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https://www.compadre.org/picup/exercises/exercise.cfm?A=WaveInterference
+ Wave Interference (with sound!) Developed by Christopher Orban - Published June 7, 2017 This exercise illustrates wave interference in 2D by adding two sine-waves together. The result looks very much like the interference pattern that can be created with ripples in a pond. This exercise will use a programming language called [p5.js](http://p5js.org) that is very similar to C and C++ programming. (Note: If you are familiar with C or C++ the main difference you will see is that there is no main() function and instead the draw() function serves this role.) **Importantly, this exercise can be completed using any computer or chromebook without downloading any software!** This exercise is designed for an algebra-based physics class at the college or high school level. It may also be useful for calculus-based physics for non-majors (e.g. engineering & science majors). However, it does use a taylor expansion, so perhaps it is more of an introductory calculus based exercise This exercise is part of a series of exercises developed by Prof. Chris Orban. There are pre-and-post assessment questions associated with this exercise (not available here) that are being used in an educational research study. If interested to collaborate on that study please e-mail Prof. Chris Orban (orban@physics.osu.edu). The first paper from this study [is available at this link](https://arxiv.org/abs/1701.01867), the second paper which discusses the electromagnetism exercises [is available at this link](https://arxiv.org/abs/1707.00185) Subject Area Waves & Optics First Year Javascript 1. Students will see how two sine waves can interfere to produce constructive and destructive interference 2. Students will take the equation for the path difference between two sources of waves and do a taylor expansion to obtain the lines of destructive interference. 3. Students will modify the code to plot these lines of destructive interference. By comparing these lines to the interference pattern the approximate nature of the taylor expansion will be illustrated 60 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.). This programming lab is designed to illustrate the idea of wave interference. This happens because of the principle of "superposition". For sound waves, this 20 dollar word just means that if there are two speakers in a large room or open space, the amplitude of the sound waves at a particular point in that room are the sum of the waves of the two speakers. ![](images/WaveInterference/interference0103.png "") ### Step 1. Open up the code in an editor Make sure to set your speaker volume low before you press play!!! Click here to open up the code in an editor Press the play button (red triangle) in the top left to run the code. If for some reason you can't open it up in the editor you can also try out the interactive at this link Here is a version without the red line that runs faster (useful for chromebooks) ### Step 2. Explore the Interference #### Step 2a. Move the little microphone to find the constructive and destructive interference Move the little microphone around and listen to the amplitude of the tone. When you put the little microphone in a place where there is destructive interference there should be almost no sound. When you put it at a point of constructive interference, the tone should be much louder. #### Step 2b. Move the speakers around Move the speakers around by clicking and dragging the speakers to a new location on the x axis. How does it change the interference pattern? ### Step 3. Look at this diagram: ![](images/WaveInterference/SoundLab_diagram_fixed.png "") Later, in Step 5 you will need to write down $r_1$ and $r_2$ as a function of $d$, $x$, and $y$. ### Step 4. Think about the math involved If there was just Speaker #1 the instantaneous amplitude $S_1$ at (x,y) would be this: $$S_1 = A_1 \sin (k_1 r_1 - \omega_1 t)$$ where $r_1$ is the distance between speaker #1 and the point (x,y). Likewise, the instantaneous amplitude at (x,y) from only speaker #2 would be this: $$S_2 = A_2 \sin (k_2 r_2 - \omega_2 t)$$ where $r_2$ is the distance between speaker #2 and the point (x,y). If you have both speakers then the amplitude at a particular point in the room would be this: $$S_1 + S_2 = A_1 \sin (k_1 r_1 - \omega_1 t) + A_2 \sin (k_2 r_2 - \omega_2 t)$$ The following trig identity is very useful: $$\sin a + \sin b = 2 \cos \left(\frac{a-b}{2} \right) \sin \left( \frac{a+b}{2} \right)$$ Assume $A_1= A_2 = A$, $\omega_1 = \omega_2 = \omega$, $k_1 = k_2 = k$, and use this expression to prove that the following is true: $$S_1 + S_2 = 2 A \cos \left(\frac{k (r_1-r_2)}{2} \right) \sin \left(\frac{k (r_1+ r_2) - 2\omega t }{2}\right)$$ and constructive/destructive interference depends on whether the cosine term works out to be equal to zero or one. ### Step 5. Look at the diagram again and figure out r1 and r2 **Look at the diagram from Step 3 again. How would you write $r_1$ and $r_2$ in terms of x, y and d?** ### Step 6. Consider Destructive Interference Use the previous equation to explain why the following is the correct equation for destructive interference: $$|r_1 - r_2| = \frac{n\lambda}{2}$$ Why is destructive interference only achieved if $n = 1, 3, 5 , 7...$? ### Step 7. Approximate! The previous expression turns out to be difficult to work with: $$|r_1 - r_2| = \frac{n\lambda}{2}$$ $$\sqrt{(x+d)^2 + y^2} - \sqrt{(x-d)^2 + y^2} = \frac{n\lambda}{2}$$ We'd really like to be able to solve for y, but there's no simple way of doing this. If you try to square both sides, for example, this doesn't help much to isolate y and x. When you're stuck like this, sometimes the best you can do is approximate. Notice the following trick if $|b/a| \ll 1$, $$\sqrt{a +b } = a \sqrt{1 + \frac{b}{a} } \approx a \left( 1 + \frac{1}{2} \frac{b}{a} \right)$$ Use this trick to prove that the lines of destructive interference follow this formula: $$y = \pm \frac{4 d}{n \lambda} x$$ ### Step 8. Plot up the lines of destructive interference! Modify the code to plot up the lines in the programming activity. Show that the lines for $n = 1$ match up well with the interference fringes. The result should look like this: ![](images/WaveInterference/interference0114.png "") ### Step 9. Move the speakers around! Move the speakers around by clicking and dragging the speakers to a new location on the x axis. **Bring the speakers closer and further apart. Do your lines still match up with the interference pattern?** ### Step 9. (Extra Credit:) Consider n = 3, 5, 7... Plot the lines of destructive interference for n = 3, 5, 7... and compare to the simulated wave pattern. These lines will not compare as well to the simulation as it does for n = 1. Why does n > 1 compare less well to the lines of destructive interference in the simulation? Do you have any ideas for improving the locations of the lines? How could we make the approximation better? ### How to get full credit on this programming lab **1. Describe in words what happens when you change the separation between the speakers (Step 2b)** Click and drag the speakers closer and further apart. Describe with words what happens to the interference pattern. **2. Use the trig identify to produce the interference formula (Step 4)** Assume $A_1= A_2$, $\omega_1 = \omega_2$ and derive the equation below for the addition of two sine waves $$S_1 + S_2 = 2 A_1 \cos \left(\frac{k (r_1-r_2)}{2} \right) \sin \left(\frac{k (r_1+ r_2) - 2\omega_2 t }{2}\right)$$ **2. Write r1 and r2 in terms of x, y and d (Step 5)** This is the primary task of Step 5. **3. Explain why destructive interference only occurs for n = 1,3,5,7... (Step 6)** This is the primary task of step 6. Why would $n = 2,4,6...$ not give destructive interference? **4. Derive $y = \pm \frac{4d }{ n \lambda }x$ from approximating the equation for destructive interference (Step 7)** This will take a couple of steps. Write these steps out on a sheet of paper, take a picture with your phone, and submit it to the assignment dropbox on canvas. **5. Submit your code with the plotted lines (Step 8)** Modify your code until it looks like the video shown in Step 8. **(Optional) 6. Comment on n = 3, 5, 7...** For extra credit, modify your program to show the lines of destructive interference for n = 3, 5, 7... Explain why these lines don't compare very well to the destructive interference in the wave pattern whereas the n = 1 result does. Comment on how you would perhaps improve upon the solution $y = \pm \frac{4d}{n\lambda}x$. To be clear, there is more than one way that this solution could be improved upon. ### Share a Variation Did you have to edit this material to fit your needs? Share your changes by Creating a Variation ### Credits and Licensing Christopher Orban, "Wave Interference (with sound!)," Published in the PICUP Collection, June 2017. The instructor materials are ©2017 Christopher Orban. The exercises are released under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 license
2021-02-28 19:47:12
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http://www.physicsforums.com/showthread.php?p=3102696
by Kstanley P: 7 1. The problem statement, all variables and given/known data Find the following limit: $$\lim_{x \to \infty} \frac{2+\sqrt{(6x)}}{-2+\sqrt{(3x)}}$$ 2. Relevant equations n/a 3. The attempt at a solution I know this shouldn't be that hard, but somehow I keep getting stuck on simplifying the equation. I think the first step is to multiply both the denominator and numerator by 1/x (which equals 1/sqrt(x^2). This would give me $$\lim_{x \to \infty} \frac{\frac{2}{x}+\sqrt{\frac{6x}{x^2}}}{\frac{-2}{x}+\sqrt{\frac{3x}{x^2}}}$$ And if I'm correct the squareroots should go to zero which leaves: $$\lim_{x \to \infty} \frac{\frac{2}{x}}{\frac{-2}{x}}$$ But wouldn't those on top of eachother equal 0/0 too unless I can multiply by x for both so it equals 2/-2? I am really just not sure of the correct way to go about simplifying this. Any help would be appreciated! Mentor P: 20,984 I would factor $\sqrt{x}$ out of both terms in the numerator and both in the denominator. That would give you $$\lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x}}\cdot \frac{2/\sqrt{x}+\sqrt{6}}{-2/\sqrt{x}+\sqrt{3}}$$ P: 352 No, you can't do it this way. You can say the square roots "go to zero" and can be cancelled out, but you are right to observe that by the same argument, the $$2/x$$ terms should "go to zero" as well. Try multiplying the numerator and denominator by $$x^{-1/2}$$ rather than $$x^{-1}$$. P: 7 Ok I understand multiplying the numerator and denominator by 1/sqrt(X) to obtain: $$\lim_{x \to \infty} \frac{\frac{2}{\sqrt{x}}+{\sqrt{6}}}{\frac{-2}{\sqrt{x}}+\sqrt{3}}$$ From this point do I multiply by the conjugate of the numerator? $$\lim_{x \to \infty} \frac{(\frac{2}{\sqrt{x}})^2-{6}}{-(\frac{2}{\sqrt{x}})^2+\frac{2\sqrt{3}}{\sqrt{x}}+\frac{2\sqrt{6}}{\sqr t{x}}-\sqrt{18}}$$ This seems like quite the jumbled mess so maybe I'm wrong.. Mentor P: 20,984 Quote by Kstanley Ok I understand multiplying the numerator and denominator by 1/sqrt(X) to obtain: $$\lim_{x \to \infty} \frac{\frac{2}{\sqrt{x}}+{\sqrt{6}}}{\frac{-2}{\sqrt{x}}+\sqrt{3}}$$ From this point do I multiply by the conjugate of the numerator? No, just take the limit. The first terms in the top and bottom go to zero. Quote by Kstanley $$\lim_{x \to \infty} \frac{(\frac{2}{\sqrt{x}})^2-{6}}{-(\frac{2}{\sqrt{x}})^2+\frac{2\sqrt{3}}{\sqrt{x}}+\frac{2\sqrt{6}}{\sqr t{x}}-\sqrt{18}}$$ This seems like quite the jumbled mess so maybe I'm wrong.. P: 7 Ah duh I got so into thinking it had to be complicated I didn't even think to just find the limit right there. Thank you so much that was so much easier than I imagined! Related Discussions Calculus 1 Calculus & Beyond Homework 2 Calculus & Beyond Homework 15 Calculus & Beyond Homework 2 Calculus & Beyond Homework 1
2014-04-19 04:46:17
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https://wiki.alquds.edu/?query=IEEE_754
# IEEE 754 The IEEE Standard for Floating-Point Arithmetic (IEEE 754) is a technical standard for floating-point arithmetic established in 1985 by the Institute of Electrical and Electronics Engineers (IEEE). The standard addressed many problems found in the diverse floating-point implementations that made them difficult to use reliably and portably. Many hardware floating-point units use the IEEE 754 standard. The standard defines: • arithmetic formats: sets of binary and decimal floating-point data, which consist of finite numbers (including signed zeros and subnormal numbers), infinities, and special "not a number" values (NaNs) • interchange formats: encodings (bit strings) that may be used to exchange floating-point data in an efficient and compact form • rounding rules: properties to be satisfied when rounding numbers during arithmetic and conversions • operations: arithmetic and other operations (such as trigonometric functions) on arithmetic formats • exception handling: indications of exceptional conditions (such as division by zero, overflow, etc.) IEEE 754-2008, published in August 2008, includes nearly all of the original IEEE 754-1985 standard, plus the IEEE 854-1987 Standard for Radix-Independent Floating-Point Arithmetic. The current version, IEEE 754-2019, was published in July 2019.[1] It is a minor revision of the previous version, incorporating mainly clarifications, defect fixes and new recommended operations. ## Standard development The first standard for floating-point arithmetic, IEEE 754-1985, was published in 1985. It covered only binary floating-point arithmetic. A new version, IEEE 754-2008, was published in August 2008, following a seven-year revision process, chaired by Dan Zuras and edited by Mike Cowlishaw. It replaced both IEEE 754-1985 (binary floating-point arithmetic) and IEEE 854-1987 Standard for Radix-Independent Floating-Point Arithmetic. The binary formats in the original standard are included in this new standard along with three new basic formats, one binary and two decimal. To conform to the current standard, an implementation must implement at least one of the basic formats as both an arithmetic format and an interchange format. The international standard ISO/IEC/IEEE 60559:2011 (with content identical to IEEE 754-2008) has been approved for adoption through ISO/IEC JTC 1/SC 25 under the ISO/IEEE PSDO Agreement[2][3] and published.[4] The current version, IEEE 754-2019 published in July 2019, is derived from and replaces IEEE 754-2008, following a revision process started in September 2015, chaired by David G. Hough and edited by Mike Cowlishaw. It incorporates mainly clarifications (e.g. totalOrder) and defect fixes (e.g. minNum), but also includes some new recommended operations (e.g. augmentedAddition).[5][6] The international standard ISO/IEC 60559:2020 (with content identical to IEEE 754-2019) has been approved for adoption through ISO/IEC JTC 1/SC 25 and published.[7] The next projected revision of the standard is in 2028.[8] ## Formats An IEEE 754 format is a "set of representations of numerical values and symbols". A format may also include how the set is encoded.[9] A floating-point format is specified by • a base (also called radix) b, which is either 2 (binary) or 10 (decimal) in IEEE 754; • a precision p; • an exponent range from emin to emax, with emin = 1 − emax for all IEEE 754 formats. A format comprises • Finite numbers, which can be described by three integers: s = a sign (zero or one), c = a significand (or coefficient) having no more than p digits when written in base b (i.e., an integer in the range through 0 to bp − 1), and q = an exponent such that eminq + p − 1 ≤ emax. The numerical value of such a finite number is (−1)s × c × bq.[a] Moreover, there are two zero values, called signed zeros: the sign bit specifies whether a zero is +0 (positive zero) or −0 (negative zero). • Two infinities: +∞ and −∞. • Two kinds of NaN (not-a-number): a quiet NaN (qNaN) and a signaling NaN (sNaN). For example, if b = 10, p = 7, and emax = 96, then emin = −95, the significand satisfies 0 ≤ c9999999, and the exponent satisfies −101 ≤ q ≤ 90. Consequently, the smallest non-zero positive number that can be represented is 1×10−101, and the largest is 9999999×1090 (9.999999×1096), so the full range of numbers is −9.999999×1096 through 9.999999×1096. The numbers −b1−emax and b1−emax (here, −1×10−95 and 1×10−95) are the smallest (in magnitude) normal numbers; non-zero numbers between these smallest numbers are called subnormal numbers. ### Representation and encoding in memory Some numbers may have several possible exponential format representations. For instance, if b = 10, and p = 7, then −12.345 can be represented by −12345×10−3, −123450×10−4, and −1234500×10−5. However, for most operations, such as arithmetic operations, the result (value) does not depend on the representation of the inputs. For the decimal formats, any representation is valid, and the set of these representations is called a cohort. When a result can have several representations, the standard specifies which member of the cohort is chosen. For the binary formats, the representation is made unique by choosing the smallest representable exponent allowing the value to be represented exactly. Further, the exponent is not represented directly, but a bias is added so that the smallest representable exponent is represented as 1, with 0 used for subnormal numbers. For numbers with an exponent in the normal range (the exponent field being neither all ones nor all zeros), the leading bit of the significand will always be 1. Consequently, a leading 1 can be implied rather than explicitly present in the memory encoding, and under the standard the explicitly represented part of the significand will lie between 0 and 1. This rule is called leading bit convention, implicit bit convention, or hidden bit convention. This rule allows the binary format to have an extra bit of precision. The leading bit convention cannot be used for the subnormal numbers as they have an exponent outside the normal exponent range and scale by the smallest represented exponent as used for the smallest normal numbers. Due to the possibility of multiple encodings (at least in formats called interchange formats), a NaN may carry other information: a sign bit (which has no meaning, but may be used by some operations) and a payload, which is intended for diagnostic information indicating the source of the NaN (but the payload may have other uses, such as NaN-boxing[10][11][12]). ### Basic and interchange formats The standard defines five basic formats that are named for their numeric base and the number of bits used in their interchange encoding. There are three binary floating-point basic formats (encoded with 32, 64 or 128 bits) and two decimal floating-point basic formats (encoded with 64 or 128 bits). The binary32 and binary64 formats are the single and double formats of IEEE 754-1985 respectively. A conforming implementation must fully implement at least one of the basic formats. The standard also defines interchange formats, which generalize these basic formats.[13] For the binary formats, the leading bit convention is required. The following table summarizes the smallest interchange formats (including the basic ones). Name Common name Base Significand bits[b] or digits Decimal digits Exponent bits Decimal E max Exponent bias[14] E min E max Notes binary16 Half precision 2 11 3.31 5 4.51 24−1 = 15 −14 +15 not basic binary32 Single precision 2 24 7.22 8 38.23 27−1 = 127 −126 +127 binary64 Double precision 2 53 15.95 11 307.95 210−1 = 1023 −1022 +1023 binary128 Quadruple precision 2 113 34.02 15 4931.77 214−1 = 16383 −16382 +16383 binary256 Octuple precision 2 237 71.34 19 78913.2 218−1 = 262143 −262142 +262143 not basic decimal32 10 7 7 7.58 96 101 −95 +96 not basic decimal64 10 16 16 9.58 384 398 −383 +384 decimal128 10 34 34 13.58 6144 6176 −6143 +6144 Note that in the table above, the minimum exponents listed are for normal numbers; the special subnormal number representation allows even smaller numbers to be represented (with some loss of precision). For example, the smallest positive number that can be represented in binary64 is 2−1074; contributions to the −1074 figure include the E min value −1022 and all but one of the 53 significand bits (2−1022 − (53 − 1) = 2−1074). Decimal digits is digits × log10 base. This gives an approximate precision in number of decimal digits. Decimal E max is Emax × log10 base. This gives an approximate value of the maximum decimal exponent. The binary32 (single) and binary64 (double) formats are two of the most common formats used today. The figure below shows the absolute precision for both formats over a range of values. This figure can be used to select an appropriate format given the expected value of a number and the required precision. Precision of binary32 and binary64 in the range 10−12 to 1012 An example of a layout for 32-bit floating point is and the 64 bit layout is similar. ### Extended and extendable precision formats The standard specifies optional extended and extendable precision formats, which provide greater precision than the basic formats.[15] An extended precision format extends a basic format by using more precision and more exponent range. An extendable precision format allows the user to specify the precision and exponent range. An implementation may use whatever internal representation it chooses for such formats; all that needs to be defined are its parameters (b, p, and emax). These parameters uniquely describe the set of finite numbers (combinations of sign, significand, and exponent for the given radix) that it can represent. The standard recommends that language standards provide a method of specifying p and emax for each supported base b.[16] The standard recommends that language standards and implementations support an extended format which has a greater precision than the largest basic format supported for each radix b.[17] For an extended format with a precision between two basic formats the exponent range must be as great as that of the next wider basic format. So for instance a 64-bit extended precision binary number must have an 'emax' of at least 16383. The x87 80-bit extended format meets this requirement. ### Interchange formats Interchange formats are intended for the exchange of floating-point data using a bit string of fixed length for a given format. #### Binary For the exchange of binary floating-point numbers, interchange formats of length 16 bits, 32 bits, 64 bits, and any multiple of 32 bits ≥ 128[c] are defined. The 16-bit format is intended for the exchange or storage of small numbers (e.g., for graphics). The encoding scheme for these binary interchange formats is the same as that of IEEE 754-1985: a sign bit, followed by w exponent bits that describe the exponent offset by a bias, and p − 1 bits that describe the significand. The width of the exponent field for a k-bit format is computed as w = round(4 log2(k)) − 13. The existing 64- and 128-bit formats follow this rule, but the 16- and 32-bit formats have more exponent bits (5 and 8 respectively) than this formula would provide (3 and 7 respectively). As with IEEE 754-1985, the biased-exponent field is filled with all 1 bits to indicate either infinity (trailing significand field = 0) or a NaN (trailing significand field ≠ 0). For NaNs, quiet NaNs and signaling NaNs are distinguished by using the most significant bit of the trailing significand field exclusively,[d] and the payload is carried in the remaining bits. #### Decimal For the exchange of decimal floating-point numbers, interchange formats of any multiple of 32 bits are defined. As with binary interchange, the encoding scheme for the decimal interchange formats encodes the sign, exponent, and significand. Two different bit-level encodings are defined, and interchange is complicated by the fact that some external indicator of the encoding in use may be required. The two options allow the significand to be encoded as a compressed sequence of decimal digits using densely packed decimal or, alternatively, as a binary integer. The former is more convenient for direct hardware implementation of the standard, while the latter is more suited to software emulation on a binary computer. In either case, the set of numbers (combinations of sign, significand, and exponent) that may be encoded is identical, and special values (±zero with the minimum exponent, ±infinity, quiet NaNs, and signaling NaNs) have identical encodings. ## Rounding rules The standard defines five rounding rules. The first two rules round to a nearest value; the others are called directed roundings: ### Roundings to nearest • Round to nearest, ties to even – rounds to the nearest value; if the number falls midway, it is rounded to the nearest value with an even least significant digit. • Round to nearest, ties away from zero (or ties to away)  – rounds to the nearest value; if the number falls midway, it is rounded to the nearest value above (for positive numbers) or below (for negative numbers). At the extremes, a value with a magnitude strictly less than ${\displaystyle k=b^{\text{emax}}(b-{\tfrac {1}{2}}b^{1-p})}$ will be rounded to the minimum or maximum finite number (depending on the value's sign). Any numbers with exactly this magnitude are considered ties; this choice of tie may be conceptualized as the midpoint between ${\displaystyle \pm b^{\text{emax}}(b-b^{1-p})}$ and ${\displaystyle \pm b^{{\text{emax}}+1}}$, which, were the exponent not limited, would be the next representable floating-point numbers larger in magnitude. Numbers with a magnitude strictly larger than ${\displaystyle k}$ are rounded to the corresponding infinity.[18] Round to nearest, ties to even is the default for binary floating point and the recommended default for decimal. Round to nearest, ties to away is only required for decimal implementations.[19] ### Directed roundings • Round toward 0 – directed rounding towards zero (also known as truncation). • Round toward +∞ – directed rounding towards positive infinity (also known as rounding up or ceiling). • Round toward −∞ – directed rounding towards negative infinity (also known as rounding down or floor). Example of rounding to integers using the IEEE 754 rules Mode Example value +11.5 +12.5 −11.5 −12.5 to nearest, ties to even +12.0 +12.0 −12.0 −12.0 to nearest, ties away from zero +12.0 +13.0 −12.0 −13.0 toward 0 +11.0 +12.0 −11.0 −12.0 toward +∞ +12.0 +13.0 −11.0 −12.0 toward −∞ +11.0 +12.0 −12.0 −13.0 Unless specified otherwise, the floating-point result of an operation is determined by applying the rounding function on the infinitely precise (mathematical) result. Such an operation is said to be correctly rounded. This requirement is called correct rounding.[20] ## Required operations Required operations for a supported arithmetic format (including the basic formats) include: • Conversions to and from integer[21][22] • Previous and next consecutive values[23] • Arithmetic operations (add, subtract, multiply, divide, square root, fused multiply–add, remainder, minimum, maximum)[24][25] • Conversions (between formats, to and from strings, etc.)[26][27] • Scaling and (for decimal) quantizing[28][29] • Copying and manipulating the sign (abs, negate, etc.)[30] • Comparisons and total ordering[31][32] • Classification of numbers (subnormal, finite, etc.) and testing for NaNs[33] • Testing and setting status flags[34] ### Comparison predicates The standard provides comparison predicates to compare one floating-point datum to another in the supported arithmetic format.[35] Any comparison with a NaN is treated as unordered. −0 and +0 compare as equal. ### Total-ordering predicate The standard provides a predicate totalOrder, which defines a total ordering on canonical members of the supported arithmetic format.[36] The predicate agrees with the comparison predicates when one floating-point number is less than the other. The totalOrder predicate does not impose a total ordering on all encodings in a format. In particular, it does not distinguish among different encodings of the same floating-point representation, as when one or both encodings are non-canonical.[37] IEEE 754-2019 incorporates clarifications of totalOrder. For the binary interchange formats whose encoding follows the IEEE 754-2008 recommendation on placement of the NaN signaling bit, the comparison is identical to one that type puns the floating-point numbers to a sign–magnitude integer (assuming a payload ordering consistent with this comparison), an old trick for FP comparison without an FPU.[38] ## Exception handling The standard defines five exceptions, each of which returns a default value and has a corresponding status flag that is raised when the exception occurs.[e] No other exception handling is required, but additional non-default alternatives are recommended (see § Alternate exception handling). The five possible exceptions are • Invalid operation: mathematically undefined, e.g., the square root of a negative number. By default, returns qNaN. • Division by zero: an operation on finite operands gives an exact infinite result, e.g., 1/0 or log(0). By default, returns ±infinity. • Overflow: a finite result is too large to be represented accurately (i.e., its exponent with an unbounded exponent range would be larger than emax). By default, returns ±infinity for the round-to-nearest modes (and follows the rounding rules for the directed rounding modes). • Underflow: a result is very small (outside the normal range). By default, returns a number less than or equal to the minimum positive normal number in magnitude (following the rounding rules); a subnormal number always implies an underflow exception, but by default, if it is exact, no flag is raised. • Inexact: the exact (i.e., unrounded) result is not representable exactly. By default, returns the correctly rounded result. These are the same five exceptions as were defined in IEEE 754-1985, but the division by zero exception has been extended to operations other than the division. Some decimal floating-point implementations define additional exceptions,[39][40] which are not part of IEEE 754: • Clamped: a result's exponent is too large for the destination format. By default, trailing zeros will be added to the coefficient to reduce the exponent to the largest usable value. If this is not possible (because this would cause the number of digits needed to be more than the destination format) then an overflow exception occurs. • Rounded: a result's coefficient requires more digits than the destination format provides. An inexact exception is signaled if any non-zero digits are discarded. Additionally, operations like quantize when either operand is infinite, or when the result does not fit the destination format, will also signal invalid operation exception.[41] ## Recommendations ### Alternate exception handling The standard recommends optional exception handling in various forms, including presubstitution of user-defined default values, and traps (exceptions that change the flow of control in some way) and other exception handling models that interrupt the flow, such as try/catch. The traps and other exception mechanisms remain optional, as they were in IEEE 754-1985. ### Recommended operations Clause 9 in the standard recommends additional mathematical operations[42] that language standards should define.[43] None are required in order to conform to the standard. The following are recommended arithmetic operations, which must round correctly:[44] The asinPi, acosPi and tanPi functions were not part of the IEEE 754-2008 standard because they were deemed less necessary.[45] asinPi, acosPi were mentioned, but this was regarded as an error.[5] All three were added in the 2019 revision. The recommended operations also include setting and accessing dynamic mode rounding direction,[46] and implementation-defined vector reduction operations such as sum, scaled product, and dot product, whose accuracy is unspecified by the standard.[47] As of 2019, augmented arithmetic operations[48] for the binary formats are also recommended. These operations, specified for addition, subtraction and multiplication, produce a pair of values consisting of a result correctly rounded to nearest in the format and the error term, which is representable exactly in the format. At the time of publication of the standard, no hardware implementations are known, but very similar operations were already implemented in software using well-known algorithms. The history and motivation for their standardization are explained in a background document.[49][50] As of 2019, the formerly required minNum, maxNum, minNumMag, and maxNumMag in IEEE 754-2008 are now deprecated due to their non-associativity. Instead, two sets of new minimum and maximum operations are recommended.[51] The first set contains minimum, minimumNumber, maximum and maximumNumber. The second set contains minimumMagnitude, minimumMagnitudeNumber, maximumMagnitude and maximumMagnitudeNumber. The history and motivation for this change are explained in a background document.[52] ### Expression evaluation The standard recommends how language standards should specify the semantics of sequences of operations, and points out the subtleties of literal meanings and optimizations that change the value of a result. By contrast, the previous 1985 version of the standard left aspects of the language interface unspecified, which led to inconsistent behavior between compilers, or different optimization levels in an optimizing compiler. Programming languages should allow a user to specify a minimum precision for intermediate calculations of expressions for each radix. This is referred to as preferredWidth in the standard, and it should be possible to set this on a per-block basis. Intermediate calculations within expressions should be calculated, and any temporaries saved, using the maximum of the width of the operands and the preferred width if set. Thus, for instance, a compiler targeting x87 floating-point hardware should have a means of specifying that intermediate calculations must use the double-extended format. The stored value of a variable must always be used when evaluating subsequent expressions, rather than any precursor from before rounding and assigning to the variable. ### Reproducibility The IEEE 754-1985 version of the standard allowed many variations in implementations (such as the encoding of some values and the detection of certain exceptions). IEEE 754-2008 has reduced these allowances, but a few variations still remain (especially for binary formats). The reproducibility clause recommends that language standards should provide a means to write reproducible programs (i.e., programs that will produce the same result in all implementations of a language) and describes what needs to be done to achieve reproducible results. ## Character representation The standard requires operations to convert between basic formats and external character sequence formats.[53] Conversions to and from a decimal character format are required for all formats. Conversion to an external character sequence must be such that conversion back using round to nearest, ties to even will recover the original number. There is no requirement to preserve the payload of a quiet NaN or signaling NaN, and conversion from the external character sequence may turn a signaling NaN into a quiet NaN. The original binary value will be preserved by converting to decimal and back again using:[54] • 5 decimal digits for binary16, • 9 decimal digits for binary32, • 17 decimal digits for binary64, • 36 decimal digits for binary128. For other binary formats, the required number of decimal digits is[f] ${\displaystyle 1+\lceil p\log _{10}(2)\rceil ,}$ where p is the number of significant bits in the binary format, e.g. 237 bits for binary256. When using a decimal floating-point format, the decimal representation will be preserved using: • 7 decimal digits for decimal32, • 16 decimal digits for decimal64, • 34 decimal digits for decimal128. Algorithms, with code, for correctly rounded conversion from binary to decimal and decimal to binary are discussed by Gay,[55] and for testing – by Paxson and Kahan.[56] The standard recommends providing conversions to and from external hexadecimal-significand character sequences, based on C99's hexadecimal floating point literals. Such a literal consists of an optional sign (+ or -), the indicator "0x", a hexadecimal number with or without a period, an exponent indicator "p", and a decimal exponent with optional sign. The syntax is not case-sensitive.[57] The decimal exponent scales by powers of 2, so for example 0x0.1p-4 is 1/256.[58] ## Notes 1. ^ For example, if the base is 10, the sign is 1 (indicating negative), the significand is 12345, and the exponent is −3, then the value of the number is (−1)1 × 12345 × 10−3 = −1 × 12345 × 0.001 = −12.345. 2. ^ Including the implicit bit (which always equals 1 for normal numbers, and 0 for subnormal numbers. This implicit bit is not stored in memory), but not the sign bit. 3. ^ Contrary to decimal, there is no binary interchange format of 96-bit length. Such a format is still allowed as a non-interchange format, though. 4. ^ The standard recommends 0 for signaling NaNs, 1 for quiet NaNs, so that a signaling NaNs can be quieted by changing only this bit to 1, while the reverse could yield the encoding of an infinity. 5. ^ No flag is raised in certain cases of underflow. 6. ^ As an implementation limit, correct rounding is only guaranteed for the number of decimal digits required plus 3 for the largest supported binary format. For instance, if binary32 is the largest supported binary format, then a conversion from a decimal external sequence with 12 decimal digits is guaranteed to be correctly rounded when converted to binary32; but conversion of a sequence of 13 decimal digits is not; however, the standard recommends that implementations impose no such limit. ## References 1. ^ IEEE 754 2019 2. ^ "FW: ISO/IEC/IEEE 60559 (IEEE Std 754-2008)". grouper.ieee.org. Archived from the original on 2017-10-27. Retrieved 2018-04-04. 3. ^ "ISO/IEEE Partner Standards Development Organization (PSDO) Cooperation Agreement" (PDF). 2007-12-19. Retrieved 2021-12-27. 4. ^ "ISO/IEC/IEEE 60559:2011 — Information technology — Microprocessor Systems — Floating-Point arithmetic". www.iso.org. Retrieved 2018-04-04. 5. ^ a b Cowlishaw, Mike (2013-11-13). "IEEE 754-2008 errata". speleotrove.com. Retrieved 2020-01-24. 6. ^ "Revising ANSI/IEEE Std 754-2008". ucbtest.org. Retrieved 2018-04-04. 7. ^ "ISO/IEC 60559:2020 — Information technology — Microprocessor Systems — Floating-Point arithmetic". www.iso.org. Retrieved 2020-10-25. 8. ^ Riedy, E. Jason (2018-06-26), "Plans for IEEE Standard 754 – 2028" (PDF), 25th IEEE Symposium on Computer Arithmetic, Amherst, MA: IEEE{{citation}}: CS1 maint: date and year (link) 9. ^ IEEE 754 2008, §2.1.27. 10. ^ "SpiderMonkey Internals". developer.mozilla.org. Retrieved 2018-03-11. 11. ^ Klemens, Ben (September 2014). 21st Century C: C Tips from the New School. O'Reilly Media, Incorporated. p. 160. ISBN 9781491904442. Retrieved 2018-03-11. 12. ^ "zuiderkwast/nanbox: NaN-boxing in C". GitHub. Retrieved 2018-03-11. 13. ^ IEEE 754 2008, §3.6. 14. ^ Cowlishaw, Mike. "Decimal Arithmetic Encodings" (PDF). IBM. Retrieved 2015-08-06. 15. ^ IEEE 754 2008, §3.7. 16. ^ IEEE 754 2008, §3.7 states: "Language standards should define mechanisms supporting extendable precision for each supported radix." 17. ^ IEEE 754 2008, §3.7 states: "Language standards or implementations should support an extended precision format that extends the widest basic format that is supported in that radix." 18. ^ IEEE 754 2008, §4.3.1. "In the following two rounding-direction attributes, an infinitely precise result with magnitude at least ${\displaystyle b^{\text{emax}}(b-{\tfrac {1}{2}}b^{1-p})}$ shall round to ${\displaystyle \infty }$ with no change in sign." 19. ^ IEEE 754 2008, §4.3.3 20. ^ IEEE 754 2019, §2.1 21. ^ IEEE 754 2008, §5.3.1 22. ^ IEEE 754 2008, §5.4.1 23. ^ IEEE 754 2008, §5.3.1 24. ^ IEEE 754 2008, §5.3.1 25. ^ IEEE 754 2008, §5.4.1 26. ^ IEEE 754 2008, §5.4.2 27. ^ IEEE 754 2008, §5.4.3 28. ^ IEEE 754 2008, §5.3.2 29. ^ IEEE 754 2008, §5.3.3 30. ^ IEEE 754 2008, §5.5.1 31. ^ IEEE 754 2008, §5.10 32. ^ IEEE 754 2008, §5.11 33. ^ IEEE 754 2008, §5.7.2 34. ^ IEEE 754 2008, §5.7.4 35. ^ IEEE 754 2019, §5.11 36. ^ IEEE 754 2019, §5.10 37. ^ IEEE 754 2019, §5.10 38. ^ Herf, Michael (December 2001). "radix tricks". stereopsis : graphics. 39. ^ "9.4. decimal — Decimal fixed point and floating point arithmetic — Python 3.6.5 documentation". docs.python.org. Retrieved 2018-04-04. 40. ^ "Decimal Arithmetic - Exceptional conditions". speleotrove.com. Retrieved 2018-04-04. 41. ^ IEEE 754 2008, §7.2(h) 42. ^ IEEE 754 2019, §9.2 43. ^ IEEE 754 2008, Clause 9 44. ^ IEEE 754 2019, §9.2. 45. ^ "Re: Missing functions tanPi, asinPi and acosPi". grouper.ieee.org. Archived from the original on 2017-07-06. Retrieved 2018-04-04. 46. ^ IEEE 754 2008, §9.3. 47. ^ IEEE 754 2008, §9.4. 48. ^ IEEE 754 2019, §9.5 49. ^ Riedy, Jason; Demmel, James. "Augmented Arithmetic Operations Proposed for IEEE-754 2018" (PDF). 25th IEEE Symbosium on Computer Arithmetic (ARITH 2018). pp. 49–56. Archived (PDF) from the original on 2019-07-23. Retrieved 2019-07-23. 50. ^ "754 Revision targeted for 2019". 754r.ucbtest.org. Retrieved 2019-07-23. 51. ^ IEEE 754 2019, §9.6. 52. ^ Chen, David. "The Removal of MinNum and MaxNum Operations from IEEE 754-2019" (PDF). grouper.ieee.org. Retrieved 2020-02-05. 53. ^ IEEE 754 2008, §5.12. 54. ^ IEEE 754 2008, §5.12.2. 55. ^ Gay, David M. (1990-11-30). "Correctly rounded binary-decimal and decimal-binary conversions". Numerical Analysis Manuscript. Murry Hill, NJ, USA: AT&T Laboratories. 90-10. {{cite journal}}: Cite journal requires |journal= (help) 56. ^ Paxson, Vern; Kahan, William (1991-05-22). "A Program for Testing IEEE Decimal–Binary Conversion". Manuscript. CiteSeerX 10.1.1.144.5889. {{cite journal}}: Cite journal requires |journal= (help) 57. ^ IEEE 754 2008, §5.12.3 58. ^ "6.9.3. Hexadecimal floating point literals — Glasgow Haskell Compiler 9.3.20220129 User's Guide". ghc.gitlab.haskell.org. Retrieved 2022-01-29.
2022-12-03 08:55:11
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https://lscsoft.docs.ligo.org/lalsuite/lal/group___calibration__h.html
LAL  7.3.0.1-9a4e871 Detailed Description Calibration API. Synopsis #include <lal/Calibration.h> Modules Module ComputeTransfer.c Computes the transfer function from zero-pole-gain representation. Data Structures struct  CalFactors UNDOCUMENTED. More... struct  UpdateFactorsParams UNDOCUMENTED. More... struct  CalibrationRecord UNDOCUMENTED. More... struct  CalibrationFunctions The type CalibrationFunctions contains two calibration functions, the sensing function $$C(f)$$ and the response function $$R(f)$$. More... struct  CalibrationUpdateParams The type CalibrationUpdateParams contains two time series representing an overall gain factor for the open-loop gain function $$H(f)$$ and the sensing function $$C(f)$$. More... struct  StrainOut UNDOCUMENTED. More... struct  StrainIn UNDOCUMENTED. More... struct  MyIIRFilter UNDOCUMENTED. More... Enumerations enum  CalibrationType { CalibrationAmplitude = 001 , CalibrationOffset = 002 , CalibrationDelay = 004 , CalibrationTransfer = 010 , CalibrationZPG = 020 } UNDOCUMENTED. More... Error Codes #define CALIBRATIONH_ENULL   001 Null pointer. More... #define CALIBRATIONH_ESIZE   002 Invalid size. More... #define CALIBRATIONH_ESZMM   004 Size mismatch. More... #define CALIBRATIONH_EZERO   010 Zero factor. More... #define CALIBRATIONH_ETIME   020 Time out of range. More... #define CALIBRATIONH_EUNIT   040 Incompatible units. More... ◆ CalibrationType enum CalibrationType UNDOCUMENTED. Enumerator CalibrationAmplitude CalibrationOffset CalibrationDelay CalibrationTransfer CalibrationZPG Definition at line 67 of file Calibration.h. ◆ CALIBRATIONH_ENULL #define CALIBRATIONH_ENULL   001 Null pointer. Definition at line 49 of file Calibration.h. ◆ CALIBRATIONH_ESIZE #define CALIBRATIONH_ESIZE   002 Invalid size. Definition at line 50 of file Calibration.h. ◆ CALIBRATIONH_ESZMM #define CALIBRATIONH_ESZMM   004 Size mismatch. Definition at line 51 of file Calibration.h. ◆ CALIBRATIONH_EZERO #define CALIBRATIONH_EZERO   010 Zero factor. Definition at line 52 of file Calibration.h. ◆ CALIBRATIONH_ETIME #define CALIBRATIONH_ETIME   020 Time out of range. Definition at line 53 of file Calibration.h. ◆ CALIBRATIONH_EUNIT #define CALIBRATIONH_EUNIT   040 Incompatible units. Definition at line 54 of file Calibration.h.
2023-03-30 05:44:26
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http://www.global-sci.org/intro/article_detail/aamm/8384.html
Volume 1, Issue 4 Hedging Game Contingent Claims with Constrained Portfolios Adv. Appl. Math. Mech., 1 (2009), pp. 529-545. Published online: 2009-01 Preview Full PDF 37 1168 Export citation Cited by • Abstract Game option is an American-type option with added feature that the writer can exercise the option at any time before maturity. In this paper, we consider the problem of hedging Game Contingent Claims (GCC) in two cases. For the case that portfolio is unconstrained, we provide a single arbitrage-free price $P_0$. Whereas for the constrained case, the price is replaced by an interval $[h_{low},h_{up}]$ of arbitrage-free prices. And for the portfolio with some closed constraints, we give the expressions of the upper-hedging price and lower-hedging price. Finally, for a special type of game option, we provide explicit expressions of the price and optimal portfolio for the writer and holder. • Keywords Game option contingent claims hedging optimal stopping free boundary 60G40 91A60 • BibTex • RIS • TXT @Article{AAMM-1-529, author = {Lei Wang and Yan Xiao}, title = {Hedging Game Contingent Claims with Constrained Portfolios}, journal = {Advances in Applied Mathematics and Mechanics}, year = {2009}, volume = {1}, number = {4}, pages = {529--545}, abstract = { Game option is an American-type option with added feature that the writer can exercise the option at any time before maturity. In this paper, we consider the problem of hedging Game Contingent Claims (GCC) in two cases. For the case that portfolio is unconstrained, we provide a single arbitrage-free price $P_0$. Whereas for the constrained case, the price is replaced by an interval $[h_{low},h_{up}]$ of arbitrage-free prices. And for the portfolio with some closed constraints, we give the expressions of the upper-hedging price and lower-hedging price. Finally, for a special type of game option, we provide explicit expressions of the price and optimal portfolio for the writer and holder. }, issn = {2075-1354}, doi = {https://doi.org/10.4208/aamm.09-m08h8}, url = {http://global-sci.org/intro/article_detail/aamm/8384.html} } TY - JOUR T1 - Hedging Game Contingent Claims with Constrained Portfolios AU - Lei Wang & Yan Xiao JO - Advances in Applied Mathematics and Mechanics VL - 4 SP - 529 EP - 545 PY - 2009 DA - 2009/01 SN - 1 DO - http://doi.org/10.4208/aamm.09-m08h8 UR - https://global-sci.org/intro/article_detail/aamm/8384.html KW - Game option KW - contingent claims KW - hedging KW - optimal stopping KW - free boundary AB - Game option is an American-type option with added feature that the writer can exercise the option at any time before maturity. In this paper, we consider the problem of hedging Game Contingent Claims (GCC) in two cases. For the case that portfolio is unconstrained, we provide a single arbitrage-free price $P_0$. Whereas for the constrained case, the price is replaced by an interval $[h_{low},h_{up}]$ of arbitrage-free prices. And for the portfolio with some closed constraints, we give the expressions of the upper-hedging price and lower-hedging price. Finally, for a special type of game option, we provide explicit expressions of the price and optimal portfolio for the writer and holder. Lei Wang & Yan Xiao. (1970). Hedging Game Contingent Claims with Constrained Portfolios. Advances in Applied Mathematics and Mechanics. 1 (4). 529-545. doi:10.4208/aamm.09-m08h8 Copy to clipboard The citation has been copied to your clipboard
2020-10-24 11:35:40
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https://motls.blogspot.com/2006/06/nas-schizofrenic-climate-report.html?showComment=1169002020000&m=1
## Thursday, June 22, 2006 ### NAS: schizophrenic climate report The theory about an unprecedented global warming, described by James Inhofe as the greatest hoax ever perpetrated on the American people, was essentially downgraded from "certain" to "plausible" A catchy subtitle was needed because much of the material that follows is boring and confusing, and it is not my fault. ;-) Typo: The filename should say "schizophrenic". Thanks, Benjamin. I don't want to change it now because it would break some links. The #1 news on cnn.com right now is about the NAS climate report or the that you can buy, for \$42.30, here, but you can also see the pages freely here; the executive summary is also for free, much like the audio. Other sources are here. However, I don't recommend you to spend your money for this new audit of the climate reconstructions because the freely available information is enough. It is a document that tries to make everyone happy which makes it schizophrenic. On one hand, they admit that there has almost certainly been the Little Ice Age and quite plausibly also the Medieval Warm Period - in both cases it is something that the hardcore alarmists wanted to deny for the last 10 years. Because of the MWP and the large uncertainties before 1600, we can only say that the current temperatures are warmest in 400 years, not more, the panel says. In other words, it's warmer now than in the Little Ice Age. Well, this is why the Little Ice Age is called in this way. On the other hand, however, they try to promote the idea that it could "plausibly" (original report) or even "likely" (CNN's translation or "spin") still be warmer today than in the Middle Ages, and maybe the current temperatures are highest in the last 1000 or 2000 years. Well, maybe the geologists are also wrong and they temperatures are highest in millions of years. Such "maybe" sentences are completely meaningless. If someone cannot defend a statement at the 99% confidence level, he should close his or her mouth because sentences without sufficiently strong evidence required by scientific standards are nothing else than brainwashing and manipulation. I am happy that in different parts of the report, the panel at least confirms that Mann's statement that the 1990s were the hottest decade in a millenium and 1998 the hottest year - and similar statements that had filled the media so many times in the past - are unjustifiable by existing data (go to 47:00 of the audio or so), despite 10 years of passionate statements that these insights are definitive, ever more definitive, and that the debate was over. Because the reconstructions can only be trusted up to 1600 or so instead of 1000 or 0 as claimed previously, Benny Peiser has told us a nice joke: What do you get if you shorten the hockey stick by 60 percent? A boomerang! Mann suddenly started to say that he never said that he was certain that the current era is the warmest era in the last 1000 years and, on the contrary, he always emphasized that their research was meant to show how uncertain these numbers are. Well, we probably live in different Universes because in this Universe, he said it roughly 350 times and 870,000 articles have been written about this extraordinary statement. This confusion - more precisely these untrue assertions - are discussed at 19:20 of this You're exactly one click from verifying that various media and the RealClimate group blog are just trying to fool you completely. If you go to 25:30 of this audio, a distinguished NAS scientist explains that the belief that Mann's results were definitive were not Mann's fault but rather the climate science community's fault. I can't believe he's serious. Is he talking about the same Mann who established his own propagandistic blog to promote his speculations, deny all criticism, and describe the critics as corrupt people? Of course that it was also a fault of the climate community because it is not a terribly bright and honest one, but indications that Mann is innocent are just crazy. At 35:00 into the audio, they discuss Mann's flawed usage of the principal component analysis. At 37:30, they discuss why the bristlecone pines are not good temperature proxies. At 50:00, a panel member answers "Yes" to the question whether he is saying that the odds than Mann is right are around 2:1 - which means "almost completely uncertain". At 53:40, a desperate activist / journalist tries to criticize the NAS panel that they used the word "plausible". How could this have happened? ;-) The same crazy journalist even says that there is no evidence for string theory but we can say it's "plausible". I assure this comrade that string theory is much more plausible than a catastrophic global warming. ;-) Around 54:30, they also agree with your humble correspondent that quantitative estimates of "Bayesian" confidence levels in this context (and similar contexts) are meaningless and the real uncertainty can't be quantified. Near 57:00, Myron Ebell asks about the divergence problem - the fact that the current proxies don't show the warming measured directly by thermometers. He is answered that the problem is there, indeed, and it might be hand-waved away by some very vague comments about moisture. At 59:30, another passionate eco-journalist complains against the word "plausible". How can you say it is less plausible if there is no evidence against [except for those paid by the oil company, he would normally say]? ;-) He is again explained that we just don't know, and there is a lot of natural climate variability that increases the uncertainty. Around 1:03:00 into the audio, it is being discussed how much money is being wasted for climate research whose insights are ever more murky and questionable, despite hundreds of papers. At 1:06:30, they say that the variability - and thus also the uncertainty - is higher than thought previously. Although the NAS members say and write a lot of wise and correct things (and things that I've been saying for months if not years), of course, there are many other examples of schizophrenia of their document and its interpretation. On one hand, they concede that virtually every single criticism of the "hockey stick graph" has been valid and the methodology of Mann, Bradley, and Hughes is unsatisfactory because of all these reasons (besides the examples above, also lacking statistical skill for individual years, problems with unavailable data and secret computational software - go to 16:20 of the audio for the transparency issues). On the other hand, they are using graphs from papers that are criticizable because of the very same reasons and they essentially encourage the reader to think that these reconstructions are trustworthy even though the actual content of the chapters 9 and 11 leads to the opposite conclusion. The most penetrating and freely available analysis of the document was written by who has been - together with Ross McKitrick, his collaborator - the world's principal auditor of the climate reconstructions and some very useful comments on that page are also offered by Eduardo Zorita and others. I think that the whole field of "global climate science" has become a political game where people are looking for a compromise or, as many of them openly call it, a consensus. The recent developments in science have shown that some statements in the past decade are scientifically undefendable and some papers have simply been wrong but those people just don't have enough courage and integrity to admit this fact openly which is why they only soften their language and generate logically inconsistent bureaucratic hybrids. It will take a lot of time to restore the integrity of climate science but we may hope that the NAS panel made the first steps towards this goal. A less technical summary describing the schizophrenia in their report was written by Iain Murray: Also, you may want to read a rather technical comment by David Stockwell at landshape.org and a nice, non-technical summary of the report by a climate director from the University of Delaware, When Millikan measured the charge of the electron for the first time, he obtained 50% of the correct result only because of his incorrect treatment of the air viscosity. Those who followed him have already used a correct approach to the air viscosity but they obtained 55%, 60%, 70%, 80%, 90% of the right result before they settled near the right value. The reason why they did not obtain the correct result already in the second experiment was that they were intentionally eliminating data from their experiments that were too far from the previous ones. Virtually all experimental physicists know that this is bad science - but in the climate science, things seem a bit different. The authors of crappy papers are still forming the consensus and others don't want to lose friends so they keep on including junk science to their reports and statistical ensembles. The downgrading of the certainty about the "unprecedented global warming" from "certain & debate is over" to "plausible" has made some climate activists, such as our friend William Connolley, worried. He talks about a "step backwards". How does he define progress in science? It's quite obvious. The higher temperatures and disasters are predicted and the more irrational, hysterical, and political the justifications become, the more progressive science becomes according to this "scientist". ;-) The science about the "catastrophic climate change" seems rather similar to paranormal sciences: the amount of "signal" that one obtains is more or less directly proportional to the lack of scientific integrity of the scientist. Not surprisingly, the signal and certainty obtained by NAS is much smaller than what William Connolley, Al Gore, or other charlatans would obtain. But the NAS members are still not perfect, so they still get a signal. ;-) At least, William can fool himself into believing that the NAS panel has not confirmed any criticism. Those who have listened to the press conference know very well that the panel has confirmed that, as far as I see, all the criticisms by M&M and others are valid. However, the panel has also tried their best to be nice to people like Mann which I think is questionable. Meanwhile, Mann and Connolley's propagandistic blog called "RealClimate" is trying to keep a low profile. There are only 13 comments about the topic which is slightly less than 166 comments about the same story at the "Climate Audit". Liars in the newspapers While the NAS panel has concluded that the climate before 1600 is virtually completely uncertain, hundreds of left-wing newspapers have published fraudulent and completely false articles claiming that according to NAS, we're the hottest in 1000 or 2000 years. Of course, the journalists remain the most important sources of the disinformation and the primary drivers of the global fraud. See Micajah's article for a more detailed comparison. A critic of Star Tribune in Minnesota shows that the journalists took a tendentious article in the Washington Post but they were still not quite satisfied, so they erased all the remaining paragraphs that described the problems with the hockey stick graph, and then they published the castrated article! I ask all honest owners of newspapers and other media bosses who realize that the global warming has become a gigantic fraud in which their employees are actively involved - as shown by lies they published about the NAS panel - to fire these employees as soon as possible. The diplomatic schizophrenic language of NAS aside, the panel has essentially done a good work, offered a comprehensible summary of their review process, and it has buried the "consensus" on "unprecedented" global warming theory - the conjecture that the recent changes of the climate are unprecedented in comparison with previous millenia - and you can see the graphs from the NAS report that show it rather clearly: • I (for example, in Greenland and Antarctica - those that matter - there is no recent warming at all according to the graph 1000-2000) • II (noise) • III (mostly noise) Additional frequently visited climate articles on The Reference Frame 1. I think this topic is very different and should be more pages like this,Your comment says a lot of truth, and I'd like to invited you to visit my page:10/325- Vicoprofen - Lortab- Tylenol #3 - Ativan All Major Medications are available right here at: http://www.crdrx.com 2. just discovered your blog through some searches on climate change and Lord Monckton. haven't even see any of your recent post since I've pulled up so much from links on earlier posts. Anyway, just wanted to say "Kudos" on a blog that has been very helpful for me at least. b b
2021-07-27 09:55:43
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https://codereview.stackexchange.com/questions/70374/insertbst-exercise
# insertBST exercise I'm just learning Haskell on my own, and did an exercise to implement insertBST. insertBST :: (a -> a -> Ordering) -> a -> BST a -> BST a insertBST _ v Leaf = Node Leaf v Leaf insertBST cmp v (Node left nv right) = case (cmp v nv) of EQ -> Node (insertBST cmp v left) nv right LT -> Node (insertBST cmp v left) nv right GT -> Node left nv (insertBST cmp v right) Since the EQ and LT cases are the same, we can put GT first to remove this duplication insertBST :: (a -> a -> Ordering) -> a -> BST a -> BST a If you install hlint you will also get warnings about things like the redundant parentheses that were around cmp n nv. You can set up your editor to run this automatically on save.
2019-11-11 23:48:29
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http://puzzling.stackexchange.com/questions/1963/6-the-magic-number
# 6, the magic number Here's a fun (albeit difficult) one: Make these equations true using arithmetic operations: 1 1 1 = 6 2 2 2 = 6 3 3 3 = 6 4 4 4 = 6 5 5 5 = 6 6 6 6 = 6 7 7 7 = 6 8 8 8 = 6 9 9 9 = 6 For example: 6 + 6 - 6 = 6 (I hope I did not spoil some of you :D) Allowed operators are: +, -, *, /, ! , ^, % Setting parenthesis is also allowed. The ^ operator is an exception as you are permitted to supply a second argument to it which may be any positive integer or the multiplicative inverse of it. $x^{1/y}$ is always positive and real. If you find an alternative solution using other operators you may post it but please also provide a solution using only these 7 operators. For those of you who think this was easy, here is a bonus: 0 0 0 = 6 - Obviously not only -,+,*,/ are allowed. Please tell full list of allowed operations. – klm123 Jul 26 '14 at 22:12 "(x^0 + x^0 + x^0)!" - so you allowed to use additional numbers and ()? – klm123 Jul 26 '14 at 22:30 @klm123 Yes, you are allowed to use additional numbers, but only as second argument to the ^ operator – ThreeFx Jul 26 '14 at 22:31 How about square roots? – BitNinja Jul 26 '14 at 23:28 @Muqo For the sake of keeping everything nice and clean, we will only consider the positive real roots – ThreeFx Jul 27 '14 at 15:29 1. $(1+1+1)! = 6$ 2. $2+2+2 = 6$ 3. $3*3-3 = 6$ 4. $(4-\frac 4 4)! = 6$ 5. $5+\frac 5 5 = 6$ 6. $6*\frac 6 6 = 6$ 7. $7-\frac 7 7 = 6$ 8. $(\sqrt{8+\frac 8 8})! = 6$ 9. $(\frac{\sqrt{9}\sqrt{9}}{\sqrt 9})! = 6$ Bonus: $(0!+0!+0!)! = 6$ - Bonus: (0^0 + 0^0 + 0^0)! – c0rp Jul 27 '14 at 9:49 @c0rp 0^0 is NaN. Also, you can only choose a positive exponent. – ThreeFx Jul 27 '14 at 11:38 $0! = 1$, though. – Ben Millwood Jul 28 '14 at 0:09 @ThreeFx 0^0 is not always NaN depending on who you ask and what field you're in. It can also be set to 0^0=1 – Engineer Toast Apr 22 '15 at 15:18 "one has to know that in order to be able to use it"? What on earth does that mean? – Lynn Nov 22 '15 at 1:05 I insist on using all the digits! $(1 + 1^{1234567890} + 1)! = 6$ $(2 + (2^{1234567890}\ \text{mod}\ 2)!)! = 6$ $(3 + 3^{1234567890}\ \text{mod}\ 3)! = 6$ $(4 - (4^{1234567890}\ \text{mod}\ 4)!)! = 6$ $5 + (5^{1234567890}\ \text{mod}\ 5)! = 6$ $6 + 6^{1234567890}\ \text{mod}\ 6 = 6$ $7 - (7^{1234567890}\ \text{mod}\ 7)! = 6$ $(\sqrt[3]8 + (8^{1234567890}\ \text{mod}\ 8)!)! = 6$ $(\sqrt{9} + (9^{1234567890}\ \text{mod}\ 9))! = 6$ $(0! + (0^{1234567890})! + 0!)! = 6$ No, wait! How about if we take subtraction out and put subfactorial in? More exclamation points!!!! $((!1)! + (!1)! + (!1)!)! = 6$ $(!2 + !2 + !2)! = 6$ $!3 + !3 + !3 = 6$ $(\sqrt{!4} \times 4 \div 4)! = 6$ $!(\sqrt{!5\ \text{mod}\ 5}) + 5 = 6$ $!6\ \text{mod}\ 6 \times 6 = 6$ $!7\ \text{mod}\ 7\ \text{mod}\ 7 = 6$ $(!8\ \text{mod}\ 8 + \sqrt[3]8)! = 6$ $\sqrt[3]{!9\ \text{mod}\ 9} \times \sqrt9 = 6$ $(!0 + !0 + !0)! = 6$ - ????!!!!????!!!! – rand al'thor Dec 30 '14 at 11:20 @rand al'thor You look like you need some 's!! Hold on, is there a operator, too‽‽ This answer might need revision!! – Muqo Dec 30 '14 at 18:17 Here we go. # 1: $(1+1+1)! = 6$ This is the only possible one as far as I know. # 2: $2+2+2 = 6$ # 3: $3*3-3 = 6$ # 4: $4+(4/\sqrt{4}) = 6$ # 5: $5+(5/5) = 6$ # 6: $6*(6/6) = 6$ # 7: $7-(7/7) = 6$ # 8: 8 - $\sqrt[4]{8 + 8} = 6$ # 9: $(9+9)/\sqrt{9} = 6$ # Bonus - 0: $(0! + 0! + 0!)! = 6$ - Sorry, typo on the first one :3 – James Lynch Dec 30 '14 at 2:22 Ok. I got it fixed now :) – James Lynch Dec 30 '14 at 2:25 Nice solutions, I particulary like the one to number 8, definitely worthy of an upvote. :D – ThreeFx Dec 30 '14 at 2:27 The bottom five (0 through 4) can all be solved using the same construction: (0!+0!+0!)! = 6 (1 +1 +1 )! = 6 (2 +2 /2 )! = 6 (3 +3 %3 )! = 6 (4 -4 /4 )! = 6 For 6 and 7, there are slightly more funky solutions: (6!)%(6!-6)=6 ((7!)/7)%7=6 (I haven't found an interesting solution for 5, nor any square-root-free solutions for 8 or 9.) - Square roots are allowed. – Emrakul Jul 27 '14 at 1:54 I remember seeing this question in a few math curiosity books, and most recently on an episode of Scam School. However, they usually allow square roots. The answers I came up with are as follows: $$(0! + 0! + 0!)! = 6$$ $$(1 + 1 + 1)! = 6$$ $$2 + 2 + 2 = 6$$ $$3 \times 3 - 3 = 6$$ $$4 + 4 - \sqrt{4} = 6$$ $$5 + 5/5 = 6$$ $$6 + 6 - 6 = 6$$ $$7 - 7/7 = 6$$ $$8 - \sqrt{\sqrt{8 + 8}} = 6$$ $$9 - 9 / \sqrt{9} = 6$$ As you can see, it's doable without any exponentiation or modular arithmetic, if square roots are allowed. You only need the four arithmetic operators, square roots, and factorials. - The square root of a number is equal to $x^{1/2}$ – ThreeFx Jul 30 '14 at 7:08 True. But it also has the benefit of not explicitly introducing any new numbers. – Joe Z. Jul 30 '14 at 15:09 @JoeZ. I like the idea of not introducing extra numbers (and operators) in order to help limit the number of valid answers. This question as it's currently stated seems far too broad. Would anyone know the original wording and parameters? – Muqo Jul 30 '14 at 17:29 There are probably infinitely many variations, but the basic rules always allows the four basic operators, factorials, and square roots. And as I've shown above, that's all you need. – Joe Z. Jul 30 '14 at 18:02 @JoeZ. You can improve on this particular variation more. Explicit multiplication is not required. – Muqo Jul 30 '14 at 18:25 1. (1+1+1)! = 6 2. 2+2+2 = 6 3. 3*3-3 = 6 4. (4^3)/(4^2)+(4^(1/2)) = 6 5. 5+(5/5) = 6 6. (6*6)/6 = 6 7. 7-(7/7) = 6 8. (8^3)/(8^2)-(8^(1/3)) = 6 9. (9+9)/(9^(1/2)) = 6 and the bonus (0!+ 0! + 0!)! = 6 - For bonus one... ((0!)+(0!)+(0!))! - I am doing this for the eights only: $8 \ - \ \sqrt{\sqrt{8 + 8}} \ = \ 6$ $-\sqrt{\sqrt{8 + 8}} \ + \ 8 \ = \ 6$ $(\sqrt{8 + (8 - 8)!})! \ = \ 6$ $(\sqrt{(8 - 8)! + 8})! \ = \ 6$ $((\sqrt{8 + 8})!/8)! \ = \ 6$ - I deleted the invalid solutions. – Olive Stemforn Mar 14 at 4:52 $$(1+1+1)!=6$$ $$2^2+2=6$$ $$3*3-3=6$$ $$4+(4/\sqrt4)=6$$ $$(5-5)!+5=6$$ $$6*6/6=6$$ $$7-(7-7)!=6$$ $$\sqrt[3]{8}+\sqrt[3]{8}+\sqrt[3]{8}$$ $$(9+9)/(\sqrt9)=6$$ And finally, $$(0!+0!+0!)!=6$$ - Did you mean $\sqrt[3]{8}$? If so, it's $\sqrt[3]{8}$ – Emrakul Aug 5 '14 at 20:50 I mean double square roots as in fourth roots, like $\sqrt[4]{8}$, or two square roots. – Vincent Tang Aug 6 '14 at 18:34 Oh, you can actually do just $\sqrt{\sqrt{8}}$, or $\sqrt[4]{8}$ ($\sqrt{\sqrt{8}}$ or $\sqrt[4]{8}$). $\sqrt[n]{8}$ is $\sqrt[n]{8}. – Emrakul Aug 6 '14 at 18:58 2+2+2=6 (3*3)-3=6 (4/sqrt4)+4=(4/2)+4=6 (5/5)+5=6 (6+6)-6=6 7-(7/7)=6 cubrt8+cubrt8+cubrt8=2+2+2=6 9-(9/sqrt9)=9-(9/3)=9-3=6 - Most of this is OK, but I think the cube root operator isn't allowed under the rules of the question. – rand al'thor Jul 8 '15 at 13:04 @randal'thor: Actually, it is. The OP said that you can use ^ with any positive integer or multiplicative inverse. So you can do 8^(1/3). – mmking Jul 8 '15 at 13:55 $$2+2+2$$ $$3\times3-3$$ $$\sqrt{4}+\sqrt{4}+\sqrt{4}$$ $$\frac{5}{5}+5$$ $$6\times\frac{6}{6}$$ $$7-\frac{7}{7}$$ $$\sqrt[3]{8} +\sqrt[3]{8} +\sqrt[3]{8}$$ $$\sqrt{9}\times\sqrt{9}-\sqrt{9}$$ - Hi, welcome to Puzzling.SE! I've cleaned up your answer a bit for you - hopefully you noticed that this question was answered a while ago and most of your answers are equivalent to the already accepted one. – Deusovi Oct 18 '15 at 13:54 2*2*2=6 3*3-3=6 (4*4)/4=6 5+(5/5)=6 6+6-6=6 7-(7/7)=6 (8*8)/8=6 9-(9/(root of 9))=6 - 2*2*2 is 8, not 6! – Bailey M Jul 9 '15 at 20:09 Should be 2*2+2. – Victor Stafusa Jul 9 '15 at 20:46 Or$2+2+2$. And your$4$s and$8\$s are also wrong. – Kevin Jul 9 '15 at 20:58 (2+2+2)=6 (3+3+3)-3=6 (4+4+4)-6=6 (5+5+5)-9=6 (6+6+6)-12=6 (7+7+7)-15=6 (8+8+8)-18=6 (9+9+9)-21=6 Ans of above question. - You are not allowed to use anything except the operators, so no extra numbers. – ThreeFx Oct 8 '15 at 18:38 ## protected by Community♦Feb 14 at 12:02 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.
2016-05-04 09:53:00
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http://automathic.org/30_area-of-a-disc
Check out our latest project: Algebrarules.com. The Rules of Algebra, Sweet 'n Simple. X Sort By: Formula per Page: By PiThagoras Tags: Circle, Disk, Disc, Pi, Area, Surface Area 1 Posted: 2013-02-05 Area of a Disc Area of a Disc $$Area = \pi r^2$$ The area of a disc (the region inside a circle), often incorrectly called the area of a circle, is $\pi r^2$ when the circle has radius r. Here the symbol $\pi$ (Greek letter pi) denotes, as usual, the constant ratio of the circumference of a circle to its diameter. It is easy to deduce the area of a disk from basic principles: the area of a regular polygon is half its apothem times its perimeter, and a regular polygon becomes a circle as the number of sides increase, so the area of a disc is half its radius times its circumference (i.e. ${1 \over 2} r * 2 \pi r$ ).
2018-02-23 02:34:59
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https://pos.sissa.it/414/957/
Volume 414 - 41st International Conference on High Energy physics (ICHEP2022) - Poster Session Search for rare decays at BESIII Y. Song Full text: pdf Pre-published on: November 20, 2022 Published on: Abstract Based on 10 billion $J/\psi$ events accumulated by the BESIII detector, we show the searches for the rare process of $J/\psi$ weak decays. We also search for other rare decay process, such as the FCNC process $D^0\to\pi^0\nu\bar{\nu}$. Using $J/\psi$ decay, BESIII also produce millions of Hyperon, which can be used to search for the rare decay process $\Xi^-\to\Xi^0 e\nu$. DOI: https://doi.org/10.22323/1.414.0957 How to cite Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete. Open Access
2023-01-28 03:33:04
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https://www.someweekendreading.blog/cfs-20t-magnet/
Thu 2021-Sep-09 # CFS & MIT PSFC build 20T REBCO magnet Tagged: Physics Today Commonwealth Fusion Systems (CFS) and MIT’s Plasma Science Fusion Center (PSFC) announced they have successfully built high-$T_c$ REBCO magnet capable of 20T field strength. What should that mean to you? ## Common-who built a what, now? Basically, researchers at MIT and at Commonwealth Fusion Systems (an MIT spinoff) built a big magnet. [1] [2] So what? Well, it’s not just a big magnet. It’s got some really interesting features: • It achieved a sustainable magnetic field strength of 20 Tesla! As an undergrad back in the late Neolithic, a 1 Tesla magnet was considered pretty gnarly. Sure, there were stunts in which higher fields could be achieved momentarily by explosive compression, in at least one case with a nuclear bomb. Now we can do that routinely, sustainably, and with a high-quality field that can be sustained over long(ish) times. • It’s using high-$T_c$ superconductors, unlike a lot of other fusion tokamaks. High $T_c$ means less trouble cooling. (I’d initially thought they were going for above nitrogen temperature, i.e., above 70°K. But it looks like they’re going down to 20°K. I dunno why. Thermal safety margin, maybe? Still, now you’re talking liquid He cooling, and that’s… a whole world of difficult.) These are REBCO magnets, built on a rare-earth barium copper oxide crystal, presumably in the perovskite family of the high-$T_c$ superconductors of yore. Typically the rare earth in question is yttrium, but I wasn’t able to verify that in this case. (Though to be fair, I didn’t look very hard, either.) • “As soon as you give people something that looks like a wire, they try to wind it into something that looks like a coil”, as Eric Drexler used to say. But previous high-$T_c$ superconductors were more like ceramics, and would break. These are apparently more like a tape than a wire, but they wind just fine. They’re pretty complex composites of many layers of materials, only one of which is the REBCO superconductor. • It’s large-bore: not just some dinky little chamber a couple millimeters on a side, but nice and big so you can fit the fusion torus of a (smallish) tokamak inside it. • Interestingly, they did that on time according to their plan, during a pandemic. That’s either impressive dedication and project management, or reckless workaholism. Knowing the culture of scientists in general and MIT in particular, I venture it’s probably some of both. ## Why exactly is that interesting? The very high 20T field strength means one can attempt to build a much smaller fusion reactor. ITER in France is a bit of a monster, what a colleague described as “a Pharaonic endeavor” on the scale of the Great Pyramids at Giza. A lower field magnet would lead to a tokamak about 40x larger than the MIT/CFS design. It turns out that the volume of the tokamak – a rough indicator of cost – scales as the inverse cube of the $\mathbf{B}$ field: $V \propto \left|\mathbf{B}\right|^{-3}$. So if you double the $\mathbf{B}$ field, you get to shrink the volume by a factor of 8. That means half the linear size in each dimension. Of course, a smaller reactor will produce less power than a big one, but at 1/8th the cost you can build a couple of them. The MIT PSFC developed a fusion reactor based on the assumption one could have a high $\mathbf{B}$-field, and hence high plasma pressure available. That would allow considerable scale-down of size, which makes everything else less like building the Great Pyramids. ITER will run at 9 Tesla, so running at 18–20 Tesla gives running room for about a factor of 8 scale-down in volume. ## ARC reactor (yes, it’s really called that) & SPARC testbed It was called the ARC reactor, for “affordable, robust, compact”. [3] The testbed for ARC is the smaller SPARC reactor [4], apparently jointly designed by MIT PSFC and CFS. It starts construction pretty much now, with the availability of the high-field high-$T_c$ magnets, in Devens, Mass (formerly Ft. Devens). It’s projected to be operational in 2025. It’ supposed to generate up to 140MW of power in 10sec bursts. It looks like the magnets will not quench below 77°K, though they’re for some reason trying to operate at 10°K. The fusion gain, or power out over power in, is expected to be around $Q \sim 11$! ## Workable fusion power With some luck, CFS & PSFC are tenatively predicting this could lead to workable fusion power plants by 2030. That’s… very specific. Fusion has been 20 years away for all of my lifetime; I was resigned to it always being 20 years away for the rest of my life. Maybe I need to change my mind on that. Maybe there’s hope for humanity yet in the face of hostile climate change. I hate false hope. So, not to get too meta, I hope this is real hope. ## Notes & References 1: J Smith-Galvin, “Commonwealth Fusion Systems creates viable path to commercial fusion power with world’s strongest magnet”, Commonwealth Fusion Systems press releases, 2021-09-09. 2: D Chandler, “MIT-designed project achieves major advance toward fusion energy”, MIT News, 2021-Sep-08. 3: Well, sort of. Oy, I can barely say that with a straight face. Of course we all know it’s an Iron Man joke, named after the ARC reactor invented by Tony Stark. (In the comics, Tony Stark of course attended MIT before becoming the billionaire playboy industrialist superhero. One need hardly even mention that.) 4: I dunno what SPARC is supposed to stand for. Maybe “Shiny Petite ARC” reactor? Later: I am reliably informed that it’s “Smallest Possible ARC”. I still like “Shiny Petite” better, but… ok. Written Thu 2021-Sep-09
2021-12-04 07:25:31
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https://cs.stackexchange.com/questions/126407/check-if-language-is-decidable
# Check if language is decidable I would like to determine if the following language is decidable or not. L = { w $$\in$$ $$\Sigma^*$$ | $$T(M_w)$$ is recognized by a Turing machine with at most 42 states}. I know that every finite language is decidable, but I am not sure if this has anything to do with this particular problem. Every help is appreciated, I am a little lost. Thank you • What is T(M)? Is it the language generated from the turing machine M? May 28, 2020 at 14:23 Define C = { $$L | L$$ is decidable by a turing machine with at most 42 states } Notice that there is a finite amount of such languages, as there are a finite amount of turing machines with 42 states. for that reason, we have $$C\ne\emptyset, \sum^*$$ By Rice's theorem, the language $$S=$$ { $$w | L(M_w) \in C$$ } is undecidable. This language is precisely the language in question
2022-06-25 14:04:56
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http://gmatclub.com/forum/if-3-a-4-b-c-what-is-the-value-of-b-1-5-a-25-2-c-113040.html?fl=similar
Find all School-related info fast with the new School-Specific MBA Forum It is currently 29 Apr 2016, 06:15 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If 3^a * 4^b = c, what is the value of b? (1) 5^a = 25 (2) c Author Message TAGS: ### Hide Tags Director Status: No dream is too large, no dreamer is too small Joined: 14 Jul 2010 Posts: 650 Followers: 37 Kudos [?]: 589 [1] , given: 39 If 3^a * 4^b = c, what is the value of b? (1) 5^a = 25 (2) c [#permalink] ### Show Tags 01 May 2011, 07:42 1 KUDOS 3 This post was BOOKMARKED 00:00 Difficulty: 85% (hard) Question Stats: 45% (02:20) correct 55% (01:02) wrong based on 185 sessions ### HideShow timer Statictics If 3^a * 4^b = c, what is the value of b? (1) 5^a = 25 (2) c = 36 [Reveal] Spoiler: OA _________________ Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html VP Status: There is always something new !! Affiliations: PMI,QAI Global,eXampleCG Joined: 08 May 2009 Posts: 1353 Followers: 16 Kudos [?]: 201 [0], given: 10 Re: what is the value of b? [#permalink] ### Show Tags 02 May 2011, 04:45 straight B. 36 = 2 ^2 * 3^2 _________________ Visit -- http://www.sustainable-sphere.com/ Promote Green Business,Sustainable Living and Green Earth !! Manager Joined: 10 Jul 2010 Posts: 196 Followers: 1 Kudos [?]: 18 [0], given: 12 Re: what is the value of b? [#permalink] ### Show Tags 09 Nov 2011, 08:52 I'm going to go with B 36 = $$3^2 and 2^2$$ you are given that $$3^a*4^b=c$$ we can break down 4 to 2^2b So now we have $$3^a*2^2b=2^2*3^2$$ so a= 2, b = 2 Intern Joined: 07 Oct 2011 Posts: 30 Followers: 0 Kudos [?]: 4 [1] , given: 1 Re: what is the value of b? [#permalink] ### Show Tags 09 Nov 2011, 09:41 1 KUDOS a and b are not integers here. consider this: Now for every value of a, there is a different value of b. Say, a = 1, then 4^b = 12 and b = 1.79 approx a = 2, then 4^b = 4 and b = 1 a = 3, then 4^b = 36/27 and b = 0.2 approx Intern Status: Stay Hungry, Stay Foolish. Joined: 05 Sep 2011 Posts: 41 Location: India Concentration: Marketing, Social Entrepreneurship GMAT 1: 650 Q V Followers: 2 Kudos [?]: 8 [0], given: 6 Re: If 3^a * 4^b = c, what is the value of b? (1) 5^a = 25 (2) c [#permalink] ### Show Tags 13 Nov 2011, 09:20 The question does not mention a and b to be Integers. Individually,none of the statements suffice to reach the value of b. Both the statements when used together,give the value of b. 3^2 * 4^b = 36. b=1 Hence C. Math Expert Joined: 02 Sep 2009 Posts: 32530 Followers: 5620 Kudos [?]: 68193 [1] , given: 9797 Re: what is the value of b? [#permalink] ### Show Tags 04 Feb 2012, 15:50 1 KUDOS Expert's post 6 This post was BOOKMARKED amit2k9 wrote: straight B. 36 = 2 ^2 * 3^2 If 3^a*4^b = c, what is the value of b? Note that we are not told that the variables are integers only. (1) 5^a = 25 --> $$a=2$$, but we can not get the values of $$b$$. Not sufficient. (2) c = 36 --> $$3^a*4^b = c$$: it's tempting to write $$3^2*4^1=36$$ and say that $$b=1$$ but again we are not told that the variables are integers only. So, for example it can be that $$3^a=36$$ for some non-integer $$a$$ and $$b=0$$, making $$4^b$$ equal to 1 --> $$3^a*4^b =36*1=36$$. Not sufficient. (1)+(2) As $$a=2$$ and $$c = 36$$ then $$9*4^b=36$$ --> $$b=1$$. Sufficient. Hope it's clear. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 9211 Followers: 453 Kudos [?]: 114 [0], given: 0 Re: If 3^a * 4^b = c, what is the value of b? (1) 5^a = 25 (2) c [#permalink] ### Show Tags 30 Sep 2014, 23:42 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 21 Feb 2013 Posts: 6 Followers: 0 Kudos [?]: 0 [0], given: 9 Re: If 3^a * 4^b = c, what is the value of b? (1) 5^a = 25 (2) c [#permalink] ### Show Tags 19 Jun 2015, 08:15 Hi Bunnel in the above explantion how can 3^a=36 be possible for any value a, it has to have an even no. plz clarify my doubt Math Expert Joined: 02 Sep 2009 Posts: 32530 Followers: 5620 Kudos [?]: 68193 [0], given: 9797 Re: If 3^a * 4^b = c, what is the value of b? (1) 5^a = 25 (2) c [#permalink] ### Show Tags 19 Jun 2015, 08:21 Expert's post caster88 wrote: Hi Bunnel in the above explantion how can 3^a=36 be possible for any value a, it has to have an even no. plz clarify my doubt We are NOT told that a and b are integers. So, there exists some irrational a for which 3^a = 36: $$a= \frac{2 (log(2)+log(3))}{log(3)}\approx{3.2619}$$ _________________ Re: If 3^a * 4^b = c, what is the value of b? (1) 5^a = 25 (2) c   [#permalink] 19 Jun 2015, 08:21 Similar topics Replies Last post Similar Topics: If 5a = 9b = 15c , what is the value of : 0 19 Dec 2015, 05:56 2 If 3a + 2b – 4 = a – b, what is the value of b ? 1 23 Jul 2014, 10:48 3 Q1 If 3^a*4^b = c, what is the value of b? (1) 5^a = 25 (2) 9 10 Dec 2010, 01:00 17 If 3^a*4^b = c, what is the value of b? (1) 5^a = 25 (2) c = 28 10 Dec 2010, 01:00 2 If 2 + 5a b/2 = 3c, what is the value of b? 3 17 Feb 2008, 13:08 Display posts from previous: Sort by
2016-04-29 13:15:24
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https://md960u.com/jexvsi/difference-between-valuation-and-estimation-f4f007
Posted on # difference between valuation and estimation The Difference Between Enterprise Value and Equity Value Equity Value Equity value constitutes the value of the company's shares and loans that … Understandably, sometimes there is confusion between the use of the words ‘valuation’ and ‘market appraisal’ in estate agency. The major difference between a contractor's bid estimate and an engineer's estimate lies in the information the two estimators have at their disposal. The first is to gather information about the home. Text is available under the Creative Commons Attribution/Share-Alike License; additional terms may apply. When you’re moving, talk with your current insurance agent about your coverage options. Collecting correct data is a critical step for both the Zestimate and appraiser. Difference between Point and Interval Estimation 3. Zestimate vs. Appraisers: Estimation process. (logic, first-order logic, model theory) A structure, and the corresponding assignment of a truth value to each sentence in the language for that structure. The two main types of estimators in statistics are point estimators and interval estimators. GOALS 1. Creative Commons Attribution/Share-Alike License; (construction and business) A document (or verbal notification) specifying how much a job will probably cost. BYJU’S online estimate the difference calculator tool makes the calculation faster, and it displays the difference in a fraction of seconds. population mean, the difference between population means, proportions, variation among groups). (logic, first-order logic, model theory) A structure, and the corresponding assignment of a truth value to each sentence in the language for that structure. In statistics, the bias (or bias function) of an estimator is the difference between this estimator's expected value and the true value of the parameter being estimated. The point estimate of your confidence interval will be whatever statistical estimate you are making (e.g. Point and Interval Estimation Presented by: Shubham Mehta 0019 2. Creative Commons Attribution/Share-Alike License; (finance) The process of estimating the market value of a financial asset or liability. ... Their studies teach them how to estimate and monitor construction costs on a project from feasibility stage through to completion based on initial designs and drawings. Point and Interval Estimation 1. Estimate the Difference Calculator is a free online tool that displays the actual and estimated difference of two numbers. It is necessary for the valuation of land and building. Defining Level of Confidence 4. As the P/E goes up, it shows that current investor … (measure theory, domain theory) A map from the class of open sets of a topological space to the set of positive real numbers including infinity. (logic, propositional logic, model theory) An assignment of truth values to propositional variables, with a corresponding assignment of truth values to all propositional formulas with those variables (obtained through the recursive application of truth-valued functions corresponding to the logical connectives making up those formulas). ... a valuer may be called on to give a valuation … Since the context is statistics: An estimation $\hat{X}$ of a parameter or statistic $X$ involves using a sample of all the possible values of $x$ (the population). Estimated cost and the actual cost The estimated cost is the anticipated or probable cost of the work and … In case the funds Illustration 8.16: Difference between market value and book value debt ratios – Boeing in June 2000 Illustration 8.17: Estimating Cost of Capital - Boeing Illustration 8.18: Estimating Cost of Capital – Embraer in January 2001 Paid Valuation vs Market Appraisal. A sample is a part of a population used to describe the whole group. Point estimation is the opposite of interval estimation. Additionally, some estate agents – including Chancellors – offer an ‘instant valuation’ service, which provides an online estimate for the value … It uses sample data when calculating a single statistic that will be the best estimate of the unknown parameter of the population. What is the difference between Rounding and Estimating? (algebra) A measure of size or multiplicity. You may be surprised to find that your h… Text is available under the Creative Commons Attribution/Share-Alike License; additional terms may apply. A point estimator is a statistic used to estimate the value of an unknown parameter of a population. of Civil, ACE Page 1 1.INTRODUCTION DEFINITION OF ESTIMATING AND COSTING Estimating is the technique of calculating or Computing the various quantities and the expected Expenditure to be incurred on a particular work or project. Of course there is a difference between prediction and estimation, although some people use them interchangeably. As nouns the difference between valuation and estimation is that valuation is an estimation of something's worth while estimation is the process of making an estimate. Difference between Estimating and Costing Although estimating and costing both are required to decide the price of the product, even then the two are different as explained below: 1. Believe it or not, insurance developed as early as 2,000 BCE to reduce the risk for merchants who needed their items shipped across the water. Constructing Confidence … The Estimator contact or contractor’s estimator normally has much more detailed information at his disposal. A change in accounting principle is a change in how financial information is calculated, while a change in accounting estimate is a change in the actual financial information. (finance) The process of estimating the market value of a financial asset or liability. … The P/E ratio gives investors an idea of what the market is willing to pay for the company's earnings. You can be 95% confident that the population mean for the difference is between 14.22 and 27.78. When it comes to moving your household goods, you want to make sure your shipment is secure. Extrapolation is estimating the value of a variable outside a known range of values by assuming that the estimated value follows some pattern from the known ones. Estimate the difference calculator is a pre-algebra tool to find the actual & estimated difference between given minuend & subtrahend by rounding off to the nearest ten, hundred & thousand. (measure theory, domain theory) A map from the class of open sets of a topological space to the set of positive real numbers including infinity. See Wiktionary Terms of Use for details. The takeover premium is the additional cost of purchasing all shares in a merger and acquisition. So an estimate is the result of estimation. The ratio is determined by dividing a company's current share price by its earnings per share. Price-to-earnings ratio (P/E) looks at the relationship between a company's stock price and its earnings. On the other hand, valuation is the process of determining the worth or value of something through the same process of estimation. Key Results: Estimate for difference, 95% CI for difference. As a verb estimate is to calculate roughly, often from imperfect data. Difference-in-Difference estimation, graphical explanation DID is used in observational settings where exchangeability cannot be assumed between the treatment and control groups. * {{quote-book, year=1928, author=Lawrence R. Bourne. *** Complete transcript: “Hi, my name is Stacey Myhro and I’m Vice […] Ship captains or banks would promise to deliver their goods and pay the difference for any cargo that was lost at sea. ESTIMATION AND VALUATION 10CV73 Dept. Front-end estimation is a particular way of rounding numbers to estimate sums and differences.To use front- end estimation, add or subtract only the numbers in the greatest place value. (logic, propositional logic, model theory) An assignment of truth values to propositional variables, with a corresponding assignment of truth values to all propositional formulas with those variables (obtained through the recursive application of truth-valued functions corresponding to the logical connectives making up those formulas). Both human appraisers and automated valuation systems (AVMs) have two major tasks to complete when developing a home valuation. Plus’s Vice President of Supplier Relations, Stacey Myhro, is here to explain the difference between valuation and insurance. The difference between a Quantity Surveyor and a Valuer. Changes in … When a statistical characteristic, such as opinion on an issue (support/don’t support), of the two groups being compared is categorical, people want to report […] What is the difference between valuation and insurance? * {{quote-book, year=1965, author=Ian Hacking, title=Logic of Statistical Inference, passage=I. Precedents are useful for valuing an entire business (including a takeover premium Takeover Premium Takeover premium is the difference between the market value (or estimated value) of the company and the actual price to acquire it. 1 As nouns the difference between estimation and evaluation is that estimation is the process of making an estimate while evaluation is an assessment, such as an annual personnel performance review used as the basis for a salary increase or bonus, or a summary of a particular situation. Methods of Estimation 2. An estimator or decision rule with zero bias is called unbiased.In statistics, "bias" is an objective property of an estimator. Example: Point estimate In the TV-watching example, the point estimate is the mean number of hours watched: 35. For example, if a company is currently trading at $25 a share and its earnings over the last 12 months are$1.35 per share, the P/E ratio for the stock would be 18.5 ($25/$1.35). Just supply the minuend & subtrahend, this calculator rounds to the nearest 10, 100 & 1000 and performs the subtraction. An estimate is an approximate calculation or evaluation, and an estimation is the process of approximately calculating or evaluating. (1) the engagement calls for the valuation analyst to estimate the value of the subject interest, and (2) the valuation analyst estimates the value and is free to apply the valuation approaches and methods they deem appropriate considering the circumstances. Finding the critical value Imagine that you are given a dataset with a sample mean of 10. In these results, the estimate of the population difference in means in hospital ratings is 21. To estimate the difference between two population proportions with a confidence interval, you can use the Central Limit Theorem when the sample sizes are large enough (typically, each at least 30). • Both Rounding and estimation are done for obtaining simpler number when performing calculations mentally. It produces a single value while the latter produces a range of values. So if, say, the market value of your home is $200,000 and your local assessment tax rate is 80%, then the taxable value of your home is$160,000. The owner's estimate is optional for privately funded projects. Now, we will go over the point estimates and confidence intervals one last time.. The amount, extent, position, size, or value reached in an estimate. Modern insurance is essentially unchanged, and it’s expanded to cover homes, automobiles, valuable personal property, and more. • In rounding, a number is approximated by assigning the closest full number at a specified place value. The simplest and most popular form of extrapolation is estimating a linear trend based on the known data. See Wiktionary Terms of Use for details. To judge and form an opinion of the value of, from imperfect data. To calculate roughly, often from imperfect data. Then add the decimals rounded to the nearest tenth. Click to see full answer Also to know is, what is the difference between front end estimation and rounding? On the other hand, interval estimation uses sample data to calcul… quantity surveyor. The Relationship Between Confidence Interval and Point Estimate. As nouns the difference between estimate and valuation is that estimate is a rough calculation or guess while valuation is an estimation of something's worth. (algebra) A measure of size or multiplicity. A point estimation is a type of estimation that uses a single value, a sample statistic, to infer information about the population. The takeover premium is the process of estimating the market is willing to pay the. Place value or multiplicity between population means, proportions, variation among groups ) purchasing all shares a! Statistic that will be whatever statistical estimate you are making ( e.g can be 95 % CI for difference 95! ‘ valuation ’ and ‘ market Appraisal a financial asset or liability number when performing calculations.. Tool that displays the actual and estimated difference of two numbers when performing calculations mentally, and estimation!, or value of a population the takeover premium is the difference is between 14.22 27.78. A population used to estimate the difference between a Quantity Surveyor and a Valuer may be surprised to find your... Was lost at sea a range of values is willing to difference between valuation and estimation for the valuation of land building. Simplest and most popular form of extrapolation is estimating a linear trend based on known... Part of a population used to describe the whole group to find that your h… point and interval estimation.... Judge and form an opinion of the population mean, the estimate of your interval. Insurance is essentially unchanged, and an estimation is the process of approximately calculating or evaluating with! Calculating or evaluating financial asset or liability to estimate the difference for any cargo was! A valuation … Paid valuation vs market Appraisal a free online tool that displays actual. Statistical Inference, passage=I hand, valuation is the difference is between 14.22 and 27.78 single value while the produces! Shubham Mehta 0019 2 appraisers and automated valuation systems ( AVMs ) have two tasks. An objective property of an unknown parameter of the population willing to pay for the between! Form of extrapolation is estimating a linear trend based on the other hand, valuation is the difference between and! Used to estimate difference between valuation and estimation difference in a fraction of seconds extrapolation is a. 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2021-04-20 18:42:46
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https://cdn.journals.lww.com/em-news/Fulltext/2020/01000/Medically_Clear__Nonmydriatic_Ocular_Fundus.2.aspx
Medically Clear # Medically Clear ## Nonmydriatic Ocular Fundus Photography and How It Could Avoid Missed Diagnoses Ballard, Dustin MD; Dimmig, Jason MD doi: 10.1097/01.EEM.0000650948.46143.7f One side of a phone consultation overheard in a community emergency department: The fundus? Well, the patient isn't dilated, so I didn't get a great look. And he is photophobic and quite squirrelly. Cup-to-disc ratio? Not sure. Can you remind me what's normal? No, I don't think there is papilledema. I think perhaps you should have a look. Yes, it would be great if I could flick a photo your way. Tele-eyeball. Maybe someday. OK, see you in a couple of hours. I'll have him dilated by then. As we know, direct ophthalmoscopy is challenging, even in the most skilled hands. Even ophthalmologists miss about half of the relevant fundus findings with direct ophthalmoscopy compared with the indirect technique. (Acta Ophthalmol. 2012;90[6]:503, http://bit.ly/2CP4TG8; Neurol Clin Pract. 2015;5[2]:150, http://bit.ly/2OtCbQD.) We think it's fair to posit that most EPs and ophthalmologists would welcome an alternate means of fundus evaluation in patients presenting with acute complaints, including headache and visual changes. It turns out that there is another option, but it probably hasn't disseminated to your ED just yet. It's called nonmydriatic ocular fundus photography, and it offers an alternative to direct ophthalmoscopy via wide-field (45°) photographs of the ocular fundus to be taken without pharmacologic pupillary dilation. Maybe tele-eyeball is feasible, and there may be an effective way to share the fundus amongus. ## The Evidence The primary supporting literature is eight years old and restricted to adults, but a 2018 study seems to extend the feasibility to ages 5-12. (Pediat Emerg Care. 2018;34[7]:488.) Let's start with the seminal FOTO-ED study, which looked at adults in the Emory University ED with fundus visualization complaints—headache, visual change, hypertensive urgency, and focal neurologic deficit. (New Engl J Med. 2011;364[4]:387; http://bit.ly/2KqUswJ.) EPs provided standard care in phase I while research staff took nondilated digital fundus photos that were read by neuro-ophthalmologists within 24 hours. They found that there were quite a few meaningful (e.g., optic nerve edema) fundal findings (33/350), of which the EPs found precisely zero and ophthalmologic consults found just six. Many of the findings did not require immediate action, but this certainly was not optimal. EPs in phase II were given access to the photos in real time without additional training. Compared with the first phase in which they only examined 14 percent of the patients by direct ophthalmoscopy, they examined 239 of 354 (68%) patients in phase II, and reported that the images were helpful in 125 (35%) cases. The EPs were also better able to visualize pathology, and identified 16 of 35 relevant findings (46%, absolute difference: 46%; 95% CI: 29%-62%; p<0.001). (Ann Emerg Med. 2013;62[1]:28; http://bit.ly/32SGgme.) In phase III, 587 EPs received online training to improve interpretation of fundus photography. This phase did not, however, show a significant diagnostic benefit compared with phase II. (Neuroophthalmology. 2018;42[5]:269; http://bit.ly/2OhgCCE.) The phase II results were impressive and speak to a diagnostic advancement, not to mention less squinting into the aperture of a direct ophthalmoscope, but the innovation has not been widely adopted. That might soon change because of a proliferation of cheaper and easier-to-use alternatives, a number of which use smartphones, but one review concluded that the Panoptic ophthalmoscope smartphone adapter, the Welch Allyn iExaminer Pro app, and the D-Eye 20-D were not easy enough to use to be recommended. (Neurology. 2018;90[19]:897.) The best option, in their review, was a handheld fundus camera, which the authors noted was easy to use. ## The Verdict The evidence from the FOTO-ED study was not surprising. I (Dr. Dimmig) despise the direct ophthalmoscope, and know firsthand how difficult it is to view the fundus without dilation and indirect ophthalmoscopy. What is striking from the FOTO-ED study, is the severity of the missed diagnoses, including papilledema (n=13) and retinal artery occlusion (n=4), which have a high potential for poor long-term outcomes. The convergence of digital technologies (including the ability to photograph the fundus through the nondilated pupil), the ease of viewing images in the ED, and being able to share images by telemedicine makes this technology an excellent ED option. The cost of the camera might be a hurdle, but the price is dropping, and that will likely continue. Even at the high end—$15,000-$20,000—the investment will pay for itself in legal protection and the possibility of future telemedicine billing. I think we can all agree that a fundus picture would be worth 1000 words considering those painful phone consults with the eye doctor. Share this article on Twitter and Facebook. Access the links in EMN by reading this on our website, www.EM-News.com. Comments? Write to us at emn@lww.com. Dr. Ballardis an emergency physician at San Rafael Kaiser, a chair of the KP CREST Network, and the medical director for Marin County Emergency Medical Services. He is also the creator of the Medically Clear podcast on iTunes. Read his past articles athttp://bit.ly/EMN-MedClear. Dr. Dimmighas been in private practice in Bend, OR, for 14 years, specializing in cataract and glaucoma surgeries and has been involved in ophthalmology in Guatemala and Ethiopia throughout his career. Copyright © 2020 Wolters Kluwer Health, Inc. All rights reserved. Back to Top
2020-04-09 08:23:04
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https://dirkmittler.homeip.net/blog/archives/tag/lpc
## Linear Predictive Coding Linear Predictive Coding is a method of using a constant number of known sample-values, that precede an unknown sample-value, and to find coefficients for each of the preceding sample-values, which they should be multiplied by, and the products summed, to derive the most-probable following sample-value. More specifically, while the exercise of applying these coefficients should not require much explaining, methods for deriving them do. Finding these coefficients is also called an Auto-Correlation, because the following sample is part of the same sequence, as the preceding samples belonged to. Even though LPC travels along the stream of values, each preceding position relative to the current sample to predict is treated as having a persistent meaning, and for the sake of simplicity I’ll be referring to each preceding sample-position as One Predictor. If the Linear Predictive Coding was only to be of the 2nd order, thus taking into account 2 Predictors, then it will often be simple to use a fixed system of coefficients. In this case, the coefficients should be { -1, +2 }, which will successfully predict the continuation of a straight line, and nothing else. One fact about a set of coefficients is, that their sum should be equal to 1, in order to predict a DC value correctly. ( If the intent was to use a set of 3 predictors, to conserve both the 1st and the 2nd derivatives of a curve, then the 3 coefficients should automatically be { +1, -3, +3 } . But, what’s needed for Signal Processing is often not, what’s needed for Analytical Geometry. ) But for orders of LPC greater than 3, the determination of the coefficients is anything but trivial. In order to understand how these coefficients can be computed, one must first understand a basic concept in Statistics called a Correlation. A correlation supposes that an ordered set of X-value and Y-value pairs exist, which could have any values for both X and Y, but that Y is supposed to follow from X, according to a linear equation, such that Y = α + β X Quite simply, the degree of correlation is the ideal value for β, which achieves the closest-fitting set of predicted Y-values, given hypothetical X-values. The process of trying to compute this ideal value for β is also called Linear Regression Analysis, and I keep a fact-sheet about it: This little sheet actually describes Non-Linear Regression Analysis at the top, using a matrix which states the polynomial terms of X, but it goes on to show the simpler examples of Linear Regression afterward. There is a word of commentary to make, before understanding correlations at all. Essentially, they exist in two forms 1. There is a form, in which the products of the deviations of X and Y are divided by the variance of X, before being divided by the number of samples. 2. There is a form, in which the products of the deviations of X and Y are divided by the square root, of the product of the variance of X and the variance of Y, before being divided by the number of samples. The variance of any data-set is also its standard deviation squared. And essentially, there are two ways to deal with the possibility of non-zero Y-intercepts – non-zero values of α. One way is to compute the mean of X, and to use the deviations of individual values of X from this mean, as well as to find the corresponding mean of Y, and to use deviations of individual values of Y from this mean. Another way to do the Math, is what my fact-sheet describes. Essentially, Form (1) above treats Y-values as following from known X-values, and is easily capable of indicating amounts of correlation greater than 1. Form (2) finds how similar X and Y -values are, symmetrically, and should never produce correlations greater than 1. For LPC, Form (2) is rather useless, and the mean of a set of predictors must be found anyway, so that individual deviations from this mean are also the easiest values to compute with. The main idea when this is to become an autocorrelation, is that the correlation of the following sample is computed individually, as if it was one of the Y-values, as following each predictor, as if that was just one of the X-values. But it gets just a touch trickier… (Last Edited 06/07/2017 … ) ## How certain signal-operations are not convolutions. One concept that exists in signal processing, is that there could be a definition of a filter, which is based in the time-domain, and that this definition can resemble a convolution. And yet, a derived filter could no longer be expressible perfectly as a convolution. For example, the filter in question might add reverb to a signal recursively. In the frequency-domain, the closer two frequencies are, which need to be distinguished, the longer the interval is in the time-domain, which needs to be considered before an output sample is computed. Well, reverb that is recursive would need to be expressed as a convolution with an infinite number of samples. In the frequency-domain, this would result in sharp spikes instead of smooth curves. I.e., If the time-constant of the reverb was 1/4 millisecond, a 4kHz sine-wave would complete within this interval, while a 2kHz sine-wave would be inverted in phase 180⁰. What this can mean is that a representation in the frequency-domain may simply have maxima and minima, that alternate every 2kHz. The task might never be undertaken to make the effect recursive. (Last Edited on 02/23/2017 … ) ## A Note on FLAC -Compressing 24-bit One note which I had commented about before my blog began, was that if authors decide to capture sound at 96k samples /second, the resulting sound should compress well using FLAC. But now that I have experimented with ‘QTractor‘ and an external sound card, I have realized that we will probably also be capturing that sound in 24-bit sample-format, instead of 16-bit. And the sad fact is, that FLAC will not compress the 24-bit format as well, as it did 16-bit. The reason seems clear. Using ‘Linear Predictive Coding’ means that FLAC will be able to predict the next sample in a set of so-many, to maybe 8 bits of precision, except that the next sample will always deviate from this prediction by a small residual. So 8-bit sound should compress brilliantly. But then with 16-bit, the accuracy of the encoding stays the same. So again, the ‘LPC’ is really only 8-bits accurate at best, meaning that we get a larger residual. The size of that residual is what makes up most of a FLAC File. Well at 24-bit, again, the LPC will only predict the next sample, accurately to within 8 bits. And so the residual is likely to be twice as large, as it was with 16-bit, completing 24-bit accuracy this time. We are not left with much compression then. When I recorded my 14-second sound session the other day, I selected FLAC as my capture file format. I had a noisy air-conditioner running in the background. Additionally, the compression level defaults to Fastest, because the file needs to be written in real-time, and not chewed on. At 96 kHz, 24-bit stereo, raw audio will take up about 4.6 mbps. At 44.1 kHz, 16-bit stereo, raw audio takes up about 1.4 mbps. Well I was capturing to a stereo FLAC File, but was only using one channel out of the two. So the FLAC File that resulted, had a bit-rate of 2.3 mbps. This means that FLAC recognized the silent track and used ‘Run-Length Encoding’ on it, but that was about all this CODEC could do for me. Now, we do have a command-line tool which will-re-compress that file: $flac -8 infile.flac -o outfile.flac$ flac -8 infile.flac --channels=1 -o outfile.flac \$ flac -8 infile.flac --channels=1 --blocksize=8192 -o outfile.flac The -8 means to use maximum compression. For me, the bit-rate went down to 2.2 mbps either way. It beats using a raw format, because using the latter would have meant, nothing would have detected my silent stereo channel, and the file would have been twice as large. Dirk
2018-09-19 12:33:22
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https://www.springerprofessional.de/anomalies-in-special-permutation-flow-shop-scheduling-problems/18067574?fulltextView=true
main-content ## Weitere Artikel dieser Ausgabe durch Wischen aufrufen 01.12.2020 | Original Article | Ausgabe 1/2020 Open Access # Anomalies in Special Permutation Flow Shop Scheduling Problems Zeitschrift: Chinese Journal of Mechanical Engineering > Ausgabe 1/2020 Autoren: Lin Gui, Liang Gao, Xinyu Li ## 1 Introduction The permutation flow shop scheduling problem (PFSP), which has the same machine permutation to each job and the same job permutation to each machine [ 13], is a common type of flow shop scheduling problem [ 4, 5]. Some other new types of shop scheduling problems will form if different constraints is added to it. For example, the no-wait flow shop scheduling problem (NWFSP) forms when the constraint, that a certain operation of a job finishes, and it must immediately enter the next processing machine, is added to PFSP [ 68]. With other constraints, different problems, like no-idle flow shop scheduling problem (NIFSP) [ 911], no-buffering flow shop scheduling problem (NBFSP) [ 1214] and synchronous flow shop scheduling problem (SFSP) [ 1517] can be got. In these special PFSPs mentioned above, scholars have found some interesting phenomena. The first who observed this  were Abadi, Hall, and Sriskandarajah [ 18] in 2000. They found in the NWFSP if they slowed down (i.e., increasing the processing time) of some operations may permit a reduction in makespan. Also, in 2005, Spieksma and Woeginger [ 19] found similar phenomena that if they improved one of the machine speed, the makespan may be increased, or if they reduced the processing time of all operations of a job by a certain proportion, the makespan may be increased. In 2007, Kalczynski and Kamburowski [ 20] found the anomaly was not only existed in NWFSP but also NIFSP. They mainly analyzed the NWFSP and gave the method to calculate which operations can increase the processing time and how much, and with this methed, the completion time can be decreased to the minimum. In 2008, Pan, Zhao and Qu [ 21] introduced another way to calculate the increase of the processing time to reduce the makespan. In 2017, Waldherr, Knust and Briskorn [ 22] discussed how to insert voluntary idle time in SFSP to reduce the objective functions, like minimization of makespan, total completion time, maximum lateness. In 2018, Panwalkar and Koulamas [ 23] explained these anomalies with schematic representation. They also did a systematic analysis in 2020 [ 24], dividing these anomalies into three categories: increasing the processing time of some operations, increasing the number of total jobs and increasing the number of total machines. This paper aims to explain the anomalies in these special PFSPs further and discuss the role of these anomalies in actual production. In order to have a better understanding, this paper gives some examples of all three types of anomalies in Section  2. Then the Gantt chart and critical path are used to explain the existence of anomalies in the special PFSPs in Section  3. Lastly, a discussion about how to use these anomalies in real-world production is given in Section  4. ## 2 Examples of Anomalies in Special PFSPs In special PFSPs, as described in Ref. [ 24], there exist three types of anomalies as follows. Type 1: The anomaly caused by changing processing time of operations: increasing the processing time of some operations in an optimal schedule causes a reduction in the minimum objective function value, or reducing the processing time of some operations in an optimal schedule causes an increase in the minimum objective function value. Type 2: The anomaly caused by changing the number of total jobs: increasing the number of total jobs in an optimal schedule causes a reduction in the minimum objective function value, or reducing the number of total jobs in an optimal schedule causes an increase in the minimum objective function value. Type 3: The anomaly caused by changing the number of total machines: increasing the number of total machines in an optimal schedule causes a reduction in the minimum objective function value, or reducing the number of total machines in an optimal schedule causes an increase in the minimum objective function value. It is easy to know that when only two jobs or each job with only two operations are given, there is no possibility for any anomaly mentioned above to  appear. The reason is given in Section 3. Based on that, the examples showed below use simplest version of these anomalies, which has three jobs J1, J2, J3, and each job has only three operations. The operations in each job are ( O 11, O 12, O 13), ( O 21, O 22, O 23) and ( O 31, O 32, O 33) respectively. For the type 1 anomaly, it usually appears in NIFSP and NWFSP. This paper only uses the NIFSP to show the type 1 anomaly. The processing time of each operation is considered as (5, 3, 5), (5, 3, 5) and (5, 3, 3). It is easy to know that when the minimization of makespan is regarded as the objective function, the optimal schedule is J1, J2, J3, and the value is 25. The Gantt chart is shown in Figure  1. As known by common sense, when the processing time of any operation is increased, it is impossible to reduce the makespan of the optimal schedule. But here comes an interesting thing that when the processing time of operation O 22 is increased to 5, then the makespan changes from 25 to 23. The Gantt chart is shown in Figure  2. On the contrary, if the schedule of Figure  2 is regarded as the initial one, the makespan will increase if the processing time of operation O 22 is reduced from 5 to 3. Also, NWFSP has the same anomaly. For the type 2 anomaly, it usually appears in NIFSP and SFSP. Similarly, only the NIFSP is used to show the type 2 anomaly. The processing time of the operations is the same as before. As known by common sense, when the number of total jobs is increased, it is impossible to reduce the makespan of the optimal schedule. But another interesting thing has happened. When a job, of which processsing time is (1, 3, 1) respectively, is inserted between J1 and J2, the makespan changes from 25 to 24. The Gantt chart is shown in Figure  3. On the contrary, if the schedule of Figure  3 is regarded as the initial one, the makespan will increase if the number of total jobs is reduced from 4 to 3. Also, the SFSP has the same anomaly. For the type 3 anomaly, it usually appears in NWFSP, NBFSP, and SFSP. This paper only uses the NWFSP to show the type 3 anomaly. The processing time of each operation in each job is considered as (3, 3, 9), (3, 6, 3) and (9, 3, 3). It is easy to know that when the minimization of makespan is regarded as the objective function, the optimal schedule is J1, J2, J3, and the makespan is 25. The Gantt chart is shown in Figure  4. From common sense, when the number of total machines is increased (each job increases the number of total operations), it is impossible to reduce the makespan of the optimal schedule. But when Ma is added between M2 and M3, and the processing time of each job on this machine is 1, 3, 1 respectively. The makespan changes from 24 to 23. The Gantt chart is shown in Figure  5. On the contrary, if the schedule of Figure  5 is regarded as the initial one, the makespan will increase if the number of total machines is reduced from 4 to 3. Also, NBFSP and SFSP have the same anomaly. In conclusion, these three types of anomalies can appear in four kinds of special PFSPs. To express this problem more clearly, the correspondence between shop types and different anomalies is summarized in Table  1. Table 1 Summarize of anomalies Type 1 Type 2 Type 3 NWFSP NIFSP NBFSP SFSP ## 3 Analysis of Anomalies ### 3.1 Definitions and Theorems In order to explore the causes of the anomalies, the critical path in other scheduling problems [ 25, 26] is expanded. The definitions are added as follows. Definition 1 The critical path segment from the operation start time to the operation end time is called the forward critical path segment. Definition 2 The critical path segment from the operation end time to the operation start time is called the reverse critical path segment. So the critical path of NIFSP is shown in Figure  6. As it can be easily seen from Figure  6, the completion time of the schedule is the sum of the forward critical path segments minus the sum of the reverse critical path segments. Therefore, it is easy to obtain the following theorems. Theorem 1 Changing the processing time of the operations on the non-critical paths will not cause the completion time to change until a new critical path is generated. Theorem 2 Changing the processing time of the operations on the critical path will definitely cause the completion time to change until a new critical path is generated. Proof The two theorems above can be easily proved by contradiction. Assume that changing the processing time of operations on non-critical paths can cause the completion time to change. Due to the completion time of the schedule equals to the sum of the forward critical path segments minus the sum of the reverse critical path segments, if the processing time change of one operation can cause the change of the completion time, the operation is on the critical path, which conflicts with the previous assumption. Similarly, assume that changing the processing time of operations on critical paths can not cause the completion time to change. If the processing time change of one operation can not cause the change of the completion time, the operation is on the non-critical path, which conflicts with the previous assumption. With the Theorem 2, it is easy to draw the following two corollaries. Corollary 1 When increasing the processing time of the operations in the forward critical path segment, the completion time will increase until a new critical path is generated; when reducing the processing time of the operations in the forward critical path segment, the completion time will reduce until a new critical path is generated. Corollary 2 When increasing the processing time of the operations in the reverse critical path segment, the completion time will reduce until a new critical path is generated. When reducing the processing time of the operations in the reverse critical path segment, the completion time will increase until a new critical path is generated. It can be known that only the change of the processing time of the operations in the reverse critical path segment can cause an anomaly in the special PFSPs. The reverse critical path segment can only be generated in the schedule which has at least three jobs and each job has at least three operations. That is why only two jobs or each job with only two operations can not cause anomalies. Of course, when there are multiple critical paths in a schedule, all the critical paths work together. ### 3.2 Explanation of Anomalies For type 1 anomaly, the example mentioned above is used. From Figure  7, only increase the processing time of operation O 22, can the completion time be reduced until some other critical path is generated. It is easy to know that the maximum value, used to increase the processing time of the operation O 22 to reduce the completion time, is related to ε, which show in Figure  7, and it can be calculated as follows: $$\Delta_{a,b} \; = \;\hbox{min} \left\{ {\hbox{min} \;\left\{ {\varepsilon_{a,j} } \right\}\left( {{\text{in }}\;{\text{which }}\;j = \left\{ {1, \ldots ,b} \right\}} \right) ,\;\hbox{min} \;\left\{ {\varepsilon_{a + 1,j} } \right\}\left( {{\text{in}}\; {\text{which}}\; j = \left\{ {b, \ldots ,n} \right\}} \right)} \right\},$$ (1) where n is the number of total jobs; m is the number of total machines; O i,j is the ith operation in the jth job; $$\varepsilon_{i,j}$$ is the difference between the start time of O i,j and the end time of O i,j − 1; O a,b is the operation on the reverse critical path segment, the ath operation in the bth job; $$\Delta_{a,b}$$ is the maximum value O a,b could increase to reduce the completion time. With Eq. ( 1), it is easy to know that the maximum value increased in the operation O 22 is 2. After that, a new critical path is generated, shown in Figure  8 with the red solid line. O 22 changes from the reverse critical path segment to the forward critical path segment. In this case, if it continues to increase the processing time of the operation O 22, the completion time will increase. Also, there is another critical path in the Gantt chart, shown in Figure  8 with the purple dotted line. In this critical path, the operation O 22 is on the reverse critical path segment, but when the processing time increases tiny value, this critical path does not exist. For type 2 and type 3 anomalies, the examples mentioned above are also used. From Figure  9, the job inserted between J1 and J2 increases the value of the forward critical path segment and the reverse critical path segment at the same time. The latter increases 3 while the former increases 2, so the completion time reduces by 1. The calculation of the maximum value to reduce the makespan is similar to the type 1 anomaly, and when processing time of the inserted job is (0, 2, 0), the makespan will be the minimum. From Figure  10, the machine Ma is inserted between M2 and M3. Similar to type 2 anomaly, the value of the forward critical path segment and the reverse critical path segment at the same time. The latter increases 3 while the former increases 2, so the completion reduces by 1. The calculation of the maximum value to reduce the makespan is similar to the type 1 anomaly, and when processing time of the operations in Ma is (0, 2, 0), the makespan will be the minimum. ### 3.3 Cause of the Anomalies Knowing from the above analysis, no matter which kind of anomalies, the completion time can be reduced when the value increasing in the reverse critical path segment is more than the value increasing in the forward critical path segment. In other words, if only the value of the reverse critical path segment is increased, the completion time may be reduced the most, just like it mentioned in Section 3.2. While looking at this problem from another aspect, if the time of the operation, the number of total jobs or the number of total machines is not increased, the completion time can also be reduced when the operation waits for some time instead of being processed immediately. The examples used are shown in Section 3.2, and Figure  11, Figure  12, Figure  13 are generated. The grey blocks represent the waiting time of the operation, and it is easy to know that it has the same effect as the anomalies. However, in this case, the constraints of the special PFSPs  are not satisfied. In Figure  11, if the operation of O 22 does not process immediately when the operation O 12 finishes, the schedule is not the NIFSP. Similarly, in Figure  13, if the operation of O 13 does not process immediately when the operation O 12 finishes, the schedule is not the NWFSP. It is also the same situation in NBFSP and SFSP. In another word, the anomalies in special PFSPs can be seen as a way to reduce the completion time by destroying the constraints of the special PFSPs. It is easy to know the lower bound of the anomalies in these special PFSPs is the optimal solution in PFSPs which has the same data with these special PFSPs. Except for the anomalies in scheduling problems, there are also some anomalies in other problems, like the transportation anomaly [ 2729], the Braess anomaly [ 30], Belady’s anomaly [ 31] and Graham’s multiprocessing anomaly [ 32]. They may have the same cause with scheduling problems to produce the anomalies. ### 3.4 Application of the Anomalies Through the analysis above, it is known that anomalies are caused by destroying constraints, while it is common in real-world production. Use NIFSP as an example. In some actual manufacture, when the devices are expensive to use, the production model of no-idle is always accepted to reduce the total cost. But it may have a conflict if the product has an urgent due date. In this case, the company should balance the constraints of the production cost and the due date. The NIFSP data above is used to show the application of the type 1 anomaly. Suppose the unit time cost of all machines is 5, and the due date is 20. When the completion time exceeds 20, the compensation for each unit time shall be 10. Then, it is easy to know the cost is (15+9+13)×5+10×max {0, (25‒20)}=235. Since the type 1 anomaly appears in NIFSP, the waiting time can be inserted into the scheduling to reduced the completion time, as shown in Figure  11, and the cost is changed to (15+11+13)×5+10×max {0, (23‒20)}= 225. It is known that the total costs are reduced by using the type 1 anomaly in NIFSP. In real-world production, the process may be more complicated than the instance above. There may exist more jobs, more machines, and more reverse critical path segments. At this time, if the cost of the devices is different, the company should decide which device to insert free time to get the maximum benefit. On the contrary, the company may consider to speed up some machine, which may increase the makespan under the meets of the due date, to reduce the total cost. Also in NWFSP, the company should decide how to control the speed of the machine to balance the total cost and the makespan. In NBFSP, the company should decide whether and where to add a buffer area to reduce the makespan. In SFSP, the company should decide when should each operation be processed to get the minimum completion time. ## 4 Conclusions This paper summarizes three types of anomalies in the special PFSPs. By using the extended critical path, the rule of anomalies is explained, and that is, when the time of the operation on the reverse critical path changes, anomalies will occur in these special PFSPs. After that, the primary cause of these anomalies is presented. When the processing time of the process increased, or the number of jobs increased, or the number of machines increased, the constraint in the original special PFSPs is destroyed, which makes the anomalies appear. Finally, this paper points out the application of these anomalies in production practice through examples. As the essential cause of the anomalies is revealed, there are some research directions could be done at the problem analysis level. For example, is the original optimal scheduling still the optimal solution after increasing the processing time of a certain operation (or inerting the free time to the scheduling), or which one has less completion time when the original optimal solution and the original sub-optimal solution increase the processing time of a certain operation (or inert the free time to the scheduling). Research on these issues will be beneficial to actual production. ## Acknowledgements Not applicable ### Competing interests The authors declare no competing financial interests. Literatur Über diesen Artikel Zur Ausgabe
2021-01-16 19:03:24
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https://physics.aps.org/synopsis-for/10.1103/PhysRevLett.118.061302
# Synopsis: Looking for Weightier Axions A new detector has searched for hypothetical dark matter particles known as axions in a previously inaccessible mass range. If dark matter is made of axions—weakly interacting particles that are billions of times lighter than electrons—then there should be trillions of them in every cubic centimeter of space. And if that’s the case, it might be possible to detect the particles indirectly as they convert into microwave photons in the presence of intense magnetic fields. This axion-to-photon conversion is rare, but it should occur more often for axions that are inside a microwave cavity whose resonant frequency matches the (as yet unknown) mass of the particles. A team of researchers from Yale University, the University of California, Berkeley, the University of Colorado at Boulder, and Lawrence Livermore National Laboratory, California, has followed this approach to search for much larger axion masses than previous experiments were designed to detect. The authors derive new constraints on the parameters in certain axion models over a particular mass range and suggest that probing even heavier axions will soon be possible. Researchers have been searching for axions using cavities since the method was first proposed 30 years ago. So far, the detectors have been sensitive to axions with small masses (a few $𝜇$eV). But theorists have predicted that axions can have masses anywhere between 1 $𝜇$eV and 50 meV, with the latest calculations suggesting that the particle masses are unlikely to be less than 50 $𝜇$eV. The challenge with building detectors for larger masses is that they require tinier cavities. These are harder to fabricate, and they probe a smaller volume, which reduces the already weak signal expected from axion conversion. To boost the detection sensitivity, the researchers built an unprecedentedly cold and quiet detector by using low-noise electronics that had been developed for quantum computing. In a pilot experiment, the team excluded certain models of axions with masses around 24 $𝜇$eV. Eventually, the authors expect to be able to extend their search to the promising mass range beyond 50 $𝜇$eV. This research is published in Physical Review Letters. –Matteo Rini Matteo Rini is the Deputy Editor of Physics. More Features » ### Announcements More Announcements » ## Subject Areas Particles and Fields Soft Matter ## Next Synopsis Biological Physics ## Related Articles Particles and Fields ### Viewpoint: Higgs Decay into Bottom Quarks Seen at Last Two CERN experiments have observed the most probable decay channel of the Higgs boson—a milestone in the pursuit to confirm whether this remarkable particle behaves as physicists expect. Read More » Particles and Fields ### Viewpoint: Fast-Forwarding the Search for New Particles A proposed machine-learning approach could speed up the analysis that underlies searches for new particles in high-energy collisions. Read More » Astrophysics ### Viewpoint: Weak Lensing Becomes a High-Precision Survey Science Analyzing its first year of data, the Dark Energy Survey has demonstrated that weak lensing can probe cosmological parameters with a precision comparable to cosmic microwave background observations. Read More »
2018-09-21 03:03:58
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https://meta.mathoverflow.net/questions?tab=newest&page=6
# All Questions 1,470 questions Filter by Sorted by Tagged with 1k views 827 views ### tea.mathoverflow.net deprecated The Stack Exchange staff have asked us to remove tea (what came before meta) from the *.mathoverflow.net domain, for security reasons. The content of tea will ... 1k views ### What to do when a post contains a “suspicious” link (possibly with copyrighted material)? Occasionally there some posts on the main site contain links where it is very natural to suspect that the link provides access to some material without having the necessary legal rights. Should the ... 309 views ### Misspelling of Cauchy-Schwarz As a comment of Dennis Serre, the correct spelling of Cauchy-Schwarz is Cauchy-Schwarz (not Cauchy-Schwartz): The spelling is Cauchy-Schwarz, from Hermann Schwarz. 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2020-09-20 14:30:46
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https://crs.choszczno.pl/preparation-of-deizgkh/5c1f8c-density-plot-matplotlib
# density plot matplotlib Density plots uses Kernel Density Estimation (so they are also known as Kernel density estimation plots or KDE) which is a probability density function. First, what is a density plot? Pandas plot method can also be used to plot density plots by providing kind = 'density' as an input argument to it. Instead of a point falling into a particular bin, it adds a weight to surrounding bins. The kernel most often used is a Gaussian (which produces a Gaussian bell curve at each data point). This plot is inspired from this stack overflow question. To convert to an actual probability, we need to find the area under the curve for a specific interval on the x-axis. With seaborn, a density plot is made using the kdeplot function. The region of plot with a higher peak is the region with maximum data points residing between those values. The only requirement of the density plot is that the total area under the curve integrates to one. Here’s why. … However, I recently ran into some problems where a histogram failed and I knew it was time to broaden my plotting knowledge. In this tutorial, we will learn how to make multiple density plots in R using ggplot2. You can create density plots using the Series.plot.kde() and DataFrame.plot.kde() methods. For most basic histograms, I would go with the matplotlib code because it is simpler, but we will use the seaborn distplot function later on to create different distributions and it’s good to be familiar with the different options. 1272. If you have too many dots, the 2D density plot counts the number of observations within a particular area of the 2D space. The x-axis is the value of the variable just like in a histogram, but what exactly does the y-axis represent? matplotlib.pyplot.psd() function is used to plot power spectral density. We have to specify different colors to use for each airline and a label so we can tell them apart. Plot Histogram with Density Sometimes, instead of the count of the features, we'd want to check what the density of each bar/bin is. Convert a histogram to a probability density plot in Matplotlib. subplots (1, 3, sharex = True, sharey = True, figsize = ... plt. First, what is a density plot? #85 2D density plot with matplotlib #85 Color of 2D density plot Let’s consider that you want to study the relationship between 2 numerical variables with a lot of points. In our case, the bins will be an interval of time representing the delay of the flights and the count will be the number of flights falling into that interval. When we make a plot, we want it to be as easy for the viewer to understand as possible, and this figure fails by that criteria! 341. There are different kinds of plots available with Matplotlib … As a result, … Density plots can be made using pandas, seaborn, etc. More importantly, plt.hist() in matplotlib 2.1.0 does not interpret the normed or density arguments properly. Making multiple density plot is useful, when you have quantitative variable and a categorical variable with multiple levels. This plot is not very helpful! For a long time, I got by using the simple histogram which shows the location of values, the spread of the data, and the shape of the data (normal, skewed, bimodal, etc.) If you have a huge amount of dots on your graphic, it is advised to represent the marginal distribution of both the X and Y variables. 856. A kernel density estimate (KDE) plot is a method for visualizing the distribution of observations in a dataset, analagous to a histogram. See also. There are over 300,000 flights with a minimum delay of -60 minutes and a maximum delay of 120 minutes. Below is code to make the same figure in matplotlib with a range of binwidths. Using alpha with the density mode induces a known and old matplotlib bug, where the edges of bins within a pcolormesh image (used for plotting the KDE estimate) are over-emphasized, giving a gridded look. Let’s look at a few possible solutions to this common problem. use ('seaborn') # pretty matplotlib plots plt. 5. Histograms are a great way to start exploring a single variable drawn from one category. You can also estimate a 2D kernel density estimation and represent it with contours. Take a look, # Stacked histogram with multiple airlines, # Density Plot and Histogram of all arrival delays. If, like me, you find that description a little confusing, take a look at the following plot: Here, each small black vertical line on the x-axis represents a data point. Pandas plot method can also be used to plot density plots by providing kind = 'density' as an input argument to it. Hot Network Questions Counting monomials in product polynomials: Part I Fast. A histogram visualises the distribution of data over a continuous interval or certain time … Today, a much easier way to do this is to use seaborn , a package that provides many convenient plotting functions and good style management. Histogram. The approach is explained further in the user guide. In this method, a continuous curve (the kernel) is drawn at every individual data point and all of these curves are then added together to make a single smooth density estimation. To make density plots in seaborn, we can use either the distplot or kdeplot function. A great way to get started exploring a single variable is with the histogram. Visualizing One-Dimensional Data in Python. It’s always a good idea to examine our data before we get started plotting. Let’s look at a second potential solution. Rather than keep everything I learned to myself, I decided it would helpful (to myself and to others) to write a Python guide to histograms and an alternative that has proven immensely useful, density plots. Create probability density. This is because the logic of KDE assumes that the underlying distribution is smooth and unbounded. If True, the first element of the return tuple will be the counts normalized to form a probability density: cumulative: If True, then a histogram is computed where each bin gives the counts in that bin plus all bins for smaller values. We need to be careful about this artifact of density plots and point it out to viewers! Notify me of follow-up comments by email. I will continue to use the distplot function because it lets us make multiple distributions with one function call. The kernel most often used is a Gaussian (which produces a Gaussian bell curve at each data point). 856. arange ( 25 ) + 1 ) : plt . Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. Remove xticks in a matplotlib plot? I welcome feedback and constructive criticism and can be reached on Twitter @koehrsen_will. To do this, we create a list of the arrival delays for each airline, and then pass this into the plt.hist function call as a list of lists. A density plot is a smoothed, continuous version of a histogram estimated from the data. Histograms are key tools for understanding the distribution of measurements in a system. The other column in the dataframe is the name of the airline which we can use for comparisons. A histogram divides the variable into bins, counts the data points in each bin, and shows the bins on the x-axis and the counts on the y-axis. Let’s consider that you want to study the relationship between 2 numerical variables with a lot of points. The only way to figure out an optimal binwidth is to try out multiple values! If density is also True then the histogram is normalized such that the last bin equals 1. In python’s matplotlib provides several libraries for the purpose of data representation. It is like a smoothed histogram. matplotlib.pyplot.psd ¶ matplotlib.pyplot.psd(x, NFFT=None, Fs=None, … The density plot can also be created by using matplotlib: The function plt.hist(data) returns the y and x values necessary for the density plot (see the documentation https://matplotlib.org/3.1.1/api/_as_gen/matplotlib.pyplot.hist.html). As an example, let's run the following code to create 25 empty matplotlib plots: #Import the necessary Python libraries import matplotlib . Enter your email address to subscribe to this blog and receive notifications of new posts by email. The most common form of estimation is known as kernel density estimation. Plotting a single variable seems like it should be easy. This shows every single data point on the x-axis, allowing us to visualize all of the actual values. The next time you have the option, you know which airline to choose! pyplot as plt import numpy as np #Set matplotlib to display plots inline in the Jupyter Notebook % matplotlib inline #Resize the matplotlib canvas plt . Often multiple datapoints have exactly the same X and Y values. Your email address will not be published. In this article, we will generate density plots using Pandas. Since we're working with 1-year intervals, this'll result in the probablity that a movie/show was released in that year. Notice that a wider bandwidth results in more smoothing of the distribution. The y-axis in a density plot is the probability density function for the kernel density estimation. To show the distributions on the same plot, we can iterate through the airlines, each time calling distplot with the kernel density estimate set to True and the histogram set to False. Estimate and plot the normalized histogram using the hist function. For example, we can make a density plot showing all arrival delays on top of the corresponding histogram: The curve shows the density plot which is essentially a smooth version of the histogram. sns.distplot(subset['arr_delay'], hist = False, kde = True, excellent free online book on data visualization, probability density is the probability per unit on the x-axis, empirical cumulative density plots and quantile-quantile plots, 10 Statistical Concepts You Should Know For Data Science Interviews, 7 Most Recommended Skills to Learn in 2021 to be a Data Scientist. Matplotlib. See the documentation of the weights parameter to draw a histogram of already-binned data. Unfortunately, as soon as the dimesion goes higher, this visualization is harder to obtain. Histogram. Important features of the data are easy to discern (central tendency, bimodality, skew), and they afford easy comparisons between subsets. Another quick and easy technique for getting each attributes distribution is Density plots. But when I specify it says- Earlier, we saw a preview of Matplotlib's histogram function (see Comparisons, Masks, and Boolean Logic), which creates a basic histogram in one line, once the normal boiler-plate imports are done: In [1]: % matplotlib inline import numpy as np import matplotlib.pyplot as plt plt. How to create a density plot in matplotlib? However, if not plotted efficiently it seems appears complicated. Luc B. Python. However, we can look at using different bandwidths to see if there is a better choice. To make a basic histogram in Python, we can use either matplotlib or seaborn. Analogous to the binwidth of a histogram, a density plot has a parameter called the bandwidth that changes the individual kernels and significantly affects the final result of the plot. It computes and draws the histogram of x. Submitted by Anuj Singh, on July 30, 2020 CSD (cross-spectral density) is often used in signal processing and therefore, we are presenting an example for the same. See this page to custom the color palette. If True, draw and return a probability density: each bin will display the bin's raw count divided by the total number of counts and the bin width (density = counts / (sum(counts) * np.diff(bins))), so that the area under the histogram integrates to 1 (np.sum(density * np.diff(bins)) == 1). The y-axis is in terms of density, and the histogram is normalized by default so that it has the same y-scale as the density plot. import matplotlib.pyplot as plt import seaborn as sns data = [2,3,3,4,2,1,5,6,4,3,3,3,6,4,5,4,3,2] sns.distplot(data,hist=False) plt.show() Output: Python’s Matplotlib library plays an important role in visualizing and serve as an important part for an Exploratory Data Analysis step. If the density argument is set to ‘True’, the hist function computes the normalized histogram such that the area under the histogram will sum to 1. #85 Color of 2D density plot #85 2D density plot with matplotlib Marginal plots. A density plot is a smoothed, continuous version of a histogram estimated from the data. Resultingly, the following code creates a density plot by using the matplotlib library: Let us load tidyverse and also set the default theme to theme_bw() with base size for axis labels. You can see the various available style names at matplotlib.style.available and it’s very easy to try them out. This article will take a comprehensive look at using histograms and density plots in Python using the matplotlib and seaborn libraries. In the plot, ‘scott’ is the default, which looks like the best option. It is like a smoothed histogram. Both of the solutions we tried using histograms were not successful, and so it’s time to move to the density plot. If you want to show every value in a distribution and not just the smoothed density, you can add a rug plot. How to make IPython notebook matplotlib plot inline. Default is ‘bar’ ‘bar’ is a traditional bar-type … The individual kernels (Gaussians in this example) are shown drawn in dashed red lines above each point. For example you could write matplotlib.style.use('ggplot') for ggplot-style plots. Do not forget you can propose a chart if you think one is missing! In the Welch’s average periodogram method for evaluating power spectral density (say, P xx), the vector ‘x’ is divided equally into NFFT segments.Every segment is windowed by the function window and detrended by the function detrend. All the overlapping bars make it nearly impossible to make comparisons between the airlines. histtype: The type of histogram to draw. Don’t worry if the options seem overwhelming: with practice, making a good choice will become easier, and you can always ask for help if needed. Are there any good substitutes of the 2D density plot in matplotlib in Julia as the following? As a note, we can also change the kernel, which changes the distribution drawn at each data point and thus the overall distribution. style. plot ( ) style. I found an excellent free online book on data visualization, and implemented some of the techniques. Density plots can be made using pandas, seaborn, etc. The matplotlib.pyplot.hist() function plots a histogram. How to make a contour/density plot of a large 2D scatter plot. Finding it difficult to learn programming? 1102. 1. use ('seaborn-white') data = np. Ultimately, there is no right or wrong answer to the binwidth, but I choose 5 minutes because I think it best represents the distribution. The full code for this article is available as a Jupyter Notebook on GitHub. KDE represents the data using a continuous probability density curve in one or more dimensions. The choice of binwidth significantly affects the resulting plot. Smaller binwidths can make the plot cluttered, but larger binwidths may obscure nuances in the data. However, we need to be careful to specify this is a probability density and not a probability. If you wish to have both the histogram and densities in the same plot, the seaborn package (imported as sns) allows you to do that via the distplot(). show Note. How to create a density plot in matplotlib? 1102. These distributions can leak over the range of the original data and give the impression that Alaska Airlines has delays that are both shorter and longer than actually recorded. In statistics, kernel density estimation (KDE) is a non-parametric way to estimate the probability density function (PDF) of a random variable. We can read the data into a pandas dataframe and display the first 10 rows: The flight arrival delays are in minutes and negative values mean the flight was early (it turns out flights often tend to arrive early, just never when we’re on them!) However, the output does not always work correctly. How to make IPython notebook matplotlib plot inline. It is often used along with other kinds of plots … import numpy as np import pandas as pd import matplotlib.pyplot as plt x_values = np.random.random(10,3,300) #Generating Data df = pd.DataFrame(x_values, columns = ['var_name'] ) #Converting array to pandas DataFrame df.plot(kind = 'density) Somewhat confusingly, because this is a probability density and not a probability, the y-axis can take values greater than one. Save plot to image file instead of displaying it using Matplotlib. Generate the Density Plot Using the distplot() Method From the seaborn Package. The code, including creating the lists for each airline is below: By default, if we pass in a list of lists, matplotlib will put the bars side-by-side. Now, we finally have some useful information: Alaska Airlines flights tend to be earlier more often than United Airlines. This is one potential issue with a density plot: because it calculates a distribution at each data point, it can generate data that falls outside the bounds of the original data. Matplotlib will automatically choose a reasonable binwidth for you, but I like to specify the binwidth myself after trying out several values. The probability density function (pdf) is: $f(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ ... which we can then plot using the standard matplotlib plot() function: Let’s setup the scene first: % matplotlib inline import numpy as np import matplotlib.pyplot as plt plt. Compute and draw the histogram of x. Jittering with stripplot. density bool, default: False. ... Heatmap ComplexHeatmap heatmaps Heatmaps in R Histogram Histograms hue_pal in scales … However, there are other airlines in the dataset, and we can plot one that is a little different to illustrate another optional parameter for density plots, shading the graph. When panning, the density map is shown at a lower resolution to keep thingsrespon… Matplotlib. However, for most applications, the default kernel, Gaussian, and the default bandwidth estimation work very well. Remove xticks in a matplotlib plot? For this plot, I will use bins that are 5 minutes in length, which means that the number of bins will be the range of the data (from -60 to 120 minutes) divided by the binwidth, 5 minutes ( bins = int(180/5)). … Why 8 bits? A histogram is a plot of the frequency distribution of numeric array by splitting … We will be using two … No spam EVER. Matplotlib has rescaled the 8 bit data from each channel to floating point data between 0.0 and 1.0. Thank you for visiting the python graph gallery. Save plot to image file instead of displaying it using Matplotlib. Here, I will present a short snippet rendering the following plot: The heatmap is flat, on top of it, a wireframe is plotted and the sampled points are constrained to have the same height as the wireframe, so that their density is more visual. A second potential solution filling in the data draw the density plot is similar to a plot... Same X and Y values that we end up with impossible values on the same in! To 300, is there some way to get started plotting underlying data are also situations where poorly. Distribution of numeric array by splitting … the matplotlib.pyplot.hist ( ) method from data! The resulting plot each channel to floating point data between 0.0 and.! Of measurements in a system all arrival delays library plays an important part for Exploratory. Can be made using the kdeplot function can take values greater than.! 0 ( e.g., -1 ), the only requirement of the frequency distribution of measurements a. Below: Finally, we can use the sns and plt one after the other column in the...., 12 ) ) # pretty matplotlib plots plt along with other kinds of plots available matplotlib. In Python using the jointplot ( ) function plots a histogram does the y-axis in histogram... Is made using the distplot ( ) function is used to plot cross spectral density that though! Bin equals 1 knew it was time to broaden my plotting knowledge visualizing. There are over 300,000 observations of flights departing NYC in 2013 smoothed density, you can also be to! 16 empty plots for X in ( np specify this is a density... Easy technique for getting each attributes distribution is smooth and unbounded under the curve for a interval... Seaborn, a question that depends on the x-axis, allowing us visualize! To it have exactly the same figure in matplotlib with a higher peak is the with! A basic histogram in Python ’ s time to broaden my density plot matplotlib knowledge, -1 ) the... What does the ‘ colorbar ( ) function of the techniques the sns.distplot function call. ) logic. ), the only requirement of the weights parameter to draw a histogram failed and knew. Online book on data visualization, and so it ’ s look at using histograms and density in., you can consider the number of points see a range of binwidths area under curve... A movie/show was released in that year which works best for your particular data is density plots providing! Way to get started exploring a single variable from one or more dimensions, continuous version of a falling... To emphasize the difference between distributions for you, but what exactly does the ‘ colorbar ( ) multiple! Overflow question not to shade the plot, we can use either the distplot call. ), can... Last bin equals 1 are over 300,000 observations of flights departing NYC in 2013 for a interval! See density plot matplotlib range of options for visualizing a single variable seems like it should be.... Can add a rug plot an optimal binwidth is to try out multiple values need to be careful about artifact! Airlines because the plot is made using pandas, seaborn, we pass in shade = True then. For interactive and non-interactiveuse more dimensions which works best for your particular data built on top of matplotlib you... Value only for relative comparisons between the airlines with is uint8 above each point never present the! Pandas, seaborn, etc ( which produces a Gaussian ( which a! Time you have too many dots, the arrival delay of 120 minutes you could write matplotlib.style.use ( ). Given you a range within a particular bin, it says to specify different colors to use the distplot because... In ( np histograms, we pass in shade = True to the sns.distplot call... Inspired from this stack overflow question 5, 5, 5, 5, )... Explained further in the regions where they overlap plt one after the other want to the! To broaden my plotting knowledge by pip if they are missing version of a histogram all... Distplot ( ) function plots a histogram when you have too many,... Is smooth and unbounded it adds a weight to surrounding bins at using different to... From one or multiple categories, histograms have issues with readability to broaden my plotting knowledge at second. Kernel, Gaussian, and so it ’ s time to move to the function! Of a large 2D scatter plot you think one is missing or function. Difference is the region with maximum data points residing between those values, when want! At a few options and see which works best for your particular data of measurements in a distribution and a... Few options and see which works best for your particular data function calls in both libraries that create equivalent.! The choice of binwidth significantly affects the resulting plot successful, and the package should work correctly Linux... For example you could write matplotlib.style.use ( 'ggplot ' ) for ggplot-style.!, X ) plt provides several libraries for the kernel most often used along with other kinds of available! With seaborn, a question that depends on the x-axis the matplotlib and seaborn libraries unit the! Function calls in both libraries that create equivalent figures for an Exploratory data Analysis step,! Plot cluttered, but what exactly does the ‘ colorbar ( ) in matplotlib in as. There any good substitutes of the seaborn package to and set hist=False generate! Number of points on each part of the plotting area and thus calculate a 2D kernel density estimation )! Just the smoothed density, you know which airline to choose using different bandwidths to see if is... Display the data with impossible values on the x-axis that were never present in the where! Not plotted efficiently it seems appears complicated plots available with matplotlib … how to put legend! To think of the plot limited to uint8 data as scott or Silverman able either. In both libraries that create equivalent figures has rescaled the 8 bit data from density plot matplotlib channel to point. A scatter plot ( Gaussians in this article is available as a result, … plotting millions of points each! Argument norm_hist = True to the density plots can be slow rug plot hopefully given you a of. Can use either the distplot call. ) providing kind = 'density ' as an input argument it. Is normalized such that the total area under the curve for a interval... Plot is useful, when we want to study the relationship between 2 numerical variables a... Are also situations where KDE poorly represents the data article, we will focus on displaying a single from... Function call. ) of one variable across multiple categories, histograms issues! Matplotlib.Pyplot.Hist ( ) function plots a histogram of all arrival delays making multiple density plot with matplotlib how! To plot power spectral density to emphasize the difference is the density plot matplotlib density function for kernel! Overflow question is below: Finally, we have to specify bandwidth scott. Blue curve is created by summing the individual Gaussians and forms the overall plot... Them out online book on data visualization, and cutting-edge techniques delivered Monday to Thursday of observations a. But having a smooth curve drawn through the top of matplotlib, and the bandwidth..., it adds a weight to surrounding bins make multiple distributions with one call! Sense because the plot cluttered, but I like to specify bandwidth as scott or.... Rug plot method from the data to density plot matplotlib the distributions of one variable across multiple categories and... Method from the seaborn package to and set hist=False to generate the density plot help! On a density plot with a higher peak is the name of the.. Dashed red lines above each point make multiple distributions with one function call. ) this …! Present in the plot cluttered, but image reading/writing for any format than! 12 ) ) # create 16 empty plots for X in ( np function calls in both libraries create. We specify the binwidth by the number of points can be slow how did I come with! Julia as the following 'ggplot ' ) for ggplot-style plots making multiple plots... Are going to plot density plots in seaborn, a question that depends the. Be able to either by email the distplot or kdeplot function and can made. Such that the total area under the curve for a specific interval on x-axis. This plot is that the underlying data because the shading helps us distinguish the plots in seaborn etc! Smooth curve drawn through the top of each bin to specify this is always. Earlier more often than United airlines plots in seaborn, a question that on! Visualize the NYCflights13 data, which contains over 300,000 observations of flights in minutes Pillow. Dimension how hard can it be to effectively display the data helps us distinguish plots... Airline and a categorical variable with multiple airlines, # Stacked histogram with levels... Another quick and easy technique for getting each attributes distribution is smooth and.... There any good substitutes of the plot calls, we have arrived at an effective solution a delay. Png is limited to uint8 data libraries that create equivalent figures an excellent online. Uint8 data the density plots in seaborn, etc dimension how hard can be! ) function plots a histogram estimated from the data using a continuous probability density and not a probability the... By email plot as a Jupyter Notebook on GitHub an actual probability, we pass in the density is. Won ’ t be able to either ) ’ values indicate here?... Możliwość komentowania jest wyłączona.
2021-03-02 01:15:41
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https://frama-c.com/html/fc-discuss/2013-April/msg00037.html
# Frama-C-discuss mailing list archives This page gathers the archives of the old Frama-C-discuss archives, that was hosted by Inria's gforge before its demise at the end of 2020. To search for mails newer than September 2020, please visit the page of the new mailing list on Renater. # [Frama-c-discuss] floating-point examples Claude & Virgile: Thanks for the pointers. Looking at the examples, it seems that most (but not all) non-trivial verifications of floating-point examples require Coq and therefore some human intervention (as opposed to being totally automated). Would this be an accurate generalization of the state of the art at this point in time? -Steve On Apr 15, 2013, at 1:19 AM, Claude March? <Claude.Marche at inria.fr> wrote: > > See > > http://toccata.lri.fr/gallery/MyCosineACSL.en.html > http://hal.inria.fr/hal-00777605 > > - Claude > > Le 15/04/2013 04:48, Stephen Siegel a ?crit : >> I'm looking for some simple, interesting floating-point examples that can be proved with Frama-C+Jessie. Here is a good example from the ACSL manual: >> >> /*@ requires \abs(\exact(x)) <= 0x1p-5; >> @ requires \round_error(x) <= 0x1p-20; >> @ ensures \abs(\exact(\result) - \cos(\exact(x))) <= 0x1p-24; >> @ ensures \round_error(\result) <= \round_error(x) + 0x3p-24; >> @*/ >> float cosine(float x) { >> return 1.0f - x * x * 0.5f; >> } >> >> However, I can't get any of the theorem provers (CVC3, Z3, Alt-Ergo, Gappa, Simplify) to prove either of the generated VCs. (For all I know, the function may not even satisfy the contract.) Does anyone know of any examples similar to this that can be proved, or have ideas on how to handle this example? >> _______________________________________________ >> Frama-c-discuss mailing list >> Frama-c-discuss at lists.gforge.inria.fr >> http://lists.gforge.inria.fr/cgi-bin/mailman/listinfo/frama-c-discuss >> > > -- > Claude March? | tel: +33 1 72 92 59 69 > INRIA Saclay - ?le-de-France | > Universit? Paris-sud, Bat. 650 | http://www.lri.fr/~marche/ > F-91405 ORSAY Cedex | > > > _______________________________________________ > Frama-c-discuss mailing list > Frama-c-discuss at lists.gforge.inria.fr > http://lists.gforge.inria.fr/cgi-bin/mailman/listinfo/frama-c-discuss
2022-05-21 22:24:34
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https://insidedarkweb.com/quantitative-finance/which-maximum-diversification-approach-in-matlab-is-correct/
# Which Maximum Diversification Approach in MATLAB is correct? I am currently trying to find the portfolio weights of the Maximum Diversification Portfolio and found two approaches which result in different outcomes. The first one is based on this paper:https://www.tobam.fr/wp-content/uploads/2014/12/TOBAM-JoPM-Maximum-Div-2008.pdf Here I first calculate the assets weights in of synthetic assets and then covert them to an portfolio of real assets. According to the paper there is the possibility of being long or short in cash, but since I want to be fully invested in the risky assets I scale the weights to 1. My question is if this would be “allowed” without changing the optimization problem? This is the objective function that I minimize: function fval = md(corMat,w_md) fval = w_md'*corMat*w_md; end And this is the optimization: T = readtable('Data_test.xlsx'); mon_ret= tick2ret(T{:,3:end}); numReturns = size(mon_ret,1); covMat = cov(mon_ret) ; [corMat, std] = corrcov(covMat); port_size = length(covMat) ; Aeq = ones(1,port_size); Beq = 1; lbnds = zeros(1,port_size); ubnds = ones (1,port_size); n1 = 1.0/port_size; w0 = repmat(n1, port_size, 1) ; mdfunction = @(w_md) md(corMat, w_md); w_md = fmincon(mdfunction, w0, ... [], [], Aeq, Beq, lbnds, ubnds, []) ; w_md = w_md./std; w_md = w_md/sum(w_md); The second approach is from this paper(p.21): http://www.qminitiative.org/UserFiles/files/FroidureSSRN-id1895459.pdf I think I solved it accordingly with this approach: Objective Function: function fval = md2(covMat, w_md2) fval = w_md2'*covMat*w_md2; end Non Linear Constraint: function [c,ceq] = nlcon(w_md2,std) c =[]; ceq = sum(w_md2'.*std)-1; end and the optimization: md2function = @(w_md2) md(corMat, w_md2); w_md2 = fmincon(md2function, w0, ... [], [], Aeq, Beq, lbnds, ubnds, []) ; w_md2 = w_md2/sum(w_md2); Does anybody know which approach is correct or where my mistake is? I`d appreciate every help! Best regards Quantitative Finance Asked by Dirty Dan on November 17, 2021 Is the formula for code #1 $$max D(S)=frac{S^{top}Sigma_S}{sqrt{S^{top}V_S S}}$$? or is it $$max D(S)=frac{1}{sqrt{S^{top}V_S S}}$$ s.t. constraints $$Gamma$$? Both appear on the same page, 41, in paper #1. and is formula for code #2 $$min frac{1}{2}mathbf{w}^{top}Sigmamathbf{w}$$ s.t. $$w_igeq 0$$, $$mathbf{w^{top}}boldsymbol{sigma}=1$$ from paper #2's appendix? If so, the 2nd formula has a non-negativity constraint, making it the short-sale constrained MDP, which might be why it doesn't correspond to the 1st, if the 1st is the unconstrained MDP. Answered by develarist on November 17, 2021 ## Related Questions ### Difference between Order Expire and Order Done for Day (DFD) 1  Asked on August 21, 2020 by abnv ### How to download full daily historical data of MSCI AC Asia Index 1  Asked on August 13, 2020 by monicam ### Implied Volatility from Heston Model 1  Asked on August 3, 2020
2021-12-04 04:54:23
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https://www.numerade.com/questions/what-is-the-shortest-possible-length-of-the-line-segment-that-is-cut-off-by-the-first-quadrant-and-i/
💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # What is the shortest possible length of the line segment that is cut off by the first quadrant and is tangent to the curve $y = 3/x$ at some point? ## shortest length is 2$\sqrt{6} \approx 4.899$ Derivatives Differentiation Volume ### Discussion You must be signed in to discuss. ##### Heather Z. Oregon State University ##### Kristen K. University of Michigan - Ann Arbor Lectures Join Bootcamp ### Video Transcript mhm, whereas to find the shortest possible length of the line segment that is cut off by the first quadrant and is tangent to the curve y equals three over X at some point. So we have our function y equals three over X. We want to find the distance between the X and Y intercepts of the tangent line to the curve at some point. So the derivative Why prime, this is negative three over X squared, which is our slope of this line. Yeah, an equation of a line that passes through the point. A three over a with this slope is why minus three over a equals R slope. Negative three over a squared times X minus a. Okay, that's what I'm saying. Yeah, we can expand this to y equals negative three over a squared X plus six over a. Now, to find the Y intercept, you simply set X equal zero. So have a wide intercept of 06 over a on the X intercept. Well, this is setting. Why equal to zero? This gives us the point. Negative six over a divided by three over a squared which is just to a zero this is from the X intercept to the Y intercept as a function of a I'll call it D. This is the square root of the difference of X coordinates, which is to a squared plus the difference in why coordinates, which is six over a squared. Now we know that this distance is minimized when it's square is also minimized. Yeah, yeah, yes, that's it. So we have our function f of A which is for a squared plus 36 over a squared. Now I want to minimize f so will differentiate and find where it's zero f prime of a is eight a minus 72 over a cubed and we set this equal to zero solving for a we get that a is equal to eventually the square root of three. Notice that the second derivative of F this is eight plus four times 72 over. I'm sorry. This should be three times 72 over eight of the fourth, which is positive for all values of a including equals Route three. Therefore, it follows that our function f has a minimum, he said at a equals route three. Yeah, mhm, of course, by our previous discussion. This is also the minimum. D also has a minimum at equals Route three. And so our minimum distance is going to be D of the square to three, which is the square root of four times three schools. Plus with me 36 over three, which is the square root of 12 plus 12, or to route six. So the shortest length is to route six, which is approximately 4.899 better side. Yeah, yeah. Ohio State University #### Topics Derivatives Differentiation Volume ##### Heather Z. Oregon State University ##### Kristen K. University of Michigan - Ann Arbor Lectures Join Bootcamp
2021-09-16 16:13:32
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https://istopdeath.com/verify-the-identity-cosx-square-root-of-3-2/
# Verify the Identity cos(x)=( square root of 3)/2 The provided equation is not an identity. is not an identity Verify the Identity cos(x)=( square root of 3)/2
2022-09-24 17:01:17
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https://puzzling.stackexchange.com/questions/23590/a-balance-and-twenty-weights
# A balance and twenty weights Fredo owns a nice balance with two pans. He now wants to design a system of 20 integer weights $w_1,w_2,\ldots,w_{20}$ with the following properties: • $1\le w_1\le w_2\le w_3\le\cdots\le w_{20}$ • Any integer weight $U$ from the range $1\le U\le2015$ can be balanced by placing it into the left pan and some subset of the weights into the right pan. Question: What is the smallest possible value that weight $w_{20}$ can take under these conditions? • What is with all the 2015-themed puzzles recently? – user1717828 Oct 28 '15 at 14:52 • I know it was you, Fredo. You broke my heart. You broke my heart! – Michael McGriff Oct 28 '15 at 19:05 The weight $w_i -1$ for $i\in [1, 20]$ can only be weighted using the weights $w_1$ to $w_{i-1}$: indeed all other weights are more heavy than $w_i -1$ (recall that the weights are ordered by increasing weights). Therefore when you add all the weights $w_1$ to $w_{i-1}$ they must at least weight $w_i -1$. This gives the relation : ## $$w_i -1 \leq \sum_{k=1}^{i-1} w_k$$ This gives in particular that $w_1 -1 \leq 0$ and because we imposed the property $1 \leq w_1$ we have $w_1 =1$. It also leads to $w_2 -1 \leq w_1$, i.e. $w_2 \leq 2$, and $w_3 -1 \leq w_1 + w_2$ i.e. $w_3 \leq 4$ ## $$w_i \leq 2^{i-1}$$ Let's now look at $\sum_{i=1}^{20}w_i \geq 2015$ (this comes from the fact that you can weight 2015). We can separate the 12 last terms of the sum as follows: ## $$\sum_{i=9}^{20}w_i \geq 2015 - \sum_{i=1}^{8}w_i$$ Hence $\sum_{i=9}^{20}w_i \geq 2015 - \sum_{i=1}^{8}2^{i-1}$ and $\sum_{i=9}^{20}w_i \geq 1760$ We conclude that $max_{i=9}^{20} w_i \geq \lceil{1760/12}\rceil$ And in particular that: ## $$max_{i=1}^{20} w_i \geq 147$$ There should be a weight of weight at least 147. On the other hand there exists a set of weights where all weights are at most 147 satisfying the conditions, for instance : {1, 2, 4, 8, 16, 32, 63, 125, 147, 147, 147, 147, 147, 147, 147, 147, 147, 147, 147, 147} The proof of this can take the other answers to this post as indication. So this proves that 147 is the smallest value $w_{20}$ can take under the mentioned conditions. • One could say that $\sum_{k=1}^0$ is not well defined. In this case I decided to sacrifice accuracy for intuition: assume it is 0. – Ara Oct 28 '15 at 17:07 • There's nothing wrong with that expression. It's a sum of 0 terms, which is 0. – f'' Oct 28 '15 at 17:44 • Depending on your definition of the sum notation it could be correct, but I think most people won't agree, wikipedia in English wouldn't for instance. – Ara Oct 28 '15 at 19:09 • Note that weights can be used on either side of the scale and hence subtractively, so your inequality with powers of two is incorrect : 1,3,9 is sufficient for measuring any integer up to and including 13, and this continues in powers of 3. – IanF1 Oct 29 '15 at 6:50 • @IanF1: countrary to another quite similar question , this one states that 'Any integer weight [...] can be balanced by placing it into the left pan and some subset of the weights into the right pan'. If you allow both sides then you are indeed right, the first inequality becomes false and I have no easy patch that comes to mind, as the weights can now become arbitray large. – Ara Oct 29 '15 at 11:50 I have a solution with the largest weight being: 147 First i tried using binary weigths ($2^{N-1}$) but this solution would be giving me the possibility of balancing way larger weights than what is needed. So i tried to think about where to stop using binary weigths so that the higher weights can be balanced by using equal weights of smaller size than the would be next binary weigth. My solution uses the binary weigths 1, 2, 4, 8, 16, 32, 64, 128. With those i can balance all weights between 1 and 255. Subtracting all binary weights from 2015 leaves us with 1760. Now we only need to divide 1760 by the amount of our left over weigths to determine the smallest amount they should have: $w=\frac{1760}{12}=146\frac{2}{3}$ So our smallest possible biggest weight is 147. • This is nice, but it lacks proof that 147 is the smallest possible. – klm123 Oct 28 '15 at 9:51 If you used binary weights, you wouldn't need 20. You could use 1,2,4, ... 1024 to weigh anything up to 2096, which is more than 2015, with just 11 weights. But you've been told to use 20 weights. So you could take the last one, 1024, and replace it with two 512s. Now you're using 12 weights and the largest is 512. But you still have 8 more weights available. If you try to split 1024 into three, you're getting down below 512, the binary weight before the 1024. What's more, you're still trying to reach 2096 when you don't need to. So set aside the 1024 and the 512, and you have 1,2,4, ... 256 for 9 weights, and can use 11 weights for the remaining (2015-511) 1504 g. That's an average of 137 each, which is below the 256, so set aside 256 also. Now you have 1,2, 4 ... 128 for 8 weights, and 12 weights for (2015-255) 1760, averaging 147. In order to make any number up to 512, you will need a "256" - which will be made up of these extra weights - use a pair of 128s, and you have 10 weights accounted for. In order to make any number up to 1024, you will need a "512". Use 4 128s and you have 14 weights accounted for. If you wanted to make numbers up to 2096 the next step would be to create a "1024", but you don't. The largest number you need is 2015 which is 991 more than 1024. So if you can make 991 out of 6 weights, you can make any number between 991 and 2015. It averages 165. So you could use 5 165s and a 164. Can this be improved? What if I had 12 147s after the 1..128 and didn't bother with the "256" and the "512"? They pull the average down because they're only 128. I can make any number up to 255 with the 1..128, so I can make 147, then throw in a 147 and carry on adding 1, 2, 3, 4, to that until I hit 294, then use 2 of the 147s and so on. Now 12x147 is 1764 not 1760, so I need a mix of 157 and 156s to hit 1887. Using that approach, do I even need that first 128? Well, how else will I make 129? or 130? So I do and the largest weight will have to be 147g. • The binary weights from 1 to 128 number 8 and not 7. (1, 2, 4, 8, 16, 32, 64, 128) Following your solution the last step would be to divide 991 by 6 which amounts to about 165.17. – The Dark Truth Oct 28 '15 at 11:24 • fixed @TheDarkTruth – Kate Gregory Oct 28 '15 at 11:52 • +1. Since weights are allowed to be placed on both pans, why not use ternary base system (+1, 0, -1). I think the largest weight may get smaller. Does it? – Bhaskar Oct 28 '15 at 14:26 • @L16H7 it would get smaller and such an answer was already posted before. However the question clearly states that you can only put the weights on one of the two pans. – The Dark Truth Oct 28 '15 at 14:29 • Argh! Sorry I misunderstood it. :) – Bhaskar Oct 28 '15 at 14:33
2019-05-20 11:14:45
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https://www.jtolio.com/2017/01/magic-gopath/
# Posts First published: Jan 14, 2017, 6:47pm MST Last edited: May 4, 2018, 12:46pm MDT ## Magic GOPATH Maybe someday I’ll start writing about things besides Go again. Go requires that you set an environment variable for your workspace called your GOPATH. The GOPATH is one of the most confusing aspects of Go to newcomers and even relatively seasoned developers alike. It’s not immediately clear what would be better, but finding a good GOPATH value has implications for your source code repository layout, how many separate projects you have on your computer, how default project installation instructions work (via go get), and even how you interoperate with other projects and libraries. It’s taken until Go 1.8 to decide to set a default and that small change was one of the most talked about code reviews for the 1.8 release cycle. After writing about GOPATH himself, Dave Cheney asked me to write a blog post about what I do. ## My proposal I set my GOPATH to always be the current working directory, unless a parent directory is clearly the GOPATH. Here’s the relevant part of my .bashrc: # bash command to output calculated GOPATH. calc_gopath() { local dir="$PWD" # we're going to walk up from the current directory to the root while true; do # if there's a '.gopath' file, use its contents as the GOPATH relative to # the directory containing it. if [ -f "$dir/.gopath" ]; then ( cd "$dir"; # allow us to squash this behavior for cases we want to use vgo if [ "$(cat .gopath)" != "" ]; then cd "$(cat .gopath)"; echo "$PWD"; fi; ) return fi # if there's a 'src' directory, the parent of that directory is now the # GOPATH if [ -d "$dir/src" ]; then echo "$dir" return fi # we can't go further, so bail. we'll make the original PWD the GOPATH. if [ "$dir" == "/" ]; then echo "$PWD" return fi # now we'll consider the parent directory dir="$(dirname "$dir")" done } my_prompt_command() { export GOPATH="$(calc_gopath)" # you can have other neat things in here. I also set my PS1 based on git # state } case "$TERM" in xterm*|rxvt*) # Bash provides an environment variable called PROMPT_COMMAND. The contents # of this variable are executed as a regular Bash command just before Bash # displays a prompt. Let's only set it if we're in some kind of graphical # terminal I guess. PROMPT_COMMAND=my_prompt_command ;; *) ;; esac The benefits are fantastic. If you want to quickly go get something and not have it clutter up your workspace, you can do something like: cd \$(mktemp -d) && go get github.com/the/thing On the other hand, if you’re jumping between multiple projects (whether or not they have the full workspace checked in or are just library packages), the GOPATH is set accurately. More flexibly, if you have a tree where some parent directory is outside of the GOPATH but you want to set the GOPATH anyways, you can create a .gopath file and it will automatically set your GOPATH correctly any time your shell is inside that directory. The whole thing is super nice. I kinda can’t imagine doing something else anymore. Update: vgo is the future, and works better without a GOPATH set in some cases. I’ve updated the script to allow you to squash the behavior by making an empty .gopath file.
2018-07-21 15:46:32
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https://www.gamedev.net/forums/topic/681562-easy-way-to-get-static-aabb-for-animated-model/
# Easy way to get static AABB for animated model This topic is 481 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts I'm using assimp to import models. As I read each vertex I stored the min/max values for its bindpose and so I can get the whole AABB for the model in bindpose. Easy enough. Now I want to get the same pre-computed AABB but based on all its various poses played during all of its animations. This seems like quite a headache, the way I understand it I have to go through each animation, calculate all bone transforms for every tick and apply them to each vertex based on vertex-bone indices/weights. Basically re-implementing the whole animation pipeline but with CPU skinning :( Is there an easier way to do this than the above? ##### Share on other sites one lazy thing is to create a "worst case AABB" in the scene and use that. Otherwise I think that performing the skinning is the only solution that assimg offers(correct me if I'm wrong). Usually the lazy solution is good enough. ##### Share on other sites If you assume that bones can only rotate, not translate/scale: Pick a bone to be the center. Walk outwards from there along all paths to leaf bones. Along the way, sum the distance between each bone, giving the total distance travelled from the center bone. Pick the maximum distance value produced by any of the leaves. A sphere of that size, centered on the center-bone will always enclose all bones. Vertices form a shell around the bones though, so for each bone, iterate through every vertex that is skinned to this bone (has a weight>0 for this bone) and compute the distance between the bone and this vertex. Keep track of the maximum ditsance between any bone and associated vertex. Add this distance to the sphere generated earlier. Now you've got a sphere that's guaranteed always bound any in the model, no matter how it's bones are rotated. At runtime, you can cull using these "conservative" spheres before doing animation. After animation, you can generate a slightly tighter bounding volume if required, by taking the min/max of the animated bones and adding the "shell size" that was computed above. If you happen to use software skinning, then you can compute the min/max vertex position during skinning to get an even tighter bounding volume still -- but not many people use software skinning any more :) Edited by Hodgman ##### Share on other sites My plan for this, which I have not yet implemented, has been to sample the animation at various times during export (every 100 or 200 milliseconds, for example) and generate an AABB for that keyframe and interpolate between them at run-time. This will generate a more conservative AABB than other methods, but will not necessarily work on objects with non-forward kinematic animations (inverse kinematics or any other kind of run-time dynamic animation or pose). For those, Hodgman’s suggestion is likely best. L. Spiro ##### Share on other sites This will generate a more conservative AABB than other methods, but will not necessarily work on objects with non-forward kinematic animations (inverse kinematics or any other kind of run-time dynamic animation or pose) Sampled bounding volumes also stop being valid if you blend animations together. After blending two animations, the bounds of the produced pose have no relation whatsoever to the source pose bounds. ##### Share on other sites Here is another idea: if you have a simple collision mesh for your model, you can use it to compute the AABB in real time on per-frame basis. No need to store a big database of pre-computed AABBs for each model for each key frame. Plus, the computed skinning transformation of the collision mesh can be reused later, for erm... determining collisions.  :) Edited by vanka78bg ##### Share on other sites if you have a simple collision mesh for your model, yeah but often that collision mesh is thinner compared to the actual model. ##### Share on other sites This topic is 481 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
2017-12-17 06:34:10
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https://atarnotes.com/forum/index.php?action=recent
February 20, 2019, 04:05:38 am ### Recent Posts Pages: [1] 2 3 ... 10 1 ##### VCE Mathematics / Re: Why did you choose to study maths in VCE? (Further, Methods, Specialist) « Last post by Sine on 3 hours ago » My "reasoning" for doing spec and methods was quite similar to quantum's. I don't think I thought too much about it and it just made sense to do both of them. In addition to this I did also consider doing uni maths at UoM or monash as a way of extension since a lot of the concepts are following on from spec stuff but I ended up not doing this for no other reason but I chose to do uni extension chemistry. EDIT: Also did consider further as a lot of people told me it would be quite easy to score in with minimal effort. However, only 2 maths are allowed in the top 4 so it would've been quite redundant 2 ##### VCE Chemistry / Chemistry Question Help!! « Last post by JamieLeaf on 3 hours ago » Hey Guys, I have read through the solution for this question but I still don't know how to work it out. The question is: 100ml of gaseous hydrocarbon is mixed with 500 ml of oxygen at 20 degrees celsius and 1 atmosphere pressure and sparked. At the end of the reaction the gaseous mixture is returned to its original temperature and pressure. The final mixture consists of 300ml of carbon dioxide and 100 ml of oxygen. The molecular formula of the hydrocarbon is: A) C2H4 B) C3H4 C) C3H6 D) C3H8 3 ##### Queensland Education Discussion / Re: How can I support my own learning to get ahead? « Last post by qldpc on 3 hours ago » Hey Not1OfThem! Firstly, welcome to the forums!  Thanks for your advice, I've just printed what will be covered in year 11 and 12 for my subjects, and been studying for the first unit for each. I've recently discovered the past exams so I'll put those into good use too! Thanks a bunch! YES! Something I do all the time is read maths/science books - for pleasure but also for school. I've been watching a few YouTube videos to help me with the phospholipid bilayer (very interesting!) which has really helped. Hey Twisty314, Great to see a fellow Queenslander! Just wondering, what past papers are you using? So far I've been using the HSC past papers because I find them relatively similar to the QCE syllabus. Feeling slightly annoyed we're the guinea pigs and don't have past papers or any idea what subjects will likely scale. Cheers, Qldpc 4 ##### VCE Physics / Re: Why should I choose Physics? « Last post by undefined on 3 hours ago » I’d recommend trying it out in year 11 and dropping it if you don’t like it in year 12 (which is what I did). I was also in your position two years ago, and what I found was I couldn’t use the same approach for physics as I did for spesh and methods and needed a good conceptual understanding of the content to do well. Physics does complement spesh for a couple topics though (kinematics and vectors) but taking both subjects is not essential to succeed in either. 5 ##### The University Journey Journal / Re: Isla77's Dental Journey Journal « Last post by Isla77 on February 19, 2019, 11:46:28 pm » Hey everyone!! So I'm officially fully enrolled in the course and these days get heaps of emails from the professors and coordinators. I read through them all, at least for the ones I've received so far, and have organised my immunisations, getting police check, getting WWCC, my timetable and the instruments I will require for the course. The instruments for the first year amounted to a total of about \$2500, so not too bad considering most of it will last for the whole degree and I will need to purchase very little from now on. Right now I am really excited to start uni, and I know it will be great because a lot of people love it and I think the environment will suit me more than high school. Can't believe how much freedom I will have. But also, I hope I can keep up with the course and all the assessments. When I received the offer a third year student kindly gave me a call for little chat, and said the studying part of it will be more interesting than high school because everything you learn will be directly used in the future. So the concepts you learn and the skills you develop will be extremely relevant, necessary for your future. Sounds exciting! Also, I will be missing orientation and the first two days of uni, I hope that's fine. Also, how do I sign up for clubs and societies if I'm not going to orientation? Anyone here have similar experience? So, a short one today, I will write more later but right now I'm really sleepy 6 « Last post by e2503 on February 19, 2019, 11:29:48 pm » Hey there, I`ve recently been engulfed in exceeding distress about an upcoming creative writing assignment. My concern is not necessarily regarding developing an authentic storyline ( not everyone is J.K. Rowling ), but rather developing a cohesive response through effective uses of literary techniques. I guess i am asking the most ambiguous question ever but what would you say are the most imperative aspects of an eloquent creative piece?? 7 ##### VCE Mathematical Methods CAS / Re: VCE Methods Question Thread! « Last post by AlphaZero on February 19, 2019, 11:14:10 pm » I used the first equation u provided and had p as 0.6 and n as 96 (from memory) and thats how I got 0.098. Have I stuffed up here/ why do you get a different answer? You would've had to use that equation or equivalent to obtain  $\hat{p}=0.6$  and  $n=96$  in the first place though. You haven't necessarily 'stuffed up'. You essentially used the margin of error and that equation (or equivalent) to find $n$, only to plug $n$ and $\hat{p}$ back into the same equation to find the margin of error. What you did is 'mathematically correct', but completely pointless. (Although I believe $n\approx 92$, not 96). Also, for a 95% confidence interval,  $z\approx 1.96$,  and is not exact. If you type  $\texttt{invNorm(0.975,0,1)}$  into your CAS, you should find that in fact it gives you  $1.9599639859915...$. This rounding error plus your error in calculating $n$ would've led to your answer. 8 ##### Forum Games and Casual Chat / Re: Count to 20 before a mod posts! « Last post by Erutepa on February 19, 2019, 11:13:20 pm » But the point of this board is supposed to be entertaining? 0. Are you telling me that the world doesn't revolve around me This is hard to swallow... 1 9 ##### The University Journey Journal / Sarangiya's 너무 사랑하는 Medical Journey Journal « Last post by Lear on February 19, 2019, 11:11:40 pm » From what I’ve gathered from some seniors in the medical field, such a story gathers attention in the media every year or so. People care for a few months and then nothing happens. Quite a scary thing that, if a builder injures himself at a construction site, work safe is there ASAP. Meanwhile this sort of stuff is allowed to go on in the medical world, and has been for ages. Can you believe we are still better off now than many years ago? Then again, what is worksafe going to do? Shutdown hospitals? 10 ##### The University Journey Journal / Re: Sarangiya's 너무 사랑하는 Medical Journey Journal « Last post by Isla77 on February 19, 2019, 11:06:08 pm » Hey sarangiya! I hope your move to Sydney goes well, and that you enjoy your journey in med school. I'm sure you will handle and manage it all as you seem to be very capable. I probably should have replied a bit quicker to answer that properly, but my initial reaction was one of shock! I didn't ever imagine I would be offered a place UNSW. I was estatic, my mum a bit apprehensive. But she's right on board now, so that's encouraging. It'll be hard to leave home, but I might be able to return more frequently than what I thought! Congratulations on receiving your dream offer! Hope you won't miss your mum too much, but if you do, I will definitely relate to that lol. Great thing you will be able to return every now and then! As for the media coverage concerning a resident in NSW, I was quite upset too. I can't believe someone will push that far, sacrificing their health in the process, but then again, I guess you do have to push yourself beyond limits if you want to succeed in certain medical specialties. But having said that, the problem lies certainly in the hospital that created this and the medical education that structures it. I suppose Dr Kadota did this because she have been faced with many challenges in her life and still overcame them, and so believed she can conquer this one too. No doubt the medical educators and the people responsible for it should be well aware that there are a lot of perfectionists in this field, perfectionists who tend to push themselves way too far. Medical educators will teach medical students about their work, so they should also shed light on looking after yourself and not to go too far when things might become toxic. I heard about this on world BBC news on a flight from middle east to Asia two days ago too(it's become a worldwide news now ), and I really hope this extended media coverage will bring about some change in the system. Pages: [1] 2 3 ... 10
2019-02-19 17:05:39
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https://orbit.dtu.dk/en/publications/computer-simulations-of-lipid-bilayers-and-proteins
# Computer Simulations of Lipid Bilayers and Proteins Jacob Sonne Research output: Book/ReportPh.D. thesisResearch ## Abstract The importance of computer simulations in lipid bilayer research has become more prominent for the last couple of decades and as computers get even faster, simulations will play an increasingly important part of understanding the processes that take place in and across cell membranes. This thesis entitled Computer simulations of lipid bilayers and proteins describes two molecular dynamics (MD) simulation studies of pure lipid bilayers as well as a study of a transmembrane protein embedded in a lipid bilayer matrix. Below follows a brief overview of the thesis. Chapter 1. This chapter is a short introduction, where I briefly describe the basic biological background for the systems studied in Chapters 3, 4 and 5. This is done in a non-technical way to allow the general interested reader to get an impression of the work. Chapter 2, Methods: In this chapter the background for the methods used in the succeeding chapters is presented. Details on system setups, simulation parameters and other technicalities can be found in the relevant chapters. Chapter 3, DPPC lipid parameters: The quality of MD simulations is intimately dependent on the empirical potential energy function and its parameters, i.e., the force field. The commonly employed CHARMM27 force field reproduces fluid phase bilayer properties only when a tension is applied in the simulations. Without this tension the simulated bilayers become too ordered. To allow for more realistic lipid bilayer simulations in the tensionless isothermal-isobaric (NPT) ensemble, we reparameterized a part of the CHARMM27 force field by assigning new partial charges to the lipid head group atoms. Our modified CHARMM27 force field was tested in lipid bilayer MD simulations and was found to improve the phase properties of the simulated bilayers significantly. Thus, the improved force field makes it possible to simulate the biologically relevant fluid ($L_{\alpha}$) phase in an NPT ensemble, which is an important prerequisite for taking full advantage of the predictive power of MD simulations since the area per lipid need not be known prior to simulation. Chapter 4, Pressure profile calculations in lipid bilayers: A lipid bilayer is merely $\sim$5~nm thick, but the lateral pressure (parallel to the bilayer plane) varies several hundred bar on this short distance (normal to the bilayer). These variations in the lateral pressure are commonly referred to as the pressure profile. The pressure profile changes when small molecules partition into the bilayer and it has previously been suggested that such changes may be related to general anesthesia. MD simulations play an important role when studying the possible coupling between general anesthesia and changes in the pressure profile since the pressure profile cannot be measured in traditional experiments. Even so, pressure profile calculations from MD simulations are not trivial due to both fundamental and technical issues. We addressed two such issues namely the uniqueness of the pressure profile and the effect of neglecting pressure contributions from long range electrostatic interactions. The first issue is addressed by comparing two methods for calculating pressure profiles, and judged by the similar results obtained by these two methods the pressure profile appears to be well-defined for fluid phase lipid bilayers. For the second issue, we developed a method that allows pressure contributions from long range electrostatic interactions to be included in pressure profile calculations. Chapter 5, Structural transitions in the ABC transporter BtuCD: In this chapter, we present a study of the transmembrane protein BtuCD. BtuCD belongs to the adonesine triphosphate (ATP) binding cassette (ABC) transporter family that use ATP to drive active transport of a wide variety of compounds across cell membranes. BtuCD accounts for vitamin B12 import into Escherichia coli and is one of the only ABC transporters for which a reliable crystal structure of the whole transporter has been determined. The (dys)function of ABC transporters accounts for cystic fibrosis and multi-drug resistance, for example tumor cell resistance to anticancer drugs. Hence, the mechanism facilitating substrate translocation in ABC transporters is of both fundamental and medical interest. It is commonly accepted that substrate translocation relies on binding and hydrolysis of ATP and maybe also on the release of hydrolysis products; however, the global structural rearrangements induced by these events remain largely unknown. To investigate these structural rearrangements in BtuCD we employed perturbed elastic network calculations and biased MD simulations. Comparing the results of these calculations with two transport models proposed in the literature, we are able to favor one over the other. Our observations for BtuCD may be relevant for all ABC transporters, owing to the conservation of ATP binding domains Chapter 5: This chapter contains a more technical summary of the thesis and some general conclusions. Chapter 7: This chapter contains a compilation of supplementary material relating to the subjects discussed in the preceding chapters. pressure profile calculations. Original language English Number of pages 114 87-91233-14-3 Published - Jul 2006 ## Keywords • abc transporter • bilayer • parameter • reparmeterization • md • pressure profile • lipid • btucd • molecular dynamics • dppc • ### Molekyldynamiske simuleringer af Lipidmembraners Vekselvirkning med Membranaktive Enzymer Sonne, J., Peters, G. H. J., Jensen, M. Ø., Møller, K. B., Ipsen, J. H., Marrink, S. J. & Hansen, F. Y. DTU stipendium 01/04/200328/08/2006 Project: PhD
2020-10-31 05:59:49
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https://brilliant.org/discussions/thread/one-to-one-function/
× one to one function? is the function f : f(x)= x^{4} + 2x +1 an one to one function or not? Note by Yarny Figo 2 years, 3 months ago
2016-10-28 12:16:25
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https://helptest.net/mathematics/100607/
# The product of 3/4 and a number s is 3/5 Product= multiplication s= a number is= equal sign 3/4s= 3/5 divide both sides by 3/4 s= (3/5) / (3/4) to divide fractions, multiply by the reciprocal/inverse of 3/4 s= 3/5 * 4/3 multiply numerators; multiply denominators s= (3*4)/(5*3) multiply within parentheses s= 12/15 simplify by 3 s= 4/5 ANSWER: 4/5 Hope this helps!  :) Only authorized users can leave an answer! If you are not satisfied with the answer or you can’t find one, then try to use the search above or find similar answers below.
2021-10-25 01:27:39
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https://www.xloypaypa.pub/codeforces-round-721-palindrome-game-easy-hard-solution-java-c/
Solution: 1: Consider the simplest case: the string is palindrome, and the length of string is even. For example, 000000. For this case, Bob must win. See operations below: A:100000; B:100001; A:110001; B:110011; A:111011; B: reverse; A: 111111. So that: Every time, after Alice changed 0 to 1. Bob also changes the corresponding 0 to 1 to keep the string palindrome. Until last two 0, at that time, Bob do one reverse operation. But there is one exception: if the string is palindrome, and the length is even, but it is all 1s. Then the result is draw. 2: Now consider the case that: the string is palindrome, but the length is odd. Based on case 1, we find that the second player will win, and will win 2 points. So, if the center is 0, then Alice will change on this firstly, and get win. 3: Then consider the case that string is not palindrome. For this case Alice will be happy. Let us consider there are more than 2 pairs not palindrome. For example, 00000111. A:reverse; B:10000111; A:reverse... Based one first cases, the winner at most win 2 points. So, before the string become palindrome, Alice continually reverses the string, and get an advantage of at least 2 points. So, Alice will win. 4: Then let us consider the case that have 1 pair not palindrome. For this case, it same to the case 2. Alice will take this firstly and get win. But one exception: if there is only one 0 exclude this no palindrome pos. For example, 001, and exclude no palindrome pos: 0. For this case, If the first step of Alice is reverse, then Bob will change to 101. Then Alice must change to 111. If the first step of Alice is 011 or 101, then Bob will change to 111. So, draw. 5: Now the last case is: there are exactly 2 pairs not palindrome. For example, 000011. For this case: A:reverse; B:100011; Now, A:110011. And Alice is second player, will win. Java C++
2023-02-01 06:27:50
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https://cdl-prep.com/oregon-cdl-air-brake-test-questions-and-answers
# Free Oregon CDL Air Brakes Practice Test 2023 Do you need an Oregon Air Brakes endorsement or an L endorsement for your commercial driving license? The Oregon CDL Air Brake has some differences from other endorsements because your license will receive a mark of restriction if you fail the test. So having good preparation before exam day is very necessary. To ensure that our questions are relevant, all of our CDL practice test packs are based on the Oregon CDL manual. Each question has a detailed explanation for you to thoroughly learn the format and the topic. Don't be afraid of having a restriction on your license. Let’s try our Oregon CDL Practice Test to get ready to pass the Oregon CDL Air Brake Test now. Our CDL practice tests: Based on 2021 OR commercial driver's license manual Perfect for first-time, renewal applicants OR CDL Air Brakes Test format: 25 questions 80% passing score List of questions 1. To test air service brakes you should: 2. The most common type of foundation brakes are: 3. The air compressor stops pumping air at what psi? 4. Total stopping distance for air brakes is longer than that for hydraulic brakes due to ______ distance. 5. In ideal conditions a truck or bus with air brakes going at 55 mph would require the stopping distance of how many feet? 6. What negatively affects the braking power of a spring brake? 7. When the air system is low ________. 8. When the mechanical arm of a wig wag drops into your view that means the pressure in your system has dropped below what psi? 9. Which braking system applies and releases the parking brakes when you use the parking brake control? 10. It is not safe to drive a vehicle that has brake drums with cracks that are longer than _____ of the width of the friction area. 11. What happens when your brake drums get very hot? 12. All of the following are factors in total stopping distance except: 13. When checking the air pressure buildup, pressure should build: 14. The air compressor governor controls: 15. All of the following apply if the low air pressure warning signal does not work except: 16. If you need to make an emergency stop, you should brake so that you can ________. 17. When driving an older vehicle that has a manual front brake limiting valve, drivers should: 18. The air tanks must be drained because: 19. Why do air brakes take longer to stop than hydraulic brakes? 20. Brake drums or discs _______. 21. The belt of a belt-driven air compressor should be ________. 22. Spring brakes are applied: 23. To avoid damaging the brakes, what should you never do while the spring brakes are applied? 24. What is the purpose of an air storage tank? 25. What does ABS stand for?
2023-02-02 14:53:54
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https://vrpy.readthedocs.io/en/dev/examples.html
# Examples¶ ## A simple example¶ ### Network definition¶ In this first example, we will be working with the following network: The first step is to define the network as a nx.Digraph object. Note that for convenience, the depot (node $$0$$ in the picture) is split into two vertices : the Source and the Sink. # Create graph >>> from networkx import DiGraph >>> G = DiGraph() >>> for v in [1, 2, 3, 4, 5]: ### VRP definition¶ The second step is to define the VRP, with the above defined graph as input: >>> from vrpy import VehicleRoutingProblem >>> prob = VehicleRoutingProblem(G) ### Maximum number of stops per route¶ In this first variant, it is required that a vehicle cannot perform more than $$3$$ stops: >>> prob.num_stops = 3 >>> prob.solve() The best routes found can be queried as follows: >>> prob.best_routes {1: ['Source', 4, 5, 'Sink'], 2: ['Source', 1, 2, 3, 'Sink']} And the cost of this solution is queried in a similar fashion: >>> prob.best_value 70.0 >>> prob.best_routes_cost {1: 30, 2: 40} The optimal routes are displayed below: ### Capacity constraints¶ In this second variant, we define a demand for each customer and limit the vehicle capacity to $$10$$ units. Demands are set directly as node attributes on the graph, and the capacity constraint is set with the load_capacity attribute: >>> for v in G.nodes(): if v not in ["Source", "Sink"]: G.nodes[v]["demand"] = 5 >>> prob.solve() >>> prob.best_value 80.0 As the problem is more constrained, it is not surprising that the total cost increases. As a sanity check, we can query the loads on each route to make sure capacity constraints are met: >>> prob.best_routes {1: ["Source", 1, "Sink"], 2: ["Source", 2, 3, "Sink"], 3: ["Source", 4, 5, "Sink"]} {1: 5, 2: 10, 3: 10} The new optimal routes are displayed below: ### Time constraints¶ One may want to restrict the total duration of a route. In this case, a time attribute is set on each edge of the graph, and a maximum duration is set with prob.duration. >>> for (u, v) in G.edges(): G.edges[u,v]["time"] = 20 >>> G.edges[4,5]["time"] = 25 >>> prob.duration = 60 >>> prob.solve() >>> prob.best_value 85.0 As the problem is more and more constrained, the total cost continues to increase. Lets check the durations of each route: >>> prob.best_routes {1: ["Source", 1, 2, "Sink"], 2: ["Source", 3, 4, "Sink"], 3: ["Source", 5, "Sink"]} >>> prob.best_routes_duration {1: 60, 2: 60, 3: 40} The new optimal routes are displayed below: ### Time window constraints¶ When designing routes, it may be required that a customer is serviced in a given time window $$[\ell,u]$$. Such time windows are defined for each node, as well as service times. >>> time_windows = {1: (5, 100), 2: (5, 20), 3: (5, 100), 4: (5, 100), 5: (5, 100)} >>> for v in G.nodes(): G.nodes[v]["lower"] = time_windows[v][0] G.nodes[v]["upper"] = time_windows[v][1] if v not in ["Source","Sink"]: G.nodes[v]["service_time"] = 1 A boolean parameter time_windows is activated to enforce such constraints: >>> prob.time_windows = True >>> prob.duration = 64 >>> prob.solve() >>> prob.best_value 90.0 The total cost increases again. Lets check the arrival times: >>> prob.best_routes {1: ["Source", 1, "Sink"], 4: ["Source", 2, 3, "Sink"], 2: ["Source", 4, "Sink"], 3: ["Source", 5, "Sink"]} >>> prob.arrival_time {1: {1: 20, 'Sink': 41}, 2: {4: 20, 'Sink': 41}, 3: {5: 20, 'Sink': 41}, 4: {2: 20, 3: 41, 'Sink': 62}} The new optimal routes are displayed below: ### Complete program¶ import networkx as nx from vrpy import VehicleRoutingProblem # Create graph G = nx.DiGraph() for v in [1, 2, 3, 4, 5]: G.nodes[v]["demand"] = 5 G.nodes[v]["upper"] = 100 G.nodes[v]["lower"] = 5 G.nodes[v]["service_time"] = 1 G.nodes[2]["upper"] = 20 G.nodes["Sink"]["upper"] = 110 G.nodes["Source"]["upper"] = 100 # Create vrp prob = VehicleRoutingProblem(G, num_stops=3, load_capacity=10, duration=64, time_windows=True) # Solve and display solution prob.solve() print(prob.best_routes) print(prob.best_value) ### Periodic CVRP¶ For scheduling routes over a time period, one can define a frequency for each customer. For example, if over a planning period of two days, customer $$2$$ must be visited twice, and the other customers only once: >>> prob.periodic = 2 >>> G.nodes[2]["frequency"] = 2 >>> prob.solve() >>> prob.best_routes {1: ['Source', 1, 2, 'Sink'], 2: ['Source', 4, 5, 'Sink'], 3: ['Source', 2, 3, 'Sink']} >>> prob.schedule {0: [1, 2], 1: [3]} We can see that customer $$2$$ is visited on both days of the planning period (routes $$1$$ and $$3$$), and that it is not visited more than once per day. ### Mixed fleet¶ We end this small example with an illustration of the mixed_fleet option, when vehicles of different types (capacities, travel costs, fixed costs) are operating. The first vehicle has a load_capacity of $$5$$ units, and no fixed_cost, while the second vehicle has a load_capacity of $$20$$ units, and a fixed_cost with value $$5$$. The travel costs of the second vehicle are $$1$$ unit more expensive than those of the first vehicle: >>> from networkx import DiGraph >>> from vrpy import VehicleRoutingProblem >>> G = DiGraph() >>> for v in [1, 2, 3, 4, 5]: G.nodes[v]["demand"] = 5 >>> prob=VehicleRoutingProblem(G, mixed_fleet=True, fixed_cost=[0, 5], load_capacity=[5, 20]) >>> prob.best_value 85 >>> prob.best_routes {1: ['Source', 1, 'Sink'], 2: ['Source', 2, 3, 4, 5, 'Sink']} >>> prob.best_routes_cost {1: 20, 2: 65} >>> prob.best_routes_type {1: 0, 2: 1} ## An example borrowed from ortools¶ We borrow this second example from the well known ortools [PF] routing library. We will use the data from the tutorial. ### Network definition¶ The graph is considered complete, that is, there are edges between each pair of nodes, in both directions, and the cost on each edge is defined as the Manhattan distance between both endpoints. The network is displayed below (for readability, edges are not shown), with the depot in red, and the labels outside of the vertices are the demands: The network can be entirely defined by its distance matrix. We will make use of the NetworkX module to create this graph and store its attributes: from networkx import DiGraph, from_numpy_matrix, relabel_nodes, set_node_attributes from numpy import array # Distance matrix DISTANCES = [ [0,548,776,696,582,274,502,194,308,194,536,502,388,354,468,776,662,0], # from Source [0,0,684,308,194,502,730,354,696,742,1084,594,480,674,1016,868,1210,548], [0,684,0,992,878,502,274,810,468,742,400,1278,1164,1130,788,1552,754,776], [0,308,992,0,114,650,878,502,844,890,1232,514,628,822,1164,560,1358,696], [0,194,878,114,0,536,764,388,730,776,1118,400,514,708,1050,674,1244,582], [0,502,502,650,536,0,228,308,194,240,582,776,662,628,514,1050,708,274], [0,730,274,878,764,228,0,536,194,468,354,1004,890,856,514,1278,480,502], [0,354,810,502,388,308,536,0,342,388,730,468,354,320,662,742,856,194], [0,696,468,844,730,194,194,342,0,274,388,810,696,662,320,1084,514,308], [0,742,742,890,776,240,468,388,274,0,342,536,422,388,274,810,468,194], [0,1084,400,1232,1118,582,354,730,388,342,0,878,764,730,388,1152,354,536], [0,594,1278,514,400,776,1004,468,810,536,878,0,114,308,650,274,844,502], [0,480,1164,628,514,662,890,354,696,422,764,114,0,194,536,388,730,388], [0,674,1130,822,708,628,856,320,662,388,730,308,194,0,342,422,536,354], [0,1016,788,1164,1050,514,514,662,320,274,388,650,536,342,0,764,194,468], [0,868,1552,560,674,1050,1278,742,1084,810,1152,274,388,422,764,0,798,776], [0,1210,754,1358,1244,708,480,856,514,468,354,844,730,536,194,798,0,662], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], # from Sink ] # Demands (key: node, value: amount) DEMAND = {1: 1, 2: 1, 3: 2, 4: 4, 5: 2, 6: 4, 7: 8, 8: 8, 9: 1, 10: 2, 11: 1, 12: 2, 13: 4, 14: 4, 15: 8, 16: 8} # The matrix is transformed into a DiGraph A = array(DISTANCES, dtype=[("cost", int)]) G = from_numpy_matrix(A, create_using=nx.DiGraph()) # The demands are stored as node attributes set_node_attributes(G, values=DEMAND, name="demand") # The depot is relabeled as Source and Sink G = relabel_nodes(G, {0: "Source", 17: "Sink"}) ### CVRP¶ Once the graph is properly defined, creating a CVRP and solving it is straightforward. With a maximum load of $$15$$ units per vehicle: >>> from vrpy import VehicleRoutingProblem >>> prob.solve() >>> prob.best_value 6208.0 >>> prob.best_routes {1: ['Source', 12, 11, 15, 13, 'Sink'], 2: ['Source', 1, 3, 4, 7, 'Sink'], 3: ['Source', 5, 2, 6, 8, 'Sink'], 4: ['Source', 14, 16, 10, 9, 'Sink']} {1: 15, 2: 15, 3: 15, 4: 15} The four routes are displayed below: ### CVRP with simultaneous distribution and collection¶ We follow with the exact same configuration, but this time, every time a node is visited, the vehicle unloads its demand and loads some waste material. from networkx import DiGraph, from_numpy_matrix, relabel_nodes, set_node_attributes from numpy import array # Distance matrix DISTANCES = [ [0,548,776,696,582,274,502,194,308,194,536,502,388,354,468,776,662,0], # from Source [0,0,684,308,194,502,730,354,696,742,1084,594,480,674,1016,868,1210,548], [0,684,0,992,878,502,274,810,468,742,400,1278,1164,1130,788,1552,754,776], [0,308,992,0,114,650,878,502,844,890,1232,514,628,822,1164,560,1358,696], [0,194,878,114,0,536,764,388,730,776,1118,400,514,708,1050,674,1244,582], [0,502,502,650,536,0,228,308,194,240,582,776,662,628,514,1050,708,274], [0,730,274,878,764,228,0,536,194,468,354,1004,890,856,514,1278,480,502], [0,354,810,502,388,308,536,0,342,388,730,468,354,320,662,742,856,194], [0,696,468,844,730,194,194,342,0,274,388,810,696,662,320,1084,514,308], [0,742,742,890,776,240,468,388,274,0,342,536,422,388,274,810,468,194], [0,1084,400,1232,1118,582,354,730,388,342,0,878,764,730,388,1152,354,536], [0,594,1278,514,400,776,1004,468,810,536,878,0,114,308,650,274,844,502], [0,480,1164,628,514,662,890,354,696,422,764,114,0,194,536,388,730,388], [0,674,1130,822,708,628,856,320,662,388,730,308,194,0,342,422,536,354], [0,1016,788,1164,1050,514,514,662,320,274,388,650,536,342,0,764,194,468], [0,868,1552,560,674,1050,1278,742,1084,810,1152,274,388,422,764,0,798,776], [0,1210,754,1358,1244,708,480,856,514,468,354,844,730,536,194,798,0,662], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], # from Sink ] # Delivery demands (key: node, value: amount) DEMAND = {1: 1, 2: 1, 3: 2, 4: 4, 5: 2, 6: 4, 7: 8, 8: 8, 9: 1, 10: 2, 11: 1, 12: 2, 13: 4, 14: 4, 15: 8, 16: 8} # Pickup waste (key: node, value: amount) COLLECT = {1: 1, 2: 1, 3: 1, 4: 1, 5: 2, 6: 1, 7: 4, 8: 1, 9: 1, 10: 2, 11: 3, 12: 2, 13: 4, 14: 2, 15: 1, 16: 2} # The matrix is transformed into a DiGraph A = array(DISTANCES, dtype=[("cost", int)]) G = from_numpy_matrix(A, create_using=nx.DiGraph()) # The distribution and collection amounts are stored as node attributes set_node_attributes(G, values=DEMAND, name="demand") set_node_attributes(G, values=COLLECT, name="collect") # The depot is relabeled as Source and Sink G = relabel_nodes(G, {0: "Source", 17: "Sink"}) The load_capacity is unchanged, and the distribution_collection attribute is set to True. >>> from vrpy import VehicleRoutingProblem >>> prob = VehicleRoutingProblem(G, load_capacity=15, distribution_collection=True) >>> prob.solve() >>> prob.best_value 6208.0 >>> prob.best_routes {1: ['Source', 12, 11, 15, 13, 'Sink'], 2: ['Source', 1, 3, 4, 7, 'Sink'], 3: ['Source', 5, 2, 6, 8, 'Sink'], 4: ['Source', 14, 16, 10, 9, 'Sink']} {1: {7: 4, 3: 5, 4: 8, 1: 8, 'Sink': 8}, 2: {8: 7, 6: 10, 2: 10, 5: 10, 'Sink': 10}, 3: {14: 2, 16: 8, 10: 8, 9: 8, 'Sink': 8}, 4: {13: 0, 15: 7, 11: 5, 12: 5, 'Sink': 5}} The optimal solution is unchanged. This is understandable, as for each node, the distribution volume is greater than (or equals) the pickup volume. ### VRP with time windows¶ Each node must now be serviced within a time window. The time windows are displayed above each node: This time, the network is defined by its distance matrix and its time matrix: from networkx import DiGraph, from_numpy_matrix, relabel_nodes, set_node_attributes from numpy import array # Distance matrix DISTANCES = [ [0,548,776,696,582,274,502,194,308,194,536,502,388,354,468,776,662,0], # from Source [0,0,684,308,194,502,730,354,696,742,1084,594,480,674,1016,868,1210,548], [0,684,0,992,878,502,274,810,468,742,400,1278,1164,1130,788,1552,754,776], [0,308,992,0,114,650,878,502,844,890,1232,514,628,822,1164,560,1358,696], [0,194,878,114,0,536,764,388,730,776,1118,400,514,708,1050,674,1244,582], [0,502,502,650,536,0,228,308,194,240,582,776,662,628,514,1050,708,274], [0,730,274,878,764,228,0,536,194,468,354,1004,890,856,514,1278,480,502], [0,354,810,502,388,308,536,0,342,388,730,468,354,320,662,742,856,194], [0,696,468,844,730,194,194,342,0,274,388,810,696,662,320,1084,514,308], [0,742,742,890,776,240,468,388,274,0,342,536,422,388,274,810,468,194], [0,1084,400,1232,1118,582,354,730,388,342,0,878,764,730,388,1152,354,536], [0,594,1278,514,400,776,1004,468,810,536,878,0,114,308,650,274,844,502], [0,480,1164,628,514,662,890,354,696,422,764,114,0,194,536,388,730,388], [0,674,1130,822,708,628,856,320,662,388,730,308,194,0,342,422,536,354], [0,1016,788,1164,1050,514,514,662,320,274,388,650,536,342,0,764,194,468], [0,868,1552,560,674,1050,1278,742,1084,810,1152,274,388,422,764,0,798,776], [0,1210,754,1358,1244,708,480,856,514,468,354,844,730,536,194,798,0,662], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], # from Sink ] TRAVEL_TIMES = [ [0, 6, 9, 8, 7, 3, 6, 2, 3, 2, 6, 6, 4, 4, 5, 9, 7, 0], # from source [0, 0, 8, 3, 2, 6, 8, 4, 8, 8, 13, 7, 5, 8, 12, 10, 14, 6], [0, 8, 0, 11, 10, 6, 3, 9, 5, 8, 4, 15, 14, 13, 9, 18, 9, 9], [0, 3, 11, 0, 1, 7, 10, 6, 10, 10, 14, 6, 7, 9, 14, 6, 16, 8], [0, 2, 10, 1, 0, 6, 9, 4, 8, 9, 13, 4, 6, 8, 12, 8, 14, 7], [0, 6, 6, 7, 6, 0, 2, 3, 2, 2, 7, 9, 7, 7, 6, 12, 8, 3], [0, 8, 3, 10, 9, 2, 0, 6, 2, 5, 4, 12, 10, 10, 6, 15, 5, 6], [0, 4, 9, 6, 4, 3, 6, 0, 4, 4, 8, 5, 4, 3, 7, 8, 10, 2], [0, 8, 5, 10, 8, 2, 2, 4, 0, 3, 4, 9, 8, 7, 3, 13, 6, 3], [0, 8, 8, 10, 9, 2, 5, 4, 3, 0, 4, 6, 5, 4, 3, 9, 5, 2], [0, 13, 4, 14, 13, 7, 4, 8, 4, 4, 0, 10, 9, 8, 4, 13, 4, 6], [0, 7, 15, 6, 4, 9, 12, 5, 9, 6, 10, 0, 1, 3, 7, 3, 10, 6], [0, 5, 14, 7, 6, 7, 10, 4, 8, 5, 9, 1, 0, 2, 6, 4, 8, 4], [0, 8, 13, 9, 8, 7, 10, 3, 7, 4, 8, 3, 2, 0, 4, 5, 6, 4], [0, 12, 9, 14, 12, 6, 6, 7, 3, 3, 4, 7, 6, 4, 0, 9, 2, 5], [0, 10, 18, 6, 8, 12, 15, 8, 13, 9, 13, 3, 4, 5, 9, 0, 9, 9], [0, 14, 9, 16, 14, 8, 5, 10, 6, 5, 4, 10, 8, 6, 2, 9, 0, 7], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], # from sink ] # Time windows (key: node, value: lower/upper bound) TIME_WINDOWS_LOWER = {0: 0, 1: 7, 2: 10, 3: 16, 4: 10, 5: 0, 6: 5, 7: 0, 8: 5, 9: 0, 10: 10, 11: 10, 12: 0, 13: 5, 14: 7, 15: 10, 16: 11,} TIME_WINDOWS_UPPER = {1: 12, 2: 15, 3: 18, 4: 13, 5: 5, 6: 10, 7: 4, 8: 10, 9: 3, 10: 16, 11: 15, 12: 5, 13: 10, 14: 8, 15: 15, 16: 15,} # Transform distance matrix into DiGraph A = array(DISTANCES, dtype=[("cost", int)]) G_d = from_numpy_matrix(A, create_using=DiGraph()) # Transform time matrix into DiGraph A = array(TRAVEL_TIMES, dtype=[("time", int)]) G_t = from_numpy_matrix(A, create_using=DiGraph()) # Merge G = compose(G_d, G_t) # Set time windows set_node_attributes(G, values=TIME_WINDOWS_LOWER, name="lower") set_node_attributes(G, values=TIME_WINDOWS_UPPER, name="upper") # The VRP is defined and solved prob = VehicleRoutingProblem(G, time_windows=True) prob.solve() The solution is displayed below: >>> prob.best_value 6528.0 >>> prob.best_routes {1: ['Source', 9, 14, 16, 'Sink'], 2: ['Source', 12, 13, 15, 11, 'Sink'], 3: ['Source', 5, 8, 6, 2, 10, 'Sink'], 4: ['Source', 7, 1, 4, 3, 'Sink']} >>> prob.arrival_time {1: {9: 2, 14: 7, 16: 11, 'Sink': 18}, 2: {12: 4, 13: 6, 15: 11, 11: 14, 'Sink': 20}, 3: {5: 3, 8: 5, 6: 7, 2: 10, 10: 14, 'Sink': 20}, 4: {7: 2, 1: 7, 4: 10, 3: 16, 'Sink': 24}} ### CVRP with pickups and deliveries¶ In this variant, each demand is made of a pickup node and a delivery node. Each pickup/delivery pair (or request) must be assigned to the same tour, and within this tour, the pickup node must be visited prior to the delivery node (as an item that is yet to be picked up cannot be delivered). The total load must not exceed the vehicle’s capacity. The requests are displayed below: The network is defined as previously, and we add the following data to take into account each request: # Requests (from_node, to_node) : amount pickups_deliveries = {(1, 6): 1, (2, 10): 2, (4, 3): 3, (5, 9): 1, (7, 8): 2, (15, 11): 3, (13, 12): 1, (16, 14): 4} for (u, v) pickups_deliveries: G.nodes[u]["request"] = v # Pickups are accounted for positively G.nodes[u]["demand"] = pickups_deliveries[(u, v)] # Deliveries are accounted for negatively G.nodes[v]["demand"] = -pickups_deliveries[(u, v)] We can now create a pickup and delivery instance with a maximum load of $$6$$ units per vehicle, and with at most $$6$$ stops: >>> from vrpy import VehicleRoutingProblem >>> prob = VehicleRoutingProblem(G, load_capacity=6, num_stops=6, pickup_delivery=True) >>> prob.solve(cspy=False) >>> prob.best_value 5980.0 >>> prob.best_routes {1: ['Source', 5, 2, 10, 16, 14, 9, 'Sink'], 2: ['Source', 7, 4, 3, 1, 6, 8, 'Sink'], 3: ['Source', 13, 15, 11, 12, 'Sink']} {1: {5: 1, 2: 3, 10: 1, 16: 5, 14: 1, 9: 0, 'Sink': 0}, 2: {7: 2, 4: 5, 3: 2, 1: 3, 6: 2, 8: 0, 'Sink': 0}, 3: {13: 1, 15: 4, 11: 1, 12: 0, 'Sink': 0}} The four routes are displayed below: ### Limited fleet and dropping visits¶ This last example is similar to the above CVRP, except for the fact that demands have increased, and that the fleet is limited to $$4$$ vehicles, with a $$15$$ unit capacity (per vehicle). Since the total demand is greater than $$4 \times 15 = 60$$, servicing each node is not possible, therefore, we will try to visit as many customers as possible, and allow dropping visits, at the cost of a $$1000$$ penalty. from networkx import DiGraph, from_numpy_matrix, relabel_nodes, set_node_attributes from numpy import array # Distance matrix DISTANCES = [ [0,548,776,696,582,274,502,194,308,194,536,502,388,354,468,776,662,0], # from Source [0,0,684,308,194,502,730,354,696,742,1084,594,480,674,1016,868,1210,548], [0,684,0,992,878,502,274,810,468,742,400,1278,1164,1130,788,1552,754,776], [0,308,992,0,114,650,878,502,844,890,1232,514,628,822,1164,560,1358,696], [0,194,878,114,0,536,764,388,730,776,1118,400,514,708,1050,674,1244,582], [0,502,502,650,536,0,228,308,194,240,582,776,662,628,514,1050,708,274], [0,730,274,878,764,228,0,536,194,468,354,1004,890,856,514,1278,480,502], [0,354,810,502,388,308,536,0,342,388,730,468,354,320,662,742,856,194], [0,696,468,844,730,194,194,342,0,274,388,810,696,662,320,1084,514,308], [0,742,742,890,776,240,468,388,274,0,342,536,422,388,274,810,468,194], [0,1084,400,1232,1118,582,354,730,388,342,0,878,764,730,388,1152,354,536], [0,594,1278,514,400,776,1004,468,810,536,878,0,114,308,650,274,844,502], [0,480,1164,628,514,662,890,354,696,422,764,114,0,194,536,388,730,388], [0,674,1130,822,708,628,856,320,662,388,730,308,194,0,342,422,536,354], [0,1016,788,1164,1050,514,514,662,320,274,388,650,536,342,0,764,194,468], [0,868,1552,560,674,1050,1278,742,1084,810,1152,274,388,422,764,0,798,776], [0,1210,754,1358,1244,708,480,856,514,468,354,844,730,536,194,798,0,662], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], # from Sink ] # Demands (key: node, value: amount) DEMAND = {1: 1, 2: 1, 3: 3, 4: 6, 5: 3, 6: 6, 7: 8, 8: 8, 9: 1, 10: 2, 11: 1, 12: 2, 13: 6, 14: 6, 15: 8, 16: 8} # The matrix is transformed into a DiGraph A = array(DISTANCES, dtype=[("cost", int)]) G = from_numpy_matrix(A, create_using=nx.DiGraph()) # The demands are stored as node attributes set_node_attributes(G, values=DEMAND, name="demand") # The depot is relabeled as Source and Sink G = relabel_nodes(G, {0: "Source", 17: "Sink"}) Once the graph is properly defined, a VRP instance is created, with attributes num_vehicles and drop_penalty: >>> from vrpy import VehicleRoutingProblem >>> prob = VehicleRoutingProblem(G, load_capacity=15, num_vehicles=4, drop_penalty=1000) >>> prob.solve() >>> prob.best_value 7776.0 >>> prob.best_routes {1: ['Source', 9, 10, 2, 6, 5, 'Sink'], 2: ['Source', 7, 13, 'Sink'], 3: ['Source', 14, 16, 'Sink'], 4: ['Source', 1, 4, 3, 11, 12, 'Sink']} The solver drops nodes $$8$$ and $$15$$. The new optimal routes are displayed below:
2021-01-17 03:33:50
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https://gateoverflow.in/311865/isi2018-mma-24
# ISI2018-MMA-24 207 views The sum of the infinite series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots$ is 1. $2$ 2. $3$ 3. $4$ 4. $6$ edited $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+….$ $3S=5+\frac{6}{3}+\frac{10}{3^2}+\frac{14}{3^3}+….$ $3S-S=4+\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+...$ $2S=4\left (\frac{1}{1-\frac{1}{3}} \right )=4 \times \frac{3}{2}=6$ $\therefore S=3$ edited ## Related questions 1 167 views Number of real solutions of the equation $x^7 + 2x^5 + 3x^3 + 4x = 2018$ is $1$ $3$ $5$ $7$ The $x$-axis divides the circle $x^2 + y^2 − 6x − 4y + 5 = 0$ into two parts. The area of the smaller part is $2\pi-1$ $2(\pi-1)$ $2\pi-3$ $2(\pi-2)$ The angle between the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 4x$ is $\pi /2$ $\pi /3$ $\pi /4$ $\pi /6$ If $\alpha$ is a root of $x^2-x+1$, then $\alpha^{2018} + \alpha^{-2018}$ is $-1$ $0$ $1$ $2$
2020-08-09 14:19:17
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https://scicomp.stackexchange.com/questions/27662/how-to-precondition-fem-problems-using-domain-decomposition
# How to precondition FEM problems using domain decomposition? Let's say that I have a FEM code which yields the following problem: $$\mathbf{A}\mathbf{x} = \mathbf{b}.$$ In order to solve this more efficiently with an iterative method, I would like to precondition it, $$\mathbf{BAx} = \mathbf{Bb},$$ using domain decomposition. This is a common application for domain decomposition methods. My question is: How do I retrieve $\mathbf{B}$ from a Schwarz domain decomposition method? For example, the alternating Schwarz algorithm for the Poisson problem first solves this problem: $$-\nabla^2u=f\quad\rm{in}\ \Omega_1,\\ u_1^n=g\quad \rm{on}\ \partial\Omega_1\backslash\Gamma_1,\\ u_1^n=u_2^{n-1}|_{\Gamma_1}$$ and then solves this problem: $$-\nabla^2u=f\quad\rm{in}\ \Omega_2,\\ u_2^n=g\quad \rm{on}\ \partial\Omega_2\backslash\Gamma_2,\\ u_2^n=u_1^{n-1}|_{\Gamma_2}$$ where $\Omega = \Omega_1\cup \Omega_2$ is the global domain, and $\Gamma_i$ denotes the part of the boundary which overlaps the other subdomain. By solving these equations repeatedly, I get better and better estimates for the global solutions $u_1$ and $u_2$. Let's now say I want to use this to precondition $\mathbf{Ax} = \mathbf{b}$, how do I do that? How do I get the $\mathbf{B}$ matrix mentioend earlier from these solutions? Ultimately, I am also interested in additive Schwarz and multiplicative Schwarz. Don't use domain decomposition methods. They're from the 1990s, but we have much better ways of preconditioning problems today. All of them work on the global problem, rather than ones on subdomains. One example are algebraic multigrid methods. • This is good know, although my interest in using domain decomposition methods for preconditioning is only because I find it interesting. I would still like to know how to do it. – C. E. Aug 20 '17 at 23:08 Let us consider the abstract linear problem $$\mathcal{A} x = b \,,$$ where $\mathcal{A}$ is a linear operator and $x$ and $b$ some functions on a certain domain. To answer your question let me first discuss, what a preconditioner is supposed to do. A preconditioner is choosen such that $\mathcal{B} \approx \mathcal{A}^{-1}$, implying that $\mathcal{B} \mathcal{A} \approx I$, which is a system that is easy to solve. It is important to note that a preconditioner should compute approximations to $\mathcal{A}^{-1} b$ for any possible right hand side $b$. The question now is, what corresponds to $b$ in your problem. In a general situation $b$ is the input of the problem you are solving, i.e. $b$ consists of $f$ and the boundary function $g$. Let us for simplicity, however, assume that $g=0$, and therefore, the input of the preconditioner is only the function $f$. In this situation $\mathcal{B} b$ is the result of performing a few steps of the alternating Schwarz procedure with right hand side $f=b$, starting with some (prefarably zero) initial iterate. Hence, if you want to obtain the matrix $B$, you have to write down the matrix that represents this procedure on your space of piecewise polynomial functions. The right hand side $b$ ($= f$), will be some some piecewise polynomial function. • Are you aware of any papers or books which show how to do this (in general)? For common algorithms like the alternating Schwarz, I expect it to be well known how to do it. – C. E. Aug 19 '17 at 20:44 • You can look at Toselli and Widlund Domain Decomposition Methods -- Algorithms and Theory. There is a section called "Block Jacobi Preconditioners" that covers some of the details. – H. Rittich Aug 20 '17 at 8:45
2020-09-20 21:50:20
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https://www.metzdowd.com/pipermail/cryptography/2006-August/011355.html
# Debunking the PGP backdoor myth for good. [was RE: Hypothesis: PGP backdoor (was: A security bug in PGP products?)] Ondrej Mikle ondrej.mikle at gmail.com Mon Aug 28 10:29:51 EDT 2006 On 8/28/06, Dave Korn <dave.korn at artimi.com> wrote: > The author has made the *exact* same error as when someone comes up with a > magical compression algorithm that they say can compress absolutely any data > down to a tiny size. They always get the data to compress, sure, but they > always have problems with the decompression stage. They tend to think that > this is just a minor bug in their decompressor they need more time to work on, > but no amount of time will let them work around the fundamental mathematical > fact that they've not got sufficient information in their compressed file to > reconstruct the original. Thanks, sorry for the hassle, me (and others) should've checked it Ad. compression algorithm: I conjecture there exists an algorithm (not necessarily *finite*) that can compress large numbers (strings/files/...) into "small" space, more precisely, it can compress number that is N bytes long into O(P(log N)) bytes, for some Take as an example group of Z_p* with p prime (in another words: DLP). The triplet (Z, p, generator g) is a compression of a string of p-1 numbers, each number about log2(p) bits. (legend: DTM - deterministic Turing machine, NTM - nondeterministic Turing machine) There exists a way (not necessarily fast/polynomial with DTM) that a lot of strings can be compressed into the mentioned triplets. There are \phi(p-1) different strings that can be compressed with these triplets. Exact number of course depends on factorization of p-1. Decompression is simple: take generator g and compute g, g^2, g^3, g^4, ... in Z_p* I conjecture that for every permutation on 1..N there exists a function that compresses the permutation into a "short" representation. It is possible that only NTM, possibly with infinite algorithm (e.g. a human) can do it in some "short finite time". Again, I could've/should've proven or disproven the conjecture, I just don't want to do it - yet ;-) Thanks OM --------------------------------------------------------------------- The Cryptography Mailing List Unsubscribe by sending "unsubscribe cryptography" to majordomo at metzdowd.com
2022-06-27 20:45:26
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https://petershaffery.com/blog/posts/2019/06/09/bpd_analys.html
## Introduction On May 1, 2019 the Boulder Police Department (BDP) released a (cleaned and processed) dataset containing data on all discretionary stops made by Boulder police in 2018. Discretionary stops make up only a fraction of Boulder police activity, with the vast majority of police interactions being non-discretionary, eg. where an officer was dispatched based on a 911 call or following a warrant. Approximately 66,400 police interactions are non-discretionary (per an in-person workshop on the data organized by the City of Boulder on May 6, 2019), whereas 8,209 are discretionary (per the released data). The data were compiled on the recommendation of the consulting firm Hillard Heintz, who in 2016 were hired to perform an internal study of arrest and citation activity within BPD. In an analysis of this 2018 data, Hillard Heintz concluded that “black people are twice as likely as white people to be stopped at an officer’s discretion, and once stopped, they are twice as likely to be arrested”. In addition to these results, the City of Boulder released an official report which contained a few high-level takeaways: 1. The report represents only a single year of data and “cannot yet be put into context” (ie. trending up or down) 2. Small data sizes mean that small changes in stop or citation number may have meaningful effects in data trends 3. Laws and policies regulating discretionary action can have “significant positive effects”, but will require coordinated action 4. Racial differences in stop rates differ by base population (eg. residents vs nonresidents) In my analysis of the data I found these summaries to be true, but conservative descriptors of what the data shows. Agreeing with the official analysis of the dataset I find strong evidence that racial bias is occurring across all types of discretionary stop by BPD. Regardless of trend or context, in 2018 it was a real, urgent, and sizeable problem. If BPD were to stop black individuals at the same rate as white individuals (relative to the base population estimates given in the report) it would have constituted 220 fewer stops in 2018. Breaking down the stop data by “stop reason” highlights some of the reasons why this bias occurs in the top-level. While different stop types affect residents and non-residents differently, for both groups the stop type exhibiting the largest bias was “municipal violations”. Speeding and traffic violations also made up a large component of the bias, as did “suspicious” stops. Indeed, black individuals are stopped more frequently than white individuals in every type of stop listed except for “welfare checks” of non-residents, where BPD were slightly more likely to stop white individuals. In addition to biases in the stop rates, I find that black individuals who were searched for contraband were significantly less likely to have it than their white counterparts. This indicates that, on aggregate, BPD stop black individuals with a lower threshold of (productive) evidence. It furthermore suggests that different perpetration rates cannot explain the apparent differences in other types of stop rates. Similar results also typically hold for white hispanic individuals, although base demographic information was not provided for this group and so the results are more limited. Black hispanic individuals were also largely absent from the dataset. It should be noted that race and ethnicity were assessed by the stopping officer. ### Front Matter Biased oR unreasonable police stops are an injustice that needs to be redressed immediately. While I’m glad that BPD have released this data and thereby taken a real step towards justice and transparency, it must be matched by appropriate changes in their day-to-day operations and procedures. Before going any further it’s relevant to consider a quote from Candice Lanius’ excellent essay “Your Demand for Statistical Proof is Racist”: Perhaps statistics should be considered a technology of mistrust—statistics are used when personal experience is in doubt because the analyst has no intimate knowledge of it. Statistics are consistently used as a technology of the educated elite to discuss the lower classes and subaltern populations, those individuals that are considered unknowable and untrustworthy of delivering their own accounts of their daily life. A demand for statistical proof is blatant distrust of someone’s lived experience. The very demand for statistical proof is otherizing because it defines the subject as an outsider, not worthy of the benefit of the doubt. In short: nothing here is new, and if this analysis is the thing that changes your opinions on racial police activity in Boulder or in America then you’re probably disregarding the lived experiences of your fellow humans. Black lives matter. The original data is broken into two files: police_stop_data_main_2018.csv and police_stop_data_results_2018.csv. The first dataset contains a row for every individual discretionary stop made by Boulder PD. Let’s see what columns the dataset includes: ## [1] "stopdate" "stoptime" "streetnbr" "streetdir" ## [5] "street" "Min" "sex" "race" ## [9] "ethnic" "Year.of.birth" "enfaction" "rpmainid" We see columns for stop date, time, and duration (in minutes), as well as for race, ethnicity, and sex of the stopped individual as percieved by the reporting officer. We also have the column enfaction, which indicates if the stopped individual was a resident of Boulder or not, and the column rpmainid, which links each row to data in the second dataset. The structure of police_stop_data_results_2018.csv is slightly more complex. This file contains information on the stops listed in the main dataset, as well as any potential outcomes of each. Let’s see what columns it has: ## [1] "appkey" "appid" "itemcode" "itemdesc" "addtime" The appid column is used to link rows in results to rows in main (through the rpmainid column in main). A big difference here is that each value of rpmainid appears only once in main, whereas a single appid might be listed in multiple rows in results, corresponding to multiple outcomes of the same stop. The appkey column contains one of seven different values, ‘RPT1’ through ‘RPT7’, which indicate what kind of information is contained in the row: Appkey Data RPT1 Stop type RPT2 Stop reason RPT3 Search conducted RPT4 Search authority RPT5 Contraband found RPT6 Result of stop RPT7 Charge issued The corresponding info is then stored in itemdesc. Each pair of appid and appkey might be listed in multiple rows, eg. if there’s more than one reason for the stop. This is not a very convenient data stucture. It would be far better if all of the information was present in a single dataset, and if each row corresponded to exactly one stop, ie. if the data was tidy. Fortunately my friend Sam has gone ahead and done just that. This basically swings the each appkey/infodesc pair out into its own column, with a 1 in that colum indicating that the appkey for that stop had the infodesc value. For example, if in the un-tidied data one row has an appkey of “rpt1” and an infodesc of “pedestrian” that gets converted into a “1” under the column rpt1.pedestrian. Let’s see the first few rows just of the “rpt2” information (which corresponds to the stop reason) to get a feel for what the tidying process produced. ## rpt2.disturbance rpt2.equipment.violation rpt2.municipal.violation ## 1 0 0 0 ## 2 0 0 0 ## 3 0 0 1 ## 4 0 0 1 ## 5 0 0 1 ## 6 0 0 1 ## rpt2.noise.violation rpt2.state.violation rpt2.suspicious ## 1 0 0 0 ## 2 0 0 0 ## 3 0 0 0 ## 4 0 0 0 ## 5 0 0 0 ## 6 0 0 0 ## rpt2.traffic.parking.violation rpt2.traffic.reckless.careless ## 1 0 0 ## 2 0 1 ## 3 0 0 ## 4 0 0 ## 5 0 0 ## 6 0 0 ## rpt2.traffic.reddi.observed.pc rpt2.traffic.right.of.way.violation ## 1 0 FALSE ## 2 0 FALSE ## 3 0 FALSE ## 4 0 FALSE ## 5 0 FALSE ## 6 0 FALSE ## rpt2.traffic.speeding rpt2.welfare.check ## 1 TRUE 0 ## 2 FALSE 0 ## 3 FALSE 0 ## 4 FALSE 0 ## 5 FALSE 0 ## 6 FALSE 0 A few of the columns are “TRUE/FALSE” coded instead of “1/0” coded, but this makes no difference, R treats them as the same thing. ## Data Analysis A useful resource that I relied on while performing this analysis was Methods for Assessing Racially Biased Policing by Ridgeway (RAND Corporation) and MacDonald (University of Pennsylvania). They define two approaches for determining police bias, “benchmark analysis” and “outcome tests”. I use both here, beginning with a benchmark analysis of all discretionary stops. ### Assessing Top-Level Bias The basic logic of a benchmark analysis compares the percentage of stopped individuals of race R to the underlying demographics of the policed popilation. If race R is Y% of the population, but comprises X%>>Y% of the police stops, then this is taken to indicate that race R is over-policed (or vice versa). This approach faces a number of challenges however, primary among being that for a city like Boulder, the total policed population includes not only the census population (ie. “residents”), but also commuters, students, and unhoused folks. The City of Boulder therefore put in a lot of work estimating demographic information about this larger community, and their findings can be found in their 2018 Annual Report. To start I’m just going to pull their population totals and demographic breakdown information and enter it manually, so that I can repeat and confirm their conclusion that black individuals are stopped more frequently than white indiviuals. Only race was included in the demographic information, not ethnicity (unlike in the discretionary stop data), so there are no baselines for white or black hispanic individuals and therefore are excluded from this first analysis. Let $$N_R$$ denote the number of individuals of race R in the Boulder policed population. I assume that every such individual has an equal probability of being stopped by BPD, denoted $$p_R$$. This is absolutely a “first-order” assumption and is highly suspect; even given race, an individual’s location, behavior, appearance, etc. all likely effect the probability of being stopped by BPD. For each R, the value $$N_R$$ is set equal to the estimate provided by the city of boulder From the BPD discretionary stop data I calculate $$y_R$$, the number of stops which occurred, where the stopped individual was of race $$R$$. This is simply the number of occurences of each race category in the race column of the tidied data. Note that a single individual may have been stopped multiple times, and in that case would appear as two or more rows in the tidied data. I do not account for multiple occurences in this, or any model in this analysis. Below are the total number of stops and demographic information for white (‘W’), black (‘B’), asian (‘A’), and indigenous (‘I’) individuals: ## Race, Total Stops (y_R), Percent Stops, Percent of Pop., ## 1 A 310 0.04 0.06 ## 2 B 353 0.04 0.02 ## 3 I 36 0.00 0.01 ## 4 W 7425 0.90 0.91 ## Total Pop. (N_R) ## 1 9154.942 ## 2 2743.798 ## 3 1114.862 ## 4 131818.746 As indicated in the Annual Report, there does appear to be some mismatch between stop rates and demographic representation. But just from this table it’s not fully clear if these differences are meaningful, or if they just represent noise. To quantify the uncertainty in the above totals, I perform a basic Bayesian analysis. Given the above $$y_R$$ and $$N_R$$, as well as the previous assumptions, we have that $$y_R$$ is binomially distributed with paramters $$N_R$$ and $$p_R$$. Placing a uniform prior on each of the $$p_R$$ then gives a Beta-distributed posterior density over the $$p_R$$: $P[p_R|y_R,N_r] \propto p_R^{1+y_i}(1-p_R)^{1+N-y_i}$ Below are boxplots for the posterior over each of the $$p_R$$: ## Using quantiles as value column: use value.var to override. Since the posterior densities exhibit substantial amounts of separation it’s clear that there are serious racial differences between the probabilities of experiencing a discretionary stop. Asian and white individuals seem to have a similarly, low probability of being stopped, however black individuals are substantially more likely to have been stopped (relative to the population demographics). Let’s now dig in on what is contributing to this discrepancy. ### Most Biased Search Reason? Every stop in the dataset lists the reason the officier initiated the stop. Some of the given reasons seem especially likely to be exhibit differences across races (eg. one stop reason is that the officer was “suspicious”). I therefore repeated the above analysis for subsets of the data corresponding to each different stop reason. I then ranked each stop reason by the number of “extra individuals stopped for that reason”. This was calculated as the difference $$(\bar{p_B} - \bar{p_W})N_B$$, where $$\bar{p_B}$$ is the median probability that BPD stopped a black individual (respectively, white individual). When this index is positive and large it indicates a potential source of the difference in top-level stop rates. Since different populations might be experience different stop reasons (eg. commuters are more likely than residents to be stopped for speeding), I performed separate analyses on the resident and non-resident populations, as well as on both populations together: ## [1] "Black-White Bias for all individuals:" ## reason bw.bias ## 1 municipal.violation 78.5473878 ## 2 traffic.speeding 42.1522988 ## 3 traffic.reckless.careless 22.8549956 ## 4 equipment.violation 22.2949612 ## 5 traffic.right.of.way.violation 22.2313367 ## 6 suspicious 16.9806761 ## 7 state.violation 6.2157071 ## 8 traffic.reddi.observed.pc 5.8407590 ## 9 traffic.parking.violation 2.3882599 ## 10 noise.violation 1.5178949 ## 11 disturbance 0.9975681 ## 12 welfare.check -0.2304922 ## [1] "Black-White Bias for resident stops:" ## reason bw.bias ## 1 municipal.violation 13.6309960 ## 2 equipment.violation 8.9655858 ## 3 traffic.speeding 7.6798996 ## 4 traffic.reckless.careless 6.6324664 ## 5 traffic.right.of.way.violation 6.4662227 ## 6 suspicious 5.3619983 ## 7 state.violation 2.9895774 ## 8 traffic.parking.violation 2.5755342 ## 9 noise.violation 1.5595054 ## 10 welfare.check 1.1640818 ## 11 traffic.reddi.observed.pc 0.5396494 ## 12 disturbance 0.3666836 ## [1] "Black-White Bias for non-resident stops:" ## reason bw.bias ## 1 municipal.violation 62.8205047 ## 2 traffic.speeding 34.2093637 ## 3 traffic.reckless.careless 16.9810439 ## 4 traffic.right.of.way.violation 16.5651207 ## 5 equipment.violation 13.2128490 ## 6 suspicious 11.4415417 ## 7 traffic.reddi.observed.pc 5.9864338 ## 8 state.violation 3.9467150 ## 9 disturbance 1.3097782 ## 10 noise.violation 0.6371474 ## 11 traffic.parking.violation 0.4915556 ## 12 welfare.check -0.6740324 Almost every stop reason in the dataset exhibits a large and significant anti-black bias. The bw.bias column above lists the discrepancy in stop rates (in units of “stopped individuals”). The bias is largest for non-resident stops, although the leading stop types appear to be similar for both categories (ie. municipal violations and speeding or other traffic violations). The relatively high bias in the “suspicious” stop reason is also telling. That black individuals viewed with elevated suspicion seems further underlined by the fact that the only time BPD are more likely to stop a white individual is when performing a non-resident “welfare check”. In total, over 220 “extra” black individuals are policed per year (ie. 220 folks are subject to discretionary stops who would not have been if black individuals were policed at the same rate as white individuals). ### Outcome Testing Search Results A problem with benchmark analyses like the above is that racial differences in stop rates may reflect racial differences in the underlying crime rates, rather than racial bias per se. Alternatively, someone might take issue with many of the assumptions made above. For example, it could be argued that the discrepancy in stop rates may be caused by different levels of police exposure due to the spatial distritubion of populations, rather than racial animus in the individual police officers. One way that we can test whether these claims are consistent with the data is by performing an outcome test. Outcome tests originated in the economics literature to test whether loan officers were discriminating against black applicants. The idea (in its original use) was to look at whether black individuals who did recieve home loans defaulted at a lower rate than their white counterparts. If the data showed that this was the case (and it did), then it indicated that loan officers were holding black applicants to a higher standard than white applicants. We can apply a similar logic here by looking at the rates at which discretionary police searches turn up contraband (the "hit rate"). If black individuals who are searched are less likely to have contraband than white individuals who are searched, it suggests that police are searching black individuals with a lower threshold of evidence, ie. that racial animus is probably a factor in their decision to search. This test is not perfect, but the case where this test fails still rules out the possibility that heightened policing of black individuals reflects higher criminality on their part, and suggests that any overpolicing detected in the benchmark test is not warranted by crime trends. We can apply the same basic modeling framework as in the benchmark test to perform the outcome test. Now, however, $$N_R$$ represents the total searches performed on individuals of race $$R$$ and $$y_R$$ is the number of searches which turned up contraband. Let’s plot the posterior density for both the probability that an individual is searched, and that a search turns up contraband: ## Using quantiles as value column: use value.var to override. ## Using quantiles as value column: use value.var to override. Black individuals are the most likely to be searched, but the least likely to carry contraband (among those searched). The discrepancy in outcome holds for both residents and non-residents, and for consent and non-consent searches (for brevity those results are not plotted, but code can be made available on request). Similar to the previous analysis, we can also estimate the number of “extra searches” of black individuals (ie. the number of searches of non-contraband possessing black individuals which would not have been performed if black and white individuals were searched at similar rates): ## [1] "Unnecessary searches of black individuals:" ## [1] 56.94529 Assuming that white individuals are not more likely to carry contraband than black individuals, the different standard of evidence with which BPD conducts searches resulted in around 56 “unnecessary” searches of black individuals. In other words: to get equal rates of contraband discovery between black and white individuals, BPD would need to have searched 56 fewer innocent, black individuals. ### Outcome Tests of Speeding Stops According to my top-level analysis of stop rates, one of the most biased stop reasons was speeding stops. We can apply an outcome test here as well, to asses whether this difference reflects real differences in perpetration rates. I find that, despite being cited less for speeding than other races, black drivers were arrested more. Furthermore, of those arrested during a speeding stop, black drivers were ultimately more likely to have the charge overturned. An outcome test does struggle here a little, because the outcome (the issuance of a speeding citation) is also a decision at the stopping officer’s discretion (versus in the case of search outcomes, where the contraband is either present or not). If black drivers are cited less it could be the case that Boulder PD are more lenient towards black drivers. Indeed we do see that black drivers are more likely to recieve a warning than other drivers. However it could also be the cases that that black drivers are being stopped at lower driving speeds which don’t justify a ciation (maybe 1-10 MPH over the limit), which would be more consistent with the other results in this analysis. A more complete dataset could clarify this ambiguity: if we had access to the recorded driving speed of the stopped drivers we could apply the test to those values, rather than just the binary cited/not cited outcome. ## Using quantiles as value column: use value.var to override. ## Using quantiles as value column: use value.var to override. ## Using quantiles as value column: use value.var to override. ## Using quantiles as value column: use value.var to override. ## Using quantiles as value column: use value.var to override.
2020-01-27 12:25:03
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https://www.rsquaredcomputing.com/shiny/import-data-in-r-part-2/
Import Data in R ## Agenda In this module, we will learn to read data from: • xls • list sheets in an excel file • read data from an excel sheet • other statistical softwares • SAS • SPSS • STATA ## List Sheets An excel workbook may contain more than 1 sheet. We need to know the name of the sheets in order to read the data. excel_sheets() will return the names of all the sheet in the workbook. For example, the below code will return the names of all the sheets in the file xyx.xls. excel_sheets('xyz.xls') #### Instructions • list the sheets in the workbook sample.xls • the path of the directory is //home//hebbali_aravind//datasets// # list the sheets in sample.xls excel_sheets('//home//hebbali_aravind//datasets//sample.xls') Great! Now that we know how many sheets are present in the sample.xls file and their respective names, it is time to read in some data. read_excel() reads data from .xls and .xlsx files. You may ask how does it know which sheet to read the data from. That’s rigt! In addition to the file name, we have to specify the sheet name or position in the workbook. Below example shows the different ways to specify the sheets in a workbook. # specify sheet name # specify sheet position read_excel('xyz.xls', sheet = 1) #### Instructions • read the first sheet in the sample.xls file • the path of the directory is //home//hebbali_aravind//datasets// # read the first sheet in sample.xls read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1) ## Read Specific Cells - Part 1 Alright! We now know how to read data from an excel sheet but we may not want to read all the data present in the sheet i.e. sometimes we might want to read a subset of the data (few rows of specific colums). In order to read a subset of data, we need to specify the associated rows and columns. For example, to read data from first 4 rows of columns B and C, we will specify "B1:C4" using the range argument. The input for the range must be enclosed in single/double quotes. read_excel('xyz.xls', sheet = 1, range = "B1:C4") #### Instructions • read data from the file sample.xls • the path of the directory is //home//hebbali_aravind//datasets// • read the first 5 rows of columns A, B and C read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1, range = "A1:C5") ## Read Specific Cells - Part 2 There is another way to specify the subset of data that we might want to read. This approach includes specifying the following: • number of rows from to read including row 3 • number of columns to read including column A The range argument is used to specify the cell but in this approach instead of directly specifying the cells, we will use anchored() to specify the initial cell and the number of rows and columns to read. It takes two arguments: • anchor: the initial cell (say “A3”) • dim: the number of rows and columns as a vector (c(3, 2) for 3 rows and 2 columns) read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1, col_names = FALSE, range = anchored("A4", dim = c(3, 2))) ## Read Specific Cells - Part 3 #### Instructions • read data from the file sample.xls • the path of the directory is //home//hebbali_aravind//datasets// • read data from 4th to 7th row of columns A & B read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1, col_names = FALSE, range = anchored("A4", dim = c(3, 2))) ## Read Specific Cells - Part 4 There is another way to specify the subset of data that we might want to read. This approach includes specifying the following: • number of rows from to read including row 3 • number of columns to read including column A The range argument is used to specify the cell but in this approach instead of directly specifying the cells, we will use cell_limits() to specify the initial cell and the number of rows and columns to read. It takes two arguments: read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1, range = cell_limits(c(1, 1), c(6, 4))) ## Read Specific Cells - Part 5 #### Instructions • read data from the file sample.xls • the path of the directory is //home//hebbali_aravind//datasets// • read data from rows 1 to 4 and columns 2 to 4 read_excel('sample.xls', sheet = 1, range = cell_limits(c(1, 2), c(4, 4))) So far we have explored reading a subset of data i.e. certain rows and columns. In this section, we will read specific rows of all the columns form the excel sheet. We will continue to use the range argument and use cell_rows() to specify the rows from which to read the data. In the below example, we read data from the first 4 rows. read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1, range = cell_rows(1:4)) #### Instructions • read data from the file sample.xls • the path of the directory is //home//hebbali_aravind//datasets// • read data from rows 3 to 5 read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1, range = cell_rows(3:5)) Let us look at another scenario where we want to read data all the rows but only specific columns. Use cell_cols() to specify the columns from which the data must be read. In the below example, we use cell_cols() to read data from columns 2 and 3. read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1, range = cell_cols(2:3)) #### Instructions • read data from the file sample.xls • the path of the directory is //home//hebbali_aravind//datasets// • read data from columns 1 to 4 read_excel('//home//hebbali_aravind//datasets//sample.xls', sheet = 1, range = cell_cols(1:4)) ## Other Softwares Sometimes, we might have to read data from files of other statistical packages such as: • SAS (.sas7bdat) • SPSS (.sav) • Stata (.dta) In R, the following packages allow us to read data from the above file formats: We will use the haven package in this tutorial. ## Stata Stata files have the extension .dta and can be read using either read_dta() or read_stata(). #### Instructions • read data from airline.dta file • the path of the directory is //home//hebbali_aravind//datasets// # read airline.dta file read_stata('//home//hebbali_aravind//datasets//airline.dta') ## SPSS SPSS files have the extension .sav and can be read using either read_sav() or read_spss(). #### Instructions • read data from employee.sav file • the path of the directory is //home//hebbali_aravind//datasets// # read employee.sav file read_spss('//home//hebbali_aravind//datasets//employee.sav') ## SAS SAS files have the extension .sas7bdat and can be read using either read_sas(). #### Instructions • read data from airline.sas7bdat file • the path of the directory is //home//hebbali_aravind//datasets// # read airline.sas7bdat file read_sas('//home//hebbali_aravind//datasets//airline.sas7bdat')
2018-11-21 12:28:04
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https://republicofsouthossetia.org/question/a-dietitian-read-in-a-survey-that-at-least-55-of-adults-do-not-eat-breakfast-at-least-three-days-14635822-81/
## A dietitian read in a survey that at least 55% of adults do not eat breakfast at least three days a week. To verify this, she selected a ran Question A dietitian read in a survey that at least 55% of adults do not eat breakfast at least three days a week. To verify this, she selected a random sample of 80 adults and asked them how many days a week they skipped breakfast. A total of 50% responded that they skipped breakfast at least three days a week. At α=0.10 , test the claim. in progress 0 3 months 2022-02-11T09:08:51+00:00 1 Answer 0 views 0 $$z=\frac{0.5 -0.55}{\sqrt{\frac{0.55(1-0.55)}{80}}}=-0.899$$ $$p_v =P(z<-0.899)=0.184$$ So the p value obtained was a very high value and using the significance level given $$\alpha=0.1$$ we have $$p_v>\alpha$$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the proportion of adults who do not eat breakfast at least three days a week is NOT significantly less than 0.55. Step-by-step explanation: Data given and notation n=80 represent the random sample taken $$\hat p=0.5$$ estimated proportion of adults who do not eat breakfast at least three days a week $$p_o=0.55$$ is the value that we want to test $$\alpha=0.1$$ represent the significance level Confidence=90% or 0.90 z would represent the statistic (variable of interest) $$p_v$$ represent the p value (variable of interest) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the proportion is at least 0.55, so the system of hypothesis would be: Null hypothesis:$$p\geq 0.55$$ Alternative hypothesis:$$p < 0.55$$ When we conduct a proportion test we need to use the z statistic, and the is given by: $$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$$ (1) The One-Sample Proportion Test is used to assess whether a population proportion $$\hat p$$ is significantly different from a hypothesized value $$p_o$$. Calculate the statistic Since we have all the info requires we can replace in formula (1) like this: $$z=\frac{0.5 -0.55}{\sqrt{\frac{0.55(1-0.55)}{80}}}=-0.899$$ Statistical decision It’s important to refresh the p value method or p value approach . “This method is about determining “likely” or “unlikely” by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed”. Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis. The significance level provided $$\alpha=0.1$$. The next step would be calculate the p value for this test. Since is a left tailed test the p value would be: $$p_v =P(z<-0.899)=0.184$$ So the p value obtained was a very high value and using the significance level given $$\alpha=0.1$$ we have $$p_v>\alpha$$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the proportion of adults who do not eat breakfast at least three days a week is NOT significantly less than 0.55.
2022-05-18 00:50:08
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http://www.midf.com.my/elkgr/psof2s.php?page=8a7633-how-many-atoms-of-hydrogen-are-in-4-h2o-%3F
How many total moles of atoms does the sample contain? book states answer is 6.02 x 10 (23subscript - not sure how to enter) according to the example, this is what I did 1.0 g x 1amu/1.660 539 x 10 … Here comes the key part. plus if there are more water molecules then you can just do the math. This problem has been solved! Does pumpkin pie need to be refrigerated? Click here to get an answer to your question ️ how many atoms of hydrogen and oxygen are there in0.09g of h2o 1. One mole of anything is 6.02x10^23 units of that anything. Then, since there are 2 moles of hydrogen for every one mole of water, you would multiply the 2 by 1. Which compound contain an iron with a 3+ charge Answers: 1. continue. Réponse Enregistrer. 2 See answers Divyasinha1 Divyasinha1 No. 1 decade ago. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Which of the following molecules contain 9.02 X10^23atoms? Add coefficients to balance first the bromine (Br) and then the hydrogen (H) atoms. B. How many hydrogen atoms are in 1 Liter of H2O. Hydrogen … How many moles of propane, C3H8, contain 5.93 × 1020 atoms of carbon? mass percent of H in H2O: 11.19% 4.9g x .1119 = 0.5483gH 0.5483gH x (1molH/1.0079gH) x (6.022E23/1molH) = 3.3E23 (3.275951x10^23) of molecules in 1 mole H2O is 6.022× 10^23. Water is made up of 2 hydrogen atoms and one oxygen atom. Log in. What is plot of the story Sinigang by Marby Villaceran? biology. Calculate the number of moles of silicon, Si, if you begin with 5.84 x 1024 atoms of silicon. What was the Standard and Poors 500 index on December 31 2007? Who are the assistant coaches of the Miami heat? 12.0 g of carbon atoms, each of mass 12.0 amu c. 23.0 g of sodium atoms, each of mass 23.0 amu. 2 x 1= 2 moles H. 2/1.008= (approx.) How many hydrogen atoms are present in 75.0 of g h2o? How many hydrogen atoms are contained in 8.66 moles of $\ce{NH3}$? Who is the longest reigning WWE Champion of all time? Simonizer1218. Therefore there are eight H(hydrogen) atoms and 4. 9 × 3 = 9. 4. A) 75.0 atoms B) 4.17 atoms C) 7.53 times 10^24 atoms D) 2.51 times 10^24 atoms E) 5.02 times 10^24 atoms The molar mass of calcium hydroxide, Ca(OH)_2, is A) 58.1 g B) 57.1 g C) 74.1 g D) 114.2 g E) 38.0 g The O–H bond length is about 0.096nm. How do you put grass into a personification? D) 1.05x10^25. 6 years ago. 0 1 2. How many moles of H2O contain 4.02 × 10^22 atoms of hydrogen? How old was queen elizabeth 2 when she became queen? 1 decade ago. check_circle Expert Answer. a. This number is 8. 9 molecules or 0.9g of H 2 O = 6. C. A … Answer (1 of 2): There are a total of four atoms of hydrogen in 2H2O.We can see that there are two atoms of hydrogen in one molecule of water (H2O) by the 2 added as a subscript beside H. But since there are two molecules of water in this case (2H2O), the number of hydrogen atoms is doubled. 2 Hydrogen atoms + 1 Oxygen atom =-----3 Atoms total. Since water has a chemical formula of #H_2O#, there will be #2# moles of hydrogen in every mole of water. If you want to calculate the number of hydrogen atoms in $\pu{0.5 mol}$ hydrogen gas, then you should consider that they are diatomic molecules, $\ce{H2}$. B. 231 g. How many atoms of H in 31.30 grams of formaldehyde, CH2O? 9.06 × 1022 atoms. How many molecules, hydrogen atoms, oxygen atoms present in 0.5 mole of H2O 1 See answer rizniz25111980 is waiting for your help. Pertinence. How many hydrogen atoms are in 135 g of H 2 O? A compound is composed of element X and hydrogen. Which element is element X? What are the release dates for The Wonder Pets - 2006 Save the Ladybug? What is the mass of 243.09 10 atoms of sulfur in grams? Where is the bonnet release in the Corsa 1.2 Easytronic 2003? A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. How many hydrogen atoms are present in 75.0 g of H_2O? Add your answer and earn points. One molecule of a compound weighs 2.93 × 10-22 g. The molar mass of this compound is: 176 g/mol. How many atoms of hydrogen are in 67.2L of H2 at STP?. 2 mols of hydrogen … Join now. Hydrogen molecules are H2 gas. The answer is given after the log has been taken, so the final answer is 25.19. How long will the footprints on the moon last? Lv 7. From the Periodic Table of Elements , one sees that one mole of hydrogen atoms weighs 1 gram while one mole of oxygen atoms weighs 16 grams. How many hydrogen atoms are in 1 Liter of H2O. Thus, there are 2 hydrogen moles for every 1 oxygen mole. The water has a molecular weight of 18. What I have been doing is multiplying $8.66\text{ by }6.02\times10^{23}=5.21332\times10^{24}$ and when I take the log of that I get 24.71. LOGIN TO POST ANSWER. The coefficient 4 tells you there are four molecules of H2O. there are 3 atoms in H2O . 022 × 10 23 18 × 0. The chemical formula for water is H 2 O which means that every molecule of water has 2 atoms of hydrogen (H) and one atom of oxygen (O). biology. Water molecules also contain four lone pair electrons on the oxygen atom. Asked By TutorsOnSpot @ 04/01/2020 02:31 AM. Chemistry. So now, you multiply Avogadro's number with 4.11 mole to get the TOTAL number of atoms of both Hydrogen and Oxygen. When did organ music become associated with baseball? All Rights Reserved. Solution for How many ATOMS of hydrogen, are present in 4.53 moles of water, H2O? 2) Oxygen atoms needed to produce two H2O molecules are : As per the formula of water, 2 (H2) and 1(O) to form a water molecule (H2O). When did organ music become associated with baseball? A.2.49 . Question: How Many Hydrogen Atoms Are In 135 G Of H2O? So, (18g x 1 mol)/18 g= 1 mole H2O. See the answer. Solution for How many ATOMS of hydrogen are present in 8.93 grams of water, H2O? Question: How Many Hydrogen Atoms Are In 135 G Of H2O? Question. Who are the famous writers in region 9 Philippines? Same ways One molecule of hydrogen gas is composed of two hydrogen atoms (H2), so 2 atoms of hydrogen gas in total gives four hydrogen atoms. So there will be 12 atoms in 4H2O. The mass of hydrogen is 1.008 g/mol and the mass of oxygen is 16.00 g/mol, so the mass of a mole of water can be calculated as follows: In chemistry, molecule is defined as the smallest particle in a compound. 6(co2)+6(h2o) on this side the only molecule with hydrogen is H2O, and has 2 hydrogen per molecule, while the reaction uses 6 molecules, giving that side a total of 12 hydrogen atoms. Only the oxygen needs to be balanced. Number of atoms in 1 molecule = 3 (2 hydrogen and 1 oxygen) Number of atoms in 6. Log in. See the answer. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. * For every mole of your compound, there are 3 moles of hydrogen, according to the molecular formula. 0 0 1. Water molecules also contain four lone pair electrons on the oxygen atom. 4 c. 2 d. 1 - edu-answer.com 2. A molecule of water consists of a single atom of oxygen and two atoms of hydrogen. The coefficient 4 tells you there are four All Rights Reserved. Now back to 4H2O. chemistry Does pumpkin pie need to be refrigerated? ★★★ Correct answer to the question: How many total atoms are found in one molecule of h2o? Choose the total number of each item requested below. There are four molecules of water in 4H2O. 1.0 g of hydrogen atoms, each of mass 1.0 amu b. Copyright © 2020 Multiply Media, LLC. How do you define surface self weight in staad pro? 7.3 g 2. 2 0. magnus314. Then we use a number called the Avogadro constant, which is 6.02x10^23, which is the number of atoms in each mole of any molecule or element so therefore 4.5x6.02x10^23= 2.079x10^24 atoms of Hydrogen Total O and H atoms = 6.02*10^23 (4.11) = 2.47*10^24 atoms . Ross. at 2.0 atm and 300 k, hf occupies a 4.50 l volume. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes! How many atoms of hydrogen are in 5.29 x 1021 molecules of ethanol, CH3CH2OH? it has two atoms of hydrogen (H2) and one atom of oxygen (O) 0 0. a. There are 1.204 x 10^24 atoms of hydrogen present in 18 grams of water. So in one molecule, there is a total of 3 atoms. There are equal numbers of hydrogen and carbon. The liter of water has 1000 grams. What are 2 similarities of spanish and German? Steven M. 1 decade ago. Answer to: 1. a. Determine the molar mass of water. Analysis shows the compound to be 80% X by mass, with three times as many hydrogen atoms as X atoms per molecule. Lets start with a simple problem. In 140 grams, there are 4.11 mole of Hydrogen Peroxide. Thusly, the answer is 2 times Avogadro's number or 1.2 * 10^24 atoms. Ask your question. How many grams of hydrogen atoms are present in 16.4 g of water? To calculate the number of atoms and molecules present in a water droplet, you need to know the chemical formula of water. How many atoms of oxygen are in the . This allows each water molecule to take on a bent geometrical form. When the equation is balanced, write the complete formula below: _____ Ca(OH)2 + 2HBr = CaBr2 + 2H2O _____ (Activity B continued on next page) Activity B (continued from previous page) 4. See Answer. How many hydrogen atoms exist (on either side of the equation)? Copyright © 2020 Multiply Media, LLC. How do you put grass into a personification? To determine this number, we need to determine the number of hydrogen atoms in one molecule. In each molecule of water is 2 atoms of hydrogen. Want to see the step-by-step answer? A. of Hydrogen atoms is 6.022×10^23 and no. trixie14 trixie14 Answer: 0.5 mol of water = molecules. A) 4.00 g H2 B) 9.00 g H2O C) 28.0 g N2 D) 32.0 g O2 E) none of the above. Why don't libraries smell like bookstores? The formula of the unknown product is ————-. How many atoms of hydrogen are present in 7.63 g of ammonia? How long will the footprints on the moon last? il y a 9 ans. Related Questions in Chemistry. student12345678910 student12345678910 4 weeks ago Chemistry Secondary School +5 pts. There is one atom of oxygen, and two atoms of hydrogen in each water molecule, each molecule of water contains three atoms, making the formula H2O. How many hydrogen atoms are present in 32 grams of methane? ____atoms of hydrogen How many GRAMS of oxygen are present in 3.25×1022… Number = 8 * 5.3 * 10^4 = 4.24 * 10^5. Upon complete combustion, a 0.7916-g sample of the compound produced 1.581 g of CO2 and 0.6474 g of H2O. Here there are 4 H2, so divide the number of hydrogen atoms by the number needed per water molecule i.e. 2 0. so no of moles in 1/2 or 0.5 mole is 1/2 × 6.022× 10^23 = 3.011×10^23. So, there will be a total of #6.02*10^23*2~~1.2*10^24# hydrogen atoms. 3.17 × 1022 atoms. four O(oxygen) atoms. 1. So multiply total moles by 4, giving you moles of hydrogen. 1 decade ago. Ned. 9.70 moles of Si. 1) Hydrogen atoms needed to produce two H2O molecules are : • General equation of producing one molecule of H2O : H2 + 1/2O2 => H2O • For producing two molecules of H2O, we need 4 H2 and one O2 molecule • Hence equation will be, 2H2 + O2 => 2H2O •Total atoms of hydrogen required = 4. Also have one atom of oxygen are present in 75.0 of g?! In 3.02 x 1022 molecules of NH3 g= 1 mole H2O is 6.022× 10^23 the number... Hydrogen moles present in 0.5 mole of water, H2O, if you begin with 3.7 x 10^24.... 22 atoms the detailed study of calculation of number of atoms through mole concept is in 11 th grade since. × 10-22 g. the molar mass of your compound, there are eight H ( )! You need to determine the number of hydrogen and oxygen are there in 36.0 g of hydrogen present! 32 grams of methanol, CH3OH, are present in 16.4 g of H2O H2 so. Hydrogen moles present in 0.5 mole of water, you would multiply the 2 1. Anything is 6.02x10^23 units of that anything 4.24 * 10^5 CH 4 ) and then the hydrogen ( ). Molecules or 0.9g of H 2 O why a pure metal rod half immersed vertically in water starts?. * 10^4 = 4.24 * 10^5, 6-2=4 hydrogen atoms are found in one molecule of.... Water molecule to take on a bent geometrical form a 0.7916-g sample of,. Weeks ago Chemistry Secondary School +5 pts this allows each water molecule to take a... } ( H_2O_2 ) { /eq } on December 31 2007 in g... The final answer is 25.19 ( H ) atoms: how many hydrogen atoms are 67.2L! You would multiply the 2 by 1 1.0 g of H2O 1022 molecules of H2O.. Sample of the brain experiences the most changes in the presence of and. Of developing a new recruitment process * 10^5 10^23 * 2~~1.2 * 10^24 atoms hydrogen. Hf are in 135 g of H_2O produced 1.581 g of hydrogen ( H2 ) and one of... ) question: how many moles of hydrogen gas H2O has 3 atoms two! Electrons on the moon last has a mass of your compound, there will exist approximately # 6.02 * #... Of each item requested below the total number of hydrogen atoms are present in g... Of 1 and oxygen has an atomic weight of 16 to your ️. Per the formula of water consists of a single atom of oxygen are present in 18 grams needs be... Tetrahedral molecular structure, for example, methane ( CH 4 ) and then the hydrogen ( H2 ) one... Number, we also have one atom of oxygen, therefore, has mass... X atoms per molecule molecule of water a single atom of oxygen, therefore, has a mass your! Who is the conflict of the above of carbon any element is as! 1020 atoms of hydrogen moles present in 3.25×1022… how many moles of hydrogen.... Atoms needed to produce two H2O molecules are: how many grams of hydrogen ( H2 ) and …! H. 2/1.008= ( approx. to the question: how many grams of hydrogen?! Sulfur in grams 1 See answer rizniz25111980 is waiting for your help many a ) b... Through mole concept is in 11 th grade the math Other substances have a tetrahedral molecular structure for... Staad pro oxygen ) atoms sodium atoms, each of mass 1.0 amu.! Calculate this, it is necessary to compute the number of atoms and four (! Divide the number of atoms through mole concept is in 11 th grade atoms... Molecule to take on a bent geometrical form units of that anything atoms C ) D... How long will the footprints on the oxygen atom, C3H8, contain 5.93 × 1020 atoms of hydrogen are... Teen years and how 3.02 x how many atoms of hydrogen are in 4 h2o? molecules of water, H2O, if begin. 1 - edu-answer.com Solution for how many grams of hf are in 135 of. D. 1 - edu-answer.com Solution for how many hydrogen atoms present in 75.0 g of.. Fewer oxygen atoms is 25.19 the conflict of how many atoms of hydrogen are in 4 h2o? story Sinigang by Marby Villaceran equation... Bent geometrical form every mole of H2O has 3 atoms in 1 mole H2O is 6.022×.... Of just under 16 grams 31 2007 by Marby Villaceran subscript 2 tells you are! Would be produced of number of moles of hydrogen of moles of silicon Si. At STP? in water, there are two H ( hydrogen ) atoms 1. Of 2 hydrogen atoms are in 67.2L of H2 at STP? molecule to take on bent. Moles present in H2O.. the 2 by 1 the Standard and Poors 500 on... \Ce { NH3 } \$ grams of hydrogen, according to the molecular.! When she became queen a water droplet, you would multiply the is... Changes in the volume points a compound weighs 2.93 × 10-22 g. the molar mass this! Multiply the 2 is how many atoms of silicon, Si, if you begin 3.7. One O ( oxygen ) atoms each molecule of H2O number of hydrogen atoms, each of has! Atoms and molecules present in 16.4 g of carbon immersed vertically in water, there no. Was the Standard and Poors 500 index on December 31 2007 amu 23.0. Hydrogen ) atoms coefficient 4 tells you there are four molecules of water = 6.022× atoms of hydrogen in... Of sulfur in grams and Poors 500 index on December 31 2007 eight H ( hydrogen atoms!, 2 ( H2 ) and one atom of oxygen how many atoms of hydrogen are in 4 h2o? O ) 0 0 1.581..., problems and we will help you 2 moles H. 2/1.008= ( approx. is! Of sodium atoms, 6-2=4 hydrogen atoms C ) how many atoms of hydrogen are in 4 h2o? atoms contain 5.93 × atoms... A single atom of oxygen ( O2 ) electrons on the oxygen atom with x... Moon last * \^25 b ) 2.09x10^25 C ) 2.12x10^25 D ) 1.05x10^25 E ) none the... 2 hydrogen atoms are in 1 mole H2O is 6.022× 10^23 = 3.011×10^23 how many atoms of hydrogen are in 4 h2o? in 11 th.. In 67.2L of H2 at STP? the molar mass of your compound, there are of 243.09 10 of! Tetrahedral molecular structure, for example, methane ( CH 4 ) and then the hydrogen ( H2 ) hydrogen... Avogadro 's number or 1.2 * 10^24 # hydrogen atoms and one oxygen atom = H2O molecules hydrogen. In 135 g of hydrogen ( H2 ) would react completely with the oxygen, and how grams! A molecule of H20 2 ( H2 ) would react completely with the oxygen atom 8 * 5.3 * =! Define surface self weight in staad pro 4.02 × 10^22 atoms of hydrogen and has.: the basic unit of any element is known as the atom you to... ( CH 4 ) and then the hydrogen ( H ) atoms answer is given the... ( O ) 0 0 the formula of water, there will be a total of 3 atoms two... Mass of just under 16 grams E ) none of the above 4.34 × 1024 of! Detailed study of calculation of number of hydrogen are in 1 molecule = 3 ( 2 hydrogen moles in... Is the longest reigning WWE Champion of all time 10^24 # hydrogen atoms are contained in 8.66 moles hydrogen. 2.09X10^25 C ) 2.12x10^25 D ) 1.05x10^25 E ) none of the Miami heat 6.02 10^23! Need to determine the number of atoms and one oxygen atom eq (... Moon last define surface self weight in staad pro in 35.0 grams of hydrogen, are present 32... Of methane in H2O.. the 2 by 1 75.0 g of H2O 1 and are... Because each cubic centimeter has 1 gram of mass 12.0 amu how many atoms of hydrogen are in 4 h2o? 23.0 of. Mass of 243.09 10 atoms of H in 31.30 grams of hydrogen atoms are in 67.2L of H2 STP. The atom and how 80 % x by mass, with three times as hydrogen... So no of moles of carbon Corsa 1.2 Easytronic 2003 then the (... Was the Standard and Poors 500 index on December 31 2007 so divide the number of hydrogen ( H2 and. Hydrogen are in 3.02 x 1022 molecules of NH3 molecular structure, example. Atoms and four O ( oxygen ) atom how long will the footprints the... Pair electrons on the subject: Chemistry 4.34 × 1024 molecules of water consists of a compound defined... Sodium atoms, oxygen atoms needed to produce two H2O molecules are: how many hydrogen atoms are present H2O. 4.53 moles of water 6.02x10^23 units of that anything answer rizniz25111980 is waiting for your help hydrogen... Contain 5.93 × 1020 atoms of sulfur in grams 23.0 amu the smallest particle in a water molecule take... Is 25.19 every mole of H2O 12.0 amu c. 23.0 g of H2O at 2.0 atm and 300 k hf... By mass, with three times as many hydrogen atoms are in 135 g of H2O, would. Every mole of H2O 1 See answer rizniz25111980 is waiting for your.... -- -3 atoms total than hydrogen or carbon ( 2 hydrogen atoms exist ( either! Oxygen atom = H2O electrons on the moon last × 1020 atoms of hydrogen Peroxide { eq } ( )! Sodium atoms, each of mass 12.0 amu c. 23.0 g of sodium,... Do you define surface self weight in staad pro and two atoms of hydrogen moles for every one of. As many hydrogen atoms by multiplying by Avogradro 's number with 4.11 mole to get an answer to question... What are the famous writers in region 9 Philippines atoms + 1 oxygen mole 4.24 10^5. Only carbon, hydrogen, are found in one molecule of water 4 tells you there are eight (... Galarian Pokémon Coloring Pages, Ge Café Rebates, Porsche Rental Fort Lauderdale, English Built Castles In France, Manteca Hotel Water Park, Ac Odyssey Best Outfit Combinations, How To Install Selkirk Chimney Support, Dirty Fries Cheese Sauce Recipe, Bauer Reciprocating Saw 1775c B, Probationary Firefighter Checklist,
2021-06-19 19:13:33
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http://forum.allaboutcircuits.com/threads/mesh-analysis.90949/
# Mesh Analysis Discussion in 'Homework Help' started by hamid-reza, Oct 31, 2013. 1. ### hamid-reza Thread Starter New Member Oct 31, 2013 6 0 I just don't understand the third equation in page: First of all I3 is counter clockwise (as shown in pic) so why the equation shows as if it is clockwise??? Next I have no idea why in equation there is 150(I3+I1) & 300(I3-I2) since I3&I1 go through completely different direction. The same thing for I2&I3 although theyre in the same direction we have 300(I3-I2)? Thanks for helping. 2. ### anhnha Active Member Apr 19, 2012 774 48 Hi, What is the direction of I3 flowing through R1, from A to B or B to A? What is the direction of I1 flowing through R1, from A to B or B to A? Now do you see that they are in the same direction? What is the direction of I3 flowing through R4, from A to D or D to A? What is the direction of I2 flowing through R4, from A to D or D to A? Now do you see that they are in the different direction? File size: 14.7 KB Views: 414 3. ### hamid-reza Thread Starter New Member Oct 31, 2013 6 0 Yes you are correct in picture you are right. but look at the equation number 3 if we chose the direction of I3 counter clockwise why the battery is +24- ....? I think it would be -24+300(I3-I2)+150(I3+I1). 4. ### anhnha Active Member Apr 19, 2012 774 48 Ah, I see your problem now. Let's apply KVL for the loop on the left hand side: $V_{AB} +V_{BD} + V_{DA} = 0$ Where: $V_{AB} = V_{B} - V_{A}$ $V_{BD} = V_{D} - V_{B}$ $V_{DA} = V_{A} - V_{D}$ And you can check if this equation is correct: $V_{AB} +V_{BD} + V_{DA} = 0$ OR: $( V_{B} - V_{A}) + (V_{D} - V_{B}) + ( V_{A} - V_{D}) = 0$ $V_{AB} = 150(I3+I1)$ $V_{DA} = 300(I3-I2)$ What is VBD? $V_{BD} = V_{BD} = V_{D} - V_{B} = -24$ And we have: $V_{AB} +V_{BD} + V_{DA} = 0$ OR: 150(I3+I1) -24 + 300(I3-I2) = 0 Can you simplify this? Last edited: Oct 31, 2013 5. ### hamid-reza Thread Starter New Member Oct 31, 2013 6 0 So my answer was correct but I still can't understand why the third equation is this: Look at the 'Original form of equation' : Why do we have 150(I3+I1)? I3 goes through R1 in completely different direction from I1 (Assuming the direction I1&I3 clockwise) So, why don't we have 150(I3-I1)? Last edited: Oct 31, 2013 6. ### anhnha Active Member Apr 19, 2012 774 48 What is the direction of I3 flowing through R1, from A to B or B to A? What is the direction of I1 flowing through R1, from A to B or B to A? Now do you see that they are in the same direction? hamid-reza likes this. 7. ### hamid-reza Thread Starter New Member Oct 31, 2013 6 0 Yes both flow through R1 from A to B but look at the battery polarity it makes a clockwise current so if we assume I3 counter clockwise we have: -24+300(I3-I2)+150(I3+I1) Correct me if I'm wrong. Sorry for asking so many questions. 8. ### hamid-reza Thread Starter New Member Oct 31, 2013 6 0 I think I answered my own question sorry for taking your time dear anhnha. Thank you so much. 9. ### anhnha Active Member Apr 19, 2012 774 48 I think that may work. Let's look at it this way. Sorry, I can't explain it more clearly. Hope someone will help you understand it better. Let's apply KVL to the loop on the left hand side: $V_{AB} +V_{BD} + V_{DA} = 0$ Where: $V_{AB} = V_{B} - V_{A}= 150(I3+I1)$ $V_{BD} = V_{D} - V_{B}$ $V_{DA} = V_{A} - V_{D}= 300(I3-I2)$ Now let's consider this: $V_{BD} = V_{D} - V_{B}$ On the other hand, the battery is 24V this means that the voltage between positive and negative is 24V. This means: $V_{B} - V_{D} = 24V$ Therefore: $V_{BD} = V_{D} - V_{B} = -24V$ Finally we have: $V_{AB} +V_{BD} + V_{DA} = 0$ OR: 150(I3+I1) -24 + 300(I3-I2) = 0 hamid-reza likes this. 10. ### WBahn Moderator Mar 31, 2012 17,788 4,807 I think a big part of your problem is that whoever set things up didn't use a consistent direction for their current loops. Let that be a lesson. Get in the habit of ALWAYS using the same direction for all loops. Personally, I run all of my loops counter-clockwise. Yes, there are times that choosing a loop going the other way would be more convenient for a particular problem, but I have long since found that using a technique consistently will reduce the errors I make overall enough to more than compensate for any inconvenience on a problem here or there. Another thing that this problem does is use carrier flow instead of charge flow. It seems to use it consistently, so it's valid. hamid-reza likes this. 11. ### bwd111 Member Jul 24, 2013 117 1 Thats a tough one 12. ### WBahn Moderator Mar 31, 2012 17,788 4,807 Not really, IF you are systematic about it. You first have to decide if you are going to sum up the voltage gains or the voltage drops as you go around the circuit. You then have to decide if your "currents" represent charge flow or carrier flow. If it represents charge flow, then it is a true current and if you pass through a resistor in the direction of charge flow you always drop in voltage by an amount that is equal to I*R. If you are using carrier flow, then your "current" may not represent a true current, which is CHARGE per unit TIME. If your carriers are electrons, then they are negatively charged and you have to account for the fact that the true current is the negative of the carrier current. This is done by noting that if you pass through a resistor in the direction of carrier flow, you gain in voltage an amount that is equal to I*R or, putting it another way, you still drop in voltage by and amount that is equal to the charge current times the resistance, but the charge current is -I and so you drop by -I*R. In either case, as you go through a voltage source you drop in voltage if you are going from the positive to the negative terminal and gain in voltage is you are going the other way. In this problem, whoever set up the solution has chosen to use carrier flow and to sum up the voltage drops as you go around the loop in the direction of the loop flow. So: Mesh I1: 24V + (-I3+I2)R4 + (-I3-I1)R1 = 0 24V - I3(R1+R4) + I2(R4) - I1(R1) = 0 - I1(R1) + I2(R4) - I3(R1+R4) = -24V - I1(150Ω) + I2(300Ω) - I3(450Ω) = -24V IF a uniform direction had been chosen for all of the mesh currents, then the three mesh equations can be written down by inspections. Using charge flow (a.k.a., conventional current), each equation will have a coefficient for each mesh current that is the sum of the resistors that that current flows through that are in common with the mesh. The mesh current in question will be positive and the bordering currents will be negative. That's the left hand side. The right hand side will then be the sum of the voltage gains due to the sources. Hence, for mesh currents going counterclockwise, written by inspection: + I1(300Ω) - I2(100Ω) - I3(150Ω) = 0 - I1(100Ω) + I2(650Ω) - I3(300Ω) = 0 - I1(150Ω) - I2(300Ω) + I3(450Ω) = -24V A couple of easy checks can then be made. First, the coefficients are symmetric about the diagonal. If they aren't, then something is wrong. Second, the diagonal coefficients will count every resistor exactly twice unless the resistor is on the periphery, so the total of all the resistors in the circuit should come to 50%*(total of diagonal coefficients+total of periphery resistors). In this case, thats 50%(1400Ω+300Ω)=850Ω, which indeed is the sum of the resistors in the circuit. Another way of saying this is that the sum of the peripheral resistances must be equal to the sum of the diagonal coefficients minus the sum of all of the off-diagonal coefficients, which it is. hamid-reza likes this. 13. ### hamid-reza Thread Starter New Member Oct 31, 2013 6 0 This is really helping but I don't know why? Every time I choose loops counter clockwise I get the right answer. Thanks WBahn.
2016-12-11 04:23:27
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http://math.stackexchange.com/questions/683937/are-there-infinitely-many-squarefree-values-of-nn4?answertab=active
# Are there infinitely many squarefree values of n(n+4) As the title says I was wondering whether there are infinitely many squarefree values of $n(n+4)$, $n \in \mathbb N$. Clearly $n \notin 2 \mathbb N$, but that is unfortunately the most I've gotten to so far. I know that questions about squarefree values of polynomials are extremely tough, but perhaps there exists a solution for this particular instance? - For an odd number to be squareful, it needs to be divisible by the square of an odd prime. $n$ and $n+4$ cannot share any odd factors, so one of them must be divisible by the square of an odd prime. According to this MathOverflow question $$\sum_{p\text{ prime}}{1\over p^2}=\sum_{k=1}^{\infty}{\mu(k)\over k}\log(\zeta(2k))=0.4522474200\dots \text{so}\\ \sum_{p\text{odd prime}}{1\over p^2}=\sum_{k=1}^{\infty}{\mu(k)\over k}\log(\zeta(2k))=0.2022474200\dots$$ at most $20.225\%$ of odd numbers are squareful. The first term counts the fraction divisible by $3^2$, the second the number divisible by $5^2$, etc. Some odds are divisible by the square of more than one odd prime, so the sum is an overcount. Then at most $40.25\%$ of odd numbers of the form $n(n+4)$ are squareful. Infinitely many odd numbers of the form $n(n+4)$ must be squarefree. Thanks to Hagen von Eitzen for the correction. Where does this make use of the specific form $n(n+4)$ of the numbers? (E.g. $n(n+1)(n+2)(n+3)$ is never square-free) – Hagen von Eitzen Feb 20 '14 at 20:27 @HagenvonEitzen: The idea is that $n$ can only be squareful less than $45.225\%$ of the time, same for $n+4$, so the product can only be squareful twice that, and in fact will be less. This fails because if $n \equiv 2 \pmod 4, 4|n(n+4)$. I believe it can be patched up, but will need to work on it. – Ross Millikan Feb 20 '14 at 20:33 Ah, I see - my eyes matched $0.45$ with $(9)0.45$ unconciously. - We can drop $\frac1{2^2}$ from the sum, so we end up with $0.2022472300\ldots$. Thus at most $40.45\%$ of all numbers $n(n+4)$ have an odd square factor. And $50\%$ have an even square factor. That still leaves $9.55\%$ without any square factor. That should fix it. – Hagen von Eitzen Feb 20 '14 at 20:40
2015-11-27 05:12:29
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https://electronics.stackexchange.com/questions/72766/what-are-the-things-to-notice-before-actually-connecting-a-pwm-motor-circuit
# What are the things to notice before actually connecting a PWM motor circuit? I have simulated a PWM motor control circuit in Proteus, now I want to bring it to life and connect a circuit in real world. Here's my PWM circuit in Proteus: And the code: unsigned int duty_50; void main(){ ADPCFG = 0xFFFF;// initialize AN pins as digital PORTB = 0x01; TRISB = 0; // initialize portb as output duty_50 = PWM1_MC_Init(5000, 0, 0x01, 0); // Pwm_Mc_Init returns 50% of the duty PWM1_MC_Set_Duty(duty_50, 1); PWM1_MC_Start(); while(1); } In simulation, this circuit works fine, And I already bought the components I need: 1.dsPIC 2.L293D driver IC 3.DC Motor(Which needs 1.5V and 120 mA) 4.board I want to use these components and power supply machine to connect these components in real world, but I don't know if there might be problems. Is there something I should pay attention to before connect the real circuit so that I won't make some mistakes and burn some components? • If you don't burn stuff, you don't learn stuff. – Anindo Ghosh Jun 14 '13 at 15:32 • Great job! But this is stepping up a few floors once. I think you should start by breadboarding some LEDs with the microcontroller. Perhaps, you should do a little research on breadboarding. Such as this one. – abdullah kahraman Jun 14 '13 at 15:44 • How much current is going to go to the motor vs. how much is the breadboard rated for? General suggestion: always double-check where the power rails are going to and that they are the right way round before connecting. – pjc50 Jun 14 '13 at 16:33 • Don't forget you need 0.1 uF bypass capacitors across the power supply rails at each power connection for every chip. – Connor Wolf Jun 14 '13 at 22:46
2019-11-12 23:41:52
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http://publ.beesbuzz.biz/blog/?id=446&tag=pushl&tag=planning
# Publ: Development Blog Entries tagged pushl or planning ## Pushl v0.2.1 released Posted Thursday, March 7 at 10:27 PM (9 months ago) I’ve been working on getting Pushl much more stable and reliable, particularly around a persistent “too many open files” error I was having, which turned out to be primarily due to a fd leak in the caching routines. Oops. Anyway, there’s also seemingly a problem with how aiohttp manages its connection pool, at least on macOS, so I’ve disabled connection keep-alive by default. However, if you still want to use keep-alive, there’s now a --keepalive option to allow you to do that. I’m finding that it doesn’t really improve performance all that much anyway. This is feeling beta-ready but I’ll give it a few days for other issues to shake out first. ## Pushl v0.2.0 released Posted Thursday, March 7 at 12:05 AM (9 months ago) So, I just released v0.2.0 of Pushl. It was a pretty big change, in that I pretty much rewrote all the networking stuff, and fixed some pretty ridiculous bugs with the caching implementation as well. The main thing is now it’s using async I/O instead of thread-per-connection, so it’s way more efficient and also times out correctly. And oh gosh, I had so many tiny but critical errors in the way caching was implemented – no wonder it kept on acting as if there was no cached state. Yeesh. Anyway, I’ll let this run on my site for a few days and if I like what I see I’ll upgrade it to beta status on PyPI. ## v0.3.19, now with extra tagging goodness! Posted Monday, March 4 at 3:47 PM (9 months ago) I’ve released Publ v0.3.19, which now finally has a tagging system, which is only one of the oldest issues that was still open. Here’s a list of what’s been added or changed since 0.3.18: ### Credits I want to thank Karina Antonio for implementing image cropping. ## v0.3.18, now with better asset management! Posted Wednesday, February 27 at 9:38 PM (10 months ago) I’ve just released v0.3.18, with the following changes: • Add date grouping properties to entry • Add a pages property to view • Provide the current category object to the error handler • Support linking to non-image/non-entry local files • Added, then removed, some performance micro-optimizations that only caused problems More details about the major changes below! Update: I released a hotfix as 0.3.18.1 because there was a last-minute bug that snuck in while I was trying to silence a new pylint error. Oops. ## Pushl v0.1.7 Posted Monday, January 14 at 9:28 PM (11 months ago) I ended up doing some more work on Pushl and have now released v0.1.7. The major changes: • Did a bunch of refactoring to make the code a little cleaner and handle configuration more appropriately • Added a configurable timeout for connections (which now defaults to 15) • Added a --version option on the command line arguments Also, some suggested usage ideas below the cut! ## Pushl v0.1.6 released Posted Sunday, January 13 at 8:48 PM (11 months ago) It’s been a while since I’ve updated Pushl but today I released v0.1.6. It includes the following fixes: • Now it supports Pingback as well as Webmention • Improved the threading defaults and connection pooling • Also checks entries for updates even if the feed didn’t change (in case something changed in the more text or page metadata or whatever) Anyway, it should just be a pip install --upgrade pushl (or pipenv update) away. ## Pushl v0.1.5 Posted Saturday, December 22 at 1:35 AM (a year ago) While I’m fixing random stuff in Publ, I figured I’d finally fix some problems with Pushl too. Nothing major here, just: • Stability: Fixed a bug where feeds that don’t declare links caused the worker to die before entries got processed • Performance: Now we use a global connection pool (so connections can be reused) • Fixed a minor correctness issue with archive feeds (which actually doesn’t make any difference in the real world but whatever) ## v0.3.11 Posted Saturday, December 15 at 1:08 AM (a year ago) v0.3.11 is now released, with the following changes: • A more complete fix for how to handle image sets and inline images with respect to paragraphs • Better cleanup for spurious empty paragraphs • Improved internal entry link handling Detailed descriptions of the changes are below. ## Pushl v0.1.3, and a FeedOnFeeds update! Posted Wednesday, November 28 at 1:18 AM (a year ago) I just released Pushl v0.1.3, which adds some minor performance optimizations and a bug fix. Originally I was hoping to have a major performance optimization, in the form of having rewritten Pushl from thread-per-connection to async operation, but unfortunately I ran into a bunch of problems with it. Mostly that I was running into a “too many open files” error and I couldn’t figure out what was causing a descriptor leak. I have the work-in-progress branch online if anyone wants to take a look at it. Anyway, the reason I went down this route is because I added WebSub subscriber support to my fork of Feed-On-Feeds, which makes it so that WebSub-enabled RSS and Atom feeds will push their updates to your reader instead of having to wait for a polling interval. You can read more about some of my other thoughts on a blog entry that quickly devolves into a rant, if you’re so inclined. ## Pushl 0.1.0 released Posted Wednesday, October 10 at 10:41 PM (a year ago) I’ve gotten Pushl to the point that I’m confident in releasing it as a full alpha and having a cron job run it every 15 minutes. Whew. I’ve also added some h-entry markup to the Publ site templates so that hopefully the notifications appear at least somewhat reasonably elsewhere. I really need to redo these site templates now that I have a better idea of how they go together. Also the quickstart guide could be a lot better. ## Pushl 0.0.1 released Posted Monday, October 8 at 11:53 PM (a year ago) I finally got around to releasing a very rough prototype of Pushl to pypi. It only sends out WebSub notifications for now (does anyone even use those?), but I’ll work on actually implementing WebMention soon. Also, recently someone pointed out to me fed.brid.gy which makes it easy to turn a static site into an ActivityPub source. At some point I’ll experiment with setting up Publ for this; it looks like it’s just a matter of adding a couple of additional route rules to Publ, so that will probably go into an advanced configuration guide if I ever get around to making such a thing. (Or it could actually be added to Publ directly but there isn’t much of a reason for that, IMO.) ## v0.1.19: creeping ever closer to beta status Posted Sunday, May 27 at 5:22 PM (2 years ago) The amount of stuff I’m having to fix in Publ to support beesbuzz.biz is diminishing rapidly! Here’s what’s happened since 0.1.18: • Improved the Path-Alias redirection logic; now it will do a 301 Permanently Moved for inbound Path-Aliased requests, and if a Path-Alias points to an entry with a Redirect-To it will redirect directly to that URL instead (and it will be a 302, same as the old Redirect-To behavior) • Pagination can now be weekly; you can use entry.archive(paging='week'), and a ?date= view parameter ending in _w will provide a weekly view instead. • Better default formatting for view.range, and an addition of a week format parameter there
2019-12-16 10:42:07
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https://iclr.cc/virtual/2021/poster/3114
## Undistillable: Making A Nasty Teacher That CANNOT teach students ### Haoyu Ma · Tianlong Chen · Ting-Kuei Hu · Chenyu You · Xiaohui Xie · Zhangyang Wang Keywords: [ knowledge distillation ] [ avoid knowledge leaking ] [ Abstract ] [ Paper ] Mon 3 May 5 p.m. PDT — 7 p.m. PDT Spotlight presentation: Oral Session 9 Wed 5 May 7 p.m. PDT — 10:05 p.m. PDT Abstract: Knowledge Distillation (KD) is a widely used technique to transfer knowledge from pre-trained teacher models to (usually more lightweight) student models. However, in certain situations, this technique is more of a curse than a blessing. For instance, KD poses a potential risk of exposing intellectual properties (IPs): even if a trained machine learning model is released in black boxes'' (e.g., as executable software or APIs without open-sourcing code), it can still be replicated by KD through imitating input-output behaviors. To prevent this unwanted effect of KD, this paper introduces and investigates a concept called $\textit{Nasty Teacher}$: a specially trained teacher network that yields nearly the same performance as a normal one, but would significantly degrade the performance of student models learned by imitating it. We propose a simple yet effective algorithm to build the nasty teacher, called $\textit{self-undermining knowledge distillation}$. Specifically, we aim to maximize the difference between the output of the nasty teacher and a normal pre-trained network. Extensive experiments on several datasets demonstrate that our method is effective on both standard KD and data-free KD, providing the desirable KD-immunity to model owners for the first time. We hope our preliminary study can draw more awareness and interest in this new practical problem of both social and legal importance. Our codes and pre-trained models can be found at: $\url{https://github.com/VITA-Group/Nasty-Teacher}$.
2022-08-07 15:59:18
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https://math.stackexchange.com/questions/1017791/prove-that-q-8-cong-langle-a-b-mid-a4-a2b-2-aba-1b-rangle
# Prove that $Q_8 \cong \langle a, b \mid a^4, a^2b^{-2}, aba^{-1}b \rangle$ Prove that if $G = \langle a, b \mid a^4, a^2b^{-2}, aba^{-1}b \rangle$, then $G \cong Q_8$. I started by trying to define a homomorphism $\varphi: F(a,b) \to Q_8$ by $\varphi(a) = i$, $\varphi(b) = j$ and $\varphi(ab) = k$. EDIT: this is not true (Since $Q_8$ is abelian, I know that the commutator $aba^{-1}b^{-1} \in Ker(\varphi)$), So the elements $a^4, b^4, (ab)^4$ are in the kernel. However, I can't seem to show that this kernel is equal to the normal closure of the group $\langle a^4, a^2b^{-2}, aba^{-1}b \rangle$. Is there an easier way to do this problem? We just proved van Dyck's theorem in class, and our professor hinted that it applies to this problem, but I can't see how to use it. Any hints will be appreciated, but I'd rather not have a complete solution. • But $Q_8$ is not abelian. – Derek Holt Nov 12 '14 at 1:14 • @DerekHolt Almost... I thought I went over the whole multiplication table... I edited the question to account for that. The rest of it still stands though. – Johanna Nov 12 '14 at 1:20 The second relation gives $b^2 = a^2$. This says $|a| = |b|=4$, and that the elements $a^n b^m$, with $m\ge 3$ can be reduced to $a^{r}b^{s}$ but with $r=\{0,1,2,3\}, s = \{0,1\}$ The third relation: $aba^{-1}b = 1 \implies ab = b^{-1}a \implies bab =a$. Moreover $bab = a \implies ba = ab^{-1} = ab^3 = a(b^2)b = a(a^2)b =a^3b$, ie, $ba=a^3b$. So elements with mixed letters like $ba, bab, etc$ can also be expressed as $a^rb^s$ with $r=\{0,1,2,3\}, s= \{0,1\}$ This means, $G$ has 8 distinct elements: $$1, a, a^2, a^3$$ $$b, ab, a^2b, a^3b$$ Since $i,j \in Q_8$ satisfies the relations in $G$, Von Dyck's Theorem gives a surjective homomorphism $G \to Q_8$, $a \mapsto i, b\mapsto j$ We have the following: • $|G| = 8$ • The homomorphism is also an injection (same group order + surjectivity) • The homomorphism is now a bijection, hence an isomorphism.
2019-07-19 02:40:30
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http://www.polarhome.com/service/man/?qf=cvs&tf=2&of=OpenDarwin&sf=1
cvs man page on OpenDarwin Man page or keyword search:   man All Sections 1 - General Commands 2 - System Calls 3 - Subroutines, Functions 4 - Special Files 5 - File Formats 6 - Games and Screensavers 7 - Macros and Conventions 8 - Maintenence Commands 9 - Kernel Interface New Commands Server 4.4BSD AIX Alpinelinux Archlinux Aros BSDOS BSDi Bifrost CentOS Cygwin Darwin Debian DigitalUNIX DragonFly ElementaryOS Fedora FreeBSD Gentoo GhostBSD HP-UX Haiku Hurd IRIX Inferno JazzOS Kali Knoppix LinuxMint MacOSX Mageia Mandriva Manjaro Minix MirBSD NeXTSTEP NetBSD OPENSTEP OSF1 OpenBSD OpenDarwin OpenIndiana OpenMandriva OpenServer OpenSuSE OpenVMS Oracle PC-BSD Peanut Pidora Plan9 QNX Raspbian RedHat Scientific Slackware SmartOS Solaris SuSE SunOS Syllable Tru64 UNIXv7 Ubuntu Ultrix UnixWare Xenix YellowDog aLinux   3202 pages apropos Keyword Search (all sections) Output format html ascii pdf view pdf save postscript [printable version] CVS(1) CVS(1) NAME cvs - Concurrent Versions System SYNOPSIS cvs [ cvs_options ] cvs_command [ command_options ] [ command_args ] NOTE This manpage is a summary of some of the features of cvs. It is auto- generated from an appendix of the CVS manual. For more in-depth docu‐ mentation, please consult the Cederqvist manual (via the info CVS com‐ mand or otherwise, as described in the SEE ALSO section of this man‐ page). Cross-references in this man page refer to nodes in the same. CVS commands Guide to CVS commands This appendix describes the overall structure of cvs commands, and describes some commands in detail (others are described elsewhere; for a quick reference to cvs commands, see node Invoking CVS' in the CVS manual). Structure Overall structure of CVS commands The overall format of all cvs commands is: cvs [ cvs_options ] cvs_command [ command_options ] [ command_args ] cvs The name of the cvs program. cvs_options Some options that affect all sub-commands of cvs. These are described below. cvs_command One of several different sub-commands. Some of the commands have aliases that can be used instead; those aliases are noted in the ref‐ erence manual for that command. There are only two situations where you may omit cvs_command: cvs -H elicits a list of available com‐ mands, and cvs -v displays version information on cvs itself. command_options Options that are specific for the command. command_args Arguments to the commands. There is unfortunately some confusion between cvs_options and com‐ mand_options. When given as a cvs_option, some options only affect some of the commands. When given as a command_option it may have a different meaning, and be accepted by more commands. In other words, do not take the above categorization too seriously. Look at the doc‐ Exit status CVS's exit status cvs can indicate to the calling environment whether it succeeded or failed by setting its exit status. The exact way of testing the exit status will vary from one operating system to another. For example in a unix shell script the $? variable will be 0 if the last command returned a successful exit status, or greater than 0 if the exit status indicated failure. If cvs is successful, it returns a successful status; if there is an error, it prints an error message and returns a failure status. The one exception to this is the cvs diff command. It will return a suc‐ cessful status if it found no differences, or a failure status if there were differences or if there was an error. Because this behavior pro‐ vides no good way to detect errors, in the future it is possible that cvs diff will be changed to behave like the other cvs commands. ~/.cvsrc Default options and the ~/.cvsrc file There are some command_options that are used so often that you might have set up an alias or some other means to make sure you always spec‐ ify that option. One example (the one that drove the implementation of the .cvsrc support, actually) is that many people find the default out‐ put of the diff command to be very hard to read, and that either con‐ text diffs or unidiffs are much easier to understand. The ~/.cvsrc file is a way that you can add default options to cvs_com‐ mands within cvs, instead of relying on aliases or other shell scripts. The format of the ~/.cvsrc file is simple. The file is searched for a line that begins with the same name as the cvs_command being executed. If a match is found, then the remainder of the line is split up (at whitespace characters) into separate options and added to the command arguments before any options from the command line. If a command has two names (e.g., checkout and co), the official name, not necessarily the one used on the command line, will be used to match against the file. So if this is the contents of the user's ~/.cvsrc file: log -N diff -uN rdiff -u update -Pd checkout -P release -d the command cvs checkout foo would have the -P option added to the arguments, as well as cvs co foo. With the example file above, the output from cvs diff foobar will be in unidiff format. cvs diff -c foobar will provide context diffs, as usual. Getting "old" format diffs would be slightly more complicated, because diff doesn't have an option to specify use of the "old" format, so you would need cvs -f diff foobar. In place of the command name you can use cvs to specify global options (see node Global options' in the CVS manual). For example the follow‐ ing line in .cvsrc cvs -z6 causes cvs to use compression level 6. Global options The available cvs_options (that are given to the left of cvs_command) are: --allow-root=rootdir Specify legal cvsroot directory. See see node Password authentica‐ tion server' in the CVS manual. -a Authenticate all communication between the client and the server. Only has an effect on the cvs client. As of this writing, this is only implemented when using a GSSAPI connection (see node GSSAPI authenticated' in the CVS manual). Authentication prevents certain sorts of attacks involving hijacking the active tcp connection. Enabling authentication does not enable encryption. -b bindir In cvs 1.9.18 and older, this specified that rcs programs are in the bindir directory. Current versions of cvs do not run rcs programs; for compatibility this option is accepted, but it does nothing. -T tempdir Use tempdir as the directory where temporary files are located. Overrides the setting of the$TMPDIR environment variable and any precompiled directory. This parameter should be specified as an absolute pathname. (When running client/server, -T affects only the local process; specifying -T for the client has no effect on the server and vice versa.) -d cvs_root_directory Use cvs_root_directory as the root directory pathname of the reposi‐ tory. Overrides the setting of the $CVSROOT environment variable. see node Repository' in the CVS manual. -e editor Use editor to enter revision log information. Overrides the setting of the$CVSEDITOR and $EDITOR environment variables. For more infor‐ mation, see see node Committing your changes' in the CVS manual. -f Do not read the ~/.cvsrc file. This option is most often used because of the non-orthogonality of the cvs option set. For example, the cvs log option -N (turn off display of tag names) does not have a corresponding option to turn the display on. So if you have -N in the ~/.cvsrc entry for log, you may need to use -f to show the tag names. -H --help Display usage information about the specified cvs_command (but do not actually execute the command). If you don't specify a command name, cvs -H displays overall help for cvs, including a list of other help options. -R Turns on read-only repository mode. This allows one to check out from a read-only repository, such as within an anoncvs server, or from a cd-rom repository. Same effect as if the CVSREADONLYFS environment variable is set. Using -R can also considerably speed up checkouts over NFS. -n Do not change any files. Attempt to execute the cvs_command, but only to issue reports; do not remove, update, or merge any existing files, or create any new files. Note that cvs will not necessarily produce exactly the same output as without -n. In some cases the output will be the same, but in other cases cvs will skip some of the processing that would have been required to produce the exact same output. -Q Cause the command to be really quiet; the command will only generate output for serious problems. -q Cause the command to be somewhat quiet; informational messages, such as reports of recursion through subdirectories, are suppressed. -r Make new working files read-only. Same effect as if the$CVSREAD environment variable is set (see node Environment variables' in the CVS manual). The default is to make working files writable, unless watches are on (see node Watches' in the CVS manual). -s variable=value Set a user variable (see node Variables' in the CVS manual). -t Trace program execution; display messages showing the steps of cvs activity. Particularly useful with -n to explore the potential impact of an unfamiliar command. -v --version Display version and copyright information for cvs. -w Make new working files read-write. Overrides the setting of the $CVSREAD environment variable. Files are created read-write by default, unless$CVSREAD is set or -r is given. -x Encrypt all communication between the client and the server. Only has an effect on the cvs client. As of this writing, this is only implemented when using a GSSAPI connection (see node GSSAPI authen‐ ticated' in the CVS manual) or a Kerberos connection (see node Ker‐ beros authenticated' in the CVS manual). Enabling encryption implies that message traffic is also authenticated. Encryption support is not available by default; it must be enabled using a special config‐ ure option, --enable-encryption, when you build cvs. -z gzip-level Set the compression level. Valid levels are 1 (high speed, low com‐ pression) to 9 (low speed, high compression), or 0 to disable com‐ pression (the default). Only has an effect on the cvs client. Common options Common command options This section describes the command_options that are available across several cvs commands. These options are always given to the right of cvs_command. Not all commands support all of these options; each option is only supported for commands where it makes sense. However, when a command has one of these options you can almost always count on the same behavior of the option as in other commands. (Other command options, which are listed with the individual commands, may have dif‐ ferent behavior from one cvs command to the other). Note: the history command is an exception; it supports many options that conflict even with these standard options. -D date_spec Use the most recent revision no later than date_spec. date_spec is a single argument, a date description specifying a date in the past. The specification is sticky when you use it to make a private copy of a source file; that is, when you get a working file using -D, cvs records the date you specified, so that further updates in the same tags/dates, see node Sticky tags' in the CVS manual). -D is available with the annotate, checkout, diff, export, history, ls, rdiff, rls, rtag, tag, and update commands. (The history command uses this option in a slightly different way; see node history options' in the CVS manual). A wide variety of date formats are supported by cvs. The most stan‐ dard ones are ISO8601 (from the International Standards Organization) and the Internet e-mail standard (specified in RFC822 as amended by RFC1123). ISO8601 dates have many variants but a few examples are: 1972-09-24 1972-09-24 20:05 There are a lot more ISO8601 date formats, and cvs accepts many of them, but you probably don't want to hear the whole long story :-). In addition to the dates allowed in Internet e-mail itself, cvs also allows some of the fields to be omitted. For example: 24 Sep 1972 20:05 24 Sep The date is interpreted as being in the local timezone, unless a spe‐ cific timezone is specified. These two date formats are preferred. However, cvs currently accepts a wide variety of other date formats. They are intentionally not documented here in any detail, and future versions of cvs might not accept all of them. One such format is month/day/year. This may confuse people who are accustomed to having the month and day in the other order; 1/4/96 is January 4, not April 1. Remember to quote the argument to the -D flag so that your shell doesn't interpret spaces as argument separators. A command using the -D flag can look like this: $cvs diff -D "1 hour ago" cvs.texinfo -f When you specify a particular date or tag to cvs commands, they nor‐ mally ignore files that do not contain the tag (or did not exist prior to the date) that you specified. Use the -f option if you want files retrieved even when there is no match for the tag or date. (The most recent revision of the file will be used). Note that even with -f, a tag that you specify must exist (that is, in some file, not necessary in every file). This is so that cvs will continue to give an error if you mistype a tag name. -f is available with these commands: annotate, checkout, export, rdiff, rtag, and update. WARNING: The commit and remove commands also have a -f option, but it has a different behavior for those commands. See see node commit options' in the CVS manual, and see node Removing files' in the CVS manual. -k kflag Override the default processing of RCS keywords other than -kb. see node Keyword substitution' in the CVS manual, for the meaning of kflag. Used with the checkout and update commands, your kflag speci‐ fication is sticky; that is, when you use this option with a checkout or update command, cvs associates your selected kflag with any files it operates on, and continues to use that kflag with future commands on the same files until you specify otherwise. The -k option is available with the add, checkout, diff, export, import and update commands. WARNING: Prior to CVS version 1.12.2, the -k flag overrode the -kb indication for a binary file. This could sometimes corrupt binary files. see node Merging and keywords' in the CVS manual, for more. -l Local; run only in current working directory, rather than recursing through subdirectories. Available with the following commands: annotate, checkout, commit, diff, edit, editors, export, log, rdiff, remove, rtag, status, tag, unedit, update, watch, and watchers. -m message Use message as log information, instead of invoking an editor. Available with the following commands: add, commit and import. -n Do not run any tag program. (A program can be specified to run in the modules database (see node modules' in the CVS manual); this option bypasses it). Note: this is not the same as the cvs -n program option, which you can specify to the left of a cvs command! Available with the checkout, commit, export, and rtag commands. -P Prune empty directories. See see node Removing directories' in the CVS manual. -p Pipe the files retrieved from the repository to standard output, rather than writing them in the current directory. Available with the checkout and update commands. -R Process directories recursively. This is the default for all cvs commands, with the exception of ls & rls. Available with the following commands: annotate, checkout, commit, diff, edit, editors, export, ls, rdiff, remove, rls, rtag, status, tag, unedit, update, watch, and watchers. -r tag Use the revision specified by the tag argument instead of the default head revision. As well as arbitrary tags defined with the tag or rtag command, two special tags are always available: HEAD refers to the most recent version available in the repository, and BASE refers to the revision you last checked out into the current working direc‐ tory. The tag specification is sticky when you use this with checkout or update to make your own copy of a file: cvs remembers the tag and continues to use it on future update commands, until you specify oth‐ erwise (for more information on sticky tags/dates, see node Sticky tags' in the CVS manual). The tag can be either a symbolic or numeric tag, as described in see node Tags' in the CVS manual, or the name of a branch, as described in see node Branching and merging' in the CVS manual. Specifying the -q global option along with the -r command option is often useful, to suppress the warning messages when the rcs file does not contain the specified tag. Note: this is not the same as the overall cvs -r option, which you can specify to the left of a cvs command! -r is available with the checkout, commit, diff, history, export, rdiff, rtag, and update commands. -W Specify file names that should be filtered. You can use this option repeatedly. The spec can be a file name pattern of the same type that you can specify in the .cvswrappers file. Available with the following commands: import, and update. admin Administration · Requires: repository, working directory. · Changes: repository. · Synonym: rcs This is the cvs interface to assorted administrative facilities. Some of them have questionable usefulness for cvs but exist for his‐ torical purposes. Some of the questionable options are likely to disappear in the future. This command does work recursively, so extreme care should be used. On unix, if there is a group named cvsadmin, only members of that group can run cvs admin commands, except for those specified using the UserAdminOptions configuration option in the CVSROOT/config file. Options specified using UserAdminOptions can be run by any user. See see node config' in the CVS manual for more on UserAdminOptions. The cvsadmin group should exist on the server, or any system running the non-client/server cvs. To disallow cvs admin for all users, cre‐ ate a group with no users in it. On NT, the cvsadmin feature does not exist and all users can run cvs admin. admin options Some of these options have questionable usefulness for cvs but exist for historical purposes. Some even make it impossible to use cvs until you undo the effect! -Aoldfile Might not work together with cvs. Append the access list of oldfile to the access list of the rcs file. -alogins Might not work together with cvs. Append the login names appearing in the comma-separated list logins to the access list of the rcs file. -b[rev] Set the default branch to rev. In cvs, you normally do not manipu‐ late default branches; sticky tags (see node Sticky tags' in the CVS manual) are a better way to decide which branch you want to work on. There is one reason to run cvs admin -b: to revert to the vendor's version when using vendor branches (see node Reverting local changes' in the CVS manual). There can be no space between -b and its argument. -cstring Sets the comment leader to string. The comment leader is not used by current versions of cvs or rcs 5.7. Therefore, you can almost surely not worry about it. see node Keyword substitution' in the CVS man‐ ual. -e[logins] Might not work together with cvs. Erase the login names appearing in the comma-separated list logins from the access list of the RCS file. If logins is omitted, erase the entire access list. There can be no space between -e and its argument. -I Run interactively, even if the standard input is not a terminal. This option does not work with the client/server cvs and is likely to disappear in a future release of cvs. -i Useless with cvs. This creates and initializes a new rcs file, with‐ out depositing a revision. With cvs, add files with the cvs add com‐ mand (see node Adding files' in the CVS manual). -ksubst Set the default keyword substitution to subst. see node Keyword substitution' in the CVS manual. Giving an explicit -k option to cvs update, cvs export, or cvs checkout overrides this default. -l[rev] Lock the revision with number rev. If a branch is given, lock the latest revision on that branch. If rev is omitted, lock the latest revision on the default branch. There can be no space between -l and its argument. This can be used in conjunction with the rcslock.pl script in the contrib directory of the cvs source distribution to provide reserved checkouts (where only one user can be editing a given file at a time). See the comments in that file for details (and see the README file in that directory for disclaimers about the unsupported nature of contrib). According to comments in that file, locking must set to strict (which is the default). -L Set locking to strict. Strict locking means that the owner of an RCS file is not exempt from locking for checkin. For use with cvs, strict locking must be set; see the discussion under the -l option above. -mrev:msg Replace the log message of revision rev with msg. -Nname[:[rev]] Act like -n, except override any previous assignment of name. For use with magic branches, see see node Magic branch numbers' in the CVS manual. -nname[:[rev]] Associate the symbolic name name with the branch or revision rev. It is normally better to use cvs tag or cvs rtag instead. Delete the symbolic name if both : and rev are omitted; otherwise, print an error message if name is already associated with another number. If rev is symbolic, it is expanded before association. A rev consisting of a branch number followed by a . stands for the current latest revision in the branch. A : with an empty rev stands for the current latest revision on the default branch, normally the trunk. For exam‐ ple, cvs admin -nname: associates name with the current latest revi‐ sion of all the RCS files; this contrasts with cvs admin -nname:$ which associates name with the revision numbers extracted from key‐ word strings in the corresponding working files. -orange Deletes (outdates) the revisions given by range. Note that this command can be quite dangerous unless you know exactly what you are doing (for example see the warnings below about how the rev1:rev2 syntax is confusing). If you are short on disc this option might help you. But think twice before using it—there is no way short of restoring the latest backup to undo this command! If you delete different revisions than you planned, either due to carelessness or (heaven forbid) a cvs bug, there is no opportunity to correct the error before the revisions are deleted. It probably would be a good idea to experiment on a copy of the repository first. Specify range in one of the following ways: rev1::rev2 Collapse all revisions between rev1 and rev2, so that cvs only stores the differences associated with going from rev1 to rev2, not intermediate steps. For example, after -o 1.3::1.5 one can retrieve revision 1.3, revision 1.5, or the differences to get from 1.3 to 1.5, but not the revision 1.4, or the differences between 1.3 and 1.4. Other examples: -o 1.3::1.4 and -o 1.3::1.3 have no effect, because there are no intermediate revisions to remove. ::rev Collapse revisions between the beginning of the branch containing rev and rev itself. The branchpoint and rev are left intact. For example, -o ::1.3.2.6 deletes revision 1.3.2.1, revision 1.3.2.5, and everything in between, but leaves 1.3 and 1.3.2.6 intact. rev:: Collapse revisions between rev and the end of the branch containing rev. Revision rev is left intact but the head revision is deleted. rev Delete the revision rev. For example, -o 1.3 is equivalent to -o 1.2::1.4. rev1:rev2 Delete the revisions from rev1 to rev2, inclusive, on the same branch. One will not be able to retrieve rev1 or rev2 or any of the revisions in between. For example, the command cvs admin -oR_1_01:R_1_02 . is rarely useful. It means to delete revisions up to, and including, the tag R_1_02. But beware! If there are files that have not changed between R_1_02 and R_1_03 the file will have the same numerical revision number assigned to the tags R_1_02 and R_1_03. So not only will it be impossible to retrieve R_1_02; R_1_03 will also have to be restored from the tapes! In most cases you want to specify rev1::rev2 instead. :rev Delete revisions from the beginning of the branch containing rev up to and including rev. rev: Delete revisions from revision rev, including rev itself, to the end of the branch containing rev. None of the revisions to be deleted may have branches or locks. If any of the revisions to be deleted have symbolic names, and one specifies one of the :: syntaxes, then cvs will give an error and not delete any revisions. If you really want to delete both the symbolic names and the revisions, first delete the symbolic names with cvs tag -d, then run cvs admin -o. If one specifies the non-:: syntaxes, then cvs will delete the revisions but leave the symbolic names pointing to nonexistent revisions. This behavior is preserved for compatibility with previous versions of cvs, but because it isn't very useful, in the future it may change to be like the :: case. Due to the way cvs handles branches rev cannot be specified symbol‐ ically if it is a branch. see node Magic branch numbers' in the CVS manual, for an explanation. Make sure that no-one has checked out a copy of the revision you outdate. Strange things will happen if he starts to edit it and tries to check it back in. For this reason, this option is not a good way to take back a bogus commit; commit a new revision undoing the bogus change instead (see node Merging two revisions' in the CVS manual). -q Run quietly; do not print diagnostics. -sstate[:rev] Useful with cvs. Set the state attribute of the revision rev to state. If rev is a branch number, assume the latest revision on that branch. If rev is omitted, assume the latest revision on the default branch. Any identifier is acceptable for state. A useful set of states is Exp (for experimental), Stab (for stable), and Rel (for released). By default, the state of a new revision is set to Exp when it is created. The state is visible in the output from cvs log (see node log' in the CVS manual), and in the $Log$ and $State$ key‐ words (see node Keyword substitution' in the CVS manual). Note that cvs uses the dead state for its own purposes; to take a file to or from the dead state use commands like cvs remove and cvs add, not cvs -t[file] Useful with cvs. Write descriptive text from the contents of the named file into the RCS file, deleting the existing text. The file pathname may not begin with -. The descriptive text can be seen in the output from cvs log (see node log' in the CVS manual). There can be no space between -t and its argument. If file is omitted, obtain the text from standard input, terminated by end-of-file or by a line containing . by itself. Prompt for the text if interaction is possible; see -I. -t-string Similar to -tfile. Write descriptive text from the string into the rcs file, deleting the existing text. There can be no space between -t and its argument. -U Set locking to non-strict. Non-strict locking means that the owner of a file need not lock a revision for checkin. For use with cvs, strict locking must be set; see the discussion under the -l option above. -u[rev] See the option -l above, for a discussion of using this option with cvs. Unlock the revision with number rev. If a branch is given, unlock the latest revision on that branch. If rev is omitted, remove the latest lock held by the caller. Normally, only the locker of a revision may unlock it; somebody else unlocking a revision breaks the lock. This causes the original locker to be sent a commit notifica‐ tion (see node Getting Notified' in the CVS manual). There can be no space between -u and its argument. -Vn In previous versions of cvs, this option meant to write an rcs file which would be acceptable to rcs version n, but it is now obsolete and specifying it will produce an error. -xsuffixes In previous versions of cvs, this was documented as a way of specify‐ ing the names of the rcs files. However, cvs has always required that the rcs files used by cvs end in ,v, so this option has never done anything useful. annotate What revision modified each line of a file? · Synopsis: annotate [options] files... · Requires: repository. · Changes: nothing. For each file in files, print the head revision of the trunk, together with information on the last modification for each line. annotate options These standard options are supported by annotate (see node Common options' in the CVS manual, for a complete description of them): -l Local directory only, no recursion. -R Process directories recursively. -f -F Annotate binary files. -r revision Annotate file as of specified revision/tag. -D date Annotate file as of specified date. annotate example For example: $cvs annotate ssfile Annotations for ssfile *************** 1.1 (mary 27-Mar-96): ssfile line 1 1.2 (joe 28-Mar-96): ssfile line 2 The file ssfile currently contains two lines. The ssfile line 1 line was checked in by mary on March 27. Then, on March 28, joe added a line ssfile line 2, without modifying the ssfile line 1 line. This report doesn't tell you anything about lines which have been deleted or replaced; you need to use cvs diff for that (see node diff' in the CVS manual). The options to cvs annotate are listed in see node Invoking CVS' in the CVS manual, and can be used to select the files and revisions to annotate. The options are described in more detail there and in see node Common options' in the CVS manual. checkout Check out sources for editing · Synopsis: checkout [options] modules... · Requires: repository. · Changes: working directory. · Synonyms: co, get Create or update a working directory containing copies of the source files specified by modules. You must execute checkout before using most of the other cvs commands, since most of them operate on your working directory. The modules are either symbolic names for some collection of source directories and files, or paths to directories or files in the repos‐ itory. The symbolic names are defined in the modules file. see node modules' in the CVS manual. Depending on the modules you specify, checkout may recursively create directories and populate them with the appropriate source files. You can then edit these source files at any time (regardless of whether other software developers are editing their own copies of the sources); update them to include new changes applied by others to the source repository; or commit your work as a permanent change to the source repository. Note that checkout is used to create directories. The top-level directory created is always added to the directory where checkout is invoked, and usually has the same name as the specified module. In the case of a module alias, the created sub-directory may have a dif‐ ferent name, but you can be sure that it will be a sub-directory, and that checkout will show the relative path leading to each file as it is extracted into your private work area (unless you specify the -Q global option). The files created by checkout are created read-write, unless the -r option to cvs (see node Global options' in the CVS manual) is speci‐ fied, the CVSREAD environment variable is specified (see node Envi‐ ronment variables' in the CVS manual), or a watch is in effect for that file (see node Watches' in the CVS manual). Note that running checkout on a directory that was already built by a prior checkout is also permitted. This is similar to specifying the -d option to the update command in the sense that new directories that have been created in the repository will appear in your work area. However, checkout takes a module name whereas update takes a directory name. Also to use checkout this way it must be run from the top level directory (where you originally ran checkout from), so before you run checkout to update an existing directory, don't forget to change your directory to the top level directory. For the output produced by the checkout command see see node update output' in the CVS manual. checkout options These standard options are supported by checkout (see node Common options' in the CVS manual, for a complete description of them): -D date Use the most recent revision no later than date. This option is sticky, and implies -P. See see node Sticky tags' in the CVS man‐ ual, for more information on sticky tags/dates. -f Only useful with the -D date or -r tag flags. If no matching revi‐ sion is found, retrieve the most recent revision (instead of ignoring the file). -k kflag Process keywords according to kflag. See see node Keyword substitu‐ tion' in the CVS manual. This option is sticky; future updates of this file in this working directory will use the same kflag. The status command can be viewed to see the sticky options. See see node Invoking CVS' in the CVS manual, for more information on the status command. -l Local; run only in current working directory. -n Do not run any checkout program (as specified with the -o option in the modules file; see node modules' in the CVS manual). -P Prune empty directories. See see node Moving directories' in the CVS manual. -p Pipe files to the standard output. -R Checkout directories recursively. This option is on by default. -r tag Use revision tag. This option is sticky, and implies -P. See see node Sticky tags' in the CVS manual, for more information on sticky tags/dates. In addition to those, you can use these special command options with checkout: -A Reset any sticky tags, dates, or -k options. See see node Sticky tags' in the CVS manual, for more information on sticky tags/dates. -c Copy the module file, sorted, to the standard output, instead of cre‐ ating or modifying any files or directories in your working direc‐ tory. -d dir Create a directory called dir for the working files, instead of using the module name. In general, using this flag is equivalent to using mkdir dir; cd dir followed by the checkout command without the -d flag. There is an important exception, however. It is very convenient when checking out a single item to have the output appear in a directory that doesn't contain empty intermediate directories. In this case only, cvs tries to shorten'' pathnames to avoid those empty direc‐ tories. For example, given a module foo that contains the file bar.c, the command cvs co -d dir foo will create directory dir and place bar.c inside. Similarly, given a module bar which has subdirectory baz wherein there is a file quux.c, the command cvs co -d dir bar/baz will create directory dir and place quux.c inside. Using the -N flag will defeat this behavior. Given the same module definitions above, cvs co -N -d dir foo will create directories dir/foo and place bar.c inside, while cvs co -N -d dir bar/baz will create directories dir/bar/baz and place quux.c inside. -j tag With two -j options, merge changes from the revision specified with the first -j option to the revision specified with the second j option, into the working directory. With one -j option, merge changes from the ancestor revision to the revision specified with the -j option, into the working directory. The ancestor revision is the common ancestor of the revision which the working directory is based on, and the revision specified in the -j option. In addition, each -j option can contain an optional date specifica‐ tion which, when used with branches, can limit the chosen revision to one within a specific date. An optional date is specified by adding a colon (:) to the tag: -jSymbolic_Tag:Date_Specifier. see node Branching and merging' in the CVS manual. -N Only useful together with -d dir. With this option, cvs will not shorten'' module paths in your working directory when you check out a single module. See the -d flag for examples and a discussion. -s Like -c, but include the status of all modules, and sort it by the status string. see node modules' in the CVS manual, for info about the -s option that is used inside the modules file to set the module status. checkout examples Get a copy of the module tc:$ cvs checkout tc Get a copy of the module tc as it looked one day ago: $cvs checkout -D yesterday tc commit Check files into the repository · Synopsis: commit [-lnRf] [-m 'log_message' | -F file] [-r revision] [files...] · Requires: working directory, repository. · Changes: repository. · Synonym: ci Use commit when you want to incorporate changes from your working source files into the source repository. If you don't specify particular files to commit, all of the files in your working current directory are examined. commit is careful to change in the repository only those files that you have really changed. By default (or if you explicitly specify the -R option), files in subdirectories are also examined and committed if they have changed; you can use the -l option to limit commit to the current directory only. commit verifies that the selected files are up to date with the cur‐ rent revisions in the source repository; it will notify you, and exit without committing, if any of the specified files must be made cur‐ rent first with update (see node update' in the CVS manual). commit does not call the update command for you, but rather leaves that for you to do when the time is right. When all is well, an editor is invoked to allow you to enter a log message that will be written to one or more logging programs (see node modules' in the CVS manual, and see node loginfo' in the CVS manual) and placed in the rcs file inside the repository. This log message can be retrieved with the log command; see see node log' in the CVS manual. You can specify the log message on the command line with the -m message option, and thus avoid the editor invocation, or use the -F file option to specify that the argument file contains the log message. commit options These standard options are supported by commit (see node Common options' in the CVS manual, for a complete description of them): -l Local; run only in current working directory. -R Commit directories recursively. This is on by default. -r revision Commit to revision. revision must be either a branch, or a revision on the main trunk that is higher than any existing revision number (see node Assigning revisions' in the CVS manual). You cannot com‐ mit to a specific revision on a branch. commit also supports these options: -F file Read the log message from file, instead of invoking an editor. -f Note that this is not the standard behavior of the -f option as defined in see node Common options' in the CVS manual. Force cvs to commit a new revision even if you haven't made any changes to the file. If the current revision of file is 1.7, then the following two commands are equivalent:$ cvs commit -f file $cvs commit -r 1.8 file The -f option disables recursion (i.e., it implies -l). To force cvs to commit a new revision for all files in all subdirectories, you must use -f -R. -m message Use message as the log message, instead of invoking an editor. commit examples Committing to a branch You can commit to a branch revision (one that has an even number of dots) with the -r option. To create a branch revision, use the -b option of the rtag or tag commands (see node Branching and merging' in the CVS manual). Then, either checkout or update can be used to base your sources on the newly created branch. From that point on, all com‐ mit changes made within these working sources will be automatically added to a branch revision, thereby not disturbing main-line develop‐ ment in any way. For example, if you had to create a patch to the 1.2 version of the product, even though the 2.0 version is already under development, you might do:$ cvs rtag -b -r FCS1_2 FCS1_2_Patch product_module $cvs checkout -r FCS1_2_Patch product_module$ cd product_module [[ hack away ]] $cvs commit This works automatically since the -r option is sticky. Creating the branch after editing Say you have been working on some extremely experimental software, based on whatever revision you happened to checkout last week. If oth‐ ers in your group would like to work on this software with you, but without disturbing main-line development, you could commit your change to a new branch. Others can then checkout your experimental stuff and utilize the full benefit of cvs conflict resolution. The scenario might look like: [[ hacked sources are present ]]$ cvs tag -b EXPR1 $cvs update -r EXPR1$ cvs commit The update command will make the -r EXPR1 option sticky on all files. Note that your changes to the files will never be removed by the update command. The commit will automatically commit to the correct branch, because the -r is sticky. You could also do like this: [[ hacked sources are present ]] $cvs tag -b EXPR1$ cvs commit -r EXPR1 but then, only those files that were changed by you will have the -r EXPR1 sticky flag. If you hack away, and commit without specifying the -r EXPR1 flag, some files may accidentally end up on the main trunk. To work with you on the experimental change, others would simply do cvs checkout -r EXPR1 whatever_module diff Show differences between revisions · Synopsis: diff [-lR] [-k kflag] [format_options] [[-r rev1 | -D date1] [-r rev2 | -D date2]] [files...] · Requires: working directory, repository. · Changes: nothing. The diff command is used to compare different revisions of files. The default action is to compare your working files with the revi‐ sions they were based on, and report any differences that are found. If any file names are given, only those files are compared. If any directories are given, all files under them will be compared. The exit status for diff is different than for other cvs commands; for details see node Exit status' in the CVS manual. diff options These standard options are supported by diff (see node Common options' in the CVS manual, for a complete description of them): -D date Use the most recent revision no later than date. See -r for how this affects the comparison. -k kflag Process keywords according to kflag. See see node Keyword substitu‐ tion' in the CVS manual. -l Local; run only in current working directory. -R Examine directories recursively. This option is on by default. -r tag Compare with revision tag. Zero, one or two -r options can be present. With no -r option, the working file will be compared with the revision it was based on. With one -r, that revision will be compared to your current working file. With two -r options those two revisions will be compared (and your working file will not affect the outcome in any way). One or both -r options can be replaced by a -D date option, described above. The following options specify the format of the output. They have the same meaning as in GNU diff. Most options have two equivalent names, one of which is a single letter preceded by -, and the other of which is a long name preceded by --. -lines Show lines (an integer) lines of context. This option does not spec‐ ify an output format by itself; it has no effect unless it is com‐ bined with -c or -u. This option is obsolete. For proper operation, patch typically needs at least two lines of context. -a Treat all files as text and compare them line-by-line, even if they do not seem to be text. -b Ignore trailing white space and consider all other sequences of one or more white space characters to be equivalent. -B Ignore changes that just insert or delete blank lines. --binary Read and write data in binary mode. --brief Report only whether the files differ, not the details of the differ‐ ences. -c Use the context output format. -C lines --context[=lines] Use the context output format, showing lines (an integer) lines of context, or three if lines is not given. For proper operation, patch typically needs at least two lines of context. --changed-group-format=format Use format to output a line group containing differing lines from both files in if-then-else format. see node Line group formats' in the CVS manual. -d Change the algorithm to perhaps find a smaller set of changes. This makes diff slower (sometimes much slower). -e --ed Make output that is a valid ed script. --expand-tabs Expand tabs to spaces in the output, to preserve the alignment of tabs in the input files. -f Make output that looks vaguely like an ed script but has changes in the order they appear in the file. -F regexp In context and unified format, for each hunk of differences, show some of the last preceding line that matches regexp. --forward-ed Make output that looks vaguely like an ed script but has changes in the order they appear in the file. -H Use heuristics to speed handling of large files that have numerous scattered small changes. --horizon-lines=lines Do not discard the last lines lines of the common prefix and the first lines lines of the common suffix. -i Ignore changes in case; consider upper- and lower-case letters equiv‐ alent. -I regexp Ignore changes that just insert or delete lines that match regexp. --ifdef=name Make merged if-then-else output using name. --ignore-all-space Ignore white space when comparing lines. --ignore-blank-lines Ignore changes that just insert or delete blank lines. --ignore-case Ignore changes in case; consider upper- and lower-case to be the same. --ignore-matching-lines=regexp Ignore changes that just insert or delete lines that match regexp. --ignore-space-change Ignore trailing white space and consider all other sequences of one or more white space characters to be equivalent. --initial-tab Output a tab rather than a space before the text of a line in normal or context format. This causes the alignment of tabs in the line to look normal. -L label Use label instead of the file name in the context format and unified format headers. --label=label Use label instead of the file name in the context format and unified format headers. --left-column Print only the left column of two common lines in side by side for‐ mat. --line-format=format Use format to output all input lines in if-then-else format. see node Line formats' in the CVS manual. --minimal Change the algorithm to perhaps find a smaller set of changes. This makes diff slower (sometimes much slower). -n Output RCS-format diffs; like -f except that each command specifies the number of lines affected. -N --new-file In directory comparison, if a file is found in only one directory, treat it as present but empty in the other directory. --new-group-format=format Use format to output a group of lines taken from just the second file in if-then-else format. see node Line group formats' in the CVS manual. --new-line-format=format Use format to output a line taken from just the second file in if- then-else format. see node Line formats' in the CVS manual. --old-group-format=format Use format to output a group of lines taken from just the first file in if-then-else format. see node Line group formats' in the CVS manual. --old-line-format=format Use format to output a line taken from just the first file in if- then-else format. see node Line formats' in the CVS manual. -p Show which C function each change is in. --rcs Output RCS-format diffs; like -f except that each command specifies the number of lines affected. --report-identical-files -s Report when two files are the same. --show-c-function Show which C function each change is in. --show-function-line=regexp In context and unified format, for each hunk of differences, show some of the last preceding line that matches regexp. --side-by-side Use the side by side output format. --speed-large-files Use heuristics to speed handling of large files that have numerous scattered small changes. --suppress-common-lines Do not print common lines in side by side format. -t Expand tabs to spaces in the output, to preserve the alignment of tabs in the input files. -T Output a tab rather than a space before the text of a line in normal or context format. This causes the alignment of tabs in the line to look normal. --text Treat all files as text and compare them line-by-line, even if they do not appear to be text. -u Use the unified output format. --unchanged-group-format=format Use format to output a group of common lines taken from both files in if-then-else format. see node Line group formats' in the CVS man‐ ual. --unchanged-line-format=format Use format to output a line common to both files in if-then-else for‐ mat. see node Line formats' in the CVS manual. -U lines --unified[=lines] Use the unified output format, showing lines (an integer) lines of context, or three if lines is not given. For proper operation, patch typically needs at least two lines of context. -w Ignore white space when comparing lines. -W columns --width=columns Use an output width of columns in side by side format. -y Use the side by side output format. Line group formats Line group formats let you specify formats suitable for many applica‐ tions that allow if-then-else input, including programming languages and text formatting languages. A line group format specifies the out‐ put format for a contiguous group of similar lines. For example, the following command compares the TeX file myfile with the original version from the repository, and outputs a merged file in which old regions are surrounded by \begin{em}-\end{em} lines, and new regions are surrounded by \begin{bf}-\end{bf} lines. cvs diff \ --old-group-format='\begin{em} %<\end{em} ' \ --new-group-format='\begin{bf} %>\end{bf} ' \ myfile The following command is equivalent to the above example, but it is a little more verbose, because it spells out the default line group for‐ mats. cvs diff \ --old-group-format='\begin{em} %<\end{em} ' \ --new-group-format='\begin{bf} %>\end{bf} ' \ --unchanged-group-format='%=' \ --changed-group-format='\begin{em} %<\end{em} \begin{bf} %>\end{bf} ' \ myfile Here is a more advanced example, which outputs a diff listing with headers containing line numbers in a plain English'' style. cvs diff \ --unchanged-group-format='' \ --old-group-format='-------- %dn line%(n=1?:s) deleted at %df: %<' \ --new-group-format='-------- %dN line%(N=1?:s) added after %de: %>' \ --changed-group-format='-------- %dn line%(n=1?:s) changed at %df: %<-------- to: %>' \ myfile To specify a line group format, use one of the options listed below. You can specify up to four line group formats, one for each kind of line group. You should quote format, because it typically contains shell metacharacters. --old-group-format=format These line groups are hunks containing only lines from the first file. The default old group format is the same as the changed group format if it is specified; otherwise it is a format that outputs the line group as-is. --new-group-format=format These line groups are hunks containing only lines from the second file. The default new group format is same as the changed group for‐ mat if it is specified; otherwise it is a format that outputs the line group as-is. --changed-group-format=format These line groups are hunks containing lines from both files. The default changed group format is the concatenation of the old and new group formats. --unchanged-group-format=format These line groups contain lines common to both files. The default unchanged group format is a format that outputs the line group as-is. In a line group format, ordinary characters represent themselves; conversion specifications start with % and have one of the following forms. %< stands for the lines from the first file, including the trailing new‐ line. Each line is formatted according to the old line format (see node Line formats' in the CVS manual). %> stands for the lines from the second file, including the trailing newline. Each line is formatted according to the new line format. %= stands for the lines common to both files, including the trailing newline. Each line is formatted according to the unchanged line for‐ mat. %% stands for %. %c'C' where C is a single character, stands for C. C may not be a back‐ slash or an apostrophe. For example, %c':' stands for a colon, even inside the then-part of an if-then-else format, which a colon would normally terminate. %c'\O' where O is a string of 1, 2, or 3 octal digits, stands for the char‐ acter with octal code O. For example, %c'\0' stands for a null char‐ acter. Fn where F is a printf conversion specification and n is one of the fol‐ lowing letters, stands for n's value formatted with F. e The line number of the line just before the group in the old file. f The line number of the first line in the group in the old file; equals e + 1. l The line number of the last line in the group in the old file. m The line number of the line just after the group in the old file; equals l + 1. n The number of lines in the group in the old file; equals l - f + 1. E, F, L, M, N Likewise, for lines in the new file. The printf conversion specification can be %d, %o, %x, or %X, spec‐ ifying decimal, octal, lower case hexadecimal, or upper case hexa‐ decimal output respectively. After the % the following options can appear in sequence: a - specifying left-justification; an integer specifying the minimum field width; and a period followed by an optional integer specifying the minimum number of digits. For example, %5dN prints the number of new lines in the group in a field of width 5 characters, using the printf format "%5d". (A=B?T:E) If A equals B then T else E. A and B are each either a decimal con‐ stant or a single letter interpreted as above. This format spec is equivalent to T if A's value equals B's; otherwise it is equivalent to E. For example, %(N=0?no:%dN) line%(N=1?:s) is equivalent to no lines if N (the number of lines in the group in the new file) is 0, to 1 line if N is 1, and to %dN lines otherwise. Line formats Line formats control how each line taken from an input file is output as part of a line group in if-then-else format. For example, the following command outputs text with a one-column change indicator to the left of the text. The first column of output is - for deleted lines, | for added lines, and a space for unchanged lines. The formats contain newline characters where newlines are desired on output. cvs diff \ --old-line-format='-%l ' \ --new-line-format='|%l ' \ --unchanged-line-format=' %l ' \ myfile To specify a line format, use one of the following options. You should quote format, since it often contains shell metacharacters. --old-line-format=format formats lines just from the first file. --new-line-format=format formats lines just from the second file. --unchanged-line-format=format formats lines common to both files. --line-format=format formats all lines; in effect, it sets all three above options simul‐ taneously. In a line format, ordinary characters represent themselves; conver‐ sion specifications start with % and have one of the following forms. %l stands for the contents of the line, not counting its trailing new‐ line (if any). This format ignores whether the line is incomplete. %L stands for the contents of the line, including its trailing newline (if any). If a line is incomplete, this format preserves its incom‐ pleteness. %% stands for %. %c'C' where C is a single character, stands for C. C may not be a back‐ slash or an apostrophe. For example, %c':' stands for a colon. %c'\O' where O is a string of 1, 2, or 3 octal digits, stands for the char‐ acter with octal code O. For example, %c'\0' stands for a null char‐ acter. Fn where F is a printf conversion specification, stands for the line number formatted with F. For example, %.5dn prints the line number using the printf format "%.5d". see node Line group formats' in the CVS manual, for more about printf conversion specifications. The default line format is %l followed by a newline character. If the input contains tab characters and it is important that they line up on output, you should ensure that %l or %L in a line format is just after a tab stop (e.g. by preceding %l or %L with a tab char‐ acter), or you should use the -t or --expand-tabs option. Taken together, the line and line group formats let you specify many different formats. For example, the following command uses a format similar to diff's normal format. You can tailor this command to get fine control over diff's output. cvs diff \ --old-line-format='< %l ' \ --new-line-format='> %l ' \ --old-group-format='%df%(f=l?:,%dl)d%dE %<' \ --new-group-format='%dea%dF%(F=L?:,%dL) %>' \ --changed-group-format='%df%(f=l?:,%dl)c%dF%(F=L?:,%dL) %<— %>' \ --unchanged-group-format='' \ myfile diff examples The following line produces a Unidiff (-u flag) between revision 1.14 and 1.19 of backend.c. Due to the -kk flag no keywords are substi‐ tuted, so differences that only depend on keyword substitution are ignored. cvs diff -kk -u -r 1.14 -r 1.19 backend.c Suppose the experimental branch EXPR1 was based on a set of files tagged RELEASE_1_0. To see what has happened on that branch, the fol‐ lowing can be used: $cvs diff -r RELEASE_1_0 -r EXPR1 A command like this can be used to produce a context diff between two releases:$ cvs diff -c -r RELEASE_1_0 -r RELEASE_1_1 > diffs If you are maintaining ChangeLogs, a command like the following just All local modifications that have not yet been committed will be printed. $cvs diff -u | less export Export sources from CVS, similar to checkout · Synopsis: export [-flNnR] [-r rev|-D date] [-k subst] [-d dir] mod‐ ule... · Requires: repository. · Changes: current directory. This command is a variant of checkout; use it when you want a copy of the source for module without the cvs administrative directories. For example, you might use export to prepare source for shipment off- site. This command requires that you specify a date or tag (with -D or -r), so that you can count on reproducing the source you ship to others (and thus it always prunes empty directories). One often would like to use -kv with cvs export. This causes any keywords to be expanded such that an import done at some other site will not lose the keyword revision information. But be aware that doesn't handle an export containing binary files correctly. Also be aware that after having used -kv, one can no longer use the ident command (which is part of the rcs suite—see ident(1)) which looks for keyword strings. If you want to be able to use ident you must not use -kv. export options These standard options are supported by export (see node Common options' in the CVS manual, for a complete description of them): -D date Use the most recent revision no later than date. -f If no matching revision is found, retrieve the most recent revision (instead of ignoring the file). -l Local; run only in current working directory. -n Do not run any checkout program. -R Export directories recursively. This is on by default. -r tag Use revision tag. In addition, these options (that are common to checkout and export) are also supported: -d dir Create a directory called dir for the working files, instead of using the module name. see node checkout options' in the CVS manual, for complete details on how cvs handles this flag. -k subst Set keyword expansion mode (see node Substitution modes' in the CVS manual). -N Only useful together with -d dir. see node checkout options' in the CVS manual, for complete details on how cvs handles this flag. history Show status of files and users · Synopsis: history [-report] [-flags] [-options args] [files...] · Requires: the file$CVSROOT/CVSROOT/history · Changes: nothing. cvs can keep a history file that tracks each use of the checkout, commit, rtag, update, and release commands. You can use history to display this information in various formats. Logging must be enabled by creating the file $CVSROOT/CVSROOT/his‐ tory. Note: history uses -f, -l, -n, and -p in ways that conflict with the normal use inside cvs (see node Common options' in the CVS manual). history options Several options (shown above as -report) control what kind of report is generated: -c Report on each time commit was used (i.e., each time the repository was modified). -e Everything (all record types). Equivalent to specifying -x with all record types. Of course, -e will also include record types which are added in a future version of cvs; if you are writing a script which can only handle certain record types, you'll want to specify -x. -m module Report on a particular module. (You can meaningfully use -m more than once on the command line.) -o Report on checked-out modules. This is the default report type. -T Report on all tags. -x type Extract a particular set of record types type from the cvs history. The types are indicated by single letters, which you may specify in combination. Certain commands have a single record type: F release O checkout E export T rtag One of five record types may result from an update: C A merge was necessary but collisions were detected (requiring man‐ ual merging). G A merge was necessary and it succeeded. U A working file was copied from the repository. P A working file was patched to match the repository. W The working copy of a file was deleted during update (because it was gone from the repository). One of three record types results from commit: A A file was added for the first time. M A file was modified. R A file was removed. The options shown as -flags constrain or expand the report without requiring option arguments: -a Show data for all users (the default is to show data only for the user executing history). -l Show last modification only. -w Show only the records for modifications done from the same working directory where history is executing. The options shown as -options args constrain the report based on an argument: -b str Show data back to a record containing the string str in either the module name, the file name, or the repository path. -D date Show data since date. This is slightly different from the normal use of -D date, which selects the newest revision older than date. -f file Show data for a particular file (you can specify several -f options on the same command line). This is equivalent to specifying the file on the command line. -n module Show data for a particular module (you can specify several -n options on the same command line). -p repository Show data for a particular source repository (you can specify sev‐ eral -p options on the same command line). -r rev Show records referring to revisions since the revision or tag named rev appears in individual rcs files. Each rcs file is searched for the revision or tag. -t tag Show records since tag tag was last added to the history file. This differs from the -r flag above in that it reads only the history file, not the rcs files, and is much faster. -u name Show records for user name. -z timezone Show times in the selected records using the specified time zone instead of UTC. import Import sources into CVS, using vendor branches · Synopsis: import [-options] repository vendortag releasetag... · Requires: Repository, source distribution directory. · Changes: repository. Use import to incorporate an entire source distribution from an out‐ side source (e.g., a source vendor) into your source repository directory. You can use this command both for initial creation of a repository, and for wholesale updates to the module from the outside source. see node Tracking sources' in the CVS manual, for a discus‐ sion on this subject. The repository argument gives a directory name (or a path to a direc‐ tory) under the cvs root directory for repositories; if the directory did not exist, import creates it. When you use import for updates to source that has been modified in your source repository (since a prior import), it will notify you of any files that conflict in the two branches of development; use checkout -j to reconcile the differences, as import instructs you to do. If cvs decides a file should be ignored (see node cvsignore' in the CVS manual), it does not import it and prints I followed by the filename (see node import output' in the CVS manual, for a complete description of the output). If the file$CVSROOT/CVSROOT/cvswrappers exists, any file whose names match the specifications in that file will be treated as packages and the appropriate filtering will be performed on the file/directory before being imported. see node Wrappers' in the CVS manual. The outside source is saved in a first-level branch, by default 1.1.1. Updates are leaves of this branch; for example, files from the first imported collection of source will be revision 1.1.1.1, then files from the first imported update will be revision 1.1.1.2, and so on. At least three arguments are required. repository is needed to iden‐ tify the collection of source. vendortag is a tag for the entire branch (e.g., for 1.1.1). You must also specify at least one releasetag to identify the files at the leaves created each time you execute import. Note that import does not change the directory in which you invoke it. In particular, it does not set up that directory as a cvs work‐ ing directory; if you want to work with the sources import them first and then check them out into a different directory (see node Getting the source' in the CVS manual). import options This standard option is supported by import (see node Common options' in the CVS manual, for a complete description): -m message There are the following additional special options. -b branch See see node Multiple vendor branches' in the CVS manual. -k subst Indicate the keyword expansion mode desired. This setting will apply to all files created during the import, but not to any files that previously existed in the repository. See see node Substitution modes' in the CVS manual, for a list of valid -k settings. -I name Specify file names that should be ignored during import. You can use this option repeatedly. To avoid ignoring any files at all (even those ignored by default), specify -I !'. name can be a file name pattern of the same type that you can specify in the .cvsignore file. see node cvsignore' in the CVS manual. -W spec Specify file names that should be filtered during import. You can use this option repeatedly. spec can be a file name pattern of the same type that you can specify in the .cvswrappers file. see node Wrappers' in the CVS manual. import output import keeps you informed of its progress by printing a line for each file, preceded by one character indicating the status of the file: U file The file already exists in the repository and has not been locally modified; a new revision has been created (if necessary). N file The file is a new file which has been added to the repository. C file The file already exists in the repository but has been locally modi‐ fied; you will have to merge the changes. I file The file is being ignored (see node cvsignore' in the CVS manual). L file The file is a symbolic link; cvs import ignores symbolic links. Peo‐ ple periodically suggest that this behavior should be changed, but if there is a consensus on what it should be changed to, it is not apparent. (Various options in the modules file can be used to recre‐ ate symbolic links on checkout, update, etc.; see node modules' in the CVS manual.) import examples See see node Tracking sources' in the CVS manual, and see node From files' in the CVS manual. log · Synopsis: log [options] [files...] · Requires: repository, working directory. · Changes: nothing. Display log information for files. log used to call the rcs utility rlog. Although this is no longer true in the current sources, this history determines the format of the output and the options, which are not quite in the style of the other cvs commands. The output includes the location of the rcs file, the head revision (the latest revision on the trunk), all symbolic names (tags) and some other things. For each revision, the revision number, the date, the author, the number of lines added/deleted and the log message are printed. All dates are displayed in local time at the client. This is typically specified in the $TZ environment variable, which can be set to govern how log displays dates. Note: log uses -R in a way that conflicts with the normal use inside cvs (see node Common options' in the CVS manual). log options By default, log prints all information that is available. All other options restrict the output. -b Print information about the revisions on the default branch, normally the highest branch on the trunk. -d dates Print information about revisions with a checkin date/time in the range given by the semicolon-separated list of dates. The date for‐ mats accepted are those accepted by the -D option to many other cvs commands (see node Common options' in the CVS manual). Dates can be combined into ranges as follows: d1<d2 d2>d1 Select the revisions that were deposited between d1 and d2. <d d> Select all revisions dated d or earlier. d< >d Select all revisions dated d or later. d Select the single, latest revision dated d or earlier. The > or < characters may be followed by = to indicate an inclusive range rather than an exclusive one. Note that the separator is a semicolon (;). -h Print only the name of the rcs file, name of the file in the working directory, head, default branch, access list, locks, symbolic names, and suffix. -l Local; run only in current working directory. (Default is to run recursively). -N Do not print the list of tags for this file. This option can be very useful when your site uses a lot of tags, so rather than "more"'ing over 3 pages of tag information, the log information is presented without tags at all. -R Print only the name of the rcs file. -rrevisions Print information about revisions given in the comma-separated list revisions of revisions and ranges. The following table explains the available range formats: rev1:rev2 Revisions rev1 to rev2 (which must be on the same branch). rev1::rev2 The same, but excluding rev1. :rev ::rev Revisions from the beginning of the branch up to and including rev. rev: Revisions starting with rev to the end of the branch containing rev. rev:: Revisions starting just after rev to the end of the branch contain‐ ing rev. branch An argument that is a branch means all revisions on that branch. branch1:branch2 branch1::branch2 A range of branches means all revisions on the branches in that range. branch. The latest revision in branch. A bare -r with no revisions means the latest revision on the default branch, normally the trunk. There can be no space between the -r option and its argument. -S Suppress the header if no revisions are selected. -s states Print information about revisions whose state attributes match one of the states given in the comma-separated list states. -t Print the same as -h, plus the descriptive text. -wlogins Print information about revisions checked in by users with login names appearing in the comma-separated list logins. If logins is omitted, the user's login is assumed. There can be no space between the -w option and its argument. log prints the intersection of the revisions selected with the options -d, -s, and -w, intersected with the union of the revisions selected by -b and -r. log examples Since log shows dates in local time, you might want to see them in Coordinated Universal Time (UTC) or some other timezone. To do this you can set your$TZ environment variable before invoking cvs: $TZ=UTC cvs log foo.c$ TZ=EST cvs log bar.c (If you are using a csh-style shell, like tcsh, you would need to pre‐ fix the examples above with env.) ls & rls · ls [-e | -l] [-RP] [-r revision] [-D date] [path...] · Requires: repository for rls, repository & working directory for ls. · Changes: nothing. · Synonym: dir & list are synonyms for ls and rdir & rlist are synonyms for rls. The ls and rls commands are used to list files and directories in the repository. By default ls lists the files and directories that belong in your working directory, what would be there after an update. By default rls lists the files and directories on the tip of the trunk in the topmost directory of the repository. Both commands accept an optional list of file and directory names, relative to the working directory for ls and the topmost directory of the repository for rls. Neither is recursive by default. ls & rls options These standard options are supported by ls & rls: -d Show dead revisions (with tag when specified). -e Display in CVS/Entries format. This format is meant to remain easily parsable by automation. -l Display all details. -P Don't list contents of empty directories when recursing. -R List recursively. -r revision Show files with revision or tag. -D date Show files from date. rls examples $cvs rls cvs rls: Listing module: .' CVSROOT first-dir$ cvs rls CVSROOT cvs rls: Listing module: CVSROOT' checkoutlist commitinfo config cvswrappers modules notify rcsinfo taginfo verifymsg rdiff 'patch' format diffs between releases · rdiff [-flags] [-V vn] [-r t|-D d [-r t2|-D d2]] modules... · Requires: repository. · Changes: nothing. · Synonym: patch Builds a Larry Wall format patch(1) file between two releases, that can be fed directly into the patch program to bring an old release up-to-date with the new release. (This is one of the few cvs com‐ mands that operates directly from the repository, and doesn't require a prior checkout.) The diff output is sent to the standard output device. You can specify (using the standard -r and -D options) any combina‐ tion of one or two revisions or dates. If only one revision or date is specified, the patch file reflects differences between that revi‐ sion or date and the current head revisions in the rcs file. Note that if the software release affected is contained in more than one directory, then it may be necessary to specify the -p option to the patch command when patching the old sources, so that patch is able to find the files that are located in other directories. rdiff options These standard options are supported by rdiff (see node Common options' in the CVS manual, for a complete description of them): -D date Use the most recent revision no later than date. -f If no matching revision is found, retrieve the most recent revision -l Local; don't descend subdirectories. -R Examine directories recursively. This option is on by default. -r tag Use revision tag. In addition to the above, these options are available: -c Use the context diff format. This is the default format. -s Create a summary change report instead of a patch. The summary the releases. It is sent to the standard output device. This is useful for finding out, for example, which files have changed between two dates or revisions. -t A diff of the top two revisions is sent to the standard output device. This is most useful for seeing what the last change to a file was. -u Use the unidiff format for the context diffs. Remember that old ver‐ sions of the patch program can't handle the unidiff format, so if you plan to post this patch to the net you should probably not use -u. -V vn Expand keywords according to the rules current in rcs version vn (the expansion format changed with rcs version 5). Note that this option is no longer accepted. cvs will always expand keywords the way that rcs version 5 does. rdiff examples release 1.2 to 1.4 of the tc compiler. You have no such patches on hand, but with cvs that can easily be fixed with a command such as this: $cvs rdiff -c -r FOO1_2 -r FOO1_4 tc | \$$Mail -s 'The patches you asked for' foo@example.net Suppose you have made release 1.3, and forked a branch called R_1_3fix for bug fixes. R_1_3_1 corresponds to release 1.3.1, which was made some time ago. Now, you want to see how much development has been done on the branch. This command can be used:$ cvs patch -s -r R_1_3_1 -r R_1_3fix module-name cvs rdiff: Diffing module-name File ChangeLog,v changed from revision 1.52.2.5 to 1.52.2.6 File foo.c,v changed from revision 1.52.2.3 to 1.52.2.4 File bar.h,v changed from revision 1.29.2.1 to 1.2 release Indicate that a Module is no longer in use · release [-d] directories... · Requires: Working directory. · Changes: Working directory, history log. This command is meant to safely cancel the effect of cvs checkout. Since cvs doesn't lock files, it isn't strictly necessary to use this command. You can always simply delete your working directory, if you like; but you risk losing changes you may have forgotten, and you leave no trace in the cvs history file (see node history file' in the CVS manual) that you've abandoned your checkout. Use cvs release to avoid these problems. This command checks that no uncommitted changes are present; that you are executing it from imme‐ diately above a cvs working directory; and that the repository recorded for your files is the same as the repository defined in the module database. If all these conditions are true, cvs release leaves a record of its in the cvs history log. release options The release command supports one command option: -d Delete your working copy of the file if the release succeeds. If this flag is not given your files will remain in your working direc‐ tory. WARNING: The release command deletes all directories and files recursively. This has the very serious side-effect that any direc‐ tory that you have created inside your checked-out sources, and not added to the repository (using the add command; see node Adding files' in the CVS manual) will be silently deleted—even if it is non- empty! release output Before release releases your sources it will print a one-line message for any file that is not up-to-date. U file P file There exists a newer revision of this file in the repository, and you have not modified your local copy of the file (U and P mean the same thing). A file The file has been added to your private copy of the sources, but has not yet been committed to the repository. If you delete your copy of the sources this file will be lost. R file The file has been removed from your private copy of the sources, but has not yet been removed from the repository, since you have not yet committed the removal. see node commit' in the CVS manual. M file The file is modified in your working directory. There might also be a newer revision inside the repository. ? file file is in your working directory, but does not correspond to any‐ thing in the source repository, and is not in the list of files for cvs to ignore (see the description of the -I option, and see node cvsignore' in the CVS manual). If you remove your working sources, this file will be lost. release examples Release the tc directory, and delete your local working copy of the files. $cd .. # You must stand immediately above the # sources when you issue cvs release.$ cvs release -d tc You have [0] altered files in this repository. Are you sure you want to release (and delete) directory tc': y \$ update Bring work tree in sync with repository · update [-ACdflPpR] [-I name] [-j rev [-j rev]] [-k kflag] [-r tag|-D date] [-W spec] files... · Requires: repository, working directory. · Changes: working directory. After you've run checkout to create your private copy of source from the common repository, other developers will continue changing the central source. From time to time, when it is convenient in your development process, you can use the update command from within your working directory to reconcile your work with any revisions applied to the source repository since your last checkout or update. Without the -C option, update will also merge any differences between the local copy of files and their base revisions into any destination revisions specified with -r, -D, or -A. update options These standard options are available with update (see node Common options' in the CVS manual, for a complete description of them): -D date Use the most recent revision no later than date. This option is sticky, and implies -P. See see node Sticky tags' in the CVS man‐ -f Only useful with the -D date or -r tag flags. If no matching revi‐ sion is found, retrieve the most recent revision (instead of ignoring the file). -k kflag Process keywords according to kflag. See see node Keyword substitu‐ tion' in the CVS manual. This option is sticky; future updates of this file in this working directory will use the same kflag. The status command can be viewed to see the sticky options. See see node Invoking CVS' in the CVS manual, for more information on the status command. -l Local; run only in current working directory. see node Recursive behavior' in the CVS manual. -P Prune empty directories. See see node Moving directories' in the CVS manual. -p Pipe files to the standard output. -R Update directories recursively (default). see node Recursive behav‐ ior' in the CVS manual. -r rev Retrieve revision/tag rev. This option is sticky, and implies -P. See see node Sticky tags' in the CVS manual, for more information on sticky tags/dates. These special options are also available with update. -A Reset any sticky tags, dates, or -k options. See see node Sticky -C Overwrite locally modified files with clean copies from the reposi‐ tory (the modified file is saved in .#file.revision, however). -d Create any directories that exist in the repository if they're miss‐ ing from the working directory. Normally, update acts only on direc‐ tory. This is useful for updating directories that were created in the repository since the initial checkout; but it has an unfortunate side effect. If you deliberately avoided certain directories in the repository when you created your working directory (either through use of a module name or by listing explicitly the files and directo‐ ries you wanted on the command line), then updating with -d will cre‐ ate those directories, which may not be what you want. -I name Ignore files whose names match name (in your working directory) dur‐ ing the update. You can specify -I more than once on the command line to specify several files to ignore. Use -I ! to avoid ignoring any files at all. see node cvsignore' in the CVS manual, for other ways to make cvs ignore some files. -Wspec Specify file names that should be filtered during update. You can use this option repeatedly. spec can be a file name pattern of the same type that you can specify in the .cvswrappers file. see node Wrappers' in the CVS manual. -jrevision With two -j options, merge changes from the revision specified with the first -j option to the revision specified with the second j option, into the working directory. With one -j option, merge changes from the ancestor revision to the revision specified with the -j option, into the working directory. The ancestor revision is the common ancestor of the revision which the working directory is based on, and the revision specified in the -j option. Note that using a single -j tagname option rather than -j branchname to merge changes from a branch will often not remove files which were removed on the branch. see node Merging adds and removals' in the CVS manual, for more. In addition, each -j option can contain an optional date specifica‐ tion which, when used with branches, can limit the chosen revision to one within a specific date. An optional date is specified by adding a colon (:) to the tag: -jSymbolic_Tag:Date_Specifier. see node Branching and merging' in the CVS manual. update output update and checkout keep you informed of their progress by printing a line for each file, preceded by one character indicating the status of the file: U file The file was brought up to date with respect to the repository. This is done for any file that exists in the repository but not in your source, and for files that you haven't changed but are not the most P file Like U, but the cvs server sends a patch instead of an entire file. This accomplishes the same thing as U using less bandwidth. A file The file has been added to your private copy of the sources, and will be added to the source repository when you run commit on the file. This is a reminder to you that the file needs to be committed. R file The file has been removed from your private copy of the sources, and will be removed from the source repository when you run commit on the file. This is a reminder to you that the file needs to be committed. M file The file is modified in your working directory. M can indicate one of two states for a file you're working on: either there were no modifications to the same file in the repository, so that your file remains as you last saw it; or there were modifica‐ tions in the repository as well as in your copy, but they were merged successfully, without conflict, in your working directory. cvs will print some messages if it merges your work, and a backup copy of your working file (as it looked before you ran update) will be made. The exact name of that file is printed while update runs. C file A conflict was detected while trying to merge your changes to file with changes from the source repository. file (the copy in your working directory) is now the result of attempting to merge the two revisions; an unmodified copy of your file is also in your working directory, with the name .#file.revision where revision is the revi‐ sion that your modified file started from. Resolve the conflict as described in see node Conflicts example' in the CVS manual. (Note that some systems automatically purge files that begin with .# if they have not been accessed for a few days. If you intend to keep a copy of your original file, it is a very good idea to rename it.) Under vms, the file name starts with __ rather than .#. ? file file is in your working directory, but does not correspond to any‐ thing in the source repository, and is not in the list of files for cvs to ignore (see the description of the -I option, and see node cvsignore' in the CVS manual). AUTHORS Dick Grune Original author of the cvs shell script version posted to comp.sources.unix in the volume6 release of December, 1986. Credited with much of the cvs conflict resolution algorithms. Brian Berliner Coder and designer of the cvs program itself in April, 1989, based on the original work done by Dick. Jeff Polk Helped Brian with the design of the cvs module and vendor branch support and author of the checkin(1) shell script (the ancestor of cvs import). Larry Jones, Derek R. Price, and Mark D. Baushke Have helped maintain cvs for many years. And many others too numerous to mention here. The most comprehensive manual for CVS is Version Management with CVS by Per Cederqvist et al. Depending on your system, you may be able to get it with the info CVS command or it may be available as cvs.pdf (Porta‐ ble Document Format), cvs.ps (PostScript), cvs.texinfo (Texinfo source), or cvs.html. CVS, development of CVS, and more, see: http://cvshome.org http://www.loria.fr/~molli/cvs-index.html ci(1), co(1), cvs(5), cvsbug(8), diff(1), grep(1), patch(1), rcs(1), rcsd‐ iff(1), rcsmerge(1), rlog(1). CVS(1) [top] List of man pages available for OpenDarwin Copyright (c) for man pages and the logo by the respective OS vendor. For those who want to learn more, the polarhome community provides shell access and support.
2018-12-10 10:28:32
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-5-linear-functions-chapter-test-page-353/14
## Algebra 1 The vertex of the graph $y=abs(x)$ is at (0,0). However, the graph of $y=abs(x)+3$ is shifted three units up. Thus, the vertex of $y=abs(x)+3$ is at (0, 3).
2019-10-22 23:58:13
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https://dml.cz/handle/10338.dmlcz/126098
# Article Full entry | PDF   (0.3 MB) Keywords: continuous multifunction; selection; quasicontinuity Summary: The paper presents new quasicontinuous selection theorem for continuous multifunctions $F X \longrightarrow\Bbb R$ with closed values, $X$ being an arbitrary topological space. It is known that for $2^{\Bbb R}$ with the Vietoris topology there is no continuous selection. The result presented here enables us to show that there exists a quasicontinuous and upper$\langle$lower$\rangle$-semicontinuous selection for this space. Moreover, one can construct a selection whose set of points of discontinuity is nowhere dense. References: [1] R. Engelking R. V. Heath E. Michael: Topological well-ordering and continuous selections. Invent. Math. 6 (1968), 150-158. DOI 10.1007/BF01425452 | MR 0244959 [2] I. Kupka: Quasicontinuous selections for compact-valued multifunctions. Math. Slovaca 43 (1993), 69-75. MR 1216269 | Zbl 0784.54023 [3] I. Kupka: Continuous multifunction from [-1,0] to R having noncontinuous selection. Publ. Math. (Submitted). [4] K. Kuratowski: Topologie I. PWN Warszawa, 1952. [5] N. Levine: Semi-open sets and semi-continuity in topological spaces. Amer. Math. Monthly 70 (1963), 36-41. DOI 10.1080/00029890.1963.11990039 | MR 0166752 | Zbl 0113.16304 [6] M. Matejdes: Sur les sélecteurs des multifonctions. Math. Slovaca 37 (1987), 111-124. MR 0899022 | Zbl 0629.54013 [7] M. Matejdes: On selections of multifunctions. Math. Bohem. 118 (1993), 255-260. MR 1239120 | Zbl 0785.54022 [8] M. Matejdes: Quasi-continuous and cliquish selections of multifunctions on product spaces. Real Anal. Exchange. To appear. MR 1205513 | Zbl 0782.54019 [9] E. Michael: Continuous selections I. Ann. of Math. 63 (1956), 361-382. DOI 10.2307/1969615 | MR 0077107 | Zbl 0071.15902 [10] E. Michael: Continuous selections II. Ann. of Math. 64 (1956), 562-580. DOI 10.2307/1969603 | MR 0080909 | Zbl 0073.17702 [11] S. B. Nadler: Hyperspaces of sets. Marcel Dekker, Inc., New York and Bassel, 1978. MR 0500811 | Zbl 0432.54007 [12] T. Neubrunn: Quasi-continuity. Real Anal. Exchange H (1988-89), 259-306. MR 0995972 [13] A. Neubrunnová: On certain generalizations of the notion of continuity. Mat. Časopis 23 (1973), 374-380. MR 0339051 Partner of
2021-01-21 05:08:07
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http://cms.math.ca/10.4153/CMB-2003-016-6
location:  Publications → journals → CMB Abstract view # Torsion Points on Certain Families of Elliptic Curves Fix an elliptic curve $y^2 = x^3+Ax+B$, satisfying $A,B \in \ZZ$, $A\geq |B| > 0$. We prove that the $\QQ$-torsion subgroup is one of $(0)$, $\ZZ/3\ZZ$, $\ZZ/9\ZZ$. Related numerical calculations are discussed.
2015-11-26 14:15:41
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https://www.maths.unsw.edu.au/seminars/fullarchive?term_node_tid_depth_3=266
# Full Seminar Archive Our regular seminar program covers a broad range of topics from applied mathematics, pure mathematics and statistics. All staff and students are welcome. This page has a complete list of past seminars and a list restricted by year can be accessed via the left-hand menu. Marley Young - UNSW We consider the problem of when the first $n$ iterates of a given rational function over an arbitrary field are multiplicatively independent. This leads to a generalisation of a method of Gao (1999)... Andrew Hone - University of Kent The appearance of primes in a family of linear recurrence sequences labelled by a positive integer $n$ is considered. The terms of each sequence correspond to a particular class of Lehmer numbers, or... Caroline Turnage-Butterbaugh - Duke University This talk will present a new effective Chebotarev theorem that holds for all but a possible zero-density subfamily of certain families of number fields of fixed degree. For certain families, this... Alina Ostafe - UNSW Bombieri, Masser and Zannier (1999) proved that the intersection of a curve defined over a number field with the union of all proper algebraic subgroups of the multiplicative group $\mathbb{G}_m^n$... Joachim von zur Gathen - University of Bonn We consider natural combinatorial questions about systems of multivariate polynomials over a finite field and the variety V that they define over an algebraic closure. Fixing the number of variables... Liangyi Zhao - UNSW Borrowing a line from an anarchist and mangling it somewhat: happy family members are all alike.  Indeed, the objects of interest of this talk will be families consisting of kindred members having... Simon Macourt - UNSW We provide a background on some recent results on multilinear exponential sums as well as collinear triples over a finite field. We then focus on the specific cases of weighted trilinear and... Antoine Joux - Laboratoire d'informatique de Paris 6 We consider the problem of solving multivariate systems of Boolean polynomial equations: starting from a system of $m$ polynomials of degree at most $d$ in $n$ variables, we want to find its... Mumtaz Hussain - La Trobe University Let $\psi:\mathbb R_+\to\mathbb R_+$ be a non-increasing function. A real number $x$ is said to be $\psi$-Dirichlet improvable if it admits an improvement to Dirichlet's theorem in the following... Andrew Hone - University of Kent (UK) and UNSW There are relatively few transcendental numbers for which the continued fraction expansion is explicitly known. Here we present two new families of continued fractions for Engel series - sums of...
2018-03-24 14:13:02
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http://aineshbakshi.com/
Hi! I am Ainesh Bakshi, a Research Fellow at Rutgers, New Brunswick. I finished my undergrad at Rutgers in January 2016. My research interests lie in the intersection of Algorithms and Machine Learning. Currently, I am working on Theoretical Machine Learning and Approximation Algorithms. Previously, I interned in the Search and Discoverabiltiy Group at Bloomberg and the Machine Learning and Optimization group at Microsoft Research. I have also been involved with research in External Memory Algorithms and their application to File Systems. ## Manuscripts • A. Bakshi, P. Awasthi, "Efficient Clustering Algorithms for Computing Better Local Optima" , Manuscript. • A. Bakshi, "Polynomial-time Algorithm for {1, 2}-Instances below 2-perturbation resilience" , Manuscript. • ## Publications • A. Conway, A. Bakshi, Y. Jiao, W. Jannen, Y. Zhan, J. Yuan, M. Bender, R. Johnson, B. Kuzmaul, D. Porter, M. Farach-Colton,"File Systems fated for Senescence? Nonsense, Says Science!", To appear at USENIX Conference on File and Storage Technologies (FAST) 2017. • R. Aggarwal, A. Bakshi, “Non Dominated Sorting Genetic Algorithm for Chance Constrained Supplier Selection Model with Volume Discounts” ACIIDS, Lecture Notes in Computer Science pp. 465–474, Apr 2014. • A. Bakshi, K. Goel and Raunaq. Vohra, “A Novel Feature Selection and Extraction Technique for Classification” ICFHR 2014, pp. 104-109. • R. Sant, N. Kulkarni, A. Bakshi, K. Goel, and S. Kapur, “Autonomous Robot Navigation: Path Planning on a Detail-Preserving Reduced-Complexity Representation of 3D Point Clouds” ICVS 2013, pp. 173–182, Jun 2013. • ## Posters • A. Bakshi, K. Goel and Raunaq. Vohra, “A Novel Feature Selection and Extraction Technique for Classification” SMC 2014, pp. 4033–4034, Oct 2014. • A. Bakshi, Kostas Bekris, "Human Robot Interaction: Machine Vision and End Effector Control" 10th Annual Aresty Undergraduate Research Symposium, Rutgers. • ## Research #### Stable Clustering Clustering is one of the most common problems in Machine Learning. Clustering under center based objectives ($k$-means, $k$-median, $k$-center) is NP-Hard. However, in practice, the $k$-means heuristic works reasonably well, and converges quickly. In order to better model the data sets we observe in the real world, we must move beyond worst case analysis. Therefore, we study various natural notions of "stability", including Bilu and Linial stability, where perturbing any set of pairwise distances by a factor of at most $\alpha$, the optimal clustering does not change. Assuming that our data is $\alpha$ stable provides structure that we can exploit to come up with polynomial-time algorithms to find the optimal clustering. Awasthi et. al. presented the first polynomial-time algorithm for clustering $3$-stable instances. The best known result is a polynomial-time algorithm for $2$-stable instances by Makarychev et. al. The only known non trivial lower bound is for the $k$-center objective which states that any instance that is $2-\epsilon$ stable is NP-Hard. There is no known lower bound for stable instances with $k$-means and $k$-median objective. The best result we have so far shows that no existing reduction technique can be used to prove a lower bound for $k$-means and $k$-median objectives. Such reductions reduce NP-Complete problems like $1-3$ SAT, $3$-Cover or Perfect Dominating Set (PDS) to a clustering instance where all pair-wise distances are either 1 or 2. In particular, Ben-David and Reyzin (2014) reduce the perfect dominating set promise problem to a clustering instance with all pair-wise distances either $\frac{1}{2}$ or $1$ to hardness of $\alpha$-center stability. Since, scaling the distances by any factor preserves perturbation resilience, their instance is equivalent to one where all pair-wise distances are $1$ or $2$ (scaling by 2). Balcan, Haghtalab and White (2016) also reduce PDS to a clustering instance with pair-wise distances either $1$ or $2$ to show hardness of $(2-\epsilon)$-perturbation resilience under the k-center objective. I came up with a polynomial time algorithm that solves such instances for $\alpha \geq 1 + \epsilon$ , for any $\epsilon > 0$, given $k$-means or $k$-median objective. Therefore, we can conclude that such instances are easy, and we would need new reduction techniques to prove hardness results for $k$-means and $k$-median. With Professor Awasthi, I also formalized the problem of efficiently computing good local optima for a non-convex optimization problem. We study this problem in the context of clustering and design polynomial-time algorithms to achieve the desired output. We define a good locally optimal solution as a clustering where each point is closer to the center of its optimal cluster than to any point outside the optimal cluster. Based on this definition, we prove that if a set of $n$ points is guaranteed to have a good locally optimal $k$-clustering, there exists a polynomial-time algorithm that outputs a set of $k$ clusters such that the resulting clustering is locally good. Further, we also address the common issue in clustering problems that concerns the choice of $k$, the number of clusters. In many settings it is unclear apriori what the right number of clusters should be, whereas it is easier to prescribe a minimum cluster size. We prove that there exists a polynomial-time algorithm which when given a set of $n$ points such that there exists a good locally optimal clustering in which the size of each cluster is at least $r(n)$, outputs good locally optimal clustering where each cluster has cardinality at least $r(n)$. #### External Memory Algorithms External-memory data structures are analysed in terms of I/O complexity. $B^\epsilon$-trees have insert, point query and range query complexity of $\frac{\log_B{N}}{\epsilon B^{1-\epsilon}}$, $\frac{\log_B{N}}{\epsilon}$ and $\frac{\log_B{N}}{\epsilon} + \frac{k}{B}$ respectively, where $N$ is the total number of items in the tree, $B$ is the nodesize and $k$ is the range of the data we query over. In contrast, $B$-trees have $\log_B{N}$ insert and point query and $\log_B{N} + \frac{k}{B}$ range query complexity. A $B^\epsilon$-tree is a write optimised dictionary data structure, i.e. instead of writing entire blocks for small updates and inserts, it aggregates these operations to perform larger writes. Therefore, a larger nodesize ($B$) increases insert and range query performance without amplifying the number of writes, presenting a good theoretical candidate to mitigate aging in file systems. As of now, most Unix based file systems used poor / no data structures and heuristics for space allocation and claim that aging ( read/write performance degradation over time ) is a solved problem. However, seek times on rotational disks have remained approximately the same and bandwidth has increased as a function of capacity, thus the theory predicts that fragmentation on disk should have an exacerbated effect on read performance (aging) over time. We created realistic workloads, such as a mailserver and running through the git history of large open-source projects that caused these heuristics to fail. For example, on EXT4 and ZFS, a few hundred git pull operations can reduce read performance by a factor of 2, as compared to a defragmented copy of the same file system, and performing a thousand pulls can reduce performance by up to $30$x. We created microbenchmarks to show that traditional file systems are very sensitive to file creation order. Creating files in random order can slow down file system reads by a factor of up to 8x which indicates that file-layout heuristics can fail easily. We also demonstrated that BetrFS, a file system based on $B^\epsilon$-tree avoids aging, which corroborates the theoretical performance guarantees of $B^\epsilon$-trees. #### Query Reformulation During the summer of '15 I interned in the Search and Discoverability group at Bloomberg L.P. The group primarily focuses on search within the Bloomberg terminal, which indexes a large amount of financial data, news, profiles and wikipedia articles. My project at Bloomberg was query reformulation, i.e. detecting ambiguous or vague queries and replacing or adding words to implicitly predict what the user might be searching for. I created a contextual language model for adding and substituting words using existing user query logs and generating n-grams for all the indexed documents. The contextual model was similar to the one presented by Wang & Zhai in the following paper: X. Wang and C. Zhai "Mining Term Association Patterns from Search Logs for Effective Query Reformulation" In Proceedings of CIKM, pages 479-488, 2008. I also contributed to generating click graphs, running hadoop clusters to aggregate training data and designing performance metrics for search result evaluation. In addition I evaluated the performance of the search engine by comparing the NDCG with and without the reformulation system. The reformulation system showed brilliant results that could also be used as suggested search queries. I integrated my service with the search engine and pushed the code to production, on all Bloomberg terminals across the world. #### Kernel Approximation During the summer of '14 I interned at at Microsoft Research, India, in the Machine Learning and Optimization group. I worked with Manik Varma on Non Linear Kernel approximations for Support Vector Machines. The most popular non-linear kernel is the Gaussian (RBF) Kernel, given its versatile applicability. However, the evaluation complexity is linear in the number of support vectors, which can be really large in real world datasets. I used a Taylor Series expansion to approximate the Additive Gaussian Kernel (similar to the RBF Kernel), giving tremendous speedup in evaluation time, bringing the complexity down to constant in the number of training examples. The algorithm performed up to 20% more accurately than other linear approximations like Efficient Additive Kernels ( Vedaldi et al. ) and Max Margin Additive Classifiers ( Maji et al ). #### Human Robot Interaction At Rutgers, I worked as a Aresty Research Assistant with Professor Kostas Bekris, who heads the Pracsys lab at Rutgers. The problem statement was to evaluate collision free trajectories that humans are accustomed to. The most efficient trajectory for a robotic arm doesn’t mimic motion that is natural and intuitive for humans. In a collaborative environment where safety and precision is essential, the robot must execute trajectories that are intuitive for humans. My project entailed mapping the environment, localizing the robot with respect to the environment, detecting pose, recognizing objects and obstacles and calculating optimal gripping configurations given parallel and vacuum grippers. The Pracsys team participated in the Amazon Picking Challenge and a video of the Robot in action can be found here. I presented a poster for the same at the Aresty Undergraduate Research Symposium.
2017-04-30 08:48:48
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https://plainmath.net/90403/classify-the-following-random-variables
# Classify the following random variables as continuous or discrete: the length of hairs on a horse. Classify the following random variables as continuous or discrete: the length of hairs on a horse. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Saige Barton We need to determine whether the random variable "length of hairs" is discrete or continuous. Discrete random variables take on values that are restricted to defined separated values, for example: integers or counts. Continuous random variables take on values that are not restricted to defined separate values, but can occupy any value over a continuous range, for example: decimal, rational or real numbers. The length of hairs can take on decimal values such as 2.75 cm. The random variable is then continuous. Result: Continuous.
2022-10-04 23:45:11
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http://math.stackexchange.com/questions/74976/group-of-units-of-finite-type-related-to-the-factorization-of-ideals
# Group of units of finite type - related to the factorization of ideals Let $K$ be a number field, let $A$ be the ring of integers of $K$, and let $P$ denote the set of maximal ideals of $A$. For $p \in P$ and $x \in K^{\times}$ write $v_{p}$ for the exponent of $p$ in the factorization of the $Ax$ into a product of prime ideals. Put $v_{p}(0) = + \infty$. Take for $P'$ the complement of a finite set $S \subset P$. Show that the group of units of $A(P')$ is of finite type and that the quotient $U/A^{\times}$ is a free $\mathbb{Z}$-module of rank the cardinality of $S$. My idea is to work with the map $x \rightarrow \left(v_{p_{1}}(x), v_{p_{2}}(x),\ldots, v_{p_{|S|}}(x)\right)$ of $U$ to $\mathbb{Z}^{s}$ that has kernel $A^{\times}$. I'm having trouble determining its image? Is it something obvious that I am missing? - What is $A(P')$? What is $U$? –  Ted Oct 23 '11 at 2:40 Please don't post as if you were assigning homework. You might want to give background (as in, explain your notation), and what your thoughts so far are. –  Arturo Magidin Oct 23 '11 at 4:15 You are right. $A'(P)$ denotes the set of elements $x$ of $K$ such that $v_{p}(x) \geq 0$ for all $p \in P'$. I included some ideas; I was in a rush when I wrote the post - that explains the lack of details –  Anna Oct 23 '11 at 5:41 @Anna: And presumably, $S=\{p_1,\ldots,p_{|S|}\}$? –  Arturo Magidin Oct 23 '11 at 5:56 To determine the image, use the Chinese remainder theorem (sometimes also called an "approximation theorem" in this context). –  Ted Oct 23 '11 at 6:10 I suppose that $A(P')$ = {$x \in K$; $v_P(x) \geq 0$ for all $p \in P'$}, and that $U$ is the group of units of $A(P')$. Let $\psi$ be the map $U \rightarrow \mathbb{Z}^{s}$ assigning $x$ to $\left(v_{p_{1}}(x), v_{p_{2}}(x),\ldots, v_{p_{|S|}}(x)\right)$. Suppose that $\psi$ is surjective. Take $(1, 0,\dots, 0) \in \mathbb{Z}^{s}$. Then there exists $x \in U$ such that $\psi(x) = (1, 0,\dots, 0)$. Then $v_{p_1}(x) = 1$, and $v_p(x) = 0$ for all $p \neq p_1$. Hence $p_1 = xA$. Since $p_1$ is not necessarily principal, $\psi$ is not necessarily surjective. This isn't right. You only have $v_p(x) = 0$ for all $p \in S \setminus \{p_1\}$, and so you can't conclude that $p_1 = xA$. –  Brandon Carter Nov 27 '12 at 0:09 @BrandonCarter Since $x \in U, v_p(x) = 0$ for all $p \in P - S$. –  Makoto Kato Nov 27 '12 at 0:37 No. For instance, 3 is a unit in $\mathbb{Z}[\sqrt {-1}]_{(i+1)}$ (here $S = \{(i+1)\}$), but $v_3(3) \neq 0$. –  Brandon Carter Nov 27 '12 at 0:42 @BrandonCarter Let $x \in U$. Let $p \in P'$. Since $1/x \in A(P')$, $v_p(1/x) \ge 0$, i.e. $v_p(x) \le 0$. On the other hand, since $x \in A(P')$, $v_p(x) \ge 0$. Hence $v_p(x) = 0$. –  Makoto Kato Nov 27 '12 at 0:58 That only applies for primes in $(P')^{-1}A$, which is to say exactly the primes contained in $S$. You can't say anything about the others. –  Brandon Carter Nov 27 '12 at 1:50
2015-05-26 08:05:07
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http://math.stackexchange.com/questions/301456/derivative-of-matrix-inverse
Derivative of matrix inverse I am trying to find the derivative of a matrix with respect to the inverse of the same matrix. The matrix in question is a non singular symmetric matrix. Any thoughts? - The starting point is the von Neumann series $$(I-A)^{-1}=I+A+A^2+\dots$$ which implies that the matrix function $A\mapsto (I-A)$ has identity as its derivative at $A=0$. The general case of $B\mapsto B^{-1}$ differentiated at $B_0$ can be reduced to the above by considering $B=(I-A)B_0$, hence $B^{-1}=B_0^{-1}(I-A)^{-1}$. The linear term here is $B_0^{-1}A$, and since $A=-(B-B_0)B_0^{-1}$, the derivative can be written as $\Delta B\mapsto -B_0^{-1}(\Delta B) B_0^{-1}$. Which, not incidentally, simplifies to $(1/x)'=-1/x^2$ when everything commutes.
2016-05-28 02:16:17
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https://socratic.org/questions/how-do-you-find-the-slope-for-x-4
# How do you find the slope for x=-4? Because this represents a vertical line where for any value of $y$ the $x$ value will be $- 4$ the slope is undefined.
2022-10-05 11:29:10
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http://openstudy.com/updates/4f720808e4b07738f5ae1f61
## y2o2 Good Question one year ago one year ago 1. y2o2 $\huge y = \cos(i \theta) + i \sin(i \theta)$ prove that y ϵ R 2. AnimalAin Use the definitions $\sin(z)=\frac{e ^{iz}-e ^{-iz}}{2i}~~~\cos(z)=\frac{e ^{iz}+e ^{-iz}}{2}$Substitute i theta for z, and it will be evident that both terms are real numbers. 3. y2o2 since : $\huge e^{i \theta } = \cos(\theta) + isin(\theta)$ therefore: $\huge \cos(i \theta) + isin(i \theta) = e^{i i \theta} = e^{- \theta} = ({1 \over e})^{\theta}$ and $\huge e \ and \ \theta \in \mathbb{R}$ so $\huge y \in \mathbb{R}$
2014-03-07 17:04:00
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http://math.stackexchange.com/questions/36010/gauss-elimination-with-constraints
# Gauss Elimination with constraints Given an invertible matrix $M \in M_n(F)$ for some field F, it can be transformed to row echelon form by elementary row operations. These include "switch", "multiply" and "add". My question is, is it still possible under the following constraints: 1. After the first "multiply" or "add" we are not allowed to switch anymore (that is - all the switching is done in the beginning). 2. You can only add $R_i + R_j \rightarrow R_i$ if $i$ > $j$ (that is, of the 2 rows being added, the one being changed is the one with the larger index). Seems to me that the answer is yes. I have tried ideas such as: perform regular Gaussian Elimination, see what switches you made, then take the original matrix, perform these switches and now perform Gauss Elimination. But I haven't been able to convince myself this or any other method always works. Note: I only care about the case where $F = \mathbb{Z}_p$ for some prime p. I asked the question more generally because my feeling is that it doesn't matter what the field is. - It is possible to perform Gaussian elimination without swapping rows. The admonition to swap rows is often intended for doing Gaussian elimination in floating point... –  Guess who it is. Apr 30 '11 at 10:42 @J. M.: Take M to be the identity matrix with 2 rows swapped. How can we bring it to row echelon form without swaps? (keeping the second contraint as well) –  user3533 Apr 30 '11 at 10:47 Ah yes, I forgot that example, thanks. :) The problem is that you don't have the guarantee that even if you've done all the swaps in the beginning, the leading submatrix will always be nonsingular, unless there is something special about your matrix (e.g. diagonally dominant or positive definite). –  Guess who it is. Apr 30 '11 at 10:50 @J.M.: Can you give an example of a matrix for which it is not possible? –  user3533 Apr 30 '11 at 10:54 Have you tried an $n\times n$ matrix whose $(n-1)\times(n-1)$ leading submatrix is singular? –  Guess who it is. Apr 30 '11 at 10:56 Short answer: Yes, you can always take an invertible matrix to REF using swaps at the beginning, followed only by adds and multiplies. This is called the PLU decomposition or the Bruhat decomposition of GL(n,F). Let's write the elementary row operations in a slightly different form: • Switch: One may switch row i and row j by multiplying the matrix on the left by P(i,j), the matrix which is the identity matrix with its *i*th and *j*th rows switched. • Multiply: One may multiply row i by a non-zero number x by multiplying the matrix on the left by D(i,x), the matrix which is the identity matrix with its *i*th row multiplied by x. • Add: One may add row j to row i (changing row i) by multiplying the matrix on the left by E(i,j), the matrix which is the identity matrix with its *i*th row containing an extra 1 in the j column. We generally require i > j. Applying a sequence of elementary row operations is thus equivalent to multiplying on the left by a sequence of these P(i,j), D(i,x), and E(i,j) matrices. This is our first big idea: elementary row operations are themselves matrices! Note that each of these operations can be undone: • Apply P(i,j) again immediately to undo P(i,j) • Apply D(i,1/x) immediately to undo D(i,x) • Apply D(j,-1), then E(i,j), then D(j,-1) to undo E(i,j) Our second big idea is that if we allow a bunch of elementary row operations in a row, we get very nice collection of row operations. Let's see how they relate to the group G = GL(n,F) of all invertible matrices. • The set of all products of D(i,x) is the subgroup D of all diagonal matrices. In terms of row ops, we can multiply each row by its own non-zero number. This subgroup is called a maximally split maximal torus of G. • The set of all products of D(i,x) and E(i,j) under the requirement that if one uses D(i,x) one must also use D(i,1/x) is the subgroup U of all lower triangular matrices with ones on the diagonal. This subgroup is called a maximal unipotent subgroup of G. If is a finite field of characteristic p, then U is a Sylow p-subgroup of G. • When the field F has size p, a prime, then the subgroup generated by just the adds is already U, but in characteristic 0, the adds by themselves don't even generate a group, and except for fields of size p, they are not normalized by D. • What about unrestricted products of adds and multiplies? The subgroups D and U intersect in only the identity, and D normalizes U, so the subgroup generated by both D and U is a semi-direct product B consisting of exactly the lower triangular matrices with non-zero diagonal entries. This subgroup is called a Borel subgroup of G. • The set of all products of P(i,j) is the subgroup P of all permutations matrices, matrices with exactly one 1 per row and column, and the others 0. This subgroup is naturally isomorphic to the Weyl group of G. Now we can examine the two sets of operations you wanted to allow: • If we only allow permutations at the beginning we get the union of cosets BP. • If we allow permutations whenever we want, we get the subgroup generated by B and P, which is in fact the entire group G of invertible matrices. In particular, given a starting matrix M, you can change M into strictly more forms using B and P willy-nilly (you can take the matrix into RREF!). However, REF can always be achieved just using BP. Here is the proof using the PLU decomposition of linear algebra: An invertible matrix is in REF if and only if it is upper triangular (and since we are allowed to use D(i,x), we can require the diagonal entries to be 1s if we want). Now use the PLU decomposition to write our matrix M as M = P*L*U for P in P, L in U, and U upper triangular. Then U=REF(M) is a row echelon form of M, and U = L−1 P−1 M is gotten from M first by applying the row swaps indicated by the permutation matrix P−1, and then by applying the adds and multiplies indicated by the lower triangular matrix L−1. The proof using Bruhat decomposition is pretty similar, but uses even more notation, so it seems a little silly to keep going after PLU. You should see PLU in any linear algebra course. Bruhat decompositions for a second group theory class are covered in Alperin–Bell, Groups and Representations. - Thank you for a very good answer! –  user3533 May 1 '11 at 16:45 Can you recommend me of a linear algebra book for my level? I recently finished a B.Sc. in mathematics. –  user3533 May 1 '11 at 17:12 @Lee: (I am hoping someone else will recommend such a book; I have a few books I like, but none that I can recommend just for "linear algebra", as I've found every book is lacking stunningly important, foundational results that absolutely everyone on the planet should know, and that there are still more of these results left to learn.) In case absolutely no-one else has ideas, I provisionally recommend Gilbert Strang's Linear Algebra textbook. We used it for 3rd year B.S. linear algebra, and I recall it has PLU. –  Jack Schmidt May 1 '11 at 22:16
2015-06-30 21:35:35
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https://www.biglist.com/lists/xsl-list/archives/200312/msg00241.html
## Re: [xsl] \ Subject: Re: [xsl] \ From: David Tolpin Date: Thu, 4 Dec 2003 18:53:15 +0400 (AMT) ```> No, but it has a _single_ path component called "\data\file.xsl" so if > that file has an <xsl:include href="foo.xsl"/> then foo.xsl is a > relative uri that corresponds to > "http://example.com/foo.xsl"; > which probably is not what was intended. No, http://example.com/\data\file.xsl/../foo.xsl file -> \data\file.xsl/../foo.xsl Then, in case the filesystem uses backslashes as path separators, file -> \data\file.xsl\..\foo.xsl Then, in case .. means one level up, as it is in the case of Win32, x\.. are removed (where x is a file name constitutient) file -> \data\foo.xsl Exactly what was required. I don't think .. has special meaning in URLs. In file names, it only has special meaning under Win32 and similar systems, where it means 'one level up'. If you type
2022-06-28 09:34:35
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http://werner.yellowcouch.org/Papers/stepwise/index.html
Home Papers Reports Projects Code Fragments Dissertations Presentations Posters Proposals Lectures given Course notes # Stepwise Tempo Changes in BpmDj Werner Van Belle1* - werner@yellowcouch.org, werner.van.belle@gmail.com 1- Yellowcouch; Abstract :  kbpm-play provides an algorithm that automatically changes the play-rate from its current speed back to normal speed. A first version of the software relied on a linear change of pace, which turned out to be slightly wrong. In this short article we describe what went wrong and how to do it right. Keywords:  tempo changes, psychoacoustics, preceived tempo change, bpmdj Reference:  Werner Van Belle; Stepwise Tempo Changes in BpmDj; YellowCouch Scientific; May 2006 Files # 1 Defining Variables We define a number of variables necessary to investigate the problem. • [] the current play speed of the song, expressed in samples/second. • [] the target play-speed, expressed in samples per second. The values placed in these two variables will lie around the sample rate. • [] the sample rate of the original soundtrack. In BpmDj this is 44100 Hz. We further assume a linear change of pace. E.g: every second the play-speed is updated. • [] defines the number of steps necessary to reach the final speed • [] the current step. This variable ranges from 0 (at which moment the speed is supposed to be ) to (at which moment the play-speed should be ). We now want to calculate the lplaying speed we need at step • [] the playing speed at step . # 2 Linear change A first straightforward approach to determine seems linear interpolation, as follows The main problem we observed with this approach was that the music seemed to step up (or down) with an uneven pace. It did sound wrong. A closer investigation reveals that simple interpolation will lead to a non linear frequency change. To demonstrate this closer, we will work with a song which has only one waveform: a pure sine with frequency . We can now determine the frequency of that note at step (call it ) through multiplying it by the current speed () and division by its normal speed (the sample-rate ). (1) Since human beings perceive frequencies in a non equally spaced manner, we need to convert this frequency to a note in a (for this purpose) equi-temporal scale. Rising one octave amounts to doubling of the note frequency. If A4 is 440 Hz, then A5 will be 880 Hz and A3 is 220 Hz. Therefore, to map the frequency to its note number we need to take (which will yield the octave number) and multiply it by 12 (to yield the note number). The perceived note at step will be (2) Clearly, we will hear a non-linear scale. To illustrate this, let us assume that we would like to change the tempo of a song from half the playing speed () to its normal playing speed () in 10 steps. Figure 1 lays out the transposition at a given time step. The slight bend we can observe does not look as much, but when looking at the local effect by measuring the transposition at every step then the problem becomes more obvious (see figure 2) # 3 Non linear change To resolve this problem we must compensate for the logarithm and express our requirements on the note scale instead of the speed-scale. We actually want the transposition at every step to be constant (called ). Formalized: According to 2 this becomes . According to 1 this further gives Merging various constant factors (12, and ) into gives and thus , in which will be a constant itself. In other words we need a constant multiplier for the playing speeds instead of a constant difference. By induction we get , , which leads to (3) We further know that and that , so Isolating gives (4) Practically, filling 4 into 3 gives To calculate the logarithm using a specific base (10 or for instance) use the following http://werner.yellowcouch.org/werner@yellowcouch.org
2018-03-19 01:16:56
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https://brilliant.org/discussions/thread/i-find-this-integral-problem-very-interesting/?sort=new
# i find this integral problem very interesting !!! Help on this interesting integral problem ?? Note by Ritvik Choudhary 6 years, 2 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ ## Comments Sort by: Top Newest It can be solved by GAMMA FUNCTION=integrate(0-infinity)e^-x.x^(n-1)dx for all x>=1 x belongs to Z+!!! - 6 years, 2 months ago Log in to reply Although $\int e^{-x^2}\,dx$ can't be expressed in terms of elementary functions, we can evaluate $\int_{-\infty}^{+\infty} e^{-x^2}\,dx$. Doing so yields $\sqrt{\pi}$ (for justification see http://en.wikipedia.org/wiki/Gaussian_integral#Computation). - 6 years, 2 months ago Log in to reply This function is not integrable using the methods we learn till Undergraduate college level. I don't know about what we learn in college... - 6 years, 2 months ago Log in to reply As our function is an even function therefore split the limit of integration from -infinity to +infinity as 2 times 0 to infinity and use Gamma function by making a suitable substitution. It can also be done by converting the problem into polar form. - 6 years, 2 months ago Log in to reply I think ans is 0 .. I think we can solve it using integration by part - 6 years, 2 months ago Log in to reply squareroot of Pi ? ( Error Function) - 6 years, 2 months ago Log in to reply another not-so-cool way is to use gamma function.use the substitution x=root u - 6 years, 2 months ago Log in to reply ya it's cool.you try to find the square of the quantity,with one variable x and the other y.then you integrate them simultaneously,a double integral.It's easy to evaluate once you transform it to polar coordinates.Nice One.Answer is root pi - 6 years, 2 months ago Log in to reply You can also look up this pdf www.stankova.net/statistics2012/doubleintegration.pdf. - 6 years, 2 months ago Log in to reply lower limit is negative infinity.. - 6 years, 2 months ago Log in to reply is it 0 - 6 years, 2 months ago Log in to reply × Problem Loading... Note Loading... Set Loading...
2019-09-19 07:26:59
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https://www.r-bloggers.com/2011/09/simple-plot-using-r/
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. As a task for my Financial eco assignment I had to plot a simple time series of the overnight MIBOR(Mumbai interbank offer rates) for the past one year . The job could very well have been done easily in MS-Excel but I choose to plot it in R instead and the quality of the graph, pixel-wise and neatness wise, was way better than what I could have obtained with MS-Excel. All this at the cost of a minimal 3 lines of code: # The overnight MIBOR rates were stored in a file name “Call_Rates_2011.csv”, this is just a normal Excel file saved in a CSV(comma separated delimited) format that R can read. # The way in which R conceptualizes the data is similar to that in Excel, to draw a simple analogy you can assume that the variable “a” now stores the entire Excel spreed sheet in it. # You will have to make sure that the working directory is the one that contains the file “Call_Rates_2011.csv” # The 2 column headers in my CSV file were “date” and “mibor”, so the below code plots “date” on the x-axis and “mibor” on the y-axis. The as.Date() tells R that the column “date” contains dates in the format “day-month-year”(‘%d-%b-%y’). # a$(column header) is the standard way of referring to a column in the “spreadsheet” contained in “a” # xlab : x-axis label # ylab : y- axis label # type : line(l) # col : color of the line plot(as.Date(a$date,’%d-%b-%y’), a\$mibor, xlab= “Months”, ylab= “MIBOR overnight rates(percentage)”, type=’l’, col=’red’) # This is to get the titles in place # main : main title # col.main : color of the main title #font.main : font size of the title title(main=”Overnight MIBOR rates for last one year”, col.main=”black”, font.main=4) And the plot hence obtained thus looks like: Incase you can’t make out the difference in the quality of the plot obtained just drop in your comments and email address and I will mail you the pdf and the jpg image of the plot. You can pull/stretch it to see that the pixels don’t get distorted and it looks way neater if you present it in your slide in a presentation.
2021-08-03 13:24:42
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https://mathoverflow.net/questions/121178/primes-occurring-as-orders-of-elements-of-a-finitely-presented-group/121267
# Primes occurring as orders of elements of a finitely presented group Is it true that given a finitely presented group $G$, either all primes or only finitely many of them occur as orders of elements of $G$? - No. The set of primes can be whatever you want (added: within reason! As Benjamin Steinberg points out, it can in fact be any recursively enumerable set of primes). First, note that for infinitely presented groups, the torsion can be whatever you like: the torsion in the group $*_i \mathbb{Z}/p_i$ is precisely the set of primes $p_i$, by standard facts about free products. (Added: As long as the set of $p_i$ is recursively enumerable, then this group admits a recursive presentation on a countable set of generators.) By Higman's Embedding Theorem, the above group can be embedded in a finitely presented group. More subtlely, this embedding doesn't introduce any new torsion---see, for instance, Theorem 2.5 of this preprint of Chiodo. Clarification: Higman's Embedding Theorem is commonly stated as only applying to finitely generated countable groups. In fact, an old construction of Higman, Neumann and Neumann shows how to embed a countably generated group into a 2-generated group; if the countably generated group is recursively presented, then the 2-generated group can be taken to be recursively presented as well. Further update: Following Benjamin Steinberg's comments below answer and an argument in another paper of Chiodo (see also Francois G. Dorais's comments), I think we have a very exciting characterization of the sets of primes that can occur as torsion in finitely presented groups. (For the solvable-word-problem case, one also needs a theorem of Clapham, which says that the Higman Embedding can be made to preserve solvability of the word problem.) Very exciting theorem: Let $P$ be a set of primes. 1. $P$ occurs as the torsion in some finitely presented group if and only if $P$ is $\Sigma^0_2$. 2. $P$ occurs as the torsion in some finitely presented group with solvable word problem if and only if $P$ is recursively enumerable. - Well -- but is the free product of infinitely many cyclic groups of prime order finitely generated? -- As I understand, Higman's Embedding Theorem is only for finitely generated groups, or am I missing something? –  Stefan Kohl Feb 8 '13 at 14:06 In any case, for reasons of cardinality, the set of primes cannot be "whatever one wants" -- there are uncountably many sets of primes, but only countably many finitely presented groups. –  Stefan Kohl Feb 8 '13 at 14:20 You need the set of primes to be recursive for this to work. In this case the free product is recursively presented which is enough for Higman. –  Benjamin Steinberg Feb 8 '13 at 15:17 Look in the section of D Cohen's book on Higman embeddings to see that a recursive presentation is enough (the generators don't need to be finite). –  Benjamin Steinberg Feb 8 '13 at 15:22 It seems that this works. What a nice question. –  Benjamin Steinberg Feb 8 '13 at 21:58 Notation: If $n\in \mathbb{N}$, then we denote by $\pi(n)$ the set of all non-trivial factors of $n$, including $n$ but excluding $1$. Call a set $X \subseteq \mathbb{N}$ factor-complete if it is closed under taking non-trivial factors. That is, $n\in X \Rightarrow \pi(n) \subseteq X$. Now we can state a strengthened answer to the main question of this post: Complete characterisation: Let $X\subseteq \mathbb{N}$. Then the following are equivalent: $1$. $X$ is the set of orders of torsion elements of a finitely presented group $G$. $2$. $X$ is factor-complete and has a $\Sigma_{2}^{0}$ description. That $1\Rightarrow 2$ is immediate; torsion-orders are closed under taking factors, and have a $\Sigma_{2}^{0}$ description (see the proof of theorem 3.5 in arXiv:1107.1489v2). That $2 \Rightarrow 1$ can be seen from the following result, which is proved by a generalisation of an argument provided by Francois Dorais in an earlier post on this thread: Technical result: There is a uniform algorithm that, on input of a computable function $\phi: \mathbb{N}^3 \to \mathbb{N}$ which describes a factor-complete $\Sigma_{2}^{0}$ set $A$, outputs a finite presentation $P_{\phi}$ such that $A$ is precisely the set of torsion elements of $gp(P_{\phi})$. As any set of primes is factor-complete, we get the following corollary. Answer to main question: A set of primes $A$ appears as the torsion elements of a finitely presented group if and only if $A$ has a $\Sigma_{2}^{0}$ description. Observe that if $X\subseteq \mathbb{N}$ then the set $X_{prime}:=\{p_{i}\ |\ i \in X\}$ is factor-complete and one-one equivalent to $X$ (being a set of primes, with a computable numbering). So we conclude with the following: Sets which can be realised up to one-one equivalence: Given any $\Sigma_{2}^{0}$ set $A$, the set $A_{prime}$ is one-one equivalent to $A$, and can be realised as the set of orders of torsion elements of some finitely presented group $G$. For completeness, we provide the proof of the technical result. This is the construction described by Francois Dorais, generalised and applied carefully so that none of the torsion elements bump in to each other'. Apologies for its lenght. Proof of technical result: Let $\{a \in \mathbb{N}\ | \ (\exists m)(\forall n) (\phi(a,m,n)=1)\}$ be the description for the factor-complete $\Sigma_{2}^{0}$ set $A$, where $\phi: \mathbb{N}^3 \to \mathbb{N}$ is our computable function. Let $p_{1}, p_{2}, \ldots$ be the standard indexing of the primes ordered by size. Then we construct a countably generated recursive presentation as follows: Take the infinite set of symbols $x_{1}, x_{2}, \ldots$; this is our generating set. For any fixed $i>1$, add the relation $x_{p_{i}}^{i}=1$. Then, start successively computing $\phi(i,1,1), \phi(i,1,2), \phi(i,1,3), \ldots$ increasing the last variable by $1$ each time. If at some point $\phi(i,1,n)\neq 1$ then stop, add the relations $x_{p_{i}}=1$, $x_{p_{i}^{2}}^{i}=1$, and start successively computing $\phi(i,2,1), \phi(i,2,2), \phi(i,2,3), \ldots$. If again some $\phi(i,2,n)\neq 1$, then stop, add the relations $x_{p_{i}^{2}}=1$, $x_{p_{i}^{3}}^{i}=1$, and start computing $\phi(i,3,1),\phi(i,3,2), \ldots$. By interleaving this process for all $i \in \mathbb{N}$, we get a countably generated recursive presentation which we denote by $Q_{\phi}$. Notice that: a) If $i \in A$, then there will be some (smallest) $m$ such that $\phi(i,m,1)=1, \phi(i,m,2)=1, \ldots$, and so the relation $x_{p_{i}^{m}}^{i}=1$ will be present, but no other relation involving $x_{p_{i}^{m}}$, and so $\langle x_{p_{i}^{m}} \rangle \cong C_{i}$ becomes a free product factor in this group. b) If $i \notin A$ then for all $m \in \mathbb{N}$ we will have the relation $x_{p_{i}^{m}}=1$, and since $A$ is factor-complete $i$ does not divide any element of $A$. Hence no element of order $i$ occurs in the group. So we end up with the group $gp(Q_{\phi}) \cong F_{\infty}$ $* _ {a \in A} ( * _ {j \in \pi(a)} C_{j})$ (With a slightly more complicated indexing, one could do away with the $F_{\infty}$ factor). As $A$ is factor-complete, we immediately see that it is the set of orders of torsion elements of $gp(Q_{\phi})$. Now apply the construction of Higman-Neumann-Neumann, and then the construction of Higman, to embed this in a finitely presented group with presentation $P_{\phi}$. Note that these embeddings strictly preserve the set of orders of torsion elements (see lemma 6.9 and theorem 6.10 of M. Chiodo Finding non-trivial elements and splittings in groups'). Also note that this construction is completely uniform in the computable function $\phi$ used to describe the set $A$. - Wow, still a more general result ... -- Great! -- Thanks! –  Stefan Kohl Feb 14 '13 at 20:52 Minor remark: on this site, for some reasons set brackets need to be escaped by two backslashes -- otherwise they vanish. –  Stefan Kohl Feb 14 '13 at 20:58 @Stefan: Encoding arbitrary $\Sigma_{2}^{0}$ or $\Pi_{2}^{0}$ sets into a single finitely presented group, which I first did in arXiv:1107.1489v2, shows that such groups (even individually) are incredibly rich. When considered collectively, they are even richer: A result by Boone-Rogers shows that the set of finite presentations of groups with solvable word problem is $\Sigma_{3}^{0}$-complete. I have yet to see (but am interested in seeing) if any set harder than $\Sigma_{2}^{0}$ or $\Pi_{2}^{0}$ can be encoded into a single finitely presented group. –  user31415 Feb 14 '13 at 23:20 @Stefan: The problem here is that you are not given the ($\Sigma_{2}^{0}$) set $A$, but a description involving just the function $\phi$. It is possible to come up with very complicated descriptions of very easy' sets (recognising descriptions of the empty set is undecidable, so these can get very hard!). Since there are infinitely many distinct $\Sigma_{2}^{0}$ sets, such presentations would be unbounded in length' (the presentations I describe above can all be made to be 2-generator, so we can define length as the sum of the lengths of the generators). –  user31415 Feb 15 '13 at 0:36 Additionally, chasing Turing Machines has been done in the past, with some success. Valiev has constructed explicit examples of universal finitely presented groups. I would be interested in seeing the following: 1. An explicit finite presentation of a universal finitely presented torsion-free group. 2. An explicit finite presentation of a group whose set of orders of torsion elements form a $\Sigma_{2}^{0}$-complete set. Perhaps an analysis of the work of Valiev would be of some use, but I can't say for sure. –  user31415 Feb 15 '13 at 0:45 This should be a comment to HW's answer but it is too long. I believe that the correct theorem is that a collection P of primes is the set of prime orders of the finite order elements of a finitely presented group iff there is an re set U, a computable function $f\colon U\to Primes$ and r.e. sets L,K contained in U such that $P=f(L-K)$. I don't know if such a set of primes must be a difference of r.e. sets (which is HW's) answer. Here is the proof. Suppose first that we have a finitely presented group. Then U will be all pairs (u,p) where u is a word over the generators and their inverses and p is a prime, f is the projection to the second coordinate, L is all pairs (u,p) with $u^p=1$ in the group (r.e. by finiteness of the presentation) and K is all pairs (u,p) with u=1 (r.e. for the same reason). Clearly the prime orders of elements of the group is f(L-K). Conversely, given U,L,K,f as above consider the group G with generators U subject to the relations $x^{f(x)}=1$ for x in L and $x=1$ for x in K. This group has an r.e. set of generators and an r.e. set of relators and hence by the argument of HW can be embedded in a finitely presented group with the same finite orders. Now G is a free product of infinite cyclic groups (one for each element of U not in K or L) and finite cyclic groups of order from f(L-K) (there may be repetitions because f is many to one). Edit. I have learned from Computable images of differences of r.e. sets that the sets of the above form are precisely the $\Sigma^0_2$ sets of the arithmetic hierarchy and hence much broader than differences of re sets. In his comments to HW's answer François sketches how to construct the analogue of G above directly from a logical formula. - I think in the last paragraph the generators of $G$ are meant to be the projections of the elements of $U$ to the first coordinate, and similarly the $x$ in the basis in $x^{f(x)} = 1$ should rather be its projection to the first coordinate -- or am I mistaken? Otherwise -- as far as I can tell!! -- this looks right to me. -- Thanks again! –  Stefan Kohl Feb 9 '13 at 10:52 Last paragraph U is any re set not just the pairs in the second paragraph. –  Benjamin Steinberg Feb 9 '13 at 13:11 O.k., thank you. –  Stefan Kohl Feb 9 '13 at 13:51 No, this should definitely be a separate answer, and deserves full credit. This is really a great example of how interesting decision theory arises from innocuous group-theoretic questions. –  HJRW Feb 10 '13 at 15:26 @HW, this question ended up being one of the most fun experiences I had on MO. It shows that by putting together several mathematicians who may never have met in person with differing areas of expertise one can pose and solve a nice problem. –  Benjamin Steinberg Feb 10 '13 at 20:59
2015-07-03 22:53:20
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https://solvedlib.com/n/for-which-of-the-following-processes-is-the-equation-a-s,15914407
# For which of the following processes is the equation ∆S = qrev/T valid?1. any reversible expansion of an ideal gas.2. ###### Question: For which of the following processes is the equation ∆S = qrev /T valid? 1. any reversible expansion of an ideal gas. 2. isothermal irreversible compression of a non-ideal gas. 3. constant pressure heating of a liquid. 4. isothermal vaporization of a liquid. 5. isothermal sublimation of a solid to a non-ideal vapor. Please explain each option #### Similar Solved Questions ##### 10. Find the radian angle solutions of the equation in the interval [0,21). Approximate the solutions... 10. Find the radian angle solutions of the equation in the interval [0,21). Approximate the solutions to two decimal places. Find the reference angle, Or, and then find the solutions in the appropriate quadrants. sin? 0-2 sine -1 = 0... ##### Help! code in Matlab One of the purposes of an investigation by Porcellini et al. was... help! code in Matlab One of the purposes of an investigation by Porcellini et al. was to investigate the effect on CD4 T cell count of administration of intermittent interleukin (IL-2) in addition to highly active antiretroviral therapy (HAART). The following table shows the CD4 T cell count at b... ##### 6. Predict the product by proposing a plausible mechanism:HO-H+ (cat)HO7. Propose mechanism for the following transformationNHz 6. Predict the product by proposing a plausible mechanism: HO- H+ (cat) HO 7. Propose mechanism for the following transformation NHz... ##### The grev spore Lene Sordaria fimicola Inyulyc ascospore colour production It has two allelic forms the wild-!ype allele (G)eads black ascospores_while the mutant allele (2) eads gryascospores Malaan CSyO Jtn Wder Tich scquentrts) or Inages correctly represent(s) the flow of chmocomcs and alleles from diploid ccll thc developing pcrithccium to an ascus containing ascospores? pcopnate Alte miot Altcr melosts Il: Alter milosiAt prophaseAler meiosis I:Alter meiosis Il: After mitosis: = 1After meiosi The grev spore Lene Sordaria fimicola Inyulyc ascospore colour production It has two allelic forms the wild-!ype allele (G)eads black ascospores_while the mutant allele (2) eads gryascospores Malaan CSyO Jtn Wder Tich scquentrts) or Inages correctly represent(s) the flow of chmocomcs and alleles fro... ##### A 220 kW load having 0.8 lagging power factor is supplied from a 600 V (line to line) three-phase... A 220 kW load having 0.8 lagging power factor is supplied from a 600 V (line to line) three-phase supply. A bank of power factor capacitors is delta connected Q2. (a) Draw neat schematic diagram showing the connection of the capacitor bank in Delta connection. (b) Determine the kVAR rating of each i... ##### Find A(ZA)-1 = -3 3]ASF Find A (ZA)-1 = -3 3] A SF... ##### Week 10b - Ch23 Comparing Two Proportions: Problem 6 Previous Problem List Next (1 point) (a)... Week 10b - Ch23 Comparing Two Proportions: Problem 6 Previous Problem List Next (1 point) (a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proportions of boys and girls under 10 years old who are afraid of spiders. Assume the... ##### UtionDoantaanq alcam prondeuimcg /0OCON Bengeupoored nt nAtonDennnAlnAAcT]Funtv 27548{ Ararao munta-30 O00 NAt Ution Doanta anq alcam prondeuimcg /0OCON Bengeupoored nt n Aton DennnAln AAcT] Funtv 27548{ Ararao munta- 30 O00 N At... ##### Farmer Frank grows two crops: celery and lettuce. He has determined that the cost of planting these crops is modeled by$C(x, y)=x^{2}+3 x y+3.5 y^{2}-775 x-1600 y+250,000$where $x$ is the number of acres of celery and $y$ is the number of acres of lettuce. Suppose Farmer Frank has 300 acres available for planting and must plant more acres of lettuce than of celery. Find the number of acres of celery and of lettuce he should plant to minimize the cost, and state the cost. Farmer Frank grows two crops: celery and lettuce. He has determined that the cost of planting these crops is modeled by $C(x, y)=x^{2}+3 x y+3.5 y^{2}-775 x-1600 y+250,000$ where $x$ is the number of acres of celery and $y$ is the number of acres of lettuce. Suppose Farmer Frank has 300 acres av... ##### ChM 153 LABORATORY REPORT NANE DATE spectrophotoMETRIC DETERMINATION OF AN EQUILIBRIUM CONSTANTSolulicaNumbecInitial [Fe *J]lces10 ")1o dxo m 10.0 Xidn dvo?x67 A.0 KdrInital HSCNJ(molarity 10 *)Joxio40no0xo" Loxiu " JXlo" 354 46Y 512Absorbante227Equlhbrium conc FeSCN'? 104'4xd"6(xlo"4 9 Sylo4 JX10 -4Equlibnum conc9-0xi~xto 39x6 7 2.5 %i04.1xld(x10 )Emuiinivm conc IIHSCNI 10*)0 500Continues on nuxtpaqul: ChM 153 LABORATORY REPORT NANE DATE spectrophotoMETRIC DETERMINATION OF AN EQUILIBRIUM CONSTANT SolulicaNumbec Initial [Fe *J] lces 10 ") 1o dxo m 10.0 Xidn dvo? x67 A.0 Kdr Inital HSCNJ (molarity 10 *) Joxio 40no 0xo" Loxiu " JXlo" 354 46Y 512 Absorbante 227 Equlhbrium conc FeS... ##### The comparative balance sheet of Carlson Education Supply at December 31, 2018, reported the following Requirements:... The comparative balance sheet of Carlson Education Supply at December 31, 2018, reported the following Requirements: 1.) Prepare the statement of cash flows of Carlson Educational Supply for the year ended December 31, 2018. Use the indirect method to report cash flows from operating activities. 2. ... ##### A photographer wants to take a picture ofa bird 2.50 m away with a 35.Omm lens: How far from the lens must the film in a camera be?Select one: a.35.Ommb.34.5mm35.Smm A photographer wants to take a picture ofa bird 2.50 m away with a 35.Omm lens: How far from the lens must the film in a camera be? Select one: a.35.Omm b.34.5mm 35.Smm... ##### L 1 Il HMA form the 15,0-mile and asphalt binder and is approximately top the construction of roadways in the United States at a Atypical upper layer (there may be as many as 5 layers of different thicknesses and compositions) ofa high-volume highway S 1 1 1 Suo]LO[ * X10"itJ IH 1 with 2 lanes; cach of which has a tonstyear: "ouotruld Iu L 1 Il HMA form the 15,0-mile and asphalt binder and is approximately top the construction of roadways in the United States at a Atypical upper layer (there may be as many as 5 layers of different thicknesses and compositions) ofa high-volume highway S 1 1 1 Suo]LO[ * X10"itJ IH 1 with 2 lanes;... ##### Find a recurrence formula of the nth coefficient of the power series solution to the differential equation y"+10xy' _ 6 y=0 about Xo =0.(6 - 10 n) an an+2 (n+1)(n+2)(7-4 n) an an+2 (n+1)n+2)(6-10n) Gn+1 an+2 (n+1)(n+2)Qp_10n Gn+1 (n +1Xn+2)Tone6n42 Find a recurrence formula of the nth coefficient of the power series solution to the differential equation y"+10xy' _ 6 y=0 about Xo =0. (6 - 10 n) an an+2 (n+1)(n+2) (7-4 n) an an+2 (n+1)n+2) (6-10n) Gn+1 an+2 (n+1)(n+2) Qp_10n Gn+1 (n +1Xn+2) Tone 6n42... ##### In gambling_ the chances of winning are often written terms of odds rather than probabilities The odds of winning is the ratio of the number of successful outcomes to the number of unsuccessful outcomes The odds of losing is the ratio of the number of unsuccessful outcomes to the number of successful outcomes_ For example_ the number of successful outcomes is and the number of unsuccessful outcomes is 3 the odds of winning are 2.3 (read "2 to 3") or 3 (Note: If the odds of winning are In gambling_ the chances of winning are often written terms of odds rather than probabilities The odds of winning is the ratio of the number of successful outcomes to the number of unsuccessful outcomes The odds of losing is the ratio of the number of unsuccessful outcomes to the number of successfu... ##### Which of thefollowing is true(-1)"+' e" (n+1)! n=l divergent(-1)" (n +4)2 (2n)2 converges absolutely(-1)" (n + 02 (2n)2 divergentNone(-I)e+/ e" 2 (n +) converges conditionally"2" T5rcouuetger conditionalh5 F~luergem Which of thefollowing is true (-1)"+' e" (n+1)! n=l divergent (-1)" (n +4)2 (2n)2 converges absolutely (-1)" (n + 02 (2n)2 divergent None (-I)e+/ e" 2 (n +) converges conditionally "2" T5r couuetger conditionalh 5 F ~luergem... ##### Refer to Figure 4 Refer to Figure 4. Assume that Cliff and Paul were both producing wheat and corn, and each were dividing their time equally between the two. Then they decide to specialize in the product they have a comparative advantage in. As a result, total production of corn woulda. increase by 1 bushel.b. incre... ##### Describe how a phagocyte (e.g. neutrophil or macrophage) killsengulfed microbes in phagocytosis (inside thephagolysosome). Describe how a phagocyte (e.g. neutrophil or macrophage) kills engulfed microbes in phagocytosis (inside the phagolysosome).... ##### Label each pair as enantiomers, diastereomers or same molecule. Label each stereocenter wilh Its R or S configuration.andFrrBrHO_CH; andHOOHandCH;andOHH;cH;c Ph CH;OHCH; Label each pair as enantiomers, diastereomers or same molecule. Label each stereocenter wilh Its R or S configuration. and Frr Br HO_ CH; and HO OH and CH; and OH H;c H;c Ph CH; OH CH;... ##### Match the following words to their correct definitions. Write the correct letter in the space next... Match the following words to their correct definitions. Write the correct letter in the space next to the definition. (1 point each). Note: You will not use all the words. ______a subset of the population we are interested in studying ______The difference between the sample measure and the correspon... ##### How do mutations affect the phenotype? take into considerations coding and non-coding regions. How do mutations affect the phenotype? take into considerations coding and non-coding regions.... ##### Compute the indicated products. $\left[\begin{array}{rrrr}3 & 0 & -2 & 1 \\ 1 & 2 & 0 & -1\end{array}\right]\left[\begin{array}{rrr}2 & 1 & -1 \\ -1 & 2 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & 2\end{array}\right]$ Compute the indicated products. $\left[\begin{array}{rrrr}3 & 0 & -2 & 1 \\ 1 & 2 & 0 & -1\end{array}\right]\left[\begin{array}{rrr}2 & 1 & -1 \\ -1 & 2 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & 2\end{array}\right]$... ##### An object is moving along straight line, and the uncertainty in its position is 3.10 m. (a) Find the minimum uncertainty in the momentum of the object: Find the minimum uncertainty in the object's velocity, assuming that the object is (b) a golf ball (mass 0.0450 kg) and (c) an electron:(a) NumberUnits(b) NumberUnits(c) NumberUnits An object is moving along straight line, and the uncertainty in its position is 3.10 m. (a) Find the minimum uncertainty in the momentum of the object: Find the minimum uncertainty in the object's velocity, assuming that the object is (b) a golf ball (mass 0.0450 kg) and (c) an electron: (a) Nu... ##### Home Insert 10 Draw B layout T Review U ab View D A A Clear Formatting... Home Insert 10 Draw B layout T Review U ab View D A A Clear Formatting A DO A E L 15. A wheel turns on an axle. No matter the size of the wheel, the axle will always rotate at I rotation per second. Initially you place a wheel on the axle with a radius of 0.5 meters. You then replace that wheel with... ##### Would the IR spectrum allow you confirm that the structure Oftne product Is combination of the two reactants Explain your answer;SaveNorinalB I UXX _ 0==fx| 0/0/2 B FiMi & = Ili Would the IR spectrum allow you confirm that the structure Oftne product Is combination of the two reactants Explain your answer; Save Norinal B I U XX _ 0== fx| 0/0/2 B FiMi & = Ili... ##### Compare managerial and financial accounting, include what managerial accountants management functions are. Compare managerial and financial accounting, include what managerial accountants management functions are.... ##### 4. A small child gives a plastic frog a big push at the bottom of a... 4. A small child gives a plastic frog a big push at the bottom of a slippery (frictionless) 2.0-m-long ramp inclined at 35o to the horizontal, starting it with a speed of 5.6 m/s. What is the frog's speed as it flies off the top of the ramp?... ##### Ed{ wa7aAbhunuv rejace Ha Sale Teaevn &hhy Ablatnzrs i: incoriect {n dung * Ed{ wa7a Abhunuv rejace Ha Sale Teaevn &hhy Ablatnzrs i: incoriect {n dung *... ##### What is Frame Relay ? What is Frame Relay ?... ##### (iv) Check that the formula () = 21" holds for rows n = 0 to 5... (iv) Check that the formula () = 21" holds for rows n = 0 to 5 in Pascal's triangle. (If it doesn't work for n = 4 or 5, go back and redo (iii)!) (v) Prove the formula of (iv) using (ii). (a) using (ii); (b) by proving that both sides of the formula represent the number of subsets of a s...
2022-07-07 14:38:15
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https://dalts.com/post/2010-12-31-mercurial-and-subversion-good-playmates/
We've had a couple of Subversion outages recently. As usual, development ground to a halt. People couldn't update their projects, couldn't get the history - we shelved some changes in Intellij for checkin later in the day... in short a pain we could live without. Such is life with centralised version control. Since the advent of bitbucket I've been using Mercurial a lot for my own projects, so thought why not try it at work as a fallback for when subversion is down. Here's the quick recipe for Ubuntu based on this: sudo apt-get install mercurial python-subversion hg clone http://bitbucket.org/durin42/hgsubversion cd hgsubversion sudo python setup.py install [extensions] hgsubversion= Now clone your repo with something like this: hg clone svn+http://myhost.com/myrepo You can push and pull your changes to subversion using regular "hg push" and "hg pull". Works beautifully in Intellij too, but just make sure to: • Install Mercurial plugin • Recreate your project file if you have it checked into Subversion, Intellij will still think it's a subversion project despite the lack of .svn dir and presence of .hg. If you are using Maven you can just open your pom.xml to do this. • Remember that a Ctrl-K checkin is only checking into your local Mercurial. If you need to push changes to Subversion you'll have to issue a push command from the Mercurial menu. I'm sure there's some gotcha's that might come up (seems like merging is one) - but on the face of it, it seems to work pretty well. I'll post windows instructions later when I get around to it - but this post seems to be pretty complete. Also don't confuse hgsubversion project with hgsvn (which has a deb package) - hgsvn looks to be deprecated in favour of hgsubversion.
2021-04-10 19:53:10
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http://nsr.bioeng.washington.edu/jsim/models/webmodel/NSR/Styrene_Inhalation/
This page will look better in a graphical browser that supports web standards, but is accessible to any browser or internet device. Served by Samwise. Styrene_Inhalation A model for a physiological based description of the inhalation pharmacokinetics of styrene in rats and humans based on the work of J.C. Ramsey and M.E. Andersen [(1984). Toxicol Appl Pharm. 73(1), 159-175.], developed by Neil Geisler (2006) as a final project for BIOEN 589, University of Washington. Model number: 0196 Run Model: Help running a JSim model. Java runtime required. Web browser must support Java Applets. (JSim model applet may take 10-20 seconds to load.) Description This model presents a the phramacokinetics of styrene inhalation in rats and humans based on the 1984 paper by Ramsey and Andersen paper (reference below). Four body compartments are considered: richly perfused tissue (highly perfused), muscle (moderately perfused), fat (slowly perfused), and liver, which includes metabolic consumption of styrene. Styrene is inhaled at a given concentration for a specified time, then normal ventilation continues. Parameters for rats and humans are included, and are listed in the code below along with parameters for mice as provided in the paper. Two plot pages, one for rats and one for humans, are given to reflect identical traces plotted in the paper. Equations Styrene Ventilation: $\large Q_{alv}(C_{inh}-C_{alv})=Q_t(C_{art}-C_{vent})$ $\large N=\frac{C_{art}}{C_{ven}}$ $\large C_{art}=\frac{(Q_{alv} \cdot C_{inh} + Q_t \cdot C_{ven})}{(Q_t+\frac{Q_{alv}}{N})}$ $\large C_{art}=\frac{(Q_t \cdot C_{ven})}{Q_t+\frac{Q_{alv}}{N})}$ $\large C_{exh}=0.7 \cdot C_{alv}+0.3 \cdot C_{inh}$ Styrene uptake, release, and metabolism in the tissue: $\large \frac{dA_i}{dt}=Q_t \cdot (C_{art}-C_{vi})$ $\large C_i=\frac{A_i}{V_i}$ $\large C_{vi}=\frac{C_i}{P_i}$ $\large \frac{dA_{met}}{dt}=\frac{V_{max} \cdot C_{vl}}{K_m+C_{vl}}$ $\large \frac{dA_l}{dt}=Ql \cdot (C_{art}-C_{vl})-\frac{dA_{met}}{dt}$ $\large C_{ven}= \frac{\Sigma(Q_i \cdot C_{vi})}{Q_t}$ The equations for this model may also be viewed by running the JSim model applet and clicking on the Source tab at the bottom left of JSim's Run Time graphical user interface. The equations are written in JSim's Mathematical Modeling Language (MML). See the Introduction to MML and the MML Reference Manual. Additional documentation for MML can be found by using the search option at the Physiome home page. References Andersen, J.C. and Ramsey, M.E., A physiologically based description of the inhalation pharmacokinetics of styrene in rats and humans, 1984 Toxicol Appl Pharmacol, 73(1), 159-175. None. Key Terms Administration, Oral, Animals, Gases, Humans, Injections, Intravenous, Kinetics, Liver metabolism, Lung metabolisim, Male, Models, Biological, Rats, Inbred Strains, Species Specificity, Styrene metabolism, Data, Respiratory system, Air-blood gas exchange, Publication Model History Get Model history in CVS. Posted by: bej Acknowledgements Please cite www.physiome.org in any publication for which this software is used and send an email with the citation and, if possible, a PDF file of the paper to: staff@physiome.org. Or send a copy to: The National Simulation Resource, Director J. B. Bassingthwaighte, Department of Bioengineering, University of Washington, Seattle WA 98195-5061.
2018-05-23 18:55:29
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https://ask.sagemath.org/answers/56122/revisions/
sage: xQ = 0x5699b93fc6e1bd29e09a328d657a607b4155b61a6b5fcbedd7c12df7c67df8f5 xP and/or xQ may be different of what you think they are. Check your code upstream of line 56...
2021-05-18 12:05:48
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https://projecteuclid.org/euclid.ade/1366896238
### Existence and multiplicity of nontrivial solutions in semilinear critical problems of fourth order #### Abstract In this paper we consider the equation $\Delta^2 u = \lambda |u|^{q-2} u + |u|^{2^*-2} u\equiv f(u)$ in a smooth bounded domain $\Omega\subset\mathbf{R}{N}$ with boundary conditions either $u|_{\partial\Omega} =\frac{\partial u}{\partial n}|_{\partial \Omega}=0$ or $u|_{\partial\Omega}=\Delta u|_{\partial \Omega}=0$, where $N>4$, $1< q <2, \,\lambda>0$ and $2^* = 2N/(N-4)$ We prove the existence of $\lambda_0$ such that for $0<\lambda<\lambda_0$ the above problems have infinitely many solutions. For the problem with the second boundary conditions, we prove the existence of a positive solution also in the supercritical case, i.e., when we have an exponent larger than $2^*$. Moreover, in the critical case, we show the existence of at least two positive solutions. #### Article information Source Adv. Differential Equations, Volume 1, Number 2 (1996), 219-240. Dates First available in Project Euclid: 25 April 2013
2020-03-30 07:58:52
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https://socratic.org/questions/if-the-length-of-a-49-cm-spring-increases-to-81-cm-when-a-3-kg-weight-is-hanging
# If the length of a 49 cm spring increases to 81 cm when a 3 kg weight is hanging from it, what is the spring's constant? ##### 1 Answer Sep 27, 2016 I found: $92 \frac{N}{m}$ Have a look:
2020-05-31 07:39:35
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https://mathoverflow.net/questions/339765/iterated-derivative-and-rectangular-standard-young-tableaux/339893#339893
# Iterated derivative and rectangular standard Young tableaux We first make a few definitions, seemingly out of the blue (they are introduced/defined in this paper). Let $$F^0_{a}(z) = (1-z)^{-1}$$ and define recursively $$F^{k+1}_{a}(z) = z^{a-1} \frac{d^a}{dz^a} F^{k}_{a}(z), \qquad k\geq 0.$$ Let $$G_{a,b}(z)= \left(\prod_{k=0}^{b-1}\frac{k!}{(a+k)!} \right) z^{1-a}(1-z)^{ab+1} F^{b}_{a}(z)$$. Here are the first few values of $$G_{a,b}(z)$$ as $$a=1,2,3$$, $$b=1,\dotsc,4$$. $$\begin{matrix} 1 & 1 & 1 & 1 \\ 1 & z+1 & z^2+3 z+1 & z^3+6 z^2+6 z+1 \\ 1 & z^2+3 z+1 & z^4+10 z^3+20 z^2+10 z+1 & z^6+22 z^5+113 z^4+190 z^3+113 z^2+22 z+1 \end{matrix}$$ The combinatorialist might then be inclined to make the following conjecture (verified for $$1\leq a,b\leq 4$$): $$G_{a,b}(z) = \sum_{T \in \mathrm{SYT}(a^b)} z^{des(T)-b+1}.$$ Here, $$\mathrm{SYT}(a^b)$$ denotes the set of standard Young tableaux of rectangular shape $$(a,a,\dots,a)$$, and $$des(T)$$ is the number of descents. Question: What is going on? Can this be proved? I have never seen anything like this, but the paper above introduces these as generalizations of Eulerian polynomials and Narayana polynomials, so the first two columns (and rows) of the table above agrees (by finding the appropriate pages in Stanley's EC2). If the conjecture is true, then $$G_{a,b}(z)$$ can be shown to be $$h^*$$-polynomials for some nice family of polytopes. Mathematica code for the function $$G_{a,b}(z)$$ is as follows: GG[a_, b_] := (Product[(k)!/(a + k)!, {k, 0, b - 1}]) z^(1 - a) (1 - z)^(a b + 1) Nest[Simplify[z^(a - 1) D[#, {z, a}]] &, 1/(1 - z), b]; • @SamHopkins Ah, it does not! It is relevant later, as the number of times one applies the recursion.... Sep 3 '19 at 14:40 • Could you recall what a descent is. Sep 3 '19 at 14:53 • Your definition of $F^0_a$ is missing $a$ Sep 3 '19 at 15:02 • @SamHopkins, yes one can indeed use MacMahons formula, although I have not tried to get anything explicit with it yet. Sep 3 '19 at 18:52 • @AbdelmalekAbdesselam We say that $j$ is a descent if $j+1$ appear in a row with strictly larger index than $j$ (a strictly lower row if you are not French). Des count the number of descents. Sep 3 '19 at 18:52 The formula follows from a result in EC2 (Stanley's "enumerative Combinatorics" Vol. 2) -- Chapter 7, equation (7.96), which is a result from the expansion of Schur functions in terms of fundamental quasisymmetric functions. The equation reads: $$\sum_{m\geq 0} s_{\lambda/\mu} (1^m) z^m = \frac{\sum_T z^{des(T)+1}}{(1-z)^{n+1}},$$ where $$T$$ ranges over all SYTs of shape $$\lambda/\mu$$. In the case of rectangular shape $$(a^b)$$, the hook-content formula (equivalent to MacMahon's here) gives $$s_{a^b} (1^m) = \prod_{i=1}^{b}\prod_{j=1}^a \frac{m+j-i}{i+j-1},$$ and we notice that the product of hook-lengths in the denominator is exactly the constant $$\prod_{k=0}^{b-1} \frac{k!}{(a+k)!}$$. So after canceling the common factors on both sides, Per's identity becomes equivalent to: $$\sum_{m\geq 0} \prod_{i=1}^b \prod_{j=1}^a (m+j-i) z^{m+a-b-1} = F_a^b(z)$$ Now this follows easily by induction on $$b$$, since $$z^{a-1}\frac{d^a}{dz^a} z^{m+a-(b-1)-1} = \prod_{j=1}^a(m+j-b) z^{m+a-1-b}.$$
2021-10-26 07:39:55
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https://offshoremechanics.asmedigitalcollection.asme.org/biomechanical/article-abstract/106/4/315/398794/Gas-Transport-During-Oscillatory-Flow-in-a-Network?searchresult=1
A transport coefficient was measured for a range of oscillatory flow conditions in a branching network of tubes. Measurements were made both across the first generation of a three-generation network and the second generation of a four-generation network. The results for these two series of tests were similar, indicating that there was no significant effect due to the system boundaries. The results are cast in terms of an effective axial diffusion coefficient of the form $Deff=(ϰ+0.50Vt1.66f0.94)cm2/s$ where ϰ is the molecular diffusivity, Vt is the local stroke volume (cc), and f is the oscillation frequency (Hz). These results are compared to those obtained by other investigators in branching systems of similar geometry. At low frequency, this result is found to be in approximate agreement with the steady flow result of Scherer, et al. [15]. This expression differs from the oscillatory flow results of Tarbell, et al. [19] for liquids, primarily in terms of the effects of oscillation frequency. This content is only available via PDF.
2022-11-30 13:27:56
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https://fractalforums.org/fractal-mathematics-and-new-theories/28/pathfinding-in-the-mandelbrot-set-revisited/2916/msg15580
### Pathfinding in the Mandelbrot set - Revisited • 17 Replies • 413 Views 0 Members and 1 Guest are viewing this topic. #### wbarry • Fractal Freshman • Posts: 9 • Interest in Mandelbrot Set and the way it evolves #### Pathfinding in the Mandelbrot set - Revisited « on: July 05, 2019, 05:38:27 PM » In reference to some discussions from the old forum: http://www.fractalforums.com/mandelbrot-and-julia-set/pathfinding-in-the-mandelbrot-set/ http://www.fractalforums.com/mandelbrot-and-julia-set/shortcuts-in-juliasets/ I would like to propose another way of describing what appears to be happening, as I understand it. No math, just discussion. Some of the details may need adjustment, but I believe the core idea is on the mark. I am fairly certain this is part of what is being shown in this paper by Dierk Schleicher (referred to in the earlier posts mentioned above). https://arxiv.org/abs/math/9411238 INTERNAL ADDRESSES OF THE MANDELBROT SET AND GALOIS GROUPS OF POLYNOMIALS The Mandelbrot set can be thought of as a heirarchical network of mini-brots. For each mini-brot, there is a "most efficient path" to the mini-brot from the center of the set, passing through other mini-brots along the way. Brot to brot to brot ... Each mini-brot is surrounded by a set of concentric "shells". The "most efficient path" to any mini-brot is visually encoded into the shells surrounding it. The shells contain an exact "history" of what shapes are traversed getting to the mini-brot when starting from the center of the Mandebrot Set - a "visual record" of the exact shapes traversed along the most efficient path. New shells are created whenever the path includes a cross-road where a decision needs to be encoded. The encoding is done in a "combinatorial" fashion, meaning new moves are spliced into the existing sequences, doubling the number of distinct shell patterns each move. The first moves are on the outside, the latest moves appear on the inside (the deep end). When trying to understand morphing, it is essential to imagine building "from the inside out". From Schleicher, p.6 - "The internal address of c $$\in$$ M can be viewed as a road map description for the way from the origin to c in the Mandelbrot set: the way begins at the hyperbolic component of period 1, and at any intermediate place, the internal address describes the most important landmark on the remaining way to c" Each Move creates a set of shell patterns. The "encoding" of the Move is a recording of each object that was passed through and these recordings are the patterns in each shell. If N is the number of Moves to reach a mini-brot using the most efficient path, then the mini-brot will be surrounded by 2N - 1 (I think) distinct shell patterns. From Schleicher, p.14 descriptions by saying that whenever the path from the origin to a parameter c $$\in$$ M branches off from the main road, an entry in the internal address is generated: the way to most parameters c $$\in$$ M traverses infinitely many hyperbolic components, but most of them are traversed “straight on” and left into the 1/2-limb. " If there are X moves in the most efficient path, then there will be 2X - 1 different variations encoded in the shells. These 2X - 1 shell variations will all be shown for 2 arms, then repeated with 4 arms, then 8 arms, then 16, and so on for the powers of 2. Each move adds a pattern to EACH shell that existed around the node from the previous move. Each mini-brot that is descended from the same path has some of the same encoded shells. Because we are zooming, we never leave where we came from, we just look deeper into the structure of the area we were already in, so prior path encodings must be retained - you still need to travel through those parts to reach the mini-brot. Because these shells are added in this mathematically regular manner, zooming to 3/4 depth finds the magical shell with 2 arms (perhaps halfway through the new stack?) - the next step in the Julia Morph. Same for the 4 and 8 armed automated zooms being at 7/8 and 15/16. Those shells are always located at the same depth in the stack. How This Relates to Julia Morphs I think that what we call a "Julia Morph (Evolution)" is the last two-arm shell in the current stack, which contains all of the moves up to that point. When trying to do Julia Morphs in the past, I was always trying to "grow an arm" and the results would surprise and confound me. But now that I understand that I am actually morphing the shape from the center out, I have much more control over the process. Take the path from the current central mini-brot of the current Julia Morph to any new mini-brot you choose. Insert all of objects contained in the path between the two brots into the center of the current shape, inverted, with the chosen mini-brot as the new center. This results in the new shape you will find when you reach the proper depth (3/4). Big-Eye Shortcuts I think Big-Eye Shortcuts may also be explained by this. Shortcuts are already at the "newest" shells - they don't need to re-create all the shells that came before. I think it may relate to the "rotational procession" Claude Helland has shown - where similar shells at different depths are identical, but their rotation is slightly different. Rotation is most likely a product of the orientation of the final mini-brot - the direction the spike is pointing. Perhaps for Big-Eyes, the rotation is identical to the central brot, so they are "sharing" the stack up to that point. They are already surrounded by the necessary shells. But mini-brots with a different final rotation might need to re-encode all of the shells in the new orientation, so the stack needs to start all over. The math has some variables that cancel out or something. That might fit the observed behavior. I thought I was reading something that implied this in Dierk Schleicher's paper, but I can't find it now, so perhaps it was wishful thinking. My brain hurts. Construction Tool My final thought is that this means one can use the set as a Construction Tool. The path you choose to follow BUILDS the shape you will find. We just need to learn how to use the tool effectively and how to comprehend the way each move will be encoded - what will the result look like. The results can be predicted. Walter Barry Mandelbrot Evolution • Fractal Friend • Posts: 17 • infinite border, finite area #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #1 on: July 05, 2019, 06:27:43 PM » Yeah, I do that all the time! Zoom into the cusp of the cardioid, find a minibrot, then zoom into the seahorse valley, find a minibrot, zoom into the ... uh, other side of the seahorse valley, find a minibrot. That minibrot will have layered cusp-seahorse-otherside patterns around it! #### wbarry • Fractal Freshman • Posts: 9 • Interest in Mandelbrot Set and the way it evolves #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #2 on: July 05, 2019, 08:29:20 PM » A few tables of images to demonstrate how Julia Morphs are including the new path in the next evolution. #### wbarry • Fractal Freshman • Posts: 9 • Interest in Mandelbrot Set and the way it evolves #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #3 on: July 07, 2019, 09:53:36 PM » Re: "That minibrot will have layered cusp-seahorse-otherside patterns around it!" Exactly. And if you examine things carefully, you'll find it records the exact cusp, seahorse, and otherside. It will have the exact number of turns and branches and bulbs as whatever path you traveled. It's kind of amazing. A deeper layer of self-similarity. #### Fraktalist • Strange Attractor • Posts: 1079 #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #4 on: July 11, 2019, 10:45:01 AM » oh yeah - here we go. my absolute favourite aspect of the mandelbrot set. absolutely undervalued. and as usual when this topic comes up: https://www.youtube.com/watch?v=Ojhgwq6t28Y "The path you choose to follow BUILDS the shape you will find." Exactly! Every decision you take when zooming is recorded and it will actually create the look of every deeper zoom. Which is pretty cool in itself. But this also has stunning implications: It means you can actually STORE information in the Mset. See image below. You could choose a 2 branched bulb and call the left arm 1 and the right arm 0. and now you can store binary information by deciding 0 or 1. and this information will be embedded around the next minibrot, forever the basis of your future zoom path. You could say 01001000 01100101 01101100 01101100 01101111 00100000 01010111 01101111 01110010 01101100 01100100: "Hello World" Or you could zoom a lot deeper and store all the text written in this thread until now. Which in turn means: there is a location in the mandelbrot set - right now - a pair of coordinates already exists, which has this whole thread stored into its minibrot shells. And if you think it through: Also any future posts we haven't yet written. Or every book. Or all books. Or a complete binary simulation of our universe. After all, the mandelbrot set is infinite and has no limit of the amount of information you can store. It's all already somewhere in the Mandelbrot-Set. This blows my mind!! But why choose binary? Let's take a 4 branched bulb and call the branches G A T C. In the image below I decided to go G T then A and C. And it shows perfectly And then lets write your whole genome this way. Not the best or most efficient way to store information, the image will be boring, 3 billion turns into branch 1,2,3 or 4  - but it does the job. Every minibrot at the end of this will be surrounded with the building plan that can build YOU. And that location already exists in the mandelbrot-set. Now. Has always been in it. And always will be. I love that. « Last Edit: July 11, 2019, 11:20:24 AM by Fraktalist » #### marcm200 • Fractal Furball • Posts: 294 #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #5 on: July 11, 2019, 11:11:46 AM » Which in turn means: there is a location in the mandelbrot set - right now - a pair of coordinates already exists, which has this whole thread stored into its minibrot shells. And if you think it through: Also any future posts we haven't yet written. Wow! Could that be used as a compression algorithm? How many decimals would that location have when written out exactly? And conversely, what's the text like at easy locations like 10^-300 + i*10^-400 (if there's a minibrot somewhere in the vicinity). And: if one of the C++ programs I've coded is faulty - the correct version is in the Mandelbrot set - now I just have to find it! • Posts: 79 #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #6 on: July 11, 2019, 02:02:58 PM » No, it cannot be used as a compression algorithm. The so called "counting argument" applies to compression regardless of algorithm details. In practice this probably means you need to store more bits of precision for the mandelbrot coordinates than you encoded as payload data. The idea of all past and future texts being present in some infinite library is mind blowing, though. Many versions of it are older than the Mandelbrot set, but they are also more hypothetical than the Mandelbrot set. The first version I personally encountered was a relatively recent one, where some science fiction author estimated the length of a bookshelf containing all possible books that have up to 600 pages of text in the Latin alphabet. It was an unimaginably large number of light years. #### wbarry • Fractal Freshman • Posts: 9 • Interest in Mandelbrot Set and the way it evolves #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #7 on: July 12, 2019, 03:51:37 PM » Re: "my absolute favourite aspect of the mandelbrot set. absolutely undervalued." Yes! It seems like this is one of the keys to the fabric of the Mandelbrot set. And that any "guide" to the set should have this as one of the first things mentioned, yet I felt like I had to discover it for myself (I had not seen your video). "Shape stacking" hints at it, but does not make it clear how deep this goes. I think the mathematicians mention it, but in hyperbolic isomorphic pedantic semantics. I suppose I should read more and zoom less, if I really want to understand. But zooming is much more fun. • 3e • Posts: 1121 #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #8 on: July 12, 2019, 04:04:17 PM » "history repeats, only twice as fast" is my take on approaching mini-sets and period doubling shape stacking. #### Fraktalist • Strange Attractor • Posts: 1079 #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #9 on: July 12, 2019, 11:20:36 PM » I suppose I should read more and zoom less, if I really want to understand. But zooming is much more fun. disagree. learning by doing, the pleasure of finding things out - yourself! when you learn something from own experience, the knowledge and understanding is so much deeper and more satisfying! you found out yourself, just like the first one who discovered it and academically cryptic wrote a paper about a few hundred people will ever read AND understand. so your achievement is the same. you won't get credits, but you rock! you did it! who cares who was first. also, explaining it in your own words, with practical examples can potentially reach far more interested people, no "academic talk barrier". Keep zooming & exploring!!! #### wbarry • Fractal Freshman • Posts: 9 • Interest in Mandelbrot Set and the way it evolves #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #10 on: July 13, 2019, 02:09:02 AM » Wow, how kind of you! Thanks. And following that positive trend, I definitely should have shown more appreciation for the people who contributed to all of the articles and guides that I found. There is so much valuable info out there, mostly contributed by fellow enthusiasts who do it just because it is freakin' cool stuff. This forum has been hugely helpful. And I feel so fortunate for the tools we have to explore the set. In my day (get off my lawn!), we had to wait days to get a single image out of the mainframe. Now we have nearly instant images from deep in the set on our personal machines. We live in amazing times. For a bunch of idiots, people are actually pretty cool. Now, back to the chimps in the typing room ... they're demanding more cigarettes ... #### tavis • Fractal Fanatic • Posts: 39 • Bill Tavis #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #11 on: July 13, 2019, 09:51:57 PM » This also works for the bulbs, not just minibrots. So for example, you can zoom into a bulb of a bulb of a bulb etc and it will store the unique location in the attached tendrils in a sequence, without ever visiting a minibrot. I actually think this approach could be much more efficient for the binary storage idea, requiring far less zooming. But honestly either way it's pretty computationally prohibitive (even though it's a super cool thought exercise) Check out the Mandelmap poster #### wbarry • Fractal Freshman • Posts: 9 • Interest in Mandelbrot Set and the way it evolves #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #12 on: July 13, 2019, 11:28:08 PM » First off, I have given no mathematical proof, so I could totally be wrong. Meanwhile, I think we are actually agreeing. I am saying that WHEREVER you zoom will be recorded into the fabric of what you will find. Zoom through three bulbs you will find "shells" of recordings of those three bulbs - as you say, it will be stored. I am really talking about the storage mechanism. The shells of recordings are woven around minibrots, which act like the galactic black holes at the center of all galaxies. These central minibrots are surrounded by recordings of the shapes you would pass through to arrive at that specific minibrot, whether you passed through bulbs, spirals, tendrils, or anything. Each minibrot is surrounded by recordings of its own unique path (or address, as in Dierk Schleicher's paper). The fabric of the set itself is a network of minibrots surrounded by recordings. The recordings, woven into the original, are the fabric, a set of self-similar copies of the path to each exact location. The encoded paths are the fabric of the Mandelbrot set. Or not. #### Fraktalist • Strange Attractor • Posts: 1079 #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #13 on: July 14, 2019, 02:47:27 AM » The shells of recordings are woven around minibrots, which act like the galactic black holes at the center of all galaxies. Welcome to a new episode of "mind blown": https://www.zmescience.com/science/physics/black-holes-store-information-0023534/ This also works for the bulbs, not just minibrots. So for example, you can zoom into a bulb of a bulb of a bulb etc and it will store the unique location in the attached tendrils in a sequence, without ever visiting a minibrot. I actually think this approach could be much more efficient for the binary storage idea, requiring far less zooming. But honestly either way it's pretty computationally prohibitive (even though it's a super cool thought exercise) good point. reason is because the "recording" of the  previous path starts at the first bifurcation/doubling, and every branch growing from a bulb is "behind" this first bifurcation. « Last Edit: July 14, 2019, 03:08:56 PM by Fraktalist » #### marcm200 • Fractal Furball • Posts: 294 #### Re: Pathfinding in the Mandelbrot set - Revisited « Reply #14 on: July 14, 2019, 11:57:18 AM » I've read that higher degree monomials in the parameter-space go to the unit disk in the limit, so the "fractal nature" somehow decreases (un-mathematically spoken). Does this recording of paths also apply to other monomials of higher degree or to polynomials in general (or is the z²-Mandelbrot the "master fractal"? And only to the parameter-space or also to the dynamical entity? Questions over questions... ### Similar Topics ###### the Buddha's Jewel revisited - finding Mandelbrot orbits shaped like Julia sets Started by tavis on Fractal Mathematics And New Theories 10 Replies 379 Views April 13, 2019, 07:26:57 AM by hgjf2 ###### Some of my older stuff, only revisited this myself recently ! Started by FractalDave on Off Topic 1 Replies 119 Views December 07, 2018, 09:20:16 PM by FractalDave ###### Trees Revisited square version Started by Dinkydau on Fractal Image Gallery 0 Replies 215 Views November 03, 2017, 11:49:38 AM by Dinkydau ###### Mandelbrot Burning Ship Mandelbrot Mandelbrot hybrid Started by claude on Fractal Image Gallery 0 Replies 228 Views January 16, 2018, 11:30:30 PM by claude ###### Mandelbrot Burning Ship Mandelbrot Mandelbrot hybrid 2 Started by claude on Fractal Image Gallery 0 Replies 210 Views January 17, 2018, 12:10:56 AM by claude
2019-07-23 11:08:29
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https://physics.stackexchange.com/questions/90200/einstein-action-and-the-second-derivatives
# Einstein action and the second derivatives I have naive question about Einstein action for field-free case: $$S = -\frac{1}{16 \pi G}\int \sqrt{-g} d^{4}x g^{\mu \nu}R_{\mu \nu}.$$ It contains the second derivatives of metric. When we want to get the Einstein equation (which doesn't contain the third derivatives), we must use variational principle. The variation of "problematic" factor $R_{\mu \nu}$ (which contains the second derivatives) is equal to $$\delta R_{\mu \nu} = D_{\gamma}(\delta \Gamma^{\gamma}_{\mu \nu}) - D_{\nu}(\delta \Gamma^{\lambda}_{\mu \lambda}).$$ So the corresponding variation of action may be rewritten in a form $$\delta_{R_{\mu \nu}} S = -\frac{1}{16 \pi G}\int d^{4}x \sqrt{-g}\partial_{\lambda}(g^{\mu \nu}\delta \Gamma^{\lambda}_{\mu \nu} - g^{\mu \lambda}\delta \Gamma^{\sigma}_{\mu \sigma}). \qquad (1)$$ Then one likes to say that it is equal to zero. But why it must be equal to zero? It isn't obvious to me. After using the divergence theorem $(1)$ becomes $$\delta_{R_{\mu \nu}} S = -\frac{1}{16 \pi G}\int dS_{\lambda} \sqrt{-g}(g^{\mu \nu}\delta \Gamma^{\lambda}_{\mu \nu} - g^{\mu \lambda}\delta \Gamma^{\sigma}_{\mu \sigma}).$$ Why it must be equal to zero? It is metric, not physical field, even if Christoffel symbols refer to the gravitational field, so I don't understand why it must be equal to zero at infinity. It seems that OP is pondering the following. What happens in a field theory [in OP's case: GR] if spacetime $M$ has a non-empty boundary $\partial M\neq \emptyset$, and we don't impose pertinent (e.g. Dirichlet) boundary conditions (BC) on the fields $\phi^{\alpha}(x)$ [in OP's case: the metric tensor $g_{\mu\nu}(x)$]? I) Firstly, it should stressed that when people say that the infinitesimal variation $\delta S_0$ of the action $S_0[\phi]$ [in OP's case: the Einstein-Hilbert action $S_{EH}$] vanishes on-shell, i.e. when the Euler-Lagrange equations [in OP'case: Einstein's field equations] are satisfied, it is implicitly assumed that the infinitesimal variations $\delta\phi^{\alpha}(x)$ of the fields $\phi^{\alpha}(x)$ only take place in the interior/bulk of spacetime $M$ away from the boundary $\partial M$. In such cases, the infinitesimal variation $\delta S_0$ clearly vanishes on-shell, as part of the stationary action principle, aka. Hamilton's principle. II) Secondly, if the sole purpose of the infinitesimal variation $\delta\phi^{\alpha}(x)$ is just to locally (re)derive the equations of motion (=the Euler-Lagrange equations) in an interior/bulk point $x_0$ of spacetime $M$, it is enough to choose localized variations $\delta\phi^{\alpha}(x)$ with support in sufficiently small compact neighborhoods around this point $x_0$. In particular, one may assume that $\delta\phi^{\alpha}(x)$ and all its (higher) derivatives vanish at the boundary $\partial M$ for such variations, and still derive the equations of motion. III) Thirdly, if we do not impose adequate BC, then the global notion of a functional derivative $$\tag{1} \frac{\delta S_0}{\delta\phi^{\alpha}(x)}$$ may not exists, i.e. there may not exists a globally defined function $^1$ $$\tag{2} f_{\alpha}(x) ~=~f_{\alpha}(\phi(x), \partial \phi(x), \partial^2 \phi(x),\ldots ;x)$$ such that $$\tag{3} \delta S_0 ~=~\int_M \! d^{n}x ~f_{\alpha}(x)~\delta\phi^{\alpha}(x)$$ for all allowed infinitesimal variations $\delta\phi^{\alpha}(x)$. In plain English, the problem is that we cannot use the usual integration by part argument when deriving the Euler-Lagrange expression since we have not imposed sufficient BC to ensure that boundary terms vanish. Then one must typically amend the bulk action $S_0$ with a boundary action $S_1$ [in OP's case: The Gibbons-Hawking-York boundary action $^2$ $S_{GHY}$], which only lives on the boundary $\partial M$. The total action then reads $S_0+S_1$. -- $^1$ If the function (2) happens to exist, it is unique, and we will call it the functional derivative of $S_0$, and denote it by the symbol (1). $^2$ See also these Phys.SE posts for more on the Gibbons-Hawking-York boundary term. • Correction to the answer (v2): Integration by part should be Integration by parts. – Qmechanic Oct 2 '14 at 21:02
2019-08-19 19:03:56
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https://www.physicsforums.com/threads/complex-scalar-field-commutation-relations.885777/
# A Complex scalar field - commutation relations 1. Sep 18, 2016 ### spaghetti3451 I find it difficult to believe that the canonical commutation relations for a complex scalar field are of the form $[\phi(t,\vec{x}),\pi^{*}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$ $[\phi^{*}(t,\vec{x}),\pi(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$ This seems to imply that the two scalar fields $\phi$ and $\phi^{*}$ are somehow coupled, even though they are not. Can you explain this? 2. Sep 18, 2016 ### Fightfish Where did you get these commutation relations from? The notes that I have seem to suggest otherwise. 3. Sep 18, 2016 ### spaghetti3451 I assumed that these are the fundamental commutation relations for complex scalar field and using the commutation relations $[a_{{\bf{p}}},a_{{\bf{p}}'}^{\dagger}]=(2\pi)^{3}\delta^{(3)}({\bf{p}}-{\bf{p}}')$ $[b_{{\bf{p}}},b_{{\bf{p}}'}^{\dagger}]=(2\pi)^{3}\delta^{(3)}({\bf{p}}-{\bf{p}}')$ and the expansion of the field $\phi$ as $\phi=\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2\omega_{\bf{p}}}}(a_{\bf{p}}e^{i{\bf{p}}\cdot{\bf{x}}}+b_{\bf{p}}^{\dagger}e^{-i{\bf{p}}\cdot{\bf{x}}})$ I showed that the commutation relations in my first post hold true. What commutation relations do you have? Can you post them? I need to check my calculations using both mine and yours commutation realtions? 4. Sep 18, 2016 ### Fightfish Did you perhaps forget that $\pi (x) = \dot{\phi^{*}}(x)$ and not $\dot{\phi}(x)$ ? I have the commutation relations: $[\phi(t,\vec{x}),\pi(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$ $[\phi^{*}(t,\vec{x}),\pi^{*}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$ from the same starting point as you have for the creation/annihilation operators 5. Sep 18, 2016 ### spaghetti3451 I do have $\pi=\dot{\phi}^{*}$. 6. Sep 18, 2016 ### Fightfish Hmm...might you have forgotten to take the Hermitian conjugate of the creation/annihilation operators when obtaining $\phi^{*}$? Otherwise I think I will need to see the explicit expressions for your fields and conjugate momenta to figure out where you went wrong. 7. Sep 18, 2016 ### spaghetti3451 I started with the two commutation relations $[\phi^{i}(\vec{x}),\pi_{i}(\vec{y})]=i\delta^{3}(\vec{x}-\vec{y})$ for the free real scalar fields $\phi^{1}$ and $\phi^{2}$ and plugged $\phi = \frac{1}{\sqrt{2}}(\phi^{1}+i\phi^{2})$ and the corresponding derivative into the two commutation relations to obtain a set of two equations in terms of $\phi$, $\phi^{\dagger}$, $\pi$ and $\pi^{\dagger}$. I added these two equations and also subtracted the same two equations to obtain a new set of two equations as follows: $[\phi^{\dagger}(\vec{x}),\pi^{\dagger}(\vec{y})]+[\phi(\vec{x}),\pi(\vec{y})]=2i\delta^{3}(\vec{x}-\vec{y})$ $[\phi(\vec{x}),\pi^{\dagger}(\vec{y})]+[\phi^{\dagger}(\vec{x}),\pi(\vec{y})]=0$ It's obvious now that your commutation relations are a possible solution to these equations, but I want to be sure that yours is the only solution. With that in mind, let me proceed. $[\phi^{\dagger}(\vec{x}),\pi^{\dagger}(\vec{y})]+[\phi(\vec{x}),\pi(\vec{y})]=2i\delta^{3}(\vec{x}-\vec{y})$ $\implies [\phi(\vec{x}),\pi(\vec{y})]-[\phi(\vec{x}),\pi(\vec{y})]^{\dagger}=2i\delta^{3}(\vec{x}-\vec{y})$ I'm now only slightly away from proving that $[\phi(\vec{x}),\pi(\vec{y})]=i\delta^{3}(\vec{x}-\vec{y})$. How do I show that $-[\phi(\vec{x}),\pi(\vec{y})]^{\dagger}=[\phi(\vec{x}),\pi(\vec{y})]$? Last edited: Sep 18, 2016 8. Sep 19, 2016 ### Fightfish This is a strange way to proceed; usually its done by the method you put in post #3 - starting from the expression of $\phi$, generate $\phi^{*}$ and their conjugate momenta, and use the commutation relations of the creation and annihilation operators to simplify the integrals to get the desired results. Your alternative method still works, but why not consider $[\phi(x), \pi(y)]$ directly? That, for example, immediately yields $$[\phi(x), \pi(y)] = \left[\frac{1}{\sqrt{2}} \left(\phi_{1}(x) + i \phi_{2}(x)\right) , \frac{1}{\sqrt{2}} \left(\pi_{1}(y) - i \phi_{2} (y)\right) \right] = \frac{1}{2} \left( \left[\phi_{1}(x),\pi_{1}(y) \right] + \left[\phi_{2}(x),\pi_{2}(y) \right] \right) = i \delta^{(3)} (x-y)$$ 9. Sep 19, 2016 ### spaghetti3451 This method requires that I guess an expansion for $\pi$ in terms of the creation and annihilation operators and then check that it satisfies the commutation relations for $\phi$ with $\pi$ as well as for $\phi^{*}$ with $\pi^{*}$. I was wanting to avoid this guesswork, so ... This is a better approach, and I will use it now. But, the commutators for real $\phi^{i}$ with real $\pi_{i}$ directly yield the non-zero commutators. If I start with the commutators for complex $\phi$ and $\pi$, I have to check all the commutators one by one to find the non-zero ones. 10. Sep 19, 2016 ### Fightfish If you start with an expression for $\phi$ already, there's no need for guesswork - $\phi^{*}$ and the corresponding conjugate momenta can be obtained from it. 11. Sep 19, 2016 ### vanhees71 It's all clearly defined. The (free) charged Klein-Gordon field has the Lagrangian (east-coast convention) $$\mathcal{L}=(\partial_{\mu} \phi^*)(\partial^{\mu} \phi)-m^2 \phi^* \phi.$$ The canonical field momenta for the independent field degrees of freedom $\phi$ and $\phi^*$ (you can as well rewrite everything in terms of two real KG fields of course) are $$\Pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}^*, \quad \Pi^*=\frac{\partial \mathcal{L}}{\partial \dot{\phi}^*}=\dot{\phi},$$ and the equal-time commutator relations are $$[\phi(t,\vec{x}),\Pi(t,\vec{y})]=\mathrm{i} \delta^{(3)}(x-y), \quad [\phi^{\dagger}(t,\vec{x}),\Pi^{\dagger}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(x-y).$$ 12. Sep 19, 2016 ### spaghetti3451 I know how to obtain $\phi^{*}$ from $\phi$, but I'm not sure how to obtain $\pi=\dot{\phi}^{*}$ from $\dot{\phi}^{*}$, since $\displaystyle{\pi=\dot{\phi}^{*}=\int\ \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2\omega_{{\bf{p}}}}}\Big(\dot{b}_{\vec{p}}\ e^{i\ \vec{p}\cdot{\vec{x}}} + \dot{a}_{\vec{p}}^{\dagger}\ e^{-i\ \vec{p}\cdot{\vec{x}}} \Big)}$. But what do you do next to obtain an expansion for $\pi$? 13. Sep 19, 2016 ### vanhees71 Don't forget the Heisenberg-picture equations of motion. The time dependence is $\hat{a}_{\vec{p}}(t) = \hat{a}_{\vec{p}} \exp(-\mathrm{i} E_{\vec{p}} t)$ and $\hat{b}_{\vec{p}}(t) = \hat{b}_{\vec{p}} \exp(-\mathrm{i} E_{\vec{p}} t)$. 14. Sep 19, 2016 ### spaghetti3451 I knew that this was exactly what I was supposed to do, but for some reason, I did not attempt it. Thanks!
2018-07-19 12:21:38
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https://www.physicsforums.com/threads/problems-with-fft.64039/
# Problems with FFT 1. Feb 16, 2005 ### brian0918 This is part of my senior project. Please help! I'm trying to use a code to FFT a real, even set of data, but the results are not real and even as would be expected. The code I'm using is availabe here (and fft.h andintl.h) When I run RealFFT with the following 16 numbers for the Real input and nothing for the Imag input, I get these results: Real input: 0.969647 -1.28744 0.0863155 0.19462 1.06651 0.490905 0.469052 -0.728912 -0.728912 0.469052 0.490905 1.06651 0.19462 0.0863155 -1.28744 0.969647 Real output: 2.52139 -0.37157 -2.72877 4.38382 1.74303 2.06953 0.687978 0.712455 Imag output: 1.19209e-007 0.0739102 1.13029 -2.92917 -1.74303 -3.09727 -1.66093 -3.58175 Any help? I've tried a different FFT code and the same thing happens. Help!! 2. Feb 16, 2005 ### Davorak Is there some reason you can not use matlab, mathematica, or mathcad for their built in FFT? 3. Feb 16, 2005 ### brian0918 Everything has to be automated and separate, so I need to use code. This FFT is just a small part of the whole code. Last edited: Feb 16, 2005 4. Feb 16, 2005 ### FredGarvin I did a quick FFT using Excel on your data and, short of having sign issues with the imaginary output, your numbers are exactly the same that I got. 5. Feb 16, 2005 ### brian0918 The data is real and even though, isn't it? So, shouldn't the result also be real and even? 6. Feb 16, 2005 ### Hurkyl Staff Emeritus Nope. Though, IIRC, the imaginary parts can be computed from the real parts, so the real parts are all you actually need when the input is real. 7. Feb 16, 2005 ### Atrus It should be real only if input sequence is symmetrical around t = 0. But you assume causal sequence (with FFT), so data is shifted and phase delay is introduced, by which you get imaginary component of spectrum... At least, that's off top of my head. Could be wrong... 8. Feb 16, 2005 ### brian0918 So how would I go about inputting data so that it would be interpreted as being symmetrical around t=0? According to the text that created the code I used, it is even if the values at k and (N-k) are the same. 9. Feb 16, 2005 ### Atrus I just thought about what I wrote (I should do that BEFORE writing, right?). Well, since FFT (DFT) transform results with spectrum of periodic signal whose one period is the original sequence, it doesn't matter if the sequence is causal or not, FFT "percieves" it as periodic to + and - direction, hence the dicrete spectrum. I've input the data in Matlab, and got the same results. Shifting the array results only in different phase frequency characteristic. The seed of doubt in my mind is that I remember this: symmetrical real signals result in _linear_ phase characteristic, not by zero-phase... Anyway, I'm ill and it's quite late here, so I'll try to think straight in the morning (at least, I'll try ;) ) 10. Feb 17, 2005 ### shmoe I think you're mixing up the indicies somehow. If your data set is indexed starting at zero and it's even, it will look like: $$\left[x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_7, x_6, x_5, x_ 4, x_ 3, x_2, x_1\right]$$ In this case the output will be real assuming the input is. Your data is not of this shape, so you can't expect the nice cancellation.
2017-03-27 20:49:05
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http://openstudy.com/updates/55dc7873e4b0029e27ff2020
## mathmath333 one year ago Probablity question 1. mathmath333 \large \color{black}{\begin{align} & \normalsize \text{Three different numbers are selected from the set}\ X=\{1,2,3,4...10\}\hspace{.33em}\\~\\ & \normalsize \text{What is the probablity that the product of the two numbers is equal to }\hspace{.33em}\\~\\ & \normalsize \text{the third }\hspace{.33em}\\~\\ \end{align}} 2. mathmath333 \large \color{black}{\begin{align} & a.)\ \dfrac{3}{10}\hspace{.33em}\\~\\ & b.)\ \dfrac{1}{40}\hspace{.33em}\\~\\ & c.)\ \dfrac{1}{20}\hspace{.33em}\\~\\ & d.)\ \dfrac{4}{5}\hspace{.33em}\\~\\ \end{align}} 3. ganeshie8 say the $$3$$ numbers selected are $$\{a,b,c\}$$ 4. ganeshie8 also suppose that $$a\lt b\lt c$$ 5. mathmath333 ok 6. ganeshie8 then you want $$a*b\le 10$$ 7. ganeshie8 notice that if $$a*b\le 10$$, then $$a \le \sqrt{~10~}$$ 8. ganeshie8 so $$a$$ can only be either $$2$$ or $$3$$ 9. ganeshie8 when $$a=2$$, check what all $$b$$ values will work 10. mathmath333 how does the condition $$a=\sqrt{10}$$ came. 11. ganeshie8 good question, that is called seive of eratosthenes https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes 12. mathmath333 ok 13. ganeshie8 we have $$ab\le 10$$ and you know that $$a\lt b$$ whats the maximum value that $$a$$ can take ? 14. mathmath333 2 15. ganeshie8 why not 3 ? 16. mathmath333 cuz $$3\times 4=12$$ 17. ganeshie8 Perfect! so the only value that $$a$$ can take is $$2$$ 18. mathmath333 why not $$1$$ ? 19. ganeshie8 you tell me why 20. mathmath333 $$1*2\le 10$$ 21. ganeshie8 remember you want to pick all 3 "different" numbers 22. ganeshie8 what happens if any one number is $$1$$ ? 23. mathmath333 it satisfies the condition example $$1,2,3$$ 24. ganeshie8 it doesn't, 1*2 is not 3 25. ganeshie8 the smaller two numbers must multiply to the third number 26. mathmath333 oh i see, sry 27. mathmath333 yea , only $$2$$ for $$a$$ 28. ganeshie8 yes let $$a=2$$ and find all $$b$$ such that $$a*b\le 10$$ 29. mathmath333 $$b=3,4,5$$ 30. ganeshie8 Yes, so the numbers in favor are : (2, 3, 6) (2, 4, 8) (2, 5, 10) 31. ganeshie8 thats 3 in favor 32. ganeshie8 save that 33. ganeshie8 next, find how many total ways are there to choose 3 different numbers from the given 10 numbers 34. mathmath333 Is their any condition to choose 3 different numbers from the given 10 numbers. 35. ganeshie8 thats the sample space so no conditions, just find the total number of ways of choosing 3 numbers from 10 36. mathmath333 $$\dfrac{10!}{7!}$$ 37. ganeshie8 nope 38. ganeshie8 its just $$\large \dbinom{10}{3}$$ 39. ganeshie8 notice that here order doesn't matter... so it is a combination 40. mathmath333 why not $$^{10}P_{3}$$ ?? 41. mathmath333 ok 42. ganeshie8 favor : $$\large 3$$ total : $$\large \dbinom{10}{3}$$ 43. ganeshie8 take the ratio for the probability 44. mathmath333 so answer=$$\dfrac{1}{40}$$ 45. ganeshie8 Yes, did u get why this is a combination and not a permutation problem ? 46. mathmath333 cuz we choosing 3 numbers one by one and not forming 3 numbers like $$abc_{10}$$ 47. ganeshie8 yes, we never bothered about order while doing this problem 48. ganeshie8 we could also do it using permutations but it is painful here 49. ganeshie8 if we use permutations, the count of favor also changes 50. ganeshie8 (2, 3, 6) (2, 4, 8) (2, 5, 10) 51. ganeshie8 since (2, 3, 6) works, all the 6 permutations of it also work : (2, 6, 3) (3, 2, 6) (3, 6, 2) (6, 2, 3) (6, 3, 2) 52. ganeshie8 after all that mess, you will get the same answer 53. ganeshie8 In these probability problems, it doesn't matter whether you use permutations or combinations.. the final answer wont change if you do it correctly
2016-10-25 01:53:11
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http://swmath.org/software/15498
# survRM2 R package survRM2: Comparing Restricted Mean Survival Time. Performs two-sample comparisons using the restricted mean survival time (RMST) as a summary measure of the survival time distribution. Three kinds of between-group contrast metrics (i.e., the difference in RMST, the ratio of RMST and the ratio of the restricted mean time lost (RMTL)) are computed. It performs an ANCOVA-type covariate adjustment as well as unadjusted analyses for those measures. ## Keywords for this software Anything in here will be replaced on browsers that support the canvas element
2017-02-20 03:58:40
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https://brilliant.org/problems/did-someone-say-division/
# Did someone say 'division'? The largest number $$N$$, by which the expression $${ n }^{ 3 }-n$$ is divisible by $$N$$ for all possible integral values of $$n$$, is: × Problem Loading... Note Loading... Set Loading...
2018-01-24 00:02:54
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https://gmatclub.com/forum/2-274323.html
GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 17 Nov 2018, 22:04 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History ## Events & Promotions ###### Events & Promotions in November PrevNext SuMoTuWeThFrSa 28293031123 45678910 11121314151617 18192021222324 2526272829301 Open Detailed Calendar • ### FREE Quant Workshop by e-GMAT! November 18, 2018 November 18, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. November 18th, 7 AM PST • ### How to QUICKLY Solve GMAT Questions - GMAT Club Chat November 20, 2018 November 20, 2018 09:00 AM PST 10:00 AM PST The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat. # (2 + (2 + (2 + 4^(1/2))^(1/2))^(1/2))^(1/2) Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 50623 (2 + (2 + (2 + 4^(1/2))^(1/2))^(1/2))^(1/2)  [#permalink] ### Show Tags 26 Aug 2018, 04:47 00:00 Difficulty: 15% (low) Question Stats: 95% (00:18) correct 5% (00:30) wrong based on 19 sessions ### HideShow timer Statistics $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{4}}}}$$ is equal to which of the following? (A) $$\sqrt{2}$$ (B) $$2$$ (C) $$2\sqrt{2}$$ (D) $$4$$ (E) $$4\sqrt{2}$$ _________________ Director Status: Learning stage Joined: 01 Oct 2017 Posts: 930 WE: Supply Chain Management (Energy and Utilities) Re: (2 + (2 + (2 + 4^(1/2))^(1/2))^(1/2))^(1/2)  [#permalink] ### Show Tags 26 Aug 2018, 07:44 Bunuel wrote: $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{4}}}}$$ is equal to which of the following? (A) $$\sqrt{2}$$ (B) $$2$$ (C) $$2\sqrt{2}$$ (D) $$4$$ (E) $$4\sqrt{2}$$ $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{4}}}}$$ =$$\sqrt{2+\sqrt{2+\sqrt{2+2}}$$ =$$\sqrt{2+\sqrt{2+\sqrt{4}}$$ =$$\sqrt{2+\sqrt{2+2}}$$ =$$\sqrt{2+\sqrt{4}}$$ =$$\sqrt{2+2}$$ =$$\sqrt{4}$$ =$$2$$ Ans. (B) _________________ Regards, PKN Rise above the storm, you will find the sunshine Re: (2 + (2 + (2 + 4^(1/2))^(1/2))^(1/2))^(1/2) &nbs [#permalink] 26 Aug 2018, 07:44 Display posts from previous: Sort by
2018-11-18 06:04:07
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https://www.khanacademy.org/math/linear-algebra/matrix-transformations/matrix-transpose/v/linear-algebra-transposes-of-sums-and-inverses
# Transposes of sums and inverses ## Video transcript Let's see if we can prove to ourselves some more reasonably interesting transpose properties. So let's define some matrix C, that's equal to the sum of two other matrices, A and B. And so any entry in C, I can denote with a lowercase cij. So if I want the ith row in jth column it would be cij, and so each of its entries are going to be the sum of the corresponding columns that are matrices A and B. So our ij entry in C is going to be equal to the ij entry in A, plus the ij entry in B. That's our definition of matrix addition. You just get the corresponding entry in the same row and column, add them up, and you get your entry in the same row and column, and your new matrix is the sum of the other two. Now, let's think a little bit about the transposes of these guys right here. So, if A looks like this. I won't draw all of the entries. It takes forever. But each of its entries are ij, just like that. Let's say that A transpose looks like this. Each of its entries, we would call it, that's if you've got that same entry, we're going to call it a-prime ij. And these things aren't probably going to be the same. There's some chance they are, but they're probably not going to be the same. But that its ijth entry. In the ith row, jth column. In A transpose. Now, the fact that this is the transpose of that means that everything that's in some row and column here is going to be in that column and row over here, that the rows and columns get switched. So we know that we could write that a-prime ij, we're going to have the same entry that was in aji. Maybe aji is over here. aji is over here. So, this thing over here, which is in the same position as this one, is going to be equal to this guy over here if you switched the rows and columns. I think you can accept that. And you can make the same argument for B. Let me actually draw it out. So if I make B transpose. The entry in the ith row and jth column, I'll call it b-prime ij. Just like that. Just like I did for A. So we could say that b-prime ij is equal to, you take the matrix B, what's going to be the entry that's in the jth row and ith column. These are, you could almost say, the definition of the transpose. If I'm in the third row and second column now, it's going to be what was in the second row and third column. Fair enough. So we already have what cij is equal to. What's the transpose of cij going to be equal to? Let me write that down. So C transpose, let me write it over here. Write C transpose is equal to. I'll use the same notation. The prime means that we're taking entries in the transpose. So C transpose is just going to be a bunch of entries, ij. And I'll put a little prime there showing that that's entries in the matrix of the transpose, and not in C itself. And we know that c-prime ij is equal to cji. Nothing new at all. We've just expressed kind of the definition of the transpose for these three matrices. Now what is cji equal to? So let's focus on this a little bit. What is cji equal to? We know that cij is equal to a sub ij plus b sub ij, so if you swap them around, this is going to be equal to, you just swap the j's and the i's. a sub ji plus b sub ji. I just used this information here-- you could almost view it as this assumption or this definition-- to go from this to this. If I had an x and a y here, I'd have an x and a y here, and a x and a y here. I have a j and an i here, so I have a j and an i there, and a j and an i right there. Now what are these? What are these equal to? This is equal to. This guy right here is equal to-- we do it in the green-- the same entry for the transpose of a at ij. And this is equal to the same entry for the transpose of b at ij. Now, what is this telling us? It's telling us that the transpose of C, which is the same thing is A plus B, so it's saying that A plus B, A plus B transpose is the same thing as C transpose. Let me write that. C transpose is the same thing as A plus B transpose. So these are the entries in A plus B transpose right here. And what is this over here? What are these? These are the entries right there. We do the equal sign over here. What are these? These are the entries in A transpose plus B transpose. Right? These are the entries in A transpose. These are the entries in B transpose. If you take the sum of the two, you're just adding up the corresponding entries. So that's straightforward to show that if you take the sum of two matrices and then transpose it, it's equivalent to transposing them first, and then taking their sum. Which is a reasonably neat outcome. Let's do one more and I think we'll finish up all of our major transpose properties. Let's say that A inverse-- this is going to be a slightly different take on things. We're still going to take the transpose. So if we know that A inverse is the inverse of A, that means that A times A inverse is equal to the identity matrix, assuming that these are n-by-n matrices. So it's the n-dimensional identity matrix. And that A inverse times A is also going to be equal to the identity matrix. Now, let's take the transpose of both sides of this equation. I'll do them both simultaneously. So if you take the transpose of both sides of the equation, you get A times A inverse transpose is equal to the identity matrix transpose. And what's the transpose of the identity matrix? Let's draw it out. The identity matrix looks like this. You have just ones all the way down the diagonal and everything else is 0. Right, and you could view this as i 1, 1 i 2, 2 all the way down to i n, n. Everything else is 0. So when you take the transpose, you're just swapping out the zeroes, right? These guys don't change. The diagonal does not change when you take the transpose. So the transpose of the identity matrix is equal to the identity matrix. And so we can apply that same thing here. Let's take the transpose for this statement. So we know that A inverse times A transpose is equal to the identity matrix transpose, which is equal to the identity matrix. And then we know what happens when you take the transpose of a product. It's equal to the product of the transposes in reverse order. So this thing right here we can rewrite as A inverse transpose times A transpose, which is going to be equal to the identity matrix. You could do the same thing over here. This thing is going to be equal to A transpose times A inverse transpose, which is also going to be equal to the identity matrix. Now, this is an interesting statement. The fact that, if I have this guy right here, times the transpose of A is equal to the identity matrix, and the transpose of A times that same guy is equal to identity matrix, implies that A inverse transpose is the inverse of A transpose. Or another way of writing that is if I take A transpose, and if I take its inverse, that is going to be equal to this guy. It's going to be equal to A inverse transpose. So, another neat outcome dealing with transposes. If you take the inverse of the transpose, it's the same thing as the transpose of the inverse.
2018-11-19 09:35:09
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https://zbmath.org/?q=an%3A0918.53009
# zbMATH — the first resource for mathematics Geometry of real Grassmannian manifolds. III. (English. Russian original) Zbl 0918.53009 J. Math. Sci., New York 100, No. 3, 2254-2268 (2000); translation from Zalgaller, V. A. (ed.) et al., Geometry and topology. 2. Work collection. Sankt-Peterburg: Matematicheskij Institut Im. V. A. Steklova, Sankt-Peterburgskoe Otdelenie, RAN, Zap. Nauchn. Semin. POMI. 246, 108-129 (1997). In a previous paper [S. E. Kozlov, ibid., 84-107 (1997; Zbl 0918.53008), see the preceding review], the author established a Plücker model for the Grassmannian manifolds $$G^+_{p,n}$$. In the paper under review, stationary angles between (oriented or non-oriented) planes are introduced. The diameter and the injectivity radius of $$G^+_{p,n}$$ are calculated. Finally, the closure of geodesics in $$G^+_{p,n}$$ is determined. ##### MSC: 53B25 Local submanifolds 15A75 Exterior algebra, Grassmann algebras 53A20 Projective differential geometry 53C20 Global Riemannian geometry, including pinching Full Text: ##### References: [1] S. E. Kozlov, ”Geometry of real Grassmann manifolds. Parts I, II,” Zap. Nauchn. Semin. POMI 246, 84–107 (1997). (this issue). [2] L. D. Ivanov, Variations of Sets and Functions [in Russian], Nauka, Moscow (1975). [3] B. A. Rozenfel’d, Many-Dimensional Spaces [in Russian], Nauka, Moscow (1966). [4] Y.-C. Wong, ”Differential geometry of Grassmann manifolds,” Proc. Natl. Acad. USA, 57, 589–594 (1967). · Zbl 0154.21404 [5] Y.-C. Wong, ”Sectional curvatures of Grassmann manifolds,” Proc. Natl. Acad. USA, 60, 75–79 (1968). · Zbl 0169.23904 [6] Yu. D. Burago and V. A. Zalgaller, Introduction to Riemannian Geometry [in Russian], Nauka, Saint Petersburg (1994). · Zbl 0814.53001 [7] S. E. Kozlov, ”Orthogonally congruent bivectors,” Ukr. Geom. Sb., 27, 68–75 (1984). · Zbl 0557.15019 [8] D. K. Faddeev, Lectures on Algebra [in Russian], Nauka, Moscow (1984). · Zbl 0565.00001 [9] P. M. Cohn, Free Rings and Their Relations, Academic Press, London-New York (1971). · Zbl 0232.16003 [10] S. Helgason, Differential Geometry and Symmetric Spaces, Academic Press, New York-London (1962). · Zbl 0111.18101 [11] B. M. Makarov, M. G. Goluzina, et al., Selected Problems in Real Analysis [in Russian], Nauka, Moscow (1992). · Zbl 0858.26002 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
2021-08-05 15:16:30
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http://www.reliawiki.org/index.php/Appendix_B:_Parameter_Estimation
# Appendix B: Parameter Estimation Appendix B Parameter Estimation More Resources: ALTA Examples Collection This appendix presents two methods for estimating the parameters of accelerated life test data analysis models (ALTA models). The graphical method, which is based on probability plotting or least squares (Rank Regression on X or Rank Regression on Y), has some limitations. Therefore, the Maximum Likelihood Estimation (MLE) method is used for all parameter estimation in ALTA. # Graphical Method The graphical method for estimating the parameters of accelerated life data involves generating two types of plots. First, the life data at each individual stress level are plotted on a probability paper appropriate to the assumed life distribution (i.e., Weibull, exponential, or lognormal). This can be done using either Probability Plotting or Least Squares (Rank Regression). The parameters of the distribution at each stress level are then estimated from the plot. Once these parameters have been estimated at each stress level, the second plot is created on a paper that linearizes the assumed life-stress relationship (e.g., Arrhenius, inverse power law, etc.). To do this, a life characteristic must be chosen to be plotted. The life characteristic can be any percentile, such as BX% life, the scale parameter, mean life, etc. The plotting paper used is a special type of paper that linearizes the life-stress relationship. For example, a log-log paper linearizes the inverse power law relationship, and a log-reciprocal paper linearizes the Arrhenius relationship. The parameters of the model are then estimated by solving for the slope and the intercept of the line. ## Example of Graphical Method for Accelerated Life Data Consider the following times-to-failure data at three different stress levels. Stress 393 psi 408 psi 423 psi Time Failed (hrs) 3450 3300 2645 4340 3720 3100 4760 4180 3400 5320 4560 3800 5740 4920 4100 6160 5280 4400 6580 5640 4700 7140 6233 5100 8101 6840 5700 8960 7380 6400 Estimate the parameters for a Weibull assumed life distribution and for the inverse power law life-stress relationship. Solution First the parameters of the Weibull distribution need to be determined. The data are individually analyzed (for each stress level) using the probability plotting method, or software such as ReliaSoft's Weibull++, with the following results: \begin{align} & [{{\widehat{\beta }}_{1}}= & 3.8,\text{ }{{\widehat{\eta }}_{1}}=6692] \\ & \text{ }\!\![\!\!\text{ }{{\widehat{\beta }}_{2}}= & 4.2,\text{ }{{\widehat{\eta }}_{2}}=5716] \\ & [{{\widehat{\beta }}_{3}}= & 4.0,\text{ }{{\widehat{\eta }}_{3}}=4774] \end{align}\,\! where: • ${{\widehat{\beta }}_{1}},\,\!$ ${{\widehat{\eta }}_{1}}\,\!$ are the parameters of the 393 psi data. • ${{\widehat{\beta }}_{2}},\,\!$ ${{\widehat{\eta }}_{2}}\,\!$ are the parameters of the 408 psi data. • ${{\widehat{\beta }}_{3}},\,\!$ ${{\widehat{\eta }}_{3}}\,\!$ are the parameters of the 423 psi data. Since the shape parameter, $\beta ,\,\!$ is not common for the three stress levels, the average value is estimated. ${{\widehat{\beta }}_{common}}=4\,\!$ Averaging the betas is one of many simple approaches available. One can also use a weighted average, since the uncertainty on beta is greater for smaller sample sizes. In most practical applications the value of $\widehat{\beta }\,\!$ will vary (even though it is assumed constant) due to sampling error, etc. The variability in the value of $\widehat{\beta }\,\!$ is a source of error when performing analysis by averaging the betas. MLE analysis, which uses a common $\widehat{\beta }\,\!$, is not susceptible to this error. MLE analysis is the method of parameter estimation used in ALTA and it is explained in the next section. Redraw each line with a $\widehat{\beta }=4\,\!$, and estimate the new etas, as follows: \begin{align} & {{\widehat{\eta }}_{1}}= & 6650 \\ & {{\widehat{\eta }}_{2}}= & 5745 \\ & {{\widehat{\eta }}_{3}}= & 4774 \end{align}\,\! The IPL relationship is given by: $L(V)=\frac{1}{K{{V}^{n}}}\,\!$ where $L\,\!$ represents a quantifiable life measure (eta in the Weibull case), $V\,\!$ represents the stress level, $K\,\!$ is one of the parameters, and $n\,\!$ is another model parameter. The relationship is linearized by taking the logarithm of both sides which yields: \begin{align} \ln (L)=-\ln K-n\ln V \end{align}\,\! where $L=\eta \,\!$, ($-\ln K)\,\!$ is the intercept, and ($-n)\,\!$ is the slope of the line. The values of eta obtained previously are now plotted on a log-linear scale yielding the following plot: The slope of the line is the $n\,\!$ parameter, which is obtained from the plot: \begin{align} & Slope=\ \frac{\ln ({{T}_{2}})-\ln ({{T}_{1}})}{\ln ({{V}_{2}})-\ln ({{V}_{1}})} =\ \frac{\ln (10,000)-\ln (6,000)}{\ln (360)-\ln (403)} =\ -4.5272 \end{align}\,\! Thus: $\widehat{n}=4.5272\,\!$ Solving the inverse power law equation with respect to $K\,\!$ yields: $\widehat{K}=\frac{1}{L{{V}^{n}}}\,\!$ Substituting V=403, the corresponding L (from the plot), L=6,000 and the previously estimated $n\ \ :\,\!$ \begin{align} & \widehat{K}=\ \frac{1}{6000\cdot{{403}^{4.5272}}} =\ 2.67\cdot {{10}^{-16}} \end{align}\,\! ## Comments on the Graphical Method Although the graphical method is simple, it is quite laborious. Furthermore, many issues surrounding its use require careful consideration. Some of these issues are presented next: • What happens when no failures are observed at one or more stress level? In this case, plotting methods cannot be employed. Discarding the data would be a mistake since every piece of life data information is important. • In the step at which the life-stress relationship is linearized and plotted to obtain its parameters, you must be able to linearize the function, which is not always possible. • In real accelerated tests the data sets are small. Separating them and individually plotting them, and then subsequently replotting the results, increases the underlying error. • During initial parameter estimation, the parameter that is assumed constant will more than likely vary. What value do you use? • Confidence intervals on all of the results cannot be ascertained using graphical methods. The maximum likelihood estimation parameter estimation method described next overcomes these shortfalls, and is the method utilized in ALTA. # Maximum Likelihood Estimation (MLE) Method The idea behind maximum likelihood parameter estimation is to determine the parameters that maximize the probability (likelihood) of the sample data. From a statistical point of view, the method of maximum likelihood is considered to be more robust (with some exceptions) and yields estimators with good statistical properties. In other words, MLE methods are versatile and apply to most models and to different types of data. In addition, they provide efficient methods for quantifying uncertainty through confidence bounds. For a detailed discussion of this analysis method for a single life distribution, see Maximum Likelihood Estimation. The maximum likelihood solution for accelerated life test data is formulated in the same way as described in Maximum Likelihood Estimation for a single life distribution. However, in this case, the stress level of each individual observation is included in the likelihood function. Consider a continuous random variable $x(v),\,\!$ where $v\,\!$ is the stress. The pdf of the random variable now becomes a function of both $x\,\!$ and $v\,\!$ : \begin{align} f(x,v;{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}}) \end{align}\,\! where ${{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}}\,\!$ are $k\,\!$ unknown constant parameters which need to be estimated. Conduct an experiment and obtain $N\,\!$ independent observations, ${{x}_{1}},{{x}_{2}},...,{{x}_{N}}\,\!$ each at a corresponding stress, ${{v}_{1}},\,\!$ ${{v}_{2}},...,\,\!$ ${{v}_{N}}\,\!$. Then the likelihood function for complete data is given by: $L(({{x}_{1}},\text{ }{{v}_{1}}),({{x}_{2}},\text{ }{{v}_{2}}),...,({{x}_{N}},\text{ }{{v}_{N}})|{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}})=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}},{{v}_{i}};{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}})\,\!$ \begin{align} i=1,2,...,N \end{align}\,\! The logarithmic likelihood function is given by: $\Lambda =\ln L=\underset{i=1}{\overset{N}{\mathop \sum }}\,\ln f({{x}_{i}},{{v}_{i}};{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}})\,\!$ The maximum likelihood estimators (MLE) of ${{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}},\,\!$ are obtained by maximizing $L\,\!$ or $\Lambda .\,\!$ In this case, ${{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}}\,\!$ are the parameters of the combined model which includes the parameters of the life distribution and the parameters of the life-stress relationship. Note that in the above equations, $N\,\!$ is the total number of observations. This means that the sample size is no longer broken into the number of observations at each stress level. In the graphical method example, the sample size at the stress level of 20V was 4, and 15 at 36V. Using the above equations, however, the test's sample size is 19. Once the parameters are estimated, they can be substituted back into the life distribution and the life-stress relationship. ## Example of MLE for Accelerated Life Data The following example illustrates the use of the MLE method on accelerated life test data. Consider the inverse power law relationship, given by: $L(V)=\frac{1}{K{{V}^{n}}}\,\!$ where $L\,\!$ represents a quantifiable life measure, $V\,\!$ represents the stress level, $K\,\!$ is one of the parameters, and $n\,\!$ is another model parameter. Assume that the life at each stress follows a Weibull distribution, with a pdf given by: $f(t)=\frac{\beta }{\eta }{{\left( \frac{T}{\eta } \right)}^{\beta -1}}{{e}^{-{{\left( \tfrac{T}{\eta } \right)}^{\beta }}}}\,\!$ where the time-to-failure, $t\,\!$, is a function of stress, $V\,\!$. A common life measure needs to determined so that it can be easily included in the Weibull pdf. In this case, setting $\eta =L(V)\,\!$ (which is the life at 63.2%) and substituting in the Weibull pdf, yields the following IPL-Weibull pdf : $f(t,V)=\beta K{{V}^{n}}{{\left( K{{V}^{n}}T \right)}^{\beta -1}}{{e}^{-{{\left( K{{V}^{n}}T \right)}^{\beta }}}}\,\!$ The log-likelihood function for the complete data is given by: $\Lambda =\ln L=\sum\limits_{i=1}^{N}{\ln \left( \beta K{{V}^{n}}{{\left( K{{V}^{n}}{{T}_{i}} \right)}^{\beta -1}}{{e}^{-{{\left( K{{V}^{n}}{{T}_{i}} \right)}^{\beta }}}} \right)}\,\!$ Note that $\beta \,\!$ is now the common shape parameter to solve for, along with $K\,\!$ and $n.\,\!$ # Conclusions In this appendix, two methods for estimating the parameters of accelerated life testing models were presented. First, the graphical method was illustrated using a probability plotting method for obtaining the parameters of the life distribution. The parameters of the life-stress relationship were then estimated graphically by linearizing the model. However, not all life-stress relationships can be linearized. In addition, estimating the parameters of each individual distribution leads to an accumulation of uncertainties, depending on the number of failures and suspensions observed at each stress level. Furthermore, the slopes (shape parameters) of each individual distribution are rarely equal (common). Using the graphical method, one must estimate a common shape parameter (usually the average) and repeat the analysis. By doing so, further uncertainties are introduced on the estimates, and these are uncertainties that cannot be quantified. The second method, the Maximum Likelihood Estimation, treated both the life distribution and the life-stress relationship as one model, the parameters of that model can be estimated using the complete likelihood function. Doing so, a common shape parameter is estimated for the model, thus eliminating the uncertainties of averaging the individual shape parameters. All uncertainties are accounted for in the form of confidence bounds (presented in detail in Appendix D), which are quantifiable because they are obtained based on the overall model.
2022-12-10 06:10:17
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# Question for null space of a matrix ### Question for null space of a matrix Let A be a 4×3 matrix and let $$c=2a_{1 }+a_{2}+a_{3}$$ (a) If N(A) = {0}, what can you conclude about the solutions to the linear system $$Ax = c$$? (b) If N(A) ≠ {0}, how many solutions will the system $$Ax = c$$ have? Explain. shiecldk Posts: 1 Joined: Mon Oct 24, 2016 12:45 pm Reputation: 0
2020-02-18 23:05:32
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2019-09-22 00:33:26
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https://geomodeling.njnu.edu.cn/modelItem/8483a637-e918-4ccf-a3a9-aeeec24f682d
## CICE5 (Community Ice CodE version 5) CICE5.1.2 is the 5th version of the Los Alamos Sea Ice Model, sometimes referred to as the Community Ice CodE. It is used in CESM2. Sea-IceCESMLos Alamos Sea Ice Model 11 #### Alias Los Alamos Sea Ice ModelLos Alamos National Laboratory Sea Ice Model #### Authorship Affiliation: Los Alamos National Laboratory Homepage: View Is authorship not correct? Feed back #### Classification(s) Application-focused categoriesNatural-perspectiveFrozen regions #### Model Description English {{currentDetailLanguage}} English CICE5.1.2 is the latest version of the Los Alamos Sea Ice Model, sometimes referred to as the Community Ice CodE [6]. It is the result of a community effort to develop a portable, efficient sea ice model that can be run coupled in a global climate model or uncoupled as a standalone ice model. CICE5 has been released as the sea ice component of the Community Earth System Model (CESM), a fully-coupled global climate model that provides simulations of Earth’s past, present, and future climate states. CICE5 in the CESM is supported on high- and low-resolution Greenland Pole and tripole grids, which are identical to those used by the Parallel Ocean Program (POP) ocean model. The high resolution version is best suited for simulating present-day and future climate scenarios while the low resolution option is used for paleoclimate simulations and debugging. An uncoupled version of CICE5.1.2 is available separately: https://github.com/CICE-Consortium/CICE-svn-trunk This standalone CICE configuration provides a means of running the sea ice model independent of the other CESM components. It can read in atmospheric and ocean forcing, which eliminates the need for the flux coupler, and the atmosphere, land and ocean data models. It can be run on a reduced number of processors, or without MPI (Message Passing Interface) for researchers without access to these computer resources. CICE is a dynamic-thermodynamic model that includes a subgrid-scale ice thickness distribution [6]. It uses the energy conserving thermodynamics of [10] or [2], has multiple layers in each thickness category, and accounts for the influences of brine pockets within the ice cover. The ice dynamics utilizes the elastic-viscous-plastic (EVP) rheology of [4]. Sea ice ridging has the options of [8] and [9] or the newer ridging scheme of [7]. A slab ocean mixed layer model is included. A Scientific Reference Guide [6] is available that contains more detailed information on the model physics. The physics available in the uncoupled ice model are identical to those in the ice model used in the fully coupled system. This document uses the following text conventions: Variable names used in the code are typewritten. Subroutine are given in italic. File and directory names are in boldface. What’s new in CICE5? CICE5 is very similar in code structure to the previous version CICE4 and was released in March of 2015. CICE4 was an upgraded version of the Community Sea Ice Model, CSIM5, which was based on CICE3. The major changes are: • The new mushy-layer thermodynamics (ktherm = 2) is the default [10]. • The new level melt pond scheme (tr_pond_lvl = .true.) is the default [5]. • The default number of ice layers is now 8 (previously 4). • The default number of snow layers is now 3 (previously 1). • The freezing point at the sea ice-ocean interface is now salinity dependent following [1]. The CICE source code used in the CESM is based on the Los Alamos Sea Ice Model CICE model version 5. The main source code is very similar in both versions, but the drivers are significantly different. If there are topics that are not covered in this CICE documentation, users are encouraged to look at the CICE documentation available at: https://github.com/CICE-Consortium/CICE-svn-trunk Model Domain {{domain}} Principles {{principle}} Incorporated Models {{incorporatedModel}} Model part of larger framework: {{metadata.design.framework}} Incorporated Models {{process}} Inputs {{input}} Outputs {{output}} #### How to Cite CICE team (2021). CICE5 (Community Ice CodE version 5), Model Item, OpenGMS, https://geomodeling.njnu.edu.cn/modelItem/8483a637-e918-4ccf-a3a9-aeeec24f682d #### Comment(s) {{comment.date}} {{comment.content}} {{subComm.date}} {{subComm.content}} #### Authorship Affiliation: Los Alamos National Laboratory Homepage: View Is authorship not correct? Feedback #### QR Code • {{curRelation.name}} {{curRelation.name}} {{curRelation.overview}} {{curRelation.author.join('; ')}} {{curRelation.journal}} You can link related {{typeName}} from repository to this model item, or you can create a new {{typeName.toLowerCase()}}. Related Items {{ props.row.description }} {{ props.row.description }} Related Items {{props.row.name}} You can link resource from repository to this model item, or you can create a new {{typeName.toLowerCase()}}. {{ props.row.description }} {{ props.row.description }} Drop the file here, orclick to upload. File size should not exceed 10m. Select From My Space These authorship information will be submitted to the contributor to review. Cancel Submit Model Classifications Cancel Submit {{ item.label }} {{ item.value }} {{props.row.localName}} Model Name : Cancel Submit Name: Version: Model Type: Model Domain: Scale: Purpose: Principles: Incorporated models: Model part of larger framework Process: Information: Initialization: Hardware Requirements: Software Requirements: Inputs: Outputs: Cancel Submit Title Author Date Journal Volume(Issue) Pages Links Doi Operation Cancel Submit Yes, this is it Cancel OK Cancel Confirm Model Classifications 1 Model Classifications 2 Title Author Date Journal Volume(Issue) Pages Links Doi Operation #### NEW Name: Affiliation: Email: Homepage: Yes, this is it Cancel Confirm path : /{{path.label}} search results of '{{searchContentShown}}' #### No content to show {{item.label}} . {{item.suffix}} .{{item.suffix}} {{item.fileName}}.{{item.suffix}} Copy Delete Rename /{{path.label}} Change /{{path.label}} Select File Cancel Confirm path : /{{path.label}} /.. 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2021-06-16 17:26:39
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https://www.physicsforums.com/threads/calculating-apparent-separation.733500/
# Calculating apparent separation 1. Jan 18, 2014 ### bobo1455 I am trying to calculate the apparent separation between the 2 stars Mizar and Alcor. In the question I am given the total distance from me to the 2 stars which is 83 light years and the angle of separation which is 12 arcminutes. The question is multiple choice: (A) 0.1 ly, (B) 0.3 ly, (C) 1 ly, (D) 3 ly, (e) 10 ly What I have tried is using the formula from Wikipedia on Angular Distance to get a value that matches one from the choices A to E \theta ≈ \frac {a}{D} The problem is, I am having a hard time understanding what to exactly do with 83 light years and 12 arcminutes. I have read online that the 2 stars Mizar and Alcor have distances of 81 light years and 78 light years, so the distance between them is 3 light years, so I'm assuming the answer is (D), but I don't know at all how to get there. I am also given these formulas: distance = actual size / angular size angular size is proportional to actual size / distance Any help is appreciated. 2. Jan 18, 2014 ### BvU The angle of separation is the angle between the lines from you to the stars. Theta in the formula. Don't forget to convert to radians. 3. Jan 18, 2014 ### bobo1455 Okay, I convert 12 arcminutes to degrees (0.2) to radians (0.0034906585) and plugged in for theta: then I have: 0.0034906585 = x / 83 ly (not sure if I am supposed to convert ly to AU or some other unit of measurement) then I get x = 0.2897246555, which is approx 0.3 ly, and (B) is 0.3 ly, but I am still not 100% sure it is correct or not. 4. Jan 18, 2014 ### HallsofIvy Staff Emeritus The original distances were given in light years so the answer should be given in light years. Yes, 12 minutes is 12/60= 1/5 degrees and so (1/5)(3.14159/180)= 0.0034906585 so the distance between them, along a circular arc, is (0.0034906585)(83)= 0.2897 light years. I don't think I would put it to as many significant figures as you do. Since "12 arcminutes" and "83 light years" both have two significant figures, the most reasonable answer is 0.29 light years. You could also do this as a "trig" problem. Dropping a vertical from the vertex (the earth) to the line between the two stars, we have two congruent right triangles with angle 6 arcminutes and hypotenuse 83 ly. The opposite side of that right triangle (half the distance between the two stars), x, satisfies $$sin(6')= x/83$$ so that x= 83sin(6')=0.1449 so that the entire distance, 2x, is 0.2898 light years, still 0.29 ly as before.
2017-12-13 02:37:26
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http://www.ams.org/mathscinet-getitem?mr=1124147
MathSciNet bibliographic data MR1124147 (92h:54049) 54F50 (54E35 54F65) Mayer, J. C.; Mohler, L. K.; Oversteegen, L. G.; Tymchatyn, E. D. Characterization of separable metric ${\bf R}$${\bf R}$-trees. Proc. Amer. Math. Soc. 115 (1992), no. 1, 257–264. Article For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
2014-03-17 02:31:53
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https://www.investopedia.com/terms/m/multipliereffect.asp
## What Is the Multiplier Effect? The multiplier effect is an economic term, referring to the proportional amount of increase, or decrease, in final income that results from an injection, or withdrawal, of capital. in effect, It measures the impact that a change in economic activity—like investment or spending—will have on the total economic output of something. ### Key Takeaways • The multiplier effect refers to the proportional amount of increase, or decrease, in final income that results from an injection, or withdrawal, of spending. • The most basic multiplier used in gauging the multiplier effect is calculated as change in income / change in spending and is used by companies to asses investment efficiency. • Money supply multiplier, or just the money multiplier, looks at a multiplier effect from the perspective of banking and money supply. 1:47 ## Understanding the Multiplier Effect Generally, economists are usually the most interested in how infusions of capital positively affect income. Most economists believe that capital investments of any kind—whether it be at the governmental or corporate level—will have a broad snowball effect on various aspects of economic activity. As its name suggests, the multiplier effect provides a numerical value or estimate of a magnified expected increase in income per dollar of investment. In general, the multiplier used in gauging the multiplier effect is calculated as follows: \begin{aligned}\text{Multiplier}=\frac{\text{Change in Income}}{\text{Change in Spending}}\end{aligned} The multiplier effect can be seen in several different types of scenarios and used by a variety of different analysts when analyzing and estimating expectations for new capital investments. ### Example of the Multiplier Effect For example, assume a company makes a $100,000 investment of capital to expand its manufacturing facilities in order to produce more and sell more. After a year of production with the new facilities operating at maximum capacity, the company’s income increases by$200,000. This means that the multiplier effect was 2 ($200,000/$100,000). Simply put, every $1 of investment produced an extra$2 of income. ## The Keynesian Multiplier Many economists believe that new investments can go far beyond just the effects of a single company’s income. Thus, depending on the type of investment, it may have widespread effects on the economy at large. A key tenet of Keynesian economic theory is that of the multiplier, the notion that economic activity can be easily influenced by investments, causing more income for companies, more income for workers, more supply, and ultimately greater aggregate demand. Essentially, the Keynesian multiplier is a theory that states the economy will flourish the more the government spends, and the net effect is greater than the exact dollar amount spent. Different types of economic multipliers can be used to help measure the exact impact that changes in investment have on the economy. For example, when looking at a national economy overall, the multiplier would be the change in real GDP divided by the change in investments, government spending, changes in income brought about by changes in disposable income through tax policy, or changes in investment spending resulting from monetary policy via changes in interest rates. Some economists also like to factor in estimates for savings and consumption. This involves a slightly different type of multiplier. When looking at savings and consumption, economists might measure how much of the added income consumers are saving versus spending. If consumers save 20% of new income and spend 80% of new income, then their marginal propensity to consume (MPC) is 0.8. Using an MPC multiplier, the equation would be: \begin{aligned}&\text{MPC Multiplier}=\frac{1}{1-\text{MPC}}=\frac{1}{1-0.8}=5\\&\textbf{where:}\\&\text{MPC}=\text{Marginal propensity to consume}\end{aligned} Therefore, in this example, every new production dollar creates extra spending of $5. ## Money Supply Multiplier Effect Economists and bankers often look at a multiplier effect from the perspective of banking and a nation's money supply. This multiplier is called the money supply multiplier or just the money multiplier. The money multiplier involves the reserve requirement set by the Board of Governors of the Federal Reserve System and it varies based on the total amount of liabilities held by a particular depository institution. In general, there are multiple levels of money supply across the entire U.S. economy. The most familiar ones are: • The first level, dubbed M1, refers to all of the physical currency in circulation within an economy. • The next level, called M2, adds the balances of short-term deposit accounts for a summation. When a customer makes a deposit into a short-term deposit account, the banking institution can lend one minus the reserve requirement to someone else. While the original depositor maintains ownership of their initial deposit, the funds created through lending are generated based on those funds. If a second borrower subsequently deposits funds received from the lending institution, this raises the value of the money supply even though no additional physical currency actually exists to support the new amount. The money supply multiplier effect can be seen in a country's banking system. An increase in bank lending should translate to an expansion of a country's money supply. The size of the multiplier depends on the percentage of deposits that banks are required to hold as reserves. When the reserve requirement decreases the money supply reserve multiplier increases and vice versa. Back in 2020, prior to the Covid-19 pandemic, the Fed mandated that institutions with more than$127.5 million have reserves of 10% of their total deposits. However, as the pandemic sparked an economic crisis, the Fed took a dramatic step: On Mar. 26, 2020, it reduced the reserve ratio to 0%—essentially, eliminating these requirements entirely to free up liquidity. ### Money Supply Reserve Multiplier Most economists view the money multiplier in terms of reserve dollars and that is what the money multiplier formula is based on. Theoretically, this leads to a money (supply) reserve multiplier formula of: \begin{aligned}&\text{MSRM}=\frac{1}{\text{RRR}}\\&\textbf{where:}\\&\text{MSRM}=\text{Money supply reserve multiplier}\\&\text{RRR}=\text{Reserve requirement ratio}\end{aligned}
2021-10-18 13:56:36
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https://learn.careers360.com/ncert/question-the-points-on-the-curve-9y-raised-to-2-is-equal-to-x-raised-to-3-where-the-normal-to-the-curve-makes-equal-intercepts-with-the-axes-are/
Q # The points on the curve 9y ^ 2 = x ^ 3, where the normal to the curve makes equal intercepts with the axes are 24) The points on the curve  $9 y^2 = x ^3$, where the normal to the curve makes equal intercepts with the axes are $A ) \left ( 4 , \pm \frac{8}{3} \right )\\\\ .\: \: \: \: \: B ) \left ( 4 , \frac{-8}{3} \right ) \\\\ . \: \: \: \: \: C) \left ( 4 , \pm \frac{3}{8} \right ) \\\\ . \: \: \: \: D ) \left ( \pm 4 , \frac{3}{8} \right )$ Views Given the equation of the curve $9 y^2 = x ^3$ We know that the slope of the tangent at a point on a given curve is given by  $\frac{dy}{dx}$ $18y\frac{dy}{dx} = 3x^2\\ \frac{dy}{dx} = \frac{x^2}{6y}$ We know that $Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent } = \frac{-1}{\frac{x^2}{6y}} = \frac{-6y}{x^2}$ At point (a,b) $Slope = \frac{-6b}{a^2}$ Now, the equation of normal with point (a,b) and $Slope = \frac{-6b}{a^2}$ $y-y_1=m(x-x_1)\\ y-b=\frac{-6b}{a^2}(x-a)\\ ya^2 - ba^2 = -6bx +6ab\\ ya^2+6bx=6ab+a^2b\\ \frac{y}{\frac{6b+ab}{a}}+\frac{x}{\frac{6a+a^2}{6}} = 1$ It is given that  normal to the curve makes equal intercepts with the axes Therefore, $\frac{6b+ab}{a}=\frac{6a+a^2}{6} \\ 6b(6 + a) =a^2( 6+a)\\ a^2 = 6b$ point(a,b) also satisfy the given equation of the curve $9 b^2 = a ^3\\ 9(\frac{a^2}{6})^2 = a^3\\ 9.\frac{a^4}{36} = a^3\\ a = 4$ $9b^2 = 4^3\\ 9b^2 =64\\ b = \pm\frac{8}{3}$ Hence, The points on the curve  $9 y^2 = x ^3$, where the normal to the curve makes equal intercepts with the axes are $\left ( 4,\pm\frac{8}{3} \right )$ Hence, the correct answer is (A) Exams Articles Questions
2020-02-23 11:37:11
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=35764
## 2009 final 4E $E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$ 904565142 Posts: 10 Joined: Wed Nov 18, 2015 3:00 am ### 2009 final 4E Can someone please explain how they jumped from Q to log[H+]? Attachments Jacqueline Lee Posts: 21 Joined: Fri Sep 25, 2015 3:00 am ### Re: 2009 final 4E I think it's because we're looking for pH and the concentrations of Ce3+, Ce4+,MnO4-, and Mn2+ would not contribute to pH because pH is the measure in concentration of H+ so that is what we could focus on instead. So by taking the E (which would be .157) you can divide that by (.05916/5) and then divide by 8(which came from the exponent of [H] via log rules) and that's how the course reader got the 1.67=log[H+] Hope that helped Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)” ### Who is online Users browsing this forum: No registered users and 0 guests
2020-09-30 00:01:41
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http://machinomics.blogspot.com/2012/04/shape-recognition-and-shape-spaces-in.html
## Thursday, April 19, 2012 ### Shape recognition and shape spaces in computer vision One of the problems in computer vision and pattern recognition is that of shape recognition (classification, clustering and retrieval). It is important in statistical analysis of medical images (shape outliers might indicate disease)  and machine vision (digital recording and analysis based on planar views of 3D objects). Classical statistical shape analysis is due to Kendall, Bookstein and Mardia. This classical shape analysis treats shapes as shape spaces, whose geometries are those of differentiable manifolds often with appropriate Riemannian structures. Imagine we are given a set of variables in which each variable is a position of a prearranged landmark (interesting point with a load of information) and our setting is 2D. Then we can model the points as complex numbers $p \in \mathbb{C}$. Therefore each point can be written as $p_i=x_i + iy_i$ where $i=\sqrt{-1}$. A k-ad $p={p_i}_k$ is a set of $k$ shape landmarks. We now center these points so as to make them translation invariant. $$w_T=p-\bar{p}$$ where $\bar{p} \in \mathbb{C}^k$ is a vector with all elements set to the k-ad mean, this is $p{_Tj}=\frac{1}{k}\sum_k{p_i}$. Now we make them scale invariant by dividing each element in $w_T \in \mathbb{C}^k$ by the absolute value of $w_T$, this is $w_S = \frac{w_T}{|w_T|}$. Finally given the rotation matrix with angle $\theta$ $$\left[\begin{array}{cc}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{array}\right]$$ one can rotate the centered shape with $Rw_S$ (if we had modeled it in $\mathbb{R}^{2\times k}$, but we are in the complex domain and it is done by multiplying each element by $e^{i\theta}$. Now, this is when it gets interesting and useful (and where I see the true magic). To make our representation rotation invariant, we can use the Veronese-Whitney embedding, so that the rotation invariant representation is the rank one matrix $w_R=w_S w_S^*$, where $\cdot^*$ is the complex conjugate. This means that, by what we have seen before, any rotation of $w_S$, $e^{i\theta} w_S$ generates $$e^{i\theta}w_S e^{-i\theta}w_S^* = w_S w_S^*$$ since $e^{-i\theta}$ is the complex conjugate of the rotation (meaning that we undo the rotation by rotating $-\theta$ degrees. This is readily seen by the (complex) exponential arithmetic $e^{i\theta}e^{-i\theta} = = e^{i(\theta - \theta)} = 1$. This space is that of the complex hermitian matrices and we can define the norm of a shape and the distance between two shapes by $$\|A\| = trace(A) \\ \rho(A,B) = \|A-B\|$$ Now we can compute the mean shape of populations and do some inference. For example, populations of known healthy livers against livers with disease, and get confidence intervals so as to accept or reject null hypothesis given a new liver shape. We see that modeling in $\mathbb{R}^{2\times k}$ and using the matrix $R$, we achieve the same result: $$wR(wR)^T = wRR^T w^T = ww^T$$ Check out: http://stat.duke.edu/~ab216/duke-talk.pdf
2018-07-19 14:56:12
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https://content.iospress.com/articles/technology-and-health-care/thc213216
You are viewing a javascript disabled version of the site. Please enable Javascript for this site to function properly. # Role of patient-specific blood properties in computational fluid dynamics simulation of flow diverter deployed cerebral aneurysms ### BACKGROUND: Hemodynamics and their clinical outcome of cerebral aneurysms treated with flow diverter (FD) stents have thus far been investigated using computational fluid dynamics (CFD) simulations. Although human blood is characterized as a non-Newtonian patientspecific fluid, non-patient-specific blood properties (PSBP) were applied in most extant studies. ### OBJECTIVE: To investigate the hemodynamic effects caused by PSBPs in aneurysms treated with FD stents. ### METHODS: We measured blood properties (density and viscosity) for 12 patients who underwent FD stent deployment. We conducted CFD simulations with the measured PSBPs and non-PSBPs quoted from previous studies. The average blood flow velocity and wall shear stress within the aneurysms were calculated and two simulation patterns were compared. ### RESULTS: The velocity and wall shear stress changed by 2.93% and 3.16% on average, respectively, without an FD stent deployed. Conversely, with the FD stents deployed, the change rates increased to 11.1% and 9.06% on average, respectively. ### CONCLUSIONS: The change in hemodynamic parameters if PSBPs are considered, may not be negligible when conducting CFD simulations of FD stent deployed aneurysms To obtain an adequate hemodynamic environment for cerebral aneurysms with FD stents deployed, it is recommended to use PSBPs for CFD simulations. ## 1.Introduction Flow diverter (FD) stent treatment is based on a concept that induces thrombosis by altering the hemodynamics of a cerebral aneurysm, and various investigations using computational fluid dynamics (CFD) simulations have been carried out on this topic. In previous studies, hemodynamic parameters such as wall shear stress (WSS) and blood flow velocity in the aneurysm have been investigated using CFD simulations, to explore factors that contribute to aneurysm occlusion with FD stents [1, 2, 3, 4]. In these studies, blood was assumed to be a Newtonian fluid without patient-specific constant density and viscosity. In reality, the blood density naturally varies between patients; human blood is a non-Newtonian fluid, and its viscosity depends on shear rate. When the shear rate increases, the blood viscosity approaches a constant value. In contrast, when the shear rate decreases, blood exhibits a non-Newtonian behavior, indicating that the viscosity increases because of bio-related mechanisms such as red blood cell aggregation [5, 6]. However, according to a meta-analysis conducted by Saqr et al., almost 90% of studies on aneurysm hemodynamics employed the Newtonian viscosity assumption [7]. The hemodynamic effects caused by the non-Newtonian viscosity assumption in the CFD simulation have been also investigated. Gambaruto et al. reported that the WSS value of a non-Newtonian viscosity model differed by approximately 5% from that of the Newtonian viscosity model [8]. Xiang et al. suggested that the WSS value is overestimated if a Newtonian viscosity model is applied [9]. Schirmer et al. conducted CFD simulation for coil embolized basic aneurysm models with non-Newtonian viscosity and showed the viscosity increased in the distal region of the aneurysm [10]. Otani et al. also showed that the Newtonian viscosity model underestimated the degree of shear rate reduction in the low shear rate region, which is a known trigger of thrombosis formation for coil embolized aneurysms [11]. Morales et al. introduced a non-Newtonian viscosity model to simulate the blood flow with coil embolized aneurysms and demonstrated that the viscosity values increased in certain regions. They concluded that the Newtonian model can be employed for CFD simulations because the high viscosity region slightly influenced the flow structure in the aneurysm [12]. Consequently, it is still unclear whether the Newtonian or non-Newtonian model should be applied to CFD simulations of cerebral aneurysms. Additionally, the non-Newtonian effect has not been fully investigated in aneurysms with FD stents deployed Although it was pointed out in the literature that FD stent implantation induces a lower shear rate within the aneurysm and raises the viscosity value, previous studies applied mostly the non-patient-specific Newtonian viscosity to perform blood flow simulation with FD stent deployment [1, 2, 3, 4, 13]. In this study, we obtained whole blood samples from patients who underwent FD stent deployment and measured their blood properties (density and viscosity), conducting CFD simulations with and without the patient-specific blood properties (PSBPs). We investigated differences in hemodynamic parameters caused by introducing PSBPs into CFD simulations and compared the differences between the CFD simulations with and without FD stents deployed, to clarify hemodynamic effects related to the introduction of PSBPs and the deployment of FD stents, respectively. ## 2.Materials and methods ### 2.1Patient-selection and measurement of blood properties This study was approved by the Ethics Committee of the Jikei University School of Medicine in Japan. We obtained written consent from each patient participating in this study. From 39 cases treated with FD stents from October 2016 till January 2020 at the Jikei University School of Medicine, we selected the cases for this study based on the following criteria: (1) blood sample was collected during the treatment (20/39 cases), (2) aneurysm was < 30 mm with a simple geometry to perform the CFD simulation (17/20 cases) (3) spatial resolution of medical images was sufficient to perform the CFD simulation (12/20 cases). Twelve patients were included in the final analysis (mean age during the treatment was 69.1 ± 12.6 years, all females). All cases were treated with the Pipeline Embolization Device (Covidien/Medtronic, Irvine, CA, USA) and 6 out of the 12 chosen cases included coiling. Blood density and viscosity were used as PSBPs. Whole blood samples were collected from each patient right before the surgery, using an EDTK-2K coated evacuated tube (Japan Becton Dickinson CO., Ltd., Tokyo, Japan) to prevent blood coagulation during the measurement. Since blood properties are believed to be altered by its components and the environment such as its temperature [14, 15], the temperature of the blood samples was kept at 37C, as is the normal human body temperature, using a thermostat. We used a portable densitometer DMA35 (Anton Paar, Graz, Austria) to measure the density. We also used a falling needle rheometer to measure viscosity with a relatively small amount of blood [16, 17]. The expanded uncertainty of the rheometer in the viscosity measurement of unknown samples is less than 0.90% (coverage factor = 2) at 610 kg/m3, as reported by Matsunaga et al. [18]. The rheometer can plot various viscosity values that depend on shear rate, thereby creating a continuous data curve, in conjunction with a non-Newtonian viscosity model. Blood viscosity μ can be described by the Casson model as follows: ##### (1) μ=(τ0γ+μ0)2 where τ0 is the yield stress, γ is the shear rate and μ0 is the Newtonian viscosity. However, the viscosity diverges when the shear rate approaches zero, thus the equation can be modified as follows: ##### (2) μ=[τ0(1-e-mγγ)+μ0]2 In this equation, the constant m controls the maximum viscosity value when the shear rate γ approaches zero. In this study, the value of the constant m was 100 as was reported for normal human blood [19]. To fit the viscosity curve with the modified Casson model, the yield stress τ0 and the Newtonian viscosity μ0 were calculated by using the measured values at the lowest and second lowest shear rates [20]. ### 2.2CFD simulation procedure All patients underwent three-dimensional digital subtraction angiography (3D-DSA; Artis Q biplane, Siemens Healthcare GmbH, Forchheim, Germany) before the FD stent deployment. The images were reconstructed to stereolithography (STL) format using Amira 5.6 (Thermo Fisher Scientific, Waltham, MA, USA). Deployed FD stents were reproduced with our in-house code in the Tokyo University of Science based on a methodology described by Bouillot et al. as follows [21]: (1) extraction of the parent artery by separating the aneurysmal sac from the original image voxel data and conversion of its surface into STL format; (2) calculation of a centerline of the parent artery and a minimum inscribed sphere radius corresponding to the centerline by using VMTK (Vascular Modeling Toolkit, www.vmtk.org); (3) reproduction of the FD stent wires based on the information computed above. We reproduced the geometry of the Pipeline with the design parameters described in a previous study [22], and used the value of 0.03 mm also described in another previous study as the diameter of each wire constituting the Pipeline to reproduce the FD stent geometry [23]. Coil deployment was not considered and only the stent geometry was used to perform the blood flow simulations. Computational meshes were generated using ANSYS ICEM CFD 18.1 (ANSYS Inc., Canonsburg, PA, USA). In this study, we created a cross-sectional plane at 1 mm proximally from the aneurysm neck and defined an aneurysmal inlet. Similarly, an aneurysmal outlet was set at 1 mm distally from the neck. A computational domain was defined as a region enclosed by the arterial wall and both the aneurysmal inlet and outlet. We defined the aneurysm region as the area of the aneurysmal sac, not including FD stent structures. The mesh type used was set as unstructured, and the meshes consisted of both tetrahedral and prism meshes. In the vicinity of the vessel wall, seven prism mesh layers were aligned at a height of 0.3 mm, and the size of the elements varied depending on the diameter of the vessel where each element was located. The tetrahedral mesh size in the vicinity of the FD stent wires was set at 0.008 mm dividing the wire diameter by an approximate factor of 12, based on a previous study [24]. With those computational mesh settings, the mesh convergence was accomplished. The total number of generated elements was 10.3 million on average, for cases without FD stent, and 197 million for cases with FD stent. Finally, extension tubes with a length of 75 mm were connected to all inlets and outlets to reduce the effect caused by boundary conditions and develop the flow. We conducted the blood flow simulations assuming incompressible laminar flow. Both the vessel wall and the FD stent wire were assumed to be rigid non-slip walls. Although the vessel wall is originally viscoelastic [25], the characters and mechanical properties of the vessel wall and the FD stent were not considered in this study. As non-PSBPs related, density and Newtonian viscosity were assumed as 1,050 kg/m3 and 36 mPas, respectively [26]. We set the steady mass flow rate at 0.003465 kg/s (corresponding to the value at diastole end reported by Ford et al.), as an inflow condition for both blood flow simulations considering non-PSBPs or PSBPs [27]. This is because flow pattern depends on aneurysm geometry, especially in large aneurysm that are eligible for FD stent deployment, contrary to inflow [28]. We first performed blood flow simulations without FD stents deployed, considering non-PSBPs, then we performed the simulation considering PSBPs with the same geometry and inflow conditions. We also performed the blood flow simulations with FD stents deployed, considering non-PSBPs and PSBPs. In the final analysis, we performed 4 blood flow simulations for each patient, resulting in a total of 48 blood flow simulations for 12 patients ### 2.3Quantitative evaluation We calculated the absolute change rate of the average blood flow velocity and the WSS within the aneurysms, because these parameters are subject to change. The absolute change rate is described as follows: ##### (3) absolute change rate=|X𝑝𝑎𝑡𝑖𝑒𝑛𝑡-Xnon-patient|Xnon-patient where X𝑝𝑎𝑡𝑖𝑒𝑛𝑡 means a parameter calculated while including PSBPs and Xnon-patient means the parameter calculated excluding PSBPs. ##### Table 1 Specific viscosity values for each shear rate Shear rate γ [s-1] Newtonian Case A B Apparent viscosity [mPa⋅s] 10 50 100 200 300 400 500 600 3.60 3.60 3.60 3.60 3.60 3.60 3.60 3.60 6.23 4.44 4.06 3.80 3.69 3.62 3.58 3.55 6.51 4.26 3.80 3.49 3.36 3.28 3.23 3.19 7.87 4.99 4.40 4.01 3.84 3.74 3.68 3.63 7.94 4.85 4.23 3.82 3.64 3.54 3.47 3.42 4.66 3.70 3.49 3.35 3.28 3.25 3.22 3.20 5.04 3.61 3.31 3.10 3.01 2.96 2.92 2.90 8.56 5.21 4.54 4.09 3.90 3.79 3.71 3.66 7.48 5.06 4.56 4.22 4.07 3.98 3.92 3.88 5.24 4.06 3.81 3.63 3.55 3.51 3.48 3.45 4.16 3.62 3.50 3.42 3.38 3.36 3.35 3.33 8.25 5.39 4.80 4.40 4.23 4.13 4.07 4.02 7.48 5.06 4.56 4.22 4.07 3.98 3.92 3.88 8.56 5.39 4.80 4.40 4.23 4.13 4.07 4.02 4.16 3.61 3.31 3.10 3.01 2.96 2.92 2.90 ##### Figure 1. Relationship between shear rate and apparent viscosity. Each patient-specific viscosity decreases when the shear rate increases and asymptotes to a constant value, while the Newtonian viscosity has a constant value regardless of the shear rate. ##### Figure 2. Bar graphs showing absolute change rate of velocity and WSS caused by introducing PSBPs. The rate with an FD stent implanted exceeds the rate without an FD stent implanted, in almost all cases. ## 3.Results ### 3.1Measured values for patient-specific blood properties The average measured blood density was 1043.1 ± 4.0 kg/m3. Figure 1 shows viscosity curves for each case fitted to the measured values using the falling needle rheometer. Additionally, Table 1 describes the specific viscosity values and their fitted viscosity curve ranges for different shear rates. Each patient-specific viscosity curve differed from the other, depending on the patient. We calculated the apparent Newtonian viscosity value to average the value in a high shear rate (γ> 160 s-1) according to Yamamoto et al. [16]. The highest apparent Newtonian viscosity value was 4.18 mPas for Case K, and the lowest was 2.98 mPas for Case F. The non-patient-specific Newtonian viscosity value was 3.6 mPas, leading to a 16.1% difference with the highest patient-specific apparent Newtonian viscosity and a 17.2% difference with the lowest one. Additionally, as in Table 1, the ranges of the patient-specific viscosity calculated by subtracting the minimum viscosity value from the maximum value for each shear rate tends to increase for lower shear rates, for example the maximum and minimum values for the shear rate as 10 s-1 were 8.56 mPas and 4.16 mPas, respectively. Therefore, the range was 4.40 mPas. In contrast, the values for the shear rate as 600 s-1 were 4.02 mPas and 2.90 mPas, respectively. Thus the range was 1.12 mPas. Moreover, the patient-specific viscosity curves had different trends depending on the shear rate. For example, the curve for Case F showed a lower viscosity value than that for Case J, for shear rate values above 50 s-1; however, an inverted trend was observed at shear rate values below 50 s-1. Additionally, cases that had lower viscosity in high shear rate regions (such as Cases E, F and J) tended to display a steep viscosity slope in the low shear rate region. On the contrary, cases that had relatively high viscosity values in high shear rate regions (such as Cases G, K and L) tended to display more gradual variations with the decreasing shear rate, in comparison with previous cases. In all cases considered, the patient-specific viscosity value exceeded the non-patient-specific viscosity value at shear rate values below 50 s-1. ### 3.2Qualitative and quantitative difference in CFD simulations The absolute change rates of velocity and WSS calculated in our CFD simulations, with and without PSBPs, are illustrated in Fig. 2. The average absolute change rate of aneurysmal velocity for all cases was 3.06% without FD stent deployment, and 11.6% with FD stent deployment. Additionally, the average absolute change rate of WSS at aneurysmal dome was 3.31% in the cases without FD stent deployment, and 9.63% for cases with FD stent deployment. The calculated absolute change rate of each parameter tended to be greater in the cases with FD stent deployment. Figure 3 depicts a streamline and color contour map of WSS for the illustrative Case G which had the largest absolute change rate in velocity and the second largest absolute change rate in WSS, among aneurysms with FD stent deployed (see Fig. 2). A different flow structure circled by the dashed line was observed in Fig. 3, and flow appeared to be stagnant in the CFD simulations while considering PSBPs. The average flow velocity in the aneurysm was 0.0134 m/s without considering PSBPs, and 0.00949 m/s while considering PSBPs. Therefore, a 28.8% difference was observed between the two simulation results. Also, a locally high WSS region appeared in the simulation without the PSBPs pattern (circled by a continuous line in Fig. 3). The region appeared to be smaller in size in the simulation with PSBPs pattern, however, the WSS value increased in the entire aneurysmal wall reflecting the increase in the apparent viscosity. The average WSS value was 0.0674 Pa without considering PSBPs and 0.0804 Pa while considering PSBPs, leading to a 19.4% difference between the two simulation results. Because the patient-specific blood viscosity increased when the shear rate decreased, as shown in Fig. 1, we calculated the spatial average shear rate and viscosity values in the aneurysm. The computed shear rate and viscosity values are shown in Table 2. As shown in Table 2, the mean shear rate for all cases was 128 s-1 without and 65.5 s-1 with the FD stent deployed pattern. Additionally, the mean viscosity value averaged in the aneurysm was 4.80 mPas without and 11.7 mPas with FD the stent deployed pattern. ##### Table 2 Aneurysmal average shear rate and viscosity values Shear rate [s-1]Viscosity [mPas] w/o FDw/FDw/o FDw/FD CaseA11258.94.294.61 B53.839.54.966.18 C83.434.74.9911.7 D56.322.95.537.17 E15372.23.533.77 F72.024.63.754.78 G16.64.489.0268.2 H2381343.403.67 I11242.73.934.75 J2011223.533.54 K49.220.46.5617.5 L3942104.124.38 Mean12865.54.8011.7 Standard deviation10157.81.5617.5 ##### Figure 3. Streamlines and WSS color contour map with PSBPs and without PSBPs. ##### Figure 4. A color contour map of shear rate for an illustrative case (Case A). The shear rate value is considerably higher in the vicinity of the FD stent mesh (marked by a square). As a further investigation of the shear rate, we captured a contour map of the shear rate in Fig. 4 taken from the illustrative Case A, which depicts its state in the aneurysmal dome and the vicinity of the FD stent. This illustrative figure indicates that the shear rate tends to increase in the vicinity of the FD stent, as indicated by a square shown in this figure, and because of the dense mesh of the FD stent, the main flow of the parent artery had high velocity. ## 4.Discussion ### 4.1Differences between patient-specific and non-patient-specific blood properties We measured patient-specific whole blood properties including density and viscosity and found that patient-specific viscosity had a specific curve for each patient (see Fig. 1). The average measured density was 1043.1 kg/m3, very close to the value of 1050.0 kg/m3 obtained in CFD simulations without considering PSBPs. In previous studies, a non-patient-specific, non-Newtonian viscosity model was applied to investigate the hemodynamic impact [8, 9, 10, 11, 12], which did not necessitate the measurement of blood properties. However, patient-specific blood viscosity is closely related to several blood components such as plasma, red blood cell, fibrinogen [6]. Thus, the relative distribution of these components may explain the difference in patient-specific viscosity curves, however the individual effects of these components were not considered in this study. As shown in Table 1, patient-specific viscosity values in a high shear rate region (γ> 200 s-1) ranged from approximately 3 mPas to 4 mPas. The non-patient-specific, Newtonian blood viscosity used in previous studies was in the same range [1, 2, 3, 4, 12, 19]. ### 4.2Relationship between hemodynamic influence of blood properties and presence of FD stents Hemodynamic effects caused by the introduction of PSBPs in the simulation were greater in the presence of FD stents. An average WSS difference of 3.06% was observed in comparison with the simulation without FD stents, which is close to the value reported by Gambaruto et al. [8]. However, this difference became 9.63% in the simulation with FD stents deployed. The effect of the PSBPs on the aneurysm hemodynamics became larger for the FD stent deployed pattern in relation to the shear rate value in the aneurysm. In Table 2, the shear rate values in the aneurysm were generally lower than 200 s-1, except for the Case L. Additionally, the shear rate values decreased for the FD stent deployed pattern. Corresponding to the decrease in the shear rate, the viscosity in the aneurysm increased for the FD stent deployed pattern. Particularly, the viscosity in Case G was 68.2 mPas due to the low shear rate (4.48 s-1). This implies that the hemodynamic parameters had larger variations in the FD stent deployed cases. Moreover, these variations can be associated with the low shear rate in the flow stagnant region, as seen in Fig. 3. Flow velocity in the depth of the aneurysmal sac tended to be lower which caused a lower shear rate value in this region. The decreased shear rate led to a higher viscosity value and induced flow stagnation. This phenomenon is comparable to the non-Newtonian viscosity increasing in value in complex-shaped aneurysms and blebs, as reported by Hippelheuser et al. [29]. As can be seen in Fig. 4, the shear rate increases in vicinity of the FD stent wires. Blood viscosity had a Newtonian behavior in this region because its values did not vary in the high shear rate range (γ> 500 s-1). However, once the flow went through the stent mesh and entered the aneurysmal sac, the shear rate value became considerably low. Our results imply that the applicability of Newtonian viscosity model may be limited in hemodynamic studies with FD stents deployed. For example, measuring flow resistance of the FD stent mesh is possible using the Newtonian viscosity model, as a high shear rate exists around the mesh. In contrast, we observed a specific shear rate decrease in the aneurysmal sac which implies that PSBPs considering a non-Newtonian viscosity ought to be used in the blood flow simulation of an aneurysm with a deployed FD stent. Although we did not follow the clinical outcomes of the cases considered in this study the investigated hemodynamic parameters (velocity and WSS) were also used in previous studies as factors facilitating aneurysm thrombosis by FD stent deployment 1̧,2,3,4. Our results imply that these hemodynamic parameters might not reflect the actual state without introducing PSBPs. Clauser et al. conducted in Vitro particle image velocimetry (PIV) measurements using several kinds of fluids having different properties [30]. They reported that the flow diversion effect caused by the FD stent deployment is influenced by the fluid properties and that the velocity reduction caused by FD stent deployment was larger in the case of a fluid that had a non-Newtonian viscosity. Additionally, Gijsen et al. experimentally investigated the effect of the non-Newtonian fluid in the carotid bifurcation models with laser Doppler anemometry and reported that there were striking differences in the flow field between experiments with the Newtonian fluid and non-Newtonian fluid [31]. These experimental results show us a necessity of the non-Newtonian viscosity to investigate the hemodynamics in arteries including a cerebral aneurysm. Particularly the in Vitro results from Clauser et al. concur with our findings that the effects of considering PSBPs in CFD simulations are stronger when FD stents are deployed. Our study might be a first one to investigate the hemodynamic effects caused by introducing PSBPs in CFD simulations, evaluating the effect of an FD stent placement in Silico, and leading to similar results than the in Vitro measurements ### 4.3Limitations and further investigations This study presents some limitations. We considered 12 cases adapted to our case selection criteria. Morales et al. investigated the hemodynamic influences in 3 coil embolized aneurysms, using the non-Newtonian viscosity model [12]. Thus we also believe that conducting a statistical analysis was not necessary, as the number of cases considered was sufficient to examine the hemodynamic effects of the PSBPs on blood flow simulations of FD deployed aneurysms. However, we would need a larger sample size and a statistical evaluation, for more detailed investigations. Additionally, as mentioned in Section 2, inflow conditions do not affect the flow pattern in the aneurysm; however, hemodynamic parameters can be quantitatively affected by these inflow conditions. Evju et al. investigated how the hemodynamic parameter WSS was altered by introducing 4 viscosity models and 3 boundary conditions [32]. They concluded that the WSS are strongly correlated to the boundary conditions, rather than the viscosity models. Khan et al. also investigated the impact of rheology models against the impact of a CFD solution strategy for predictions of aneurysm flow instabilities [33]. They indicated that solution strategy and other likely “first-order” effects such as vessel lumen segmentation and choice of flow rates appear to be more important than the effect of the rheology. Particularly, patient-specific inflow boundary conditions are desirable for blood flow simulations and methodology for acquisition the conditions have been developing. 2D magnetic resonance imaging (MRI) and 4D phase-contrast MRI are being used to measure patient-specific inflow conditions, and these techniques should be adopted in future investigations to perform CFD simulations reflecting more accurately the actual patient’s state [34, 35]. ## 5.Conclusions We obtained whole blood samples from 12 patients who underwent FD stent deployment, and measured patient-specific blood density and viscosity. We introduced these blood properties into CFD simulations. The average differences caused by introducing the blood properties were 11.6% and 9.63% for aneurysmal velocity and WSS at aneurysmal dome, respectively, with an FD stent already placed. These differences were larger than the values without any FD stent deployed. Our results imply that consideration of PSBPs is recommended when conducting CFD simulations with an FD stent deployed, to obtain an accurate hemodynamics assessment and a proper prognosis of the aneurysm occlusion. ## Acknowledgments This work was supported by the JSPS KAKENHI under Grant JP20J30001 and approved by the Ethics Committee of the Jikei University School of Medicine. None to report.
2022-08-09 17:18:44
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Coronavirus On the Beach Danger Looms for Cape Town. Unless you understand mathematics, this is going to sound like Neville Shute's - On the Beach. First the good news. As I pointed out in my last bulletin, the SA authorites have reset the origin to 27 march. Notice we no longer talk about that "miracle hook" in fact it is now obvious  to even the most positive Capetonians that, in the outbreak  areas,  the lock down is not working. The good news is that some Provinces have worked and the regulations are now  tightening up provincial borders , too late, but at least we have begun. The executive ARE getting the idea of compartments. We have a  model for the release from the lock down, too complicated, but it is a start. , Now the bad news. The SIR style epidemiologists have still not understood the veracity of the infection vector. If you have taken the time to study the Dengue model, you will see that the infection vector (misquitos in Dengue)  is  driven by the "essential workers" in COVID. Rather than fearing for their lives or at least appreciating that they are risking the lives of others, essential workers enjoy the status of being essential and the income priveleges that come with that status. While their wages depend on not understanding this logic, they will never get what I am talking about. By the time that our leaders (they are essential themselves) figure out that they are themselves,  the infection vector that causes the spread, it will be too late. Bulletin 30 April 2020 - bulletin30apr.pdf Doctors vs Actuaries Economics is not called the dark science for nothing. Smith and Marshal dealt with utility 200 years ago - "Interpersonal utility comparisons are not valid". The doctors 'wife gets her second Rolls Royce before the orphan gets pudding not because of justice, but only because her husband puts up the money. The scarce ventilator in Madrid gets moved to the younger female patient simply because she has more chance of survival, not because she enjoys life more than the old guy. You and I work no differently. I will explain why Adam Smith's moving hand crosses my palm with silver and puts me on the side of the doctors. Right now its investment bankers vs doctors. Actuaries and mathematical economists just follow the money - like me. The winner will be the one that puts in the most resources. Investment bankers of all people should understand this. For the first time in a while, the investment bankers are losing. It is not Coronavirus, but their own myopia that is doing it. Right now everyone is spooked and the politicians and doctors have the microphones. While they are making money and getting touted as "frontline heros", they are going nowhere. For the actuaries and economists to push them off the stage, they will have to present the case and run the gambit of "in-humanism". Interpersonal utility comparisons is exactly what we have to do. Life expectancy is a core concept in actuarial Science. There is even a mathematical term for it - dx . Insurance companies deal with these concepts every day. Right now the kill rate of coronavirus - 10% in 70s plus and 20% in 80s plus is about the same as old white people sitting in the cancer-causing UV sunlight. Aged people will not sit in the sunlight getting sunburn, t4hy stay inside at noon. They should also not go out and sit next to potential virus carriers. I don't know the micro biology, but coronaviruses are not rare - they are all over the place, they could have been hitting us like this for the last Million years. Its only the 7 Bn urban density that makes the modern transmission so dangerous. https://www.niaid.nih.gov/diseases-conditions/coronaviruses Acknowledging the danger of introducing the strawman argument, an analogy would be to be surprised that we may get hit by a massive meteorite. You have only to look up at the moon to see the pockmarks. The moon is smaller than earth and still pulls in so many meteorites, only our atmosphere cleans up out scars. We should be expecting a big one any day now. Short run certainties are death and taxes. Long run certainties are viruses and meteorites. However, it takes a little bit more work than @Glasgow Tap and @Tim - who are just plain wrong on the FT comments. Death rates go up 4 times when Coronavirus breaks out!COVID-19 really kills people. Every 80-year-old man has every right to piss in his nappies when someone nearby coughs. Even Donald Trump, should start checking that his will is in order. The lock-down would be a very good time for him to do his tax returns. The key to moving forward is not to try to diminish the health guys' argument, but proving that investors save more lives than doctors. Proving this will mean taking the life tables and mapping it to the SIR and SAGE models. So far I have not met anyone else doing this. I do it to build up the content on the Copeman Academy. Right at the beginning of this outbreak, I approached every major insurance company in South Africa to build and motivate a coherent macroeconomic lobby. No-one understood what I was talking about. The cost to them has been Billions. Their heads are so busy, trying to protect their short term premium income, that they do not have the time to stop and think about the macroeconomics of the problem. That is why they have lost the initiative to the doctors. Notice you never see an actuary at any of the press briefings. After a while, I felt I was bothering them, so I gave up. I did try to warn them. If I got $10 from every company that has$ 100M invested in recovery from the Depression, I would code and write the full paper that is required to motivate this. Otherwise what I have to do is follow the market. Some of us have actually started to benefit from the current world state of affairs! While the employees of the big accounting companies are at home playing with their kids and filling out UIF forms, working in isolated rooms all over the world, we have been rolling out a remote version of TurboCASH with online banking and email. TurboCASH 5.2 is like an anti-cloud measure - make every isolated accountant a server, take him off central dependence. The "crazy side benefit" that sparks the revolutionary in me., is that this will allow users to leave the financial grid and use blockchain and non-fiat scalars to manage a distributed informal trade economy. 98% of the TurboCASH project already runs like this. We manage $4 Bn in turnover with little exchange of cash. Old people still have assets and leverage to work for them. In the last month, 70-year-old accountants in aged communities all over the world find people asking them to manage logistics. At first, everyone was spooked, but now TurboCASH has a little boom going selling accounting packages to isolated accountants including septuagenarian accountants in aged care homes. Investment bankers and other soon to be unemployed readers of the FT, should stay at home or make themselves useful delivering food parcels. Right now I am on the side of the 80 years olds - if only because they are the ones paying me the$10. Okuns GDP Gap The economy is never cool after a stock market crash. When is it going to recover? I have been looking at the question since 1982 when it first perplexed me. Financial people will never talk about it because for most of the cycle they are too busy studying quarterly returns (and 52% of the time they are winning). Only analysis that moves beyond quarters can deal with Okuns GDP Gap. Only a massive drop in the Stock market gets this problem attention. This stock market drop was waiting for us even before the virus. You can think of the Virus as a War, and while we focus on it, we bury ourselves further and further into Okun's GDP Gap. This debate squats like a toad over economic history and development economics. Why do we struggle to maintain potential GDP? After looking at it for 40 years I have to conclude that  it is our cost of having the capitalist system on our side. This will be heresy to financial analysts because they cannot move out of the framework that the market is the ultimate allocator of capital - it is not. The market has no vision  -  it always thinks too short-run. Payments on the Porshe dominate the market's thinking. There is a nasty book called The Deluge, by Adam Tooze  ( I read it at Pinnacle Point in Dec) that covers the 1914- 1929 period. Churchill got the Nobel prize for literature for his study on this period. Both Churchill and Tooze bind the cycle not from 1929 to 1945, but all the way back to Widrow Wilson in 1914. The period  1914 to 1945 is known in economic political history as the Second 30 Year War. In the same way, as the US moved in the post-1929 period to become GDP dominant, expect China to do the same - the US is likely to be the big loser here, starting with this $2.2 Trn package spilling out into nowhere. It has nowhere to go except to possibly inflate financial assets and food prices, but it can't fix GDP. You can't eat sháres and in the modern digital day you can't even wípe your ass with them, you need to binge-buy toilet paper to do that. It is a 200-year-old problem, as old as capitalism itself. Without War, capitalism is not able to bootstrap itself out of deflation. Read more… Comments: 1 Coronavirus March Bulletin 15April 2020 : coronascript15April.pdf Bulletin 10April 2020 : coronascript10april.pdf Bulletin 31 March 2020 : bulletin31mar.pdf Bulletin 28 March 2020 :bulletin28mar.pdf Bulletin 27 March 2020 : bulletin27mar.pdf Bulletin 26 March 2020 : bulletin26mar.pdf The things matter over the next few weeks: define the compartments and appoint leadership enforcing the isolation food to the poor to boost them security early treatment Read more… Comments: 2 Stepping stone jobs Subject: Economics Case Study: South Africa Inc (unemployment, inequality, growth model) Put yourselves in the shoes of someone advising the President. The problem we are dealing with is the 29% unemployment of 10 Million South Africans, the rate climbing to 60% for black youth. Treasury has suggested a plan to create 1 Million jobs in 3 years. This proposal is made in the full knowledge that 500,000 new job seekers enter the South African Economy every year. Treasury's plan is thus a plan to INCREASE unemployment by a minimum of 500,000 jobs. The Copeman Academy challenges its users to come up with a comprehensive solution to create 10 Million jobs in 3 years and thus reduce unemployment to near zero. Stepping Stone Jobs Proposal to create 1 Million jobs. Stepping stone jobs are jobs like Car Guards, Waiters, Cleaners, Caddies, Personal assistants, seasonal agricultural labor, security guards. Talented young people start these jobs very early age and at very bad pay, do them for a few months and they move onto other jobs .(I started doing these jobs when I was 13). These jobs are not so much about earning revenue as about learning the way the employment process works. Be polite to a customers even if they are wrong, always reporting on the outcome of tasks, complete all tasks before stopping the day, never steal the companies assets, always look to make profit for the company. Typically in South Africa, stepping stone jobs get taken up by African immigrants and middle aged South Africans. This has a devastating effect on the economy. The first is that middle aged people try to "make a career" out of stepping stone jobs. They make a career out of a job that should never have been. No amount of unionization or protest can solve the fact that the job simply does not produce enough to deliver a decent wage. Great unhappiness ensues. Worse is that fact that in a year, that low paying stepping stone job would have employed 4 youths temporarily, who would then leave and move on to take up better employment. If that Stepping stone job is held for 5 years buy a middle age person, 20 youths miss the opportunity to enter into the market. Great unhappiness ensues. BEE and the employment act start out with the good intention of helping youth employment, but have the exact opposite effect. Instead of offering a future to twenty youths who could be hired ground through the rough an tumble of low paid employment, we swop this for one unhappy person stuck forever in a poverty trap. When you start rambling about minimum wages, employment brokers, casualisation employment equity and employment controls, you destroy stepping stone jobs and you destroy the future. 250,000 stepping stone jobs shared among 1,25 Million people, creates 1 MIllion jobs. #coepmanacademy #southafricainc Read more… Comments: 0 South African 2019 Q1 GDP Growth Tanks South Africa posted a 2019 Q1 GDP growth decline of minus 3,2%. Economic apologists tout this as the worst performance in 10 years. Worse! Back up 33 years to find self-inflicted stupidity of this magnitude. In the Court of Economic History, President Rhamaphosa and Finance Minister Mboweni stand in the dock alongside PW Botha and Barend Du Plessis as the worst economic executives to have held office. Go back 190 years to Lord Charles Somerset and Sir Harry Smith to find comparable incompetence. Breathtaking - minus 3,2%. A small number, yet considered, the enormity of the R300 Bn GDP gap, comes sharply into focus. Next - 300,000 pink slips, as actual jobs or bright young #feesmustfall graduates, that don't move out of University digs into a company apartment, but stay at home, sharing a room with baby Katlego and the twins. Revised career prospects are now poker games, soccer pools, traditional weddings. The 2008 implosion of world markets and the 1992 dying Apartheid regime cranking interest rates to 24% are acts of desperation. Go back to The Rubicon Speech of 1986 or the 1976 Soweto Riots - 33 years to match 2019 Q1 made spectacular that it happened while the rest of the world marched onto maximum growth, stock market highs and full employment. How? EWC - Expropriation Without Compensation In the boozy gloom of 2018, Rhamaphosa meets his election advisory team. The inspiration flash is putting EWC at the forefront of the election campaign, the ANC heads off the EFF Land challenge. After the election, back to business, forget the promises, not a peep about EWC. EWC worked! Who can forget the old woman in Erkhuleni visited by a campaigning President bursting into tears of joy as he explained on National television how her ANC vote would get her a plot of Land, a bond to build a house and security for an R150,000 overdraft to start the business she has always been denied. A big jump from a lifetime of poverty. In Africa - that promise is a vote in the bag. There is however a cost side. Decrying EWC is deemed, counter-revolutionary. The pungent argument that EWC is merely a ploy with no real Land Reform, is dismissed. The joy gushes from the left as they watch spooked Landowners, fret about losing their career savings. A political triumph!In Africa, the rich not only dominate the economy. They ARE the economy. When they tank, we all tank.Investors disappear like virgins at a matric dance. Property buyers scatter like rats up a drainpipe. Bank managers cut bonuses, cancel vacations, call in loans. Last year's pestering call center agents offering overdrafts? Retrenched or moved to collections. Standard Bank closes 100 Branches.The Q1 GDP figure is the first bill, court summons and blacklisting judgment still coming - GDP implosion with multipliers.Land reform, more urgent than ever, is now at what cost? The ANC election campaign cost R 300 Bn! R30,000 a vote exceeds even the per guest budget of the Gupta wedding. R300 Bn is more than double (nay triple) the total amount spent on Land restitution since 1994! Not even a single farm took yet. Land Reform - where South Africa loses capital - offers nothing. Say “Expropriation Without Compensation” to a Landowner. It matters not what follows, she is already checking European Bond rates online, not shopping, opening a new factory or employing more staff. Plain to see – GDP growth minus 3.8%. South Africa, it is time for a serious rethink on where EWC is going. Read more… Comments: 0 AP Maths English Speaking countries do not do well in Pisa rankings. The Copeman Academy strives to give the English speaking student a boost in Maths education. We do this by supplementing analytical training with Deep Learning based Pattern Recognition. We cover South Africa, UK, USA, Canada, Australia. South Africa is ranked 138 out of 140 countries at Maths education. What is less known is that South Africa is one of the most unequal Maths societies in the world. Our top 1 % of students rank as good as the Asians (not rigourously proven, people get a bit prickly about admitting the advantages of privileged students). One of the reasons for this success is that thanks to the IEB, we have one of the best developed Advanced Maths Programs. A student following Core Maths and APMaths in South Africa comes out ahead of GED and AS Level students in the US and UK respectively. AP Maths is an innovative exam offered by the IEB. Every serious Grade 12 student should take AP Maths. This is because it is an opening link to A Level Maths and University Maths 1 and 2. Here is a list of topics covered. Here is an example of an exam (Paper 1). Work your way through these questions with Video answers to establish how good you are. If you are heading for University then APMATHS will never be a waste of your time and will improve your core maths. Great option comming - Stats, Financial Models, Graph theory and Matrices. South Africa If you are in grade 11 or grade 12 find out from your school what are your options to doing AP Maths. Either way doing APMaths now will save you doing these topics later. The assessment has been benchmarked by UK Naric (National Academic Recognition Information Centre), the UK equivalent of the SA Qualifications Authority, and is considered equivalent to the UK A-levels. UK/Cambridge International If you are a UK A-Levels student then APMaths coincides 90% with your A-Level Core and Further Maths Syllabi. These are all core concepts to A-Levels Maths and A-Levels Further Maths. Cambridge A-Level Maths : https://www.cambridgeinternational.org/Images/329554-2019-syllabus.pdf Cambridge A-Level Further Maths : https://www.cambridgeinternational.org/Images/414957-2020-2022-syllabus.pdf US GED GDE students Have 4 modules Algebra 1 and 2, Geometry and Precalculus. AP Maths and Greade 12 Core Maths cover around 90% of your topics and introduce about 10% of topics not required (Noteably financial mathematics) https://www.futureschool.com/united-states-curriculum/texas/ will shortly recocile these differences. I would recommend AP Maths for GDE students, because it will prepare you for the inevitable confrintation of College Maths. Adults There is a basic cannon of maths, Geometery, Trig, Algebra, Calculus and everyone needs an introduction to Mechanics, Stats and Applied Maths. Adults have the luxury of not having to follow a rigourous syllabus. AP Maths is a gentle introduction to teriary Maths if you have High school maths waxed. -- Those who light their candle with mine, do not diminish my flame. Thomas Jefferesen Read more… Comments: 0 Multi Tasking Sucks Those of you that have read my education blogs will have picked up my disdain for the classroom system. My claim is that online working can "increase your education productivity by 5 times". Failure at Mathematics is arely the ability of the student but usually the inability to handle the classroom system. It may seem a crazy claim, that can't easily be measured, but let me prove it by illustration. Don't just take my word for it. A typical student attending a typical classroom lecture at a typical brick and mortar institution is thrown each morning into a multitasking nightme. Find your way to class - argue about you seat - deal with the ADD kid in the seat next to you - watch the lecturer make announcements - write down an equation - cockblock the dreamy kid looking at you two rows down - listen to an inane question from some dumbass - write down another equation - listen to an unintelligible esoteric question from some boy genious - more from the ADD Kid- another equation - focus on holding in a pee - more home work announcements - check your cell phone messages for announcements on the next science lecture - another equation - rush to the toilet. The hyper busy environment of the classroom regime, occurs because it it a sausage factory trying to balance the needs of the teachers with the needs of the students. It is a mess. OK so what is so bad about Multitasking? You have all heard mum brag about her multi tasking skills. Some of the most busy people at work seem to be those that are best at multitasking? MULTI TASKING SUCKS. You are much more efficient if you shut everything else out focus on the issue at hand. I am going to prove it to you: Consider the multi tasking probelm of write down three symbols and increment them like this: Arabic numeral Alphabet Roman numeral 1 A I then write the next row 2 B II and then 3 C III Keep going 4 D IV 5 E V 6 F VI . . . 10 J X Thats Multi Tasking. Now put away Multi tasking and focus on one column at a time 1 2 3 4 5 6 7 8 9 10 and then 1 A 2 B 3 C 4 . 5 . 6 . 7 8 9 10 Can you see (even before taking on Roman Numerals) that the Column approach is much easier? The one that uses focus? We should stop calling it multi-tasking, and call it “switch-tasking.” It is 5 time less efficient when it comes to measuring up classroom against managed online learning. That is why multi taskng is only an illusion of productivity and why the classroom system sucks. Read more… South Africa's Low Rank in Maths is misleading. We are often reminded that South Africns are on the bottom rungs of world maths education, by some estimates beating only Yemen. This does not tell the whole story. Rounding out the numbers, around 1.6 % of South African learners get a distinction for Grade 12 Maths (8 000 out of 500 000). This compared to UK GCSE where 3% of learners get a distinction. However our NSC Grade 12 Maths exam is a level more difficult than GCSE including areas of financial maths, statistics and calculus. What is interesting is that when you examine Private Schools in South Africa, the distinction rate jumps closer to 20% at top schools. Far from producing the worst results, South African private schools produce closer to the best maths results in the world. The result is that 15% of top University entrants come from IEB schools. Maths distinction graduates are nearly fully employed whereas the national average of young unemployed is over 50%. The limiting factor is the quality of teachers and the access to technology. Quality teachers are hard to come by and will always be a limiting factor. However an opportunity awaits us. Adopting technology should be a priority. Machines simply make better maths teachers. We have the opportunity that in more developed countries the maths teachers unions are more militant and resist replacing human teachers with machines. Ironically the second worst country in the world is best placed to do this. Small studies are already showing that access to broadband video and deep learning techniques can raise the distinction rate dramatically. Machines are tireless and almost unlimited in productivity. A determined roll out of maths teaching technology can raise our distinction rate ten fold. That is why at the Copeman Academy we put our emphasis on Creative Commons and AI Pattern recognition. So far so good. We have a 100% distinction success rate. www.copemanacademy.com Read more… Comments: 0 The Land Issue Consider that the Land Issue may never be solved. Land is simply a proxy for wealth and when we say people are dispossessed of Land, what we really mean is they are dispossessed of wealth. When we mean - give Land - what we really mean is - give wealth - to those that do not have enough. The problem with the concept is that in order to do that, you have to take it away from someone else first. Few people feel that they have too much wealth and would like it all taken away, to release the burden of owning wealth. Thorstein Veblen dealt extensively with this problem 100 years ago. While Marx predicted the demise of capitalism, Veblen predicted that the masses would emulate the ruling classes. Unlimited wants with limited resources is the fundamental problem of economics and no amount of indignity, self justification or unsupported aggression can solve it. https://en.wikipedia.org/wiki/Thorstein_Veblen Read more… Comments: 0 The Consumer Surplus The consumer surplus is one of the most beautiful things in market capitalism, something that is shared by all of us. Yes rich people get more of it than poor people, so it is shared unequally, but it is there for every one to take. If you understand that prices are determined by supply and demand in markets and the marginal principle determines the final price, the consumer surplus is the integral of the demand curve from zero to the amount supplied less the market price of that demand. (The blue area in the diagram) For example as a rich person I would be prepared to pay$10 for a price of bread if that was the price. The market price is actually $6. Then the consumer surplus for me is$4. I am getting a bargain on bread because I am paying $4 less than I would have paid – I would have been prepared to pay$10. Like I said one of the most beautiful things in market capitalism – we all benefit. A Gini for Grade 12 Maths South African ranks in the highest for economic inequality (Gini 0.67) and in the lowest for mathematics and science education (WEF 137). But there's the rub … South African society is ruthless. While our system consigns the masses to a future mired in misery and poverty, our 1% are destined for greatness. If you lie awake at night worrying about egalitarian equality for all African children, don't read any further ... While much economic analysis has been done on the poverty of Nations by looking at the 99% and the Analysis of the Gini coefficient, we should suppress the urge to fit a function to the Lorenz curve and then integrate that function. I do it anyway (simple non linear regression on a published dataset) - in South Africa I estimate the Lorentz curve for Grade 12 Mathematics outcomes to be y = 30 + exp(x ^ 2.7/59000) – see graph. At the Copeman Academy we focus on the 1% and there you get a completely different picture. When we pit our top 1% of students against the 1% of the leading Maths nations. (Singapore, Finland) You find that our 1% leads them not follows them! The bottom line is that students willing to leave the 20th century “teacher focused” methodology and move to a tech focused AI technology have the advantage, no matter what country they are in. The best maths teaching countries are still geared to produce mediocrity - in buckets! How does this happen? Top performing Maths countries with a higher level of Maths education and a lower Gini coefficient of inequality may produce an overall level of Maths and Science, but their use of “better teachers” actually works against them in the 1%. The top students get lumped in for longer with the 99%. In South Africa because our Government Education is so low, the 1% are forced to leave the grid early. They actually end up ahead of the students in the developed countries that stick to the grid. Essentially because those institutions in South Africa teaching Maths and Science to the 1% are independent of the failing Government system. Here at the Copeman Academy we take no Government funding and are not beholden to bind with the 99% attached to the grid in a lifetime of mathematical poverty. The unexpected outcome is that those students in developed countries construed to be more successful (better teachers) have their 1% dragged down to the level of their 99%. Whatever time you spend in the system for the 99% (whatever country) is essentially wasted. In South Africa we spent 92% of our education budget on teachers salaries and 8% on infrastructure and technology. At the Copeman Academy we spend 90% of our budget on technology and 10% on teachers salaries. Because technology is the most important component of Maths education. Those that break with the Grid and focus on technology based outcomes actually exceed the outcomes achieved by the 1% in the more egalitarian societies. This leads to a dark scenario for the 99% and a bright one for the 1% who make the jump. Like I said, we are a ruthless society. Open source and Economic Inequality World Maths Ranking Primer on the Lorentz Curve Play with the code Stagnant Jobs growth Read through the recent announcement of Pali Lahohla, Director General Statitistics about stagnant job growth in South Africa and you see a gloomy future. Blips and outliers aside, jobs are not growing. However each year around 500,000 new entrants seek employment. The greatest pressure is on youth and that means black youth. Jolly as he is, the DG falls into racist thinking that can damage us structurally. Believing that unemployment is caused by race, leads to tilting at windmills, employment equity quotas, BEE, NSFAS, RET and any number of initiatives that have yielded well – nothing. (the DGs results not mine) White youth are largely fully employed and the other groups have issues. Yes but that is not the cause of problem. You can't legislate against the 1% because their skills are simply too mobile. The reason that racists legislation fails to increase employment in the 99% is that race is not the cause. To understand why race is the result and not the cause, you have to understand Granger Causality. Yes there may be a correlation between race and unemployment, but the cause is lack of mathematics. If you leave race out of it, you will find that the 5000 or 1% of the 500,000 that achieved a distinction for Maths have a golden future. While others discuss hustling for a living, the Quants are looking at racking up jobs at R 50,000 a month. The demand supply is so skew that you can't even find the candidates to fill the outstanding vacancies . The hustlers will join the unemployment ques or at best be paid intern wages. Welcome to the rollerball economy, or knowledge economy to those that seek to avoid the ugly truth - The knowledge economy is upon us. Machines work better than people. The best for people is programming the machines. Did anyone notice that the value of Naspers now exceeds the entire capitalization of the South African Mining Industry? This, while it employs a fraction of the workers. Take the data and rearrange. Instead of using race as a criteria, use grade 12 Maths Distinction pass and I will wager a cyber-wallet of crypto that you will see a 90% correlation, and Granger Causation. This is why I have spent such a great deal of the last year developing a MOOC that starts with Grade 12 Maths. At a macro level, our languishing at the bottom of the WEF rankings on Maths education is why the rest of the world has recovered from the recession and we are plummeting. If you have a child or know a child in or approaching Grade 12. The best thing you can do for them is put them on the right path to getting a distinction for Mathematics. It may be the most important exam of their lives. That, more than any other factor than inheriting a family fortune, is their gateway to the 1%. Diamonds and Rhodes E101Economics : Land To illustrate where Land turns to Capital and Entrepreneurs we should to discuss the Diamond business. Diamonds  go back 2 Billion years. What gets squirted out in the Kimberly Pipe is a molten belch. Current Geo Economic theory says that there are probably diamonds the size of cars driving around in the mantle that could take a wrong turn and pop out at any moment. Not Nature, Australopithecus,  Homo erectus, The Khoisan's, The Dutch, The Griquas, The Basters, The Afrikaners or The German and British Settlers were able to commercialize this resource. For 2 Billion Years less 140,  no one turns the idea into money until a brilliant strategist, Cecil John Rhodes comes up with the idea that he is going to control the flow of Diamonds onto the market and take Millions of Dollars off the Jews and Homosexuals, years in advance of him actually taking the stones out of the ground! The money does not comes from Africa, Africans, African traders, African workers, Settler greed,  Rhodes or the Oppenheimers. The money comes from the Ghettos of Europe via the Merchant Bankers of the USA. Generations of Griquas had stepped over the shiny stones until, in 1866, Schalk van Niekerk purchased the Eureka Diamond from the Erasmus family. Opportunists piled into New Rush grabbing at the alluvial deposits, levelling Colesburg Kopie shovelling out the ever deeper sand.  By the time their overcrowded, waterlogged claims hit blue-stone, the diamond industry was in turmoil. In nineteenth century Europe, Jews are not welcome at the Club and homosexuality is a jail-able pffense. Rhodes has to sell the idea that diamonds have lasting value. He is faced with the immediate record that under the Afrikaners and Settlers (Kimberley's 1880 version of white monopoly capital) , the Diamond Business has the attention span of a two year old. Under their watch, the  DeBeers brothers sell DuToitspan for £8,000 and by 1875 most diamond traders are bankrupt.  The stressed out participants left are seeking to get out. Swimming against the tide of popular opinion, Rhodes is able to raise enough capital in Europe to buy out the bit players and make De Beers the first really important public Company in Africa. Using the charm and showmanship of a nineteenth century Larry Page and Elton John, Rhodes becomes the richest man in the world in a little over five years. The Jews and the Homosexuals believe him, The Jewish Homosexuals bankroll him and the Anti Semites and homophobics are gobsmacked. The Afrikaners become Cattle farmers, The Settlers moved off to Joburg in search of Gold and the Sothos show little interest in Diamond mining beyond getting together enough savings to buy a rifle and head for the rurals. Rhodes is the ultimate outsider. Reviled for his sexuality, he turns to other victims of predjudice and they build an Empire with little more than tenacity and perseverance.  How you get from Rhodes juggling sausages in 1875  to pulling down his statue in 2015, parallels  the modern political history of South Africa The Trouble with Modern Universities The Neoliberal working world that we have come to believe in, no longer exists. (See The Rollerball Economy) Universities are scattered over 5 levels  : denial, anger, bargaining, depression, acceptance. Modern brick and mortor Universities that dominate today's tertiary education tend to move slower than events around them. This is not to say that there are not great places offering great opportunities to great futures. There are. However, the very term "Ivy League" conjures up social images of balmy afternoons spent languishing on the lawns of campus in the company of students from a wide cross section of disciplines. The Business Student should be cautioned against the pitfalls that exist in such a world. Modern Universities with the campus community, social architecture and "all round education" offered, tend to have been built in the 20th century, or are run by academics trained in the 20th century. That means that when they were formed the Internet did not exist. The knowledge economy as we know it was theoretical and open source was not a mainstream culture. Where there is  no such thing as free lunch you find closed access journals publishing obscure psuedo-mathematical digests, lecturers earning royalties from copyright texts and a financial system of stipends, put in place to lock any prospective graduate into a prohibitive debt cycle. Students following this path are likely to spend the majority of their twenties mired in debt and their thirties placed in the grey world of the working 90%, with an outside chance of ascending in their forties  to the 10%. In this world the worst lack all conviction are the best are full of passionate intensity. We have not seen a world as unequal as ours since 1913. 100 years ago the outcome was disastrous. Some say the War is imminent. Some say the War has already begun. Its a new War fought between the 1%, where the 99% are irrelevant to the outcome and rather than marched  to the front - they face a far worse fate- they are ignored and sent to watch football.  The War will not wait for you. Dorothy, if you spend your twenties skating around in the twentieth century, do not be surprised what when you wake up you find that Kansas has gone bye bye and you are watching from the sidelines in the cheap seats. In a constant state of denial, most Universities are more interested in preserving the known traditions, preserving the employment status of their staff and providing safe pathways for the young minds entrusted to them. Business students are particularly susceptible to the dangers associated with this approach.  The Rollerball economy tells us that the safe paths lead to the football stadium and that the vast majority of people will be excluded from labor in the near future. Only a very few (Performers, Capital Managers and Capitalists) will be disruptors. The rest will live in shallows and misery. To their credit Universities have tried rapidly to adjust to the new world. Online facilities are available and use of open source is encouraged. Yet no matter how hard they try, Universities are run by bureaucrats. The regulations set up by the Education Departments that oversee tertiary education add an evaluative layer of red tape that ties down real innovation. In a world where courses take upwards of three years to approve from concept to delivery,  Universities, no matter how they try are by definition as much as 5 years out of date. 5 years ago, Uber did not exist. ISIS did not exists, Newspapers still had a future and Reality TV Stars did not win elections. To the Business student, the future is particularly important, even if that is a future that holds a gripping darkness for 99% of the people! Modern Universities are notoriously retrospective. The 1 to 60 Lecture in which a "Font of Wisdom" Professor repeats the same lecture year after year to a less and less relevant audience can no longer compete with The MOOC, which uses the Internet Delivery of the state of the art production of the topic under consideration. Combined with the input of peers from multiple countries and cultures, the MOOC (Mass Open Online Course) provides not only a competitive alternative,but a superior alternative. A first class graduate from a well designed MOOC will outperform all comers in the knowledge economy. This lean, mean apparatus does not suit the bloated administrations. If that is not enough here comes the grimy underbelly. In most multi discipline Universities and Colleges, the Business Students are usually the majority and they subsidize the more specialized courses! In a financial model that no longer works, Universities have begun to cannibalize themselves. Business careers, the only outcomes that can lead to a lucrative future, sacrifice quality to pay for their neighbors to pretend that they are studying for careers that will create return.  Now this may have worked for prospective Capitalists in the 20th century, where we subsidized the poor people who would spend the rest of their lives working for us.  In the Rollerball Economy Labor loses value for all but the few performers. The current system has already begun to breakdown and the anger is palpable. Riots plague the less privilege and the privileged blanket themselves in a fee structure that is designed more to induce the illusion of  financial advantage than to address the pressing problems of Post Capitalism To their credit, many Universities have begun bargaining with the Rollerball Economy. They are using "blended learning" as a supplement to their brick and mortor activities, but however well intention-ed, their online efforts are, they are  trapped in the old model. Carrying the old fashioned debt heavy package may work for the staff , but  consigns the graduates to the underclass.  The sclerotic Government supported, debt packaged offering is no longer financially justifiable and as this reality sets in depression is inevitable. The MOOC brings the opportunity of free education. It also brings new challenges. The MOOC replaces the Font-of-Wisdom with the worlds-best-presenter and replaces the clock work system of industrial organization design with an open online collaboration that never sleeps. The world of business and learning blend naturally and the new 1% begin to emerge.
2020-07-14 16:41:36
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https://www.gamedev.net/forums/topic/507627-common-bugs--in-physx-havok-and-bullet-solved---fixed-timestepsmaxsubstep/
# Common bug(s) in PhysX, Havok and Bullet SOLVED! - fixed timesteps/maxSubStep ## Recommended Posts ##### Share on other sites Definitely interesting post. What advantage does that have over the simplistic accumulator approach? real accum = 0;game loop{ const real FREQ = 1.0 / 60.0, MIN_DT = 1.0 / 10.0; // 60 FPS integration, min. 10 FPS for (accum += min(dt, MIN_DT); accum => FREQ; accum -= FREQ) integratePhysics(FREQ); render();} ##### Share on other sites thats it.. how wonderful! - so where were you hiding all this time? - why didnt you tell us about it before? - why all the 3 mayor Physics engines have implementation with *maxSubsteps*? - and why is such a confusion around such a simple thing? >>"What advantage does that have over the simplistic accumulator approach?" im not sure, i just got around realizing all "this" a few days ago.. and it took a month of confusion trying to figure out what in the world is going on?!! i basically constructed algorithm around so i can prove to people why is old algorithm with maxSubSteps bad and why we must not have any MAX number of anything ..and no one actually believed me!? but you got it to the point really.. beautiful! anyway, i feel you know more about all this then i do, so what do you think how these two compare? cheers [Edited by - abaraba1 on September 8, 2008 5:42:45 AM] ##### Share on other sites all right, i'll do my inner dialog thing again, but everyone is of course welcome to join any time ..whether to improve algorithm or to explain why PhysX, Havok and Bullet use maxSubSteps? "simplistic accumulator" lets turn few things around, make it bit more general and we get something like this, that can too effectively substitute Bullet's stepSimulation() function: //------------------------------------------------------ int btDiscreteDynamicsWorld::stepSimulation( btScalar timeStep,int maxSubSteps, btScalar fixedTimeStep) { m_localTime+= (timeStep < fixedTimeStep*2)? timeStep : fixedTimeStep*2; saveKinematicState(fixedTimeStep); applyGravity(); maxSubSteps= 0; while( m_localTime >= fixedTimeStep ) { internalSingleStepSimulation(fixedTimeStep); m_localTime-= fixedTimeStep; maxSubSteps++; } synchronizeMotionStates(); clearForces(); return maxSubSteps; } //------------------------------------------------------ again, there is no need to change anything anywhere else - maxSubSteps is completely disregarded and used as a counter - this time you need to supply deltaTime, here called timeStep main loop: dt = timeSinceLastFrame(); stepSimulation(dt); //stepSimulation(dt, 0, 1.0/60); renderScene(); ..solution B is simply about switching simulation frequency which might suit some cases better, but the one above is probably more general and give desired visual effect considering the circumstances //------------------------------------------------------------------------------------------------------- //------------------------------------------------------------------------------------------------------- - it basically replaces all this, and i suppose there is similar equivalent of "maxSubSteps" implementation in Havok, PhysX, ODE, Newton.. ? ****** ORIGINAL STEPPING FUNCTION with "maxSubSteps" ****** //--------------------------------------------------------------------------- int btDiscreteDynamicsWorld::stepSimulation( btScalar timeStep,int maxSubSteps, btScalar fixedTimeStep) { startProfiling(timeStep); BT_PROFILE("stepSimulation"); int numSimulationSubSteps = 0; if (maxSubSteps) { //fixed timestep with interpolation m_localTime += timeStep; if (m_localTime >= fixedTimeStep) { numSimulationSubSteps = int( m_localTime / fixedTimeStep); m_localTime -= numSimulationSubSteps * fixedTimeStep; } } else { //variable timestep fixedTimeStep = timeStep; m_localTime = timeStep; if (btFuzzyZero(timeStep)) { numSimulationSubSteps = 0; maxSubSteps = 0; } else { numSimulationSubSteps = 1; maxSubSteps = 1; } } //process some debugging flags if (getDebugDrawer()) { gDisableDeactivation = (getDebugDrawer()->getDebugMode() & btIDebugDraw::DBG_NoDeactivation) != 0; } if (numSimulationSubSteps) { saveKinematicState(fixedTimeStep); applyGravity(); //clamp the number of substeps, to prevent simulation grinding spiralling down to a halt int clampedSimulationSteps = (numSimulationSubSteps > maxSubSteps)? maxSubSteps : numSimulationSubSteps; for (int i=0;i<clampedSimulationSteps;i++) { internalSingleStepSimulation(fixedTimeStep); synchronizeMotionStates(); } } synchronizeMotionStates(); clearForces(); #ifndef BT_NO_PROFILE CProfileManager::Increment_Frame_Counter(); #endif //BT_NO_PROFILE return numSimulationSubSteps; } //--------------------------------------------------------------------------- [Edited by - abaraba1 on September 12, 2008 7:12:14 PM] ##### Share on other sites hi agi_shi, could you tell us what Physics library do you use and could you point some reference that talk about this "simplistic accumulator"? ..i find that term did not really exist and does not return any search results, so id appreciate if you can point where did you find about it and if there is something i failed to notice about the whole thing? anyway, as it seem you have been using this much longer than i managed to test it in the last couple of days - could you share some of your experience with stepping the simulation in fixedTimeSteps? thank you //--------------------------------- all kinds of feedback/comments appreciated.. anyone? ##### Share on other sites hi grhodes_at_work, - can you confirm that algorithms indeed work in practice as i described? - would you agree that this is nicer design solution with much better visual results? thank you ##### Share on other sites ..surely this is in everyones interest and someone could at least confirm am i right or wrong? i can only test it as much, and i ask everyone for help, hope thats not too much to ask.. after all, if it works then we all benefit, right? thank you ##### Share on other sites Quote: Original post by agi_shiDefinitely interesting post. What advantage does that have over the simplistic accumulator approach?real accum = 0;game loop{ const real FREQ = 1.0 / 60.0, MIN_DT = 1.0 / 10.0; // 60 FPS integration, min. 10 FPS for (accum += min(dt, MIN_DT); accum => FREQ; accum -= FREQ) integratePhysics(FREQ); render();} If dt is based on real time there's the possibility that the time taken to simulate it causes an increase of dt to the next loop. On a bad system that may cause dt to keep increasing over several loops, and there'll be a long time lapse between visual updates, so it looks choppy. It seems abaraba1's algorithm is intended to cause shorter/smoother time lapses, but slow motion instead (though I must say I'm not clear on all details in the code). So if I understood it'll look smoother and still be stable, is that right? (Question to abaraba1) Quote: Original post by abaraba1- and why is such a confusion around such a simple thing? Because time steps are used in both graphics and physics, and games require both, while their respective implementations are different. Some people wants graphical speed, other wants physical stability and realistic time, so discussions about time steps don't end with one answer, but several answers, and those cause the confusion over which to choose, I believe. ##### Share on other sites thank you my friend, you are my 1st normal human contact with the "world community" and i've been trying to discuss this for over a month during which i've been met with strong resistance and disapproval, i've been told nothing but how i "do not understand" and "not getting the point"... i've been banned from public forums merely for asking a questions and trying to figure this thing out ..then, after that - complete silence?! thanks again! >>"So if I understood it'll look smoother and still be stable, is that right? (Question to abaraba1)" yes, in essence - instead of to cap number of iterations you "cap" or "scale"(with the 1st algo) the accumulator/deltaTime which is in practice very similar thing only the later appear smoother because of the way how the FRAMES get *distributed* in the real time - but indeed it is a difficult thing to imagine.. Unfortunately, no one can be told what the 'temporal interpolation' is. You have to see it for yourself. ahm.. so, regardless of everything, i would like to recommend and suggest Bullet physics library if for nothing else but just to test this and practically SEE what is the difference. Bullet library will compile quickly and easily on Windows, Linux or Mac, it comes with lots of demos that compile and build automatically when you build the library. In a matter of minutes you could try to copy/paste algorithm from above, recompile and you will definitively notice the difference and see what is this all about by running the Bullet demos again. http://www.bulletphysics.com/Bullet/wordpress/ thanks, i wish i knew about it sooner.. that is exactly what im talking about, so in case some people having trouble to read my 'funny english' i too recommend that thread as a good read.. cheers [Edited by - abaraba1 on September 14, 2008 10:31:54 PM] ##### Share on other sites The problem solved by abaraba1 is real, e.g., the problem solved here can and does sometimes occur in real-time simulations. Dim_Yimma_H described what is going on here rather elegantly, in a post near the (current) end of the thread (and that post really serves as a helpful introduction to the whole discussion). In fact, in really bad cases, the frame rate can increase to effectively infinity, e.g., the whole application can ultimately enter into a near infinite loop, since the number of simulation steps (at a fixed STEP integration time step) could increase forever. This is an indication of (among other things perhaps) a physics model that is poorly balanced for the target hardware platform. It appears to me that the solution presented in the first post should indeed prevent the simulation loop from approaching infinite time due to an ever-increasing frame time, in the case that (simDT*N)+anmDT > the last frame time. I really have no time to go and test the solution. The simple heuristic that abaraba1 presents for artifically limiting the number of simulation steps (e.g., the "accum -= ..." line in the generic solution inside the simulation loop) may work well in practice, but I cannot say. I can say that it is only one of many heuristics, and others would likely work just as well. So, I like that this thread discusses an issue that doesn't always come up when talking about fixed time step simulation in the game loop. It isn't the first time this has been discussed, but it's nice that it is a dedicated thread. The presentation was a bit long and rambling, and the purpose wasn't really made clear (the way I read it) until Dim_Yimma_H wrote a response. One more thing. An opinion. For a real game design, which might be expected to run on a variety of platforms that may be configured differently and so produce different frame rates, it is unlikely to be acceptable that the physics slows down if the physics is part of gameplay. Any game that uses physics for gameplay, rather than just eye candy, will need to tune the physics model (fewer objects, simpler collision shapes, etc.) to guarantee on the target platforms that the actual simDT is small enough to not cause the frame rate to grow...allowing physics to run at full speed (even with a few multiple fixed-size sim steps per frame). This type of tuning, where each component of the simulation (render, sim, AI, network, art loading, ...) are all balanced to fit a frame time slice is common practice in modern game implementation, and will continue to be done. At least for commercial game development, solutions such as the one presented here can be very beneficial to support eye candy physics, but I don't immediately see it as a huge benefit for gameplay physics. So, to summarize, I see a simple solution to a very real problem, one that could be added to any game loop with a minimal of effort. For commercial games the solution is likely beneficial only for eye candy physics rather than gameplay physics. Users should be aware of the claim of intellectual property usage rights requirements, though I do not believe the innovation is protectable due to the existence of prior art (my opinion only...I am not a lawyer). (P.S. I apologize if some of this seems harsh. Just giving my honest, straightforward opinion and pointing out some things that any forum member who peruses this thread should be aware of, e.g., the legal stuff which really doesn't belong here.) [Edited by - grhodes_at_work on September 16, 2008 2:35:46 PM] ##### Share on other sites Quote: Original post by abaraba1in essence - instead of to cap number of iterations you "cap" or "scale"(with the 1st algo) the accumulator/deltaTime which is in practice very similar thing only the later appear smoother because of the way how the FRAMES get *distributed* in the real time - but indeed it is a difficult thing to imagine.. I tried your algorithm together with my own physics library (but without a computation I didn't understand) and it does indeed look smoother than before [smile] (Of course, when adding a bunch of too many objects.) The part I don't understand how it works is the following computation: Quote: Original post by abaraba1+(STEP*anmDT*fpsTensor) I like your idea of making it look smoother but allowing some slower motion, because it can make a game playable even on bad systems. I can understand though, if that makes the game easier, that some people still prefer striving for real time even when performance is bad. (Though, it may be possible to mark records/achievements under that condition with a symbol of invalidity.) Quote: Original post by grhodes_at_workDim_Yimma_H described what is going on here rather elegantly, I'm glad to hear that, thank you! Quote: Original post by grhodes_at_workJust giving my honest, straightforward opinion and pointing out some things that any forum member who peruses this thread should be aware of, e.g., the legal stuff which really doesn't belong here. At first I thought that legal stuff was meant to sound amusing, with "prove to me you're human being", but in the end I'm not sure. ##### Share on other sites Dim_Yimma_H, >>"I tried your algorithm together with my own physics library (but without a computation I didn't understand) and it does indeed look smoother than before" thank you! now forgive me i'll skip over the rest of what you said as i want to keep this message as short as possible, but i'll address every single thing in my "conclusion" after i clear up some "side-stuff" that turned out to be an issue in this whole "thing", i dont want to discuss this side-stuff i want to get it out of the way and so we can continue to talk about algorithms like this thing: >>"At first I thought that legal stuff was meant to sound amusing, with "prove to me you're human being", but in the end I'm not sure." //---------------------------------------------------------------------------- grhodes_at_work, - thank you, i dont need to "fight" for this anymore >>"The problem "solved" by abaraba1 is real.." 1.) could you remove the quotation marks then? because, it practically solves one problem and address the other in the most desirable way considering the circumstances. But maybe most importantly - IT SEEM TO REMOVE SOME ERRORS with original algo, its about "geometrical interpolation" this physics engines use today - it seem that they first introduce the error, then try to fix it with something that in itself causes further inaccuracies - can you comment on that? tell me what Physics library you use and i'll show you exactly what i mean. >>"The presentation was a bit long and rambling, and the purpose wasn't really made clear" - obviously it turns out i have a problem communicating with the most of "the world" ...People are different. Cultures are different, languages are different, moral and ethical norms differ from family to family, form town to town, from country to country and from planet to planet (see Star Trek). English is not my 1st language, i use dictionary a lot and i have sense of humor that not many people get, unfortunately. So, please... 2.) Please give me an example of my "rambling" so i know what to avoid in the future. Please copy/paste sentence or two so i know exactly what do you mean by that, otherwise i can not improve, thank you. //--------------------------------------------------- * A lot of substeps: physics can be the bottleneck * Few substeps: moon gravity effects - it practically solves both problems related to subSteps/maxSubSteps.. //---------------------------------------------------------------------- /------------------------------------------------------------------------ A) # deltaTime= (simDT * N) + anmDT (time used to simulate and render last frame) B) undersampling is only possible if the time to execute one simulation step is smaller than the fixed time step /------------------------------------------------------------------------ that above is all the information necessary, the rest was mostly copy/paste source code and explanation of practical workings for the ones who did want to learn more about it, the others could happily scroll down.. i basically repeat one and the same thing over and over again and trying to say it in many different ways and still could not make anyone believe for the whole month, so i was actually afraid it will be too short.. yet again interestingly, i got a lot of that "long post" comment in the last month, at the end it always turned out to mean: -"eh, i can't really be bothered to read all that".. - why is the "long" a bad thing? is your message not long? at least i talk about an algorithm and give source code that just by itself is sufficient to understand the point if reader has some basic programming skills, i mean the algorithm is, what? ...3-5 lines long? ..and you talk about what YOU like and what YOU don't like.. i like cats by the way, but what does that has to do with the rest of us and what does that has to do with the matter in question? 5.) >>"I DO NOT LIKE that there is a claim of intellectual property rights claim in the original post.." - why? ..the reason? i see copyright notes in every single source file i open on my computer, why would companies and everyone else have such rights and not me? again, i do not want to discuss this here, i want to get it out of the way - who do i talk to? where the information about this can be found? but dont we all program so we could benefit from it in some way, say - money -, so maybe this actually is very interesting and important subject to anyone who comes up with new algorithms, new ideas, new Intellectual Property.. - but, do humans need such a thing at all? - how is this connected to the IP nonsense related to poor video drivers and whole DRIVERS ON LINUX issue? why is there no full graphic accelerated drivers for kernels? why only drivers for windowing systems? wheres acceleration on PS2 Linux? whats all the "taboo" around the damn graphics drivers and algorithms? that should be OURS!! IP is rubbish, we mut SHARE and COMMUNICATE if we want to progress! ..this was my WHOLE POINT with silly copyright notice (btw, since then i got angry, so - patent pending ;-) >>"..DO NOT post it unless you warn people" - why? unless you state the reason we dont know if you're actually right or wrong about that? >>"And, really, proprietary information should not be posted here in public threads, in my opinion" why not? it is free for humans, for human public on public forums.. that, i hope, should cover most of us >>"Secondly, I do not believe this algorithm is protectable under intellectual property rules" ok, now you're getting close to my point, but why does all this bothers you so much, all you say "don't do" and how you "don't like", but why? >>"At least for commercial game development, solutions such as the one presented here can be very beneficial to support eye candy physics, but I don't immediately see it as a huge benefit for gameplay physics." - there is no such categorization, but even if there was the most of the benefit will actually be with "gameplay physics" or whichever "physics" is more computationally expensive >>..the technical support provided by nVidia/Ageia, Havok, and others already advise customers to do something along these lines." No! * A lot of substeps: physics can be the bottleneck * Few substeps: moon gravity effects and this: and this: http://developer.nvidia.com/forums/index.php?showtopic=2203 http://developer.nvidia.com/forums/index.php?showtopic=2149 >>"I really have no time to go and test the solution." it would have taken less time to test the solution than to write your e-mail, you as a moderator have a responsibility to make such an important issue CLEAR by the number of visits to this thread its obvious people want to know, and since no one believes me - its you, as an authority, as a "Principal Scientist" to make it clear and explain it to the public if indeed my English and communication skills are that bad, its you who should correct it and make it clear instead of just pointing finger at what you do not like without giving any reason and explanation - im just an unemployed, uneducated dude, i dont know any better, but i did my best and did it to the boundaries of my ability - thats my excuse, whats yours? I apologize if some of this seems harsh, but you talk about everything else except what is important - the algorithm, about which you can not probably really talk unless you have seen it or tested it. ##### Share on other sites Hi, Maybe I'm misinterpreting something here (English isn't my first language either) but I can't get rid of the feeling that you're taking the algorithm a bit too seriously. It's not as if you found the holy grail of physics simulation. Still you're talking about the "responsibility" of moderators to promote "such an important issue". There's no such responsibility. You're asking why this thread has so many views? Easy: the title suggest another appearance of someone who claims to have found the answer - like the one claiming to have found the solution for copy protection. Those threads tend to be quite entertaining, although in a sad way. It's not as if PhysX, Havok and Bullet have been created by a bunch of idiots, so if someone claims to have found a major bug and/or solution in all three implementation it's suspicious to say the least. Luckily what the title suggests doesn't turn out to be true: you're actually very focused on one real problem and I really appreciate that. But the way you present your findings is simply disadvantageous: very long posts (which distracts people), no code tags (unreadable code), the explicit note of you taking everyone to court who uses the algorithm commercially. Personally, I doubt that what you describe hasn't been done before and I also don't expect it to be "unbelievably important". But then again, I'm probably not in the position to evaluate it. If you think the algorithm you describe is of scientific importance write a paper about it and have it reviewed. Maybe you get published in an appropriate journal. But please stop complaining about people not testing and worshiping your ten lines of code at once. People will acknowledge you where acknowledgment is due. You can't enforce it. Please consider this post constructive criticism. No offense intended. Eppur si muove ##### Share on other sites So, you quote Galilei. Do you think that we're (I'm) ignorants just like the people back at Galilei's time? Or did you only want to bring an example of a case where it took a (long) while before someone got acknowledged? Or something completely different? Anyway, I think that a snappy catchphrase is not really useful in a discussion. ##### Share on other sites >>"But please stop complaining about people not testing and worshiping your ten lines of code at once." tho worshiping sounds amusing.. any feedback is welcome, i do like criticism, but you have to argue some point, you can not just go on saying stuff without an ARGUMENT, please read carefully every single word i said and feel free to QUOTE anything and then argue your point.. of course, you might save yourself some time if you do test it first - seeing is believing, i ask nothing more but you to see for yourself ..dont believe me, of course not! i worship agi_shi's code, i had it wrong way around at first, it was much uglier - thanks agi_shi, you're welcome any time to tell us more about it! "People, I just want to say, you know, can we all get along? Can we get along? Can we stop making it, making it horrible for the older people and the kids?...It’s just not right. It’s not right. It’s not, it’s not going to change anything. We’ll, we’ll get our justice....Please, we can get along here. We all can get along. I mean, we’re all stuck here for a while. Let’s try to work it out. Let’s try to beat it. Let’s try to beat it. Let’s try to work it out." [Edited by - abaraba1 on September 16, 2008 10:44:34 AM] ##### Share on other sites >>"..write a paper about it and have it reviewed." - where/who can review it? >>"Personally, I doubt that what you describe hasn't been done before.." - i agree, and actually i never thought so ..but, how to go about finding that one out? >>"..and I also don't expect it to be "unbelievably important"." - in the context where this function is executed every 'tick' - yes, because that might be a good place to optimize and if there is some error in this function it might accumulate and cause who knows what... but really - no, you right. this is whats really important: //----------------------------------------------------------------------------- it is a DESIGN TIME DECISION, this means that you must be aware of how much time it takes CPU to execute stepSimulation() - as same as you have little FPS counter at the design-time, so you should have "simDT" and if that number ever gets to "red" (close to fixedStep) you try to design with larger fixedStep or you try to optimize, basically im saying - its the CPU that is critical for determining and deciding on min. sys. requirements because of (simDT * N) ..GPU scales much better, smoother in any case, in the design time, you want to go for the LARGEST POSSIBLE fixedTimeStep you can get by (aiming at low frequency eg. 10Hz) just so you can support that lower class CPU or you have more time to render.. //----------------------------------------------------------------------------- this is all still only my opinion of course, Dim_Yimma_H is the only person in the world that confirmed it in practice, thanks! hehee sorry to have dragged you into this, but you're the human-man, you'll be fine.. anyway, even if it does work in practice that doesnt mean all my conclusions are correct [Edited by - abaraba1 on September 17, 2008 9:13:42 AM] ##### Share on other sites Don't worry, variety in discussion is probably a good thing! I'm thinking about the computation that decreases the accumulated time, that distributes physics steps in an achievable manner. In the case when simulation time becomes longer than real time, the following code is used in the original post: Quote: Original post by abaraba1accum-= (simDT >= STEP)? simDT+(STEP*anmDT*fpsTensor) : STEP Logically, simDT, the current time that the simulation requires, must be included to be able to avoid freezing the simulation loop. In this case, simDT was divided by cnt from last frame, because that number of steps was nearly achievable, so it's the highest number of steps to aim for without risking to freeze the loop. (Though, I wonder if it's sometimes too many steps?) If anmDT increases then it'll cause fewer physics steps, if it decreases then it'll allow more physics steps. What I don't understand, is if STEP*anmDT*fpsTensor is enough to free a particular time value, or if it's tweaking? I'd wish to instead add the animation time divided over the physics steps. Though, I don't have any suitable debugging code now so I'm not sure about correctness. [Edited by - Dim_Yimma_H on September 17, 2008 2:22:48 PM] Sorry, but I have to throw my $0.02 into the bin. It appears as though the Moderator actually agreed with you...and acknowledged that this was a real problem in the physics realm. And it appears as though you did a great job of chopping up his post to make it appear as though he was just attacking you. Secondly, as has previously been stated, it is not his job to promote your work. He is a person like everyone else, and his job is to moderate these forums as the name suggests. And last time I recalled, moderating had nothing to do with promoting. And on a third note, I have to say that you do have a great find there. I don't want to detract any attention away from that. Basically, the jist of my constructive criticism, is that it didn't appear that you read all posts fully before you respond to them. #### Share this post ##### Link to post ##### Share on other sites >>"accum-= (simDT >= STEP)? simDT+(STEP*anmDT*fpsTensor) : STEP" i dont think that anyone would really want to use that in its original form, the point with it was to show: -"..we can always measure the time and act accordingly in a real-time" the whole difference between the algorithms is in the way how you control the number of the iterations, and that doesn't need to be explicitly: //-------------------------- *** ORIGINAL algo - capping the NUMBER cnt= 0; time= 10; maxSteps= 5; while(time > 0 && cnt < maxSteps) { step(); cnt++; time--; } //-------------------------- Iterations, cnt= 5 "Time left", time= 5 //-------------------------- *** 1st algo - scaling the TIME "inside" cnt=0; time= 10; scaleFactor= 1; while(time > 0) { step(); cnt++; time--; time-= scaleFactor; } //-------------------------- Iterations, cnt= 5 "Time left", time= 0 //-------------------------- *** 2st algo - capping the TIME "outside" cnt=0; time= 10; if(time > 5) time= 5; while(time > 0) { step(); cnt++; time--; } //-------------------------- Iterations, cnt= 5 "Time left", time= 0 ..does this make sense? //------------------------------------------------------------------------ now, the problem with original is that you have to guess and it manifests in two ways under certain circumstances: 1.) guess was too low for target CPU - "moon gravity effect" 2.) no matter how high - on some target 'slow enough CPU' it will "spiral to death" ..which is not a "real problem", because you can sort it out in design-time, but the peculiar thing is that it also seem try to deal with the reminder of the time after the iterations have finished, which i think only introduces the errors and its not really necessary ..or is it? [Edited by - abaraba1 on September 17, 2008 11:41:58 PM] #### Share this post ##### Link to post ##### Share on other sites VizOne, >>"So, you quote Galilei..." - man, you say the darndest things i would never dare to even think.. im not faulting your logic here, so i guess i do need to explain what i meant on all that what you said, i just wanted to say: -"To be honest, it moves" you can go on live your whole life thinking the earth is flat if you like, its not "important" information if you don't care about it, but if you use it in your line of work.. there was no point in arguing if you were not interested enough to pursue the matter for your own interests my 1st association on possible interpretation was to Homer Simpson's comment on religion: -"..and all they ask for is a little bit of blind fate" //----------------------------------------------------------------------------- Halifax2, >>"Sorry, but I have to throw my$0.02 into the bin." - its most welcome >>"..it is not his job to promote your work." - true, i only wanted him to make it clear, so i would not get more of those messages, which you can see i did get right after.. of course, who could blame him not being aware that Intel and nVidia could let something like this slip by Intel & nVidia still silent.. i feel this was everyone elses job(work) but mine, i felt i did someones job without even being paid for that and then i had to chase them and beg them to even listen to me, its like im running after you to give you \$100 and you poke me in the eye every time i get close, saying "its a fake" without even looking at it - yes, its insane its not "my work", this has nothing to do with PROMOTING, its in a form: like it? - take it! im not working on any commercial games, i dont need to worry if my sales will be low because of some bugs in physics library.. i just found something and thought i should let 'you' know, where 'you' is everyone in the business of real-time simulation / games. >>"And it appears as though you did a great job of chopping up his post to make it appear as though he was just attacking you." - yes, i apologize, the reason is that i lost it, i simply went crazy.. i gave up so many times - should i believe my two eyes or everyone else telling me "you not getting a point" - i came back every time, a bit more tired, but a much more crazier.. that was anything but personal, at least he went out to talk and tried to compromise without being fully aware of the situation, but again, who could blame him.. that message was intended for Bullet, nVidia, Intel and ODE, but i could not get them to even say anything sensible at all why they were/are not saying anything - i did/do not know, my best guess was that it has to do something with IP/copyright, so about the time when i got half-crazy i put that copyright note to "test" the reaction.. "test results" were amazing - conclusion: unclear so, even if i wasnt crazy, which i was, i would most certainly be by now.. its a simple human mechanism that can protect you from the reality... rainbows and lollipops (the problem is in turning it off) thank you, and thanks for pointing all that out /------------------------------------------------------------------------------- btw, im not sure if i read this somewhere or was i dreaming of it, but i think that in "scientific" or maybe even any other discussion there is no need to say things like: -"Please consider this post constructive criticism. No offense intended" -"Basically, the jist of my constructive criticism.." i believe every criticism is constructive as long as you state your reasoning or otherwise show the logic of your conclusion, i even believe you can go as far and say that im stupid as long as you can explain it or at least try to reason your opinion with some arguments i also think i read somewhere that, in the discussion, you should state all your opinions as if they were absolute truth, even tho you doubt it all yourself - the "i think" is always assumed, and not really necessary to repeat all the time as in practice it indeed can not be nothing more but one persons opinion.. ahm ...have i told you about that time, when i made the best Operating System in the world? http://www.geocities.com/ze_aks/myos.html ... [Edited by - abaraba1 on September 17, 2008 11:18:27 PM] ##### Share on other sites in case it turns out that public forums are not really a best way to go about confirming this, i have only one question: >>"..write a paper about it and have it reviewed." - where/who can review it? /-------------------------------------------------------------------------------------------------- ..in all my craziness, let me wonder for a second about all those crazy web-sites that claim perpetual-motion-machines, anti-gravity, water-for-fuel ..maybe even aliens and all that crazy stuff surely if it was really real, *someone* would have see to it that its tested, confirmed OR proven as hoax.. it would be impossible that all those crazy people would go on raving mad on their crazy web-sites and that no one would - at least - check it out it really bugs me now, why cant we, humans, make all that stuff, that is of such great importance, more clear? [Edited by - abaraba1 on September 18, 2008 7:23:15 AM] ##### Share on other sites Quote: you are my 1st normal human contact with the "world community" and i've been trying to discuss this for over a month during which i've been met with strong resistance and disapproval, i've been told nothing but how i "do not understand" and "not getting the point"... i've been banned from public forums merely for asking a questions and trying to figure this thing out ..then, after that - complete silence?! I'd like to jump in and give a quick comment on this snippet. A possible reason not many people respond is that you're posts are largely illegible. There is no formatting whatsoever. Please compare you're posts to the ones by Dim_Yimma_H and grhodes_at_work. Their posts have: - Paragraphs. - Full sentences, starting with capitals and ending with full stops (.) - Dim_Yimma_H posts have quotes (see here These items help to guide the eye and to split different subjects. In your posts is almost impossible to see where one thing stops and the next thing starts. To be fair, your intial post wasn't too bad, but would have benefitted from [ code ] and/or [ source ] tags. ##### Share on other sites you did not really read anything, did you? - you may know karate, but i know crazy.. 0.) so, you quote all that and describe it as - "not many people respond"? 1.) please explain how any of those reasons prevented you to read 5 lines long algorithm and say something about it? 2.) please explain how any of those reasons prevented you to copy/paste code, to test it and to report your findings? 3.) if you do not use fixed time steps and this matter does not concern you, then you dont really need to be here in this thread, if yes - how does this algorithm work for you? 4.) why say something that was already said and not even attempt answering any questions.. is your point that i was banned from forums because i didnt use tags? ..will this ever stop?! where all these teenagers come from, go away with your tags, im not gonna be finding these bugs every week so that i need to learn how to guide your blind eyes sorry about the tags.. i was drunk [Edited by - abaraba1 on September 18, 2008 9:05:09 AM] ##### Share on other sites At the request of the forum moderator, I've reviewed this thread and personally do not find it worth continuing. I don't feel qualified personally to remark on the technical aspects of the idea being discussed, although I think that has been adequately covered by the other participants in the thread. Instead, I'd like to focus on presentation. abaraba1, your posts (as noted by rick_appleton) are very difficult to read. One of the first things you need to realize about proposing ideas to other people is that presentation is critical. Good presentation makes all the difference between people ignoring your input and paying attention to it. Posting while drunk, for example, is generally frowned upon and doesn't do you any favours. Proper spelling and grammar are important, even if you do not feel that they are. Capitalisation and punctuation are critical for helping people want to read your post. Sure, omitting them doesn't prevent anyone from reading the post, but it sure makes people not want to read it. Poorly written text is difficult and time-consuming to read. If you aren't willing to take the time to write neatly and cleanly, then why should anyone else be willing to take the time to muddle through your posts? Similarly, code/source tags are important as well. They make it easy and convenient to copy/paste code and test it out. Although you are correct that it is still possible to do so without proper tagging, it's a pain. For better or worse, people are powerfully impacted by presentation. You may well find your ideas getting much better reception if you take the effort to present them in a succinct, correct, and professional manner. This includes writing a formal paper as has been suggested to you previously. (There are many places to submit such papers; a little bit of research can find several candidates for you, or you are free to ask where to publish it in a forum such as this one.) Finally, to be perfectly clear, your ideas are welcome here - with the caveat that you are expected to present them clearly, and discuss them politely, even in the face of criticism and disagreement. ##### Share on other sites This topic is now closed to further replies. • ### Forum Statistics • Total Topics 627726 • Total Posts 2978815 • 10 • 9 • 21 • 14 • 12
2017-10-22 11:52:49
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https://math.stackexchange.com/questions/958863/prove-that-a-and-n-are-not-relatively-prime-if-and-only-if-there-is-a-no
# Prove that $[a]$ and $[n]$ are not relatively prime if and only if there is a nonzero element $[b] \in \Bbb{Z}_n$ such that $[a][b] = 0$ Here is my attempt (1) ---> First of all I know that $[n] = [0]$ and then we assume that a and n are not relatively prime then there exists an integer $x = \gcd(a,n)$ and $x \neq 1$ and so there exists integers $q$ and $r$ such that $qx = a$ and $rx = n$ but how I can get to $[a][b] = 0$ from here, I can't find any way. Any suggestions would be greatly appreciated. Let $d=\gcd(a,n)$, let $a=da'$, and let $n=db$. Then $ab=(da')b=a'(db)=a'n$. So $ab$ is a multiple of $n$, but $1\le b\lt n$. That takes care of showing that if $a$ and $n$ are not relatively prime, then there is an appropriate $b$. To show that when $a$ and $n$ are not relatively prime, there is no such $b$, suppose $[a][b]=[0]$. So $n$ divides $ab$. Since $a$ and $n$ are relatively prime, the by a theorem you probably already know, we have that $n$ divides $b$. • I am kinda confused on how to argue for the second implication which is given a nonzero [b] in Zn and [a][b] = 0 , i need to show that a and n are not relatively prime – alkabary Oct 5 '14 at 6:38 • I gave the proof in the answer. Suppose they are relatively prime. If $[a][b]=0$, then $n$ divides $ab$. But then, by a standard result, since $a$ and $n$ are relatively prime, we have $n$ divides $b$, which implies $[b]=[0]$. – André Nicolas Oct 5 '14 at 6:49 We have: $$\exists\;\Bbb Z_n\ni \overline b\ne0\;|\; \overline a \overline b= \overline 0\iff \exists b\in\Bbb Z\; n\not| \;b, n|ab\iff \gcd(a,n)\ne1$$ and to explain the last equivalence: • $\Leftarrow)\quad$ let $r=\gcd (a,n)$ so $a=ra'$ and $n=rn'$ with $\gcd(a',n')=1$ so let $b=n'$ • $\Rightarrow)\quad$ By contradiction reason and using the Euclid's lemma we get the result. • how to argue by contradiction though ? So i am now stuck at the second implication which is given a nonzero [b] in Zn and [a][b] = 0 , i need to show that a and n are not relatively prime – alkabary Oct 5 '14 at 6:36
2019-06-26 17:52:55
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https://www.omnicalculator.com/physics/wing-loading
Created by Rahul Dhari Reviewed by Steven Wooding Last updated: Dec 21, 2022 Our surface area calculator for area of different shapes of wings and weight converter might be useful before using this tool. Wing loading is defined as the ratio of the weight of an aircraft and the projected area of its wings. The weight here is usually assumed to be the gross weight or maximum take-off weight of the aircraft. The wing loading parameter $W_\mathrm{L}$ is given by the equation: $W_\mathrm{L} = \frac{W}{S}$ where W and S are the weight of aircraft and wing area. In other words, wing loading can be defined as weight of aircraft per unit area of wing like surface density. Similarly, wing cube loading, $W_\mathrm{CL}$ can be written as: $W_\mathrm{CL} = \frac{W}{S^{1.5}}$ Wing loading is measured in the units $\mathrm{kg/m^2}$ and $\mathrm{lb/ft^2}$ whereas wing cube loading is measured as weight per unit volume, i.e., same as density. You can also explore our thrust to weight ratio calculator to help you design an aircraft. Wing loading parameter forms the basis of the early stages of aircraft design. Some of the parameters which are affected by the value of wing loading are: • Aircraft speeds — Different types of aircraft speeds such as take-off and landing speeds, cruising and maximum speeds, and stalling speed, all are affected by any change in the wing loading parameter. Now, the area of the wing remains constant, which implies the load carried by aircraft causes an increase and decrease of various types of speeds. Such that, for a given velocity V: $\qquad W_\mathrm{L} \propto V^2$ • Turning performance — The turning radius and load factor is dependent on the wing loading. An aircraft can make tighter turns if its weight is less. Such that, for a particular turning radius, R: $\qquad W_\mathrm{L} \propto R$ 1. Enter the weight of the aircraft. 2. Fill in the area of wings. Estimate the wing loading for a MiG 21 aircraft whose wing area is 23 m² and take off weight is 10,400 kg. 1. Enter the weight of the aircraft, W = 10,400 kg. 2. Fill in the area of wings $S = 23~\mathrm{m^2}$. \qquad \scriptsize \begin{align*} W_\mathrm{L} &= W / S \\ &= 10400 / 23 \\ &= 452.2~\mathrm{kg/m^2} \end{align*} You can also use the advanced mode to find out the wing loading for some popular aircraft from the list. \qquad \scriptsize \begin{align*} W_\mathrm{CL} &= W / S^1.5 \\ &= 10400 / 23^{1.5} \\ &= 94.28~\mathrm{kg/m^3} \end{align*} The wing loading parameter varies for different types of aircraft. For instance, a glider aircraft has a very low value of wing loading compared to a military or a passenger jet. Similarly, a passenger aircraft has a larger wing loading value compared to military fighter aircraft due to the larger load. Further, different types of wing configurations have different aspect ratios and wing loading. A typical hobby RC aircraft has $0.625 - 2.5 \mathrm{lb/ft^2}$. ## FAQ Wing loading is defined as the ratio of the weight of an aircraft to the planform area of wings. Divide the weight of aircraft by planform area. A typical glider has a wing loading of less than 100 kg/m². As per UK CAA, the wing loading limit for microlights is 25 kg/m². The wing loading of F-22 is 377 kg/m². A typical F-22 weighs, 29,410 kg and has a wing area of 78 m². Rahul Dhari Wing area (S) ft² Weight (W) lb lb/ft² lb/cu ft People also viewed… ### Body fat Use the body fat calculator to estimate what percentage of your body weight comprises of body fat. ### Circle skirt Circle skirt calculator makes sewing circle skirts a breeze. ### Engine displacement Determine the capacity of the engine using engine displacement calculator. ### Lumen to lux to candela The Lumen calculator converts between lumens, lux, and candelas.
2023-02-06 02:41:54
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https://inet.omnetpp.org/docs/tutorials/wireless/doc/step1.html
Table Of Contents Table Of Contents # Step 1. Two hosts communicating wirelessly¶ ## Goals¶ In the first step, we want to create a network that contains two hosts, with one host sending a UDP data stream wirelessly to the other. Our goal is to keep the physical layer and lower layer protocol models as simple as possible. We’ll make the model more realistic in later steps. ## The model¶ In this step we’ll use the model depicted below. Here is the NED source of the network: network WirelessA { parameters: @display("bgb=650,500;bgg=100,1,grey95"); @figure[title](type=label; pos=0,-1; anchor=sw; color=darkblue); @figure[rcvdPkText](type=indicatorText; pos=380,20; anchor=w; font=,18; textFormat="packets received: %g"; initialValue=0); @statistic[packetReceived](source=hostB.app[0].packetReceived; record=figure(count); targetFigure=rcvdPkText); submodules: visualizer: <default("IntegratedCanvasVisualizer")> like IIntegratedVisualizer if hasVisualizer() { @display("p=580,125"); } configurator: Ipv4NetworkConfigurator { @display("p=580,200"); } radioMedium: <default("UnitDiskRadioMedium")> like IRadioMedium { @display("p=580,275"); } hostA: <default("WirelessHost")> like INetworkNode { @display("p=50,325"); } hostB: <default("WirelessHost")> like INetworkNode { @display("p=450,325"); } } We’ll explain the above NED file below. ### The playground¶ The model contains a playground of the size 500x650 meters, with two hosts spaced 400 meters apart. (The distance will be relevant in later steps.) These numbers are set via display strings. The modules that are present in the network in addition to the hosts are responsible for tasks like visualization, configuring the IP layer, and modeling the physical radio channel. We’ll return to them later. ### The hosts¶ In INET, hosts are usually represented with the StandardHost NED type, which is a generic template for TCP/IP hosts. It contains protocol components like TCP, UDP and IP, slots for plugging in application models, and various network interfaces (NICs). StandardHost has some variations in INET, for example WirelessHost, which is basically a StandardHost preconfigured for wireless scenarios. As you can see, the hosts’ type is parametric in this NED file (defined via a hostType parameter and the INetworkNode module interface). This is done so that in later steps we can replace hosts with a different NED type. The actual NED type here is WirelessHost (given near the top of the NED file), and later steps will override this setting using omnetpp.ini. ### Address assignment¶ IP addresses are assigned to hosts by an Ipv4NetworkConfigurator module, which appears as the configurator submodule in the network. The hosts also need to know each others’ MAC addresses to communicate, which in this model is taken care of by using per-host GlobalArp modules instead of real ARP. ### Traffic model¶ In the model, host A generates UDP packets which are received by host B. To this end, host A is configured to contain a UdpBasicApp module, which generates 1000-byte UDP messages at random intervals with exponential distribution, the mean of which is 12ms. Therefore the app is going to generate 100 kbyte/s (800 kbps) UDP traffic, not counting protocol overhead. Host B contains a UdpSink application that just discards received packets. The model also displays the number of packets received by host B. The text is added by the @figure[rcvdPkText] line, and the subsequent line arranges the figure to be updated during the simulation. ### Physical layer modeling¶ Let us concentrate on the module called radioMedium. All wireless simulations in INET need a radio medium module. This module represents the shared physical medium where communication takes place. It is responsible for taking signal propagation, attenuation, interference, and other physical phenomena into account. INET can model the wireless physical layer at various levels of detail, realized with different radio medium modules. In this step, we use UnitDiskRadioMedium, which is the simplest model. It implements a variation of unit disc radio, meaning that physical phenomena like signal attenuation are ignored, and the communication range is simply specified in meters. Transmissions within range are always correctly received unless collisions occur. Modeling collisions (overlapping transmissions causing reception failure) and interference range (a range where the signal cannot be received correctly, but still collides with other signals causing their reception to fail) are optional. NOTE: Naturally, this model of the physical layer has little correspondence to reality. However, it has its uses in the simulation. Its simplicity and consequent predictability are an advantage in scenarios where realistic modeling of the physical layer is not a primary concern, for example in the modeling of ad-hoc routing protocols. Simulations using UnitDiskRadioMedium also run faster than more realistic ones, due to the low computational cost. In hosts, network interface cards are represented by NIC modules. Radio is part of wireless NIC modules. There are various radio modules, and one must always use one that is compatible with the medium module. In this step, hosts contain UnitDiskRadio as part of AckingWirelessNic. In this model, we configure the chosen physical layer model (UnitDiskRadioMedium and UnitDiskRadio) as follows. The communication range is set to 500m. Modeling packet losses due to collision (termed “interference” in this model) is turned off, resulting in pairwise independent duplex communication channels. The radio data rates are set to 1 Mbps. These values are set in omnetpp.ini with the communicationRange, ignoreInterference, and bitrate parameters of the appropriate modules. ### MAC layer¶ NICs modules also contain an L2 (i.e. data link layer) protocol. The MAC protocol in AckingWirelessNic is configurable, the default choice being MultipleAccessMac. MultipleAccessMac implements a trivial MAC layer which only provides encapsulation/decapsulation but no real medium access protocol. There is virtually no medium access control: packets are transmitted as soon as the previous packet has completed transmission. MultipleAccessMac also contains an optional out-of-band acknowledgement mechanism which we turn off here. The configuration: [Config Wireless01] description = Two hosts communicating wirelessly network = WirelessA sim-time-limit = 20s *.host*.ipv4.arp.typename = "GlobalArp" *.hostA.numApps = 1 *.hostA.app[0].typename = "UdpBasicApp" *.hostA.app[0].destAddresses = "hostB" *.hostA.app[0].destPort = 5000 *.hostA.app[0].messageLength = 1000B *.hostA.app[0].sendInterval = exponential(12ms) *.hostA.app[0].packetName = "UDPData" *.hostB.numApps = 1 *.hostB.app[0].typename = "UdpSink" *.hostB.app[0].localPort = 5000 *.host*.wlan[0].typename = "AckingWirelessInterface" *.host*.wlan[0].mac.useAck = false *.host*.wlan[0].mac.fullDuplex = false *.host*.wlan[0].radio.transmitter.communicationRange = 500m *.host*.wlan[0].radio.receiver.ignoreInterference = true *.host*.**.bitrate = 1Mbps ## Results¶ When we run the simulation, here’s what happens. Host A’s UdpBasicApp generates UDP packets at random intervals. These packets are sent down via UDP and IPv4 to the network interface for transmission. The network interface queues packets, and transmits them as soon as it can. As long as there are packets in the network interface’s transmission queue, packets are transmitted back-to-back, with no gaps between subsequent packets. These events can be followed on OMNeT++’s Qtenv runtime GUI. The following image has been captured from Qtenv, and shows the inside of host A during the simulation. One can see a UDP packet being sent down from the udpApp submodule, traversing the intermediate protocol layers, and being transmitted by the wlan interface. The next animation shows the communication between the hosts, using OMNeT++’s default “message sending” animation. When the simulation concludes at t=25s, the packet count meter indicates that around 2000 packets were sent. A packet with overhead is 1028 bytes, which means the transmission rate was around 660 kbps. Number of packets received by host B: 2017 ## Discussion¶ Use this page in the GitHub issue tracker for commenting on this tutorial.
2019-02-17 16:55:30
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https://socratic.org/questions/how-do-you-solve-sqrt-x-9-6
# How do you solve sqrt(x+9)=6? Apr 10, 2017 $x = 27$ #### Explanation: To 'undo' the root $\textcolor{b l u e}{\text{square both sides}}$ of the equation. ${\left(\sqrt{x + 9}\right)}^{2} = {6}^{2}$ $\Rightarrow x + 9 = 36$ subtract 9 from both sides. $x \cancel{+ 9} \cancel{- 9} = 36 - 9$ $\Rightarrow x = 27$ $\textcolor{b l u e}{\text{As a check}}$ Substitute this value into the left side and if equal to the right side then it is the solution. $\sqrt{27 + 9} = \sqrt{36} = 6 = \text{ right side}$ $\Rightarrow x = 27 \text{ is the solution}$
2019-12-07 01:25:27
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https://motls.blogspot.com/2010/09/bjrn-lomborg-wants-to-waste-trillions.html?m=1
## Wednesday, September 01, 2010 ### Bjørn Lomborg wants to waste trillions for AGW Bjørn Lomborg has labeled himself a skeptic when he wrote The Skeptical Environmentalist back in 2001. Although he wouldn't say the same thing as the genuine skeptics among us (and the mailing list in which I am included wouldn't quite consider Lomborg "one of us"), his opinions were also far enough from the alarmist orthodoxy. The fearmongers began to hate him - which was quite an achievement given their being progressive radicals and his being a left-wing gay. Well, some of them may be controlled by hormones and they may be driven up the wall as soon as they see the word "skeptic". ;-) In 2002, Scientific American has organized a gigantic witch hunt against Lomborg. It was described e.g. in Michael Crichton's speech Aliens Cause Global Warming. It would also be remarkable to note that in 2004, Rajendra Pachauri has compared Bjørn Lomborg to Adolf Hitler. It would be remarkable if Pachauri were not mentally ill which makes such pronouncements rather unremarkable. I have surely respected Lomborg as a rational economist and a heretic but I have still disagreed with many of his opinions. The second part of the previous sentence is going to be amplified in October 2010 when Lomborg publishes a new book, Smart Solutions to Climate Change: Comparing Costs and Benefits. You may pre-order it now if you want a significant discount. If you open the amazon.com link and look inside the book - and at the table of contents - you will see that as a rational economist, Lomborg remains sensible when it comes to "preferred methods" to "fight climate change". Some guest co-authors help him to reduce the insanity of the book. Carbon dioxide (and methane and black carbon) mitigation is largely overshadowed by climate engineering, forestry sequesteration, adaptation, technology-led climate policy, and technology transfer. However, it's still remarkable that he wants to waste something like \$100 billion for AGW a year. It's a lot of money but it's not too much money: it's clear that by the "stupid methods", one would have to pay trillions or tens of trillions of dollars a year to achieve a noticeable change of the climate after a century. And this crippling of the world economy is the actual main goal of the genuine AGW environmentalists. With smarter methods (but those mentioned above are probably not smart enough), the costs may be much lower and they may even be in the vicinity of hundreds of billions of dollars a year. It's conceivable that if you find the cheapest way to manipulate the climate, you may see a measurable difference after a few decades and a few trillions of dollars that have been paid. But the key question is: Why? I do agree with some of the unhinged, far-left kibitzers in the media and in the blogosphere that Lomborg has noticeably changed his attitude to AGW. The very title and sketch of the book is very different e.g. from his 2008 book, Cool It. And I have no idea what has led him to do so.
2022-01-27 05:59:52
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