherical-shape objects, these objects are usually regarded as dimensionless, point-like particles. This assumption causes an overestimate of the orbital velocity of mergers, because real bodies have finite dimensions and a part of their orbital impulse is converted to the rotation of the merger. We give a statistical estimate of this impulse and, thus, the merger's velocity vector, which is a better approximation of reality. The provided formula is simple and, therefore, suitable to be used in a robust simulation of, e.g., the planet-formation process. Full text version of this article in PostScript (600dpi) format compressed by gzip; or in PDF.	{"url":"https://www.astro.sk/caosp/Eedition/Abstracts/2011/Vol_41/No_1/pp35-38_abstract.html","timestamp":"2024-11-15T02:50:06Z","content_type":"application/xhtml+xml","content_length":"3541","record_id":"<urn:uuid:4441974b-eab4-4be4-bda4-ad6496c90eaa>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00659.warc.gz"}
aive Bayes classification model for incremental learning Train naive Bayes classification model for incremental learning Since R2021a The fit function fits a configured naive Bayes classification model for incremental learning (incrementalClassificationNaiveBayes object) to streaming data. To additionally track performance metrics using the data as it arrives, use updateMetricsAndFit instead. To fit or cross-validate a naive Bayes classification model to an entire batch of data at once, see fitcnb. Mdl = fit(Mdl,X,Y) returns a naive Bayes classification model for incremental learning Mdl, which represents the input naive Bayes classification model for incremental learning Mdl trained using the predictor and response data, X and Y respectively. Specifically, fit updates the conditional posterior distribution of the predictor variables given the data. Mdl = fit(Mdl,X,Y,'Weights',Weights) also sets observation weights Weights. Incrementally Train Model with Little Prior Information Fit an incremental naive Bayes learner when you know only the expected maximum number of classes in the data. Create an incremental naive Bayes model. Specify that the maximum number of expected classes is 5. Mdl = incrementalClassificationNaiveBayes('MaxNumClasses',5) Mdl = IsWarm: 0 Metrics: [1x2 table] ClassNames: [1x0 double] ScoreTransform: 'none' DistributionNames: 'normal' DistributionParameters: {} Mdl is an incrementalClassificationNaiveBayes model. All its properties are read-only. Mdl can process at most 5 unique classes. By default, the prior class distribution Mdl.Prior is empirical, which means the software updates the prior distribution as it encounters labels. Mdl must be fit to data before you can use it to perform any other operations. Load the human activity data set. Randomly shuffle the data. load humanactivity n = numel(actid); rng(1) % For reproducibility idx = randsample(n,n); X = feat(idx,:); Y = actid(idx); For details on the data set, enter Description at the command line. Fit the incremental model to the training data, in chunks of 50 observations at a time, by using the fit function. At each iteration: • Simulate a data stream by processing 50 observations. • Overwrite the previous incremental model with a new one fitted to the incoming observations. • Store the mean of the first predictor in the first class ${\mu }_{11}$ and the prior probability that the subject is moving (Y > 2) to see how these parameters evolve during incremental learning. % Preallocation numObsPerChunk = 50; nchunk = floor(n/numObsPerChunk); mu11 = zeros(nchunk,1); priormoved = zeros(nchunk,1); % Incremental fitting for j = 1:nchunk ibegin = min(n,numObsPerChunk(j-1) + 1); iend = min(n,numObsPerChunkj); idx = ibegin:iend; Mdl = fit(Mdl,X(idx,:),Y(idx)); mu11(j) = Mdl.DistributionParameters{1,1}(1); priormoved(j) = sum(Mdl.Prior(Mdl.ClassNames > 2)); Mdl is an incrementalClassificationNaiveBayes model object trained on all the data in the stream. To see how the parameters evolve during incremental learning, plot them on separate tiles. t = tiledlayout(2,1); axis tight ylabel('\pi(Subject Is Moving)') axis tight fit updates the posterior mean of the predictor distribution as it processes each chunk. Because the prior class distribution is empirical, $\pi$(subject is moving) changes as fit processes each Specify All Class Names Before Fitting Fit an incremental naive Bayes learner when you know all the class names in the data. Consider training a device to predict whether a subject is sitting, standing, walking, running, or dancing based on biometric data measured on the subject. The class names map 1 through 5 to an activity. Also, suppose that the researchers plan to expose the device to each class uniformly. Create an incremental naive Bayes model for multiclass learning. Specify the class names and the uniform prior class distribution. classnames = 1:5; Mdl = incrementalClassificationNaiveBayes('ClassNames',classnames,'Prior','uniform') Mdl = IsWarm: 0 Metrics: [1x2 table] ClassNames: [1 2 3 4 5] ScoreTransform: 'none' DistributionNames: 'normal' DistributionParameters: {5x0 cell} Mdl is an incrementalClassificationNaiveBayes model object. All its properties are read-only. During training, observed labels must be in Mdl.ClassNames. Mdl must be fit to data before you can use it to perform any other operations. Load the human activity data set. Randomly shuffle the data. load humanactivity n = numel(actid); rng(1); % For reproducibility idx = randsample(n,n); X = feat(idx,:); Y = actid(idx); For details on the data set, enter Description at the command line. Fit the incremental model to the training data by using the fit function. Simulate a data stream by processing chunks of 50 observations at a time. At each iteration: • Process 50 observations. • Overwrite the previous incremental model with a new one fitted to the incoming observations. • Store the mean of the first predictor in the first class ${\mu }_{11}$ and the prior probability that the subject is moving (Y > 2) to see how these parameters evolve during incremental learning. % Preallocation numObsPerChunk = 50; nchunk = floor(n/numObsPerChunk); mu11 = zeros(nchunk,1); priormoved = zeros(nchunk,1); % Incremental fitting for j = 1:nchunk ibegin = min(n,numObsPerChunk(j-1) + 1); iend = min(n,numObsPerChunkj); idx = ibegin:iend; Mdl = fit(Mdl,X(idx,:),Y(idx)); mu11(j) = Mdl.DistributionParameters{1,1}(1); priormoved(j) = sum(Mdl.Prior(Mdl.ClassNames > 2)); Mdl is an incrementalClassificationNaiveBayes model object trained on all the data in the stream. To see how the parameters evolve during incremental learning, plot them on separate tiles. t = tiledlayout(2,1); axis tight ylabel('\pi(Subject Is Moving)') axis tight fit updates the posterior mean of the predictor distribution as it processes each chunk. Because the prior class distribution is specified as uniform, $\pi$(subject is moving) = 0.6 and does not change as fit processes each chunk. Specify Observation Weights Train a naive Bayes classification model by using fitcnb, convert it to an incremental learner, track its performance on streaming data, and then fit the model to the data. Specify observation Load and Preprocess Data Load the human activity data set. Randomly shuffle the data. load humanactivity rng(1); % For reproducibility n = numel(actid); idx = randsample(n,n); X = feat(idx,:); Y = actid(idx); For details on the data set, enter Description at the command line. Suppose that the data from a stationary subject (Y <= 2) has double the quality of the data from a moving subject. Create a weight variable that assigns a weight of 2 to observations from a stationary subject and 1 to a moving subject. Train Naive Bayes Classification Model Fit a naive Bayes classification model to a random sample of half the data. idxtt = randsample([true false],n,true); TTMdl = fitcnb(X(idxtt,:),Y(idxtt),'Weights',W(idxtt)) TTMdl = ResponseName: 'Y' CategoricalPredictors: [] ClassNames: [1 2 3 4 5] ScoreTransform: 'none' NumObservations: 12053 DistributionNames: {1x60 cell} DistributionParameters: {5x60 cell} TTMdl is a ClassificationNaiveBayes model object representing a traditionally trained naive Bayes classification model. Convert Trained Model Convert the traditionally trained model to a naive Bayes classification model for incremental learning. IncrementalMdl = incrementalLearner(TTMdl) IncrementalMdl = IsWarm: 1 Metrics: [1x2 table] ClassNames: [1 2 3 4 5] ScoreTransform: 'none' DistributionNames: {1x60 cell} DistributionParameters: {5x60 cell} IncrementalMdl is an incrementalClassificationNaiveBayes model. Because class names are specified in IncrementalMdl.ClassNames, labels encountered during incremental learning must be in Separately Track Performance Metrics and Fit Model Perform incremental learning on the rest of the data by using the updateMetrics and fit functions. At each iteration: 1. Simulate a data stream by processing 50 observations at a time. 2. Call updateMetrics to update the cumulative and window minimal cost of the model given the incoming chunk of observations. Overwrite the previous incremental model to update the losses in the Metrics property. Note that the function does not fit the model to the chunk of data—the chunk is "new" data for the model. Specify the observation weights. 3. Store the minimal cost. 4. Call fit to fit the model to the incoming chunk of observations. Overwrite the previous incremental model to update the model parameters. Specify the observation weights. % Preallocation idxil = ~idxtt; nil = sum(idxil); numObsPerChunk = 50; nchunk = floor(nil/numObsPerChunk); mc = array2table(zeros(nchunk,2),'VariableNames',["Cumulative" "Window"]); Xil = X(idxil,:); Yil = Y(idxil); Wil = W(idxil); % Incremental fitting for j = 1:nchunk ibegin = min(nil,numObsPerChunk(j-1) + 1); iend = min(nil,numObsPerChunkj); idx = ibegin:iend; IncrementalMdl = updateMetrics(IncrementalMdl,Xil(idx,:),Yil(idx), ... mc{j,:} = IncrementalMdl.Metrics{"MinimalCost",:}; IncrementalMdl = fit(IncrementalMdl,Xil(idx,:),Yil(idx),'Weights',Wil(idx)); IncrementalMdl is an incrementalClassificationNaiveBayes model object trained on all the data in the stream. Alternatively, you can use updateMetricsAndFit to update performance metrics of the model given a new chunk of data, and then fit the model to the data. Plot a trace plot of the performance metrics. h = plot(mc.Variables); xlim([0 nchunk]) ylabel('Minimal Cost') The cumulative loss gradually stabilizes, whereas the window loss jumps throughout the training. Perform Conditional Training Incrementally train a naive Bayes classification model only when its performance degrades. Load the human activity data set. Randomly shuffle the data. load humanactivity n = numel(actid); rng(1) % For reproducibility idx = randsample(n,n); X = feat(idx,:); Y = actid(idx); For details on the data set, enter Description at the command line. Configure a naive Bayes classification model for incremental learning so that the maximum number of expected classes is 5, the tracked performance metric includes the misclassification error rate, and the metrics window size is 1000. Fit the configured model to the first 1000 observations. Mdl = incrementalClassificationNaiveBayes('MaxNumClasses',5,'MetricsWindowSize',1000, ... initobs = 1000; Mdl = fit(Mdl,X(1:initobs,:),Y(1:initobs)); Mdl is an incrementalClassificationNaiveBayes model object. Perform incremental learning, with conditional fitting, by following this procedure for each iteration: • Simulate a data stream by processing a chunk of 100 observations at a time. • Update the model performance on the incoming chunk of data. • Fit the model to the chunk of data only when the misclassification error rate is greater than 0.05. • When tracking performance and fitting, overwrite the previous incremental model. • Store the misclassification error rate and the mean of the first predictor in the second class ${\mu }_{21}$ to see how they evolve during training. • Track when fit trains the model. % Preallocation numObsPerChunk = 100; nchunk = floor((n - initobs)/numObsPerChunk); mu21 = zeros(nchunk,1); ce = array2table(nan(nchunk,2),'VariableNames',["Cumulative" "Window"]); trained = false(nchunk,1); % Incremental fitting for j = 1:nchunk ibegin = min(n,numObsPerChunk(j-1) + 1 + initobs); iend = min(n,numObsPerChunkj + initobs); idx = ibegin:iend; Mdl = updateMetrics(Mdl,X(idx,:),Y(idx)); ce{j,:} = Mdl.Metrics{"ClassificationError",:}; if ce{j,2} > 0.05 Mdl = fit(Mdl,X(idx,:),Y(idx)); trained(j) = true; mu21(j) = Mdl.DistributionParameters{2,1}(1); Mdl is an incrementalClassificationNaiveBayes model object trained on all the data in the stream. To see how the model performance and ${\mu }_{21}$ evolve during training, plot them on separate tiles. t = tiledlayout(2,1); hold on xlim([0 nchunk]) legend('\mu_{21}','Training occurs','Location','best') hold off xlim([0 nchunk]) ylabel('Misclassification Error Rate') The trace plot of ${\mu }_{21}$ shows periods of constant values, during which the loss within the previous observation window is at most 0.05. Input Arguments X — Chunk of predictor data floating-point matrix Chunk of predictor data to which the model is fit, specified as an n-by-Mdl.NumPredictors floating-point matrix. The length of the observation labels Y and the number of observations in X must be equal; Y(j) is the label of observation j (row) in X. If Mdl.NumPredictors = 0, fit infers the number of predictors from X, and sets the corresponding property of the output model. Otherwise, if the number of predictor variables in the streaming data changes from Mdl.NumPredictors, fit issues an error. Data Types: single \| double Y — Chunk of labels categorical array \| character array \| string array \| logical vector \| floating-point vector \| cell array of character vectors Chunk of labels to which the model is fit, specified as a categorical, character, or string array, logical or floating-point vector, or cell array of character vectors. The length of the observation labels Y and the number of observations in X must be equal; Y(j) is the label of observation j (row) in X. fit issues an error when one or both of these conditions are met: • Y contains a new label and the maximum number of classes has already been reached (see the MaxNumClasses and ClassNames arguments of incrementalClassificationNaiveBayes). • The ClassNames property of the input model Mdl is nonempty, and the data types of Y and Mdl.ClassNames are different. Data Types: char \| string \| cell \| categorical \| logical \| single \| double If an observation (predictor or label) or weight contains at least one missing (NaN) value, fit ignores the observation. Consequently, fit uses fewer than n observations to create an updated model, where n is the number of observations in X. Output Arguments • Unlike traditional training, incremental learning might not have a separate test (holdout) set. Therefore, to treat each incoming chunk of data as a test set, pass the incremental model and each incoming chunk to updateMetrics before training the model on the same data. Normal Distribution Estimators Estimated Probability for Multinomial Distribution Estimated Probability for Multivariate Multinomial Distribution Observation Weights Version History Introduced in R2021a R2021b: Naive Bayes incremental fitting functions compute biased (maximum likelihood) standard deviations for conditionally normal predictor variables	{"url":"https://fr.mathworks.com/help/stats/incrementalclassificationnaivebayes.fit.html","timestamp":"2024-11-10T18:11:30Z","content_type":"text/html","content_length":"150654","record_id":"<urn:uuid:1e855846-3ded-4825-9dcf-22c9506242ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00181.warc.gz"}
Construction of upper and lower solutions with applications to singular boundary value problemsConstruction of upper and lower solutions with applications to singular boundary value problems An upper and lower solution theory is presented for the Dirichlet boundary value problem $y^{\prime\prime}+f(t,y,y^{\prime})=0$, $0<t <1$ with $y(0)=y(1)=0$. Our nonlinearity may be singular in its dependent variable and is allowed to change sign. R.P. Agarwal D. O’Regan Radu Precup Babes-Bolyai University, Cluj-Napoca, Romania R.P. Agarwal, D.O. Regan, R. Precup, Construction of upper and lower solutions with applications to singular boundary value problems, J. Comput. Anal. Appl. 7 (2005), 205-221. MR2223477, Zbl 1085.34016 [1] R.P. Agrawal, D.O’Regan and V. Lakshmikantham, An upper and lower solution approach for nonlinear singular boundary value problems with y’ dependence, Archives of Inequalities and Applicaitons 1 (2003), 119-135. [2] R.P. Agarwal, D.O’Regan, V. Lakshmikantham and S. Leela, an upper and lower solution theory for singular Emden-Fowler equations, Nonlinear Analysis: Real World Applications, 3(2002), 275-291. [3] A. Callegari and A. Nachman, Some singular nonlinear differential equaitons arising in boundary layer theory, J. Math. Anal. Appl., 64(1978), 96-105. [4] J.A. Gatica, V. Oliker and P. Waltman, Singular nonlinear boundary value problems for second order differential equations, J. Differential Equations, 79(1989), 62-78. [5] H. Lu, D.O’Regan and C. Zhong, Existence of positive solutions for the singular equation (φ_{p}(y′))′+g(t,y,y′)=0, Nonlinear Oscillations, to appear. [6] C.D.Luning and W.L.Perry, Positive solutions of negative exponent generalized Emden-Fowler boundary value problems, SIAM J. Math. Anal., 12(1981), 874-879. [7] D.O’Regan, Theory of singular boundary value problems, World Scientific, Singapore, 1994. [8] D.O’Regan and R.P. Agarwal, Singular problems: an upper and lower solution approach, J. Math. Anal. Appl., 251(2000), 230-250.	{"url":"https://ictp.acad.ro/construction-of-upper-and-lower-solutions-with-applications-to-singular-boundary-value-problems/","timestamp":"2024-11-04T17:08:59Z","content_type":"text/html","content_length":"120301","record_id":"<urn:uuid:10f387e2-af86-4497-a03c-b3c632487d15>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00718.warc.gz"}
Concrete Bag Calculator \| D'oh!-I-Y Concrete Bag Calculator You can easily work out how many cubic feet or cubic yards of concrete you need for a project. That’s fine if you are buying ready mixed concrete. But what if you are buying bags? This calculator will convert cubic yards, or feet, into the equivalent number of forty, fifty, sixty, or eighty pound bags. Just enter the number of cubic yards or cubic feet of concrete you need for your project, to calculate how many bags you need to get the job done. The calculator rounds up to the nearest bag, so you’re sure to have enough. How many cubic feet or cubic yards of concrete do you need? The number of bags of concrete you need is: • – forty pound bags, • – fifty pound bags, • – sixty pound bags, or • – eighty pound bags.	{"url":"https://www.dohiy.com/calculators/concrete-bag-calculator/","timestamp":"2024-11-03T22:23:12Z","content_type":"text/html","content_length":"14528","record_id":"<urn:uuid:95b81db2-d13d-4d52-8410-a1abda0a1e4b>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00120.warc.gz"}
Exploring Linear Algebra: Facts, Figures, and Applications Linear algebra is a branch of mathematics that deals with linear equations, linear functions, and their representations through matrices and vectors. It is a fundamental tool in many fields, including physics, engineering, computer science, economics, and many others. In this article, we will explore some interesting facts and figures related to linear algebra. Applications of Linear Algebra: Linear algebra has numerous applications in various fields. Here are some examples: 1. Computer graphics: Linear algebra is used to represent 3D objects and their transformations in computer graphics. 2. Control systems: Linear algebra is used to analyze and design control systems in engineering. 3. Quantum mechanics: Linear algebra is used to describe the state of a quantum system in quantum mechanics. 4. Machine learning: Linear algebra is used to perform computations on large datasets in machine learning. 5. Economics: Linear algebra is used in input-output analysis and game theory in economics. Famous Matrices: Matrices are one of the primary tools used in linear algebra. Here are some famous matrices: 1. Identity matrix: The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. It is denoted by I. 2. Zero matrix: The zero matrix is a matrix where all entries are zero. 3. Diagonal matrix: A diagonal matrix is a square matrix where all the entries outside the diagonal are zero. 4. Symmetric matrix: A symmetric matrix is a square matrix that is equal to its transpose. 5. Pauli matrices: The Pauli matrices are a set of three 2×2 matrices used in quantum mechanics to represent the spin of a particle. Notable Mathematicians in Linear Algebra: There have been many mathematicians who have contributed to the development of linear algebra. Here are some notable ones: 1. Carl Friedrich Gauss: Gauss made significant contributions to the theory of matrices, determinants, and linear equations. 2. Arthur Cayley: Cayley introduced the concept of matrices and developed the theory of determinants. 3. Hermann Grassmann: Grassmann introduced the concept of vector spaces and developed the theory of linear transformations. 4. Évariste Galois: Galois developed the theory of linear equations and their solutions. 5. David Hilbert: Hilbert made significant contributions to the development of linear algebra, including the theory of infinite-dimensional vector spaces. Interesting Facts about Linear Algebra: 1. The concept of vector spaces was introduced by Hermann Grassmann in 1844. 2. The Gaussian elimination algorithm for solving linear equations was developed by Carl Friedrich Gauss. 3. The determinant of a matrix was first defined by Arthur Cayley in 1841. 4. The eigenvalue and eigenvector concept was introduced by Peter Gustav Lejeune Dirichlet in 1858. 5. Linear algebra has applications in diverse fields such as computer graphics, control systems, and quantum mechanics. Linear algebra is a fundamental branch of mathematics with numerous applications in many fields. It provides a powerful tool to represent and analyze linear equations, linear functions, and their transformations. The concept of matrices and vectors has revolutionized many fields, including computer graphics, control systems, and machine learning. Linear algebra continues to be a vital subject in undergraduate and graduate programs in mathematics and other related disciplines.	{"url":"https://mathematicslearninglink.com/2023/03/exploring-linear-algebra-facts-figures-and-applications/","timestamp":"2024-11-01T20:58:12Z","content_type":"text/html","content_length":"46733","record_id":"<urn:uuid:46ad392c-d347-45e3-804d-5036afcce3a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00628.warc.gz"}
A Paley-Wiener Theorem for All Two and Three-Step Nilpotent Lie Groups. Document Type Degree Name Doctor of Philosophy (PhD) First Advisor Leonard Richardson A Paley-Wiener Theorem for all connected, simply-connected two and three-step nilpotent Lie groups is proved. If f $\epsilon \ L\sbsp{c}{\infty}({G}),$ where G is a connected, simply-connected two or three-step nilpotent Lie group such that the operator-valued Fourier transform $\\varphi(\pi)$ = 0 for all $\pi$ in E, a subset of G of positive Plancherel measure, then it is shown that f = 0 a. e. on G. The proof uses representation-theoretic methods from Kirillov theory for nilpotent Lie groups, and uses an integral formula for the operator-valued Fourier transform $\\varphi(\pi)$. It is also shown by example that the condition that G be simply-connected is necessary. Recommended Citation Park, Robert Reeve, "A Paley-Wiener Theorem for All Two and Three-Step Nilpotent Lie Groups." (1994). LSU Historical Dissertations and Theses. 5749.	{"url":"https://repository.lsu.edu/gradschool_disstheses/5749/","timestamp":"2024-11-04T08:37:21Z","content_type":"text/html","content_length":"31742","record_id":"<urn:uuid:a1e3a74c-8296-4a57-bc19-be989cbf0b03>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00658.warc.gz"}
Inspiring Drawing Tutorials How Many Amps Does Microwave Draw How Many Amps Does Microwave Draw - Nevertheless, manufacturers recommend that you plug the microwave into at least 15 amps on its own circuit. Thus, current drawn in amps for a 900 watts microwave will be: Not many, considering how convenient it is in the kitchen and also in comparison with the energy. In some regions, it may be 110v or 220v. Web how many amps does a microwave use? What factors affect the power used by a microwave? Amps = 1350 watts / 120 v. In this guide, we’ll look at how many amps does a microwave draws, how its amperage is affected by various conditions, and how to install and use your microwave securely. Amperes (a) = watts (w) / volts (v). In some regions, it may be 110v or 220v. Importance of knowing the amp drawn by a microwave: Several factors can affect the amp draw of a microwave. Therefore, for 240 volts only half of the amperage is between 2.5 to 5 amps. Web a standard microwave uses about 5 amps of current during its operation. Web most microwaves operate at a power of 600 to 1,500 watts, which translates to an average amp draw of 5 to 15 amps when operating at full power. Given the formula for current (i = p / v) and assuming the voltage is 110v, here are the different amperages of the common microwave products available: Importance of knowing the amp drawn by a microwave: Before designing a circuit, you should know the amps connected to it. Discover the average amps a microwave oven uses in this informative article. Web microwaves differ in terms of the number of amps they use and their specific electrical requirements. Web how many amps does a microwave use? Web microwaves differ in terms of the number of amps they use and their specific electrical requirements. Amps = 1350 watts / 120 v. In a nutshell, amps help you to determine how much power your appliance requires quickly. Web how many amps does a microwave use? The correlation between the two is that one goes up with the other. The more amps, the quicker the device will cook your food. Several factors can affect the amp draw of a microwave. Find out how much electricity your microwave consumes and how it impacts your energy bills. Web how many amps does a microwave use? It is important to observe these precautionary measures to avoid mishaps or damaging your microwave. Amps = 1350 watts / 120 v. Web how many amps does a microwave use? The exact amount of amperage will depend on the wattage of the microwave. This means that the amperage will be 10 amperes. Given the formula for current (i = p / v) and assuming the voltage is 110v, here are the different amperages of the. Web how many amps does a microwave use? Therefore, instead of up to 10 amps, it may only draw 5amps for a 240v circuit. In this guide, we’ll look at how many amps does a microwave draws, how its amperage is affected by various conditions, and how to install and use your microwave securely. The ampere usage varies based on. It is important to observe these precautionary measures to avoid mishaps or damaging your microwave. Amps = 1350 watts / 120 v. Standard microwaves consume 8.3 amps on average, but other factors, like wattage, largely determine this. Discover the average amps a microwave oven uses in this informative article. To get the amp rating of a microwave, divide the wattage. Therefore, for 240 volts only half of the amperage is between 2.5 to 5 amps. The more amps, the quicker the device will cook your food. Industrial microwaves generally consume above 1800 watts, requiring around 20 amps. However, in countries where the standard household outlet voltage is 240 volts, which may result in a lower amp draw. Web microwave ovens. Microwaves come in a variety of wattages, and the amount of wattage your microwave uses will determine how many amps it uses. Therefore, for 240 volts only half of the amperage is between 2.5 to 5 amps. Before designing a circuit, you should know the amps connected to it. Now that you know how many amps your microwave uses, you. The correlation between the two is that one goes up with the other. Web typically, an average microwave with a power rating of up to 1,200 watts may use a current draw of 10 amps at a 120v circuit. The higher the wattage, the more amps the microwave requires. Web a typical microwave oven uses on average 1000 watts of. The exact amount of amperage will depend on the wattage of the microwave. Standard microwaves consume 8.3 amps on average, but other factors, like wattage, largely determine this. Therefore, for 240 volts only half of the amperage is between 2.5 to 5 amps. Web a standard microwave uses about 5 amps of current during its operation. Web a microwave uses. How Many Amps Does Microwave Draw - Microwaves come in a variety of wattages, and the amount of wattage your microwave uses will determine how many amps it uses. Industrial microwaves generally consume above 1800 watts, requiring around 20 amps. Nevertheless, manufacturers recommend that you plug the microwave into at least 15 amps on its own circuit. Thus, current drawn in amps for a 900 watts microwave will be: Several factors can affect the amp draw of a microwave. The correlation between the two is that one goes up with the other. Amperes (a) = watts (w) / volts (v). The higher the wattage, the more amps the microwave requires. This means that the amperage will be 10 amperes. Web how many amps does a microwave use? Web microwaves differ in terms of the number of amps they use and their specific electrical requirements. Therefore, instead of up to 10 amps, it may only draw 5amps for a 240v circuit. The correlation between the two is that one goes up with the other. The higher the wattage, the more amps the microwave requires. Discover the average amps a microwave oven uses in this informative Web high watt vs. Web for the usa, 120 v is the standard voltage. Find out how much electricity your microwave consumes and how it impacts your energy bills. Thus, current drawn in amps for a 900 watts microwave will be: Discover the average amps a microwave oven uses in this informative article. In some regions, it may be 110v or 220v. To get the amp rating of a microwave, divide the wattage by the voltage. Nevertheless, manufacturers recommend that you plug the microwave into at least 15 amps on its own circuit. Therefore, instead of up to 10 amps, it may only draw 5amps for a 240v circuit. The higher the wattage, the more amps the microwave requires. How To Tell How Many Amps An Appliance Uses? Web microwaves differ in terms of the number of amps they use and their specific electrical requirements. The correlation between the two is that one goes up with the other. Therefore, instead of up to 10 amps, it may only draw 5amps for a 240v circuit. Average microwaves range from 600 watts to 1200 watts, employing between 5.4 and 10 amps. Thus, Current Drawn In Amps For A 900 Watts Microwave Will Be: Web how many amps does a microwave use? Industrial microwaves generally consume above 1800 watts, requiring around 20 amps. Web a microwave uses between 5.4 amps and 10 amps depending on how much power is being drawn by the appliance at the time. In some regions, it may be 110v or 220v. Amps = 1350 Watts / 120 V. Web how to calculate microwave amp usage? This means that the amperage will be 10 amperes. However, in countries where the standard household outlet voltage is 240 volts, which may result in a lower amp draw. The exact amount of amperage will depend on the wattage of the microwave. It Is Important To Observe These Precautionary Measures To Avoid Mishaps Or Damaging Your Microwave. Web typically, an average microwave with a power rating of up to 1,200 watts may use a current draw of 10 amps at a 120v circuit. In this guide, we’ll look at how many amps does a microwave draws, how its amperage is affected by various conditions, and how to install and use your microwave securely. The higher the wattage, the more amps the microwave requires. Now that you know how many amps your microwave uses, you can figure out how much electricity it consumes in a day, month, or year.	{"url":"https://one.wkkf.org/art/drawing-tutorials/how-many-amps-does-microwave-draw.html","timestamp":"2024-11-07T17:23:30Z","content_type":"text/html","content_length":"35244","record_id":"<urn:uuid:23721d5f-aad1-4ff6-a549-c5d0ff292dad>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00557.warc.gz"}
Math Colloquia - <학부생을 위한 ɛ 강연> Convergence of Fourier series and integrals in Lebesgue spaces Convergence of Fourier series and integrals is the most fundamental question in classical harmonic analysis from its beginning. In one dimension convergence in Lebesgue spaces is fairly well understood. However in higher dimensions the problem becomes more intriguing since there is no canonical way to sum (and integrate) Fourier series (and integrals, respectively), and convergence of the multidimensional Fourier series and integrals is related to complicated phenomena which can not be understood in perspective of convergence in one dimension. The Bochner-Riesz conjecture may be regarded as an attempt to understand multidimensional Fourier series and integrals. Even though the problem is settled in two dimensions, it remains open in higher dimensions. In this talk we review developments in the Bochner-Riesz conjecture and discuss its connection to the related problems such as the restriction and Kakeya conjectures.	{"url":"http://my.math.snu.ac.kr/board/index.php?mid=colloquia&page=4&document_srl=786572&sort_index=speaker&order_type=asc&l=en","timestamp":"2024-11-03T12:45:34Z","content_type":"text/html","content_length":"43455","record_id":"<urn:uuid:15ebdc8d-624f-44ea-a869-b247e4586817>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00048.warc.gz"}
A priori estimates of the accuracy of the Galerkin method with discontinuous functions for two-dimensional parabolic problems CitationZhalnin R. V., Masyagin V. F., Peskova E. E., Tishkin V. F. ''A priori estimates of the accuracy of the Galerkin method with discontinuous functions for two-dimensional parabolic problems '' [Electronic resource]. Proceedings of the International Scientific Youth School-Seminar "Mathematical Modeling, Numerical Methods and Software complexes" named after E.V. Voskresensky (Saransk, July 16-20, 2018). Saransk: SVMO Publ, 2018. - pp. 50-51. Available at: https://conf.svmo.ru/files/2018/papers/paper15.pdf. - Date of access: 12.11.2024.	{"url":"https://conf.svmo.ru/en/archive/article?id=87","timestamp":"2024-11-12T04:15:47Z","content_type":"text/html","content_length":"11054","record_id":"<urn:uuid:a15b9c96-7280-48a3-b786-66959c8a306d>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00874.warc.gz"}
Neural network Neural network Neural networks are networks of artificial neurons or nodes used to solve AI problems. They model connections between nodes as weights, which can be positive or negative depending on the type of connection. Inputs are modified by weights and summed, then an activation function controls the output. Neural networks can be trained via a dataset and can self-learn from experience. 14 courses cover this concept This course provides a deeper understanding of robot autonomy principles, focusing on learning new skills and physical interaction with the environment and humans. It requires familiarity with programming, ROS, and basic robot autonomy techniques. An in-depth course focused on building neural networks and leading successful machine learning projects. It covers Convolutional Networks, RNNs, LSTM, Adam, Dropout, BatchNorm, Xavier/He initialization, and more. Students are expected to have basic computer science skills, probability theory knowledge, and linear algebra familiarity. This course is centered on extracting information from unstructured data in language and social networks using machine learning tools. It covers techniques like sentiment analysis, chatbot development, and social network analysis. CS 224N provides an in-depth introduction to neural networks for NLP, focusing on end-to-end neural models. The course covers topics such as word vectors, recurrent neural networks, and transformer models, among others. This comprehensive course covers various machine learning principles from supervised, unsupervised to reinforcement learning. Topics also touch on neural networks, support vector machines, bias-variance tradeoffs, and many real-world applications. It requires a background in computer science, probability, multivariable calculus, and linear algebra. A survey course on neural network implementation and applications, including image processing, classification, detection, and segmentation. The course also covers semantic understanding, translation, and question-answering applications. It's ideal for those with a background in Machine Learning, Neural Networks, Optimization, and CNNs. This introductory course focuses on machine learning, probabilistic reasoning, and decision-making in uncertain environments. A blend of theory and practice, the course aims to answer how systems can learn from experience and manage real-world uncertainties. A general introduction to computer vision, this course covers traditional image processing techniques and newer, machine-learning based approaches. It discusses topics like filtering, edge detection, stereo, flow, and neural network architectures. Stanford's CS 221 course teaches foundational principles and practical implementation of AI systems. It covers machine learning, game playing, constraint satisfaction, graphical models, and logic. A rigorous course requiring solid foundational skills in programming, math, and probability. UC Berkeley's CS 188 course covers the basic ideas and techniques for designing intelligent computer systems, emphasizing statistical and decision-theoretic modeling. By the course's end, students will have built autonomous agents that can make efficient decisions in a variety of settings. Carnegie Mellon University This course provides a comprehensive introduction to deep learning, starting from foundational concepts and moving towards complex topics such as sequence-to-sequence models. Students gain hands-on experience with PyTorch and can fine-tune models through practical assignments. A basic understanding of calculus, linear algebra, and Python programming is required. Carnegie Mellon University A comprehensive exploration of machine learning theories and practical algorithms. Covers a broad spectrum of topics like decision tree learning, neural networks, statistical learning, and reinforcement learning. Encourages hands-on learning via programming assignments. A thorough introduction to machine learning principles such as online learning, decision making, gradient-based learning, and empirical risk minimization. It also explores regression, classification, dimensionality reduction, ensemble methods, neural networks, and deep learning. The course material is self-contained and based on freely available resources. Carnegie Mellon University This course gives an expansive introduction to computer vision, focusing on image processing, recognition, geometry-based and physics-based vision, and video analysis. Students will gain practical experience solving real-life vision problems. It requires a good understanding of linear algebra, calculus, and programming.	{"url":"https://cogak.com/concept/291","timestamp":"2024-11-13T06:40:04Z","content_type":"text/html","content_length":"474147","record_id":"<urn:uuid:dd7382d0-5441-4420-9e27-c9381747c335>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00300.warc.gz"}
ezEDA: A Task-Oriented Interface for Exploratory Data Analysis Viswa Viswanathan May 25, 2020 Owing to its extensive functionality and its roots in a robust grammar of graphics (Wilkinson 2013), ggplot2 (Wickham 2016) has become very popular among R users. In using ggplot, the general approach to generating plots is to first conceive of the plot and then use our undertsanding of ggolot to create the required mappings, layers and other adjustments. The process requires us to divide our cognitive resources to the main task at hand – extracting intelligence from the data – and the process of generating the requisite plot. The guiding insight behind ezEDA is that the process of exploratory data analysis can be made more productive by increasing the proportion of congitive resources devoted to imagining the kinds of analysis we could perform (by reducing the proportion devoted to the mechanices of generating the plot). Where possible, devoting more of our mental energies to thinking about the kinds of explorations we would like to perform will likely make us more Although exploratory analysis of data differs from dataset to dataset, we can still see some recurrent themes or patterns that apply to many situations. ezEDA identifies such common patterns, and for each one, it provides a single convenience function that relieves us of the mechanices of generating the plot. With ezEDA we aim to ease the task of generating ggplot-based visualizations by allowing users to think in terms of their problem domain rather than the details of how to achieve the plot that they have in mind. This approach is particularly useful when the visualization task involves standard themes or patterns. The ezEDA package currently provides functions for twelve patterns and we aim to incorporate more in future versions. ezEDA is only beneficial in situations where the analyst is able to exploit a common pattern that ezEDA has already identified. For other situations, the analyst has to use other means like constructing a ggplot plot from the ground up. Here are some general features of ezEDA functions: • all ezEDA functions are built on top of ggplot functions • like dplyr and ggplot, ezEDA functions make interactive use easier by allowing the use of unquoted column names and support variable numbers of arguments • given that the main focus of ezEDA functions is to enable users to quickly generate visualizations for supported patterns, these functions support limited customization; they do not allow the full range of customization that a direct use of ggplot would Once they have completed their initial explorations and obtained the necessary insights, it could very well be that users will then dive into ggplot and fine-tune their plots where needed. Currently, ezEDA provides functions under five categories: • Trends • Distributions • Relationships • Tallies, and • Contributions Basic concepts behind design of the ezEDA interface ezEDA is partly motivated by the work of Stephen Few (Few 2015) who described a general framework for Exploratory Data Analysis. Such a framework would be useful when we are presented with a dataset and would like to generate questions to ask of the dataset – questions that could potentially generate interesting results or confirm what we already knew or suspected. Few’s framework is driven by the observation that any dataset has columns of four types: • categories – categorical columns representing categories with no inherent numerical meaning • measures – numerical columns • time – columns representing time (could be date columns or just integers representing sequential time units) • others – columns without much significance for data exploration (like names, etc.) Once we establish this terminology, the list below shows the patterns and the corresponding ezEDA functions: • trends of measures □ functions: measure_change_over_time_wide and measure_change_over_time_long • distributions of measures □ functions: measure_distribution, measure_distribution_by_category, measure_distribution_by_two_categories and measure_distribution_over_time • relationships between measures □ functions: two_measure_relationship and multi_measure_relationship • tallies for counts of one or two categories □ functions: category_tally and two_category_tally • contributions of different categories to a measure □ functions: category_contribution and two_category_contribution ezEDA provides functions for these tasks. The package currently has twelve functions and we hope to add more as we identify more patterns. Installing ezEDA ezEDA is available under CRAN and can be installed as usual by calling the install.packages function: ezEDA depends on many other packages like ggplot2, dplyr, tidyr and others and if any of those packages are not installed on your system, the above code will cause any of the missing packages to be installed as well. # Using ezEDA As always, it is a good idea to make the package namespace available via the library function: ezEDA functions We discuss each of the functions below. For convenience, we have divided the functions into convenient groups. If a dataset has one or more measure columns and a time column, then it could be useful to study the movement of the measure columns over time. Each of the two functions in this group help us to simultaneously visualize trends of up to 6 measures. Dependning on whether the data is in wide form (different measures in different columns) or long (all measure values in one column with another column to identify each measure) you can use a differemnt function. Simultaneously visualize the change over time of several measures (up to 6) when the data is in wide form (see above). Simultaneously visualize the change over time of several measures (up to 6) when the data is in long form (see above). ## For ggplot2::economics_long, plot the trend of population and number unemployed. ## In this dataset, all measures are in the column named value and ## the names of the measures are in the column named variable measure_change_over_time_long(ggplot2::economics_long, date, variable, value, pop, unemploy) Very often we study the distributuons of numeric columns (measures).ezEDA provides four different functions. Within this broad theme, we are often also interested in studying how the distribution changes based on one or two categories or based on time. Distribution of a measure: default is histogram. ## For ggplot2::mpg, plot the distribution of highway mileage measure_distribution(ggplot2::mpg, hwy) #> Using binwidth corresponding to 30 bins. Use bwidth argument to fix if needed. #> `stat_bin()` using `bins = 30`. Pick better value with `binwidth`. Distribution of a measure: default is histogram. By default the function uses a bin width corresponding to 30 bins. Use the bwidth argument to specify the desired value for the width of each bin. ## For ggplot2::mpg, plot the distribution of highway mileage measure_distribution(ggplot2::mpg, hwy, bwidth = 2) Distribution of a measure. Get a boxplot instead of histogram. Distribution of a measure with highlighting of different values of a single category. ## For ggplot2::diamonds, plot the distribution of price while highlighting ## the counts of diamonds of different cuts measure_distribution_by_category(ggplot2::diamonds, price, cut) #> Using binwidth corresponding to 30 bins. Use bwidth argument to fix if needed. #> `stat_bin()` using `bins = 30`. Pick better value with `binwidth`. Distribution of a measure with highlighting of different values of a single category across facets. ## For ggplot2::diamonds, plot the distribution of price showing ## the distribution for each kind of cut in a different facet measure_distribution_by_category(ggplot2::diamonds, price, cut, separate = TRUE) #> Using binwidth corresponding to 30 bins. Use bwidth argument to fix if needed. #> `stat_bin()` using `bins = 30`. Pick better value with `binwidth`. Distribution of a measure for a combination of two categories within a facet grid ## For ggplot2::diamonds, plot the distribution of price separately ## for each unique combination of cut and clarity measure_distribution_by_two_categories(ggplot2::diamonds, carat, cut, clarity) #> Using binwidth corresponding to 30 bins. Use bwidth argument to fix if needed. #> `stat_bin()` using `bins = 30`. Pick better value with `binwidth`. Study the change of distribution of a measure over time ## 50 random values of three measures for each of 1999, 2000 and 2001 h1 <- round(rnorm(50, 60, 8), 0) h2 <- round(rnorm(50, 65, 8), 0) h3 <- round(rnorm(50, 70, 8), 0) h <- c(h1, h2, h3) y <- c(rep(1999, 50), rep(2000, 50), rep(2001, 50)) df <- data.frame(height = h, year = y) measure_distribution_over_time(df, h, year) #> Using binwidth corresponding to 30 bins. Use bwidth argument to fix if needed. #> `stat_bin()` using `bins = 30`. Pick better value with `binwidth`. WHen a dataset has two or more measures, studying their pairwise relationships often yields useful insights. exEDA provides two relevant functions. Scatterplot of the relationship between two measures with optional coloring of points by a category. ## For ggplot2::mpg, plot the highway mileage against the displacement two_measures_relationship(ggplot2::mpg, displ, hwy) #> `geom_smooth()` using method = 'loess' and formula 'y ~ x' Below is a variant shopwing the relationship separately for each value of a category. The two functions in this group satisfy a common need to generate counts of categories. For example, in a dataset of students, we might want to plot the counts of students by gender; in the diamonds dataset from ggplot2, we might want to generate a plot of the number of diamonds by color, and so on. Barplot of the counts based on a category column. Often when a barplot has many bars, we run the risk of the x-axis labels running onto each other. To handle this issue, the function flips the coordinates. The first argument is the dataset to be used and the second argument is the unquoted name of the relevant category column. Barplot of one category showing its conposition in terms of another category. The arguments are: the dataset to be used and the unquoted names of the two category columns. ## For ggplot2::diamonds plot the tallies for different types of ## cut and clarity two_category_tally(ggplot2::diamonds, cut, clarity) Below is a variant with facets. ## For ggplot2::diamonds, plot the tallies for different types of ## cut and clarity and show the plots for each value of the ## second category in a separate facet two_category_tally(ggplot2::diamonds, cut, clarity, separate = TRUE) #> Warning: `guides(<scale> = FALSE)` is deprecated. Please use `guides(<scale> = #> "none")` instead. The two functions in this group satisfy a common need to analyze the extent to which specific categories contribute to specific measures. For example, in a dataset of people, we might want to plot the total income by gender; in the diamonds dataset from ggplot2, we might want to generate a plot of the contribution to the price by diamonds of various kinds of cut, and so on. Barplot of the total of a measure based on a category column. Barplot of the total of a measure based on a category column. ## For ggplot2::diamonds, plot the total price for each kind of cut ## while also showing the contribution of each kind of clarity ## within each kind of cut two_category_contribution(ggplot2::diamonds, cut, clarity, price) Below is a variant with facets. Few, Stephen. 2015. Signal: Understanding What Matters in a World of Noise. Analytics Press. Wickham, Hadley. 2016. Ggplot2: Elegrant Graphics for Data Analysis. Springer. Wilkinson, Leland. 2013. The Grammar of Graphics. Springer Science & Business Media.	{"url":"http://rsync.jp.gentoo.org/pub/CRAN/web/packages/ezEDA/vignettes/ezEDA.html","timestamp":"2024-11-06T19:05:33Z","content_type":"text/html","content_length":"721862","record_id":"<urn:uuid:2c2c5a30-86e6-4554-b6ef-591e7e68cbc0>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00605.warc.gz"}
Financial algorithms on real quantum computers With the availability of real, functioning quantum computers, quantum algorithms have transitioned from theory to practice and skyrocketed interest in practical applications. Finance is one of the many fields where quantum algorithms can improve speed and accuracy. Here, we demonstrate a prototype calculation on the Helmi quantum computer through the Finnish Quantum-Computing Infrastructure The speed-up offered by quantum computers stems from the probabilistic nature of the laws of physics at the quantum scale. For a quantum system, it is possible to exist simultaneously in a mix of multiple states, a state called quantum superposition. Quantum computing exploits this property by using quantum bits, that is, qubits. A classical bit can attain a value of 0 or 1. The quantum state of the qubit is clouded by uncertainty, allowing the qubit essentially to be both 0 and 1 simultaneously. Another quantum phenomenon, quantum entanglement, allows qubits to be coupled so that changing the state of one qubit affects the state of all entangled qubits simultaneously. An ideal quantum computer can efficiently perform calculations for large and complex systems, as each additional qubit, in principle, doubles the capacity of the quantum computer. Now is the time to identify algorithms with future potential. Many applications have already been singled out: cryptography, simulations, search algorithms, and optimization problems. In finance, the use of quantum algorithms is investigated in asset pricing, risk analysis, and portfolio optimization, to mention a few [2,3]. One promising quantum algorithm for financial analysis is called quantum amplitude estimation. It provides quadratic speed-up over currently known classical methods, a significant advantage in sufficiently large calculations. Today, anyone interested in quantum algorithms can implement them with open-source software and run simulations either on personal laptops or supercomputers. More interestingly, they can also be executed on actual quantum computers, such as Helmi, the first Finnish quantum computer. Helmi is connected to the LUMI supercomputer within the Finnish Quantum-Computing Infrastructure (FiQCI), a world-leading hybrid HPC+QC platform. FiQCI is jointly maintained, operated, and developed by VTT, Aalto University, and CSC. Quantum amplitude estimation Quantum amplitude estimation (QAE) was originally developed by Brassard et al. 1. It is a generalization of the famous Grover’s search algorithm and can be used to find the amplitude of a quantum state. It uses quadratically fewer queries than classical sampling methods. Thus, what would require a million queries using purely classical methods can ideally be done with just a thousand quantum queries. A hundred million classical queries correspond to only ten thousand quantum queries, and so on. The impatient reader, who just wants to know how all of this works in practice, can skip to the next section. The technically inclined should read on. The main idea of the QAE algorithm is to find the solutions to a problem from an unstructured set of possible answers. This set of answers is encoded into the amplitudes of quantum states in a register of quantum bits. Due to the unique properties of quantum computers, a single quantum query is enough to search the whole set of solutions, find the desired answer, and mark it by inverting the phase of the corresponding quantum state. We call this particular quantum state the “good state.” The marked phase, however, does not lead to quantum advantage on its own, as it does not show up in measurements. QAE solves this by increasing the amplitude of the “good state” through the process called amplitude amplification. When done correctly, this increases the probability of measuring the good state significantly. Converting the states into the computational basis using the inverse quantum Fourier transform allows classically mapping the measured states into the estimates of the amplitudes of the original states. Because the increased amplitude of the good state causes it to be measured more frequently, the correct answer is recovered with high probability. The efficiency of the amplitude amplification and its effect on the number of queries is the source of the quadratic increase in performance. Figure 1: A simple example of a QAE circuit based on Brassard's approach [1]. The “good state” is encoded into the objective (obj) qubit by quantum gate A. The amplitude amplification is implemented by repeatedly applying different powers of the Q operator. In this approach, the amplification operations are controlled by the register of evaluation qubits (eval) The cool thing about QAE is that adjusting the function that determines the initial amplitudes of the system allows one to calculate different moments of a probability distribution. This makes QAE an extremely versatile tool, as many problems are best approached from a probabilistic point of view. In finance, QAE can calculate, for example, expected profits and VaRs (Value at Risk). Below, we test how quantum amplitude estimation performs on real quantum hardware using examples of financial applications. The original approach to quantum amplitude estimation proposed by Brassard et al. requires too many qubits for the 5-qubit quantum processing unit (QPU) of Helmi. In addition, the approach involves a lot of two-qubit gates, making the algorithm too error-prone for current quantum computers. Here, we instead consider a variant of QAE called Maximum Likelihood Quantum Amplitude Estimation (MLQAE) 4 , which runs multiple iterations of much simpler circuits (see Figure 2). To make up for the simplified circuits, MLQAE uses additional postprocessing in the form of maximum likelihood estimation. Though not quite as mathematically rigorous as the original algorithm by Brassard, the MLQAE has been subject to much research and seems to keep up with quadratic speedup [4,5]. The good performance of the MLQAE is corroborated by Figure 3, where we present a comparison between QAE and MLQAE on Helmi. Where MLQAE achieves good accuracy, QAE suffers from a small number of available qubits in addition to a significant error arising from the large number of gates and long execution time of the circuit. On the other hand, the additional classical processing in MLQAE introduces some inaccuracy to the answer. Overall, given the advantages over Brassards QAE, MLQAE looks to be a good choice for implementing financial algorithms on near-term quantum computers. Figure 2: Using MLQAE to implement the circuit of Figure 1 consists of a batch of simpler circuits with different powers of amplification operations Q. Compared to the approach in Figure 1, the MLQAE requires no additional evaluation qubits Figure 3: Comparison between QAE and MLQAE. The amplitude error is plotted as a function of queries, which in QAE are related to the number of qubits and in MLQAE to the number of circuits. Note the different scales of y-axes The next step is to select a suitable problem. An interesting example algorithm was investigated by CSC and OP Labs in 2021 6. The algorithm estimated housing prices in Helsinki using previous years’ prices and the annual growth rates from the last ten years. The calculation itself was a simple multiplication of the growth and last year’s prices. To perform the quantum algorithm, one must bin the available data sets and encode corresponding probability distributions in two different quantum registers. The accuracy of this mapping scales exponentially with the number of qubits in the register, making it accurate with large quantum registers. The downside is that even though the algorithm technically can be run using MLQAE with a very low number of qubits, it might not reach decent accuracy. Using a simulator, the estimated relative error of this implementation is 2.5%. Of course, real-world problems go beyond simple multiplication. Option pricing is another potential application for quantum computing. An option is a contract that allows but does not obligate its owner to perform some specified transaction in the future. Setting a price for an option can be difficult, as the price depends on the future value of its underlying asset and the agreed conditions of the option itself. For certain type of options, the valuation process is computationally expensive, due to the uncertainty of future markets. We chose to implement a quantum algorithm for pricing the European call option, a simple but realistic type of an option. Despite its simplicity, the implementation can be extended for more complicated options with relative ease [2] (#references). The idea is to use MLQAE to calculate the expected value of the option’s payoff function. In other words, we use a quantum computer’s parallel properties to simultaneously evaluate the option’s different future prices, something which classically would be done separately. Here, a reality check is in order. Both housing price estimation and option pricing encounter challenges on a real QPU. The circuits required for more complex problems are too long for current noise levels. Displayed in Figure 4 are the results of the European option pricing done with MLQAE. A simulation converges faster than the classical sampling method. However, results with a real QPU fall apart due to the qubits’ loss of coherence. The good news is that these algorithms work as a concept even with quantum computers like Helmi, with only a few qubits available. Increased accuracy comes in time with more advanced hardware. Figure 4: Statistical mean of amplitude error for evaluating the European call option. Comparison between MLQAE algorithm on a simulator and the Helmi quantum computer, and a classical sampling Finance and quantum computers gather public interest Experts from different fields interested in gaining a quantum advantage in the future are now working to improve their knowledge and encouraging people to learn about quantum technologies. Last November, the Hanken School of Economics in Finland and Ultrahack organized a quantum hackathon for finance-oriented minds with the theme of sustainable finance in the quantum era. The competition saw many teams with different backgrounds tackling the challenge and coming up with innovative solutions to combine the financial sector with the capabilities of quantum computers. Taking part in supporting the event, CSC and VTT with colleagues from IQM, provided participants access to the 20-qubit Leena quantum computer through the LUMI supercomputer. It was great to see teams utilizing the Finnish quantum computer as a part of their projects. One such team, Qumpula Quantum, won 2nd place in the competition. A popular topic in this year’s hackathon was quantum machine learning, which was also implemented using Leena. Figure 5: CSC and VTT had teams mentoring at Hanken Quantum Hackathon. Pictured at the bottom left: Nikolas Klemola Tango, Olli Mukkula, Modupe Falodun and Jake Muff. Photos: Aleksi Leskinen Closing thoughts It is already possible to run straightforward quantum financial examples and see how they perform on real quantum hardware. As with quantum algorithms in general, it is vital to learn how to construct algorithms best suited for real quantum devices. Hybrid algorithms combining quantum and classical processing are promising, and machine learning models are expected to help optimize quantum circuits for noisy environments. Compared to the history of classical computing, quantum algorithms have been researched for such a short time that much remains to be discovered.It may well be that new and improved algorithms are discovered tomorrow. Research for algorithms cannot sit and wait for better quantum hardware. After all, a suitable combination of both is needed for quantum advantage. Those interested in the codes used for this blog can find the Jupyter notebooks with detailed explanations in the link below. They can be executed directly on the FiQCI infrastructure. More details Give feedback! Feedback is greatly appreciated! You can send feedback directly to fiqci-feedback@postit.csc.fi.	{"url":"https://fiqci.fi/_posts/2024-04-09-Financial-Algorithms/","timestamp":"2024-11-04T00:40:55Z","content_type":"text/html","content_length":"34599","record_id":"<urn:uuid:3b4048fd-cf65-4261-93cc-49ca699ac14f>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00892.warc.gz"}
Systems Biology and its tools - AGINGSCIENCES™ - Anti-Aging Firewalls™Systems Biology and its tools - AGINGSCIENCES™ - Anti-Aging Firewalls™ Systems Biology and its tools Posted on 13. May 2011 by Vince Giuliano Victor’s recent blog entry Living on the Brink of Chaos points to Systems Biology, a relatively new research perspective likely to be of increasing importance. Here, I introduces Systems Biology a bit more systematically and briefly characterize some of the many tools of mathematics and systems theory that may be used in it – tools traditionally considered to be useful outside the biological-life sciences. Systems Biology Systems biology characterizes an approach to understanding focused on patterns of interaction of systems components rather than the traditional reductionist research approach of focusing on one process, substance, gene or even subsystem at a time. “Proponents describe systems biology as a biology-based inter-disciplinary study field that focuses on complex interactions in biological systems , claiming that it uses a new perspective (holism instead of reduction). — An often stated ambition of systems biology is the modeling and discovery of emergent properties, properties of a system whose theoretical description is only possible using techniques which fall under the remit of systems biology(ref).” These techniques include mathematical methods for finding patterns in large diverse collections of data and approaches for building large complex computer models of biological systems. I describe several of such below. We already have many simple partial models of how things work in bodies relating to health and aging, examples being the role of microglia in neuropathic pain, longevity and the GH–IGF Axis, tumor suppression by the NRG1 gene, PGC-1alpha in the health-producing effects of exercise, how DAF-16 promotes longevity in nematodes, the cell-cycle roles of JDP2, and CETP gene longevity variants. These are a sample of mostly-qualitative models previously discussed in this blog, drawn out of a pool of thousands of such existing partial models. Some of these partial models are in themselves very complex and it is not clear whether and how how many of them fit together. Along with those simple models we have petabytes of possibly relevant data coming from association studies, genomic and other studies and next-generation sequencing technologies spewing out daily mountains of new data(ref). By the early-2000’s it was clear that there was a need for approaches to building higher-level quantitative models and develop new techniques for analyzing vast quantities of data. Thus arose the interest in Systems Biology. Another basic motivation for using Systems Biology approaches is that when it comes to considering health and disease states and aging, the relationships are far from simple and it is often not possible to say what is causing what. Very rarely can we simply and accurately state “A causes B.“ That is why genome-wide SNP-disease association studies have tended to show only disappointingly weak correlations. “Nevertheless, the inauguration of genome-wide association studies only magnifies the challenge of differentiating between the expected, true weak associations from the numerous spurious effects caused by misclassification, confounding and significance-chasing biases(ref).” Indeed, most health and disease states appear to come about through a time and sequence-dependent set of interactions among very large numbers of variables. The mTOR, SIRT1, AMPK, and IGF1 pathways all have to do with aging and longevity and themselves are incredibly complex. Yet, perturbations in any one of these pathways can affect the others as well. Thus, to discover what is going on, Systems Biology as a philosophy often draws on tools of systems modeling. The 2004 publication Search for organising principles: understanding in systems biology relates: “Due in large measure to the explosive progress in molecular biology, biology has become arguably the most exciting scientific field. The first half of the 21st century is sometimes referred to as the ‘era of biology’, analogous to the first half of the 20th century, which was considered to be the ‘era of physics’. Yet, biology is facing a crisis–or is it an opportunity–reminiscent of the state of biology in pre-double-helix time. The principal challenge facing systems biology is complexity. According to Hood, ‘Systems Biology defines and analyses the interrelationships of all of the elements in a functioning system in order to understand how the system works.’ With 30000+ genes in the human genome the study of all relationships simultaneously becomes a formidably complex problem.” The 2007 document The nature of systems biology puts it “The advent of functional genomics has enabled the molecular biosciences to come a long way towards characterizing the molecular constituents of life. Yet, the challenge for biology overall is to understand how organisms function. By discovering how function arises in dynamic interactions, systems biology addresses the missing links between molecules and physiology. Top-down systems biology identifies molecular interaction networks on the basis of correlated molecular behavior observed in genome-wide “omics” studies. Bottom-up systems biology examines the mechanisms through which functional properties arise in the interactions of known components.” Aging in particular is clearly a systems phenomenon. A search in Pubmed.org for papers relevant to “systems biology and aging” retrieves 862 entries. Shown here is a nice model of human aging, a diagrammatic network model developed by John D. Furber. A larger more-readable version of the diagram with accompanying discussion can be found here. Actually, this model is a qualitative macro-model aimed at enhancing understanding of the major aging pathways in humans. When it gets down to the molecular level and gene-epigenetics-promoter interactions, the complexity increases by orders of magnitudes. The challenge of systems biology requires the application of sophisticated modeling techniques. Effective models must handle immense amounts of data and be built so that they conform to fuzzy data sets where the exact relevancy of variables may not be known and where the variables considered may not include all those necessary to predict an effect. In many cases, dynamic modeling is needed. Time sequence of events may be critical. This is known to be the case when it comes to formation of cancers, for example. And a person’s epigenome and associated gene activation patterns evolve continuously over that person’s lifetime making what goes on age-dependent. Further, to effectively reflect what is going on in complex organisms like us, models must simultaneously function on multiple scales. The 2008 publication Multiscale modeling of biological pattern formation relates “In the past few decades, it has become increasingly popular and important to utilize mathematical models to understand how microscopic intercellular interactions lead to the macroscopic pattern formation ubiquitous in the biological world. Modeling methodologies come in a large variety and presently it is unclear what is their interrelationship and the assumptions implicit in their use. They can be broadly divided into three categories according to the spatial scale they purport to describe: the molecular, the cellular and the tissue scales. Most models address dynamics at the tissue-scale, few address the cellular scale and very few address the molecular scale. Of course there would be no dissent between models or at least the underlying assumptions would be known if they were all rigorously derived from a molecular level model, in which case the laws of physics and chemistry are very well known. However in practice this is not possible due to the immense complexity of the problem. A simpler approach is to derive models at a coarse scale from an intermediate scale model which has the special property of being based on biology and physics which are experimentally well studied.” The 2009 publication Multiscale modeling of cell mechanics and tissue organization relates “Nowadays, experimental biology gathers a large number of molecular and genetic data to understand the processes in living systems. Many of these data are evaluated on the level of cells, resulting in a changed phenotype of cells. Tools are required to translate the information on the cellular scale to the whole tissue, where multiple interacting cell types are involved. Agent-based modeling allows the investigation of properties emerging from the collective behavior of individual units. A typical agent in biology is a single cell that transports information from the intracellular level to larger scales. Mainly, two scales are relevant: changes in the dynamics of the cell, e.g. surface properties, and secreted molecules that can have effects at a distance larger than the cell diameter.” Mathematical and systems tools used in systems biology Many tools have been developed to help analyze and model situations where there are large numbers of related variables, messy data sets and fuzzy understanding of relationships. A number of these tools are based on use of sophisticated mathematical techniques like multivariate factor analysis. Others are computer-implemental simulation approaches. Such tools have been applied for decades across many disciplines such as electrical engineering, physics, economics, weather forecasting and social dynamics. Though almost all of these tools were developed outside of the biological sciences, we now have a situation where they are being embraced and used under the umbrella of Systems Biology. I mention some of the most important of these tools here. The text descriptions are mainly drawn from Wikipedia: 1. Polynomial regression – “In statistics, polynomial regression is a form of linear regression in which the relationship between the independent variable x and the dependent variable y is modeled as an nth order polynomial. Polynomial regression fits a nonlinear relationship between the value of x and the corresponding conditional mean of y, denoted E(y\|x), and has been used to describe nonlinear phenomena such as the growth rate of tissues^[1]” 2. Harmonic analysis – “Harmonic analysis is the branch of mathematics that studies the representation of functions or signals as the superposition of basic waves. It investigates and generalizes the notions of Fourier series and Fourier transforms. The basic waves are called “harmonics” (in physics), hence the name “harmonic analysis,” but the name “harmonic” in this context is generalized beyond its original meaning of integer frequency multiples. In the past two centuries, it has become a vast subject with applications in areas as diverse as signal processing, quantum mechanics, and 3. Correlation matrices – “The correlation matrix of n random variables X[1][,] …, X[n ]is the n Ã— n matrix whose i,j entry is corr(X[i,] X[j)]. If the measures of correlation used are product-moment coefficients, the correlation matrix is the same as the covariance matrix of the standardized random variables X[i ]/Ïƒ (X[i)] for i = 1, …, n. This applies to both the matrix of population correlations (in which case “Ïƒ ” is the population standard deviation), and to the matrix of sample correlations (in which case “Ïƒ ” denotes the sample standard deviation). Consequently, each is necessarily a positive-semidefinite matrix.” 4. Principal factor analysis – “Factor analysis is a statistical method used to describe variability among observed variables in terms of a potentially lower number of unobserved variables called factors. In other words, it is possible, for example, that variations in three or four observed variables mainly reflect the variations in a single unobserved variable, or in a reduced number of unobserved variables. Factor analysis searches for such joint variations in response to unobserved latent variables. The observed variables are modeled as linear combinations of the potential factors, plus “error” terms. The information gained about the interdependencies between observed variables can be used later to reduce the set of variables in a dataset.” 5. Data mining – “Data mining (the analysis step of the Knowledge Discovery in Databases process, or KDD), a relatively young and interdisciplinary field of computer science,^[1][2] is the process of extracting patterns from large data sets by combining methods from statistics and artificial intelligence with database management.^[3] ” 6. Cellular automata – “A cellular automaton (pl. cellular automata, abbrev. CA) is a discrete model studied in computability theory, mathematics, physics, complexity science, theoretical biology and microstructure modeling. It consists of a regular grid of cells, each in one of a finite number of states, such as “On” and “Off” (in contrast to a coupled map lattice). The grid can be in any finite number of dimensions. For each cell, a set of cells called its neighborhood (usually including the cell itself) is defined relative to the specified cell.” 7. Complex adaptive systems – “Complex adaptive systems are special cases of complex systems. They are complex in that they are dynamic networks of interactions and relationships not aggregations of static entities. They are adaptive in that their individual and collective behaviour changes as a result of experience.^[1] ” 8. Process calculus “– the process calculi (or process algebras) are a diverse family of related approaches to formally modelling concurrent systems. Process calculi provide a tool for the high-level description of interactions, communications, and synchronizations between a collection of independent agents or processes. They also provide algebraic laws that allow process descriptions to be manipulated and analyzed, and permit formal reasoning about equivalences between processes (e.g., using bisimulation).” 9. Computational complexity theory – “Computational complexity theory is a branch of the theory of computation in theoretical computer science and mathematics that focuses on classifying computational problems according to their inherent difficulty. In this context, a computational problem is understood to be a task that is in principle amenable to being solved by a computer (which basically means that the problem can be stated by a set of mathematical instructions). Informally, a computational problem consists of problem instances and solutions to these problem instances. 10. Fractal mathematics – “A mathematical fractal is based on an equation that undergoes iteration, a form of feedback based on recursion.^[2] There are several examples of fractals, which are defined as portraying exact self-similarity, quasi self-similarity, or statistical self-similarity. While fractals are a mathematical construct, they are found in nature, which has led to their inclusion in artwork. They are useful in medicine, soil mechanics, seismology, and technical analysis.” 11. Chaos theory – Chaos theory is a field of study in applied mathematics, with applications in several disciplines including physics, economics, biology, and philosophy. Chaos theory studies the behavior of dynamical systems that are highly sensitive to initial conditions; an effect which is popularly referred to as the butterfly effect. Small differences in initial conditions (such as those due to rounding errors in numerical computation) yield widely diverging outcomes for chaotic systems, rendering long-term prediction impossible in general.^[1] This happens even though these systems are deterministic, meaning that their future behavior is fully determined by their initial conditions, with no random elements involved.^[2] In other words, the deterministic nature of these systems does not make them predictable.^[3]^[4] This behavior is known as deterministic chaos, or simply chaos.” 12. Dynamical systems theory – “Dynamical systems theory is an area of applied mathematics used to describe the behavior of complex dynamical systems, usually by employing differential equations or difference equations. When differential equations are employed, the theory is called continuous dynamical systems. When difference equations are employed, the theory is called discrete dynamical systems. When the time variable runs over a set which is discrete over some intervals and continuous over other intervals or is any arbitrary time-set such as a cantor set then one gets dynamic equations on time scales.” Sophisticated software programs like Vensim allow dynamic modeling of systems with hundreds of variables. 14. Agent-based modeling – “Agent-based models have many applications in biology, primarily due to the characteristics of the modeling method. Agent-based modeling is a rule-based, computational modeling methodology that focuses on rules and interactions among the individual components or the agents of the system.^[1] The goal of this modeling method is to generate populations of the system components of interest and simulate their interactions in a virtual world. Agent-based models start with rules for behavior and seek to reconstruct, through computational instantiation of those behavioral rules, the observed patterns of behavior.^[1]” 16. Stochastic resonance – “Stochastic resonance (SR) is a phenomenon that occurs in a threshold measurement system (e.g. a man-made instrument or device; a natural cell, organ or organism) when an appropriate measure of information transfer (signal-to-noise ratio, mutual information, coherence, d, etc.) is maximized in the presence of a non-zero level of stochastic input noise thereby lowering the response threshold;^[1] the system resonates at a particular noise level.” 17. Coupling of models – The 2005 publication Modelling biological complexity: a physical scientist’s perspective suggests another approach, which is coupling of models. “From the perspective of a physical scientist, it is especially interesting to examine how the differing weights given to philosophies of science in the physical and biological sciences impact the application of the study of complexity. We briefly describe how the dynamics of the heart and circadian rhythms, canonical examples of systems biology, are modelled by sets of nonlinear coupled differential equations, which have to be solved numerically. A major difficulty with this approach is that all the parameters within these equations are not usually known. Coupled models that include biomolecular detail could help solve this problem. Coupling models across large ranges of length- and time-scales is central to describing complex systems and therefore to biology. Such coupling may be performed in at least two different ways, which we refer to as hierarchical and hybrid multiscale modelling. While limited progress has been made in the former case, the latter is only beginning to be addressed systematically. These modelling methods are expected to bring numerous benefits to biology, for example, the properties of a system could be studied over a wider range of length- and time-scales, a key aim of Systems Biology. Multiscale models couple behaviour at the molecular biological level to that at the cellular level, thereby providing a route for calculating many unknown parameters as well as investigating the effects at, for example, the cellular level, of small changes at the biomolecular level, such as a genetic mutation or the presence of a drug.” All of the above approaches and many more are covered under the umbrella of Computational biology. “Computational biology involves the development and application of data-analytical and theoretical methods, mathematical modeling and computational simulation techniques to the study of biological, behavioral, and social systems.^[1] The field is widely defined and includes foundations in computer science, applied mathematics, statistics, biochemistry, chemistry, biophysics, molecular biology, genetics, ecology, evolution, anatomy, neuroscience, and visualization.^[2] ” Wrapping it up Systems Biology is more of a philosophical framework for developing understanding of complex biological relationships than it is a technique or discipline. The framework emphasizes viewing biological creatures as being complex systems developing in time where all components and their properties influence all others via a large multiplicity of interacting feedback paths. Another important aspect of Systems Biology is searching for meaningful patterns in very large amounts of data such a produced by collections of whole-genome disease-association studies. Systems Biology entails the introduction of new thinking paradigms into biology, ones involving the use of sophisticated mathematics and highly technical computer modeling tools and looking for meaningful relationships through analysis of vast mountains of data. 2 Responses to Systems Biology and its tools 1. at home income says: I admire the valuable information and facts you offer inside your posts. Ill bookmark your weblog and also have my children examine up right here typically. Im fairly positive theyll discover a lot of new things here than anybody else! 2. Mahiraj Ramson says: After looking through this particular blog post I have made the decision to sign up to your rss feed. I expect your coming content will certainly turn out to be just as good. Leave a Reply Cancel reply You must be logged in to post a comment. About Vince Giuliano Being a follower, connoisseur, and interpreter of longevity research is my latest career, since 2007. I believe I am unique among the researchers and writers in the aging sciences community in one critical respect. That is, I personally practice the anti-aging interventions that I preach and that has kept me healthy, young, active and highly involved at my age, now 93. I am as productive as I was at age 45. I don’t know of anybody else active in that community in my age bracket. In particular, I have focused on the importance of controlling chronic inflammation for healthy aging, and have written a number of articles on that subject in this blog. In 2014, I created a dietary supplement to further this objective. In 2019, two family colleagues and I started up Synergy Bioherbals, a dietary supplement company that is now selling this product. In earlier reincarnations of my career. I was Founding Dean of a graduate school and a full University Professor at the State University of New York, a senior consultant working in a variety of fields at Arthur D. Little, Inc., Chief Scientist and C00 of Mirror Systems, a software company, and an international Internet consultant. I got off the ground with one of the earliest PhD's from Harvard in a field later to become known as computer science. Because there was no academic field of computer science at the time, to get through I had to qualify myself in hard sciences, so my studies focused heavily on quantum physics. In various ways I contributed to the Computer Revolution starting in the 1950s and the Internet Revolution starting in the late 1980s. I am now engaged in doing the same for The Longevity Revolution. I have published something like 200 books and papers as well as over 430 substantive.entries in this blog, and have enjoyed various periods of notoriety. If you do a Google search on Vincent E. Giuliano, most if not all of the entries on the first few pages that come up will be ones relating to me. I have a general writings site at www.vincegiuliano.com and an extensive site of my art at www.giulianoart.com. Please note that I have recently changed my mailbox to This entry was posted in Uncategorized. Bookmark the permalink.	{"url":"http://www.anti-agingfirewalls.com/2011/05/13/systems-biology-and-its-tools/","timestamp":"2024-11-03T09:59:10Z","content_type":"text/html","content_length":"138763","record_id":"<urn:uuid:534f4d44-8575-4355-aabd-354a8b7b2016>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00486.warc.gz"}
Get instant live expert help on how do i multiply in excel “My Excelchat expert helped me in less than 20 minutes, saving me what would have been 5 hours of work!” I currently have the function =COUNTIF(L11:NJ11,"O") that counts all the cells in the row containing the letter "O." Now, I want to multiply all the cells falling in Columns titled "Mon" and "Fri" by 2. How do I do this? I am trying to do a simple multiplecation formula. I want to take one number so lets say A10 and then multiply it by 1.1 and then have that number be outputted to A11 however i want to have it continually multiply the new number by 1.1 many times but I can't figure out how to get excel to put multiple numbers down. which key do I use to multiply one cell with another? How can I multiply the number of blank cells in a row by 5 using conditional formatting?	{"url":"https://www.got-it.ai/solutions/excel-chat/excel-help/how-to/do/how-do-i-multiply-in-excel","timestamp":"2024-11-02T04:55:19Z","content_type":"text/html","content_length":"337141","record_id":"<urn:uuid:1d3b2cd1-eb07-49f3-a97f-423a1a94ae13>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00496.warc.gz"}
Geometric approach to analytic marginalisation of the likelihood ratio for continuous gravitational wave searches The likelihood ratio for a continuous gravitational wave signal is viewed geometrically as a function of the orientation of two vectors; one representing the optimal signal-to-noise ratio, and the other representing the maximised likelihood ratio or F-statistic. Analytic marginalisation over the angle between the vectors yields a marginalised likelihood ratio, which is a function of the F-statistic. Further analytic marginalisation over the optimal signal-to-noise ratio is explored using different choices of prior. Monte-Carlo simulations show that the marginalised likelihood ratios had identical detection power to the F-statistic. This approach demonstrates a route to viewing the F-statistic in a Bayesian context, while retaining the advantages of its efficient computation. Dive into the research topics of 'Geometric approach to analytic marginalisation of the likelihood ratio for continuous gravitational wave searches'. Together they form a unique fingerprint.	{"url":"https://researchportalplus.anu.edu.au/en/publications/geometric-approach-to-analytic-marginalisation-of-the-likelihood-","timestamp":"2024-11-13T05:37:07Z","content_type":"text/html","content_length":"53185","record_id":"<urn:uuid:bfa4768e-955a-412b-a0dd-1139f562b0e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00311.warc.gz"}
tf.switch_case \| TensorFlow v2.15.0.post1 Create a switch/case operation, i.e. View aliases Compat aliases for migration See Migration guide for more details. branch_index, branch_fns, default=None, name='switch_case' an integer-indexed conditional. See also tf.case. This op can be substantially more efficient than tf.case when exactly one branch will be selected. tf.switch_case is more like a C++ switch/case statement than tf.case, which is more like an if/elif/ elif/else chain. The branch_fns parameter is either a dict from int to callables, or list of (int, callable) pairs, or simply a list of callables (in which case the index is implicitly the key). The branch_index Tensor is used to select an element in branch_fns with matching int key, falling back to default if none match, or max(keys) if no default is provided. The keys must form a contiguous set from 0 to len(branch_fns) - 1. tf.switch_case supports nested structures as implemented in tf.nest. All callables must return the same (possibly nested) value structure of lists, tuples, and/or named tuples. switch (branch_index) { // c-style switch case 0: return 17; case 1: return 31; default: return -1; branches = {0: lambda: 17, 1: lambda: 31} branches.get(branch_index, lambda: -1)() def f1(): return tf.constant(17) def f2(): return tf.constant(31) def f3(): return tf.constant(-1) r = tf.switch_case(branch_index, branch_fns={0: f1, 1: f2}, default=f3) # Equivalent: tf.switch_case(branch_index, branch_fns={0: f1, 1: f2, 2: f3}) branch_index An int Tensor specifying which of branch_fns should be executed. branch_fns A dict mapping ints to callables, or a list of (int, callable) pairs, or simply a list of callables (in which case the index serves as the key). Each callable must return a matching structure of tensors. default Optional callable that returns a structure of tensors. name A name for this operation (optional). The tensors returned by the callable identified by branch_index, or those returned by default if no key matches and default was provided, or those returned by the max-keyed branch_fn if no default is TypeError If branch_fns is not a list/dictionary. TypeError If branch_fns is a list but does not contain 2-tuples or callables. TypeError If fns[i] is not callable for any i, or default is not callable.	{"url":"https://tensorflow.google.cn/versions/r2.15/api_docs/python/tf/switch_case","timestamp":"2024-11-04T01:29:46Z","content_type":"text/html","content_length":"47773","record_id":"<urn:uuid:51dbbff5-56fd-4b27-9aeb-7cec3b6034fb>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00253.warc.gz"}
Column Standard Deviation in R (Step by Step) - Data Science Parichay The R programming language comes with a number of helpful functions to work with the data stored in data structures like vectors, lists, dataframes, etc. In this tutorial, we will look at one such function that helps us get the standard deviation of the values in a column of an R dataframe. How to get the standard deviation of an R dataframe column? You can use the built-in sd() function in R to compute the standard deviation of values in a dataframe column. Pass the column values as an argument to the function. The following is the syntax – It returns the standard deviation of the passed vector. Steps to compute the standard deviation of values in an R column Let’s now look at a step-by-step example of using the above syntax to compute the std dev of a numeric column in R. Step 1 – Create a dataframe First, we will create an R dataframe that we will be using throughout this tutorial. # create a dataframe employees_df = data.frame( "Name"= c("Jim", "Dwight", "Angela", "Tobi", "Kevin"), "Age"= c(26, 28, 29, 32, 30) # display the dataframe Team_A Team_B We now have a dataframe containing information about some employees in an office. The dataframe has two columns – “Name” and “Age”. 📚 Data Science Programs By Skill Level Introductory ⭐ Intermediate ⭐⭐⭐ Advanced ⭐⭐⭐⭐⭐ 🔎 Find Data Science Programs 👨💻 111,889 already enrolled Disclaimer: Data Science Parichay is reader supported. When you purchase a course through a link on this site, we may earn a small commission at no additional cost to you. Earned commissions help support this website and its team of writers. Step 2 – Calculate the standard deviation column values using the sd() function To calculate the standard deviation of values in a column, pass the column values as an argument to the sd() function. You can use the [[]] notation to access the values of a column. Let’s compute the standard deviation in the “Age” column. # std dev in "Age" column sd_age = sd(employees_df[["Age"]]) # display the std dev [1] 2.236068 We get the standard deviation in the “Age” column as 2.236068. Standard deviation of a column with NA values in R What if there are NA values in a column? Let’s find out. # create a dataframe employees_df = data.frame( "Name"= c("Jim", "Dwight", "Angela", "Tobi", "Kevin"), "Age"= c(26, 28, NA, 32, 30) # display the dataframe Name Age 1 Jim 26 2 Dwight 28 3 Angela NA 4 Tobi 32 5 Kevin 30 Here, we created a new dataframe such that the “Age” column now contains some NA values. Now, let’s apply the sd() function to the “Age” column. # std dev in "Age" column sd_age = sd(employees_df[["Age"]]) # display the std dev [1] NA We get NA as the standard deviation for the “Age” column. This happened because performing any mathematical operation with NA results in an NA in R. If you want to calculate the standard deviation of a column with NA values, pass na.rm=TRUE to the sd() function to skip the NA values when computing the standard deviation. # std dev in "Age" column sd_age = sd(employees_df[["Age"]], na.rm=TRUE) # display the std dev [1] 2.581989 We now get the standard deviation of the “Age” column excluding the NA values. Summary – Standard Deviation of Column Values in R In this tutorial, we looked at how to compute the standard deviation of values in a column of an R dataframe. The following is a short summary of the steps – 1. Create a dataframe (skip this step if you already have a dataframe on which you want to operate). 2. Use the sd() function to compute the standard deviation of column values. Pass the column values vector as an argument. 3. If your column contains any NA values, pass na.rm=TRUE to the sd() function to calculate the standard deviation excluding the NA values in the column. You might also be interested in – Subscribe to our newsletter for more informative guides and tutorials. We do not spam and you can opt out any time.	{"url":"https://datascienceparichay.com/article/column-standard-deviation-in-r/","timestamp":"2024-11-13T18:32:03Z","content_type":"text/html","content_length":"262254","record_id":"<urn:uuid:bbb8f25a-decd-45f8-b8d9-3aaa77161c37>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00494.warc.gz"}
In a seminar, the number of participants in Hindi, English, and Mathematics are $60,84$ and $108$ respectively. Find the minimum number of rooms required if, in each room, the same number of participants are to be seated and all of them being in the same subject. Hint: We have to find a minimum number of participants. For that, you need to find the total number of participants. Also, find the HCF of $60,84$ and $108$. Use the formula\[Number\text{ }of\text{ } rooms\text{ }required=\dfrac{total\text{ }number\text{ }of\text{ }participants}{12}\]. Complete step-by-step answer: The greatest number which divides each of the two or more numbers is called HCF or Highest Common Factor. It is also called the Greatest Common Measure(GCM) and Greatest Common Divisor(GCD). HCF and LCM are two different methods, whereas LCM or Least Common Multiple is used to find the smallest common multiple of any two or more numbers. Follow the below-given steps to find the HCF of numbers using the prime factorization method. Step 1: Write each number as a product of its prime factors. This method is called here prime factorization. Step 2: Now list the common factors of both the numbers. Step 3: The product of all common prime factors is the HCF ( use the lower power of each common factor). The largest number that divides two or more numbers is the highest common factor (HCF) for those numbers. For example, consider the numbers $36({{2}^{2}}\times {{3}^{2}}),42(2\times 3\times 7)$. $3$ is the largest number that divides each of these numbers, and hence, is the HCF for these numbers. HCF is also known as Greatest Common Divisor (GCD) To find the HCF of two or more numbers, express each number as a product of prime numbers. The product of the least powers of common prime terms gives us the HCF. In the question, the number of rooms will be minimum if each room accommodates the maximum number of participants, since in each room, the same number of participants are to be seated and all of them must be of the same subject. Therefore, the number of participants in each room must be the HCF of $60,84$ and $108$ . The prime factorizations of $60,84$ and $108$ are as under. $60={{2}^{2}}\times 3\times 5$ $84={{2}^{2}}\times 3\times 7$ $108={{2}^{2}}\times {{3}^{2}}$ So HCF for $60,84$ and$108$ is ${{2}^{2}}\times 3=12$. \[Number\text{ }of\text{ }rooms\text{ }required=\dfrac{total\text{ }number\text{ }of\text{ }participants}{12}\] \[Number\text{ }of\text{ }rooms\text{ }required=\dfrac{60+84+108}{12}\] \[Number\text{ }of\text{ }rooms\text{ }required=\dfrac{252}{12}=21\] So the minimum number of rooms is $21$. Note: Carefully read the question. You should know the concepts related to HCF. Also, you must know that HCF can be solved by two methods: prime factorization and division method. Don’t make confusion while writing the HCF. Don’t miss any of the terms. In division method divide the largest number by the smallest number of the given numbers until the remainder is zero. The last divisor will be the HCF of given numbers.	{"url":"https://www.vedantu.com/question-answer/in-a-seminar-the-number-of-participants-in-hindi-class-7-mathematics.-cbse-5ee705f147f3231af244762b","timestamp":"2024-11-04T18:47:48Z","content_type":"text/html","content_length":"155749","record_id":"<urn:uuid:e09bf1c7-c0c8-48e6-b8a1-031945c1915e>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00194.warc.gz"}
Baccarat Rules Sep 08 2018 Baccarat Policies Baccarat is played with 8 decks of cards. Cards which are of a value less than 10 are said to be at their printed value while at the same time ten, J, Q, K are 0, and A are each equal to 1. Wagers are placed on the ‘banker,’ the ‘player’ or for a tie (these aren’t actual persons; they simply portray the 2 hands to be given out). Two hands of two cards will now be dealt to the ‘banker’ and ‘player’. The score for any hand will be the sum total of the two cards, but the initial digit is removed. For example, a hand of 7 and 5 produces a tally of two (7plusfive=twelve; drop the ‘1′). A 3rd card may be given out depending on the following guidelines: - If the bettor or banker has a tally of 8 or 9, then both bettors stand. - If the player has five or lower, he hits. bettors stand otherwise. - If gambler stands, the banker hits of 5 or lesser. If the player hits, a chart will be used to judge if the banker stands or hits. Baccarat Odds The greater of the 2 scores is the winner. Successful stakes on the banker payout 19 to 20 (even money minus a 5% commission. Commission is followed closely and paid out when you leave the table so ensure that you have dollars still before you leave). Bets on the player that end up winning pay one to 1. Winning bets for tie usually pay 8 to one and occasionally 9 to 1. (This is a crazy bet as ties will happen less than one every ten hands. Avoid wagering on a tie. Nevertheless odds are far better – 9 to 1 vs. 8 to 1) Played properly, baccarat offers generally good odds, away from the tie bet of course. Baccarat Strategy As with just about all games, Baccarat has some established false impressions. One of which is quite similar to a misconception of roulette. The past is surely not an indicator of future events. Monitoring of last outcomes on a chart is definitely a waste of paper as well as an insult to the tree that gave its life for our stationary needs. The most commonly used and almost certainly most successful strategy is the one-3-two-6 technique. This schema is used to increase successes and minimizing risk. start by wagering 1 unit. If you win, add 1 more to the 2 on the table for a total of three on the second bet. If you win you will have six on the table, remove four so you have two on the 3rd bet. If you win the third wager, add 2 to the four on the table for a sum total of 6 on the fourth wager. If you lose on the first bet, you suck up a loss of one. A win on the 1st bet quickly followed by loss on the 2nd brings about a loss of 2. Wins on the 1st 2 with a loss on the 3rd gives you a profit of 2. And wins on the first 3 with a loss on the 4th mean you come out even. Arriving at a win on all four bets leaves you with 12, a profit of 10. This means you can lose the 2nd bet 5 times for every successful streak of four bets and still break even. You must be logged in to post a comment.	{"url":"http://fastplayingaction.com/2018/09/08/baccarat-rules-15/","timestamp":"2024-11-04T11:56:46Z","content_type":"application/xhtml+xml","content_length":"26325","record_id":"<urn:uuid:7004412f-99c1-4da5-9ee6-54d9fb866161>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00373.warc.gz"}
Is it possible to terminate sample_qubo when it finds a target solution ? I am using LeapHybridSampler for solving a QUBO problem by sample_qubo. What I want to do is: 1. give a target energy value to sample_qubo, and 2. terminate sample_qubo as soon as it finds a solution with the target energy value or less, because I want to evaluate the time-to-solution for the target energy. Is this possible ? If not, is there an alternative way to evaluate the time-to-solution of LeapHybridSampler ? 5 comments • Hi Koji, Unfortunately, there isn't a feature available at the moment to terminate sample_qubo when a solution with a target energy is found. To help us isolate potential alternative methods, could you please provide any steps you have taken so far to evaluate the time-to-solution for the specific problem? While it sounds like an interesting use case for a feature request, could you please tell us a bit more about what you are looking to solve with the feature? Best Regards, Comment actions • Hi, Tanjid, Thank you very much for your comments. I am now evaluating the performance of several QUBO solvers including LeapHybridSampler. I want to evaluate the time-to-solution(tts) of QUBO problems generated by several kinds of optimization problems, for which the optimal solutions are known. Since the optimal solutions of them are known, it makes sense to evaluate the tts to obtain the optimal solutions. I can obtain the tts of the other QUBO solvers, but cannot obtain that of LeapHybridSampler. Also, I think this feature is mandatory for practical use of QUBO solvers. Probably, users want to specify both (A) time limit and (B) target energy when they start a QUBO solver. A QUBO solver terminates as soon as it finds a solution less than or equal to the target energy before the time limit. If the time limit exceeds, it just outputs the best solution obtained in the time limit. For a user who needs a solution of a target energy as soon as possible and the computing time is limited, this feature is very helpful. Best regards, Comment actions • Hi Koji, Thank you for providing details around the use case. We have put in a feature request with our development team. In HSS (Hybrid Solver Services) BQM Sampler, the answer is slightly different for each time the problem is run, implying that the longer the solver runs, the better the average solution quality A feasible approach is by creating your own problems in the same problem class with an identified target energy and benchmarking your problems to obtain TTS (time-to-solution) from gathered metrics for probability of success for different time limits where the solution is found at the target energy. A hypothetical example: On an experiment with 10 attempts to run solvers for each time limit, double the time limit until the probability of success is at a rate that you are comfortable with, and you can roughly translate those data into TTS (time-to-solution) by calculating time_limit ÷ probability_success: □ 3s: 0/10 success (3s÷(0/10)=infinite) 6s: 1/10 success (6s÷(1/10)=60s) 12s: 9/10 success (12s÷(9/10)=13s) In these cases, the last set of tests at 12s is the one with the smallest TTS. Here are resources that may be beneficial for your understanding of calculating TTS: Although, these methods are a bit QPU centric, they explain the concept well enough to build an understanding of how you can derive a similar method for HSS-BQM Best Regards, Comment actions • Dear Tanjid, Thank you very much for your reply. I understand how we can estimate the TTS. However, it is too costly to get such data to estimate the TTS. It would be nice if the API accepts a target energy, we can get the TTS for it directly. Comment actions • Hi Koji, Thank you for the details. We have already submitted a feature request based on your request and included additional information regarding the cost of finding the solution. With kind regards, Comment actions Please sign in to leave a comment.	{"url":"https://support.dwavesys.com/hc/en-us/community/posts/5736876358039-Is-it-possible-to-terminate-sample-qubo-when-it-finds-a-target-solution","timestamp":"2024-11-07T10:01:56Z","content_type":"text/html","content_length":"50455","record_id":"<urn:uuid:83b75011-8891-4402-a585-db4084dfa782>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00244.warc.gz"}
Mastering Formulas In Excel: What Is The Formula For A Triangle Mastering formulas in Excel is crucial for anyone looking to efficiently analyze and manipulate data. Whether you are a student, a professional, or a business owner, having a strong grasp of formulas can significantly increase your productivity and accuracy. Today, we will take a closer look at one specific formula: the formula for a triangle. Preview: The formula for a triangle involves a straightforward calculation that can be easily implemented in Excel. Understanding this formula can be beneficial for various tasks, such as calculating the area or perimeter of a triangle within a dataset. Key Takeaways • Mastering formulas in Excel is crucial for efficient data analysis and manipulation. • Understanding the formula for a triangle can be beneficial for calculating its area or perimeter within a dataset. • The Pythagorean Theorem, SUM function, Sine function, POWER function, and SQRT function are all essential for working with triangle formulas in Excel. • Practicing and experimenting with triangle formulas in Excel can improve productivity and accuracy. • Having a strong grasp of Excel formulas can benefit students, professionals, and business owners alike. Understanding the Pythagorean Theorem Explanation of the Pythagorean Theorem: The Pythagorean Theorem is a fundamental principle in mathematics that relates to the sides of a right-angled triangle. It states that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In mathematical terms, it can be expressed as a^2 + b^2 = c^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides. Application of the theorem in Excel: In Excel, the Pythagorean Theorem can be used to calculate the length of the hypotenuse or one of the other sides of a right-angled triangle, given the lengths of the other two sides. This can be particularly useful in various fields such as engineering, physics, and architecture, where right-angled triangles are commonly encountered. Examples of using the Pythagorean Theorem in Excel: • Example 1: Imagine we have a right-angled triangle with sides of length 3 and 4. To calculate the length of the hypotenuse, we can use the formula =SQRT(3^2+4^2), which will give us the result of • Example 2: In another scenario, if we have the length of the hypotenuse (5) and one of the other sides (3), we can use the formula =SQRT(5^2-3^2) to calculate the length of the remaining side, which will give us the result of 4. Mastering Formulas in Excel: What is the formula for a triangle When it comes to mastering formulas in Excel, the SUM function is a powerful tool that can be used to perform various calculations, including those related to geometric shapes such as triangles. In this chapter, we will explore the application of the SUM function in calculating the perimeter of a triangle and provide illustrative examples to demonstrate its usage. Description of the SUM function in Excel The SUM function in Excel is used to add up the values in a range of cells. It can be applied to a single column or row, as well as to multiple columns or rows. By entering the relevant cell references or range, the SUM function will automatically calculate the total sum of the values within the specified range. How to apply the SUM function to calculate the perimeter of a triangle Calculating the perimeter of a triangle involves adding up the lengths of its three sides. Using the SUM function, this can be achieved by entering the cell references or values representing the lengths of the sides into the function. For example, if the lengths of the sides of a triangle are represented by cells A1, A2, and A3, the formula for calculating the perimeter would be =SUM(A1:A3). Illustrative examples of using the SUM function for triangle formulas To further illustrate the application of the SUM function in calculating the perimeter of a triangle, consider the following example: • Side 1: 5 units • Side 2: 7 units • Side 3: 9 units Using the SUM function, the formula for calculating the perimeter would be =SUM(5, 7, 9), which results in a perimeter of 21 units. This demonstrates how the SUM function can be utilized to efficiently perform calculations related to triangle formulas in Excel. Utilizing the Sine function The Sine function is a trigonometric function that relates the angle of a right-angled triangle to the ratio of the length of the side opposite the angle to the length of the hypotenuse. It is denoted as sin and is widely used in mathematics and Excel for various calculations. Explanation of the Sine function The Sine function is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right-angled triangle. In mathematical terms, it can be expressed as: sin(θ) = opposite/hypotenuse Where θ is the angle of the triangle, opposite is the side opposite to the angle, and hypotenuse is the longest side of the triangle which is always opposite the right angle. Application of the Sine function to calculate the area of a triangle In Excel, the Sine function can be used to calculate the area of a triangle when the length of two sides and the angle between them are known. By using the formula: Area = 0.5 * a * b * sin(θ) Where a and b are the lengths of the two sides of the triangle, and θ is the angle between them. Step-by-step guide on using the Sine function for triangle formulas Here's a step-by-step guide on how to utilize the Sine function in Excel to calculate the area of a triangle: • Step 1: Input the lengths of the two sides of the triangle into separate cells in Excel. • Step 2: Input the value of the angle between the two sides into a separate cell. • Step 3: Use the Sine function to calculate the area of the triangle by entering the formula =0.5 * A1 * B1 * SIN(RADIANS(C1)), where A1 and B1 are the cells containing the lengths of the sides, and C1 is the cell containing the angle value. • Step 4: Press enter to get the calculated area of the triangle. Incorporating the POWER function When it comes to mastering formulas in Excel, the POWER function is a powerful tool that can be used to perform calculations involving exponents. This function allows you to raise a number to a specific power, making it a valuable asset in solving mathematical equations. Description of the POWER function in Excel The POWER function in Excel is used to raise a number to a specified power. It takes two arguments: the base number and the exponent. The syntax for the POWER function is =POWER(number, power). How to use the POWER function to find the area of a triangle One way to utilize the POWER function in the context of a triangle is to calculate the area using the formula A = 0.5 * base * height. To incorporate the POWER function, you can raise the height to the power of 2 to find the area of the triangle. Examples of applying the POWER function for triangle formulas For example, if the base of a triangle is 5 units and the height is 8 units, you can use the POWER function to calculate the area as follows: A = 0.5 * 5 * POWER(8, 2) = 0.5 * 5 * 64 = 160 square Another example could involve finding the hypotenuse of a right-angled triangle using the Pythagorean theorem. By using the POWER function to square the two shorter sides and then summing them together, you can easily find the length of the hypotenuse. Exploring the SQRT function When it comes to mastering formulas in Excel, the SQRT function is an essential tool for calculating the square root of a number. It is particularly useful in geometric calculations, such as finding the height of a triangle. A. Explanation of the SQRT function The SQRT function in Excel is used to calculate the square root of a given number. It takes a single argument, which is the number for which you want to find the square root. The syntax for the SQRT function is =SQRT(number). B. How to utilize the SQRT function to find the height of a triangle One of the most common applications of the SQRT function in geometry is to find the height of a triangle. The formula for calculating the height (h) of a triangle with the base (b) and the area (A) is: h = 2A / b. In this formula, the area (A) can be calculated using the formula A = (b * h) / 2. C. Step-by-step guide on using the SQRT function for triangle formulas Step 1: Enter the known values Start by entering the known values of the base (b) and the area (A) into separate cells in your Excel worksheet. Step 2: Calculate the height using the formula Next, use the formula for calculating the height of the triangle: h = 2A / b. Enter this formula into another cell in your worksheet, using the SQRT function to calculate the square root of the Step 3: Utilize the SQRT function Use the SQRT function to find the square root of the result from the previous step. This will give you the height of the triangle. By following these steps and utilizing the SQRT function, you can easily calculate the height of a triangle in Excel, making geometric calculations a breeze. As we wrap up, it is clear that mastering formulas in Excel is crucial for any professional looking to streamline their data analysis and reporting. In this blog post, we've explored the different Excel functions for calculating triangle properties, including area and perimeter. I encourage you to practice and experiment with these formulas in Excel to familiarize yourself with their application in various scenarios. With dedication and hands-on experience, you'll be well on your way to becoming an Excel formula expert. ONLY $99 Immediate Download MAC & PC Compatible Free Email Support	{"url":"https://dashboardsexcel.com/blogs/blog/mastering-formulas-in-excel-what-is-the-formula-for-a-triangle","timestamp":"2024-11-13T01:01:11Z","content_type":"text/html","content_length":"213489","record_id":"<urn:uuid:83677352-dfdd-45aa-af89-ab66b9eb2eb2>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00139.warc.gz"}
Anna University B.E CSE PH6151 Engineering Physics Question Bank University Anna University Download Question Bank Department B.E Computer Science Subject & Subject Code PH 6151 Engineering Physics Regulation 2018 Year Apr/May 2019 Official Website https://library.annauniv.edu/index.php Anna University Question Bank Anna University B.E Computer Science PH 6151 Engineering Physics Question Bank Download All Anna University BE Computer Science Engineering Question Bank Here Download Anna University Engineering Physics Question Bank PH 6151 Engineering Physics Regulation 2018 Download Question Bank Here Anna University Engineering Physics Question Bank Part – A :(10 x 2 = 20 marks) 1. List any four factors that affect elasticity of a material. 2. Define Simple Harmonic Motion. 3. What is noise? Give an example. 4. Why does a glass bottle break when you pour hot water in it? 5. Define Diffraction. Give an example. 6. For a semiconductor laser, the bandgap is 0.9 eV. What is the wavelength of light emitted from it? 7. What is Total Internal Reflection? 8. What is a debroglie wave? 9. State Bragg’s law. 10. What is epitaxy? Part-B : (8 x 8 = 64 marks)(Answer any 8 questions) 11. State Hooks law of elasticity. Draw stress-strain diagram and discuss the behavior of aductile material under loading 12. Describe a method to find moment of inertia of a disc using torsion pendulum. 13. Explain how ultrasonic waves can be produced by piezo electric method. 14. What is acoustic grating? Explain how it can be used to measure the velocity of ultrasonic waves in liquids. 15. With neat diagram, derive an expression for thermal conductivity through a compound media connected in series. 16. Discuss using relevant diagrams about the static part in Forbe’s method 17. Explain the working of Airwedge with necessary diagram to find the thickness of a thin wire. 18. Explain the theory behind anti-reflection coating. 19. Write short notes on homo-junction and hetero-junction lasers 20. Derive the time independent Schrödinger wave equation. 21. Find the expression for acceptance angle of a fiber optic cable. 22. Sketch and explain the Fermi-Dirac statistical distribution with respect to energy at OK and above Part-C: (2 x 8 = 16 marks) 23. Derive Einsteins coefficients and write their significance. 24. What is coordination number? Find the packing factor of a FCC lattice with neat diagrams. Similar Searches: anna university question bank, anna university question bank regulation 2017, anna university question bank regulation 2013, anna university question bank mechanical engineering, anna university mba question bank, anna university dbms question bank, mba anna university question bank, anna university 1st semester question bank , discrete mathematics anna university question bank , anna university mba mcq question bank , anna university 2nd semester question bank, anna university question bank book, business research methods anna university question bank, biochemistry question bank anna university, anna university question bank for me cse regulation 2017 Have a question? Please feel free to reach out by leaving a comment below (Visited 924 times, 1 visits today) 3 thoughts on “Anna University B.E CSE PH6151 Engineering Physics Question Bank” 1. Please upload answer key. 2. Please send 2022 Physics model question paper and answer. 3. Sir Physics question paper and answer 2022 mortal instruments	{"url":"https://www.recruitmentzones.in/anna-university-b-e-cse-ph6151-engineering-physics-question-bank/","timestamp":"2024-11-05T13:02:52Z","content_type":"text/html","content_length":"155377","record_id":"<urn:uuid:4e5a1309-2b07-4451-a1de-ee2c8bf0b992>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00783.warc.gz"}
SS2 Mathematics 1st Term - eLearning \| eClasses \| Past Questions & Answers \| Solar Electricity \| Project Management Mathematics is a core subject SS2 students are required to offer in first term. The Unit of Instruction for SS2 Mathematics 1st Term is carefully developed from the Scheme of Work, which in turn is, based on NERDC current curriculum and SSCE syllabus. Approximation, Logarithms of Numbers Less than One, Circle Geometry, Quadratic Equations, Sine Rule and Cosine Rule, Bearing and Distance, Statistics: Measure of Central Tendency of Grouped Data, ETC SUB TOPICS:A. Approximation to the nearest whole numbers, decimal places and uintsB. Significant figures, ratios and proportionsC. Absolute error and relative errorD. Percentage errors and degree of LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 Approximate numbers to the nearest ten, hundred, thousand, million, billion and trillion.2 Round off numbers to the nearest tenth, thousandth, etc.3 Compare results obtained from using logarithm table and calculators.4 Calculate percentage error of a result or measurement from a given instrument.5 Solve examples of approximation in schools, health sector and social environment. SUB TOPICS:A. Revision on logarithm of numbers greater than one (SS1 1st Term)B. Logarithm with bar notation involving addition and subtractionC. Multiplication with numbers less than 1D. Division with numbers less than 1E. Revision of SSCE past questions LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 Use logarithms table to perform calculations involving numbers less than 1.2 Identify and name the two parts of a number in logarithms.3 Write a given number in standard form and compare the number with its characteristics in logarithms.4 Compare characteristics of logarithms with standard form of numbers.5 Use logarithm table to perform multiplication and division of numbers less than 1. SUB TOPICS:A. Logarithm with bar notation involving multiplication and divisionB. Powers of numbers less than 1C. Roots of numbers less than 1D. Simple equations with logarithmsE. Revision of SSCE past questions LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 Simplify problems with bar notation either by addition or subtraction.2 Use logarithm table to calculate powers of numbers less than 1.3 Use logarithm table to calculate roots of numbers less than 1.4 Apply logarithm rules to solve simple equations. SUB TOPICS:A. Definition of terms on circles, chords and arcsB. Mid-point of a chordC. Angles in a circleD. Angles in the same segmentE. Application of circle geometry LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 Define the following terms: (i) major segment and minor segment (ii) radius and diameter (iii) chords and arcs2 Prove that angle at the mid-point of a circle is right-angled (90◦).3 Prove that the angles in the same segment of a circle are equal.4 Prove that the angles which an arc subtends at the center is twice the angle it subtends at the circumference.5 Draw diagrams of circle theorems in their books and label them correctly. SUB TOPICS:A. Angles in a semi-circleB. Opposite interior and exterior angles of a cyclic quadrilateralC. Application of cyclic quadrilateral theoremD. Revision of SSCE past questions LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 Prove that the angles in a semi-circle is 90◦2 Prove that opposite interior and exterior angles of a cyclic quadrilateral are supplementary.3 Solve problems on angles in a semi-circle and opposite angles in a cyclic quadrilateral correctly. SUB TOPICS:A. Quadratic equations and methods of solving quadratic equationsB. Equations with irrational rootsC. Completing the square methodD. Derivation of quadratic formularE. Using quadratic formular to solve any quadratic equation LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 Mention the methods used in solving quadratic equations.2 Revise and recall factorization of perfect squares.3 Expand and factorize perfect squares.4 Recognise the difference between rational and irrational roots.5 Identify equations with irrational roots.6 Solve quadratic equations by method of completing the square.7 Deduce the quadratic formular from the method of completing the square.8 Apply the quadratic formular in solving quadratic equation problems. It does not require sign up or login. But a correct and valid e-mail will help the quiz machine send you the questions and answers when you click SUBMIT. Cheers! 2. Simplify log 6 + log 2 – log 12 1991 Q5 3. A chord of length 6 cm is drawn in a circle of radius 5 cm. Find the distance of the chord from the centre of the circle. 2007 Q15 4. A chord of a circle of radius 26 cm is 10 cm from the centre of the circle. Calculate the length of the chord. 1998 Q31 5. If x = 3, y = 2 and z = 4, what is the value of 3x2 – 2y + z? 2007 Q11 6. Which of the following is not a quadratic expression? 2001 Q21 7. Solve the equation: 3a + 5 = a2 1991 Q8 8. If log 2 = 0.3010 and log 2y = 1.8062, find, correct to the nearest whole number, the value of y. 2002 Q12 9. Find the number whose logarithm to base 10 is 2.6025 1991 Q6 10. Find the equation whose roots are –8 and 5. 2000 Q35 SUB TOPICS:A. Trigonometry ratiosB. Angles between 0◦ and 360◦C. Using table to find the value of Ꝋ lying between 0◦ and 360◦D. The sine rule and application to everyday lifeE. The cosine rule and application to everyday life LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 Solve problems involving use of sine, cosine and tangent in right-angled triangle.2 Find the sin, cos and tan of angles lying between 0◦ and 360◦ of the following: (i) 0.9626 (ii) 0.2826 (iii) 0.46283 Find the values of Ꝋ lying between 0◦ and 360◦ by using table.4 Derive and apply sine rule to solve some problems.5 Solve problems involving angles and sides of triangles using sine rule.6 Derive and apply cosine rule to find angles and sides of triangles. SUB TOPICS:A. Introduction to bearing and distanceB. Application of bearing and distance using sine ruleC. Application of bearing and distance using cosine ruleD. Revision of SSCE past questions LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 State the definition of bearing.2 Give the two bearing notations.3 State their own examples of the two notations of bearing.4 Define and draw 4, 8 and 16 cardinal points.5 Solve practical problems on bearing.6 Draw cardinal points measuring through clockwise direction correctly.7 Solve problems in trigonometric ratios and angles of elevation and depression. SUB TOPICS:A. Data collection and analysis of grouped dataB. Tabular presentation of data (frequency table)C. Measure of central tendency (mean)D. Measure of central tendency (median and mode)E. Practical application to population studies LEARNING OBJECTIVES: At the end of the lesson, learners should be able to:1 Collect, tabulate and present data in meaningful form.2 Construct frequency tables from given data.3 Calculate given data with the use of measure of central tendency.4 Calculate the mean, median and mode of given data of some problems in everyday life.5 List various forms of data presentation. SS 1 SS 2 SS 3 1st 2nd 1st 2nd 1st 2nd 3rd 3rd 3rd	{"url":"https://anaarm.com/subjects-and-courses/ss2-mathematics-1st-term/","timestamp":"2024-11-05T00:48:59Z","content_type":"text/html","content_length":"210475","record_id":"<urn:uuid:490cb16c-7f70-409d-9b2c-5d6791650cb1>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00203.warc.gz"}
Optimized effective potential method in current-spin-density-functional theory Current-spin-density-functional theory (CSDFT) provides a framework to describe interacting many-electron systems in a magnetic field which couples to both spin and orbital degrees of freedom. Unlike in the usual (spin-)density-functional theory, approximations to the exchange-correlation energy based on the model of the uniform electron gas face problems in practical applications. In this work, explicitly orbital-dependent functionals are used and a generalization of the optimized effective potential method to the CSDFT framework is presented. A simplifying approximation to the resulting integral equations for the exchange-correlation potentials is suggested. A detailed analysis of these equations is carried out for the case of open-shell atoms and numerical results are given using the exact-exchange energy functional. For zero external magnetic field, a small systematic lowering of the total energy for current-carrying states is observed due to the inclusion of the current in the Kohn-Sham scheme. For states without current, CSDFT results coincide with those of spin-density-functional theory. Dive into the research topics of 'Optimized effective potential method in current-spin-density-functional theory'. Together they form a unique fingerprint.	{"url":"https://cris.huji.ac.il/en/publications/optimized-effective-potential-method-in-current-spin-density-func","timestamp":"2024-11-06T17:03:18Z","content_type":"text/html","content_length":"47980","record_id":"<urn:uuid:3bd3cf21-418a-4583-a7a2-04dca7f0db81>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00062.warc.gz"}
Advanced Functionality Splitting the pilot set An important consideration for pilot designs like the stratamatch approach is the selection of the pilot set. Ideally, the individuals in the pilot set should be similarto the individuals in the treatment group, so a prognostic model built on this pilot set will not beextrapolating heavily when estimating prognostic scores on the analysis set. To more closely ensure that the selected pilot set is a representative sample of controls, one easy step is to specify a list of categorical or binary covariates and sample the pilot set proportionally based on these covariates. This can be done in one step using auto_stratify, for example: a.strat1 <- auto_stratify(mydata, "treat", prognosis = outcome ~ X1 + X2, group_by_covariates = c("C1", "B2"), size = 500) #> Constructing a pilot set by subsampling 10% of controls. #> Subsampling while balancing on: #> C1, B2 #> Fitting prognostic model via logistic regression: outcome ~ X1 + X2 Another method is to use the split_pilot_set function, which allows the user to split the pilot set (and examine the balance) before passing the result to auto_stratify to fit the prognostic score. mysplit <- split_pilot_set(mydata, "treat", group_by_covariates = c("C1", "B2")) #> Constructing a pilot set by subsampling 10% of controls. #> Subsampling while balancing on: #> C1, B2 The result, mysplit, is a list containing a pilot_set and an analysis_set, in this case partitioned while balancing on B1 and C2. At this point, we might pass the result to auto_stratify as follows: a.strat2 <- auto_stratify(mysplit$analysis_set, "treat", prognosis = outcome ~ X1 + X2, pilot_sample = mysplit$pilot_set, size = 500) #> Using user-specified set for prognostic score modeling. #> Fitting prognostic model via logistic regression: outcome ~ X1 + X2 In this case, the pilot set splitting method is the same for a.strat1 and a.strat2, so each should be qualitatively similar. Fitting the prognostic model By default, auto_stratify uses a logistic regression (for binary outcomes) or a linear regression (for continuous outcomes) to fit the prognostic model. auto_stratify is built to accomodate other prognostic score estimation approaches: rather than letting auto_stratify do the model fitting, the user can specify a pre-fit model or a vector of prognostic scores. A word of caution is advisable: Logistic regression is generally the norm when fitting propensity scores (Stuart (2010)), and most studies discussing the prognostic score have likewise focused on linear or logistic regression (Ben B. Hansen (2008), Leacy and Stuart (2014), Aikens et al. (2020)), or less commonly the lasso (Antonelli et al. (2018)). The nuances of modeling and checking the fit of the prognostic score are still understudied. In particular, prognostic models generally are fit on only control observations, meaning that they must necessarily extrapolate to the treatment group. Users should, as always, consider diagnostics for their specific model, for their stratified data set (see “Intro to statamatch”), and for their matched dataset (see, as an introduction Stuart (2010) ). Additionally, in order to maintain the integrity of the pilot set and prevent over-fitting, users interested in trying many modeling or matching schemes are encouraged to define a single pilot/ analysis set split (e.g. with split_pilot_set, above) and use the same pilot set throughout their design process. To an extent, any prognostic score stratification – even on a poor quality model – will increase the speed of matching for large datasets. In addition, if the strata are sufficiently large, the subsequent within-strata matching step may compensate for a poor-quality prognostic model. To one view, the diagnostic check that matters most is the quality of the final matching result, whereas specific prognostic modeling concerns are perhaps secondary. Nonetheless, simulation and theory results suggest that incorporating a prognostic score into a study design (in combination, generally, with a propensity score) can have favorable statistical properties, such as decreasing variance, increasing power in gamma sensitivity analyses, and decreasing dependence on the propensity score ( Stuart (2010), Aikens, Greaves, and Baiocchi (2020), Antonelli et al. (2018), Ben B. Hansen (2008)). To that end, prognostic models which produce high-quality prognostic score estimates are expected to ultimately produce higher quality designs by improving the prognostic balance of the matched set. This section contains a few examples to introduce users who may be new to the modeling space. By no means does it begin to cover all of the modeling possibilities, or even all of the nuances of any one model. This is not a tutorial on predictive modeling. Users looking to become more familiar with modeling in R more broadly may be interested in the caret package (Kuhn (2021)), which implements support for a wide variety of predictive models. Outcomes: Binary or Continuous It’s important to select a model which is appropriate to the nature of the outcome of interest. In this tutorial, our sample data has a binary outcome, so we use models appropriate to that outcome. Users with continuous outcomes should use regression models appropriate to continuous outcomes. Other types of outcomes – such as categorical – have not yet been characterized in the prognostic score A lasso The lasso is a sparsifying linear model – it is a mathematical cousin to linear regression, but it functions well when a substaintial number of the measured covariates are actually uninformative to the outcome. This may be a particularly useful and intuitive approach when there are many measured covariates which may be redundant or uninformative. The code below uses the glmnet package (Friedman, Hastie, and Tibshirani (2010)) to fit a cross-validated lasso on the pilot set based on all the measured covariates. In this example, since the outcome is binary, we will run a logistic lasso. This is done by specifying the family = "binomial" argument to cv.glmnet (although other modeling steps are simular for continuous outcomes.) The code below does some preprocessing to convert the pilot set data to the right format before passing it to cv.glmnet. glmnet expects the input data to be a model matrix rather than a data frame, and it expects outcomes (y_pilot) to be separated from the covariate data (x_pilot). #> Loading required package: Matrix #> Loaded glmnet 4.1-3 # fit model on pilot set x_pilot <- model.matrix(outcome ~ X1 + X2 + B1 + B2 + C1, data = mysplit$pilot_set) y_pilot <- mysplit$pilot_set %>% dplyr::select(outcome) %>% cvlasso <- cv.glmnet(x_pilot, y_pilot, family = "binomial") At this point, we can run diagnostics on cvlasso. The Introduction to glmnet vignette contains an accessible starting point. In this simulated data, we happen to know that the only variable that actually affects the outcome is X1. The sparsifying model often does a great job of picking out X1 as the most important variable. We can see this by printing the coefficients: #> 8 x 1 sparse Matrix of class "dgCMatrix" #> s1 #> (Intercept) 0.02870109 #> (Intercept) . #> X1 -0.64374310 #> X2 . #> B1 . #> B2 . #> C1b . #> C1c . When we are satisfied with our prognostic model, we can estimate the scores on the analysis set with predict and pass the result to auto_stratify. # estimate scores on analysis set x_analysis <- model.matrix(outcome ~ X1 + X2 + B1 + B2 + C1, data = mysplit$analysis_set) lasso_scores <- predict(cvlasso, newx = x_analysis, s = "lambda.min", type = "response") # pass the scores to auto_stratify a.strat_lasso <- auto_stratify(data = mysplit$analysis_set, treat = "treat", outcome = "outcome", prognosis = lasso_scores, pilot_sample = mysplit$pilot_set, size = 500) An elastic net An elastic net is a fairly straightforward extension of the code above – we can use the same form of the pilot set data that we used above. An additional task is to select the alpha “mixing” parameter determining the amount of L2-regularization (1 for lasso, 0 for ridge regression). Here, I set alpha = 0.2. The tutorial for glmnet also contains some advice for selecting the alpha parameter via cross-validation. As above, we specify a logistic elastic net with family = "binomial", since in this case our outcome is binary. cvenet <- cv.glmnet(x_pilot, y_pilot, family = "binomial", alpha = 0.2) enet_scores <- predict(cvenet, newx = x_analysis, s = "lambda.min", type = "response") # pass the scores to auto_stratify a.strat_enet <- auto_stratify(data = mysplit$analysis_set, treat = "treat", outcome = "outcome", prognosis = enet_scores, pilot_sample = mysplit$pilot_set, size = 500) A random forest Random forests are a popular option for both classification and regression modeling, particularly because of their strengths in modeling nonlinear relationships in the data. Below is an example which fits a random forest for our binary outcome using randomForest. A note for users with binary outcomes: randomForest will run regression by default if the outcome column is numeric. To circumvent this (e.g. for 0/1 coded data), the outcome can be cast as a factor. #> randomForest 4.7-1 #> Type rfNews() to see new features/changes/bug fixes. #> Attaching package: 'randomForest' #> The following object is masked from 'package:dplyr': #> combine forest <- randomForest(as.factor(outcome) ~ X1 + X2 + B1 + B2, data = mysplit$pilot_set) Random forests can be somewhat more opaque than linear models in terms of understanding how predictions are made. A good starting point is running importance(forest) to check on which features are weighted heavily in the model. Below, we extract a “prognostic score” from the random forest. Another note for users with binary outcomes: The predict method for random forest classifiers outputs 0/1 predictions by default. These will be useless for stratification. Instead, we need to specify type = "prob" in the call to predict and extract the probabilities for the “positive” outcome class. Alternative matching schemes “Intro to Stratamatch” covers only the default functionality of stratamatch, which is a fixed 1:k propensity score match within strata. This tutorial covers some alternative options. Note that the examples that follow all require the R package optmatch (Ben B. Hansen and Klopfer (2006)) to be installed. Distance measure: Mahalanobis Distance Users can opt to use Mahalanobis distance rather then propensity score for the within-strata matching step by specifying the “method” to strata_match. We set k = 2, so that precisely 2 control observations are matched to each treated observation. Matching procedure: Full Matching Full matching may be a particularly useful approach when the ratio of treated to control individuals varies within strata, but the researcher still would prefer to use as much of the data as possible. To do full matching, set k = "full". This can be used in combination with mahalanobis distance matching, as shown below: fullmahalmatch <- strata_match(a.strat2, model = treat ~ X1 + X2 + B1 + B2, method = "mahal", k = "full") #> Using Mahalanobis distance: treat ~ X1 + X2 + B1 + B2 #> Structure of matched sets: #> 5+:1 4:1 3:1 2:1 1:1 1:2 1:3 1:4 1:5+ #> 8 6 14 42 453 132 84 70 206 #> Effective Sample Size: 1331.8 #> (equivalent number of matched pairs). Matching with other software stratamatch doesn’t natively support all possible matching schemes. Luckily, it can be fairly straightforward to stratify a data set with auto_stratify and match the results with other software. As an example, the code below uses the optmatch package (Ben B. Hansen and Klopfer (2006)) to match within-strata using Mahalanobis distance with a propensity score caliper. #> Loading required package: survival # mahalanobis distance matrix for within-strata matching mahaldist <- match_on(treat ~ X1 + X2 + B1 + B2, within = exactMatch(treat ~ stratum, data = a.strat2$analysis_set), data = a.strat2$analysis_set) # add propensity score caliper propdist <- match_on(glm(treat ~ X1 + X2 + B1 + B2, family = "binomial", data = a.strat2$analysis_set)) mahalcaliper <- mahaldist + caliper(propdist, width = 1) mahalcaliper_match <- pairmatch(mahalcaliper, data = a.strat2$analysis_set) #> Warning in fullmatch.BlockedInfinitySparseMatrix(x = x, min.controls = controls, : At least one subproblem matching failed. #> (Restrictions impossible to meet?) #> Enter ?matchfailed for more info. #> Matches made in 8 of 10 subgroups, accounting for 3691 of 4614 total observations. #> Structure of matched sets: #> 1:0 1:1 0:1 #> 192 954 2514 #> Effective Sample Size: 954 #> (equivalent number of matched pairs). Aikens, Rachael C, Dylan Greaves, and Michael Baiocchi. 2020. “A Pilot Design for Observational Studies: Using Abundant Data Thoughtfully.” Statistics in Medicine. Aikens, Rachael C, Joseph Rigdon, Justin Lee, Michael Baiocchi, Andrew B Goldstone, Peter Chiu, Y Joseph Woo, and Jonathan H Chen. 2020. “Stratified Pilot Matching in r: The Stratamatch Package.” Statistics arXiv , January. Antonelli, Joseph, Matthew Cefalu, Nathan Palmer, and Denis Agniel. 2018. “Doubly Robust Matching Estimators for High Dimensional Confounding Adjustment.” Biometrics 74 (4): 1171–79. Friedman, Jerome, Trevor Hastie, and Rob Tibshirani. 2010. “Regularization Paths for Generalized Linear Models via Coordinate Descent.” Journal of Statistical Software 33 (1): 1. Hansen, Ben B. 2008. “The Prognostic Analogue of the Propensity Score.” Biometrika 95 (2): 481–88. Hansen, Ben B., and Stephanie Olsen Klopfer. 2006. “Optimal Full Matching and Related Designs via Network Flows.” Journal of Computational and Graphical Statistics 15 (3): 609–27. Kuhn, Max. 2021. Caret: Classification and Regression Training Leacy, Finbarr P, and Elizabeth A Stuart. 2014. “On the Joint Use of Propensity and Prognostic Scores in Estimation of the Average Treatment Effect on the Treated: A Simulation Study.” Statistics in Medicine 33 (20): 3488–508. Stuart, Elizabeth A. 2010. “Matching Methods for Causal Inference: A Review and a Look Forward.” Statistical Science: A Review Journal of the Institute of Mathematical Statistics 25 (1): 1.	{"url":"https://cran.uvigo.es/web/packages/stratamatch/vignettes/Advanced_functionality.html","timestamp":"2024-11-03T21:44:22Z","content_type":"text/html","content_length":"49395","record_id":"<urn:uuid:1883dd0b-cc52-4c36-baa6-88f0a633b3dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00060.warc.gz"}
Optimal Classification/Rypka Method/Equations/Separatory/Characteristic/Theoretical - Wikibooks, open books for an open world Theoretical Separation The general identification equation ${\displaystyle S_{j}={\frac {1-{V^{-j}}}{1-V^{-C}}}}$ , where:^[1] ☆ S[j] is the theoretical separatory value per jth characteristic, ☆ C is the number of characteristics in the group, ☆ V is the highest value of logic in the group and ☆ j is the jth characteristic index in the target set, where: j = 0..K and where: ■ K is the number of characteristics in the target set. Minimal number of characteristics to result in theoretical separation ${\displaystyle t_{min}={\frac {\log G}{\log V}}}$ , where:^[2] ☆ t[min] is the minimal number of characteristics to result in theoretical separation, ☆ G is the number of elements in the bounded class and ☆ V is the highest value of logic in the group. 1. ↑ See page 153, page 167, Fig. 2. & page 175 of the primary reference 2. ↑ See page 157, Primary Schemes footnote of the primary reference	{"url":"https://en.m.wikibooks.org/wiki/Optimal_Classification/Rypka_Method/Equations/Separatory/Characteristic/Theoretical","timestamp":"2024-11-06T01:24:43Z","content_type":"text/html","content_length":"32891","record_id":"<urn:uuid:386fb2b2-b3bf-440a-bc24-b23fa4b78aa8>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00055.warc.gz"}
Application of the Standardised Streamflow Index for Hydrological Drought Monitoring in the Western Cape Province, South Africa: A Case Study in the Berg River Catchment Department of Water and Sanitation, South African Government, Pretoria 0001, South Africa Department of Earth Sciences, University of the Western Cape, Cape Town 7535, South Africa Center for Environmental Research and Education, Duquesne University, Pittsburgh, PA 15282, USA Author to whom correspondence should be addressed. Submission received: 17 May 2023 / Revised: 28 June 2023 / Accepted: 28 June 2023 / Published: 10 July 2023 In many regions around the world, drought has been recurrent, more frequent, and more intense over time. Hence, scientific research on drought monitoring has become more urgent in recent years. The aim of this study was to test the applicability of the Standardised Streamflow Index (SSI) for hydrological drought monitoring in the Berg River catchment (BRC), Western Cape (WC) province, South Africa (SA). Using various methods described in this study, the sensitivity of the SSI to the commonly used Gamma, Log-normal, Log-logistic, Pearson Type III, and Weibull Probability Distribution Functions (PDFs) was tested. This study has found that all the tested PDFs produced comparable results for mild to severe drought conditions. The SSI calculated using the Gamma, Log-Normal, and Weibull PDFs is recommended for the BRC because it consistently identified extreme drought conditions during the 1990–2022 study period and identified the 2015–2018 droughts as the worst during the study period. Although more studies are required to test other PDFs not considered, this study has shown that the SSI can be applicable in the BRC. This study has provided a foundation for more research on the application of the SSI in the BRC and other catchments in SA. 1. Introduction Historical records indicate that in most climate zones or regions around the world, drought has been recurrent, more frequent, and more intense. There is evidence of negative environmental and socio-economic impacts because of the recurring drought events [ ]. Since 1976, the United Kingdom (UK) has experienced several severe droughts that resulted in serious water shortages on a national scale, negatively affecting mainly the agriculture and commerce industries [ ]. It was reported that the United States of America (USA) lost over 10 billion dollars in damages because of the droughts that occurred during the year 2002 alone [ ]. In South Africa (SA), the drought that occurred in 1992 was judged to be the most severe since the beginning of the 20th century. The resulting water shortages were responsible for crop and livestock losses in the agriculture and farming industries, as well as food shortages for people. During the year 2015, the droughts that occurred in the Western Cape (WC) province in SA were reportedly the most severe in just over a century. In response to these severe droughts, increased water use restrictions were enforced by the WC government [ ]. Projections on climate variability (change) indicate that most regions worldwide, including SA, will continue to experience more frequent and more intense drought events [ ]. Hence, scientific research on the development of improved drought monitoring and early warning systems has become more urgent in recent years. If well developed and applied effectively, these systems have the potential to reduce vulnerability to drought impacts and may contribute to the development and implementation of suitable policies for improved drought management in SA [ The complex nature of drought has resulted in a lack of consensus on its definition. However, it is accepted worldwide that drought occurs in four phases: meteorological drought, agricultural drought, hydrological drought, and socio-economic drought. Both of these phases of drought are generally caused by prolonged deficits in rainfall, affecting soil moisture (meteorological), crop growth (agricultural), surface and ground water storages (hydrological), and the availability of water for human consumption (socio-economic) [ ]. The description of drought according to its propagation phases has led to the development of important indices for monitoring drought around the world. Many indices have been developed and used for many years to monitor droughts and develop drought monitoring and early warning systems. They have commonly been used to characterize droughts according to their onset, duration, magnitude, frequency, end, and spatial coverage. The Standardised Precipitation Index (SPI) is one of the most widely used drought indices around the world. It is recommended by the World Meteorological Organization (WMO) as the preferred index for meteorological drought monitoring [ ]. In the UK, the SPI has been used to characterize meteorological droughts to improve understanding of the nature of droughts and their related hazards [ ]. In SA, the SPI has been thoroughly tested to characterize drought according to its duration, frequency, severity, intensity, and spatial extent in all the provinces and climatic regions [ ]. The SPI is widely used across the world and especially in SA because it uses a simple calculation procedure, requires only rainfall data to calculate, and is flexible in that it allows the use of various time scales for monitoring different types of droughts [ ]. However, the SPI has an inherent limitation that cannot be overlooked. Its use of only precipitation in its calculation procedure means that it is not capable of providing hydrological drought information that describes the direct impact of drought on surface and groundwater storages [ ]. Hence, other indices should be considered for hydrological drought monitoring in SA. To characterize hydrological drought according to its onset, duration, magnitude, frequency, and spatial coverage, the WMO has recommended the Standardised Streamflow Index (SSI), which uses the same calculation procedure as the SPI [ ]. The SSI inherits the advantages of the SPI in that it uses a simple calculation procedure. It differs from the SPI in that it uses streamflow instead of rainfall data in its calculation procedure. Although the SSI does not incorporate the impact of water use or demand, its use has increased since its introduction. It has been tested and has performed well in various regions with different catchment characteristics around the world. It has proved useful for the characterization of hydrological droughts and the development of early warning systems in Slovenia, China, the UK, Azerbaijan, and Iran [ ]. In SA, the SSI has been used to characterize hydrological drought in all the cape provinces [ ]. The calculation simplicity of the SPI is enhanced by the fact that it has been extensively used around the world to the extent that the Gamma Probability Distribution Function (PDF) has been widely accepted for its calculation. On the other hand, the SSI has not been tested thoroughly enough to be able to reach a consensus on the universal PDF for its calculation. This consensus may not easily be reached because many or all catchments possess high spatial streamflow variability, resulting in high levels of uncertainty in the PDFs that fit streamflow data best [ ]. The Gamma PDF was used to calculate the SSI and tested at varying catchments in SA, Slovenia, China, the Netherlands, and Iran [ ]. The Generalized Extreme Value (GEV), Log-Logistic, and Tweedie PDFs were recommended to calculate the SSI and were tested in various catchments in the UK [ ]. The GEV and the Log-Logistic PDFs were recommended for calculating the SSI and were tested in catchments in Spain [ ]. The Tweedie, GEV, and Log-Logistic PDFs were recommended for calculating the SSI and were tested in various catchments in Europe [ ]. The above findings show that the best-fitting PDFs may vary with varying catchments. This is supported by Li et al. (2018) who tested the Pearson Type III (PTIII), Log-Logistic, GEV, and Log-Normal and concluded that the best fitting PDFs varied with varying catchments and streamflow gauging station locations [ ]. Hence, according to the above studies, the Gamma PDF, as used by Botai et al. (2021) and others, may not be the most suitable candidate for calculating the SSI in some SA catchments [ ]. Hence, other PDFs should be tested in SA catchments. Although it is evident that drought affects surface water supply systems such as rivers, there are very few research studies on the use of the SSI to characterize droughts in SA [ ]. Consequently, there is no consensus on the accepted PDFs for SSI calculation in the various catchments in SA. Hence, the aim of this study is twofold: to evaluate the applicability of the SSI for hydrological drought monitoring in SA and to test the sensitivity of the SSI to different PDFs at a selected catchment in the WC province of SA, potentially leading to recommendations or guidelines for the selection of the most suitable or best-fitting PDFs in other SA catchments with varying geo-hydro-climatic zones. The WMO has recommended the SSI for hydrological drought monitoring, but the SSI needs to be thoroughly tested at various catchments in SA. Therefore, the results from this study will contribute to the provision of tested scientific knowledge on the effective application of the SSI for hydrological drought monitoring in SA. Given the few studies that have been conducted on the application of the SSI in SA, the overall outcomes from this study will provide a foundation and a basis for the application of the SSI in SA. This may ultimately aid in the improvement of drought monitoring and early warning systems in SA. The remainder of the paper is organized as follows: Section 2 presents the case study area and the data used in the study. Section 3 briefly describes the methods and approaches used. Section 4 presents key results and the discussion. Section 5 presents the main conclusions of this study. 2. Study Area and Data 2.1. Study Area This study was carried out on the approximately 7700 km Berg River Catchment (BRC) ( Figure 1 ), one of the two catchments in the Berg-Olifants Water Management Area (WMA). The BRC supplies water to parts of the WC province in SA. As shown in Figure 1 , the Berg River in the BRC forms at the Franschhoek mountains and flows northwards, where it is joined by the Klein Berg, and then flows westwards until it discharges into the Atlantic Ocean. With a total length of approximately 285 km, the Berg River has up to nine major and seven minor tributaries. Six of these minor tributaries, which include the Klein Berg River, are perennial [ ]. Surface water is a major water source in the BRC. The Mean Annual Runoff (MAR) in the upper Berg River and its tributaries is approximately 277 × 106 m , approximately 263 × 106 m at the upper middle Berg River and its tributaries, approximately 288 × 106 m at the lower middle Berg River, 97 × 106 m at the lower Berg River and its tributaries, and approximately 17 × 106 m at the flood plain and estuary [ ]. Thus, monitoring hydrological drought using streamflow is crucial in the BRC. The WC province experiences both winter, summer, and all-year rainfall. The annual rainfall in the WC ranges between 300 mm and 900 mm. The BRC is situated in the winter rainfall zone of the WC province, with a maximum rainfall of approximately 30 mm in June [ 2.2. Streamflow Data The monthly mean streamflow data from 1990 to 2022 used in this study to calculate the SSI were obtained from the South African National Department of Water and Sanitation (DWS) ( Figure 1 ). The streamflow data considered were assumed to be from near-natural flow rivers. In this study, the authors used data from three streamflow gauging stations that met the minimum required record of 30-years for calculating the SSI and had minimal missing data or gaps. The G1H020 is in the upper Berg River, with a MAR of approximately 277 × 106 m . The G1H013 is in the upper middle Berg River, with a MAR of approximately 263 × 10 . The G1H008 is in the Klein Berg River on the lower middle Berg River, with a relatively low MAR of 263 × 10 . The Klein Berg River is a tributary of the Berg River ( Figure 1 Table 1 ). An assessment of the historical data obtained from G1H020, G1H013, and G1H008 indicates that those located on the middle and upper Berg River record relatively higher flows than those located on the Klein Berg River ( Figure 2 ). Hence, in this study, to test the sensitivity of the SSI to various PDFs, streamflow time series were acquired from three streamflow gauging sites: G1H008, located on the low flow Klein Berg River; G1H013 located on the relatively high-flow lower part of the Berg River; and G1H020 also located on the relatively high-flow upper part of the Berg River. 2.3. Rainfall Data The monthly mean rainfall data from 1980 to 2021 were used in this study to calculate the SPI was obtained from the South African Weather Service (SAWS). Although all three stations contained rainfall data that met the requirement of a minimum of a 30-year record, only the Franschoek and Stellenbosch stations were used in this study because they had more recent data ( Figure 3 ). Hence, in this study, the rainfall data used were obtained from two gauging stations located in the Franschoek and Stellenbosch towns ( Figure 1 ) because the data met the minimum required record of 30 years for calculating the SPI. 3. Methods 3.1. SSI Calculation In this study, the SSI was computed using streamflow data obtained from the G1H020, G1H013, and G1H008 streamflow gauging stations located in the BRC, as well as various commonly used PDFs. The resultant SSI obtained from the PDF that best fitted the streamflow data was used to characterize hydrological drought in the BRC. The simple SPI-based generic procedure was followed for calculating the SSI: (1) monthly streamflow time series were averaged over 12 months’ time scales; (2) a PDF was fitted to the streamflow time series using 12 months’ time scales; (3) the PDF’s parameters were determined using the streamflow data; (4) cumulative distribution functions were established for the streamflow and used to calculate the cumulative probability of the observed values of the variables; and (5) the inverse normal cumulative distribution function with a mean of zero and variance of one was applied to generate the SSI12 time series. In the SSI12 series, the zero values are equivalent to the mean streamflow. Negative values indicate dryer than average conditions, while positive values indicate wetter than average conditions [ ]. The SSI is therefore calculated using Equation (1) [ $S S I = W − C 0 + C 1 W + C 2 W 2 1 + d 1 W + d 2 W 2 + d 3 W 3$ $W = − 2 ln P$ ≤ 0.5. is the probability of exceeding a determined x value, and = 1 − F(x). If > 0.5, is replaced by 1 − , and the sign of the resultant is reversed. C = 2.515517; C = 0.802853; C = 0.010328; d = 1.432788; d = 0.189269; and d = 0.001308 are constants. If the PDF, F(x), is suitable for fitting the monthly streamflow series, the average value of the SSI and the standard deviation must equal 0 and 1, respectively [ Drought classification using the SSI may differ for various studies. The drought classification used in this study is described in Table 2 3.2. PDFs Considered for SSI Calculation To evaluate the applicability of the SSI for hydrological drought monitoring in the BRC, the sensitivity of the SSI to various commonly used PDFs was tested. Five PDFs, i.e., Gamma, Log-normal, Pearson Type III (PTIII), Log-Logistic, and Weibull (Equations (2)–(6)), were fitted to the streamflow time series obtained from G1H008, G1H013, and G1H020 in the BRC ( Table 3 ). The Gamma PDF was selected because it is the commonly preferred method for calculating the SPI and its performance for calculating the SSI has not been thoroughly tested in SA catchments. The only record that could be found on the SSI application in SA using the Gamma PDF is by Botai et al. (2021) [ ]. The Log-normal, PTIII, Log-Logistic, and Weibull PDFs were selected because they have been tested in European and other catchments but not in SA catchments. No records were found on the SSI application in SA using the Log-normal, PTIII, Log-Logistic, and Weibull PDFs. It is recommended that other PDFs be tested in future research. The Log-normal and Gamma are two-parameter PDFs, while the PTIII, Log-Logistic, and Weibull are three-parameter PDFs. Only the Lognormal and Gamma are bound below zero. Following the approaches by Tijdeman et al. (2020) and Stagge et al. (2015), the L moments (Lmom) were used to estimate parameters of the PDFs [ ]. Alternative parameter estimation methods may be considered for future studies. The selected PDFs and L moments have been commonly used and thoroughly tested for SPI and SSI calculation in many regions around the world, but not in RSA, especially for SSI calculation. The use of SSI is relatively new in RSA, so the focus of this study was to introduce the SSI and test it using currently commonly used and relatively easy-to-apply PDFs and parameter selection methods. Follow up studies should consider other PDFs not tested in this study as well as other parameter selection methods. The Log-logistic, Log-Normal, PTIII, Weibull, and Gamma PDFs are determined using Equations (2)–(6), as shown in Table 3 Table 3. Probability Distribution Functions used to calculate the SSI in the BRC [ Probability Distribution Function (PDF) PDF Equations Used for SSI Calculation in the BRC Log-logistic $F x = 1 + α x − γ β − 1$ (2) [40,46] $β = 2 w 1 − w 0 6 w 1 − w 0 − 6 w 2$, $α = w − 2 w 1 β Γ 1 + 1 β Γ 1 − 1 β$, $γ = w 0 − α Γ 1 + 1 β Γ 1 − 1 β$ $F x = θ l n x − a − μ σ$ (3) Log-Normal θ ≈ standard normal cumulative distribution function. [40,46] $σ = 0.999281 z − 0.006118 z 2 + 0.000127 z 5$ such that $z = 8 3 θ − 1 1 + τ 3 2$. $μ = l n ε 2 erf σ 2 − σ 2 2 e r f$ is the Gauss error function such that $erf σ 2 = 2 θ σ 2 2 − 1$ and $a = ε 1 − e μ + σ 2 2$. Pearson Type III $F x = 1 α Γ β ∫ γ x x − γ α β − 1 e − x − γ α$ (4) [40,46] If $τ 3 ≥ 1 3$, then $τ m = 1 − τ 3$, leading to $β = 0.36067 τ m − 0.5967 τ m 2 + 0.25361 τ m 3 1 − 2.78861 τ m + 2.56096 τ m 2 − 0.77045 τ m 3$ If $τ 3 < 1 3$, then $τ m = 3 π τ 3 2$; such that $β = 1 + 0.2906 τ m τ m + 0.1882 τ m 2 + 0.0442 τ m 3$, $α = π ε 2 Γ β Γ β + 1 2$ and $γ = ε 1 − α$ Weibull $F x = 1 − e − x − m a b$ (5) [40,46] $b = 1 7.859 C + 2.9554 C 2$, $C = 2 3 − τ 3 − 0.6309$, $a = ε 2 G 1 + 1 b 1 − 2 − 1 b$$m = ε 1 − a Γ 1 + 1 b$ Gamma $g x , α , β = 1 β α Γ α x α − 1 e − x β$ (6) [34,38] α > 0 and β > 0 are the estimated shape and scale parameters, x > 0 is the streamflow (m^3/s), and Γ (α) is the Gamma PDF such that, $Γ α = ∫ 0 ∞ x α − 1 e − x d x$. 3.3. SSI Computation Using R Software Package The SSI calculations for G1H008, G1H013, and G1H020 streamflow time series were carried out using the R software package. The SSI time series computation using the Gamma, Log-logistic, and PTIII PDFs was carried out in R-Studio software using the SPEI version 1.7 package. The manual for the SPEI version 1.7 package was obtained from (accessed on 9 January 2023). The SSI time series computation using Weibull and Log-Normal PDFs was carried out in R-Studio software using the SCI version 1.0–2 package. The manual for the SCI version 1.0–2 package was obtained from (accessed on 9 January 2023). 3.4. Evaluation of Best Fitting PDFs for SSI Computation According to Svensson et al. (2017), the S-W test has been found to be the most powerful test for normality, closely followed by the Anderson-Darling and the Kolmogorov-Smirnov tests [ ]. Thus, the Shapiro-Wilk (S-W) goodness-of-fit or normality test was used in this study to evaluate the sensitivity of the SSI to the selected PDFs. As used in the study by Svensson et al. (2017), the significance level chosen for this study is 95% ( -value = 0.05). The S-W test was applied to test the null hypothesis (H0) that the SSI time series is normally distributed. Thus, if the value is less than or equal to 0.05, the null hypothesis is rejected, and the time series is not normally distributed. If the value is greater than 0.05, the null hypothesis is not rejected, and the time series used is normally distributed [ ]. The S-W test helps to assess how well the considered PDFs fit the streamflow time series, resulting in an SSI time series that closely resemble the expected standard normal distribution. 3.5. Evaluation of the Correlation between the SSI Computed Using the Selected PDFs Correlation coefficients are descriptive statistics used to describe the magnitude and direction of the relationship between variables. The correlation coefficients vary from −1 to +1, whereby the sign describes the direction of the relationship (positive or negative). When the coefficient is 0 or close to 0, there is little to no relationship. However, the closer the coefficient of correlation is to −1 or +1, the stronger the relationship is between the variables [ ]. In this study, the Pearson’s correlation coefficient described by Sedgwick (2012) is used to determine the linear relationship between the different probability distribution functions [ 4. Results 4.1. SSI Calculation Using the Selected PDFs The results of the SSI calculation using the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs are shown in Figure 4 Figure 5 Figure 6 . In Figure 4 (G1H008), it can be observed that the SSI12 time series computed using the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs produced drought events with similar onset and end times but with different intensities and magnitudes. As shown in Table 4 , between November 2004 and June 2005, the SSI12 computed using Gamma, Log-Normal, and Weibull PDFs produced severe (−1.6) drought conditions, while the SSI12 computed using Log-Logistic and PTIII PDFs produced moderate (−1.4) drought conditions. As shown in Table 5 , between December 2015 and April 2016, the SSI12 computed using Gamma, Log-Normal, and Weibull PDFs produced extreme and severe (−2.2, −2.3, and −2.0, respectively) drought conditions, while the SSI12 computed using Log-Logistic and PTIII PDFs produced severe (−1.6) drought conditions. Hence, from Figure 4 as well as Table 4 Table 5 , it is apparent that the variability of drought intensity produced by the SSI12 computed using the Gamma, Log-Logistic, Log-Normal, PTIII and Weibull PDFs increases as drought conditions increase from moderate to extreme. Figure 5 (G1H013), it can be observed that the SSI12 time series computed using the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs produced drought events with similar onset and end times but with different intensities and magnitudes. This is a similar outcome to Figure 4 (G1H008). As shown in Table 6 , between November 2003 and May 2005, the SSI12 calculated using Gamma, Log-Normal, Log-Logistic, PTIII, and Weibull PDFs all produced moderate (−1.2) drought conditions. As shown in Table 7 , between August 2017 and May 2018, the SSI12 computed using Gamma, Log-Normal, PTIII, and Weibull PDFs produced extreme (−2.6, −3.2, −2.1, and −2.3, respectively) drought conditions, while the SSI12 computed using the Log-Logistic PDF produced severe (−1.9) drought conditions. Hence, from Figure 5 as well as Table 6 Table 7 , it is apparent that the variability of drought intensity produced by the SSI12 computed using the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs increases as drought conditions increase from moderate to extreme. This is similar to the results obtained from station G1H008. Figure 6 (G1H020), it can be observed that the SSI12 time series computed using the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs produced drought events with similar onset and end times but with different intensities and magnitudes. This is a similar outcome to Figure 4 (G1H008) and Figure 5 (G1H013). As shown in Table 8 , between July 2003 and June 2004, the SSI12 computed using Gamma, Log-Normal, Log-Logistic, PTIII, and Weibull PDFs all produced moderate (−1.2 for Gamma, Log-Normal, and Log-Logistic and −1.1 for PTIII and Weibull PDFs) drought conditions. As shown in Table 9 , between July 2017 and May 2018, the SSI12 computed using Gamma, Log-Normal, PTIII and Weibull PDFs produced extreme (−2.4, −28, −2.1, and −2.0, respectively) drought conditions, while the SSI12 computed using the Log-Logistic PDF produced severe (−1.9) drought conditions. Hence, from Figure 6 as well as Table 8 Table 9 , it is apparent that the variability of drought intensity produced by the SSI12 computed using the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs increases as drought conditions increase from moderate to extreme. This is similar to the results obtained from the G1H008 and G1H013 stations. 4.2. The S-W Test for Normality on the SSI Calculated Using the Selected PDFs Thus far, the SSI12 results for the three selected streamflow gauging stations (G1H008, G1H013, and G1H020) have shown that the Gamma, Log-Normal, Log-Logistic, PTIII, and Weibull PDFs produced comparable results for mild (0.00 to −0.99) to moderate (−1.00 to −1.49) drought conditions. For severe (−1.5 to −1.99) to extreme (≤−2.0), there remains uncertainty on the choice of a suitable PDF for SSI calculation due to the increased variability in drought intensity produced by the different PDFs. To determine which PDF is most suitable for SSI computation for severe to extreme drought conditions, the S-W test for normality was used. The aim was to use the S-W test for normality to assess and select the best-fitting PDF between the Gamma, Log-Normal Log-Logistic, PTIII, and Weibull PDFs. As shown in Table 10 , none of the PDFs met the S-W condition for normality. Hence, the S-W normality test results were inconclusive for the considered PDFs. 4.3. Visual Inspection of the SSI Calculated Using the Selected PDFs In the absence of conclusive S-W results to aid in the selection of the most suitable PDFs for SSI calculation in the BRC, a visual inspection of the SSI calculated using the selected PDFs was carried out to identify any obvious systematic differences or similarities. This visual inspection approach is not uncommon in this type of research; it was employed by Svensson et al. (2017) [ ]. As shown in the example in Figure 7 , it was observed visually that the Gamma, Log-Normal, and Weibull PDFs were the only PDFs that were able to identify extreme (≤2.0) drought conditions for the G1H008 between July 2015 and July 2018. Visual inspection of Figure 8 Figure 9 Figure 10 shows that the SSI calculated using Gamma, Log-Normal, and Weibull PDFs were the only PDFs that were able to consistently identify extreme (≤2.0) drought conditions for the G1H008, G1H013, and G1H020. On the other hand, the SSI calculated using PTIII and Log-Logistic failed to consistently identify extreme drought conditions for G1H008, G1H013, and G1H020. 4.4. Evaluation of the Correlation between the SSI Computed Using the Selected PDFs A correlation statistical test was performed on the SSI time series computed using the Gamma, Log-Normal, PTIII, Log-Logistic, and Weibull PDFs to assess the extent of the similarities or differences amongst them. As shown in Table 11 , the correlation statistics show that they produced a positive correlation, mostly above 99%. Hence, they produced relatively similar SSI time series. The positive correlation is an indication they all produced similar major and minor drought conditions. In Table 11 , the correlation statistics for the SSI12 time series for G1H008, G1H013, and G1H020 computed using the Gamma PDF and SPI12 for Franschoek produced a positive correlation, mostly between 75% and 83%. The positive correlation is an indication that they all produced similar major drought conditions. 4.5. Comparison of the SSI with SPI Results The credibility of the SSI12 time series obtained using both the Gamma, Lo-Logistic, PTIII, Log-Normal, and Weibull PDFs was tested by comparing it with an SPI12 time series obtained using the Gamma PDF for both the Franschoek and Stellenbosch stations. As shown in Figure 11 , the Comparison between the SPI12 time series for Franschoek and the SPI12 time series for Stellenbosch reveals that both SPI time series are closely similar. Hence, since Franschoek is located within the BRC, the Franschoek SPI 12 time series was used to test the credibility of the SSI12 for G1H008, G1H013, and G1H020. The SSI12 time series for G1H008, G1H013, and G1H020 computed using the Gamma PDF were used for comparisons with the SPI 12 for Franschoek. For instance, both the SSI12 and SPI 12 identify the extreme drought conditions that occurred during the 2015–2018 period. Figure 12 Figure 13 Figure 14 , it can be observed that both the SSI12 for G1H008, G1H013, and G1H020 computed using the Gamma PDF are closely similar to the SPI12 for Franschoek in that they both produced or identified all the major drought events that occurred between the years 1990 and 2022. 4.6. Drought Assessment Using the SSI Calculated Using the Gamma, Log-Normal and Weibull PDFs As shown in Figure 8 Figure 9 Figure 10 Table 12 Table 13 Table 14 , historical drought assessment using the SSI12 time series calculated using the Gamma, Log-Normal, and Weibull PDFs in the BRC indicates that for both G1H008, G1H013, and G1H020, drought events have been occurring with more intensity between the years 1990 and 2022; the most severe were during the 2015–2018 extreme drought condition. 5. Discussion Despite the urgent need to improve drought monitoring using streamflow-based indices, insufficient studies have been conducted to test the applicability of the SSI for hydrological drought monitoring in SA. The few studies that exist have not tested the sensitivity of the SSI to various commonly used PDFs to recommend the most suitable PDFs [ ]. Thus, this study has investigated the applicability of the SSI as well as its sensitivity to the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs in the BRC, located in the WC province of SA. Streamflow time series spanning more than 30 years were acquired and used to compute the SSI, accumulated over a 12-month period (SSI12). The study has found that all the SSI12 computed using all the selected PDFs consistently produced drought events with similar onset and end times but with different intensities and magnitudes. The variability in drought intensity was more evident in the severe to extreme drought conditions. While the SSI calculated using the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs were able to detect all the mild to severe droughts during the study period, only the SSI computed using the Gamma, Log-Normal, and Weibull PDFs could detect the extreme drought conditions in the BRC. This is evidence that not all PDFs are suitable for SSI calculation in the BRC, and possibly in many other catchments in SA. Studies such as Botai et al. (2021) used the Gamma PDF to compute the SSI for drought monitoring in the WC province, on the basis that the Gamma is most commonly used [ ]. The SSI results from this study have demonstrated that it is essential that various PDFs be tested before being accepted and used for SSI computation and application in SA catchments. The Gamma PDF cannot be universally accepted for SSI calculation as it is for SPI calculation. To propose the most suitable PDF for SSI calculation and application in the BRC, the S-W test for normality was employed. However, the S-W test produced inconclusive results, so the visual inspection approach was resorted to. Through visual inspection of the SSI computed using all the selected PDFs, it was observed that only the SSI computed using the Gamma, Log-Normal, and Weibull could detect the only extreme drought conditions (2015–2018) that occurred during the 1990–2022 study period. Therefore, this study discourages the use of the Log-Logistic and PTII PDFs to calculate the SSI for the BRC catchment due to their failure to detect the 2015–2018 extreme drought conditions. Furthermore, using the SSI computed using the Gamma, Log-Normal, and Weibull, this study has found that droughts have been occurring with increased intensity and that the detected 2015–208 extreme drought events are the worst streamflow-based hydrological droughts in the BRC during the 1990–2022 study period. This agrees with Botai et al. (2021), who, from their study in the WC province, concluded that the duration and severity of drought conditions over the WC province have increased during the 1985–2020 period, and identified the 2015–2020 drought as the worst during the 1985–2020 period. Botai attributed the increasing drought conditions in the WC to reduced streamflow, influenced by reduced precipitation or a shift in seasonal precipitation, coupled with increased temperature [ ]. The study results also agree with other drought reports in the WC. For instance, Brühl and Visser (2021) reported the 2016–2018 WC drought as the worst drought in 100 years. It led to the anticipation of the so-called ‘day zero’, described as a situation in which the city of Cape Town would be left with only 10% of available water for human consumption [ ]. Hence, this study recommends the Gamma, Log-Normal, and Weibull PDFs for computation and application of the SSI in the BRC catchment. Since the selected streamflow gauging stations are well spatially distributed across the BRC, the SSI based on the Gamma, Log-Normal, and Weibull PDFs is recommended for application throughout the BRC catchment. The results from the evaluation of the correlation amongst the SSI computed using the selected PDFs and with the SPI showed that there are good similarities amongst the SSIs as well as between the SSI and the SPI. Comparison of the SSI with the SPI has also shown that both the SSI and SPI identify all the major droughts, including the 2015–2018 extreme drought event. Hence, this study has shown that the SSI is capable of characterizing streamflow-based hydrological drought in the BRC. The close correlation between the SSI12 and the SPI12 is an indication that the streamflow-based hydrological drought may be caused by climate factors such as precipitation deficit, in concurrence with Botai et al. (2021) [ ]. Studies have shown the importance of using the SSI and the SPI to obtain the Propagation Threshold (PT) from meteorological drought to hydrological drought. This helps to provide early warning information for hydrological drought, which is vital for drought preparedness and mitigation [ ]. This study thus proposes that future research should focus on more investigations using the SSI and the SPI to study the propagation of drought from meteorological to hydrological in the BRC and other catchments in SA. 6. Conclusions To contribute to the provision of tested scientific knowledge on the effective application of the SSI in SA, this study has investigated the application of the SSI for hydrological drought monitoring in the BRC and WC provinces of SA. Using more than 30-year records of streamflow data (G1H008, G1H013, and G1H020) from the BRC, as well as five PDFs (PTIII, Log-Normal, Log-Logistic, Weibull, and Gamma), 12 months’ SSI (SSI12) time series were computed and analyzed. The study has found that all the SSI time series computed using all the selected PDFs detected mild to extreme drought conditions with varying intensities and magnitudes during the 1990–2022 study period. It is therefore recommended that different PDFs, including those not tested in this study, always be tested before they are accepted and used for SSI computation and application in all SA catchments. On the basis that only the SSI time series computed using the Gamma, Log-Normal, and Weibull PDFs detected the 2015–2018 extreme droughts events, the Gamma, Log-Normal, and Weibull PDFs are recommended for SSI computation and application in the BRC. The 2015–2018 extreme drought in the BRC has been reported by other studies to have been the worst drought in almost 100 years. Comparison of the SSI12 (Gamma) with the SPI12 (Gamma) has provided evidence that the SSI is credible and is applicable for hydrological drought monitoring in the BRC. Both the SSI and the SPI identified all the major drought events during the study period, including the 2015–2018 extreme drought event. Based on these outcomes, it is recommended that further studies be conducted to investigate the propagation and evolution of drought from meteorological to hydrological drought using the SSI and SPI. These investigations will result in improved early warning information for hydrological drought in the BRC and other catchments in SA. When comparing the study outcomes with those of other studies, it can be concluded that the detected droughts during the 1990–2022 study period are caused largely by climate-related factors such as precipitation deficit and increased evaporation. Thus, in anticipation of more frequent and intense droughts due to climate change factors, it is recommended that water resource managers take proactive action in searching for strategies to improve water resource management and drought preparedness, mitigation, and response in the WC province of SA. The application of the SSI for hydrological drought monitoring is relatively new in SA. Hence, this study has provided a foundation for more research on the application of the SSI in the WC and other catchments in SA. Author Contributions Conceptualization and methodology: M.B.M.; Software: N.S.M.; validation, formal analysis, and data curation: M.B.M., T.K., D.K. and N.S.M.; Original draft preparation: M.B.M.; Review and editing: M.B.M.; Supervision: T.K. and D.K. All authors have read and agreed to the published version of the manuscript. This research was funded by the Department of Water and Sanitation, South Africa. Informed Consent Statement Informed consent was obtained from all subjects involved in the study. Data Availability Statement Not applicable. The Department of Water and Sanitation, South Africa is acknowledged for providing funding to publish the article. Conflicts of Interest The authors declare no conflict of interest. Figure 2. Historical streamflow patterns (1990–2022) at gauging stations G1H008, G1H013, and G0H020 on the Berg and Klein Berg rivers in the Berg River Catchment (m^3/s ≅ cumecs). Figure 3. Historical rainfall (mm) (1980–2021) patterns in Franschoek, Stellenbosch, and Malmesbury towns, located in and around the Berg River Catchment. Figure 4. SSI12 results for G1H008 streamflow time series computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull Probability Distribution Functions. Figure 5. SSI12 results for G1H013 Streamflow time series computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull Probability Distribution Functions. Figure 6. SSI12 results for G1H020 streamflow time series computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull Probability Distribution Functions. Figure 7. Example of visual inspection of the SSI12 calculated using the Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull for the G1H008 streamflow gauging station; from July 2015 to July 2018. Figure 8. SSI12 time series for G1H008 computed using Gamma, PTIII (PearsonIII), Log-Normal, Log-Logistic, and Weibull PDFs in the Berg River Catchment. Figure 9. SSI12 time series for G1H013 calculated using Gamma, PTIII (PearsonIII), Log-Normal, Log-Logistic, and Weibull PDFs in the Berg River Catchment. Figure 10. SSI12 time series for G1H020 computed using Gamma, PTIII (PearsonIII), Log-Normal, Log-Logistic, and Weibull PDFs in the Berg River Catchment. Figure 11. Comparison between SPI12 time series for Franschoek and SPI12 time series for Stellenbosch. Table 1. Streamflow gauging stations that were used to obtain river discharge data for SSI calculations in the Berg River Catchment. Streamflow Gauging Station Identity River Location Coordinates (Latitude: Longitude) Period (Years) G1H008 Klein Berg −33.313889:19.074722 1990 to 2022 (32 Years) G1H013 Berg −33.130833:18.862778 1990 to 2022 (32 Years) G1H020 Berg −33.707778:18.991111 1990 to 2022 (32 Years) SPI/SSI Values Drought Classification ≥2.00 Extremely Wet 1.50 to 1.99 Severely Wet 1.00 to 1,49 Moderately Wet 0.00 to 0.99 Mildly Wet 0.00 to −0.99 Mild Drought −1.00 to −1.49 Moderate Drought −1.5 to −1.99 Severe Drought ≤−2.00 Extreme Drought Table 4. SSI12 computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs for the G1H008 station between November 2004 and June 2005. Month-Year SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought Gamma Classification Log-Logistic Classification Log-Normal Classification PTIII Classification Weibull Classification November 2004 −1.6 Severe −1.4 Moderate −1.5 Severe −1.3 Moderate −1.5 Severe December 2004 to April 2005 −1.6 Severe −1.4 Moderate −1.6 Severe −1.4 Moderate −1.6 Severe May 2005 −1.6 Severe −1.4 Moderate −1.6 Severe −1.4 Moderate −1.5 Severe June 2005 −1.6 Severe −1.4 Moderate −1.5 Severe −1.3 Moderate −1.5 Severe Average −1.6 Severe −1.4 Moderate −1.6 Severe −1.4 Moderate −1.6 Severe Table 5. SSI12 computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs for the G1H008 station between December 2015 and April 2016. Month-Year SSI12 Drought SSI12 Drought SSI1 Drought SSI12 Drought SSI12 Drought Gamma Classification Log-Logistic Classification Log-Normal Classification PTIII Classification Weibull Classification December 2015 to March 2016 −2.2 Extreme −1.6 Severe −2.3 Extreme −1.6 Severe −2.0 Extreme April 2016 −2.1 Extreme −1.6 Severe −2.2 Extreme −1.6 Severe −1.9 Severe Average −2.2 Extreme −1.6 Severe −2.3 Extreme −1.6 Severe −2.0 Extreme Table 6. SSI12 computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs for the G1H013 station between November 2003 and May 2005. Month-Year SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought Gamma Classification Log-Logistic Classification Log-Normal Classification PTIII Classification Weibull Classification November 2003 −1.2 Moderate −1.2 Moderate −1.3 Moderate −1.2 Moderate −1.2 Moderate December 2003 to May 2004 −1.3 Moderate −1.3 Moderate −1.3 Moderate −1.3 Moderate −1.3 Moderate June 2004 −1.1 Moderate −1.2 Moderate −1.3 Moderate −1.2 Moderate −1.1 Moderate July 2004 −1.1 Moderate −1.2 Moderate −1.2 Moderate −1.1 Moderate −1.1 Moderate August 2004 −1.0 Moderate −1.0 Moderate −1.0 Moderate −1.0 Moderate −1.0 Moderate September 2004 −1.1 Moderate −1.2 Moderate −1.1 Moderate −1.1 Moderate −1.1 Moderate October 2004 −1.1 Moderate −1.1 Moderate −1.1 Moderate −1.1 Moderate −1.1 Moderate November 2004 −1.0 Moderate −1.1 Moderate −1.0 Moderate −1.0 Moderate −1.0 Moderate December 2004 to April 2005 −1.2 Moderate −1.2 Moderate −1.1 Moderate −1.1 Moderate −1.1 Moderate May 2005 −1.2 Moderate −1.1 Moderate −1.0 Moderate −1.1 Moderate −1.1 Moderate Average −1.2 Moderate −1.2 Moderate −1.2 Moderate −1.1 Moderate −1.2 Moderate Table 7. SSI12 computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs for the G1H013 station between August 2017 and May 2018. Month-Year SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought Gamma Classification Log-Logistic Classification Log-Normal Classification PTIII Classification Weibull Classification August 2017 −2.3 Extreme −1.9 Severe −2.9 Extreme −2.2 Extreme −2.1 Extreme September 2017 to October 2017 −2.5 Extreme −1.8 Severe −3.0 Extreme −1.9 Severe −2.2 Extreme November 2017 −2.4 Extreme −1.8 Severe −3.0 Extreme −2.0 Extreme −2.2 Extreme December 2017 −2.6 Extreme −1.9 Severe −3.1 Extreme −2.0 Extreme −2.3 Extreme January 2018 −2.7 Extreme −1.9 Severe −3.2 Extreme −2.0 Extreme −2.3 Extreme February 2018 −2.8 Extreme −1.9 Severe −3.3 Extreme −2.1 Extreme −2.4 Extreme March 2018 to April 2018 −2.9 Extreme −1.9 Severe −3.4 Extreme −2.1 Extreme −2.5 Extreme May 2018 −2.7 Extreme −1.9 Severe −3.3 Extreme −2.1 Extreme −2.4 Extreme Average −2.6 Extreme −1.9 Severe −3.2 Extreme −2.1 Extreme −2.3 Extreme Table 8. SSI12 computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs for the G1H020 station between July 2003 and June 2004. Month-Year SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought Gamma Classification Log-Logistic Classification Log-Normal Classification PTIII Classification Weibull Classification July 2003 −1.1 Moderate −1.2 Moderate −1.2 Moderate −1.1 Moderate −1.0 Moderate August 2003 −1.2 Moderate −1.2 Moderate −1.3 Moderate −1.2 Moderate −1.2 Moderate September 2003 −1.1 Moderate −1.1 Moderate −1.1 Moderate −1.1 Moderate −1.1 Moderate October 2003 −1.2 Moderate −1.2 Moderate −1.2 Moderate −1.2 Moderate −1.2 Moderate November 2003 −1.2 Moderate −1.2 Moderate −1.2 Moderate −1.1 Moderate −1.1 Moderate December 2003 to May 2004 −1.2 Moderate −1.2 Moderate −1.2 Moderate −1.2 Moderate −1.1 Moderate June 2004 −1.0 Moderate −1.0 Moderate −1.0 Moderate −1.0 Moderate −0.9 Mild Average −1.2 Moderate −1.2 Moderate −1.2 Moderate −1.1 Moderate −1.1 Moderate Table 9. SSI12 computed using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs for the G1H020 station between July 2017 and May 2018. Month-Year SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought SSI12 Drought Gamma Classification Log-Logistic Classification Log-Normal Classification PTIII Classification Weibull Classification July 2017 −1.9 Severe −1.8 Severe −2.3 Extreme −2.1 Extreme −1.7 Severe August 2017 −2.2 Extreme −1.8 Severe −2.6 Extreme −2.0 Extreme −1.9 Severe September 2017 −2.3 Extreme −1.9 Severe −2.7 Extreme −2.0 Extreme −2.0 Extreme October 2017 −2.5 Extreme −1.9 Severe −2.9 Extreme −2.1 Extreme −2.1 Extreme November 2017 −2.4 Extreme −2.0 Severe −2.8 Extreme −2.2 Extreme −2.0 Extreme December 2017 to January 2018 −2.4 Extreme −1.9 Severe −2.9 Extreme −2.1 Extreme −2.1 Extreme February 2018 −2.6 Extreme −2.0 Extreme −3.0 Extreme −2.2 Extreme −2.1 Extreme March 2018 −2.7 Extreme −2.0 Extreme −3.1 Extreme −2.2 Extreme −2.2 Extreme April 2018 −2.6 Extreme −2.0 Extreme −3.0 Extreme −2.2 Extreme −2.2 Extreme May 2018 −2.4 Extreme −1.9 Severe −2.9 Extreme −2.2 Extreme −2.1 Extreme Average −2.4 Extreme −1.9 Severe −2.8 Extreme −2.1 Extreme −2.0 Extreme Table 10. Shapiro-Wilk Normality Test results for SSI−12 time series calculated using Gamma, Log-Logistic, Log-Normal, PTIII, and Weibull PDFs on streamflow gauging stations G1H008, G1H013 and Shapiro-Wilk Test for Normality Gamma Log-Logistic PTIII Log-Normal Weibull G1H008 W = 0.97464 W = 0.96785 W = 0.97958 W = 0.94277 W = 0.94277 p-value = 3.06× 10^−6 p-value = 1.83× 10^−7 p-value = 3.062× 10^−5 p-value = 5.416× 10^−11 p-value = 5.416× 10^−11 G1H013 W = 0.9717 W = 0.95804 W = 0.97103 W = 0.94802 W = 0.97389 p-value = 8.386× 10^−7 p-value = 5.186× 10^−9 p-value = 6.346× 10^−7 p-value = 2.293× 10^−10 p-value = 2.14× 10^−6 G1H020 W = 0.99019 W = 0.98243 W = 0.99018 W = 0.97238 W = 0.98534 p-value = 0.01188 p-value = 0.0001341 p-value = 0.01177 p-value = 1.186× 10^−6 p-value = 0.0006527 Table 11. Correlation statistics for the SSI time series computed using the Gamma, Log-Normal, PTIII, Log-Logistic, and Weibull. SSI Gamma SSI log-Logistic SSI log-Normal SSI PTIII SSI Weibull Franschoek SPI12 (Gamma) SSI Gamma 1 SSI log-Logistic 0.98223 1 SSI log-Normal 0.984512 0.947839 1 SSI PTIII 0.986172 0.998936 0.953691 1 SSI Weibull 0.991315 0.98774 0.977377 0.990254 1 Franschoek SPI12 (Gamma) 0.741493 0.750306 0.715999 0.752611 0.754136 1 SSI Gamma SSI log-Logistic SSI log-Normal SSI PTIII SSI Weibull Franschoek SPI12 (Gamma) SSI Gamma 1 SSI log-Logistic 0.989213 1 SSI log-Normal 0.991976 0.965913 1 SSI PTIII 0.994384 0.998255 0.976084 1 SSI Weibull 0.995609 0.995802 0.977003 0.998044 1 Franschoek SPI12 (Gamma) 0.803445 0.81042 0.782366 0.811022 0.810208 1 SSI Gamma SSI log-Logistic SSI log-Normal SSI PTIII SSI Weibull Franschoek SPI12 (Gamma) SSI Gamma 1 SSI log-Logistic 0.994149 1 SSI log-Normal 0.994499 0.980957 1 SSI PTIII 0.997659 0.997659 0.987417 1 SSI Weibull 0.992818 0.993924 0.975533 0.996049 1 Franschoek SPI12 (Gamma) 0.817348 0.827468 0.804458 0.82031 0.816487 1 Table 12. Drought assessment during the period between 1990 and 2022 using the SSI12 calculated using the recommended Gamma, Log-Normal, and Weibull PDFs for the G1H008 streamflow gauging station. Streamflow Gauging Station Drought Period Average SSI12 Drought Classification Gamma Log-Normal Weibull June 2000 to June 2001 −0.7 −0.7 −0.8 Mild Drought September 2004 to May 2005 −1.6 −1.6 −1.5 Severe Drought G1H008 September 2015 to April 2016 −2.1 −2.2 −2.0 Extreme Drought August 2017 to May 2018 −2.6 −2.9 −2.3 Extreme Drought September 2019 to July 2020 −1.3 −1.3 −1.2 Moderate Drought Table 13. Drought assessment during the period between 1990 and 2022 using the SSI12 calculated using the recommended Gamma, Log-Normal, and Weibull PDFs for the G1H013 streamflow gauging station. Streamflow Gauging Station Drought Period Average SSI12 Drought Classification Gamma Log-Normal Weibull August 2000 to June 2001 −0.6 −0.6 −0.7 Mild Drought August 2003 to May 2005 −1.2 −1.2 −1.2 Moderate Drought G1H013 October 2011 to July 2012 −1.0 −1.0 −1.0 Moderate Drought August 2015 to June 2016 −1.4 −1.4 −1.3 Moderate Drought July 2017 to June 2018 −2.5 −3.0 −2.2 Extreme Drought July 2018 to February 2019 −1.1 −1.1 −1.1 Moderate Drought Table 14. Drought assessment during the period between 1990 and 2022 using the SSI12 calculated using the recommended Gamma, Log-Normal, and Weibull PDFs for the G1H020 streamflow gauging station. Streamflow Gauging Station Drought Period Average SSI12 Drought Classification Gamma Log-Normal Weibull August 2000 to June 2001 −0.4 −0.4 −0.5 Mild Drought July 2003 to June 2004 −1.2 −1.2 −1.1 Moderate Drought G1H020 August 2011 to July 2012 −1.3 −1.3 −1.2 Moderate Drought August 2015 to January 2016 −1.1 −1.1 −1.1 Moderate Drought July 2017 to June 2018 −2.4 −2.8 −2.0 Extreme Drought Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https: Share and Cite MDPI and ACS Style Mukhawana, M.B.; Kanyerere, T.; Kahler, D.; Masilela, N.S. Application of the Standardised Streamflow Index for Hydrological Drought Monitoring in the Western Cape Province, South Africa: A Case Study in the Berg River Catchment. Water 2023, 15, 2530. https://doi.org/10.3390/w15142530 AMA Style Mukhawana MB, Kanyerere T, Kahler D, Masilela NS. Application of the Standardised Streamflow Index for Hydrological Drought Monitoring in the Western Cape Province, South Africa: A Case Study in the Berg River Catchment. Water. 2023; 15(14):2530. https://doi.org/10.3390/w15142530 Chicago/Turabian Style Mukhawana, Mxolisi Blessing, Thokozani Kanyerere, David Kahler, and Ndumiso Siphosezwe Masilela. 2023. "Application of the Standardised Streamflow Index for Hydrological Drought Monitoring in the Western Cape Province, South Africa: A Case Study in the Berg River Catchment" Water 15, no. 14: 2530. https://doi.org/10.3390/w15142530 Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details Article Metrics	{"url":"https://www.mdpi.com/2073-4441/15/14/2530?utm_campaign=releaseissue_waterutm_medium=emailutm_source=releaseissueutm_term=titlelink26","timestamp":"2024-11-08T11:46:27Z","content_type":"text/html","content_length":"593088","record_id":"<urn:uuid:3dd4cde9-e180-4b4b-a496-9a394e56227f>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00640.warc.gz"}
Computability is the study of a massively important question: do there exist any problems that are impossible for a computer to solve? The Halting Problem # It turns out that the above question itself is impossible to solve: in other words, there cannot exist a program HALT which determines if a program can halt in finite time given a particular input. This was originally proposed by Alan Turing- he proved the nonexistence by attempting to feed the Halting Problem into itself: if the Halting Problem doesn’t halt, then it is supposed to output an answer. That means that the Halting Problem would state that the Halting Problem halts, even though it didn’t. This paradox led to demonstrating that the Halting Problem simply cannot be solved. Reductions # Reducing a problem A to another problem B means that we can solve problem A if we know how to solve problem B. For instance, we might be able to write psuedocode that has a whole bunch of known components, but rely on the output of problem B in order to determine the final output. One application of reduction is to show that a program cannot exist if it requires the Halting Problem as a component.	{"url":"https://bigdatajunction.com/hive/discrete-math/computability/","timestamp":"2024-11-09T22:10:24Z","content_type":"text/html","content_length":"28267","record_id":"<urn:uuid:f8085aec-148a-4af9-ab3e-17e14f510dab>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00071.warc.gz"}
Conference "Approximation and discretization" The conference will be held from August 30 to September 3 in the Moscow hotel "Golden Ring", in the room "Serguiev Posad". 29 August is an arrival day, 4 September is a departure day. We plan to have morning sessions (10.00-13.30, plenary talks for 50 minutes) and evening sessions (15.00-18.20, evening talks for 25 minutes). Participant registration: on Monday, August 30, in the room "Serguiev Posad", from 9.00 to 9.50. The sessions on Thursday, September 2, are dedicated to the 70th anniversary of the academician B.S. Kashin. Plenary talks: S.V. Astashkin, P. Berna, P.A. Borodin, A.V. Dereventsov, R. DeVore, M.I. Dyachenko, G. Garrigos, G.A. Karagulyan, K. Kazarian, S.V. Konyagin, E. Kopecká, A.M. Olevskii, E.M. Semenov, M.A. Skopina, М.М. Skriganov, A.Yu. Shadrin, V.N. Temlyakov, S.Yu. Tikhonov. Evening talks: M.Sh. Burlutskaya, L.Sh. Burusheva, M.A. Chahkiev, V.V. Galatenko, S.L. Gogyan, B.I. Golubov, D.V. Gorbachev, M.G. Grigoryan, A.B. Kalmynin, N.N. Kholshchevnikova, E.D. Kosov, V.V. Lebedev, I.V. Limonova, S.Ya. Novikov, A.S. Orlova, M.G. Plotnikov, K.V. Runovskii, S.P. Sidorov, K.S. Shklyaev, P.A. Terekhin, A.I. Tyulenev, Al.R. Valiullin, Ar.R. Valiullin, M.A. Valov, K.S. Vishnevetskii. More details: http://approx-lab.math.msu.su/conf_kashin70.html	{"url":"https://mathcenter.ru/en/conference-approximation-and-discretization-2021","timestamp":"2024-11-09T14:31:13Z","content_type":"text/html","content_length":"43902","record_id":"<urn:uuid:a35159df-4e8d-466f-9b70-06d1cdf068aa>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00756.warc.gz"}
Mach Altitude Calculator - Super Avionics Mach Altitude Calculator Mach Altitude Calculator (Knot & Foot Units) Mach Altitude Calculator Take the guesswork out of high-altitude flight with our innovative Mach Altitude Calculator. This essential tool goes beyond simply calculating Mach number. It factors in the critical influence of air temperature on the speed of sound at different altitudes. Why is this important? • Optimized Flight Planning: Ensure your aircraft operates within its ideal speed range, maximizing fuel efficiency and safety at every stage of your flight. • Precise Data for Design and Testing: Whether you’re developing cutting-edge aircraft or conducting rigorous flight tests, our calculator provides the accurate Mach number data you need for • Unparalleled Training Scenarios: Elevate your pilot training with realistic simulations that incorporate precise Mach number calculations based on altitude variations. Beyond Basic Calculations: This calculator leverages atmospheric data to deliver the most accurate Mach numbers possible. No more relying on generic estimates – this tool empowers you with the knowledge to make informed decisions in the air. Formula for Mach Altitude Calculator The formula for calculating the Mach number (M) based on altitude (h) involves a few steps. Here’s how you can determine the Mach number at a given altitude: Determining Speed of Sound at a Specific Altitude The speed of sound (a) varies with altitude and temperature. For standard atmospheric conditions, the speed of sound can be approximate by: The speed of sound (a) can be calculated using the following equation: a = √(γ * R * T) • a – Speed of sound (m/s) • γ (gamma) – Ratio of specific heats (approximately 1.4 for air) • R – Gas constant for air (approximately 287.05 J/KgK) • T – Absolute temperature (Kelvin) Note that this formula requires the absolute temperature (Kelvin) at the specific altitude you’re interested in. Estimating Temperature at A Specific Altitude The temperature (T) at a specific altitude can be estimate using the International Standard Atmosphere (ISA) model. For altitudes up to 11 km, the temperature can be calculate as: T = T0 + L h • · T0 is the standard temperature at sea level (288.15 K), • · L is the temperature lapse rate(-0.0065 K/m), • · h is the altitude in meters. Calculating Mach Number At a Specific Altitude Once you have the speed of sound, the Mach number can be calculate using the formula: M = V / a • · V is the velocity of the object, • · a is the speed of sound at the given altitude. General Terms and Useful Conversions Altitude (meters) Temperature (K) Speed of Sound (m/s) 0 288.15 340.3 1000 281.65 336.4 2000 275.15 332.5 3000 268.65 328.6 4000 262.15 324.6 5000 255.65 320.5 6000 249.15 316.4 7000 242.65 312.2 8000 236.15 308.0 9000 229.65 303.7 10000 223.15 299.5 Example of Mach Altitude Calculation Let’s calculate the Mach number for an object traveling at 400 m/s at an altitude of 6000 meters. 1. 1. Determine the temperature at 6000 meters: T = 288.15 + (-0.0065 * 6000) T = 288.15 – 39 T = 249.15 K • 2. Calculate the speed of sound at 6000 meters: a = sqrt(1.4 * 287.05 * 255.65) a ≈ 320.5 m/s a = √(γ * R * T) a = √(1.4 * 287.05 * 249.15) a ≈ 316.42 m/s) • 3. Determine the Mach number: M = 400 / 316.42 M ≈ 1.26 The Mach number is approximately 1.26 FAQs About Mach Number in Aviation: 1. What is the Mach number? The Mach number (denoted by M) is a dimensionless unit that expresses the speed of an object relative to the speed of sound in the medium it’s traveling through. In simpler terms, it tells you how many times faster an object is moving compared to the speed sound travels in that specific environment. Here’s the formula for Mach number: M = V / a • M: Mach number (unitless) • V: Object’s speed (usually in meters per second or kilometers per hour) • a: Speed of sound in the medium (usually in meters per second or kilometers per hour) 2. Why is the Mach number important in aviation? The Mach number is critical in aviation for several reasons: • Aerodynamic effects: As an aircraft approaches the speed of sound (Mach 1), it encounters significant aerodynamic changes. These include increased drag, turbulence, and control difficulties. Understanding the Mach number helps pilots avoid these challenges by adjusting their speed strategically. • Transonic regime: The region around Mach 1 is called the transonic regime. It’s a challenging zone where the airflow over different parts of the aircraft can be subsonic (below Mach 1) and supersonic (above Mach 1) simultaneously. This can lead to buffeting, shock waves, and other phenomena that can affect aircraft stability and performance. Knowing the Mach number helps pilots navigate this zone safely. • Supersonic flight: For aircraft designed for supersonic travel (exceeding Mach 1), the Mach number becomes an essential parameter for monitoring performance and efficiency. 3. How does altitude affect the Mach number? Altitude significantly affects the Mach number because the speed of sound itself varies with altitude. Here’s the connection: • Temperature and Speed of Sound: The speed of sound is directly related to the temperature of the surrounding air. Colder air molecules move slower, resulting in a lower speed of sound. • Altitude and Temperature: As you climb to higher altitudes, the air temperature generally decreases. • Impact on Mach Number: Because the speed of sound decreases with altitude, an aircraft flying at a constant speed will reach a lower Mach number at higher altitudes. For example, an airplane flying at 340 meters per second (756 mph) would have a Mach number of 1 (speed of sound) at sea level, but it would only have a Mach number of 0.85 (85% the speed of sound) at 10,000 meters altitude (assuming the temperature difference between these altitudes). This is why pilots need to consider both their airspeed and altitude when referencing the Mach number to understand the true behavior of the aircraft.	{"url":"https://superavionics.com/mach-altitude-calculator/","timestamp":"2024-11-06T23:09:26Z","content_type":"text/html","content_length":"349659","record_id":"<urn:uuid:a28bd62a-666e-4935-800a-2ebd98475b98>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00211.warc.gz"}
Solve Speed Questions Easily \| PSLE Math Simplified \| Practicle Solve speed questions easily When it comes to preparing for PSLE math, speed always comes up as one of the most challenging Primary 6 topics according to most students and parents. In this simple Practicle guide on speed, let’s look at the different kind of speed questions that can appear in your PSLE math paper and learn how to solve them. Here’s what we’ll cover: The only speed formula you’ll need Let’s start with the Speed questions you’ll see in Paper 1. These Math problems usually test you on your basics. This means that you can solve them with the Distance Speed Time formula, the Speed Time Distance formula or the Time Speed Distance formula. Let’s look at each ingredient of these formulas: • Distance – Distance measures how far the object travels • Speed – Speed measures how fast an object travels from one place to another. (usually in m/s or km/h) • Time – Time is how long the object takes to travel If you are driving a car at a speed of 60 km/h, this means that the car is going to travel 60 km every hour. When you drive it for 2 hours, you would have covered a total distance of 60 km/h x 2 h which is 120 km. This is why distance = speed * time. Knowing this relationship between speed, time and distance is key to solving any speed question. Speed Questions in Math Paper 1 Now let’s see how a speed question in your PSLE Paper 1 might look like. These speed questions will provide you with 2 known values out of the 3 ingredients most of the time. The missing third value that you are asked to solve can either be the speed, time or distance. For example, you might be asked to solve a question about speed like this: Alvin was walking at an average speed of 70 m/min. At this speed, how long did he take to walk a distance of 910 m? [Source: St Nicholas Girls Primary, Primary 6 Prelim Exam Paper] As you can see, we have both the speed and distance in this Math problem. To solve for the time, we need to use the distance speed time formula (refer to the DST triangle above) and move the terms around to that it looks like this: Time = Distance / Speed. When we replace the known values into this formula, we can easily calculate the time Alvin took by dividing the distance of 910 m by his speed of 70 m/min. Did you get an answer of 13 min? Next, let’s look at the speed questions that may appear in your PSLE Paper 2. This is where things start to get a bit more complicated. Speed Questions in Math Paper 2 For PSLE speed questions in Paper 2, you might start seeing word problems that involve 1 object moving and stopping along its journey, 2 objects that starts from the same point and move in different directions or 2 objects that are trying to catch up with one another etc. Yes, that can be VERY confusing to our Primary 6 students. Let’s look at an example of one such challenging speed question in Practicle which is similar to what you might see in your exams. Aladdin and Jasmine were having a race on their magic flying carpets. They left the same starting point for the market at 13 00. Both of them did not change their speed throughout the journey. Aladdin travelled at a constant speed of 60 km/h. When he reached the market at 13 30, Jasmine was 3 km from the market. What is Jasmine’s speed? Now while reading this Math problem, many kids would already start thinking to themselves “This sounds super hard! Who’s travelling where?!?!” Luckily, there’s a way to simplify this entire complicated-sounding speed problem! Let’s jump to the next section where to learn a useful problem-solving technique called the “draw a diagram” strategy. This will be the life-saver you need when it comes to solving speed questions. How drawing speed diagrams can save your day Whenever you come across a complicated Math problem on speed, draw the speed diagram to see what’s happening. It’ll make things so much clearer! As you draw the speed diagram, you break down the problem step by step. This is important to get rid of the confusion! As you identify the key people/objects involved, their locations and their travel paths, you’ll understand their relationship better. Not to mention, the more time to focus on the Mathematical and problem solving aspects of the problem. Check out this Practicle Math video to learn the steps to draw a speed diagram with an example question on speed! Try drawing speed diagrams to represent the problems on speed today and see how it helps you solve these problems faster. Alright, now that you’ve learnt how to solve the different kinds of speed questions that you might see in your Primary 6 PSLE Math exam, how confident are you in getting that AL1? If you need more speed questions to practice on, check out Practicle where we have unlimited exam questions to help you polish up your Math skills, along with video explanations from school teachers!	{"url":"https://practicle.sg/primary-6-math/speed-questions/","timestamp":"2024-11-08T11:24:55Z","content_type":"text/html","content_length":"364736","record_id":"<urn:uuid:54e5d469-50e6-4acd-8ee1-7dd6c352ad7e>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00577.warc.gz"}
SRGS grammars, especially when combined with SISR, allow for the creation of very complex and flexible documents. The underlying logic of rule expansions and rule references makes writing SRGS grammars almost like developing small applications (and by adding JavaScript with SISR this becomes even more true). There are a number of traps new grammar writers often fall into that can be avoided with a few best practices. Consider a grammar designed to capture a number with two digits: root $TwoDigits; $TwoDigits = {out=0} ([$TensDigit {out+=rules.latest()}] [$OnesDigit {out+=rules.latest()}] \| $TeensDigit{out+=rules.latest()} \| [$OnesDigit {out+=rules.latest()*10}] [$OnesDigit {out+=rules.latest()}]) {out=out.toString()}; $TensDigit = ten {out=10} \| twenty {out=20} \| thirty {out=30} \| forty {out=40} \| fifty {out=50} \| sixty {out=60} \| seventy {out=70} \| eighty {out=80} \| ninety {out=90}; $TeensDigit = eleven {out=11} \| twelve {out=12} \| thirteen {out=13} \| fourteen {out=14} \| fifteen {out=15} \| sixteen {out=16} \| seventeen {out=17} \| eighteen {out=18} \| nineteen {out=19}; $OnesDigit = zero {out=0} \| one {out=1} \| two {out=2} \| three {out=3} \| four {out=4} \| five {out=5} \| six {out=6} \| seven{out=7} \| eight {out=8} \| nine {out=9}; Avoid Ambiguity Any input to a grammar should only have one valid parse (the Grammar Editor tool, included with the LumenVox Speech Tuner, can show how many parses a grammar returns for a given input). The more parses a grammar has, the longer it takes the Engine to decode an utterance against it. It also decreases accuracy. As the number of valid parses increases, decode time can increase In the grammar above, the grammar is capable of correctly handling parses such as "two one" or "twenty one." But if a caller says just "one", it allows for two valid parses, as the last part of the root rule allows two optional $OnesDigit rule matches. In this case, each parse has a different interpretation: the first $OnesDigit match multiplies the interpretation by 10, returning a result of 10, while the second one returns a result of 1. This sort of ambiguity not only increases decode time while decreasing accuracy, it also makes it harder for your application to correctly handle results. You would probably not expect a caller saying "one" to return a result of "10", but that is precisely one thing this grammar allows for. Eliminate Unwanted Parses You would obviously want to not allow the above example, where "one" has a valid parse that returns as "ten." But even allowing "one" to be a valid parse is quite possibly a bad idea if all you want to capture are two digit strings. The grammar allows for other parses such as "twenty zero"(it returns with an interpretation of "20"), or "ten two" (returning with an interpretation of "12"). Even a null input is a valid parse ("" returns with an interpretation of "0"). Unwanted parses slow down decodes and reduce accuracy. It's pretty unlikely that a caller would ever say "twenty zero" or that a developer would want to allow for that sort of input. Accounting for these sorts of unlikely cases increases the probability that a caller behaving appropriately will be misrecognized. E.g. somebody who says "twenty two" might get mistaken for the unreasonable "twenty Keep Rules Compact The larger and more complex rules are, the longer it takes to compile a grammar or decode against it. One good trick to keeping rules short is to combine rules with common words, where possible. For instance, the following rule: $name = James Anderson \| Jim Anderson \| Jimmy Anderson \| James \| Jim \| Jimmy; Can be combined into: $name = (James \| Jim \| Jimmy) [Anderson]; While it is a relatively small savings for one rule, across large grammars this sort of compactness can add up, decreasing load and decode times. Be Careful with Recursion SRGS allows for recursive rules, that is rules with references to themselves. Any time you work with recursion, you must be careful to avoid infinite loops. Since the LumenVox grammar parser parses from left to right, you should always avoid doing left-hand recursion. For instance, the following rule will match the word "foo" any number of times: $rule = foo ($rule \| $NULL); If the input is "foo foo," the Engine parses the rule, expanding the reference $rule each time until it matches $NULL and terminates. On the other hand, if your rule is written: $rule = ($rule \| $NULL) foo; The parser will get caught in an infinite loop. The first thing it will attempt to do is to expand the $rule reference, only to expand it again, and again, ad infinitum.	{"url":"https://www.lumenvox.com/knowledgebase/index.php?/article/AA-01066/0","timestamp":"2024-11-13T16:42:17Z","content_type":"application/xhtml+xml","content_length":"112351","record_id":"<urn:uuid:6cd2f88e-af5d-4962-9254-67385e5dd08a>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00869.warc.gz"}
Is cardiovascular fitness associated with structural brain integrity in midlife? Evidence from a population-representative birth cohort study \| Aging Figure 4. Cortical surface area (cm2) at age 45 and rate of decline in VO2Max scores (mL/min/Kg) from age 26 to 45. (A) Surface area. (B) Graph showing the correlation between total surface area (cm2, y-axis) and the rate of decline in VO2Max (average decrease in mL/min/kg between each wave of data collection; x-axis). Study members with VO2Max scores declining at a faster rate over time (i.e., larger slopes) did have slightly smaller total surface area (β = -0.09, 95% CI = -0.17 to -0.004, p=0.04). There were no regionally-specific associations. cm3 = centimeters cubed; VO2Max = volume of maximum oxygen uptake; mL/min/kg = milliliters per minute per kilogram; β = standardized coefficient; CI = confidence interval.	{"url":"https://static-site-aging-prod2.impactaging.com/figure/104112/f4","timestamp":"2024-11-02T12:10:07Z","content_type":"application/xhtml+xml","content_length":"11478","record_id":"<urn:uuid:eb648a3b-5a25-4f80-b62f-938d5e088949>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00386.warc.gz"}
Kamal and Monika appeared for an interview for two vacancies. The probability of Kamal’s selection is $\ The probability of both getting selected means multiplying the events of each person whereas the probability of selecting only one at a time requires addition of the events of both the persons. Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. Complete step by step solution: The probability of Kamal’s selection is $\dfrac{1}{3}$ And, the probability of Kamal’s selection is $\dfrac{1}{5}$ Let P(A) be the probability of Kamal gets selected $P(A) = \dfrac{1}{3}$ And, P(B) be the probability of Monika gets selected $P(B) = \dfrac{1}{5}$ Now, the probability of both getting selected means Kamal and Monika both getting selected. Therefore, \[P\left( {Both{\text{ }}get{\text{ }}selected} \right){\text{ }} = {\text{ }}P\left( A \right){\text{ }} \times P\left( B \right)\] $ = \dfrac{1}{3} \times \dfrac{1}{5}$ $ = \dfrac{1}{{15}}$ Therefore, the P (Both getting selected) = $\dfrac{1}{{15}}$ One can also approach by choosing both cases individually and subtracting 1 from it that is $\dfrac{8}{{15}} + \dfrac{8}{{15}} = \dfrac{{16}}{{15}}$ $\dfrac{{16}}{{15}} - 1 = \dfrac{1}{{15}}$ [Which is P (Both getting selected)] $\dfrac{8}{{15}}$ is nothing but P(A) + P(B) and two times $\dfrac{8}{{15}}$ is the condition of choosing both one at a time and subtracting 1 means choosing both Kamal and Monika at same time. But, approaching with the first method is advised here. The second method is less explanatory. In this type of question students often get confused while finding the probability of both. They usually add the two instead of multiplying. Do not make this mistake. Just to simplify, remember AND means multiply and OR means addition.	{"url":"https://www.vedantu.com/question-answer/kamal-and-monika-appeared-for-an-interview-for-class-12-maths-cbse-5f895dba2331d1505c8b6f9a","timestamp":"2024-11-14T10:32:31Z","content_type":"text/html","content_length":"180421","record_id":"<urn:uuid:e512ad91-0884-488a-aa9b-d50c54fccf3e>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00224.warc.gz"}
Non Uniform Random Walks Non Uniform Random WalksArticle Given $\epsilon _i ∈ [0,1)$ for each $1 < i < n$, a particle performs the following random walk on $\{1,2,...,n\:\}$par If the particle is at $n$, it chooses a point uniformly at random (u.a.r.) from $\{1,...,n-1\}$. If the current position of the particle is $m (1 < m < n)$, with probability $\epsilon _m$ it decides to go back, in which case it chooses a point u.a.r. from $\{m+1,...,n\}$. With probability $1-\epsilon _m$ it decides to go forward, in which case it chooses a point u.a.r. from $\{1,...,m-1\}$. The particle moves to the selected point. What is the expected time taken by the particle to reach 1 if it starts the walk at $n$? Apart from being a natural variant of the classical one dimensional random walk, variants and special cases of this problemarise in Theoretical Computer Science [Linial, Fagin, Karp, Vishnoi]. In this paper we study this problem and observe interesting properties of this walk. First we show that the expected number of times the particle visits $i$ (before getting absorbed at 1) is the same when the walk is started at $j$, for all $j > i$. Then we show that for the following parameterized family of $\epsilon 's: \epsilon _i = \frac {n-i}{n-i+ α · (i-1)}$,$1 < i < n$ where $α$ does not depend on $i$, the expected number of times the particle visits $i$ is the same when the walk is started at $j$, for all $j < i$. Using these observations we obtain the expected absorption time for this family of $\epsilon 's$. As $α$ varies from infinity to 1, this time goes from $Θ (log n) to Θ (n)$. Finally we studythe behavior of the expected convergence timeas a function of $\epsilon$ . It remains an open question to determine whether this quantity increases when all $\epsilon 's$ are increased. We give some preliminary results to this effect. Volume: DMTCS Proceedings vol. AC, Discrete Random Walks (DRW'03) Section: Proceedings Published on: January 1, 2003 Imported on: May 10, 2017 Keywords: Non uniform random walk,[INFO.INFO-DS] Computer Science [cs]/Data Structures and Algorithms [cs.DS],[INFO.INFO-DM] Computer Science [cs]/Discrete Mathematics [cs.DM],[MATH.MATH-CO] Mathematics [math]/Combinatorics [math.CO],[INFO.INFO-CG] Computer Science [cs]/Computational Geometry [cs.CG]	{"url":"https://dmtcs.episciences.org/3330","timestamp":"2024-11-04T21:17:34Z","content_type":"application/xhtml+xml","content_length":"54830","record_id":"<urn:uuid:0d561b5f-c854-4f89-9274-9b1ae05a796c>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00648.warc.gz"}
Find the Zeros of the Function Calculator \| Fast Solutions This tool helps you find the zeros of any given function quickly and easily. How to Use the Find the Zeros of the Function Calculator This calculator helps you find the zeros (roots) of a quadratic function of the form ax² + bx + c = 0. To use the calculator, enter the coefficients ‘a’, ‘b’, and ‘c’ into the respective input fields and click the “Calculate” button. How It Calculates the Results The calculator uses the quadratic formula: x = [-b ± sqrt(b² – 4ac)] / 2a. It first calculates the discriminant (b² – 4ac). Based on the value of the discriminant: • If the discriminant is greater than zero, there are two real roots. • If the discriminant is equal to zero, there is one real root. • If the discriminant is less than zero, there are no real roots. This calculator only works for quadratic equations (degree 2). Ensure that the coefficients entered are valid numbers. If you enter non-numeric values or leave any field empty, the calculator will prompt you to enter valid numbers.	{"url":"https://madecalculators.com/find-the-zeros-of-the-function-calculator/","timestamp":"2024-11-09T17:26:07Z","content_type":"text/html","content_length":"142496","record_id":"<urn:uuid:5fb45d80-2f72-49cc-9c30-3aa65a96c920>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00844.warc.gz"}
Kilowatt-Hours to Watt-Hours Conversion (kWh to Wh) Kilowatt-Hours to Watt-Hours Converter Enter the energy in kilowatt-hours below to convert it to watt-hours. Do you want to convert watt-hours to kilowatt-hours? How to Convert Kilowatt-Hours to Watt-Hours To convert a measurement in kilowatt-hours to a measurement in watt-hours, multiply the energy by the following conversion ratio: 1,000 watt-hours/kilowatt-hour. Since one kilowatt-hour is equal to 1,000 watt-hours, you can use this simple formula to convert: watt-hours = kilowatt-hours × 1,000 The energy in watt-hours is equal to the energy in kilowatt-hours multiplied by 1,000. For example, here's how to convert 5 kilowatt-hours to watt-hours using the formula above. watt-hours = (5 kWh × 1,000) = 5,000 Wh How Many Watt-Hours Are in a Kilowatt-Hour? There are 1,000 watt-hours in a kilowatt-hour, which is why we use this value in the formula above. 1 kWh = 1,000 Wh Kilowatt-hours and watt-hours are both units used to measure energy. Keep reading to learn more about each unit of measure. What Is a Kilowatt-Hour? A kilowatt-hour is a measure of electrical energy equal to one kilowatt, or 1,000 watts, of power over a one hour period. Kilowatt-hours are a measure of electrical work performed over a period of time, and are often used as a way of measuring energy usage by electric companies. Kilowatt-hours are usually abbreviated as kWh, although the formally adopted expression is kW·h. The abbreviation kW h is also sometimes used. For example, 1 kilowatt-hour can be written as 1 kWh, 1 kW·h, or 1 kW h. In formal expressions, the centered dot (·) or space is used to separate units used to indicate multiplication in an expression and to avoid conflicting prefixes being misinterpreted as a unit Learn more about kilowatt-hours. What Is a Watt-Hour? The watt-hour is a measure of electrical energy equal to one watt of power over a one hour period. Watt-hours are usually abbreviated as Wh, although the formally adopted expression is W·h. The abbreviation W h is also sometimes used. For example, 1 watt-hour can be written as 1 Wh, 1 W·h, or 1 W Learn more about watt-hours. Kilowatt-Hour to Watt-Hour Conversion Table Table showing various kilowatt-hour measurements converted to watt-hours. Kilowatt-hours Watt-hours 0.001 kWh 1 Wh 0.002 kWh 2 Wh 0.003 kWh 3 Wh 0.004 kWh 4 Wh 0.005 kWh 5 Wh 0.006 kWh 6 Wh 0.007 kWh 7 Wh 0.008 kWh 8 Wh 0.009 kWh 9 Wh 0.01 kWh 10 Wh 0.02 kWh 20 Wh 0.03 kWh 30 Wh 0.04 kWh 40 Wh 0.05 kWh 50 Wh 0.06 kWh 60 Wh 0.07 kWh 70 Wh 0.08 kWh 80 Wh 0.09 kWh 90 Wh 0.1 kWh 100 Wh 0.2 kWh 200 Wh 0.3 kWh 300 Wh 0.4 kWh 400 Wh 0.5 kWh 500 Wh 0.6 kWh 600 Wh 0.7 kWh 700 Wh 0.8 kWh 800 Wh 0.9 kWh 900 Wh 1 kWh 1,000 Wh 1. Bureau International des Poids et Mesures, The International System of Units (SI), 9th edition, 2019, https://www.bipm.org/documents/20126/41483022/SI-Brochure-9-EN.pdf More Kilowatt-Hour & Watt-Hour Conversions	{"url":"https://www.inchcalculator.com/convert/kilowatt-hour-to-watt-hour/","timestamp":"2024-11-07T13:06:31Z","content_type":"text/html","content_length":"73255","record_id":"<urn:uuid:119a0d11-8f6a-4570-a6ad-2117c7efdd9d>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00582.warc.gz"}