url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://dsp.stackexchange.com/questions/15624/matlabs-fft-is-wrong-dont-think-so/15626 | 1,642,359,909,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00530.warc.gz | 307,277,870 | 33,912 | # matlab's FFT is wrong? don't think so
i'm trying to get an FFT of a signal
my FFT has 1024 samples
i added two sin functions one of freq 86,13Hz (2*44100/1024) and one of freq 129,19Hz (3*44100/1024)
the WAV looks like that (for 1 period) :
i then use this code to inspect the FFT
FFTSize = 1024;
current_window = zeros(FFTSize, 1);
for k = 1:length(Y)/(FFTSize)
for m = 1:FFTSize
current_window(m, 1) = Y((k-1)*(FFTSize)+m, 1);
end;
plot(current_window)
a = abs(Ydft);
p = angle(Ydft);
end;
the bins 2 and 3 have the maximum amplitude of 256 which is right as i used two sin of frequency 2 and 3
but when i look at the phases they are -pi/2 which is weird as they should be 0 no ?
here are the phases :
so you see they are -pi/2, why is that ?
and here is the WAV file i used so you can try yourself: http://www.khaelis.com/tmp/2sins.zip
thanks for help
Jeff
• @ionone, neither. It decomposes into complex exponentials $\exp(i \omega) = \cos(\omega) + i \sin(\omega)$ Apr 14 '14 at 11:08 | 327 | 1,011 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-05 | latest | en | 0.905069 |
http://solareviews.net/1235/solar-power-cost-effective-calculation/ | 1,368,918,349,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382920/warc/CC-MAIN-20130516092622-00047-ip-10-60-113-184.ec2.internal.warc.gz | 241,609,102 | 5,793 | Saturday, May 18, 2013
You are here: Home > Solar Power > Solar Power: Solar Power Cost Effective Calculation? (7/27/2011)
# Solar Power: Solar Power Cost Effective Calculation? (7/27/2011)
An average householder uses 160 litres day-1 hot water at a temperature of 55oC, and the specific heat of water is 4.1868 kJ kg-1oC-1 and the inlet temperature of the water to the house is 10°C. Estimate the total energy requirement for hot water supply. If a 2 m2 solar collector is installed to save electricity for hot water supply, the efficiency of the solar panel is 55%. Assuming the mean availability of solar energy throughout the year is 115 Wm-2, the cost of electricity is 5.87p kWh-1, the total cost of the installation is £2000, and it has a life span of 20 years, is the scheme cost effective at a 0% discount rate?
I got the total energy required to heat the water as 11.47GJ/Annum and that its not cost effective as the solar panles will provide 34.7% of the heating requirement which is equilivent to £1300 over 20 years, thus as it cost £2000 to install, its not worth it.
Is this correct? I feel i have done it wrong
More Pages:
Want to Help July 27, 2011 at 9:50 am
Yes I think you may have done something wrong. The company that installed my solar panels helped us figure it out and in the end we realized it really would be worth it. We decided to take the plunge and do it and now 6 years later, we realize how much money we are saving ourselves. Also a good company should be able to get you some good rebates, which really helps a lot. I think you should look into it further. Good Luck!
roderick_young July 27, 2011 at 10:39 am
My calculations show the panels producing 7.25 GJ/year, and 11.00 GJ as the original electrical requirement. That means the panels would produce about 66% of the heating requirement (oddly, this is the complement of 34%).
1 kWh = 3.6 MJ, so the annual energies are
solar 2013 kWh/y
electric 3056 kWh/y
The value of the solar heating is therefore 11800p / year. I’m American, but if p are like cents, and pounds like dollars, then that’s £118 / year, or £2360 over 20 years, a bare win over the system price.
It sounds like these calculations are for a place that is right on the cusp of where solar makes sense. In Hawaii, electricity would be 3-4x that price, the system price 2/3 of that (simple panels – freezing not an issue), the insolation 2x that, and the water inlet temperature higher, making the energy requirement 2/3 of that. | 653 | 2,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2013-20 | latest | en | 0.948042 |
http://mathoverflow.net/questions/124758/another-fibration-with-a-given-singular-fiber-class | 1,469,765,153,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829970.64/warc/CC-MAIN-20160723071029-00003-ip-10-185-27-174.ec2.internal.warc.gz | 162,728,417 | 14,045 | # Another fibration with a given singular fiber class.
Let $f:X\rightarrow C$ be a fibration of complex manifold over a smooth curve $C$. Let $D$ be an irreducible component of singular fibers. How can one prove that $X$ does not admit any fibration with general fiber class $D$? Or any counterexample?
I think this is intuitively true because $D$ is an irreducible component of singular fibers, so it cannot $move$. But I don't know how to prove it.
I would appreciate it if you can provide me a proof or reference.
-
Without further hypothesis, I don't see how you would prove something like this. Consider the following example. Suppose $g:Y\to C$ is a surjective map with general fibre $D$, blow up $Y$ along a point on $D$ to get $X$. Now $D$ appears as a component of singular fibre of $X$, yet all the other smooth fibres are deformation equivalent to $D$. – Donu Arapura Mar 17 '13 at 15:24
Correction: I guess I was thinking of $Y$ as a surface. In general, you would need to blow up along a smooth codim 2 subset of $X$ supported on $D$. – Donu Arapura Mar 17 '13 at 15:26 | 286 | 1,086 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-30 | latest | en | 0.926183 |
https://brilliant.org/problems/all-hail-euler-the-ruler/ | 1,490,449,805,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188926.39/warc/CC-MAIN-20170322212948-00258-ip-10-233-31-227.ec2.internal.warc.gz | 773,505,852 | 12,709 | # All Hail Euler the ruler!
Geometry Level 4
Consider a triangle$$\Delta ABC$$ whose circumcenter is at the origin. If in $$\Delta ABC$$, the coordinates of the centroid $$G$$ are $$\left(x_G,y_G\right)$$ and the coordinates of the orthocenter $$H$$ are $$\left(x_H,y_H\right)$$.
Find the ratio $$\dfrac{x_Hy_H}{x_Gy_G}$$.
Bonus: Don't forget the lonely incenter!
× | 112 | 370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-13 | longest | en | 0.652832 |
https://www.polytechforum.com/metalworking/concentric-bolt-circles-443937-.htm | 1,695,683,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510100.47/warc/CC-MAIN-20230925215547-20230926005547-00462.warc.gz | 1,067,209,979 | 10,596 | # Concentric bolt circles?
• posted
I need to make a couple of adapter plates for the new 56HP Yanmar to fit the CV joint to the prop shaft. The gear box flange has four M10x1.5 threaded holes on a 78mm pitch circle and the CV joint has six M10x1.5 holes on a 93mm pitch circle.
At anything other than a 15 degree offset between the two circles the holes interfere. At 15 degrees I have 3mm between the holes. I am limited to
22mm in thickness so I can't drill blind holes on either side. Is 3mm between holes sufficient or is there another way to lay out the bolt circles?
• posted
Hi Glenn,
Surely the least metal you can have between the holes is 5 mm? The outer edge of the holes on the 78 mm circle will be at a radius of 83 mm, and the inner edge of those on the 93 mm circle will be at a radius of 88 mm. So I figure there's a minimum of 5 mm between them. Someone correct me if I'm wrong here...
I reckon 5 mm will be okay, but I'd be inclined to make the adaptor plate the full 22 mm thick.
There is a mathematical way of figuring out an optimum hole spacing, but right now I can't think of a simple mathematical way. But I'll have a go if you like.
Best wishes,
Chris
• posted
Having thought about it a little more, I think 15 degrees may very well be the optimum offset. I don't think there is much to be gained by doing a load of maths :-).
Chris
• posted
It's customary to specify bolt circles as the diameter, not radius. It's a tight fit.
I have no answers for the OP, but I will say this. It might be better to use only four of the six mount holes rather than compromise all four of the other set.
• posted
If that's the case, it's a very tight fit. Mike's idea of only omitting a couple of bolts could be a good one. Perhaps other people can add their thoughts?
Chris
• posted
Without looking at the flanges, it's hard to make any recommendations, but I'd be tempted to plug the holes, weld an "extension ring" (if it's needed) on the small flange to match its outside diameter with the larger one and redrill the holes, or make a new flange (although the thought of creating splines in the inside of the new flange would give me pause...).
Peter
• posted
3 mm will be pushing it. Have you looked at LHSHCS? (Low-Head Socket- Head Cap Screws). Check:
Head height is only 6 mm, leaving you 15.5 mm of material thickness without a washer (0.5 cbore clearance), subtract another 1.25 mm if you use a serrated cone (bellville style) lockwasher.
Anthony
• posted
"Mike Young" wrote
I got a look at a "stock adaptor for a 100mm PCD adaptor. Only 4 holes are drilled for the 6 hole side so that looks like the solution.
Never realized how much I have saved with my mill and lathe. Took the drawings down to a local machine shop and got a price of \$240 each for the adaptors. They are just a 4.5" diameter round 7/8" thick bored 1.75" with a 3"x 1/4" relief on one face and a 1/8"x1/8" spigot around the bore on the other. I can turn them both out in half a day on my antique 10K. In making parts for Rutu I figure I have paid for both the mill and the lathe 5 or 6 times. Of course, I bill my time at \$.01/hour after the satisfaction discount. :-)
• posted
I can't imagine why it would be a problem with a reasonably strong and malleable material (e.g., mild steel or 300 series SS). Consider that the wall between the holes at the thinnest point is only 0.5mm less than the wall of a standard M10 nut.
Is there a specific failure mode you're concerned about?
Ned Simmons
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners. | 941 | 3,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.951451 |
https://brainly.in/question/112036 | 1,484,779,129,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280364.67/warc/CC-MAIN-20170116095120-00469-ip-10-171-10-70.ec2.internal.warc.gz | 795,712,543 | 10,073 | # ? +? +?=30 Solve this fill the boxes useing ( 1.3.5.7.9.11.13.15) you can repeat the numbers
2
by premKata361
2015-05-14T18:34:47+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
as it is not possible to get an odd number by adding odd no.s odd no.of times, we could do it by using factorials,for example:
11+13+3! -=30
[ by this way,you may also find other combinations ]
2015-05-14T20:17:12+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
The sum of three odd numbers gives an odd number
So, the sum of 3 given odd numbers doesnt results 30
So, it is an impossible factor
here we can't use 3! too
because 3! [3×2×1] results a even number 6
so we cant use it.
If there will be an allowance of factorials, then the answers will be
3! + 15 + 9 =30
3! + 11 + 13 = 30 | 356 | 1,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-04 | latest | en | 0.918664 |
https://convertertoolz.com/m-to-feet/21-m-to-feet/ | 1,726,456,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00732.warc.gz | 150,130,410 | 11,897 | # 21 m to feet
## Heading 1: Understanding the Conversion from Meters to Feet
Understanding the conversion from meters to feet is essential for individuals working with measurements in various industries and areas of expertise. The process of converting meters to feet involves the transformation of a metric unit of length into an imperial unit of length. While the metric system is widely used around the world, the imperial system, which includes feet as a unit of measurement, is commonly used in the United States and a few other countries. Therefore, being able to convert between meters and feet is crucial for international trade, construction, and engineering projects, among other disciplines.
The conversion from meters to feet is based on a simple formula. One meter is equal to approximately 3.28 feet. By multiplying the given measurement in meters by this conversion factor, the corresponding value in feet can be obtained. For example, if a length is given as 10 meters, the conversion to feet would be calculated as follows: 10 meters multiplied by 3.28 feet per meter equals 32.8 feet. This formula allows for quick and accurate conversion between the two units, facilitating seamless communication and collaboration across different measurement systems. The ability to convert between meters and feet is a fundamental skill that ensures precision and consistency in various professional fields.
## Heading 2: The Importance of Unit Conversion in Everyday Life
Unit conversion is an essential skill that holds great significance in our daily lives. Whether we realize it or not, we encounter situations where converting between different units is necessary. From cooking recipes to construction projects, unit conversion allows us to effectively communicate measurements and ensure accurate results.
Without proper unit conversion, confusion and errors can easily arise. Imagine a recipe that calls for 500 grams of flour, but you only have a kitchen scale that measures in ounces. Converting the units ensures you add the correct amount and achieve the desired outcome. In sectors such as engineering, medicine, and technology, where precise measurements are critical, having a strong grasp of unit conversion is indispensable. It enables professionals to interpret and convey data accurately, ultimately contributing to the success and safety of various projects and operations.
## Heading 2: A Brief Explanation of Meters and Feet
Meters and feet are both units of measurement used to quantify distance or length. Meters, denoted by the symbol “m”, is the standard unit of measurement in the International System of Units (SI), commonly used in most countries worldwide. It is used to measure various things such as the length of an object, the height of a building, or the distance between two points.
On the other hand, feet, denoted by the symbol “ft”, is a unit of measurement predominantly used in the United States, the United Kingdom, and other countries that follow the imperial system. Initially derived from the length of a human foot, a foot is divided into 12 inches. While meters provide a more universally accepted means of measurement, feet continue to be widely used in everyday life, especially in construction, architecture, and sports.
These two units of measurement, meters and feet, can sometimes create confusion due to their differing scales. The conversion between meters and feet allows for a seamless transition between the two systems and ensures accurate communication and understanding of distances and dimensions. Understanding this conversion formula is essential to effectively work with or compare measurements expressed in these two distinct units.
## Heading 2: The Conversion Formula for Meters to Feet
To convert a measurement from meters to feet, a simple formula is used. The conversion formula is quite straightforward: 1 meter is equal to 3.28084 feet. Therefore, to convert any given measurement in meters to feet, simply multiply the value by 3.28084. For example, if you have a measurement of 5 meters, you would multiply 5 by 3.28084, resulting in 16.4042 feet.
It is essential to remember that the conversion formula is based on a precise conversion factor. This factor ensures accuracy in converting between the two units of measurement. By utilizing this formula, individuals can easily and accurately convert distances, heights, or any other measurements from meters to feet. Understanding and utilizing this conversion formula, we can seamlessly navigate between the metric and imperial systems of measurement, allowing for effective communication and understanding across various contexts.
## Heading 2: Common Applications of Meters to Feet Conversion
Common Applications of Meters to Feet Conversion.
One of the most common applications of converting meters to feet is in the field of construction and architecture. In these industries, accurate measurement of distance is crucial for ensuring that buildings, structures, and other elements are designed and built to the correct specifications. Many countries, such as the United States, primarily use feet as the unit of measurement in construction projects. Therefore, professionals in these fields often need to convert measurements given in meters to feet to ensure that they are working with the correct values.
Another common application of meters to feet conversion is in the realm of sports. Many international sports competitions, including track and field events and swimming races, use meters as the standard unit of distance. However, in countries where feet are the primary unit of measurement, such as the United States, conversions from meters to feet are necessary to communicate the distances to athletes and spectators. This ensures that everyone understands the distances being covered during the competition and allows for accurate comparison of performances across different measurements systems.
## Heading 2: Examples of Converting Meters to Feet in Real-Life Scenarios
In real-life scenarios, the conversion from meters to feet is a common practice, especially when it comes to measuring the height of objects or buildings. For example, in the field of architecture and construction, architects and engineers often need to convert measurements between these two units to ensure accurate designs and constructions. By converting meters to feet, they can easily communicate and understand the dimensions of buildings or components in a common unit that is widely used in the industry.
Another example where converting meters to feet is important is in athletics and sports. Track and field events, such as long jump or high jump, often require measurements in feet rather than meters. This is because the values are easily relatable to viewers and fans who are more familiar with feet as a unit of measurement. By converting meters to feet, athletes, coaches, and spectators can better understand the impressive distances or heights achieved by the athletes, contributing to the overall excitement and enjoyment of the sport. | 1,291 | 7,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.935283 |
http://www.algebra.com/algebra/college/linear/Linear_Algebra.faq?hide_answers=1&beginning=585 | 1,369,112,595,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699684236/warc/CC-MAIN-20130516102124-00054-ip-10-60-113-184.ec2.internal.warc.gz | 319,080,404 | 8,929 | # Questions on Algebra: Linear Algebra (NOT Linear Equations) answered by real tutors!
Algebra -> Algebra -> College -> Linear Algebra -> Questions on Algebra: Linear Algebra (NOT Linear Equations) answered by real tutors! Log On
Algebra: Linear Algebra (NOT Linear Equations) Solvers Lessons Answers archive Quiz In Depth
Question 37978: find f(g(x))when f(x)is x-1 and g(x) is 2x squared Click here to see answer by fractalier(2101)
Question 38555: Solve the linear system: x+2y= 1 -x+y= -1 Click here to see answer by lyra(94)
Question 38081: How are zeros,x-intercenpts,or soluton relate to the line of symmertry? Click here to see answer by rapaljer(4667)
Question 36646: This is my second and last question. This is not in a textbook. a.)Let T be a linear operator on V. SHow that if (Tu|v)=0 for every u,v in V, then T is the zero operator on V. b.)Prove the Generalized Pythagorean Theorem: Suppose {v1,v2,...,vn} is an orthogonal set of vectors. Then ||v1+v2+...+vn||^2=||v1||^2+...||v2||^2+...+||vn||^2 For (a.) I think you fix u and get it, but I don't understand how. For (b.) Do I only have to do this: ||v1+v2+...+vn||^2=(v2+v2+...vn|v1+v2+...+vn) =(v1|v1)+(v2|v2)+...(vn|vn) since the vectors are orthog. =||v1||^2+...||v2||^2+...+||vn||^2 Is this all i have to do or is there more steps in there? Click here to see answer by venugopalramana(3286)
Question 39022: Can you help me solve this problem: 3,520 ft =_____miles Click here to see answer by fractalier(2101)
Question 37810: Given f(x)=4x-3 and g(x)=2x(to the power of 2)-5 find f[g(-1)] and g[f(-1)]. Click here to see answer by Nate(3500)
Question 39282: Find an equation of the line through the point (4,5) and perpendicular to the line passing through the points (-1,3) and (2,9). Click here to see answer by stanbon(57282)
Question 39279: Find the slope of the line containing the given points. (-a-b,p+q) and (a+b,p-q) Click here to see answer by longjonsilver(2297)
Question 39551: I have to cnovert a linear distance from a fraction to a decimal and a decimal to a fraction. 57'- 11 7/16" what is this as a decimal? 14.81' what is this as a fraction? Click here to see answer by fractalier(2101)
Question 39551: I have to cnovert a linear distance from a fraction to a decimal and a decimal to a fraction. 57'- 11 7/16" what is this as a decimal? 14.81' what is this as a fraction? Click here to see answer by stanbon(57282)
Question 39593: Question; Determine whether or not W is a subspace of R3, with justification (general proof). W={(x,y,x+y); x and y are real)}. I have a trouble in proving(in general, not specific numbers) the closure axioms(addition and scalar multiplication). Please help! Click here to see answer by cininnatus(12)
Question 39572: Proof of a singular matrix: I need to show (in general) that either matrix A is "singular" or "A^2 = A^(-1)". A is an n*n square matrix such that A^4 = A. I want to know how to do this problem and what happens if the exponent values are different.(for example, odd number exponents?) Click here to see answer by venugopalramana(3286)
Question 40449: I posted a question just a few minutes ago and Nate answered it but he didn't exactly answer what I am looking for. My slope is 2/3 but how do I get this into y=mx+b format. The question is Line 2 is perpendicular to a line that passes through the points (4,-3) and (-2,6). I know that the slope is 2/3 but how do I get this into y=mx+b? Click here to see answer by smik(40)
Question 40572: How does regression relate to linear algebra? What are some limitations of linear regression? What are the risks? Click here to see answer by stanbon(57282)
Question 40520: Last one: Let T:R^n --> R^m be a linear transformation, and let {v1, v2, v3} be a linearly dependent set in R^n. Prove that the set {T(v1), T(v2), T(v3)} is linearly dependent. I have proved it closed under addition and scalar multiplication, but am not sure if this is correct? Should I be looking at the trivial solution? Thank you! Click here to see answer by kev82(148)
Question 40518: I am struggling with this proof, please help! Prove that the row vectos of an n x n intertible matrix A form a basis for R^n Thank you! Glenna Click here to see answer by kev82(148)
Question 40403: Proof of Hermitian matrices: If A and B are Hermitian matrices, I need to show that BA = AB iff AB is Hermitian, but I can't figure out how... Click here to see answer by venugopalramana(3286)
Question 42104: What would be a good project for Algebra 2, sections 1-5? Click here to see answer by fractalier(2101)
Question 42594: what do points on the line parallel on the y-axis in 3 units to the left of it have in common? Click here to see answer by rapaljer(4667)
Question 42593: what do point on the line parallel onthe y-axis in 3-units to the left of it have in common? Click here to see answer by rapaljer(4667)
Question 42596: what do points on the line parallel on the y-axis in 3 units to the left of it have in common?with a solution. Click here to see answer by rapaljer(4667)
Question 42604: what do points on the line parallel to the y-axis in 3 units to the left of it have in common?with a summary. Click here to see answer by AnlytcPhil(1276)
Question 42661: identify the vertex of the parabola y = x squared + 3 also y = 3x squared + 30x + 77 Click here to see answer by Nate(3500)
Question 42660: Give the equation or other information on for a parabola, find the matching discription or graph. f(x)-a(x-h)2+k the 2 after ) is square a is less than 0 h is equal to 0 choices: a. the graph of f(x) does not intersect x-axis b. you can not tell from the information given c. the graph of f(x) interescts the x-axis at one point d. the graph of f(x) intersects the x-axis at two points Click here to see answer by Nate(3500)
Question 42829: in what quadrant would a point lie if its abscissa and ordinate are equal? Click here to see answer by ichudov(499)
Question 42834: what do points on the line parallel to the y-axis and 5-units to the left of it have in common?with a solution. Click here to see answer by josmiceli(9664)
Question 42833: what is the distance from x-axis to (7,-2)?from the y-axis?with a solution. Click here to see answer by josmiceli(9664)
Question 42821: The perimeter of a rectangle is given by p=2w+2h where w is the width and h is the height. Find the height if the perimeter is 52 inches and the width is 14 inches. I have worked the question out and came up with 13in. Would like to know if this is right. Thanks again for any help. Click here to see answer by josmiceli(9664)
Question 42839: what can be said of points with equal abscissa?with equal ordinates? Click here to see answer by AnlytcPhil(1276)
Question 42839: what can be said of points with equal abscissa?with equal ordinates? Click here to see answer by psbhowmick(529)
Question 42826: what can be said of points with equal abscissas?with eqal ordinates. Click here to see answer by psbhowmick(529)
Question 42831: what is the distance from x-axis to (7,-2)?from the y-axis? Click here to see answer by fractalier(2101)
Question 42832: what is the distance from x-axis to (7,-2)?from the y-axis?with solution. Click here to see answer by fractalier(2101)
Question 42830: what is the distance from x-axis to (7,-2)?from the y-axis? Click here to see answer by fractalier(2101)
Question 42828: what do points on the parallel to the y-axis and 5-units to the left of it have in common? Click here to see answer by fractalier(2101)
Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510 | 2,675 | 8,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2013-20 | latest | en | 0.840614 |
https://math.stackexchange.com/questions/1370599/can-you-add-new-functions-to-the-set-of-elementary-functions-such-that-every-fun | 1,561,523,653,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000164.31/warc/CC-MAIN-20190626033520-20190626055520-00459.warc.gz | 507,384,520 | 35,635 | # Can you add new functions to the set of elementary functions such that every function has an anti-derivative?
Its fairly well known that not every elementary function has an elementary anti-derivative. The common examples of this are $\exp(-x^2)$ and $\sin(x)/x$. The general workaround to this problem is to create new functions like $\mathrm{Erf}(X)$ and $\mathrm{Si}(x)$ to go alongside the more elementary functions. I was wondering if there is a way to add "enough" non-elementary functions so that every function has an anti-derivative that can be expressed via other functions. The typical solution to this question is to introduce Riemann integration as an operator on functions, and then declare that the set of functions is closed under this operator. That definitely works, but I'm looking for something that is more enumerable. Like a simple countable set of functions that I can build everything else out of. Has anyone come up with a set of functions like this, or proved that making one is impossible?
• en.wikipedia.org/wiki/Hypergeometric_function – Jack D'Aurizio Jul 22 '15 at 21:52
• If your set of functions in analytic, then yes, just use Laurent series, truncated to finite order. – Alex R. Jul 22 '15 at 22:28
• @AlexR.: I don't see how Laurent series give a countable set of functions that meet Joel Turnblade's requirements. – Rob Arthan Jul 22 '15 at 23:38
• @RobArthan: Isn't $\{x^k\}_{k=0}^\infty$ countable? I'm assuming OP wants antiderivatives to be generated as (possibly infinite) linear combinations of such a set. – Alex R. Jul 22 '15 at 23:41
• I was assuming the OP was looking for a countable set of functions that would generate in the algebraic (finitistic) sense an an algebra of functions closed under antiderivation. Over to the OP to clarify. – Rob Arthan Jul 22 '15 at 23:55 | 454 | 1,824 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2019-26 | latest | en | 0.928973 |
https://community.tes.com/threads/interpreting-carol-dwecks-motivation-questionairre.453843/page-6 | 1,600,616,593,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198213.25/warc/CC-MAIN-20200920125718-20200920155718-00201.warc.gz | 344,242,097 | 22,930 | # Interpreting Carol Dweck's Motivation Questionairre
Discussion in 'Mathematics' started by mature_maths_trainee, Dec 12, 2010.
1. ### weebecka
I'll have another go at responding to this one, since I messed it up last time.
Scentless_apprentice,
Even with a very simple concept like multiplication, there is substantial variation over time and confusion now as to what it actually is. I can explain this further if you like (it helps if you spend sometime thinking about what you personally think multiplication is as part of this), or we can move on to a different theme.
2. ### DMNew commenter
Yes, but you only have to enter a room to enchant them by your genius weebee. What about the rest of us?
3. ### DMNew commenter
Happy Xmas by the way.
4. ### weebecka
That's right of course DM, I just walked in and all was sweetness and light. The children devised Euclid's algorithms for themselves while I sat back, relaxed, sipping my cuppa.
Nothing to do with using the specific strategies I've been taught by the many gifted teachers I've worked with and am happy to talk about here (combined with a ton of resiliance and a great deal of peer and personal support).
5. ### DMNew commenter
I knew it. Well, fair play to you - if you've got it, flaunt it.
6. ### bgy1mm
We had a thread recently on how to teach -2 * -2 = 4.
7. ### weebecka
Oh you don't have to get that complicated, just try 7x3. What's that?
8. ### weebecka
& how and why? what's going on when we multiply?
9. ### bgy1mm
It's the lucky numbers, 3 and 7 makes 21. I was taught that by my mother and still think of it that way.
Of course I can add 14 to 7 in my head, but that's a bit of an effort, so I don't do it like that. I just treat it as a fact.
10. ### weebecka
That's nice. Each result can be a remembered fact.
anyone else?
11. ### Betamale
Multiplication and division facts should be that for most. No real depth,m simply a parrott fashion response
Teacher "7x8"
Every student "56"
Once this is sroted, fluent for all operations then an understanding of why can be explored.
Anything beyond this is not required for many students who slug their way through 11 years of foral education to come out with a C or D grade still unable to recall basic number facts.
Numeracy is not maths, maths relies on numeracy.
Teach them as two subjects, the latter being a progression from the former for those fluent.
Why teach angle facts, area of strange shapes and scatter graphs year in year out if the pupils cant work out the monthly cost of their gas bill if they have been handed the yearly one?
12. ### MathsMANew commenter
But my problem is just what students these are, as looking at your CV/Profile I struggle to see where you have stayed long enough or had the time to implement these wonderful strategies (apart from the one school of choice in Cumbria) with a range of classes.
Your last three schools spanned 3 years in total, a period of which you have mentioned you were actually not at one of the schools. Your last year is somewhat of a mystery as we have to still to understand what you have been doing and what your link is with MMU?
So just where have you had the time, capacity and longevity to make such grandoise claims about the merits and success of these "methods"?
Where is the emperical research, where are we to find a detailed explanation of the methods, the setting, the outcomes etc?
I'm not entering in to an ad hominen fallacy, but I think it would be fair to say that an impartial observer would find ir rather difficult to square your claims with your experince to date.
13. ### weebecka
I'm happy to provide context for the purpose of enhancing a discussion, but clearly that's not happening here.
I could defend myself on each point one at a time,
for example here I was teaching Jordanian students in state schools in front of their teachers, who afterwards got to critically analyse what I was doing (and boy were they sharp). Following this we would spend time together, getting to know each other and chatting about the realities of their teaching, their constraints and their aspirations.
I find this kind of teaching more illuminating that working for a while in an expat school.
But the negative twisting and the accusations are coming too thick and fast to be compatible with an interesting thread MathsMA.
So to abridge my need to work with them all, could I perhaps instead suggest that you celebrate Christmas by having an anal encounter with your festive tree, which I hope has a nice sharp pointy star on top.
14. ### weebecka
i.e. attack the argument or the topic, not the person.
15. ### MathsMANew commenter
In general, when you have managed to respond on a topic/argument (as opposed to waffling off on a tangent and/or posting your "facilitating", read condescending, questions), people have responded (often in horror) at some of your suggestions.
Additionally, as I have mentioned above, I'm not attempting an ad hominem attack here, instead I am merely trying to gain an understanding of where your experiences have been and how robust they are particularly when you hold them up to be a sort of panacea of teaching and learning.
I don't think it is at all unreasonable to question your experience or experiences, particualry when they relate directly to some of your more boastful claims. You continually refer to "my students" or "my classes" and I am merely trying to ascertain just where and what these classes were, because looking at your profile I have to say I'm not overly confident that you have been around anywhere long enough to be speaking with such certainty and confidence.
Your experiences form a central plank of your posts on here, its just that I have my doubts about the validity of your experiences and your refusal to expand on them speaks volumes to me. I'm particularly curious with regards your last three years of teaching and your work at MMU.
PS If Boaler blustered on to this forum in the way you have with stories about the benefits of Mixed Ability Teaching or Williams on Formative Assessment, I would probably be just as miffed at their poor communication skills and their patronising attitude, but at least they have a plethora of experience and hard edged research to back up their opinions which might placate me somewhat. Instead judging by your profile (with its incorrect lengths of tenure at various places) you have nothing to back up your bluster.
16. ### weebecka
Good good. Look MathsMA. I just can't give you the level of details you're demanding here. It's inappropriate. I'm happy to meet your or others and discuss this kind of thing in an appropriate way. I'm not covering anything up - I think in the first thread here I was very clear that I'd been HOD in a small school. I've not been misrepresenting things. Again, like Jordan I chose it deliberately because being head of a small dept. give you a great deal of freedom to change things rapidly and for other reasons we could discuss face to face.
If you don't want to meet me, stop having a go because I can't give the level of detail you want here, so it's pointless. Yes my CV is slighty fragmented, but there are sensible and appropriate reasons for that.
Heaven knows why on earth you think my profile is incorrect. It's far more detailed and open than anyone else here and it's just as it is. Again, it's just totally inappropriate to write things like that. If there's one specific thing that doesn't seem right then mention it, or better still PM me so that it doesn't spam up the thread and I can reassure you.
I did invite Jo Boaler up to do observations at Ehenside, but sadly it just wasn't possible before it shut. She's a long way from here and she was working with a different focus at the time. It was a shame because I don't think she really understands the impact braodband has had on the practicalities of integrating curriculum-orientated and process-orientated curricula.
I'm not looking to win an argument with you based on my back-record MathsMA or to overwhelm you with my credibility. My resume isn't there to do that, just give people an idea of my range of experience so I don't have to start from scratch whenever someone new joins the discussion. On most forums you're required to provide this sort of stuff because discussion is so much easier if you just click on a poster to get a good idea of 'where they're coming from'.
I've no interest whatsoever in arguments being won because one poster 'is better' than the other. I'm just interested in exploring the issues - with anyone who wants to talk about them.
17. ### Maths_MikeNew commenter
Beecka - please dont take it too personally I just think that lots of us are fed up with people - many of whom have limited (either in time or type) teaching experience telling us what to do and why we are **** at our jobs and they are so much better.
I mean anyone from Ed balls (or who ever his replacement is) to Carol Vorderman to my postman things they would be a bettre treacher than I am.
They may well be right but I dont tell them how to do their job!
18. ### scentless_apprentice
My sentiments entirely. I'm all for a debate about the priniciples of Mathematics education.
But a lot of the time, we've got to look at the problems we have working at 'the coal face' - i.e. in the class!
19. ### weebecka
Er - I do get this.
Back on the thread - when we had training on Carol Dweck, the first half of it was fine - a nice gentle interactive coverage of what she was all about.
Then I/we wanted to talk about and share and develop how we were doing a lot of what she was talking about already, it's just we'd never expressed it in her way.
Instead Barry Hymer just picked the pace up and up, going on with more and more fervour and conviction about how wonderful she was and that was it.
If anyone ever catches me doing a conference session or any training in that way, please scream and shout and stop me. I don't ever want to be like that.
On-line conversations are challenging, especially when they are talking about deep issues like philosophy. I'm genuinely interested in listening to everyone but it takes some working at. I think part of the problem is people assume I'm not listening so they don't bother to work at it.
The coal face bit just means people drop in and out of the deeper conversations as and when they feel like it and that's absolutely fine. The deeper stuff is better taken on bite size chunks anyway. | 2,346 | 10,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-40 | latest | en | 0.966079 |
https://zuoyedaixie.net/data-structure%E4%BB%A3%E5%81%9A-algorithm-%E4%BB%A3%E5%81%9Aquiz-%E4%BB%A3%E5%86%99java-%E4%BB%A3%E5%86%99python-comp90038-algorithms-and-complexity/ | 1,723,414,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00811.warc.gz | 848,205,206 | 20,105 | # Data structure代做 | Algorithm | 代做quiz | 代写java | 代写Python – COMP90038 Algorithms and Complexity
### COMP90038 Algorithms and Complexity
Data structure代做 | Algorithm | 代做quiz | 代写java | 代写Python – 这是一个Data structure面向对象设计的practice, 考察Data structure的理解, 是比较有代表性的Data structure/Algorithm/java/Python等代写方向, 该题目是值得借鉴的quiz代写的题目
### Semester 1 Assessment 2020
Writing Time3 hours.
Marks Available:70 marks.
Instructions to Students:
The exam consists of two parts:
• Part A: 3 questions worth 30 marks
• You must prepare answers to these questions in a PDF document. Please use a text editor. Handwritten answers areNOTacceptable.
• Submit your solutions using the ModulesFinal ExamPart A submission link within the allocated 3-hour time window.
• Your algorithms must be written in pseudocode using a format consistent with the algorithms introduced in the lectures or plain English where appropriate. Python and/or java code isNOTacceptable.
• Plagiarism detection software will be used to check all submitted solutions. If you include pseudocode from any internet site, your solution may be penalised.
• Full marks will be given if your Algorithm is correct, with appropriate time complexity and is unambiguous the examiner/marker should not have to second guess what you mean.
• Solutions that somehow arrive at what appears to be a correct solution with obvious mistakes and/or ambiguous lines of pseudocode will be penalised.
• Part B: 26 quiz questions worth 40 marks
• The quiz questions can be accessed via the ModulesFinal ExamPart B
• You must complete the quiz questions in the allocated 3-hour time window
### Part A
Question 1 (10 marks).
Design arecursive algorithmto evaluate arithmetic expressions that consist of integer numbers, opening and closing parentheses and the operators +, -, * and /. Here / is integer division.
A well-defined arithmetic expression is defined as:
• a digitd{ 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }is a well-formed arithmetic expression.
• If X and Y are well-formed arithmetic expressions, then the four expressions (X + Y), (X – Y), (X * Y) and (X / Y) are also well-formed arithmetic expressions.
Examples of well-formed arithmetic expressions are ((8+7)*3), (4-(9-2)) and 0. However, the expressions 3+) 2 (()), (9 + ()), -4, 5-8, 108 or (8) are not well-formed arithmetic expressions.
Your algorithm should check whether an input string is a well-formed arithmetic expression. If it is not well-formed then your algorithm must return a constant NOTWELLFORMED, otherwise it must evaluate the expression and return the integer result. Your algorithm does not have to deal with division by 0.
You may assume that the input string is an array or list ofncharacters. You should make sure that your pseudocode matches any assumptions that you make.
Your algorithm must be written in unambiguous pseudocode using a style consistent with the pseudocode introduced on the lecture slides. Python and/or Java code areNOTacceptable for this question.
### Part A
Question 2 (10 marks).
Consider the problem of processing a sequence of real numbersX= (x 1 ,… , xn)of lengthn.
You are required to design anaugmented data structurethat provides (implements) twoO(log n) time operations:
``````INJECT(y, i): inserts itemybetweenxiandxi+1in the data structure
``````
``````SUM(i, j): computes the sum
``````
``````j
``````
``````t=i
``````
``````xtitems in the data structure
``````
``````a. Briefly describe the augmented Data structure based only on a data structure discussed in
the COMP90038 lectures that meets the criteria above.
``````
``````b. Explain what happens when the INJECT() operation is called.
You do not have to write an algorithm in pseudocode to answer part (b) of this question. We
are expecting that you write a couple of sentences or a short list of bullet points describing
the important steps of the INJECT() function.
``````
``````c. Explain what happens when the SUM() operation is called.
You do not have to write an algorithm in pseudocode to answer part (c) of this question. We
are expecting that you write a couple of sentences or a short list of bullet points describing
the important steps of the SUM() function.
``````
### Part A
Question 3 (10 marks).
You are part of a task force put together to trace the evolution of SARS-CoV-2, the virus causing the COVID-19 pandemic. The main goal of your team is to understand the differences between thousands of genomes (represented by large character strings) of strains of the virus collected worldwide to better guide the development of a new drug.
As a COMP90038 student, you realise that a string processing algorithm can be used to help analyse viral evolution. That is, an algorithm can be used to compare and quantify differences between genomes that are represented as a long string of four characters:A,C,G&U.
Over the course of genome evolution, the long character strings change via multiplemutations, which are modifications to the the characters in a given string accumulated over time. Each of these modifications may include one of the following operations (or mutations):
• Substitution of a character with a different character (eg.,Ais replaced by aU).
• Insertion or deletion of one character (eg., given the stringGAUCGthe random deletion of the characterUresults in a new stringGACG). Thus, an insertion or deletion results in variable length genomes.
In order to compare genomes, your challenge is to develop an algorithm tosuperimposeany two genomes in a way that maximises a similarity score based on simulating mutations (or the types of modifications described above). The algorithm uses a 2D-superimposition matrixSto record the similarity scores. That is, an entry in the superimposition matrixS(i,j), denotes the optimal superimposition between theifirst characters of one genome and thejfirst characters of the other.
When maximisingS(i,j), you must consider the following possibilities and corresponding scores:
• MatchesorMismatches
• If the corresponding characters from each of the strings match, a score of+1is at- tributed. This score means that the genome characters did not change during evolution.
• If the corresponding characters from each of the strings do not match, a score of- is attributed. This score means that the at least one of the genome characters was substituted (mutated) during evolution.
• Insertionsordeletions
• Your algorithm can also simulate an insertion or deletion on one of the genomes, with a score of-2attributed.
Two examples are provided to illustrate the superimposing process and the corresponding similar- ity score.
Example 1: providing the stringsCAAGACGandAAGAACGas input, your algorithm should identify the following superposition(by tracing back theSmatrix) to maximise the score.
``````C A A G A - C G
``````
• A A G A A C G
Here, there were 2 insertion/deletions represented by the-character (scoring -4) and 6 matches (scoring 6) resulting in an optimal total score of 2.
Example 2: providing the stringsAACTGAandACTGTas input, your algorithm should identify the following superposition (by tracing back theSmatrix) to maximise the score.
``````A A C T G A
A - C T G T
``````
Here, there was 1 insertion/deletion (scoring -2), 1 mismatch (scoring -1) and 4 matches (scoring 4), resulting in an optimal total score of 1.
Design your algorithm usingdynamic programming. Assuming that the input to your algorithm are two valid character strings (genomes), your algorithm should fill the entries in the matrix S(i,j)and return the optimal score based on the superimposing process describe above.
Your algorithm must be written in unambiguous pseudocode using a style consistent with the pseudocode introduced on the lecture slides. Python and/or Java code areNOTacceptable for this question. | 1,803 | 7,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.754228 |
http://www.convertit.com/Go/weighing-systems/Measurement/Converter.ASP?From=actus&To=height | 1,653,612,026,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00691.warc.gz | 70,230,374 | 4,249 | weighing-systems.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Roman actus = 35.47872 length (length) ``` Related Measurements: Try converting from "actus" to cable length, caliber (gun barrel caliber), digitus (Roman digitus), earth to moon (mean distance earth to moon), ell, fathom, football field, Greek span, m (meter), mil, nautical league, palm, ri (Japanese ri), Roman cubit, Roman foot, Roman mile, sazhen (Russian sazhen), span (cloth span), stadia (Greek stadia), yard, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: actus = 4,190.4 barleycorn, 1.76 chain (surveyors chain), .388 city block (informal), 310.4 cloth finger, 9.23E-08 earth to moon (mean distance earth to moon), 201,893.47 en (typography en), 1.16 engineers chain, 19.4 fathom, 116.4 foot, 106,436.16 French, 3.75E-15 light yr (light year), 35.48 m (meter), .00084083 marathon, .01915698 nautical mile, 8,380.8 pica (typography pica), 100,946.74 point (typography point), 119.88 Roman foot, .02397528 Roman mile, .388 soccer field, .03325714 verst (Russian verst).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 475 | 1,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-21 | latest | en | 0.676526 |
https://www.gradesaver.com/textbooks/math/geometry/CLONE-68e52840-b25a-488c-a775-8f1d0bdf0669/chapter-8-section-8-2-perimeter-and-area-of-polygons-exercises-page-360/9 | 1,723,456,629,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00163.warc.gz | 615,498,280 | 12,312 | ## Elementary Geometry for College Students (6th Edition)
84 $in^{2}$
Given sides of a triangle 13 in, 14 in, 15 in Using Heron's formula if three sides of a triangle have length a,b,c then area A of a triangle is A = $\sqrt s(s-a)(s-b)(s-c)$ where s = $\frac{a+b+c}{2}$ Lets take a=13 b=14 and c=15 s = $\frac{13+14+15}{2}$ = 21 in A = $\sqrt 21(21-13)(21-14)(21-15)$ =$\sqrt 21*8*7*6$ =84 $in^{2}$ | 151 | 400 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.701733 |
https://www.edplace.com/worksheet_info/maths/keystage4/year10/topic/1246/7118/solve-real-life-proportion-questions | 1,726,308,480,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00028.warc.gz | 706,289,298 | 13,047 | # Solve Real Life Proportion Questions
In this worksheet, students will solve real life proportion problems with two or more variables using the methods of applying and manipulating the information provided in the question.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR,
Curriculum topic: Ratio, Proportion and Rates of Change
Curriculum subtopic: Ratio, Proportion and Rates of Change Direct and Inverse Proportion
Difficulty level:
#### Worksheet Overview
One of the most common uses for proportion in maths is to solve real-world problems.
In these, we will be given a fact and asked to use it to find another amount.
Let's look at this in action now.
e.g. If it takes 2 hours to cut 10 m2 of grass, how long does it take to cut 25 m2 of grass?
Step 1: Write out the proportion we already know.
2 hours = 10 m2
Step 2: Find the Highest Common Factor (HCF) of the proportion we already know and the element we are trying to find.
2 hours = 10 m2
1 hour = 5 m2
Note: Whatever we do to one side, we need do to the other which, in this case, this is ÷ 2.
Step 3: Use the information we have found to answer the question.
1 hour = 5 m2
5 hours = 25 m2
So it takes 5 hours to mow 25 m2 of grass, based on the current proportion provided.
e.g. 4 people take 3 hours to clean 60 cars. How long will it take 2 people to clean 80 cars?
This one is a little more complicated as we have three variables present.
Our first step is to write out what we know as an equation:
4 people = 3 hours = 60 cars
The key step to apply here is that we can only change two of the three quantities at once.
We need to keep testing and making changes to two elements at a time, until we have the cars as 80 and the people as 2.
Change 1:
4 people = 3 hours = 60 cars ÷ Both by 3
So if the same amount of people are cleaning one-third of the cars, it will take one-third of the time:
4 people = 1 hour = 20 cars
Change 2:
4 people = 1 hour = 20 cars × Both by 4
So if the same amount of people are cleaning 4 times as many cars, it will take them 4 times as long:
4 people = 4 hours = 80 cars
Change 3:
4 people = 4 hours = 80 cars ÷ People by 2, doubles the hours required
So if half as many people are cleaning the same amount of cars, it will take twice as long:
2 people = 8 hours = 80 cars
So based on the proportion given, it will take 2 people 8 hours to clean 80 cars.
In this activity, we will solve real life proportion problems with two or more variables using the methods shown above of applying and manipulating the information provided in the question.
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started | 772 | 3,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-38 | latest | en | 0.93335 |
https://mail.haskell.org/pipermail/beginners/2013-August.txt | 1,448,716,676,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-48/segments/1448398452560.13/warc/CC-MAIN-20151124205412-00332-ip-10-71-132-137.ec2.internal.warc.gz | 839,436,873 | 47,191 | From voldermort at hotmail.com Thu Aug 1 17:00:59 2013 From: voldermort at hotmail.com (vold) Date: Thu, 1 Aug 2013 15:00:59 +0000 (UTC) Subject: [Haskell-beginners] maybe in IO Message-ID: I've defined a function similar to check x assoc = let found = lookup x assoc in when (isJust found) $putStrLn$ "found " ++ fromJust found which I've used several times from within the IO monad. Is there a more compact way of doing this? From gesh at gesh.uni.cx Thu Aug 1 17:09:11 2013 From: gesh at gesh.uni.cx (Gesh hseG) Date: Thu, 1 Aug 2013 18:09:11 +0300 Subject: [Haskell-beginners] Shorten this code In-Reply-To: References: Message-ID: On Wed, Jul 24, 2013 at 4:18 PM, Henk-Jan van Tuyl wrote: > On Tue, 23 Jul 2013 12:41:32 +0200, Nadav Chernin > wrote: > >> Hi, all >> Below is my solution to SPOJ->Polybius >> square >> >> >> Please try to shorten it: >> >> *import Data.Maybe* >> *d=[1..5]* >> *f s=unwords$map(\c->fromJust$lookup c((' >> ',""):('J',"24"):zip(['A'..'I']++['K'..'Z'])[show(x+10*y)|y<-d,x<-d]))s* >> *main=getLine>>(interact$unlines.map f.lines)* > d="12345" > f s = unwords$ map (\c -> let (Just x) = lookup c ((' ',"") : ('J',"24") : > zip (['A'..'I'] ++ ['K'..'Z']) [y : [x] | y <- d, x <- d]) in x) s > main = getLine >> (interact $unlines . map f . lines) Note that you can point-freeify that solution a little and sacrifice totality, yielding an even shorter solution. (I didn't remove the lambda expression, as it ends up being shorter than flipping lookup) Inline the definitions and remove redundant spaces and newlines in the code below: d = "12345" f = unwords . map$ \c -> fromJust $lookup c ((' ',"") : ('J',"24") : zip (['A'..'Z']\\"J") [[y,x] | y <- d, x <- d]) main = interact$ unlines . map f . drop 1 . lines Just my two cents. Gesh From gtener at gmail.com Thu Aug 1 17:34:15 2013 From: gtener at gmail.com (=?UTF-8?Q?Krzysztof_Skrz=C4=99tnicki?=) Date: Thu, 1 Aug 2013 17:34:15 +0200 Subject: [Haskell-beginners] maybe in IO In-Reply-To: References: Message-ID: I'd rather use pattern matching in this case: check x assoc = case lookup x of Just found -> putStrLn ("found " ++ found) _ -> Not really shorter but I think it's cleaner this way. On Thu, Aug 1, 2013 at 5:00 PM, vold wrote: > I've defined a function similar to > > check x assoc = let found = lookup x assoc in > when (isJust found) $putStrLn$ "found " ++ fromJust > found > > which I've used several times from within the IO monad. Is there a more > compact way of doing this? > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: From gtener at gmail.com Thu Aug 1 17:35:22 2013 From: gtener at gmail.com (=?UTF-8?Q?Krzysztof_Skrz=C4=99tnicki?=) Date: Thu, 1 Aug 2013 17:35:22 +0200 Subject: [Haskell-beginners] maybe in IO In-Reply-To: References: Message-ID: Of course there is "return ()" missing. check x assoc = case lookup x of Just found -> putStrLn ("found " ++ found) _ -> return () On Thu, Aug 1, 2013 at 5:34 PM, Krzysztof Skrz?tnicki wrote: > I'd rather use pattern matching in this case: > > check x assoc = case lookup x of > Just found -> putStrLn ("found " ++ found) > _ -> > > Not really shorter but I think it's cleaner this way. > > > > > On Thu, Aug 1, 2013 at 5:00 PM, vold wrote: > >> I've defined a function similar to >> >> check x assoc = let found = lookup x assoc in >> when (isJust found) $putStrLn$ "found " ++ fromJust >> found >> >> which I've used several times from within the IO monad. Is there a more >> compact way of doing this? >> >> >> _______________________________________________ >> Beginners mailing list >> Beginners at haskell.org >> http://www.haskell.org/mailman/listinfo/beginners >> > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From gesh at gesh.uni.cx Thu Aug 1 17:39:22 2013 From: gesh at gesh.uni.cx (Gesh hseG) Date: Thu, 1 Aug 2013 18:39:22 +0300 Subject: [Haskell-beginners] maybe in IO In-Reply-To: References: Message-ID: On Thu, Aug 1, 2013 at 6:00 PM, vold wrote: > I've defined a function similar to > > check x assoc = let found = lookup x assoc in > when (isJust found) $putStrLn$ "found " ++ fromJust found > > which I've used several times from within the IO monad. Is there a more > compact way of doing this? Firstly, note that it is more idiomatic to move the IO code to other functions, which would allow check to be pure. Secondly, one can use the fact that Maybe is a Functor to rewrite check as follows: check x db = maybeToIO $(putStrLn . ("found"++)) <$> x lookup db where maybeToIO = fromMaybe (return ()) The refactoring of maybeToIO isn't strictly necessary, but I found it clearer. HTH, Gesh From michael at orlitzky.com Thu Aug 1 17:45:02 2013 From: michael at orlitzky.com (Michael Orlitzky) Date: Thu, 01 Aug 2013 11:45:02 -0400 Subject: [Haskell-beginners] maybe in IO In-Reply-To: References: Message-ID: <51FA827E.8070609@orlitzky.com> On 08/01/2013 11:00 AM, vold wrote: > I've defined a function similar to > > check x assoc = let found = lookup x assoc in > when (isJust found) $putStrLn$ "found " ++ fromJust found > > which I've used several times from within the IO monad. Is there a more > compact way of doing this? > maybe (return ()) (putStrLn . ("found " ++)) found is the best I could do. From hjgtuyl at chello.nl Fri Aug 2 00:18:14 2013 From: hjgtuyl at chello.nl (Henk-Jan van Tuyl) Date: Fri, 02 Aug 2013 00:18:14 +0200 Subject: [Haskell-beginners] Shorten this code In-Reply-To: References: Message-ID: On Thu, 01 Aug 2013 17:09:11 +0200, Gesh hseG wrote: > On Wed, Jul 24, 2013 at 4:18 PM, Henk-Jan van Tuyl > wrote: >> d="12345" >> f s = unwords $map (\c -> let (Just x) = lookup c ((' ',"") : >> ('J',"24") : >> zip (['A'..'I'] ++ ['K'..'Z']) [y : [x] | y <- d, x <- d]) in x) s >> main = getLine >> (interact$ unlines . map f . lines) > > Note that you can point-freeify that solution a little and sacrifice > totality, > yielding an even shorter solution. (I didn't remove the lambda > expression, > as it ends up being shorter than flipping lookup) > Inline the definitions and remove redundant spaces and newlines > in the code below: > d = "12345" > f = unwords . map $> \c -> fromJust$ lookup c ((' ',"") : ('J',"24") : zip > (['A'..'Z']\\"J") [[y,x] | y <- d, x <- d]) > main = interact $unlines . map f . drop 1 . lines There are imports necessary for fromJust and \\, this makes it longer again. Regards, Henk-Jan van Tuyl -- Folding at home What if you could share your unused computer power to help find a cure? In just 5 minutes you can join the world's biggest networked computer and get us closer sooner. Watch the video. http://folding.stanford.edu/ http://Van.Tuyl.eu/ http://members.chello.nl/hjgtuyl/tourdemonad.html Haskell programming -- From chris at christopher-schneider.com Mon Aug 5 06:05:47 2013 From: chris at christopher-schneider.com (Chris Schneider) Date: Sun, 4 Aug 2013 22:05:47 -0600 Subject: [Haskell-beginners] Threading (lack of) interleaving Message-ID: Hi, I'm working on a program that should launch, spawn two threads, each with a zeromq connection, where those threads ping the zmq channel repeatedly. The main thread just sits and waits. The code is here: https://gist.github.com/cschneid/3862b33c29a803be7848 If I launch one thread, or the other (lines 13 & 14), and comment the other one out, it runs as expected. But if I attempt to launch both, both build themselves, and run one call of themselves, but then it seems that one or the other thread "wins", and the other never gets to run again. Most of the time it's the first one launched that wins, but not always. See the output for an example. Adding explicit yield calls, using forkOS vs forkIO, and moving around of where I make the ZMQ context & socket (into and out of the threads) don't seem to help. What am I missing? How can I get these threads to play nice with each other? Thanks! -------------- next part -------------- An HTML attachment was scrubbed... URL: From chris at christopher-schneider.com Mon Aug 5 17:11:04 2013 From: chris at christopher-schneider.com (Chris Schneider) Date: Mon, 5 Aug 2013 09:11:04 -0600 Subject: [Haskell-beginners] Threading (lack of) interleaving In-Reply-To: References: Message-ID: Ok, I've figured out my problem - it wasn't threading related at all - instead it was the second thread dying off due to a ZMQ problem. Specifically I was reusing the node name, which apparently isn't allowed. Tweaking it such that it uses a different name for each fixes the issue. On Sun, Aug 4, 2013 at 10:05 PM, Chris Schneider < chris at christopher-schneider.com> wrote: > Hi, I'm working on a program that should launch, spawn two threads, each > with a zeromq connection, where those threads ping the zmq channel > repeatedly. The main thread just sits and waits. > > The code is here: https://gist.github.com/cschneid/3862b33c29a803be7848 > > If I launch one thread, or the other (lines 13 & 14), and comment the > other one out, it runs as expected. > > But if I attempt to launch both, both build themselves, and run one call > of themselves, but then it seems that one or the other thread "wins", and > the other never gets to run again. Most of the time it's the first one > launched that wins, but not always. See the output for an example. > > Adding explicit yield calls, using forkOS vs forkIO, and moving around of > where I make the ZMQ context & socket (into and out of the threads) don't > seem to help. > > What am I missing? How can I get these threads to play nice with each > other? > > Thanks! > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From martin.drautzburg at web.de Mon Aug 5 21:37:18 2013 From: martin.drautzburg at web.de (martin) Date: Mon, 05 Aug 2013 21:37:18 +0200 Subject: [Haskell-beginners] How to construct complex string Message-ID: <51FFFEEE.5050209@web.de> Hello all, recently I was working on a haskell program, which does some calculations and the produces a "plot", i.e. a set of Tikz (a LaTeX thing) commands. While this was not a problem in itself, it lacked composability. While all my plots show a grid, a caption and some data points, some of them additionally show a line with a legend, some of them show two. Now I clearly do not want to duplicate the entire function just to add another line. I'd rather construct the Tikz commands (which at the end of the day is just a String) in a stepwise fashion. When constructing the set of Tikz commands, I cannot rely that additional commands are placed at the end. It is typically more an inside-out thing. The last thing I need to do is wrap the whole thing in \begin and \end directives. Additionally there may be "global" values (like the scale) which need to be known at more than one step. I had a brief look at the "Diagrams" package, which MUST have similar issues, and they "do everything with monoids", but I fail to see the light. Could anyone point me in the right direction? From stephen.tetley at gmail.com Tue Aug 6 19:01:45 2013 From: stephen.tetley at gmail.com (Stephen Tetley) Date: Tue, 6 Aug 2013 18:01:45 +0100 Subject: [Haskell-beginners] How to construct complex string In-Reply-To: <51FFFEEE.5050209@web.de> References: <51FFFEEE.5050209@web.de> Message-ID: Hi Martin If you are building syntax as a sequence of commands (or statements) a common idiom is to collect them with a Writer monad. Andy Gill's Dotgen on Hackage is a nice realization of this idiom. See the how example included in the package has acceptably nice syntax without obliging the implementation to use complex facilities such as Template Haskell. In the source code, the data type GraphElement represents the main statement types in GraphViz's "dot" language. The data type Dot is the monad - here a combination of a Writer monad and a State monad, with the State monad supplying unique integers for fresh identifiers. The monad is written as a direct combination of State and Writer (it is perhaps more common to rely on _monad transformers_ but once you know what you are looking at, Andy's code is simple and clear). http://hackage.haskell.org/package/dotgen Best wishes Stephen -------------- next part -------------- An HTML attachment was scrubbed... URL: From voldermort at hotmail.com Wed Aug 7 09:23:34 2013 From: voldermort at hotmail.com (vold) Date: Wed, 7 Aug 2013 07:23:34 +0000 (UTC) Subject: [Haskell-beginners] Why is my GHC bigger than the distribution Message-ID: I've built GHC 7.6.3 from source using the default "perf" settings, and all my programs are about 4 Mb bigger after linking than if I used the pre-built binary. What could be causing this? (x64 Linux with gold linker.) From voldermort at hotmail.com Wed Aug 7 09:33:10 2013 From: voldermort at hotmail.com (vold) Date: Wed, 7 Aug 2013 07:33:10 +0000 (UTC) Subject: [Haskell-beginners] Why is my GHC bigger than the distribution References: Message-ID: vold hotmail.com> writes: Sorry, that wasn't completely clear. It's both the GHC binary itself and everything build by GHC which is a few MB bigger. From voldermort at hotmail.com Wed Aug 7 10:00:15 2013 From: voldermort at hotmail.com (vold) Date: Wed, 7 Aug 2013 08:00:15 +0000 (UTC) Subject: [Haskell-beginners] Why is my GHC bigger than the distribution References: Message-ID: vold hotmail.com> writes: Now I've noticed something even weirder - if I replace the GHC binary from the binary distribution with the one I built, the executables I produce with it are the original size. From chris at christopher-schneider.com Wed Aug 7 19:04:15 2013 From: chris at christopher-schneider.com (Chris Schneider) Date: Wed, 7 Aug 2013 11:04:15 -0600 Subject: [Haskell-beginners] Showing a value, when there are many approaches? Message-ID: Hi, I'm writing a module that contains a data type for a MacAddress. Given that, I can do things like validate and format. But there are several ways you can display a MacAddress: 'aabbccddeeff' or '00:11:22:33:AA:BB' or 'aa:bb:cc:00:11:22'. Different external systems want the mac addresses in different formats, and I'd like to centralize that logic. ** So the design question: I can't just use a show function directly to do this, since there are options I need to pass. I could use separate functions that return a string. I could use newtype to wrap each display style of mac address in a type that has a specific show instance that spits out the right format. So a Pretty and Lower and similar types. --- Is there a best-practice here? Am I totally missing a better way to model this idea? -------------- next part -------------- An HTML attachment was scrubbed... URL: From toad3k at gmail.com Wed Aug 7 19:35:03 2013 From: toad3k at gmail.com (David McBride) Date: Wed, 7 Aug 2013 13:35:03 -0400 Subject: [Haskell-beginners] Showing a value, when there are many approaches? In-Reply-To: References: Message-ID: Show is not meant to necessarily be human readable. It is meant to be the exact opposite of Read. You show it, then you read it back in and it is unambiguous. Having different ways to show it kind of defeats the purpose of show. Most people just make a custom function that takes a macaddress and some other parameter to determine which format you want it to be output as and then just use that. On Wed, Aug 7, 2013 at 1:04 PM, Chris Schneider < chris at christopher-schneider.com> wrote: > Hi, I'm writing a module that contains a data type for a MacAddress. > > Given that, I can do things like validate and format. > > But there are several ways you can display a MacAddress: > > 'aabbccddeeff' or '00:11:22:33:AA:BB' or 'aa:bb:cc:00:11:22'. > > Different external systems want the mac addresses in different formats, > and I'd like to centralize that logic. > > ** So the design question: > > I can't just use a show function directly to do this, since there are > options I need to pass. > > I could use separate functions that return a string. > > I could use newtype to wrap each display style of mac address in a type > that has a specific show instance that spits out the right format. So a > Pretty and Lower and similar types. > > --- > > Is there a best-practice here? Am I totally missing a better way to model > this idea? > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From mike at fpcomplete.com Wed Aug 7 20:02:39 2013 From: mike at fpcomplete.com (Mike Meyer) Date: Wed, 7 Aug 2013 13:02:39 -0500 Subject: [Haskell-beginners] Looking for judges for haskell competition Message-ID: In order to foster commercial use of Haskell, FP Complete is running a competition for monthly cash prizes (details https://www.fpcomplete.com/blog/2013/07/competition-announcement). We are looking for judges from the community. People helping beginners would seem to be a good fit for this. If you're interested, could you please drop me a note along with pertinent information? Thanks, Mike -------------- next part -------------- An HTML attachment was scrubbed... URL: From mike at fpcomplete.com Wed Aug 7 20:16:44 2013 From: mike at fpcomplete.com (Mike Meyer) Date: Wed, 7 Aug 2013 13:16:44 -0500 Subject: [Haskell-beginners] Enterprise-quality XML library? Message-ID: I've been looking for an enterprise-grade XML library. My standard for comparison is Python's lxml library (a wrapper around libxml2 and libxslt). The only thing I've found that's even close is hxt, and it falls a bit short. Specifically, when I feed it invalid XML, I get this error: error: The content of element "school" must match ( "author"* ). Element "author" expected, but element "auther" found. error: Element "auther" not declared in DTD. Whereas from lxml I get: lxml.etree.XMLSyntaxError: No declaration for attribute name of element auther, line 24, column 19 This is shorter and clearer, and identifies the location of the error. It does have better error messages for ill-formed XML. Is there a library with a validating parser that provides better quality error messages? It also seems like mapping an XML Schema into a collection of Haskell types should be a function of the library. Is there on that does that, or is this completely untenable? -------------- next part -------------- An HTML attachment was scrubbed... URL: From allbery.b at gmail.com Wed Aug 7 20:51:15 2013 From: allbery.b at gmail.com (Brandon Allbery) Date: Wed, 7 Aug 2013 14:51:15 -0400 Subject: [Haskell-beginners] Showing a value, when there are many approaches? In-Reply-To: References: Message-ID: On Wed, Aug 7, 2013 at 1:04 PM, Chris Schneider < chris at christopher-schneider.com> wrote: > I can't just use a show function directly to do this, since there are > options I need to pass. > And you shouldn't use show for this, because if it prints your MacAddress as anything other than an unambiguous representation of the Haskell MacAddress type then it has lost information useful for debugging. More to the point: if I cannot see the difference between your MacAddress and a String, then it's an unacceptable show. show should reflect the Haskell data structure so I can verify that my program is doing the right thing; if it shows what looks like a String, but I'm expecting to see a MacAddress, then I will suspect a type error somewhere. show is like Perl's Data::Dumper or Python's repr. I don't want pretty, I want accuracy. -- brandon s allbery kf8nh sine nomine associates allbery.b at gmail.com ballbery at sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net -------------- next part -------------- An HTML attachment was scrubbed... URL: From byorgey at seas.upenn.edu Wed Aug 7 21:15:12 2013 From: byorgey at seas.upenn.edu (Brent Yorgey) Date: Wed, 7 Aug 2013 15:15:12 -0400 Subject: [Haskell-beginners] How to construct complex string In-Reply-To: <51FFFEEE.5050209@web.de> References: <51FFFEEE.5050209@web.de> Message-ID: <20130807191512.GA7929@seas.upenn.edu> On Mon, Aug 05, 2013 at 09:37:18PM +0200, martin wrote: > Hello all, > > recently I was working on a haskell program, which does some > calculations and the produces a "plot", i.e. a set of Tikz (a LaTeX > thing) commands. > > While this was not a problem in itself, it lacked composability. While > all my plots show a grid, a caption and some data points, some of them > additionally show a line with a legend, some of them show two. > > Now I clearly do not want to duplicate the entire function just to add > another line. I'd rather construct the Tikz commands (which at the end > of the day is just a String) in a stepwise fashion. > > When constructing the set of Tikz commands, I cannot rely that > additional commands are placed at the end. It is typically more an > inside-out thing. The last thing I need to do is wrap the whole thing in > \begin and \end directives. Hi Martin, I think the key thing you are missing is that a String is just a particular concrete representation of a TikZ program. That is not what a TikZ program really *is*. Instead, TikZ programs are trees. String is nice for storing a program in a file, but it is a terrible representation to use when working with programs, whether generating them, type-checking them, or whatever. As you have discovered, when working with Strings you have to insert stuff at various arbitrary places (at the beginning, the end, the middle), which is quite annoying and difficult (especially the middle!), whereas those same operations are simple on a tree representation of the program. To do what you are trying to do in a nice way would require (1) making an algebraic data type to represent the structure of TikZ programs (2) changing your program so that it generates values of this type, instead of Strings, and (3) making a single function to convert a value of this type into a String. This can be quite a bit of work up front, but well worth it if you are going to do more than just a bit of TikZ generation. With that said, however, did you know there is a TikZ backend for diagrams? [1] It should be possible for you to describe your plots using diagrams and then output them to TikZ (or any other format for which diagrams has a backend---SVG, PDF, PS, or PNG). I have not looked at it recently but would be happy to support it if you run into any problems (I will probably want it myself sooner or later). [1] http://hackage.haskell.org/package/diagrams%2Dtikz -Brent > > Additionally there may be "global" values (like the scale) which need to > be known at more than one step. I had a brief look at the "Diagrams" > package, which MUST have similar issues, and they "do everything with > monoids", but I fail to see the light. > > Could anyone point me in the right direction? > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners From chris at christopher-schneider.com Wed Aug 7 21:47:55 2013 From: chris at christopher-schneider.com (Chris Schneider) Date: Wed, 7 Aug 2013 13:47:55 -0600 Subject: [Haskell-beginners] Showing a value, when there are many approaches? In-Reply-To: References: Message-ID: Ok, that makes sense, I misinterpreted show since for integers & friends it is just the same as a toString in other languages. Is there a standard name for "render a human readable form of this data"? toString() in java, to_s in ruby, and similar? On Wed, Aug 7, 2013 at 12:51 PM, Brandon Allbery wrote: > On Wed, Aug 7, 2013 at 1:04 PM, Chris Schneider < > chris at christopher-schneider.com> wrote: > >> I can't just use a show function directly to do this, since there are >> options I need to pass. >> > > And you shouldn't use show for this, because if it prints your > MacAddress as anything other than an unambiguous representation of the > Haskell MacAddress type then it has lost information useful for debugging. > More to the point: if I cannot see the difference between your MacAddress > and a String, then it's an unacceptable show. show should reflect the > Haskell data structure so I can verify that my program is doing the right > thing; if it shows what looks like a String, but I'm expecting to see a > MacAddress, then I will suspect a type error somewhere. > > show is like Perl's Data::Dumper or Python's repr. I don't want pretty, > I want accuracy. > > -- > brandon s allbery kf8nh sine nomine > associates > allbery.b at gmail.com > ballbery at sinenomine.net > unix, openafs, kerberos, infrastructure, xmonad > http://sinenomine.net > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From martin.drautzburg at web.de Thu Aug 8 22:38:25 2013 From: martin.drautzburg at web.de (martin) Date: Thu, 08 Aug 2013 22:38:25 +0200 Subject: [Haskell-beginners] How to construct complex string In-Reply-To: References: <51FFFEEE.5050209@web.de> Message-ID: <520401C1.6070504@web.de> Am 08/06/2013 07:01 PM, schrieb Stephen Tetley: > Hi Martin > > If you are building syntax as a sequence of commands (or statements) a > common idiom is to collect them with a Writer monad. > > Andy Gill's Dotgen on Hackage is a nice realization of this idiom. See > the how example included in the package has acceptably nice syntax > without obliging the implementation to use complex facilities such as > Template Haskell. Thanks Stephen, exactly the nudge I was looking for, except I could not find any examples in the package. I had a brief look at the source code and it did not instantly reveal itself. Actually I already got stuck here: data Dot a = Dot { unDot :: Int -> ([GraphElement],Int,a) } Thanks Brent for you detailed explanation. From stephen.tetley at gmail.com Fri Aug 9 00:18:28 2013 From: stephen.tetley at gmail.com (Stephen Tetley) Date: Thu, 8 Aug 2013 23:18:28 +0100 Subject: [Haskell-beginners] How to construct complex string In-Reply-To: <520401C1.6070504@web.de> References: <51FFFEEE.5050209@web.de> <520401C1.6070504@web.de> Message-ID: Hi Martin The example is actually in the test directory - DotTest.hs. > data Dot a = Dot { unDot :: Int -> ([GraphElement],Int,a) } This is the monad in which you build syntax - when the code is run it is pretty printed into a String. This particular monad is a combination of the Writer monad - it accumulates a list of GraphicElement and a state monad - Int is passed around and incremented so it can generate variable names: As separate pieces the monads are: > data State a = State { unState :: Int -> (Int, a) } i.e. a function from state of type Int to a tuple of state and a polymorphic answer (Int,a). > data Writer a = Writer { unWriter :: ([GraphElement],a) } i.e. a tuple of GraphElement's accumulated in a list and a polymorphic answer. Once you have a "declaration" of a monad you also have to provide an instance of the monad class so you can use the do-notation, in the Dotgen code the instance is: instance Monad Dot where return a = Dot$ \ uq -> ([],uq,a) m >>= k = Dot $\ uq -> case unDot m uq of (g1,uq',r) -> case unDot (k r) uq' of (g2,uq2,r2) -> (g1 ++ g2,uq2,r2) This is a bit complicated as it is a combination of the monadic operation of both the State and Writer monad. With luck for just generating commands (and not generating fresh variables) you should be able to manage with only a Writer monad. There is one already in the monads package (mtl) in the Platform. The main part of the Dotgen library are the functions edge, node, (.->.) etc. which build Dot commands. In edge you can see a single command being built [GraphEdge from to attrs] in the first cell of the answer tuple: -- | 'edge' generates an edge between two 'NodeId's, with attributes. edge :: NodeId -> NodeId -> [(String,String)] -> Dot () edge from to attrs = Dot (\ uq -> ( [ GraphEdge from to attrs ],uq,())) Because no fresh variables are generated, the state uq (read as "unique") is returned unaltered in the answer tuple along with the void answer (). The other key part of the library is the so-called "run function" which evaluates monadic expressions - here the run function is called showDot: -- 'showDot' renders a dot graph as a 'String'. showDot :: Dot a -> String showDot (Dot dm) = case dm 0 of (elems,_,_) -> "digraph G {\n" ++ unlines (map showGraphElement elems) ++ "\n}\n" showDot unwraps dm from the Dot data type. As dm is a function Int -> ([GraphElement],Int,a) it needs to be applied to the initial state - here 0. This produces the answer 3-tuple, where we are only interested in the accumulated list of dot commands (the first cell). This list of commands is pretty printed into legal Dot syntax. Hope this helps somewhat. -------------- next part -------------- An HTML attachment was scrubbed... URL: From eleventynine at gmail.com Fri Aug 9 08:38:37 2013 From: eleventynine at gmail.com (Mike Ledger) Date: Fri, 9 Aug 2013 16:38:37 +1000 Subject: [Haskell-beginners] maybe in IO In-Reply-To: <51FA827E.8070609@orlitzky.com> References: <51FA827E.8070609@orlitzky.com> Message-ID: With Data.Foldable, it's quite nice: check x assoc = lookup x assoc for_ \f -> putStrLn ("found " ++ f) On Fri, Aug 2, 2013 at 1:45 AM, Michael Orlitzky wrote: > On 08/01/2013 11:00 AM, vold wrote: > > I've defined a function similar to > > > > check x assoc = let found = lookup x assoc in > > when (isJust found)$ putStrLn $"found " ++ > fromJust found > > > > which I've used several times from within the IO monad. Is there a more > > compact way of doing this? > > > > maybe (return ()) (putStrLn . ("found " ++)) found > > is the best I could do. > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: From k.bleijenberg at lijbrandt.nl Fri Aug 9 12:00:42 2013 From: k.bleijenberg at lijbrandt.nl (Kees Bleijenberg) Date: Fri, 9 Aug 2013 12:00:42 +0200 Subject: [Haskell-beginners] alternative sourcegraph Message-ID: <000001ce94e7$4f2516b0$ed6f4410$@bleijenberg@lijbrandt.nl> I want to remove 'dead' code and unnecessary exports out of modules. I tried sourceGraph with sourceGraph tbGlasDll.hs and I got: Could not parse source file ./Toetsing.hs Could not parse source file ./TestGlas.hs Could not parse source file ./TBGlasDll.hs Could not parse source file ./Glas.hs Could not parse source file ./CalcGlas.hs Could not parse source file ./calc.hs Could not parse source file ./BelastingFactoren.hs sourceGraph: Map.findMin: empty map has no minimal element The files above compile without errors with ghc. Maybe it has to do with Template Haskell code? Is there an alternative to sourceGraph? Kees -------------- next part -------------- An HTML attachment was scrubbed... URL: From byorgey at seas.upenn.edu Fri Aug 9 13:29:29 2013 From: byorgey at seas.upenn.edu (Brent Yorgey) Date: Fri, 9 Aug 2013 07:29:29 -0400 Subject: [Haskell-beginners] alternative sourcegraph In-Reply-To: <000001ce94e7$4f2516b0$ed6f4410$@bleijenberg@lijbrandt.nl> References: <000001ce94e7$4f2516b0$ed6f4410$@bleijenberg@lijbrandt.nl> Message-ID: <20130809112929.GA17085@seas.upenn.edu> On Fri, Aug 09, 2013 at 12:00:42PM +0200, Kees Bleijenberg wrote: > I want to remove 'dead' code and unnecessary exports out of modules. I tried > sourceGraph with sourceGraph tbGlasDll.hs and I got: > > Could not parse source file ./Toetsing.hs > > Could not parse source file ./TestGlas.hs > > Could not parse source file ./TBGlasDll.hs > > Could not parse source file ./Glas.hs > > Could not parse source file ./CalcGlas.hs > > Could not parse source file ./calc.hs > > Could not parse source file ./BelastingFactoren.hs > > sourceGraph: Map.findMin: empty map has no minimal element > > The files above compile without errors with ghc. > > Maybe it has to do with Template Haskell code? Is there an alternative to > sourceGraph? I don't know of any alternatives to sourceGraph. Why don't you file a bug report with the maintainer? -Brent From martin.drautzburg at web.de Fri Aug 9 19:41:57 2013 From: martin.drautzburg at web.de (martin) Date: Fri, 09 Aug 2013 19:41:57 +0200 Subject: [Haskell-beginners] How to construct complex string In-Reply-To: References: <51FFFEEE.5050209@web.de> <520401C1.6070504@web.de> Message-ID: <520529E5.9020901@web.de> Am 08/09/2013 12:18 AM, schrieb Stephen Tetley: > Hi Martin > instance Monad Dot where > return a = Dot $\ uq -> ([],uq,a) > m >>= k = Dot$ \ uq -> case unDot m uq of > (g1,uq',r) -> case unDot (k r) uq' of > (g2,uq2,r2) -> (g1 ++ g2,uq2,r2) Thanks Stephen I have a related question to the one above: The "case .. of" is just required to bind the variables left of the "->", as there is only one alternative, right? I tried to rewrite this using "where" and was bitten badly, because inside the "where" uq would not be defined. I found a way to write (a simplified version of) this using "let .. in", but I had to put the entire "let .. in" after the "->" in the lambda and it ended up equally ugly. So my question is: is "case .. of" the the "where" of lambdas? Are there alternatives? I find this confusing because "case .. of" looks like a decision between alternatives, rather than a way to bind variables. From stephen.tetley at gmail.com Fri Aug 9 20:27:44 2013 From: stephen.tetley at gmail.com (Stephen Tetley) Date: Fri, 9 Aug 2013 19:27:44 +0100 Subject: [Haskell-beginners] How to construct complex string In-Reply-To: <520529E5.9020901@web.de> References: <51FFFEEE.5050209@web.de> <520401C1.6070504@web.de> <520529E5.9020901@web.de> Message-ID: Hi Martin I think case ... of here is equivalent to let ... in the only difference is that it can avoid growing rightwards. The declaration and body of let would have to be indented to the same level as the let keyword. With case ... of the consequent expression doesn't have to be indented to the level of case provided it is on a new line. I don't think I've ever noticed this "typographical trick" of a single case _case_ in other code. -------------- next part -------------- An HTML attachment was scrubbed... URL: From martin.drautzburg at web.de Fri Aug 9 21:29:32 2013 From: martin.drautzburg at web.de (martin) Date: Fri, 09 Aug 2013 21:29:32 +0200 Subject: [Haskell-beginners] How to construct complex string In-Reply-To: References: <51FFFEEE.5050209@web.de> <520401C1.6070504@web.de> <520529E5.9020901@web.de> Message-ID: <5205431C.5040609@web.de> Am 08/09/2013 08:27 PM, schrieb Stephen Tetley: > I don't think I've ever noticed this "typographical trick" of a single > case _case_ in other code. Are there other ways of doing this? What do you guys do, when you are tempted to write "where" but you are behind the "->" of a lambda and you want to refer to the arguments? From ky3 at atamo.com Fri Aug 9 23:11:17 2013 From: ky3 at atamo.com (Kim-Ee Yeoh) Date: Sat, 10 Aug 2013 04:11:17 +0700 Subject: [Haskell-beginners] How to construct complex string In-Reply-To: <520529E5.9020901@web.de> References: <51FFFEEE.5050209@web.de> <520401C1.6070504@web.de> <520529E5.9020901@web.de> Message-ID: On Sat, Aug 10, 2013 at 12:41 AM, martin wrote: > > instance Monad Dot where > > return a = Dot $\ uq -> ([],uq,a) > > m >>= k = Dot$ \ uq -> case unDot m uq of > > (g1,uq',r) -> case unDot (k r) uq' of > > (g2,uq2,r2) -> (g1 ++ g2,uq2,r2) > > I tried to rewrite this using "where" and was bitten badly, because > inside the "where" uq would not be defined. > You could write: m >>= k = Dot f where f uq = (g1 ++ g2, uq2, r2) where (g1, uq', r) = unDot m uq (g2, uq2, r2) = unDot (k r) uq' Doubtless, there has to be a way of revealing the inner structure, which looks like some state monad with icing on top. -- Kim-Ee -------------- next part -------------- An HTML attachment was scrubbed... URL: From psismondi at arqux.com Sat Aug 10 16:22:06 2013 From: psismondi at arqux.com (Philippe Sismondi) Date: Sat, 10 Aug 2013 10:22:06 -0400 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment Message-ID: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> I am interested in tracking down discussions on how Haskell fits into an operating system environment. That is probably not a very clear question, so let me explain. One of the great obstacles I have faced in using two of my favourite languages (Common Lisp and Haskell) for anything practical is that I am constantly frustrated by their relationship with the host OS, or perhaps more accurately with the APIs provided by the OS. For example, I tend to program mostly on OS X. I have a reasonable knowledge of Cocoa, and can get quite a lot done with Objective-C. So, if I want to use Haskell on OS X, I can run off and learn about Haskell's FFI, or some hand-rolled widget package, etc. But here is the problem: Cocoa et al provide an entire ecosystem that includes data structures, error handling, and so on. So, when trying to use Haskell with a GUI, or perhaps to do some audio processing, or other things that are provided by the OS X APIs, I feel that I am layering two quite incompatible worlds together: Haskell and its host environment. I ran into the same dilemma with Common Lisp. Clozure Common Lisp provides an excellent Cocoa binding. But using it feels (to me) like communications between alien races. The OS APIs want to do things the Objective-C way, and that is (clearly) not the Haskell or Lisp way. One soon notices that books on Haskell pay virtually no attention to this issue. At least, the ones I have don't. Maybe what I am really experiencing is that OS X just "wants" to be programmed in its native language. I had something of a better feeling about using F# on Windows, because that architecture does leverage the native stuff much better. The upshot is that I remain skeptical that Haskell is actually a practical language for many kinds of development. Can anyone point me to discussions on this? I may be having a dimwitted interval here, but I don't know what to search to find this. - P - From magnus at therning.org Sat Aug 10 22:06:02 2013 From: magnus at therning.org (Magnus Therning) Date: Sat, 10 Aug 2013 22:06:02 +0200 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> Message-ID: <20130810200602.GA7446@mteis.lan> On Sat, Aug 10, 2013 at 10:22:06AM -0400, Philippe Sismondi wrote: > I am interested in tracking down discussions on how Haskell fits > into an operating system environment. That is probably not a very > clear question, so let me explain. > > One of the great obstacles I have faced in using two of my favourite > languages (Common Lisp and Haskell) for anything practical is that I > am constantly frustrated by their relationship with the host OS, or > perhaps more accurately with the APIs provided by the OS. > > For example, I tend to program mostly on OS X. I have a reasonable > knowledge of Cocoa, and can get quite a lot done with Objective-C. > So, if I want to use Haskell on OS X, I can run off and learn about > Haskell's FFI, or some hand-rolled widget package, etc. But here is > the problem: Cocoa et al provide an entire ecosystem that includes > data structures, error handling, and so on. So, when trying to use > Haskell with a GUI, or perhaps to do some audio processing, or other > things that are provided by the OS X APIs, I feel that I am layering > two quite incompatible worlds together: Haskell and its host > environment. > > I ran into the same dilemma with Common Lisp. Clozure Common Lisp > provides an excellent Cocoa binding. But using it feels (to me) like > communications between alien races. The OS APIs want to do things > the Objective-C way, and that is (clearly) not the Haskell or Lisp > way. I'm not sure there are a lot of discussions or texts on this issue, but maybe you can find some solace in me thinking that this problem pops up whenever one tries to combine components written in different languages. For instance C++ is able to call straight to C libraries, but that doesn't mean that C APIs fit well into "idiomatic C++" (or maybe I should say one of the many "idiomatic C++s"). The same goes for all other combinations I've tried out myself, Python+C, Python+C++, Java+C, Haskell+C... I'm fairly sure it is the case with *every* combination. I'm also fairly sure that the friction becomes higher the more different the languages are, i.e. combining two procedural (non-OO) languages is most likely easier than combining a procedural (non-OO) with an OO language, even if both are imperative. If the languages have different paradigms, e.g. imperative vs. functional, the friction becomes higher still. > One soon notices that books on Haskell pay virtually no attention to > this issue. At least, the ones I have don't. I believe the FFI is touched upon, but providing an idiomatic Haskell API for a C library is most likely an art (maybe a black one ;) at this point. So in short, you are right. (I'd love to be proven wrong here and be pointed to some recipe book for how to wrap C in Haskell, or C in Python, and do it well.) /M -- Magnus Therning OpenPGP: 0xAB4DFBA4 email: magnus at therning.org jabber: magnus at therning.org twitter: magthe http://therning.org/magnus I invented the term Object-Oriented, and I can tell you I did not have C++ in mind. -- Alan Kay -------------- next part -------------- A non-text attachment was scrubbed... Name: not available Type: application/pgp-signature Size: 230 bytes Desc: not available URL: From orblivion at gmail.com Sun Aug 11 02:16:58 2013 From: orblivion at gmail.com (Dan Krol) Date: Sat, 10 Aug 2013 17:16:58 -0700 Subject: [Haskell-beginners] HTTP Download -> Save File - non-strict Message-ID: Hi, I'm working on an rss file getter. I was wondering if I could get some help getting files to download and save without holding the entire file in memory in between. I chose Conduit's version of SimpleHttp only because it was recommended, and it was the quickest thing I could get to work correctly because I was eager to get started on this project, so I'd be happy to switch. Here's where I define the download and save functions: https://github.com/orblivion/feedGetter/blob/master/rss.hs#L107 And here's where I use them, getting multiple at a time with async: https://github.com/orblivion/feedGetter/blob/master/rss.hs#L208 What happens when I run this is that it outputs that it's "Getting" the file, waits a while (presumably to download the whole thing), then says it's "Saving". And I checked the file system, it's not there during the pause. I'm not entirely sure why. Is it my choice of libraries, or the way I'm using them? Perhaps something to do with async? I just tried content <- simpleHttp "http://google.com" in ghci, and it does pause for a second, so I'm guessing this is strict from the getgo. But I've done almost no I/O before. Is there a straightforward, canonical option? It seems like there perhaps should be. But if it comes down to using pipes or conduit, what the heck I'll try it out, I'd like to learn pipes eventually. Thanks a lot, Dan -------------- next part -------------- An HTML attachment was scrubbed... URL: From magnus at therning.org Sun Aug 11 08:30:30 2013 From: magnus at therning.org (Magnus Therning) Date: Sun, 11 Aug 2013 08:30:30 +0200 Subject: [Haskell-beginners] HTTP Download -> Save File - non-strict In-Reply-To: References: Message-ID: <20130811063030.GA1513@mteis.lan> On Sat, Aug 10, 2013 at 05:16:58PM -0700, Dan Krol wrote: > Hi, > > I'm working on an rss file getter. I was wondering if I could get > some help getting files to download and save without holding the > entire file in memory in between. I chose Conduit's version of > SimpleHttp only because it was recommended, and it was the quickest > thing I could get to work correctly because I was eager to get > started on this project, so I'd be happy to switch. > > Here's where I define the download and save functions: > > https://github.com/orblivion/feedGetter/blob/master/rss.hs#L107 > > And here's where I use them, getting multiple at a time with async: > > https://github.com/orblivion/feedGetter/blob/master/rss.hs#L208 > > What happens when I run this is that it outputs that it's "Getting" > the file, waits a while (presumably to download the whole thing), > then says it's "Saving". And I checked the file system, it's not > there during the pause. I'm not entirely sure why. Is it my choice > of libraries, or the way I'm using them? Perhaps something to do > with async? I just tried content <- simpleHttp "http://google.com" > in ghci, and it does pause for a second, so I'm guessing this is > strict from the getgo. But I've done almost no I/O before. > > Is there a straightforward, canonical option? It seems like there > perhaps should be. But if it comes down to using pipes or conduit, > what the heck I'll try it out, I'd like to learn pipes eventually. Michael is very good with documenting his packages, this is what I found in the docs for http-conduit (http://is.gd/WkDb7G): Note: Even though this function returns a lazy bytestring, it does not utilize lazy I/O, and therefore the entire response body will live in memory. If you want constant memory usage, you'll need to use the conduit package and http directly. /M -- Magnus Therning OpenPGP: 0xAB4DFBA4 email: magnus at therning.org jabber: magnus at therning.org twitter: magthe http://therning.org/magnus I invented the term Object-Oriented, and I can tell you I did not have C++ in mind. -- Alan Kay -------------- next part -------------- A non-text attachment was scrubbed... Name: not available Type: application/pgp-signature Size: 230 bytes Desc: not available URL: From apfelmus at quantentunnel.de Sun Aug 11 10:58:15 2013 From: apfelmus at quantentunnel.de (Heinrich Apfelmus) Date: Sun, 11 Aug 2013 10:58:15 +0200 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> Message-ID: Philippe Sismondi wrote: > I am interested in tracking down discussions on how Haskell fits into > an operating system environment. That is probably not a very clear > question, so let me explain. > > One of the great obstacles I have faced in using two of my favourite > languages (Common Lisp and Haskell) for anything practical is that I > am constantly frustrated by their relationship with the host OS, or > perhaps more accurately with the APIs provided by the OS. > > For example, I tend to program mostly on OS X. I have a reasonable > knowledge of Cocoa, and can get quite a lot done with Objective-C. > So, if I want to use Haskell on OS X, I can run off and learn about > Haskell's FFI, or some hand-rolled widget package, etc. But here is > the problem: Cocoa et al provide an entire ecosystem that includes > data structures, error handling, and so on. So, when trying to use > Haskell with a GUI, or perhaps to do some audio processing, or other > things that are provided by the OS X APIs, I feel that I am layering > two quite incompatible worlds together: Haskell and its host > environment. > > I ran into the same dilemma with Common Lisp. Clozure Common Lisp > provides an excellent Cocoa binding. But using it feels (to me) like > communications between alien races. The OS APIs want to do things the > Objective-C way, and that is (clearly) not the Haskell or Lisp way. > > One soon notices that books on Haskell pay virtually no attention to > this issue. At least, the ones I have don't. > > Maybe what I am really experiencing is that OS X just "wants" to be > programmed in its native language. I had something of a better > feeling about using F# on Windows, because that architecture does > leverage the native stuff much better. > > The upshot is that I remain skeptical that Haskell is actually a > practical language for many kinds of development. > > Can anyone point me to discussions on this? I may be having a > dimwitted interval here, but I don't know what to search to find > this. I'm not entirely sure I understand your question. The fact that, say, Objective-C uses quite a different way of speech than Haskell is not surprising, these are two different languages after all. You would have similar problems in an imaginary world where Haskell were the norm and Objective-C the exception. If you want to use both at once, you have to invest work in building a bridge. For instance, you can present Haskell in a more Objective-C-y fashion, or the other way round, or some middle ground. That's just how it is and it applies to all languages. Concerning Objective-C specifically, there was a very nice project implementing a slick FFI for Cocoa in Haskell. Unfortunately, it has been dead for years now. http://hoc.sourceforge.net/ Somewhat related, there is actually an (experimental) OS written in Haskell, see http://stackoverflow.com/a/6638207/403805 In case you're not that interested in integrating with the entire OS, but are just looking for a GUI library in Haskell anyway, I would like to toot my own horn and point to threepenny-gui, which uses the web browser as a display. https://github.com/HeinrichApfelmus/threepenny-gui Best regards, Heinrich Apfelmus -- http://apfelmus.nfshost.com From psismondi at arqux.com Sun Aug 11 17:10:48 2013 From: psismondi at arqux.com (Philippe Sismondi) Date: Sun, 11 Aug 2013 11:10:48 -0400 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> Message-ID: <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> On 2013-08-11, at 4:58 AM, Heinrich Apfelmus wrote: > Philippe Sismondi wrote: >> I am interested in tracking down discussions on how Haskell fits into > I'm not entirely sure I understand your question. Well, I admit that I am having trouble formulating my question clearly. It was probably not even clear in my own mind. However, posting and clarifying here have served a purpose for me; read on. Upon reflection, probably my real concern is not about mixing dissimilar programming languages, but about the frequently discussed issue of finding production-quality libraries for a language. This issue is a serious one for Common Lisp, Scheme, ML and OCaml, and other languages I like. It is perhaps less of a problem for languages built on a substrate such as the JVM or .NET. For most of the software that I am interested in working on, there is a vast collection of "native" stuff available in both OS X and Windows. This includes, just as an example, Core Audio in OS X. As a result of posting my (vague) question, and thinking about this, I have decided that it makes more sense to just stick to Objective-C or C when the vendor of the OS has provided a good library in their language of choice. I like Haskell and Lisp better than I like flavours of C. But for now I am going to abandon the forlorn hope of doing everything in my favourite languages when the OS ecosystem favours something else. Moreover, I am dissatisfied with the quality of Haskell libraries even for those things that are not already supplied by OS vendors. In my opinion (and I may be wrong), hackage is littered with half-baked stuff, poorly documented projects started by extremely bright grad students and then abandoned after a year or two. (Of course, there is some great stuff on there too.) As a result of this little thread I have come to another conclusion, and this is just my subjective view. Most of the software that I am interested in seems to live most comfortably with a stateful conception of the world. (The native libraries I find most useful certainly are stateful.) I am reasonably competent with monads and monad transformers in Haskell. But, to be honest, after three years of pluggin away at Haskell, I am not the least convinced that the problem of handling a changing external world in a pure functional language has been successfully solved by those techniques. I always feel as though I am using the robot arm on a space shuttle when a screwdriver would do. (Again, no need to rebut this - I may be wrong or just to stupid to use Haskell effectively - so be it.) Perhaps in the end I do not really believe that functional programming is the panacea that its devotees claim it to be. I think this post may mark the beginning of my abandonment of Haskell for many purposes. So long?. - P - From stephen.tetley at gmail.com Sun Aug 11 17:45:20 2013 From: stephen.tetley at gmail.com (Stephen Tetley) Date: Sun, 11 Aug 2013 16:45:20 +0100 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> Message-ID: I tend to see OO subtyping and heterogeneous collections as more an impediment to (pure) functional integration with the "rest-of-the-world" than state. There are a lot of quality wrappers to stateful C libraries from Haskell, but few to object oriented C++ or Objective C libraries. On 11 August 2013 16:10, Philippe Sismondi wrote: > > > As a result of this little thread I have come to another conclusion, and > this is just my subjective view. Most of the software that I am interested > in seems to live most comfortably with a stateful conception of the world. > (The native libraries I find most useful certainly are stateful.) I am > reasonably competent with monads and monad transformers in Haskell. But, to > be honest, after three years of pluggin away at Haskell, I am not the least > convinced that the problem of handling a changing external world in a pure > functional language has been successfully solved by those techniques. I > always feel as though I am using the robot arm on a space shuttle when a > screwdriver would do. (Again, no need to rebut this - I may be wrong or > just to stupid to use Haskell effectively - so be it.) > > Perhaps in the end I do not really believe that functional programming is > the panacea that its devotees claim it to be. > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From psismondi at arqux.com Sun Aug 11 18:42:08 2013 From: psismondi at arqux.com (Philippe Sismondi) Date: Sun, 11 Aug 2013 12:42:08 -0400 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> Message-ID: On 2013-08-11, at 11:45 AM, Stephen Tetley wrote: > I tend to see OO subtyping and heterogeneous collections as more an impediment to (pure) functional integration with the "rest-of-the-world" than state. There are a lot of quality wrappers to stateful C libraries from Haskell, but few to object oriented C++ or Objective C libraries. That may be true. And, state and OO types are different issues. I guess I conflated the two because the stateful OS X libraries I use are largely implemented using Cocoa or sometimes C data structures. So, do I make my own Haskell list or record or type representing something in the API, or just access the Cocoa one? etc. etc. Also, in truth I introduced a totally separate gripe, which is that I find monads and monad transformers to be a hell of a messy and difficult way to deal with state. Sometimes I get sick of all the layering in monad transformers and hand-roll my own monad with all the stuff. But it just seems baroquely difficult. Perhaps I was just to set in my ways when I came to FP, and cannot really "get it". There are things I have never looked into such as FRP, but maybe my poor old brain is just not up to it. Maybe I should have been a janitor or a car salesman. - P - > > > On 11 August 2013 16:10, Philippe Sismondi wrote: > > > As a result of this little thread I have come to another conclusion, and this is just my subjective view. Most of the software that I am interested in seems to live most comfortably with a stateful conception of the world. (The native libraries I find most useful certainly are stateful.) I am reasonably competent with monads and monad transformers in Haskell. But, to be honest, after three years of pluggin away at Haskell, I am not the least convinced that the problem of handling a changing external world in a pure functional language has been successfully solved by those techniques. I always feel as though I am using the robot arm on a space shuttle when a screwdriver would do. (Again, no need to rebut this - I may be wrong or just to stupid to use Haskell effectively - so be it.) > > Perhaps in the end I do not really believe that functional programming is the panacea that its devotees claim it to be. > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners -------------- next part -------------- An HTML attachment was scrubbed... URL: From voldermort at hotmail.com Sun Aug 11 19:45:05 2013 From: voldermort at hotmail.com (harry) Date: Sun, 11 Aug 2013 17:45:05 +0000 (UTC) Subject: [Haskell-beginners] cabal install relative to $topdir Message-ID: I would like cabal install to create paths in the package database relative to$topdir, so that I can mount it on a different machine (where the absolute path will obviously be different). Is there any way of automating this? From orblivion at gmail.com Sun Aug 11 20:44:24 2013 From: orblivion at gmail.com (Dan Krol) Date: Sun, 11 Aug 2013 11:44:24 -0700 Subject: [Haskell-beginners] HTTP Download -> Save File - non-strict In-Reply-To: <20130811063030.GA1513@mteis.lan> References: <20130811063030.GA1513@mteis.lan> Message-ID: Ah yes, the actual docs. Somehow didn't think to check that, sorry. Alright, I'll try to figure that one out, thanks. Any particular reason nobody just offers http over lazy I/O? Is it just because lazy I/O is generally discouraged? Or just particularly bad over a network? And is this an area where Conduit is better than Pipes? There doesn't seem to be a similar http for Pipes. On Sat, Aug 10, 2013 at 11:30 PM, Magnus Therning wrote: > On Sat, Aug 10, 2013 at 05:16:58PM -0700, Dan Krol wrote: > > Hi, > > > > I'm working on an rss file getter. I was wondering if I could get > > some help getting files to download and save without holding the > > entire file in memory in between. I chose Conduit's version of > > SimpleHttp only because it was recommended, and it was the quickest > > thing I could get to work correctly because I was eager to get > > started on this project, so I'd be happy to switch. > > > > Here's where I define the download and save functions: > > > > https://github.com/orblivion/feedGetter/blob/master/rss.hs#L107 > > > > And here's where I use them, getting multiple at a time with async: > > > > https://github.com/orblivion/feedGetter/blob/master/rss.hs#L208 > > > > What happens when I run this is that it outputs that it's "Getting" > > the file, waits a while (presumably to download the whole thing), > > then says it's "Saving". And I checked the file system, it's not > > there during the pause. I'm not entirely sure why. Is it my choice > > of libraries, or the way I'm using them? Perhaps something to do > > with async? I just tried content <- simpleHttp "http://google.com" > > in ghci, and it does pause for a second, so I'm guessing this is > > strict from the getgo. But I've done almost no I/O before. > > > > Is there a straightforward, canonical option? It seems like there > > perhaps should be. But if it comes down to using pipes or conduit, > > what the heck I'll try it out, I'd like to learn pipes eventually. > > Michael is very good with documenting his packages, this is what I > found in the docs for http-conduit (http://is.gd/WkDb7G): > > Note: Even though this function returns a lazy bytestring, it does > not utilize lazy I/O, and therefore the entire response body will > live in memory. If you want constant memory usage, you'll need to > use the conduit package and http directly. > > /M > > -- > Magnus Therning OpenPGP: 0xAB4DFBA4 > email: magnus at therning.org jabber: magnus at therning.org > twitter: magthe http://therning.org/magnus > > I invented the term Object-Oriented, and I can tell you I did not have > C++ in mind. > -- Alan Kay > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From mle+hs at mega-nerd.com Sun Aug 11 22:47:13 2013 From: mle+hs at mega-nerd.com (Erik de Castro Lopo) Date: Mon, 12 Aug 2013 06:47:13 +1000 Subject: [Haskell-beginners] HTTP Download -> Save File - non-strict In-Reply-To: References: <20130811063030.GA1513@mteis.lan> Message-ID: <20130812064713.b8ec626e4553316c16246d46@mega-nerd.com> Dan Krol wrote: > Ah yes, the actual docs. Somehow didn't think to check that, sorry. > > Alright, I'll try to figure that one out, thanks. Any particular reason > nobody just offers http over lazy I/O? Is it just because lazy I/O is > generally discouraged? Or just particularly bad over a network? Lazy I/O is particularly problematic when implmenting network servers and generally discouraged for networking code. > And is this an area where Conduit is better than Pipes? Conduit has been around longer and is thus more mature and complete. > There doesn't seem to be a similar http for Pipes. I believe Andrew Cowie, author of http-streams (on top of io-streams) is working on a Pipes version. HTH, Erik -- ---------------------------------------------------------------------- Erik de Castro Lopo http://www.mega-nerd.com/ From voldermort at hotmail.com Mon Aug 12 09:35:10 2013 From: voldermort at hotmail.com (vold) Date: Mon, 12 Aug 2013 07:35:10 +0000 (UTC) Subject: [Haskell-beginners] cabal logs Message-ID: If I set build-summary in cabal's config file, cabal will create duplicate logs - the one specified in build-summary and the default build-reports.log in the packages directory. There's also a log-dir option in the config file which doesn't seem to do anything. Are these bugs, or intended behaviour? From apfelmus at quantentunnel.de Mon Aug 12 11:23:14 2013 From: apfelmus at quantentunnel.de (Heinrich Apfelmus) Date: Mon, 12 Aug 2013 11:23:14 +0200 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> Message-ID: Philippe Sismondi wrote: > > Upon reflection, probably my real concern is not about mixing > dissimilar programming languages, but about the frequently discussed > issue of finding production-quality libraries for a language. [..] > > For most of the software that I am interested in working on, there is > a vast collection of "native" stuff available in both OS X and > Windows. This includes, just as an example, Core Audio in OS X. [..] > > Moreover, I am dissatisfied with the quality of Haskell libraries > even for those things that are not already supplied by OS vendors. In > my opinion (and I may be wrong), hackage is littered with half-baked > stuff, poorly documented projects started by extremely bright grad > students and then abandoned after a year or two. (Of course, there is > some great stuff on there too.) Well, there are a lot of magazines at the news kiosk, too, while only a few are of outstanding quality. That is just how a news kiosk works. Speaking of quality, what I like most about Haskell libraries, even the half-baked ones, is that they have very few bugs. I've been programming functionally for a decade now, and whenever I venture into the world of imperative languages, I always trip up bugs that just shouldn't be there. Here two recent examples: * HTML 5 drag and drop. Apart from the fact that the specification is overcomplicated, Chrome doesn't even implement the spec correctly. When the mouse enters a child element of a "dropzone'd" element, the latter receives a "dragleave" event, but will not receive a "dragenter" event again when the mouse moves away from the child element. Argh! * HTML 5 WebSockets. Chrome or Safari. After a certain amount of inactivity on the server side, the browser will close the WebSocket. However, it will only close the client side, so the client cannot send messages anymore. The connection to the server is still *open*, though, and the server can happily send data. What? Also, if you connect with a WebSocket and then reload the page and connect again, the old connection will be reused. WTF? These are just examples, this happens to me all the time. Curiously, whenever I use state, my programs start to become similarly brittle. There is no reason why state should be a fundamental element of a programming language, and as a design pattern, state is best avoided at all cost. > As a result of this little thread I have come to another conclusion, > and this is just my subjective view. Most of the software that I am > interested in seems to live most comfortably with a stateful > conception of the world. (The native libraries I find most useful > certainly are stateful.) I am reasonably competent with monads and > monad transformers in Haskell. But, to be honest, after three years > of pluggin away at Haskell, I am not the least convinced that the > problem of handling a changing external world in a pure functional > language has been successfully solved by those techniques. I always > feel as though I am using the robot arm on a space shuttle when a > screwdriver would do. (Again, no need to rebut this - I may be wrong > or just to stupid to use Haskell effectively - so be it.) > > Perhaps in the end I do not really believe that functional > programming is the panacea that its devotees claim it to be. > > I think this post may mark the beginning of my abandonment of Haskell > for many purposes. Haskell may not be easy to learn, but it's definitely worth the effort. Best regards, Heinrich Apfelmus -- http://apfelmus.nfshost.com From alanbuxton at gmail.com Mon Aug 12 22:35:30 2013 From: alanbuxton at gmail.com (Alan Buxton) Date: Mon, 12 Aug 2013 21:35:30 +0100 Subject: [Haskell-beginners] Text.XML.writeFile question Message-ID: <004001ce979b$7d5b8d00$7812a700$@gmail.com> Hi I am trying to write an XML file where the filename is created based on a timestamp. Simplified version below. This won't compile - I get this error in doWrite2 filepathtest.hs|24 col 17 error| Couldn't match expected type system-filepath-0.4.7:Filesystem.Path.Internal.FilePath' || with actual type String' || In the second argument of writeFile', namely t1' || In a stmt of a 'do' block: writeFile def t1 doc || In the expression: || do { t1 <- tsString; || writeFile def t1 doc } Somehow the String "text.xml" in doWrite1 is converted into a FilePath, but not the String t1 in doWrite2. What am I doing wrong? {-# LANGUAGE OverloadedStrings #-} module Filepathtest where import Text.XML import Data.Time.Clock.POSIX (utcTimeToPOSIXSeconds) import Data.Time.Clock (getCurrentTime) import Prelude hiding (writeFile, FilePath) tsString :: IO String tsString = do x <- getCurrentTime let x' = show$ floor $utcTimeToPOSIXSeconds x return x' doWrite1 :: Document -> IO () doWrite1 doc = writeFile def "test1.xml" doc doWrite2 :: Document -> IO () doWrite2 doc = do t1 <- tsString writeFile def t1 doc -------------- next part -------------- An HTML attachment was scrubbed... URL: From toad3k at gmail.com Mon Aug 12 22:52:06 2013 From: toad3k at gmail.com (David McBride) Date: Mon, 12 Aug 2013 16:52:06 -0400 Subject: [Haskell-beginners] Text.XML.writeFile question In-Reply-To: <004001ce979b$7d5b8d00$7812a700$@gmail.com> References: <004001ce979b$7d5b8d00$7812a700$@gmail.com> Message-ID: When you use overloadedstrings, it will type all literal strings as IsString a => a as a type. Unfortunately your tsString is not a literal, it is a run time function that always returns a string and a string is not a filepath. You should be able to do fmap fromString tsString, provided you import Data.String. On Mon, Aug 12, 2013 at 4:35 PM, Alan Buxton wrote: > Hi**** > > ** ** > > I am trying to write an XML file where the filename is created based on a > timestamp. Simplified version below. This won?t compile ? I get this error > in doWrite2 **** > > ** ** > > *filepathtest.hs|24 col 17 error| Couldn't match expected type > system-filepath-0.4.7:Filesystem.Path.Internal.FilePath' > * > > *|| with actual type String'* > > *|| In the second argument of writeFile', namely t1'* > > *|| In a stmt of a 'do' block: writeFile def t1 doc* > > *|| In the expression:* > > *|| do { t1 <- tsString;* > > *|| writeFile def t1 doc }* > > * * > > Somehow the String ?text.xml? in doWrite1 is converted into a FilePath, > but not the String t1 in doWrite2. What am I doing wrong?**** > > ** ** > > {-# LANGUAGE OverloadedStrings #-}**** > > module Filepathtest where**** > > **** > > import Text.XML**** > > import Data.Time.Clock.POSIX (utcTimeToPOSIXSeconds)**** > > import Data.Time.Clock (getCurrentTime)**** > > import Prelude hiding (writeFile, FilePath)**** > > **** > > tsString :: IO String**** > > tsString = do**** > > x <- getCurrentTime**** > > let x' = show$ floor $utcTimeToPOSIXSeconds x**** > > return x'**** > > **** > > doWrite1 :: Document -> IO ()**** > > doWrite1 doc =**** > > writeFile def "test1.xml" doc**** > > **** > > doWrite2 :: Document -> IO ()**** > > doWrite2 doc = do**** > > t1 <- tsString**** > > writeFile def t1 doc**** > > ** ** > > ** ** > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From kdamodar2000 at gmail.com Tue Aug 13 06:45:07 2013 From: kdamodar2000 at gmail.com (damodar kulkarni) Date: Tue, 13 Aug 2013 10:15:07 +0530 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> Message-ID: > Curiously, whenever I use state, my programs start to become similarly > brittle. There is no reason why state should be a fundamental element of a > programming language, and as a design pattern, state is best avoided at all > cost. Just as a curiosity, how would one avoid state in cases like protocol design? e.g. protocols specifications (like TCP/IP) do have a large element of state dependent behavior that "seems essential" to the problem. How would one deal with such cases? @Philippe Sismondi: I always feel as though I am using the robot arm on a space shuttle when a > screwdriver would do. +1 for this remark. Thanks and regards, -Damodar Kulkarni On Mon, Aug 12, 2013 at 2:53 PM, Heinrich Apfelmus < apfelmus at quantentunnel.de> wrote: > Philippe Sismondi wrote: > >> >> Upon reflection, probably my real concern is not about mixing >> dissimilar programming languages, but about the frequently discussed >> issue of finding production-quality libraries for a language. [..] >> >> >> For most of the software that I am interested in working on, there is >> a vast collection of "native" stuff available in both OS X and >> Windows. This includes, just as an example, Core Audio in OS X. [..] >> >> >> Moreover, I am dissatisfied with the quality of Haskell libraries >> even for those things that are not already supplied by OS vendors. In >> my opinion (and I may be wrong), hackage is littered with half-baked >> stuff, poorly documented projects started by extremely bright grad >> students and then abandoned after a year or two. (Of course, there is >> some great stuff on there too.) >> > > Well, there are a lot of magazines at the news kiosk, too, while only a > few are of outstanding quality. That is just how a news kiosk works. > > Speaking of quality, what I like most about Haskell libraries, even the > half-baked ones, is that they have very few bugs. I've been programming > functionally for a decade now, and whenever I venture into the world of > imperative languages, I always trip up bugs that just shouldn't be there. > Here two recent examples: > > * HTML 5 drag and drop. Apart from the fact that the specification is > overcomplicated, Chrome doesn't even implement the spec correctly. When the > mouse enters a child element of a "dropzone'd" element, the latter receives > a "dragleave" event, but will not receive a "dragenter" event again when > the mouse moves away from the child element. Argh! > > * HTML 5 WebSockets. Chrome or Safari. After a certain amount of > inactivity on the server side, the browser will close the WebSocket. > However, it will only close the client side, so the client cannot send > messages anymore. The connection to the server is still *open*, though, and > the server can happily send data. What? Also, if you connect with a > WebSocket and then reload the page and connect again, the old connection > will be reused. WTF? > > These are just examples, this happens to me all the time. Curiously, > whenever I use state, my programs start to become similarly brittle. There > is no reason why state should be a fundamental element of a programming > language, and as a design pattern, state is best avoided at all cost. > > > As a result of this little thread I have come to another conclusion, >> and this is just my subjective view. Most of the software that I am >> interested in seems to live most comfortably with a stateful >> conception of the world. (The native libraries I find most useful >> certainly are stateful.) I am reasonably competent with monads and >> monad transformers in Haskell. But, to be honest, after three years >> of pluggin away at Haskell, I am not the least convinced that the >> problem of handling a changing external world in a pure functional >> language has been successfully solved by those techniques. I always >> feel as though I am using the robot arm on a space shuttle when a >> screwdriver would do. (Again, no need to rebut this - I may be wrong >> or just to stupid to use Haskell effectively - so be it.) >> >> Perhaps in the end I do not really believe that functional >> programming is the panacea that its devotees claim it to be. >> >> I think this post may mark the beginning of my abandonment of Haskell >> for many purposes. >> > > Haskell may not be easy to learn, but it's definitely worth the effort. > > > > Best regards, > Heinrich Apfelmus > > -- > http://apfelmus.nfshost.com > > > ______________________________**_________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/**mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: From dedgrant at gmail.com Tue Aug 13 07:30:37 2013 From: dedgrant at gmail.com (Darren Grant) Date: Mon, 12 Aug 2013 22:30:37 -0700 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> Message-ID: On Mon, Aug 12, 2013 at 9:45 PM, damodar kulkarni wrote: > > Curiously, whenever I use state, my programs start to become similarly >> brittle. There is no reason why state should be a fundamental element of a >> programming language, and as a design pattern, state is best avoided at all >> cost. > > > Just as a curiosity, how would one avoid state in cases like protocol > design? e.g. protocols specifications (like TCP/IP) do have a large element > of state dependent behavior that "seems essential" to the problem. How > would one deal with such cases? > State can be achieved in pure functional languages. The difference is that the state management is not directly accessible to the programmer as it is hidden behind a highly declarative set of abstractions. This provides the benefits of consistency, and sometimes frustration when the resulting machine code doesn't perform as expected. Cheers, Darren -------------- next part -------------- An HTML attachment was scrubbed... URL: From alanbuxton at gmail.com Tue Aug 13 10:06:14 2013 From: alanbuxton at gmail.com (Alan Buxton) Date: Tue, 13 Aug 2013 09:06:14 +0100 Subject: [Haskell-beginners] Text.XML.writeFile question In-Reply-To: References: <004001ce979b$7d5b8d00$7812a700$@gmail.com> Message-ID: <01a301ce97fb$fba31ec0$f2e95c40$@gmail.com> Thanks David. Spot on. a From: Beginners [mailto:beginners-bounces at haskell.org] On Behalf Of David McBride Sent: 12 August 2013 21:52 To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Subject: Re: [Haskell-beginners] Text.XML.writeFile question When you use overloadedstrings, it will type all literal strings as IsString a => a as a type. Unfortunately your tsString is not a literal, it is a run time function that always returns a string and a string is not a filepath. You should be able to do fmap fromString tsString, provided you import Data.String. On Mon, Aug 12, 2013 at 4:35 PM, Alan Buxton wrote: Hi I am trying to write an XML file where the filename is created based on a timestamp. Simplified version below. This won't compile - I get this error in doWrite2 filepathtest.hs|24 col 17 error| Couldn't match expected type system-filepath-0.4.7:Filesystem.Path.Internal.FilePath' || with actual type String' || In the second argument of writeFile', namely t1' || In a stmt of a 'do' block: writeFile def t1 doc || In the expression: || do { t1 <- tsString; || writeFile def t1 doc } Somehow the String "text.xml" in doWrite1 is converted into a FilePath, but not the String t1 in doWrite2. What am I doing wrong? {-# LANGUAGE OverloadedStrings #-} module Filepathtest where import Text.XML import Data.Time.Clock.POSIX (utcTimeToPOSIXSeconds) import Data.Time.Clock (getCurrentTime) import Prelude hiding (writeFile, FilePath) tsString :: IO String tsString = do x <- getCurrentTime let x' = show$ floor $utcTimeToPOSIXSeconds x return x' doWrite1 :: Document -> IO () doWrite1 doc = writeFile def "test1.xml" doc doWrite2 :: Document -> IO () doWrite2 doc = do t1 <- tsString writeFile def t1 doc _______________________________________________ Beginners mailing list Beginners at haskell.org http://www.haskell.org/mailman/listinfo/beginners -------------- next part -------------- An HTML attachment was scrubbed... URL: From apfelmus at quantentunnel.de Tue Aug 13 11:33:38 2013 From: apfelmus at quantentunnel.de (Heinrich Apfelmus) Date: Tue, 13 Aug 2013 11:33:38 +0200 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> Message-ID: damodar kulkarni wrote: >> Curiously, whenever I use state, my programs start to become similarly >> brittle. There is no reason why state should be a fundamental element of a >> programming language, and as a design pattern, state is best avoided at all >> cost. > > > Just as a curiosity, how would one avoid state in cases like protocol > design? e.g. protocols specifications (like TCP/IP) do have a large element > of state dependent behavior that "seems essential" to the problem. How > would one deal with such cases? Well, in a protocol like TCP/IP, the response of a participant does depend on the history of the communication, and that history can neatly be summarized in a small amount of state. I don't think it's possible to avoid state in this case. But I meant something else, actually, and should have been more precise. Namely, it is best to avoid *mutable* state as a *design pattern*, i.e. the use of IORef and thelike. Pure functions State -> State are fine, but anything were the meaning of an expression depends heavily on the context (the program state) is prone to bugs. The problem is more about source code than it is about state. To avoid bugs, each piece of source code should be understandable in isolation, i.e. it should give the same results in all contexts ("code paths"). This way, it is a lot easier to determine its correctness. Of course, "source code" has different scales, and can include protocol design. After all, protocols are made by humans, and clever design can prevent headaches later. For instance, HTTP GET requests were designed to be stateless, and that makes the protocol a lot more robust. Best regards, Heinrich Apfelmus -- http://apfelmus.nfshost.com From lehner.lukas at gmail.com Tue Aug 13 16:46:04 2013 From: lehner.lukas at gmail.com (Lukas Lehner) Date: Tue, 13 Aug 2013 16:46:04 +0200 Subject: [Haskell-beginners] print [] Message-ID: Hello, This code runs ok http://lpaste.net/91814 but only thanks to enforcing [()] on line 12. In GHCI *Main> :t (List []) (List []) :: NestedList a and *Main> :t flatten (List []) flatten (List []) :: [a] That means ghc cannot infer the type. Is there a way how to # print flatten (List []) ? Or even more general, print [] without enforcing the type? Thank you Lukas -------------- next part -------------- An HTML attachment was scrubbed... URL: From allbery.b at gmail.com Tue Aug 13 16:59:09 2013 From: allbery.b at gmail.com (Brandon Allbery) Date: Tue, 13 Aug 2013 10:59:09 -0400 Subject: [Haskell-beginners] print [] In-Reply-To: References: Message-ID: On Tue, Aug 13, 2013 at 10:46 AM, Lukas Lehner wrote: > That means ghc cannot infer the type. > Is there a way how to # print flatten (List []) ? > Or even more general, print [] without enforcing the type? > If you turn on the ExtendedDefaultRules extension ( {-# LANGUAGE ExtendedDefaultRules #-} pragma or -X ExtendedDefaultRules ghc option), ghc will infer () for the type just as ghci does. Note that this reduces type safety a bit, since ghc will now accept programs that have what otherwise would be type errors. -- brandon s allbery kf8nh sine nomine associates allbery.b at gmail.com ballbery at sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net -------------- next part -------------- An HTML attachment was scrubbed... URL: From nadavchernin at gmail.com Tue Aug 13 17:05:15 2013 From: nadavchernin at gmail.com (Nadav Chernin) Date: Tue, 13 Aug 2013 18:05:15 +0300 Subject: [Haskell-beginners] Sorting Bank Accounts problem in Haskell Message-ID: Hi all, I learn Haskell and i try to solve questions from SPOJ. Currently, i try to solve problem *Sorting Bank Accounts* Here is my code: *import qualified Data.Map as M* * * *extractTest []=[]* *extractTest (x:xs)=(take n xs):(extractTest (drop (n+1) xs)) where* * n=read x* * * *emp=M.empty* * * *collect m []=m* *collect m (x:xs)* * |M.member x m =collect (M.insert x (amount+1) m) xs* * |otherwise=collect (M.insert x 1 m) xs* * where* * Just amount=M.lookup x m* * * *getList m=map (\(x,k)->x++" "++(show k)) (M.toList m)* *f x=unlines$(getList (collect emp x))++[""]* *main=getLine>>=(\x->interact$unlines.map f.take (read x).extractTest.lines) * * * The problem is that i can't finish it during 7 sec. So i want to know is there more quick solution/method for this problem. Thanks, Nadav -------------- next part -------------- An HTML attachment was scrubbed... URL: From lehner.lukas at gmail.com Tue Aug 13 17:08:45 2013 From: lehner.lukas at gmail.com (Lukas Lehner) Date: Tue, 13 Aug 2013 17:08:45 +0200 Subject: [Haskell-beginners] print [] In-Reply-To: References: Message-ID: Good to know extension. And without it? On Tue, Aug 13, 2013 at 4:59 PM, Brandon Allbery wrote: > On Tue, Aug 13, 2013 at 10:46 AM, Lukas Lehner wrote: > >> That means ghc cannot infer the type. >> Is there a way how to # print flatten (List []) ? >> Or even more general, print [] without enforcing the type? >> > > If you turn on the ExtendedDefaultRules extension ( {-# LANGUAGE > ExtendedDefaultRules #-} pragma or -X ExtendedDefaultRules ghc option), > ghc will infer () for the type just as ghci does. Note that this reduces > type safety a bit, since ghc will now accept programs that have what > otherwise would be type errors. > > -- > brandon s allbery kf8nh sine nomine > associates > allbery.b at gmail.com > ballbery at sinenomine.net > unix, openafs, kerberos, infrastructure, xmonad > http://sinenomine.net > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From davidleothomas at gmail.com Tue Aug 13 17:21:25 2013 From: davidleothomas at gmail.com (David Thomas) Date: Tue, 13 Aug 2013 08:21:25 -0700 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> Message-ID: It seems like it would be roughly as easy to encode TCP with continuations as with the usual imperative approaches, though I'll certainly admit I haven't tried. On Mon, Aug 12, 2013 at 9:45 PM, damodar kulkarni wrote: > > Curiously, whenever I use state, my programs start to become similarly >> brittle. There is no reason why state should be a fundamental element of a >> programming language, and as a design pattern, state is best avoided at all >> cost. > > > Just as a curiosity, how would one avoid state in cases like protocol > design? e.g. protocols specifications (like TCP/IP) do have a large element > of state dependent behavior that "seems essential" to the problem. How > would one deal with such cases? > > @Philippe Sismondi: > > > I always feel as though I am using the robot arm on a space shuttle when a >> screwdriver would do. > > > +1 for this remark. > > Thanks and regards, > -Damodar Kulkarni > > > On Mon, Aug 12, 2013 at 2:53 PM, Heinrich Apfelmus < > apfelmus at quantentunnel.de> wrote: > >> Philippe Sismondi wrote: >> >>> >>> Upon reflection, probably my real concern is not about mixing >>> dissimilar programming languages, but about the frequently discussed >>> issue of finding production-quality libraries for a language. [..] >>> >>> >>> For most of the software that I am interested in working on, there is >>> a vast collection of "native" stuff available in both OS X and >>> Windows. This includes, just as an example, Core Audio in OS X. [..] >>> >>> >>> Moreover, I am dissatisfied with the quality of Haskell libraries >>> even for those things that are not already supplied by OS vendors. In >>> my opinion (and I may be wrong), hackage is littered with half-baked >>> stuff, poorly documented projects started by extremely bright grad >>> students and then abandoned after a year or two. (Of course, there is >>> some great stuff on there too.) >>> >> >> Well, there are a lot of magazines at the news kiosk, too, while only a >> few are of outstanding quality. That is just how a news kiosk works. >> >> Speaking of quality, what I like most about Haskell libraries, even the >> half-baked ones, is that they have very few bugs. I've been programming >> functionally for a decade now, and whenever I venture into the world of >> imperative languages, I always trip up bugs that just shouldn't be there. >> Here two recent examples: >> >> * HTML 5 drag and drop. Apart from the fact that the specification is >> overcomplicated, Chrome doesn't even implement the spec correctly. When the >> mouse enters a child element of a "dropzone'd" element, the latter receives >> a "dragleave" event, but will not receive a "dragenter" event again when >> the mouse moves away from the child element. Argh! >> >> * HTML 5 WebSockets. Chrome or Safari. After a certain amount of >> inactivity on the server side, the browser will close the WebSocket. >> However, it will only close the client side, so the client cannot send >> messages anymore. The connection to the server is still *open*, though, and >> the server can happily send data. What? Also, if you connect with a >> WebSocket and then reload the page and connect again, the old connection >> will be reused. WTF? >> >> These are just examples, this happens to me all the time. Curiously, >> whenever I use state, my programs start to become similarly brittle. There >> is no reason why state should be a fundamental element of a programming >> language, and as a design pattern, state is best avoided at all cost. >> >> >> As a result of this little thread I have come to another conclusion, >>> and this is just my subjective view. Most of the software that I am >>> interested in seems to live most comfortably with a stateful >>> conception of the world. (The native libraries I find most useful >>> certainly are stateful.) I am reasonably competent with monads and >>> monad transformers in Haskell. But, to be honest, after three years >>> of pluggin away at Haskell, I am not the least convinced that the >>> problem of handling a changing external world in a pure functional >>> language has been successfully solved by those techniques. I always >>> feel as though I am using the robot arm on a space shuttle when a >>> screwdriver would do. (Again, no need to rebut this - I may be wrong >>> or just to stupid to use Haskell effectively - so be it.) >>> >>> Perhaps in the end I do not really believe that functional >>> programming is the panacea that its devotees claim it to be. >>> >>> I think this post may mark the beginning of my abandonment of Haskell >>> for many purposes. >>> >> >> Haskell may not be easy to learn, but it's definitely worth the effort. >> >> >> >> Best regards, >> Heinrich Apfelmus >> >> -- >> http://apfelmus.nfshost.com >> >> >> ______________________________**_________________ >> Beginners mailing list >> Beginners at haskell.org >> http://www.haskell.org/**mailman/listinfo/beginners >> > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > -------------- next part -------------- An HTML attachment was scrubbed... URL: From voldermort at hotmail.com Tue Aug 13 20:55:22 2013 From: voldermort at hotmail.com (harry) Date: Tue, 13 Aug 2013 18:55:22 +0000 (UTC) Subject: [Haskell-beginners] cabal binary packages Message-ID: I need to build some software on a server, but it doesn't have enough spare memory for compiling all the dependencies. I would like to create a binary distribution of e.g. snap including all of its dependencies (other than those which come with ghc), so that the server can just install them. I found http://www.haskell.org/cabal/users-guide/installing-packages.html#creating-a-binary-package, but this seems to have two deficiencies - the destination prefix needs to be know in advance (which might be OK if it can be a relative path), and it has to be done individually for each package. From bryanvick at gmail.com Wed Aug 14 11:25:02 2013 From: bryanvick at gmail.com (Bryan Vicknair) Date: Wed, 14 Aug 2013 02:25:02 -0700 Subject: [Haskell-beginners] Compiler can't deduce Bool as instance of ToField Message-ID: <20130814092502.GA4017@x220.att.net> postgresql-simple declares Bool as an instance of the ToField class. The compiler can't deduce that given this simple code however: import Database.PostgreSQL.Simple.ToField (ToField(..)) foo :: (ToField a) => a foo = True It fails with this error: Db.hs:64:7: Could not deduce (a ~ Bool) from the context (ToField a) bound by the type signature for foo :: ToField a => a at Db.hs:63:8-23 a' is a rigid type variable bound by the type signature for foo :: ToField a => a at Db.hs:63:8 In the expression: True In an equation for foo': foo = True Failed, modules loaded: none. What am I missing? From dav.vire+haskell at gmail.com Wed Aug 14 12:08:39 2013 From: dav.vire+haskell at gmail.com (David Virebayre) Date: Wed, 14 Aug 2013 12:08:39 +0200 Subject: [Haskell-beginners] print [] In-Reply-To: References: Message-ID: 2013/8/13 Lukas Lehner : > Good to know extension. And without it? Someone correct me if I'm wrong, here's how I understand it: To execute print [], ghc has to find which instance of Show to call. How could it find an instance without knowing the type ? [] could be a list of anything, and while in theory we know that in all cases it should print "[]", in practice ghc can't call that code if it doesn't know the type of the list. David. From lehner.lukas at gmail.com Wed Aug 14 12:15:11 2013 From: lehner.lukas at gmail.com (Lukas Lehner) Date: Wed, 14 Aug 2013 12:15:11 +0200 Subject: [Haskell-beginners] print [] In-Reply-To: References: Message-ID: That I understand. My question is that I have to enforce the type that specific way (don't want to use ExtendedDefaultRules) or is there some more generic way? On Wed, Aug 14, 2013 at 12:08 PM, David Virebayre < dav.vire+haskell at gmail.com> wrote: > 2013/8/13 Lukas Lehner : > > Good to know extension. And without it? > > Someone correct me if I'm wrong, here's how I understand it: > > To execute print [], ghc has to find which instance of Show to call. > How could it find an instance without knowing the type ? > [] could be a list of anything, and while in theory we know that in > all cases it should print "[]", in practice ghc can't call that code > if it doesn't know the type of the list. > > David. > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: From dav.vire+haskell at gmail.com Wed Aug 14 12:17:48 2013 From: dav.vire+haskell at gmail.com (David Virebayre) Date: Wed, 14 Aug 2013 12:17:48 +0200 Subject: [Haskell-beginners] Sorting Bank Accounts problem in Haskell In-Reply-To: References: Message-ID: You should probably start by switching strings to Data.Text. David. 2013/8/13 Nadav Chernin : > Hi all, > I learn Haskell and i try to solve questions from SPOJ. > > Currently, i try to solve problem Sorting Bank Accounts > > Here is my code: > > import qualified Data.Map as M > > extractTest []=[] > extractTest (x:xs)=(take n xs):(extractTest (drop (n+1) xs)) where > n=read x > > emp=M.empty > > collect m []=m > collect m (x:xs) > |M.member x m =collect (M.insert x (amount+1) m) xs > |otherwise=collect (M.insert x 1 m) xs > where > Just amount=M.lookup x m > > getList m=map (\(x,k)->x++" "++(show k)) (M.toList m) > f x=unlines$(getList (collect emp x))++[""] > main=getLine>>=(\x->interact$unlines.map f.take (read x).extractTest.lines) > > The problem is that i can't finish it during 7 sec. > So i want to know is there more quick solution/method for this problem. > Thanks, Nadav > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > From dav.vire+haskell at gmail.com Wed Aug 14 12:24:29 2013 From: dav.vire+haskell at gmail.com (David Virebayre) Date: Wed, 14 Aug 2013 12:24:29 +0200 Subject: [Haskell-beginners] print [] In-Reply-To: References: Message-ID: 2013/8/14 Lukas Lehner : > That I understand. My question is that I have to enforce the type that > specific way (don't want to use ExtendedDefaultRules) or is there some more > generic way? Ah, sorry, I don't know of a more generic way. But I'm not experienced enough to be sure. My experience is than in real world code, most of the time, the way you use the values gives enough info for ghc to infer a type. Sometimes, you can avoid adding type information by using aTypeOf from the Prelude. David. From ollie at ocharles.org.uk Wed Aug 14 14:16:49 2013 From: ollie at ocharles.org.uk (Oliver Charles) Date: Wed, 14 Aug 2013 13:16:49 +0100 Subject: [Haskell-beginners] Compiler can't deduce Bool as instance of ToField In-Reply-To: <20130814092502.GA4017@x220.att.net> References: <20130814092502.GA4017@x220.att.net> Message-ID: You have declared that foo is *any* type that has a ToField instance, allowing the caller of foo to determine the type at their will. However, your implementation of foo is more specific and requires a is actually Bool and nothing else. On 14 Aug 2013 10:24, "Bryan Vicknair" wrote: > postgresql-simple declares Bool as an instance of the ToField class. The > compiler can't deduce that given this simple code however: > > import Database.PostgreSQL.Simple.ToField (ToField(..)) > > foo :: (ToField a) => a > foo = True > > > It fails with this error: > > Db.hs:64:7: > Could not deduce (a ~ Bool) > from the context (ToField a) > bound by the type signature for foo :: ToField a => a > at Db.hs:63:8-23 > a' is a rigid type variable bound by > the type signature for foo :: ToField a => a at Db.hs:63:8 > In the expression: True > In an equation for foo': foo = True > Failed, modules loaded: none. > > What am I missing? > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: From giacomo at tesio.it Wed Aug 14 15:21:13 2013 From: giacomo at tesio.it (Giacomo Tesio) Date: Wed, 14 Aug 2013 15:21:13 +0200 Subject: [Haskell-beginners] Heterogeneous Lists In-Reply-To: <20130528190810.30335234@tritium.ertes.de> References: <20130528134923.GA8721@seas.upenn.edu> <20130528190810.30335234@tritium.ertes.de> Message-ID: > > I recently ran into this with a GUI application, where I needed to > > process a list of widgets that were members of the same typeclass, but > > I had to wrap them all because they were different types. > > Your misconception is likely that you need to /store/ widgets, because > you think in terms of memory. What you really need is /definitions/ of > widgets, which can very well have different types. Then you can combine > them using the usual monadic or applicative interface. > > Ertugrul, can you please expand on this? Giacomo -------------- next part -------------- An HTML attachment was scrubbed... URL: From nathan.huesken at posteo.de Wed Aug 14 17:03:48 2013 From: nathan.huesken at posteo.de (=?ISO-8859-1?Q?Nathan_H=FCsken?=) Date: Wed, 14 Aug 2013 17:03:48 +0200 Subject: [Haskell-beginners] Trouble with layout in wxHaskell Message-ID: <520B9C54.1040806@posteo.de> Hey, I am trying to understand how do a layout in wxHaskell. I am trying to make a frame with 2 text entries. One single lined at the top and one multiline below it. Both should expand with the window size horizontal. When the window is expanded vertical, only the multiline text entry should expand. Her is what I wrote: main = start mainFrame mainFrame = do f <- frame [text := "BiVision"] buffer <- textEntry f [] mainArea <- textCtrl f [] set f [layout := expand$ column 2 [hstretch $widget buffer, stretch$ widget mainArea]] This gives me a window with the desired controls. But they are not expanding. Why? I tried different combinations of expand and stretch (I do not really understand the difference?). But nothing works. I am guessing there is something simple I do not yet understand :). Thanks for any help! Nathan From michael.peternell at gmx.at Wed Aug 14 20:30:01 2013 From: michael.peternell at gmx.at (Michael Peternell) Date: Wed, 14 Aug 2013 20:30:01 +0200 Subject: [Haskell-beginners] print [] In-Reply-To: References: Message-ID: There is a way. It's print "[]" if the compiler cannot infer the type of an empty list, it's because you wrote a literal empty list []. Sorry for giving such a practical (non-theoretical) answer ;) Michael From KKelleher at homesite.com Wed Aug 14 21:04:43 2013 From: KKelleher at homesite.com (Kelleher, Kevin) Date: Wed, 14 Aug 2013 19:04:43 +0000 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <105AED40-8115-4480-B039-E58C7A013D1C@arqux.com> Message-ID: <644E83C9A0C2C54A9F40341B5F63393716B8E184@CAMBOSEXC06.camelot.local> I have had the same frustration. It feels like you're learning a foreign language that doesn't have any vocabulary for walking directions, prices, or ordering food. So, while you could discuss philosophy, say, you wouldn't be able to ask for a drink of water, or how to find the bathroom. I do believe there is a payoff to studying Haskell, and I do keep plugging away, rather inconstantly. It's difficult, because I don't really have the time now, but hopefully I can devote some of an upcoming vacation to make what I know of Haskell finally touch the ground. Kevin From: Beginners [mailto:beginners-bounces at haskell.org] On Behalf Of Stephen Tetley Sent: Sunday, August 11, 2013 11:45 AM To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Subject: Re: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment I tend to see OO subtyping and heterogeneous collections as more an impediment to (pure) functional integration with the "rest-of-the-world" than state. There are a lot of quality wrappers to stateful C libraries from Haskell, but few to object oriented C++ or Objective C libraries. On 11 August 2013 16:10, Philippe Sismondi > wrote: As a result of this little thread I have come to another conclusion, and this is just my subjective view. Most of the software that I am interested in seems to live most comfortably with a stateful conception of the world. (The native libraries I find most useful certainly are stateful.) I am reasonably competent with monads and monad transformers in Haskell. But, to be honest, after three years of pluggin away at Haskell, I am not the least convinced that the problem of handling a changing external world in a pure functional language has been successfully solved by those techniques. I always feel as though I am using the robot arm on a space shuttle when a screwdriver would do. (Again, no need to rebut this - I may be wrong or just to stupid to use Haskell effectively - so be it.) Perhaps in the end I do not really believe that functional programming is the panacea that its devotees claim it to be. This message (including any attachments) is intended only for the use of the individual or entity to which it is addressed and may contain information that is non-public, proprietary, privileged, confidential, and exempt from disclosure under applicable law or may constitute as attorney work product. If you are not the intended recipient, you are hereby notified that any use, dissemination, distribution, or copying of this communication is strictly prohibited. If you have received this communication in error, notify us immediately by telephone and (i) destroy this message if a facsimile or (ii) delete this message immediately if this is an electronic communication. Thank you. -------------- next part -------------- An HTML attachment was scrubbed... URL: From nathan.huesken at posteo.de Thu Aug 15 13:56:25 2013 From: nathan.huesken at posteo.de (=?ISO-8859-1?Q?Nathan_H=FCsken?=) Date: Thu, 15 Aug 2013 13:56:25 +0200 Subject: [Haskell-beginners] Trouble with layout in wxHaskell In-Reply-To: <520B9C54.1040806@posteo.de> References: <520B9C54.1040806@posteo.de> Message-ID: <520CC1E9.6090101@posteo.de> Ok, I need to use fill und hfill. I personally think the documentation is little confusing, but I will take this to the developers. Regards, Nathan On 08/14/2013 05:03 PM, Nathan H?sken wrote: > Hey, > > I am trying to understand how do a layout in wxHaskell. I am trying to > make a frame with 2 text entries. One single lined at the top and one > multiline below it. Both should expand with the window size > horizontal. When the window is expanded vertical, only the multiline > text entry should expand. > > Her is what I wrote: > > main = start mainFrame > > mainFrame = do > f <- frame [text := "BiVision"] > buffer <- textEntry f [] > mainArea <- textCtrl f [] > set f [layout := expand $column 2 [hstretch$ widget buffer, > stretch $widget mainArea]] > > This gives me a window with the desired controls. But they are not > expanding. > Why? > I tried different combinations of expand and stretch (I do not really > understand the difference?). But nothing works. > > I am guessing there is something simple I do not yet understand :). > > Thanks for any help! > Nathan > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners From bryanvick at gmail.com Thu Aug 15 19:49:54 2013 From: bryanvick at gmail.com (Bryan Vicknair) Date: Thu, 15 Aug 2013 10:49:54 -0700 Subject: [Haskell-beginners] Compiler can't deduce Bool as instance of ToField In-Reply-To: References: Message-ID: <20130815174954.GB3349@bry-m6300> > > Bryan Vicknair: > > postgresql-simple declares Bool as an instance of the ToField class. The > > compiler can't deduce that given this simple code however: > > > > import Database.PostgreSQL.Simple.ToField (ToField(..)) > > > > foo :: (ToField a) => a > > foo = True > > > > > > It fails with this error: > > > > Db.hs:64:7: > > Could not deduce (a ~ Bool) > > from the context (ToField a) > > bound by the type signature for foo :: ToField a => a > > at Db.hs:63:8-23 > > a' is a rigid type variable bound by > > the type signature for foo :: ToField a => a at Db.hs:63:8 > > In the expression: True > > In an equation for foo': foo = True > > Failed, modules loaded: none. > > > > What am I missing? > > > Oliver Charles > You have declared that foo is *any* type that has a ToField instance, > allowing the caller of foo to determine the type at their will. However, > your implementation of foo is more specific and requires a is actually Bool > and nothing else. > On 14 Aug 2013 10:24, "Bryan Vicknair" wrote: I see. I was confused, because the following works: bar :: (Num a) => a bar = 5 I guess it is because the literal '5' could be an Int or a Float, and the type system doesn't know. When I saw that the following does not work...:: bar :: (Show a) => a bar = False ...it made a bit more sense. It seems that if a literal can be considered part of a typeclass in more than one way, you can declare that literal to be of that typeclass. However, if there is only one way for a literal to belong to a typeclass, then you can't. Are there other literals (besides (Num a) => [a]) that don't have a concrete type like literal numbers? If I understand correctly, -XOverloadedStrings turns all literal strings into IsString instead of [Char], so that seems to be another case. I don't have a need for this, but now I'm curious: is there a way to declare that all literal numbers in a module are of a certain concrete type? Something like -XAllNumbersAreInt8? From tom.davie at gmail.com Thu Aug 15 20:11:28 2013 From: tom.davie at gmail.com (Tom Davie) Date: Thu, 15 Aug 2013 20:11:28 +0200 Subject: [Haskell-beginners] Compiler can't deduce Bool as instance of ToField In-Reply-To: <20130815174954.GB3349@bry-m6300> References: <20130815174954.GB3349@bry-m6300> Message-ID: <40F23FAC-C0F7-4C35-BAD2-6591EEF31922@gmail.com> On 15 Aug 2013, at 19:49, Bryan Vicknair wrote: >>> Bryan Vicknair: >>> postgresql-simple declares Bool as an instance of the ToField class. The >>> compiler can't deduce that given this simple code however: >>> >>> import Database.PostgreSQL.Simple.ToField (ToField(..)) >>> >>> foo :: (ToField a) => a >>> foo = True >>> >>> >>> It fails with this error: >>> >>> Db.hs:64:7: >>> Could not deduce (a ~ Bool) >>> from the context (ToField a) >>> bound by the type signature for foo :: ToField a => a >>> at Db.hs:63:8-23 >>> a' is a rigid type variable bound by >>> the type signature for foo :: ToField a => a at Db.hs:63:8 >>> In the expression: True >>> In an equation for foo': foo = True >>> Failed, modules loaded: none. >>> >>> What am I missing? >>> > >> Oliver Charles >> You have declared that foo is *any* type that has a ToField instance, >> allowing the caller of foo to determine the type at their will. However, >> your implementation of foo is more specific and requires a is actually Bool >> and nothing else. >> On 14 Aug 2013 10:24, "Bryan Vicknair" wrote: > > I see. I was confused, because the following works: > > bar :: (Num a) => a > bar = 5 > > I guess it is because the literal '5' could be an Int or a Float, and the type > system doesn't know. > > When I saw that the following does not work...:: > > bar :: (Show a) => a > bar = False > > ...it made a bit more sense. It seems that if a literal can be considered part > of a typeclass in more than one way, you can declare that literal to be of that > typeclass. However, if there is only one way for a literal to belong to a > typeclass, then you can't. That?s not it here. What you?re saying in the first case is ?no matter what numeric type you want, I can provide it?. This is true, because the compiler can decide that 5 can indeed be any numeric type. In the latter case you make the promise ?no matter what type you want, as long as you can show it, I can provide it?, but it?s false. I can say ?well okay, give me an Int? and your function can not provide it to me, despite telling me in its type signature that it can. Tom Davie From toad3k at gmail.com Thu Aug 15 23:19:16 2013 From: toad3k at gmail.com (David McBride) Date: Thu, 15 Aug 2013 17:19:16 -0400 Subject: [Haskell-beginners] Compiler can't deduce Bool as instance of ToField In-Reply-To: <20130815174954.GB3349@bry-m6300> References: <20130815174954.GB3349@bry-m6300> Message-ID: The type of 5: Prelude> :t 5 5 :: Num a => a That's why it works. The literal's type uses the type class. Other fun stuff: Prelude> :t "asdf" "asdf" :: [Char] Prelude> :set -XOverloadedStrings Prelude> :t "asdf" "asdf" :: Data.String.IsString a => a -------------- next part -------------- An HTML attachment was scrubbed... URL: From ky3 at atamo.com Thu Aug 15 23:27:55 2013 From: ky3 at atamo.com (Kim-Ee Yeoh) Date: Fri, 16 Aug 2013 04:27:55 +0700 Subject: [Haskell-beginners] Compiler can't deduce Bool as instance of ToField In-Reply-To: <20130815174954.GB3349@bry-m6300> References: <20130815174954.GB3349@bry-m6300> Message-ID: On Fri, Aug 16, 2013 at 12:49 AM, Bryan Vicknair wrote: > I guess it is because the literal '5' could be an Int or a Float, and the > type > system doesn't know. > The literal '5' is *known* to be of the *polymorphic* type (Num a => a). Ints and Floats are monomorphic instances of this polymorphic type. See http://www.haskell.org/onlinereport/basic.html#numeric-literals > When I saw that the following does not work...:: > > bar :: (Show a) => a > bar = False > > ...it made a bit more sense. It seems that if a literal can be considered > part > of a typeclass in more than one way, you can declare that literal to be of > that > typeclass. However, if there is only one way for a literal to belong to a > typeclass, then you can't. > The part "if a literal can be considered part of a typeclass in more than one way" isn't how you want to think about them (see link). Good attempt though. > I don't have a need for this, but now I'm curious: is there a way to declare that all literal numbers in a module are of a certain concrete type? Something like -XAllNumbersAreInt8? You enforce monomorphism by specifying type signatures where appropriate. In some cases, a light touch is all that's needed because type inference does the rest. -- Kim-Ee -------------- next part -------------- An HTML attachment was scrubbed... URL: From psismondi at arqux.com Sat Aug 17 23:10:26 2013 From: psismondi at arqux.com (Philippe Sismondi) Date: Sat, 17 Aug 2013 17:10:26 -0400 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> Message-ID: I am the original poster on this. I have not checked the thread for some time - I had decided to sign off on it because seemed to be veering off in the direction of a religious dispute. (That is probably my fault; my observations were perhaps too caustic.) But I seem unable to resist. Sorry if this post makes you go "TLDR". We seem to be revisiting the whole "state/mutability is evil" thing for the umpteenth time. There is an obvious dichotomy between algorithms that are stateful (but entirely internal to the program) and those that require state + IO. (There is probably some proper terminology for this difference, but I don't know what it is.) Anyway, my main frustration with pure FP is in respect of interacting with the "outside world", i.e. IO. I suppose that in part is what what drove my question of operating systems and their native libraries, all of which (in the case of OS X) rely on mutability and state. One of the commonest claims made about pure functional languages, and Haskell in particular, is that programs written in pure functional languages contain fewer bugs than?"other stuff". The claim seems largely to be about (a) avoiding mutability, and (b) type safety and static type checking. (I have no quibble with the typing part of the claim; I love this in Haskell.) In the same spirit, Haskell programmers are wont to assert that they are confident that a program that compiles is a correct program. However, these claims (as made in books on Haskell, and on mailing lists such as this) are no more than anecdotal. At the risk of pouring fuel on the fire, I would ask: Can anyone point me to scientific studies quantifying the benefits of pure functional languages? I expect they exist, but I am interested not just in lack of errors, but in overall productivity. (It's easy to avoid program bugs: just write programs that don't do anything, or don't write any software at all.) The value of being able to produce correct programs is, no doubt enormous. But that is hardly the end of the matter. My sense is that most of us who produce software are less interested in error-free software than in overall productivity. There is an enormous amount of highly useful software out there that contains all kinds of both known and as yet undiscovered bugs. In other words, I propose that the perfect is the enemy of the good here. We are all most likely exchanging our thoughts on the topic on this mailing list using buggy operating systems, buggy email software, buggy hardware, and buggy human brains. But somehow we soldier bravely on ;-) Somebody may be writing an experimental OS in Haskell, but I ain't using it at the moment, and neither are you (probably). Moreover, whenever a list appears of Haskell systems in production, it is woefully tiny. I am reminded of the same kind of lists of Lisp software: They invariably include AutoCAD scripting, emacs, Viaweb, and?and?and a few other niche thingies, and the list peters out. Because people are mostly writing software in shitty languages like Java and Objective-C and C++ and so on. The world outside your program is stateful, I think (I agree with Heraclitus on this). Any software that must deal with the "world" outside itself over time must deal with state. So even devotees of of pure FP agree that the trick is to somehow hive off or isolate state and IO, or maybe to mimic it with pure techniques in order to ensure correctness. But the problem of dealing with IO in Haskell is hardly a solved problem. Reading the history of Haskell one discovers that the use of monads was not the first attempt at dealing with IO in the language. Moreover, research papers continue to appear on other ways to handle state and IO in FPL. So it can hardly be said that, "well, we've got this problem licked, so just suck it up and learn it." It can hardly be questioned that people find monads tough to understand; some never do, and some (like me) can use them but find them a pain in the ass (so are arrows). Thus in that sense, Haskell has saved legions of programmers from writing buggy software - because it has prevented them from writing any software at all, at least in Haskell. My guess is that FP research has and will continue to provide real benefits to software development. But making that observation that other approaches are a "bad idea" is not terribly helpful (to me, anyway). I'll continue to tinker with Haskell, and when I want to write audio software or a game or something I'll use Objective-C and the massive bug-riddled but vendor-supported, documented libraries that come with OS X. - P - From allbery.b at gmail.com Sat Aug 17 23:26:50 2013 From: allbery.b at gmail.com (Brandon Allbery) Date: Sat, 17 Aug 2013 17:26:50 -0400 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> Message-ID: On Sat, Aug 17, 2013 at 5:10 PM, Philippe Sismondi wrote: > My sense is that most of us who produce software are less interested in > error-free software than in overall productivity. There is an enormous > amount of highly useful software out there that contains all kinds of both > known and as yet undiscovered bugs. In other words, I propose that the > perfect is the enemy of the good here. I have several security lists (and victims of the reported security issues) and several rather large sectors of industry (financial and medical, to name two) which are increasingly realizing that this attitude isn't viable any more. It is becoming increasingly obvious that "productivity is more important than error-free" leads to very expensive reparations down the road. -- brandon s allbery kf8nh sine nomine associates allbery.b at gmail.com ballbery at sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net -------------- next part -------------- An HTML attachment was scrubbed... URL: From apfelmus at quantentunnel.de Sun Aug 18 10:55:00 2013 From: apfelmus at quantentunnel.de (Heinrich Apfelmus) Date: Sun, 18 Aug 2013 10:55:00 +0200 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> Message-ID: Philippe Sismondi wrote: > I am the original poster on this. [..] > > One of the commonest claims made about pure functional languages, and > Haskell in particular, is that programs written in pure functional > languages contain fewer bugs than?"other stuff". The claim seems > largely to be about (a) avoiding mutability, and (b) type safety and > static type checking. (I have no quibble with the typing part of the > claim; I love this in Haskell.) In the same spirit, Haskell > programmers are wont to assert that they are confident that a program > that compiles is a correct program. However, these claims (as made in > books on Haskell, and on mailing lists such as this) are no more than > anecdotal. Note that there is a principled argument for (a) that does not rely on empirical evidence. It goes as follows: avoiding mutable state means that a piece of source code, say a function definition, can be reasoned about *without* needing to know about the possible contexts (code paths) in which it can appear. In other words, the correctness of a pure function can be checked by looking only at the function definition, the rest of the program code has no bearing on it. In contrast, mutable state usually means a "spooky action at a distance" between separate code sections. > At the risk of pouring fuel on the fire, I would ask: Can > anyone point me to scientific studies quantifying the benefits of > pure functional languages? I expect they exist, but I am interested > not just in lack of errors, but in overall productivity. (It's easy > to avoid program bugs: just write programs that don't do anything, or > don't write any software at all.) You know well that (1) such studies are hard to design for any language and (2) nobody asks for studies that quantify the benefits of imperative languages, simply because they happen to be the status quo. Still, there are some interesting case studies, for instance P Hudak and M P Jones "Haskell vs. Ada vs. C++ vs. Awk vs. ... An Experiment in Software Prototyping Productivity" http://haskell.cs.yale.edu/?post_type=publication&p=366 > The value of being able to produce correct programs is, no doubt > enormous. But that is hardly the end of the matter. > > My sense is that most of us who produce software are less interested > in error-free software than in overall productivity. There is an > enormous amount of highly useful software out there that contains all > kinds of both known and as yet undiscovered bugs. [..] I always dread the prospect of dealing with non-Haskell libraries, because I often lose several days of development time investigating how a library really behaves as opposed to how it should behave according to the specification. I have mentioned two recent examples in a previous email (HTML 5 drag & drop, WebSockets), but it has also happened to me when doing development with Cocoa. > The world outside your program is stateful, I think (I agree with > Heraclitus on this). Any software that must deal with the "world" > outside itself over time must deal with state. So even devotees of of > pure FP agree that the trick is to somehow hive off or isolate state > and IO, or maybe to mimic it with pure techniques in order to ensure > correctness. Note that source code is a *model* and we are free to choose any model we like. The "true nature" of the world being "stateful" has no bearing on this. For instance, physicists actually model the world in a stateless fashion (a particle is modeled by a function from time to position in space). > But the problem of dealing with IO in Haskell is hardly a solved > problem. Reading the history of Haskell one discovers that the use of > monads was not the first attempt at dealing with IO in the language. > Moreover, research papers continue to appear on other ways to handle > state and IO in FPL. So it can hardly be said that, "well, we've got > this problem licked, so just suck it up and learn it." It can hardly > be questioned that people find monads tough to understand; some never > do, and some (like me) can use them but find them a pain in the ass > (so are arrows). Huh? How is dealing with IO not a solved problem in Haskell? Note that there may be smarter ways to deal with "IO" (e.g. FRP), but I don't see how experiments with new models invalidate an existing model. Also note that software transactional memory (STM) is a solved problem in Haskell, something that cannot be said of any other imperative language. Best regards, Heinrich Apfelmus -- http://apfelmus.nfshost.com From aurimas.anskaitis at vgtu.lt Mon Aug 19 03:18:28 2013 From: aurimas.anskaitis at vgtu.lt (Aurimas) Date: Mon, 19 Aug 2013 04:18:28 +0300 Subject: [Haskell-beginners] Type inference in ST monad? Message-ID: <52117264.4010404@vgtu.lt> I have the following code which does not compile due to explicit type annotation (ST s Double). Error message says usual thing about "s" type variables. ---------------------------------------------------------------------------------------------------------- import Control.Monad.ST import System.Random.MWC (initialize, uniformR, Gen) import Control.Monad.Loops (whileM) import Data.Vector (singleton) import Data.Word(Word32) main :: IO () main = do print$ runSimulation 1 runSimulation :: Word32 -> [Int] runSimulation seed = runST $do gen <- initialize (singleton seed) whileM (do r1 <- uniformR (-1.0, 1.0) gen :: ST s Double -- does not compile due to this if r1 > 0.0 then return True else return False) (do r2 <- uniformR (0, 10) gen if r2 > 5 then return r2 else return 0) --------------------------------------------------------------------------------------------------------- if I rewrite runSimulation like this (below), everything is OK. --------------------------------------------------------------------------------------------------------- runSimulation :: Word32 -> [Int] runSimulation seed = runST$ do gen <- initialize (singleton seed) whileM (do r1 <- tempFun gen if r1 > 0.0 then return True else return False) (do r2 <- uniformR (0, 10) gen if r2 > 5 then return r2 else return 0) where tempFun :: Gen s -> ST s Double -- this line automatically provides required type annotation tempFun g = uniformR (-1.0, 1.0) g --------------------------------------------------------------------------------------------------------- Ca somebody explain what's wrong with the first version? Best Regards, Aurimas From toad3k at gmail.com Mon Aug 19 04:04:26 2013 From: toad3k at gmail.com (David McBride) Date: Sun, 18 Aug 2013 22:04:26 -0400 Subject: [Haskell-beginners] Type inference in ST monad? In-Reply-To: <52117264.4010404@vgtu.lt> References: <52117264.4010404@vgtu.lt> Message-ID: I'm not exactly sure why you get the error, but the easiest way to fix it is just to type it this way: runSimulation :: Word32 -> [Int] runSimulation seed = runST $do gen <- initialize (singleton seed) whileM (do r1 <- uniformR (-1.0, 1.0 :: Double) gen if r1 > 0.0 then return True else return False) (do r2 <- uniformR (0, 10 :: Int) gen if r2 > 5 then return r2 else return 0) It has something to do with the forall s in runST, although I'm not completely sure what. On Sun, Aug 18, 2013 at 9:18 PM, Aurimas wrote: > I have the following code which does not compile due to explicit type > annotation > (ST s Double). Error message says usual thing about "s" type variables. > > ------------------------------**------------------------------** > ------------------------------**---------------- > import Control.Monad.ST > import System.Random.MWC (initialize, uniformR, Gen) > import Control.Monad.Loops (whileM) > import Data.Vector (singleton) > import Data.Word(Word32) > > main :: IO () > main = do > print$ runSimulation 1 > > runSimulation :: Word32 -> [Int] > runSimulation seed = runST $do > gen <- initialize (singleton seed) > whileM (do r1 <- uniformR (-1.0, 1.0) gen :: ST s Double -- does not > compile due to this > if r1 > 0.0 then return True else return False) > (do r2 <- uniformR (0, 10) gen > if r2 > 5 then return r2 else return 0) > ------------------------------**------------------------------** > ------------------------------**--------------- > > if I rewrite runSimulation like this (below), everything is OK. > > ------------------------------**------------------------------** > ------------------------------**--------------- > runSimulation :: Word32 -> [Int] > runSimulation seed = runST$ do > gen <- initialize (singleton seed) > whileM (do r1 <- tempFun gen > if r1 > 0.0 then return True else return False) > (do r2 <- uniformR (0, 10) gen > if r2 > 5 then return r2 else return 0) > where tempFun :: Gen s -> ST s Double -- this line automatically > provides required type annotation > tempFun g = uniformR (-1.0, 1.0) g > ------------------------------**------------------------------** > ------------------------------**--------------- > > Ca somebody explain what's wrong with the first version? > > Best Regards, > Aurimas > > > ______________________________**_________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/**mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: From aurimas.anskaitis at vgtu.lt Mon Aug 19 07:36:10 2013 From: aurimas.anskaitis at vgtu.lt (Aurimas) Date: Mon, 19 Aug 2013 08:36:10 +0300 Subject: [Haskell-beginners] Type inference in ST monad? In-Reply-To: References: <52117264.4010404@vgtu.lt> Message-ID: <5211AECA.6050409@vgtu.lt> On 08/19/2013 05:04 AM, David McBride wrote: > I'm not exactly sure why you get the error, but the easiest way to fix > it is just to type it this way: > > runSimulation :: Word32 -> [Int] > runSimulation seed = runST $do > gen <- initialize (singleton seed) > whileM (do r1 <- uniformR (-1.0, 1.0 :: Double) gen > if r1 > 0.0 then return True else return False) > (do r2 <- uniformR (0, 10 :: Int) gen > if r2 > 5 then return r2 else return 0) > > It has something to do with the forall s in runST, although I'm not > completely sure what. > Thanks, this is clearly the most easy way to fix the problem. But the question remains - why the line "do r1 <- uniformR (-1.0, 1.0 :: Double) gen" cannot be annotated with ST s Double? From magnus at therning.org Mon Aug 19 10:18:11 2013 From: magnus at therning.org (Magnus Therning) Date: Mon, 19 Aug 2013 10:18:11 +0200 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> Message-ID: <20130819081811.GB1125@mteis.semcon.se> On Sun, Aug 18, 2013 at 10:55:00AM +0200, Heinrich Apfelmus wrote: > Philippe Sismondi wrote: > >I am the original poster on this. [..] [...] > >At the risk of pouring fuel on the fire, I would ask: Can anyone > >point me to scientific studies quantifying the benefits of pure > >functional languages? I expect they exist, but I am interested not > >just in lack of errors, but in overall productivity. (It's easy to > >avoid program bugs: just write programs that don't do anything, or > >don't write any software at all.) > > You know well that (1) such studies are hard to design for any > language and (2) nobody asks for studies that quantify the benefits > of imperative languages, simply because they happen to be the status > quo. > > Still, there are some interesting case studies, for instance > > P Hudak and M P Jones > "Haskell vs. Ada vs. C++ vs. Awk vs. ... > An Experiment in Software Prototyping Productivity" > http://haskell.cs.yale.edu/?post_type=publication&p=366 There are also a few articles on Erlang, and I believe there's good reason to believe the results with that language can be transferred to other "languages without assignment". - Productivity gains with Erlarng http://dl.acm.org/citation.cfm?id=1362710 slides: http://is.gd/BgGZyh (behind a paywall, I haven't found a freely available copy) - Breakthrough in software design productivity through the use of declarative programming http://www.sciencedirect.com/science/article/pii/S0925527397807669 (behind a paywall, I haven't found a freely available copy) /M -- Magnus Therning OpenPGP: 0xAB4DFBA4 email: magnus at therning.org jabber: magnus at therning.org twitter: magthe http://therning.org/magnus Perl is another example of filling a tiny, short-term need, and then being a real problem in the longer term. -- Alan Kay -------------- next part -------------- A non-text attachment was scrubbed... Name: not available Type: application/pgp-signature Size: 230 bytes Desc: not available URL: From hjgtuyl at chello.nl Mon Aug 19 12:52:20 2013 From: hjgtuyl at chello.nl (Henk-Jan van Tuyl) Date: Mon, 19 Aug 2013 12:52:20 +0200 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: <20130819081811.GB1125@mteis.semcon.se> References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <20130819081811.GB1125@mteis.semcon.se> Message-ID: On Mon, 19 Aug 2013 10:18:11 +0200, Magnus Therning wrote: > There are also a few articles on Erlang, and I believe there's good > reason to believe the results with that language can be transferred to > other "languages without assignment". > > - Productivity gains with Erlarng > http://dl.acm.org/citation.cfm?id=1362710 > slides: http://is.gd/BgGZyh > (behind a paywall, I haven't found a freely available copy) > For free: "Four-fold Increase in Productivity and Quality" Ulf Wiger, FemSYS 2001, Munich http://www.erlang.se/publications/Ulf_Wiger.pdf Regards, Henk-Jan van Tuyl -- Folding at home What if you could share your unused computer power to help find a cure? In just 5 minutes you can join the world's biggest networked computer and get us closer sooner. Watch the video. http://folding.stanford.edu/ http://Van.Tuyl.eu/ http://members.chello.nl/hjgtuyl/tourdemonad.html Haskell programming -- From allbery.b at gmail.com Mon Aug 19 15:59:28 2013 From: allbery.b at gmail.com (Brandon Allbery) Date: Mon, 19 Aug 2013 09:59:28 -0400 Subject: [Haskell-beginners] Type inference in ST monad? In-Reply-To: <5211AECA.6050409@vgtu.lt> References: <52117264.4010404@vgtu.lt> <5211AECA.6050409@vgtu.lt> Message-ID: On Mon, Aug 19, 2013 at 1:36 AM, Aurimas wrote: > Thanks, this is clearly the most easy way to fix the problem. But the > question remains - why the line > "do r1 <- uniformR (-1.0, 1.0 :: Double) gen" cannot be annotated with > ST s Double? > Maybe you need ScopedTypeVariables? But I'd be suspicious that, given the forall in the definition of ST, that such a type annotation *always* creates a new s and some other way to "force" the type is needed. -- brandon s allbery kf8nh sine nomine associates allbery.b at gmail.com ballbery at sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net -------------- next part -------------- An HTML attachment was scrubbed... URL: From aurimas.anskaitis at vgtu.lt Mon Aug 19 16:12:27 2013 From: aurimas.anskaitis at vgtu.lt (Aurimas) Date: Mon, 19 Aug 2013 17:12:27 +0300 Subject: [Haskell-beginners] Type inference in ST monad? In-Reply-To: References: <52117264.4010404@vgtu.lt> <5211AECA.6050409@vgtu.lt> Message-ID: <521227CB.30009@vgtu.lt> On 08/19/2013 04:59 PM, Brandon Allbery wrote: > On Mon, Aug 19, 2013 at 1:36 AM, Aurimas > wrote: > > Thanks, this is clearly the most easy way to fix the problem. But > the question remains - why the line > "do r1 <- uniformR (-1.0, 1.0 :: Double) gen" cannot be annotated > with ST s Double? > > > Maybe you need ScopedTypeVariables? But I'd be suspicious that, given > the forall in the definition of ST, that such a type annotation > *always* creates a new s and some other way to "force" the type is > needed. > ScopedTypeVariables does not solve the problem. As you noticed, new s is created and runST is not happy about this. It is strange that this type can be "forced" replacing given line by a function returning ST s a. -------------- next part -------------- An HTML attachment was scrubbed... URL: From paratribulations at free.fr Mon Aug 19 22:28:14 2013 From: paratribulations at free.fr (TP) Date: Mon, 19 Aug 2013 22:28:14 +0200 Subject: [Haskell-beginners] multi-parameter typeclass with default implementation Message-ID: Hi, I struggle with a dummy example using a multi-parameter typeclass containing a default implementation for a function: --------------------------------------- {-# LANGUAGE MultiParamTypeClasses #-} class Foo a b where bar :: a -> Int foobar :: a -> b -> Int foobar avalue bvalue = bar avalue instance Foo Int Int where bar i = 5 main = do print$ bar (4::Int) --------------------------------------- I obtain the following errors. I have tried various things without any success. Any help appreciated! Thanks TP PS: The errors: $runghc test.hs test.hs:8:28: Could not deduce (Foo a b1) arising from a use of bar' from the context (Foo a b) bound by the class declaration for Foo' at test.hs:(3,1)-(8,37) The type variable b1' is ambiguous Possible fix: add a type signature that fixes these type variable(s) In the expression: bar avalue In an equation for foobar': foobar avalue bvalue = bar avalue test.hs:16:9: No instance for (Foo Int b0) arising from a use of bar' The type variable b0' is ambiguous Possible fix: add a type signature that fixes these type variable(s) Note: there is a potential instance available: instance Foo Int Int -- Defined at test.hs:10:10 Possible fix: add an instance declaration for (Foo Int b0) In the second argument of ($)', namely bar (4 :: Int)' In a stmt of a 'do' block: print $bar (4 :: Int) In the expression: do { print$ bar (4 :: Int) } From bgamari.foss at gmail.com Mon Aug 19 22:59:42 2013 From: bgamari.foss at gmail.com (Ben Gamari) Date: Mon, 19 Aug 2013 16:59:42 -0400 Subject: [Haskell-beginners] multi-parameter typeclass with default implementation In-Reply-To: References: Message-ID: <87haeltn0h.fsf@gmail.com> TP writes: > Hi, > > I struggle with a dummy example using a multi-parameter typeclass containing > a default implementation for a function: > > --------------------------------------- > {-# LANGUAGE MultiParamTypeClasses #-} > > class Foo a b where > > bar :: a -> Int > The problem is in this declaration, which does not mention the type "b". This makes it impossible for the compiler to infer which instance to use when "bar" is used. This is what the compiler is trying to tell you when it says "The type variable b1' is ambiguous". As far as I know, you'd need to do something like this to accomplish what you are after, {-# LANGUAGE MultiParamTypeClasses, DefaultSignatures #-} class Bar a where bar :: a -> Int class FooBar a b where foobar :: a -> b -> Int default foobar :: Bar a => a -> b -> Int foobar avalue bvalue = bar avalue instance Bar Int where bar i = 5 instance FooBar Int Int main = do print $bar (4::Int) print$ foobar (5::Int) (2::Int) Cheers, - Ben -------------- next part -------------- A non-text attachment was scrubbed... Name: not available Type: application/pgp-signature Size: 489 bytes Desc: not available URL: From magnus at therning.org Tue Aug 20 10:11:39 2013 From: magnus at therning.org (Magnus Therning) Date: Tue, 20 Aug 2013 10:11:39 +0200 Subject: [Haskell-beginners] How Haskell Fits Into an Operating System / API Environment In-Reply-To: References: <2EEE4F93-1136-48EE-8B22-A3DC7E3C4FF6@arqux.com> <20130819081811.GB1125@mteis.semcon.se> Message-ID: <20130820081139.GA8726@mteis.semcon.se> On Mon, Aug 19, 2013 at 12:52:20PM +0200, Henk-Jan van Tuyl wrote: > On Mon, 19 Aug 2013 10:18:11 +0200, Magnus Therning > wrote: > > >There are also a few articles on Erlang, and I believe there's good > >reason to believe the results with that language can be transferred to > >other "languages without assignment". > > > >- Productivity gains with Erlarng > > http://dl.acm.org/citation.cfm?id=1362710 > > slides: http://is.gd/BgGZyh > > (behind a paywall, I haven't found a freely available copy) > > > > For free: > "Four-fold Increase in Productivity and Quality" > Ulf Wiger, FemSYS 2001, Munich > http://www.erlang.se/publications/Ulf_Wiger.pdf Thanks! I was trying to find that paper yesterday, but apparently my google-fu isn't as strong as yours ;) /M -- Magnus Therning OpenPGP: 0xAB4DFBA4 email: magnus at therning.org jabber: magnus at therning.org twitter: magthe http://therning.org/magnus Most software today is very much like an Egyptian pyramid with millions of bricks piled on top of each other, with no structural integrity, but just done by brute force and thousands of slaves. -- Alan Kay -------------- next part -------------- A non-text attachment was scrubbed... Name: not available Type: application/pgp-signature Size: 230 bytes Desc: not available URL: From paratribulations at free.fr Tue Aug 20 11:44:49 2013 From: paratribulations at free.fr (TP) Date: Tue, 20 Aug 2013 11:44:49 +0200 Subject: [Haskell-beginners] multi-parameter typeclass with default implementation References: <87haeltn0h.fsf@gmail.com> Message-ID: Ben Gamari wrote: >> {-# LANGUAGE MultiParamTypeClasses #-} >> >> class Foo a b where >> >> bar :: a -> Int >> > The problem is in this declaration, which does not mention the type > "b". This makes it impossible for the compiler to infer which instance > to use when "bar" is used. This is what the compiler is trying to tell > you when it says "The type variable b1' is ambiguous". Thanks, this works perfectly. Yet, to try to improve myself, I would like to discuss a bit on the text of the two obtained errors in my example (correct me if I am wrong below). """ test.hs:16:9: No instance for (Foo Int b0) arising from a use of bar' """ The second error is clear: indeed by calling bar on 4::Int, we don't indicate the type of b allowing to choose the right typeclass for the implementation of the bar function. The fact there is only one typeclass instance with Int as type for type variable a does not change anything: Haskell does not look at the implemented instances. In other words, ?typeclasses are open?: more typeclasses with Int as first type variable (and types different from Int for type variable b) may be added in the future, and Haskell asks to specify which one is to use now, not in the future. (Is this the right meaning for "typeclasses are open"?). """ test.hs:8:28: Could not deduce (Foo a b1) arising from a use of bar' from the context (Foo a b) bound by the class declaration for Foo' at test.hs:(3,1)-(8,37) The type variable b1' is ambiguous """ Is this the same interpretation here? That is, even in the default implementation of a typeclass function, Haskell does not suppose that the instance of bar to use is the one of the considered instance? Thanks a lot, TP From paratribulations at free.fr Tue Aug 20 12:13:28 2013 From: paratribulations at free.fr (TP) Date: Tue, 20 Aug 2013 12:13:28 +0200 Subject: [Haskell-beginners] multi-parameter typeclass with default implementation References: <87haeltn0h.fsf@gmail.com> Message-ID: <83sbea-ove.ln1@rama.universe> Ben Gamari wrote: > As far as I know, you'd need to do something like this to accomplish > what you are after, > > {-# LANGUAGE MultiParamTypeClasses, DefaultSignatures #-} > > class Bar a where > bar :: a -> Int > > class FooBar a b where > foobar :: a -> b -> Int > default foobar :: Bar a => a -> b -> Int > foobar avalue bvalue = bar avalue > > instance Bar Int where > bar i = 5 > instance FooBar Int Int > > main = do > print $bar (4::Int) > print$ foobar (5::Int) (2::Int) It seems that the "DefaultSignatures" extension is not necessary, the following version works correctly on my computer (ghc 7.6.2): ------- {-# LANGUAGE MultiParamTypeClasses #-} class Bar a where bar :: a -> Int class FooBar a b where foobar :: Bar a => a -> b -> Int foobar avalue bvalue = bar avalue instance Bar Int where bar i = 5 instance FooBar Int Int main = do print $bar (4::Int) print$ foobar (5::Int) (2::Int) ------- From twanvl at gmail.com Tue Aug 20 14:20:24 2013 From: twanvl at gmail.com (Twan van Laarhoven) Date: Tue, 20 Aug 2013 14:20:24 +0200 Subject: [Haskell-beginners] multi-parameter typeclass with default implementation In-Reply-To: <83sbea-ove.ln1@rama.universe> References: <87haeltn0h.fsf@gmail.com> <83sbea-ove.ln1@rama.universe> Message-ID: <52135F08.9040002@gmail.com> On 20/08/13 12:13, TP wrote: > {-# LANGUAGE MultiParamTypeClasses #-} > > class Bar a where > bar :: a -> Int > > class FooBar a b where > foobar :: Bar a => a -> b -> Int > foobar avalue bvalue = bar avalue > > instance Bar Int where > bar i = 5 > instance FooBar Int Int > > main = do > print $bar (4::Int) > print$ foobar (5::Int) (2::Int) It might be better to make Bar a superclass of FooBar, class Bar a => FooBar a b where foobar :: a -> b -> Int foobar a b = bar a Then the compiler knows that every instance of FooBar also requires an instance of Bar. Twan From diem at xs4all.nl Tue Aug 20 21:02:17 2013 From: diem at xs4all.nl (Dimitri Hendriks) Date: Tue, 20 Aug 2013 21:02:17 +0200 Subject: [Haskell-beginners] cabal package haskore-vintage fails to build Message-ID: <1C957484-DB86-43B2-8927-A012D4480B1A@xs4all.nl> Hi all, I'm new to this list, and know very little about haskell. I am trying to install the package haskore-vintage on mac osx version 10.6.8, using cabal, but this fails; see the log below. I have ghc version 7.6.3, and cabal-install version 1.16.0.2. ------------------------------------------------------------------------------ $cabal install haskore-vintage Resolving dependencies... Configuring haskore-vintage-0.1... Building haskore-vintage-0.1... Preprocessing library haskore-vintage-0.1... [ 1 of 16] Compiling Haskore.Monads ( src/Haskore/Monads.hs, dist/build/Haskore/Monads.o ) [ 2 of 16] Compiling Haskore.Utils ( src/Haskore/Utils.hs, dist/build/Haskore/Utils.o ) src/Haskore/Utils.hs:87:23: Not in scope: catch' src/Haskore/Utils.hs:93:23: Not in scope: catch' Failed to install haskore-vintage-0.1 cabal: Error: some packages failed to install: haskore-vintage-0.1 failed during the building phase. The exception was: ExitFailure 1$ ------------------------------------------------------------------------------ Does anyone know how to resolve this? Related problems I found: https://code.google.com/p/leksah/issues/detail?id=272 (no solution given). And here: https://github.com/haskell/cabal/issues/1137 it appears they think it is a combination of Cabal 1.10 / cabal-install 0.10 and GHC 7.6 the causes the problem with "catch", but I use cabal 1.16. Many thanks in advance, Dimitri From paratribulations at free.fr Tue Aug 20 22:12:19 2013 From: paratribulations at free.fr (TP) Date: Tue, 20 Aug 2013 22:12:19 +0200 Subject: [Haskell-beginners] multi-parameter typeclass with default implementation References: <87haeltn0h.fsf@gmail.com> Message-ID: <36vcea-pt.ln1@rama.universe> Ben Gamari wrote: > {-# LANGUAGE MultiParamTypeClasses, DefaultSignatures #-} In fact, we could try a solution using a simple parameter typeclass containing an implicit existential type b (I hope I am right): ------------------------- class Foo a where bar :: a -> Int foobar :: Foo b => a -> b -> Int foobar avalue bvalue = bar avalue instance Foo Int where bar i = 5 foobar avalue bvalue = (bar avalue) + (bar bvalue) main = do print $bar (4::Int) print$ foobar (5::Int) (3::Int) ------------------------- It works correctly: $runghc test_one_simple_parameter_typeclass.hs 5 10 But if we try to call a function external to the typeclass: ------------------------- toto :: Int -> Int toto i = 4 class Foo a where bar :: a -> Int foobar :: Foo b => a -> b -> Int foobar avalue bvalue = bar avalue instance Foo Int where bar i = 5 foobar avalue bvalue = (bar avalue) + (bar bvalue) + (toto bvalue) main = do print$ bar (4::Int) print $foobar (5::Int) (3::Int) ------------------------- We get an error message (see below) meaning that when we call "toto" with "bvalue", there is not guarantee that "bvalue" is an "Int". So, in this situation, *are we compelled to use multiparameter typeclasses*? PS: the error message yielded by the second example above:$ runghc test_one_simple_parameter_typeclass_limitation.hs test_one_simple_parameter_typeclass_limitation.hs:15:37: Could not deduce (b ~ Int) from the context (Foo b) bound by the type signature for foobar :: Foo b => Int -> b -> Int at test_one_simple_parameter_typeclass_limitation.hs:(13,5)-(15,43) b' is a rigid type variable bound by the type signature for foobar :: Foo b => Int -> b -> Int at test_one_simple_parameter_typeclass_limitation.hs:13:5 In the first argument of toto', namely bvalue' In the second argument of (+)', namely (toto bvalue)' In the expression: (bar avalue) + (bar bvalue) + (toto bvalue) From stephen.tetley at gmail.com Tue Aug 20 23:55:23 2013 From: stephen.tetley at gmail.com (Stephen Tetley) Date: Tue, 20 Aug 2013 22:55:23 +0100 Subject: [Haskell-beginners] cabal package haskore-vintage fails to build In-Reply-To: <1C957484-DB86-43B2-8927-A012D4480B1A@xs4all.nl> References: <1C957484-DB86-43B2-8927-A012D4480B1A@xs4all.nl> Message-ID: Hi Dimitri You should be able to get it to work by downloading the archive, gunzip and untarring the source and changing Utils.hs to include this line after the line "import Control.Monad": import Control.Exception hiding ( assert ) After that build with these commands from the top of the source tree: runhaskell Setup.hs configure runhaskell Setup.hs build runhaskell Setup.hs install The original problem is that catch is no longer export by Prelude in GHC 7.6 -------------- next part -------------- An HTML attachment was scrubbed... URL: From hsyl20 at gmail.com Wed Aug 21 09:49:52 2013 From: hsyl20 at gmail.com (Sylvain HENRY) Date: Wed, 21 Aug 2013 09:49:52 +0200 Subject: [Haskell-beginners] cabal package haskore-vintage fails to build In-Reply-To: References: <1C957484-DB86-43B2-8927-A012D4480B1A@xs4all.nl> Message-ID: <52147120.9000005@gmail.com> Hi, It is better to use "cabal unpack" to download and unpack the source. Cheers Sylvain Le 20/08/2013 23:55, Stephen Tetley a ?crit : > Hi Dimitri > > You should be able to get it to work by downloading the archive, > gunzip and untarring the source and changing Utils.hs to include this > line after the line "import Control.Monad": > > import Control.Exception hiding ( assert ) > > > After that build with these commands from the top of the source tree: > > runhaskell Setup.hs configure > runhaskell Setup.hs build > runhaskell Setup.hs install > > The original problem is that catch is no longer export by Prelude in > GHC 7.6 > > > > > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners -------------- next part -------------- An HTML attachment was scrubbed... URL: From diem at xs4all.nl Wed Aug 21 09:58:29 2013 From: diem at xs4all.nl (Dimitri Hendriks) Date: Wed, 21 Aug 2013 09:58:29 +0200 Subject: [Haskell-beginners] cabal package haskore-vintage fails to build In-Reply-To: References: <1C957484-DB86-43B2-8927-A012D4480B1A@xs4all.nl> Message-ID: <899939EF-27C7-4AC7-B953-8BA8EA966128@xs4all.nl> Many thanks Stephen, works like a charm! Finally, I can start composing music, can't wait! @Sylvain: why is cabal unpack better than tar zxvf ? Greetings, Dimitri On Aug 20, 2013, at 23:55 , Stephen Tetley wrote: > Hi Dimitri > > You should be able to get it to work by downloading the archive, gunzip and untarring the source and changing Utils.hs to include this line after the line "import Control.Monad": > > import Control.Exception hiding ( assert ) > > > After that build with these commands from the top of the source tree: > > runhaskell Setup.hs configure > runhaskell Setup.hs build > runhaskell Setup.hs install > > The original problem is that catch is no longer export by Prelude in GHC 7.6 > > > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners From hsyl20 at gmail.com Wed Aug 21 13:00:12 2013 From: hsyl20 at gmail.com (Sylvain HENRY) Date: Wed, 21 Aug 2013 13:00:12 +0200 Subject: [Haskell-beginners] cabal package haskore-vintage fails to build In-Reply-To: <899939EF-27C7-4AC7-B953-8BA8EA966128@xs4all.nl> References: <1C957484-DB86-43B2-8927-A012D4480B1A@xs4all.nl> <899939EF-27C7-4AC7-B953-8BA8EA966128@xs4all.nl> Message-ID: <52149DBC.5030802@gmail.com> It may be a bit quicker as it downloads and unpacks the package automatically. > cabal unpack haskore-vintage > cd haskore-vintage-0.1 > vim src/Haskore/Utils.hs > cabal install Le 21/08/2013 09:58, Dimitri Hendriks a ?crit : > @Sylvain: why is cabal unpack better than tar zxvf ? > > From paratribulations at free.fr Wed Aug 21 18:39:33 2013 From: paratribulations at free.fr (TP) Date: Wed, 21 Aug 2013 18:39:33 +0200 Subject: [Haskell-beginners] fonction in a typeclass that does not mention the type variable Message-ID: <537fea-mm5.ln1@rama.universe> Hi, I have a simple test case containing a typeclass Foo with one type variable a. The goal is to write once and for all a function symbolToInfixLevel function that combines two other functions defined in the typeclass: ----------------------- class Foo a where symbolToConstructor :: String -> ( a -> a -> a ) infixLevel :: a -> Int symbolToInfixLevel :: String -> Int symbolToInfixLevel s = infixLevel $(symbolToConstructor s) undefined undefined ----------------------- This yields an error because there is no "a" in the type signature for symbolToInfixLevel:$ runghc test_typeclass_without_typevariable.hs test_typeclass_without_typevariable.hs:1:1: The class method symbolToInfixLevel' mentions none of the type variables of the class Foo a When checking the class method: symbolToInfixLevel :: String -> Int In the class declaration for Foo' Now, if I define symbolToInfixLevel out of the typeclass: ----------------------- class Foo a where symbolToConstructor :: String -> ( a -> a -> a ) infixLevel :: a -> Int symbolToInfixLevel :: String -> Int symbolToInfixLevel s = infixLevel $(symbolToConstructor s) undefined undefined ----------------------- Now, I obtain:$ runghc test_typeclass_without_typevariable.hs test_typeclass_without_typevariable.hs:7:24: No instance for (Foo a0) arising from a use of infixLevel' In the expression: infixLevel In the expression: infixLevel $(symbolToConstructor s) undefined undefined In an equation for symbolToInfixLevel': symbolToInfixLevel s = infixLevel$ (symbolToConstructor s) undefined undefined How to get rid from this situation? Thanks in advance, TP From byorgey at seas.upenn.edu Wed Aug 21 19:01:01 2013 From: byorgey at seas.upenn.edu (Brent Yorgey) Date: Wed, 21 Aug 2013 13:01:01 -0400 Subject: [Haskell-beginners] cabal package haskore-vintage fails to build In-Reply-To: <52149DBC.5030802@gmail.com> References: <1C957484-DB86-43B2-8927-A012D4480B1A@xs4all.nl> <899939EF-27C7-4AC7-B953-8BA8EA966128@xs4all.nl> <52149DBC.5030802@gmail.com> Message-ID: <20130821170101.GA8648@seas.upenn.edu> It also uses the cached version of the tar file which is already on your disk, if you have previously installed the package, so it does not even need to download anything. -Brent On Wed, Aug 21, 2013 at 01:00:12PM +0200, Sylvain HENRY wrote: > It may be a bit quicker as it downloads and unpacks the package > automatically. > > > cabal unpack haskore-vintage > > cd haskore-vintage-0.1 > > vim src/Haskore/Utils.hs > > cabal install > > Le 21/08/2013 09:58, Dimitri Hendriks a ?crit : > >@Sylvain: why is cabal unpack better than tar zxvf ? > > > > > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > From peter.hall at memorphic.com Wed Aug 21 19:05:42 2013 From: peter.hall at memorphic.com (Peter Hall) Date: Wed, 21 Aug 2013 18:05:42 +0100 Subject: [Haskell-beginners] fonction in a typeclass that does not mention the type variable In-Reply-To: <537fea-mm5.ln1@rama.universe> References: <537fea-mm5.ln1@rama.universe> Message-ID: Maybe I don't fully understand what you are trying to achieve, but I don't think what you are trying to do makes sense. infixLevel and symbolToConstructor will have different implementations for each instance of Foo, and the compiler cannot possibly know which implementations you mean. If the implementation of symbolToInfixLevel is independent of the implementations of infixLevel and symbolToConstructor then it shouldn't need them in its definition. Peter On 21 August 2013 17:39, TP wrote: > Hi, > > I have a simple test case containing a typeclass Foo with one type variable > a. The goal is to write once and for all a function symbolToInfixLevel > function that combines two other functions defined in the typeclass: > > ----------------------- > class Foo a where > > symbolToConstructor :: String -> ( a -> a -> a ) > infixLevel :: a -> Int > > symbolToInfixLevel :: String -> Int > symbolToInfixLevel s = infixLevel $(symbolToConstructor s) undefined > undefined > ----------------------- > > This yields an error because there is no "a" in the type signature for > symbolToInfixLevel: > >$ runghc test_typeclass_without_typevariable.hs > test_typeclass_without_typevariable.hs:1:1: > The class method symbolToInfixLevel' > mentions none of the type variables of the class Foo a > When checking the class method: symbolToInfixLevel :: String -> Int > In the class declaration for Foo' > > Now, if I define symbolToInfixLevel out of the typeclass: > > ----------------------- > class Foo a where > > symbolToConstructor :: String -> ( a -> a -> a ) > infixLevel :: a -> Int > > symbolToInfixLevel :: String -> Int > symbolToInfixLevel s = infixLevel $(symbolToConstructor s) undefined > undefined > ----------------------- > > Now, I obtain: > >$ runghc test_typeclass_without_typevariable.hs > test_typeclass_without_typevariable.hs:7:24: > No instance for (Foo a0) arising from a use of infixLevel' > In the expression: infixLevel > In the expression: > infixLevel $(symbolToConstructor s) undefined undefined > In an equation for symbolToInfixLevel': > symbolToInfixLevel s > = infixLevel$ (symbolToConstructor s) undefined undefined > > How to get rid from this situation? > > Thanks in advance, > > TP > > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: From mlists at pmade.com Wed Aug 21 19:20:30 2013 From: mlists at pmade.com (Peter Jones) Date: Wed, 21 Aug 2013 11:20:30 -0600 Subject: [Haskell-beginners] fonction in a typeclass that does not mention the type variable References: <537fea-mm5.ln1@rama.universe> Message-ID: <87y57vdkpt.fsf@pmade.com> TP writes: > symbolToInfixLevel :: String -> Int > symbolToInfixLevel s = infixLevel $(symbolToConstructor s) undefined > > Now, I obtain: > >$ runghc test_typeclass_without_typevariable.hs > test_typeclass_without_typevariable.hs:7:24: > No instance for (Foo a0) arising from a use of infixLevel' > In the expression: infixLevel > In the expression: > infixLevel $(symbolToConstructor s) undefined undefined > In an equation for symbolToInfixLevel': > symbolToInfixLevel s > = infixLevel$ (symbolToConstructor s) undefined undefined > > How to get rid from this situation? Your symbolToConstructor' function is too specific for both cases you describe. You'd need to change the String' argument to a' and put it back in the type class or write it like this: symbolToInfixLevel :: (Foo a) => a -> Int symbolToInfixLevel s = infixLevel $(symbolToConstructor s) undefined -- Peter Jones, Founder, Devalot.com Defending the honor of good code From mlists at pmade.com Wed Aug 21 19:37:23 2013 From: mlists at pmade.com (Peter Jones) Date: Wed, 21 Aug 2013 11:37:23 -0600 Subject: [Haskell-beginners] fonction in a typeclass that does not mention the type variable References: <537fea-mm5.ln1@rama.universe> <87y57vdkpt.fsf@pmade.com> Message-ID: <87siy3djxo.fsf@pmade.com> Peter Jones writes: > Your symbolToConstructor' function is too specific for both cases you > describe. You'd need to change the String' argument to a' and put it > back in the type class or write it like this: > > symbolToInfixLevel :: (Foo a) => a -> Int > symbolToInfixLevel s = infixLevel$ (symbolToConstructor s) undefined Actually, the problem is with your symbolToConstructor' function. It needs to look like this: symbolToConstructor :: (Foo a) => String -> a Then symbolToInfixLevel can be: symbolToInfixLevel :: String -> Int -- Peter Jones, Founder, Devalot.com Defending the honor of good code From paratribulations at free.fr Wed Aug 21 21:44:43 2013 From: paratribulations at free.fr (TP) Date: Wed, 21 Aug 2013 21:44:43 +0200 Subject: [Haskell-beginners] fonction in a typeclass that does not mention the type variable References: <537fea-mm5.ln1@rama.universe> Message-ID: Peter Hall wrote: > Maybe I don't fully understand what you are trying to achieve, but I don't > think what you are trying to do makes sense. infixLevel and > symbolToConstructor will have different implementations for each instance > of Foo, and the compiler cannot possibly know which implementations you > mean. Indeed, my question was stupid. This is obvious in the second implementation (which instance to choose?), and it is also true in the first one because even if symbolToInfixLevel of one considered instance uses infixLevel and symbolToConstructor of this same instance (what I am not sure of), at the location where I use this function the compiler cannot known which instance I mean. > If the implementation of symbolToInfixLevel is independent of the > implementations of infixLevel and symbolToConstructor then it shouldn't > need them in its definition. No, they are not independent: it is really a function that "composes" functions infixLevel and symbolToConstructor. See my proposition in my answer to Peter Jones. From paratribulations at free.fr Wed Aug 21 21:43:33 2013 From: paratribulations at free.fr (TP) Date: Wed, 21 Aug 2013 21:43:33 +0200 Subject: [Haskell-beginners] fonction in a typeclass that does not mention the type variable References: <537fea-mm5.ln1@rama.universe> <87y57vdkpt.fsf@pmade.com> <87siy3djxo.fsf@pmade.com> Message-ID: <5shfea-vh9.ln1@rama.universe> Peter Jones wrote: > Actually, the problem is with your symbolToConstructor' function. It > needs to look like this: > > symbolToConstructor :: (Foo a) => String -> a > > Then symbolToInfixLevel can be: > > symbolToInfixLevel :: String -> Int > Sorry, I have not been able to write an example with your proposition. Might you give more informations? In the meantime, I have been able to write a working example proposing two variants: ------------------------------ {-# LANGUAGE ScopedTypeVariables #-} data Exp = Plus Exp Exp | Minus Exp Exp | Atom String deriving Show class Foo a where symbolToConstructor :: String -> ( a -> a -> a ) infixLevel :: a -> Int -- The second argument is only used to indicate the type of the -- intermediary result (i.e. the type of the result of -- symbolToConstructor). symbolToInfixLevel :: String -> a -> Int symbolToInfixLevel s u = infixLevel $(((symbolToConstructor s) undefined undefined)::a) instance Foo Exp where symbolToConstructor e = case e of "+" -> Plus "-" -> Minus infixLevel e = case e of Plus _ _ -> 6 Minus _ _ -> 7 main = do -- First variant print$ infixLevel $(((symbolToConstructor "+") undefined undefined)::Exp) print$ infixLevel $(((symbolToConstructor "-") undefined undefined)::Exp) -- Second variant print$ symbolToInfixLevel "+" (undefined::Exp) print $symbolToInfixLevel "-" (undefined::Exp) ------------------------------ From bryanvick at gmail.com Mon Aug 26 23:05:00 2013 From: bryanvick at gmail.com (Bryan Vicknair) Date: Mon, 26 Aug 2013 14:05:00 -0700 Subject: [Haskell-beginners] Lifting (Either SqlError a) in a monad stack Message-ID: <20130826210500.GA4710@bry-m6300> I'm at the beginning of the monad transformers journey, and this is stumping me. I've read "Monad Transformers Step by Step" which made me confident enough to play with simple stacks, but I'm a bit lost in the following example from work. Please excuse any incorrect transformer terminology. In a WAI web app, I want to chain these two actions inside a monad stack: > safeConnect :: ConnectInfo -> IO (Either SqlError Connection) > safeInsert :: Thing -> Connection -> IO (Either SqlError Id) Here is an example where I would like 'result' to be (Right Id) only if the DB connection *and* the insert were successful, (Left SqlError) otherwise. > add :: Request -> ResourceT IO Response > add _ = do > result <- liftIO$ safeConnect devConnInfo >>= safeInsert thing > case result of > (Left e) -> return $dbErr e > (Right _) -> return$ postRedirect > where thing = exampleThing The compiler tells me (edited for brevity): Expected type: Either SqlError Connection -> GHC.Types.IO a0 Actual type: Connection -> GHC.Types.IO (Either SqlError Id) In the return type of a call of safeInsert' In the second argument of (>>=)', namely safeInsert thing' How can I chain safeConnect and safeInsert using the (Either SqlError a) monad inside WAI's (ResourceT IO Response) monad stack? From ktvoelker at gmail.com Tue Aug 27 01:18:16 2013 From: ktvoelker at gmail.com (Karl Voelker) Date: Mon, 26 Aug 2013 16:18:16 -0700 Subject: [Haskell-beginners] Lifting (Either SqlError a) in a monad stack In-Reply-To: <20130826210500.GA4710@bry-m6300> References: <20130826210500.GA4710@bry-m6300> Message-ID: The problem is that the Either SqlError monad is not actually a part of your "monad stack". It's just something that appears in the result type of safeConnect and safeInsert. In order to put Either SqlError into your stack, you'll need a monad transformer which exhibits Either-like semantics, such as EitherT or ErrorT. Then your safeConnect would have type "ConnectInfo -> EitherT SqlError IO Connection". http://hackage.haskell.org/packages/archive/either/3.4.1/doc/html/Control-Monad-Trans-Either.html http://hackage.haskell.org/packages/archive/transformers/0.3.0.0/doc/html/Control-Monad-Trans-Error.html -Karl On Mon, Aug 26, 2013 at 2:05 PM, Bryan Vicknair wrote: > I'm at the beginning of the monad transformers journey, and this is > stumping > me. I've read "Monad Transformers Step by Step" which made me confident > enough > to play with simple stacks, but I'm a bit lost in the following example > from > work. Please excuse any incorrect transformer terminology. > > In a WAI web app, I want to chain these two actions inside a monad stack: > > > safeConnect :: ConnectInfo -> IO (Either SqlError Connection) > > safeInsert :: Thing -> Connection -> IO (Either SqlError Id) > > Here is an example where I would like 'result' to be (Right Id) only if > the DB > connection *and* the insert were successful, (Left SqlError) otherwise. > > > add :: Request -> ResourceT IO Response > > add _ = do > > result <- liftIO $safeConnect devConnInfo >>= safeInsert thing > > case result of > > (Left e) -> return$ dbErr e > > (Right _) -> return $postRedirect > > where thing = exampleThing > > The compiler tells me (edited for brevity): > > Expected type: Either SqlError Connection -> GHC.Types.IO a0 > Actual type: Connection -> GHC.Types.IO (Either SqlError Id) > In the return type of a call of safeInsert' > In the second argument of (>>=)', namely > `safeInsert thing' > > How can I chain safeConnect and safeInsert using the (Either SqlError a) > monad > inside WAI's (ResourceT IO Response) monad stack? > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners > -------------- next part -------------- An HTML attachment was scrubbed... URL: From anthony_clayden at clear.net.nz Tue Aug 27 11:46:59 2013 From: anthony_clayden at clear.net.nz (AntC) Date: Tue, 27 Aug 2013 09:46:59 +0000 (UTC) Subject: [Haskell-beginners] print [] References: Message-ID: > Lukas Lehner gmail.com> writes: > > ... My question is that I have to enforce the type that specific way (don't want to use?ExtendedDefaultRules) or is there some more generic way? > print flatten (List ([] :: Show a => [a]) Even for an empty list, we have to provide evidence that the list is showable. That's what the type signature is doing: I am a list of something/anything showable. (If you want all of your NestedLists's to be showable, it's better to put a Show constraint on the data decl. But that would take us into existential types or GADT's.) AntC From kxra at riseup.net Thu Aug 29 07:59:58 2013 From: kxra at riseup.net (=?UTF-8?B?S+G6j3Jh?=) Date: Thu, 29 Aug 2013 01:59:58 -0400 Subject: [Haskell-beginners] Haskell mentors/tutors website? Message-ID: Is there an existing website to connect haskell mentors and mentees? I'm thinking of something along the lines of railsmentors (except, of course, with haskell support). I suppose it doesn't have to be Haskell specific, as long as it's a supported language. Codementor.io seems like a really cool option along those lines, but it would also be valuable to have a network for volunteer mentors. The closest I have seen is Haskellers, which is not about mentorship, but instead only about hiring and job hunting. P.S. Are there any LGBTQIA/female/disabled people of color on this list who would be willing to teach myself and other folks from the Empowermentors Collective? http://kxra.tumblr.com/post/54139122042/please-signal-boost-the-empowermentors Thanks! K?ra -- Board of Directors, Free Culture Foundation: www.freeculture.org Web: kxra.info - StatusNet Microblog: http://identi.ca/kxra Email: kxra at freeculture.org - SMS: +1.617.340.3661 Jabber/XMPP: kxra at riseup.net - IRC: kxra @freenode @oftc @indymedia -------------- next part -------------- An HTML attachment was scrubbed... URL: From ehamberg at gmail.com Thu Aug 29 08:31:55 2013 From: ehamberg at gmail.com (Erlend Hamberg) Date: Thu, 29 Aug 2013 08:31:55 +0200 Subject: [Haskell-beginners] multi-parameter typeclass with default implementation In-Reply-To: References: Message-ID: You could also use functional dependencies here: {-# LANGUAGE FunctionalDependencies #-} class Foo a b | a -> b where bar :: a -> Int foobar :: a -> b -> Int foobar avalue bvalue = bar avalue instance Foo Int Int where bar i = 5 main = print$ bar (4::Int) Here you are saying that the type parameter b is determined by the parameter a, so GHC knows that the instance Foo Int Int is the only instance of Foo with a = Int, thus removing the ambiguity. -- Erlend Hamberg ehamberg at gmail.com -------------- next part -------------- An HTML attachment was scrubbed... URL: From shaegis at gmail.com Thu Aug 29 09:02:52 2013 From: shaegis at gmail.com (S. H. Aegis) Date: Thu, 29 Aug 2013 16:02:52 +0900 Subject: [Haskell-beginners] How can I keep delimiters in splitRegex? Message-ID: Hi. I'm newbee to Haskell. I want to keep delimiters in splitRegex like 'keepDelimsR'. I try... splitRegex (keepDelimsR (mkRegex "2013[0-9]{11}AH02")) myDataString but It doesn't work. How can I keep delimiters in splitRegex? Have a nice day. Thank you. S. Chang. -------------- next part -------------- An HTML attachment was scrubbed... URL: From nathan.huesken at posteo.de Sat Aug 31 20:05:26 2013 From: nathan.huesken at posteo.de (=?ISO-8859-1?Q?Nathan_H=FCsken?=) Date: Sat, 31 Aug 2013 20:05:26 +0200 Subject: [Haskell-beginners] Early return in IO monad Message-ID: <52223066.7010502@posteo.de> Hey, Is it somehow possible to return "early" in a do block of the IO monad? The eqivalent do in C: void doBlock() { if (some preCondition) { return; } ... } ??? Thanks! Nathan From byorgey at seas.upenn.edu Sat Aug 31 20:23:16 2013 From: byorgey at seas.upenn.edu (Brent Yorgey) Date: Sat, 31 Aug 2013 14:23:16 -0400 Subject: [Haskell-beginners] Early return in IO monad In-Reply-To: <52223066.7010502@posteo.de> References: <52223066.7010502@posteo.de> Message-ID: <20130831182316.GA23914@seas.upenn.edu> On Sat, Aug 31, 2013 at 08:05:26PM +0200, Nathan H?sken wrote: > Hey, > > Is it somehow possible to return "early" in a do block of the IO monad? > The eqivalent do in C: > > void doBlock() { > if (some preCondition) { > return; > } > ... > } > > ??? Nope. But you can say do ... when (not (some condition)) \$ do the rest of the stuff You can even indent "the rest of the stuff" directly under the 'when' if you like, though IMO that may be a bit confusing. -Brent From gnusokyo at gmail.com Sat Aug 31 22:03:35 2013 From: gnusokyo at gmail.com (Tidus Zero) Date: Sat, 31 Aug 2013 22:03:35 +0200 Subject: [Haskell-beginners] Early return in IO monad In-Reply-To: <52223066.7010502@posteo.de> References: <52223066.7010502@posteo.de> Message-ID: <262EAF29-F84B-40BE-9664-50AA29172A5D@gmail.com> You can use the EitherT monad with IO and use the 'left' function which is just 'fail' to return early if you aren't restricted to only use the IO monad. Sent from my iGNU On 31. aug. 2013, at 20.05, Nathan H?sken wrote: > Hey, > > Is it somehow possible to return "early" in a do block of the IO > monad? > The eqivalent do in C: > > void doBlock() { > if (some preCondition) { > return; > } > ... > } > > ??? > > Thanks! > Nathan > > _______________________________________________ > Beginners mailing list > Beginners at haskell.org > http://www.haskell.org/mailman/listinfo/beginners | 45,902 | 172,287 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-48 | longest | en | 0.766458 |
http://www.physicsforums.com/showthread.php?s=38a75c387398655d99b0b7c830eb4c0f&p=4172651 | 1,369,162,578,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700438490/warc/CC-MAIN-20130516103358-00010-ip-10-60-113-184.ec2.internal.warc.gz | 660,773,989 | 9,315 | ## Help with calculating vertical jump with kinematic equations
Ok, so I'm using an arduino with an led matrix and a homemade pressure sensor to figure out the height of a vertical jump and then display the info back. First I get the amount of time between when the person left the board and when they land back on it. I then divide the time in half and use the equation:
Yf = 1/2 * g * t^2 substituting g for 386.088 inches per second^2 giving
Yf = 193.044 * t^2
My problem now comes in here. I'm doing this inside a room with an 8 foot ceiling for easy reference on how high I am jumping. I'm 6' 1" so when I jump so that my head just barely touches the ceiling my vertical is roughly 23". The arduino, though, is reading a little over 16". My first thought was that the pressure pad was staying down for some time after I jumped, but I recorded a jump and it's working right starting the time right when I leave the pad. In fact I even used the time I got from comparing when I left the ground on the jump and when I landed in the video and got .58 seconds. Using the equations before that gives a little over 16 inches max height (like what the arduino measured). So, does anyone have any idea what's going on here?
PhysOrg.com physics news on PhysOrg.com >> Iron-platinum alloys could be new-generation hard drives>> Lab sets a new record for creating heralded photons>> Breakthrough calls time on bootleg booze
Recognitions: Science Advisor Your center of mass might be in different position relative to your feet when you separate from ground, and when you touch back. If you post the video and mark where you get 23" from on some screen captures, we could probably analyze it further.
Recognitions: Homework Help Science Advisor Yes, it's most likely to do with posture, but you're looking for a reason why the time recorded might be too short. So bending your legs ready to land could not account for it. Maybe it's your arms - if you raise them when descending you'll land sooner.
Recognitions:
## Help with calculating vertical jump with kinematic equations
I think it's the other way around. He was trying not to jump too high, so he might have took off with legs still bent, but landed on straight legs.
I thought about it being posture related also, but when going through the video it doesn't look like that. I'd upload it but it just a recording of my legs and a light on the arduino for comparison of where the algorithms at. On the jump I start with legs bent and jump up--legs extending and toes eventually pointing down. The arduino doesn't recognize me as leaving the ground until my toes come off the board (although I'm not pushing down anymore: this may be where the error is). I go up then land in the exact same manor, feet pointed down, until they touch the board then the rest of my body comes down to absorb the impact. When I begin to touch the board again is when the algorithm gets the total time. It takes very little force (probably about 10 pounds) to trigger the pressure pad. The measurement for comparison is the top of my head, which should be a fairly accurate recording of how high I'm getting off the ground since that would be highest point I can reach. The two errors I'm thinking of right now is that the difference between being flat footed and having feet extended is causing the error, or it's the one I mentioned earlier where I'm already decelerating before leaving the pressure pad. Thanks for the responses guys, let me know if either of those or anything else seems like a more likely cause. It just threw me off a lot when I started getting these consistently differing results. Edit: Actually looking back at the posts now, the first reason seems like what you were saying k^2
Recognitions:
Homework Help | 833 | 3,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2013-20 | latest | en | 0.932125 |
https://www.instructables.com/id/Particle-Photon-Salinity-Meter/ | 1,591,102,593,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347424174.72/warc/CC-MAIN-20200602100039-20200602130039-00230.warc.gz | 732,223,584 | 21,628 | Particle Photon Salinity Meter
754
3
1
Introduction: Particle Photon Salinity Meter
We made a measurement device to measure the salinity of water using a magnetic field and a linear hall sensor.
To make it we used a Particle Photon, but an Arduino could also be used as they work practically the same way.
To make this project you need a couple of things:
- Particle/arduino including a breadboard and some cables
- a linear hall sensor
- some magnets(we used small but strong neodymium magnets)
- a pen
- some tape
Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
Step 1: The Container
The pen will be used as a container so go ahead and take out the pin so you'll only have the plastic container.
Close the small hole with some tape, and tape the magnets near the small hole on the side of the pen.
Step 2: Connect the Particle/Arduino
Connect the particle or arduino to the breadboard. Also connect the linear hall sensor the same way as on the picture, the top pin to 3.3V, the middle pin to GND and the bottom pin to an analog input.
Step 3: The Code
On the particle photon you can just press on the pin you used as input and use the function analogRead to get the value from the hall sensor.
If you want it done automatically or if you're using an arduino you'll need a code looking something like this:
//the pin to measure from
int analogPin = A0;
//the amount of time, in milliseconds, between measurements.
//since you can not publish too many events, this has too be at least 1000
int delayTime = 5000;
//an event name so you'll recognize the measurements flowing in
String eventName = "measurement/Salinity";
String laag = "Low";
String middel = "Medium";
String hoog = "High";
void setup(){
}
void loop(){
if(measurement<=1750){
Particle.publish(eventName, laag); }
if(measurement>=1751 && measurement<=1830){
Particle.publish(eventName, middel);
}
if(measurement>=1831 && measurement<=2100){
Particle.publish(eventName, hoog);
}
if(measurement>=2101){
}
delay(delayTime);
}
Step 4: Measure!
Ofcourse the values in the code will have to be calibrated to the salinity you're using so go ahead and get 3 cups of water. Cup 1 will be just water, Cup 3 will be fully saturated with salt and Cup 2 will be somewhere in between.
Grab one of the cups and pour some of the water into the pen.
Hold the pen next to the hall sensor with the magnets sticking out the other side( so the water will be sandwiched between the magnets and the sensor)
Use the function analogRead to see the value for the water you're using and use that value in the code.
The values we measured were:
just water: 1720
Saturated with salt: 1840
somewhere in between: 1760
26 3.7K
164 12K
96 9.4K
Discussions
That's a neat setup :) | 684 | 2,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.851261 |
https://math.stackexchange.com/questions/3281619/finding-the-maxima-of-a-logarithm-with-the-form-xa-lnmxc-d | 1,575,683,097,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540491871.35/warc/CC-MAIN-20191207005439-20191207033439-00487.warc.gz | 471,523,756 | 32,474 | # Finding the maxima of a logarithm with the form: $x(a \ln(mx+c) + d)$
Currently trying to build an analytical model that requires knowing the point of maxima for the equation:
$$y = x(a \ln(mx+c)+d)$$
The parameters are such that the maxima can be graphically shown to be $$x>0$$, with $$-1.
I would like to be able to quickly know where this maxima is using the parameters, as there will be thousands of models and I would rather not have to do it by fitting.
Currently my result (as shown below) involves the Lambert function and I haven't come across that before. So my question is kind of two-fold: is my result correct and can this be simplified with the knowledge that the maxima is in a reasonable region of x?
My method:
Differentiating with the product rule I get:
$$\frac{dy}{dx} = d + \frac{amx}{c + mx} + a\ln(c + mx) = 0$$
Substituting in $$V = mx+c$$ to simplify things:
$$d + \frac{a(V-c)}V + a\ln(V) = 0$$.
Now solving for V,
$$V_{max} = \frac{c}{W(c . \exp(1 + \frac{d}{a}))}$$,
where W is the product log or Lambert function. https://en.wikipedia.org/wiki/Lambert_W_function
I've attached an example distribution with parameters of $$a=-0.15$$, $$m=0.000043$$, $$c=0.43$$ and $$d=0.31$$
• Lambert is a very beautiful function with a lot of applications. Here, in the Search bar, just type Lambert :2748 entries. Whet more precisely to you need about it ? By the way, Welcome to the site !! – Claude Leibovici Jul 3 at 8:49
• Thanks! I guess my question is whether it has an approximation that can be used to simplify it in the region of small i.e. $0<x<10$ for W(x). I'm hoping to be able to just stick the parameters quickly for thousands of curves. Maybe the answer is obvious, but it looks quite intimidating for the first time seeing it! – James Broughton Jul 3 at 9:01
• For the computation of Lambert $W(t)$, you could use Corless algorithm which is quite fast. Otherwise, the series expansions give in the Wikipedia page. – Claude Leibovici Jul 3 at 10:46
• Thanks, that's helpful! I will take a look – James Broughton Jul 3 at 11:33
As asked in comment, I had a look to what could be an approximation of $$W(x)$$ for the range $$0 \leq x \leq 10$$.
For obvious reasons, the Taylor series built around $$x=0$$ would not work over this large range since it is $$W(x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{n^{n-1}}{n!} x^n$$ showing how fact the coefficient vary.
However, and this is totally empirical, it seems that, over this range, $$x^{3/4}W(x)$$ is close to linearity; have a look at
So, the idea was to write $$W(x)=\sum_{n=1}^p a_n t^n \qquad \text{where} \qquad t=x^b$$ Using nonlinear regression for such an empirical model with $$p=5$$, what is obtained is summarized below $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a_1 & +0.7293417468 & 0.00067739 & \{+0.7280125,+0.7306710\} \\ a_2 & -0.1989488190 & 0.00084036 & \{-0.2005979,-0.1972997\} \\ a_3 & +0.0385896461 & 0.00028194 & \{+0.0380364,+0.0391429\} \\ a_4 & -0.0040728293 & 0.00004093 & \{-0.0041532,-0.0039925\} \\ a_5 & +0.0001741237 & 0.00000218 & \{+0.0001698,+0.0001784\} \\ b & +0.8488289012 & 0.00101683 & \{+0.8468335,+0.8508243\} \\ \end{array}$$ For the values gives in the example, this would lead to $$0.1300$$ for an exact value equal to $$0.1366$$. | 1,074 | 3,333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-51 | latest | en | 0.897554 |
https://www.daniweb.com/programming/software-development/threads/250175/counting-test-scores | 1,540,008,658,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512504.64/warc/CC-MAIN-20181020034552-20181020060052-00316.warc.gz | 889,899,796 | 11,540 | ## new programer
Hello all
I was looking into some textbook solution (preparing for my exam)
and I saw this program:
Question Details:
write a program that reads in a set of positive integers, representing test scores for a class, and outputs how many times a particular number appears in the list. you may assume that the data set has at most 100 numbers and -999 marks the end of the output data. the numbers must be output in increasing order. for example, for the data:
55 80 78 92 95 55 78 53 92 65 78 95 85 92 85 95 95
the output is:
test score count
53 1
55 2
65 1
78 3
80 1
85 2
``````#include<iostream>
using namespace std;
int main()
{
int numbers[101]={0},i,num;
cout<<"enter a number: ";
cin>>num;
while(num!=-999)
{
numbers[num]++;
//cout<<numbers[num]<<" "<<endl; >>this appears to be a counter
cout<<"enter a number: ";
cin>>num;
}
cout<<"test score\tcount\n";
for(i=0;i<101;i++)
if(numbers[i]!=0)
cout<<i<<"\t\t"<<numbers[i]<<endl;
system("pause");
return 0;
}``````
how does it work ?
what does "i" represent ?
## vmanes 1,165
When you want to understand what a piece of code like this does, or how it does it, it's sometimes useful to get out pencil and paper. Write down each of the variable names, and beside them keep track of their values as the program executes. Walk through it by hand.
So, you start with an array of 101 elements, all initialized to 0. (You don't need to write out all 101 0's, just use the first 10 or so as an example.)
Assume some number input to num. How does that interact with the array? As you enter other values, how does the array change?
Once the -999 is entered, what does the next loop do? If you're getting ready for any sort of exam, that should be evident. | 474 | 1,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-43 | latest | en | 0.861855 |
https://www.doubtnut.com/question-answer/in-a-triangle-abc-if-cota-2cotb-2c-cotb-2cotc-2a-and-cotc-2cota-2b-then-1-s-a-1-s-b-1-s-c-132913 | 1,631,889,774,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055645.75/warc/CC-MAIN-20210917120628-20210917150628-00663.warc.gz | 737,235,477 | 76,365 | # In a triangle ABC if cot(A/2)*cot(B/2)=c ,cot(B/2)*cot(C/2)=a and cot(C/2)*cot(A/2)=b then 1/(s-a)+1/(s-b)+1/(s-c)=
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 26-11-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
65.1 K+
3.2 K+
Image Solution
132913
3.2 K+
65.1 K+
2:44
8493384
51.4 K+
64.9 K+
3:28
130029
6.3 K+
126.1 K+
2:37
64179
10.9 K+
218.8 K+
5:00
2941131
12.7 K+
254.3 K+
2:20
41409
2.8 K+
55.1 K+
2:58
8494708
2.4 K+
48.9 K+
3:28
54849234
30.5 K+
47.9 K+
7:19
57121
4.1 K+
83.0 K+
3:22
1872348
3.3 K+
66.5 K+
2:02
1872353
6.0 K+
120.8 K+
2:02
8494485
2.9 K+
57.8 K+
2:20
6560
49.4 K+
121.2 K+
5:08
96593084
1.5 K+
30.5 K+
5:19
22004
14.8 K+
295.9 K+
3:34 | 393 | 853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-39 | latest | en | 0.406442 |
http://search.cpan.org/dist/Algorithm-Munkres/lib/Algorithm/Munkres.pm | 1,529,371,817,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00619.warc.gz | 293,809,891 | 5,847 | search.cpan.org is shutting down
Ted Pedersen > Algorithm-Munkres > Algorithm::Munkres
Algorithm-Munkres-0.08.tar.gz
Dependencies
Annotate this POD
# CPAN RT
Open 0
View/Report Bugs
Module Version: 0.08
# NAME
``` Algorithm::Munkres - Perl extension for Munkres' solution to
classical Assignment problem for square and rectangular matrices
This module extends the solution of Assignment problem for square
matrices to rectangular matrices by padding zeros. Thus a rectangular
matrix is converted to square matrix by padding necessary zeros.```
# SYNOPSIS
use Algorithm::Munkres;
``` @mat = (
[2, 4, 7, 9],
[3, 9, 5, 1],
[8, 2, 9, 7],
);```
assign(\@mat,\@out_mat);
``` Then the @out_mat array will have the output as: (0,3,1,2),
where
0th element indicates that 0th row is assigned 0th column i.e value=2
1st element indicates that 1st row is assigned 3rd column i.e.value=1
2nd element indicates that 2nd row is assigned 1st column.i.e.value=2
3rd element indicates that 3rd row is assigned 2nd column.i.e.value=0```
# DESCRIPTION
``` Assignment Problem: Given N jobs, N workers and the time taken by
each worker to complete a job then how should the assignment of a
Worker to a Job be done, so as to minimize the time taken.
Thus if we have 3 jobs p,q,r and 3 workers x,y,z such that:
x y z
p 2 4 7
q 3 9 5
r 8 2 9
where the cell values of the above matrix give the time required
for the worker(given by column name) to complete the job(given by
the row name)
then possible solutions are:
Total
1. 2, 9, 9 20
2. 2, 2, 5 9
3. 3, 4, 9 16
4. 3, 2, 7 12
5. 8, 9, 7 24
6. 8, 4, 5 17
Thus (2) is the optimal solution for the above problem.
This kind of brute-force approach of solving Assignment problem
quickly becomes slow and bulky as N grows, because the number of
possible solution are N! and thus the task is to evaluate each
and then find the optimal solution.(If N=10, number of possible
solutions: 3628800 !)
Munkres' gives us a solution to this problem, which is implemented
in this module.
This module also solves Assignment problem for rectangular matrices
(M x N) by converting them to square matrices by padding zeros. ex:
If input matrix is:
[2, 4, 7, 9],
[3, 9, 5, 1],
[8, 2, 9, 7]
i.e 3 x 4 then we will convert it to 4 x 4 and the modified input
matrix will be:
[2, 4, 7, 9],
[3, 9, 5, 1],
[8, 2, 9, 7],
[0, 0, 0, 0]```
# EXPORT
` "assign" function by default.`
# INPUT
``` The input matrix should be in a two dimensional array(array of
array) and the 'assign' subroutine expects a reference to this
array and not the complete array.
eg:assign(\@inp_mat, \@out_mat);
The second argument to the assign subroutine is the reference
to the output array.```
# OUTPUT
``` The assign subroutine expects references to two arrays as its
input paramenters. The second parameter is the reference to the
output array. This array is populated by assign subroutine. This
array is single dimensional Nx1 matrix.
For above example the output array returned will be:
(0,
2,
1)
where
0th element indicates that 0th row is assigned 0th column i.e value=2
1st element indicates that 1st row is assigned 2nd column i.e.value=5
2nd element indicates that 2nd row is assigned 1st column.i.e.value=2```
``` 1. http://216.249.163.93/bob.pilgrim/445/munkres.html
2. Munkres, J. Algorithms for the assignment and transportation
Problems. J. Siam 5 (Mar. 1957), 32-38
3. François Bourgeois and Jean-Claude Lassalle. 1971.
An extension of the Munkres algorithm for the assignment
problem to rectangular matrices.
Communication ACM, 14(12):802-804```
# AUTHOR
``` Anagha Kulkarni, University of Minnesota Duluth
kulka020 <at> d.umn.edu
Ted Pedersen, University of Minnesota Duluth
tpederse <at> d.umn.edu``` | 1,160 | 3,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-26 | latest | en | 0.806329 |
https://wiki.math.uwaterloo.ca/statwiki/index.php?title=Wavelet_Pooling_CNN&diff=prev&oldid=34076&printable=yes | 1,620,493,946,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.94/warc/CC-MAIN-20210508151721-20210508181721-00460.warc.gz | 638,932,062 | 9,725 | # Difference between revisions of "Wavelet Pooling CNN"
## Introduction
It is generally the case that Convolution Neural Networks (CNNs) out perform vector-based deep learning techniques. As such, the fundamentals of CNNs are good candidates to be innovated in order to improve said performance. The pooling layer is one of these fundamentals, and although various methods exist ranging from deterministic and simple: max pooling and average pooling, to probabilistic: mixed pooling and stochastic pooling, all these methods employ a neighborhood approach to the sub-sampling which, albeit fast and simple, can produce artifacts such as blurring, aliasing, and edge halos (Parker et al., 1983).
This paper introduces a novel pooling method based on the discrete wavelet transform. Specifically, it uses a second-level wavelet decomposition for the sub-sampling. This method, instead of nearest neighbor interpolation, uses a sub-band method that the authors claim produces less artifacts and represents the underlying features more accurately. Therefore, if pooling is viewed as a lossy process, the reason for employing a wavelet approach is to try to minimize this loss.
## Pooling Background
Pooling essentially means sub-sampling. After the pooling layer, the spatial dimensions of the data is reduced to some degree, with the goal being to compress the data rather than discard some of it. Typical approaches to pooling reduce the dimensionality by using some method to combine a region of values into one value. For max pooling, this can be represented by the equation $a_{kij} = max_{(p,q) \epsilon R_{ij}} (a_{kpq})$ where $a_{kij}$ is the output activation of the \k^th feature map at (i,j), $a_{kpq}$ is input activation at (p,q) within $R_{ij}$, and $R_{ij}\lt math\gt is the size of the pooling region. Mean pooling can be represented by the equation (EQUATION) with everything defined as before. Figure 1 provides a numerical example that can be followed. [[File:WT_Fig1.PNG|650px|center|]] The paper mentions that these pooling methods, although simple and effective, have shortcomings. Max pooling can omit details from an image if the important features have less intensity than the insignificant ones, and also commonly overfits. On the other hand, average pooling can dilute important features if the data is averaged with values of significantly lower intensities. Figure 2 displays an image of this. [[File:WT_Fig2.PNG|650px|center|]] == Wavelet Background == Data or signals tend to be composed of slowly changing trends (low frequency) as well as fast changing transients (high frequency). Similarly, images have smooth regions of intensity which are perturbed by edges or abrupt changes. We know that these abrupt changes can represent features that are of great importance to us when we perform deep learning. Wavelets are a class of functions that are well localized in time and frequency. Compare this to the Fourier transform which represents signals as the sum of sine waves which oscillate forever (not localized in time and space). The ability of wavelets to be localized in time and space is what makes it suitable for detecting the abrupt changes in an image well. Essentially, a wavelet is a fast decaying, oscillating signal with zero mean that only exists for a fixed duration and can be scaled and shifted in time. There are some well defined types of wavelets as shown in Figure 3. The key characteristic of wavelets for us is that they have a band-pass characteristic, and the band can be adjusted based on the scaling and shifting. [[File:WT_Fig3.jpg|650px|center|]] The paper uses discrete wavelet transform and more specifically a faster variation called Fast Wavelet Transform (FWT) using the Haar wavelet. There also exists a continuous wavelet transform. The main difference in these is how the scale and shift parameters are selected. == Discrete Wavelet Transform General== The discrete wavelet transform for images is essentially applying a low pass and high pass filter to your image where the transfer functions of the filters are related and defined by the type of wavelet used (Haar in this paper). This is shown in the figures below, which also show the recursive nature of the transform. For an image, the per row transform is taken first. This results in a new image where the first half is a low frequency sub-band and the second half is the high frequency sub-band. Then this new image is transformed again per column, resulting in four sub-bands. Generally, the low frequency content approximates the image and the high frequency content represents abrupt changes. Therefore, one can simply take the LL band and perform the transformation again to sub-sample even more. [[File:WT_Fig8.png|650px|center|]] [[File:WT_Fig9.png|650px|center|]] == DWT example using Haar Wavelet == Suppose we have an image represented by the following pixels: \lt math\gt \begin{bmatrix} 100 & 50 & 60 & 150 \\ 20 & 60 & 40 & 30 \\ 50 & 90 & 70 & 82 \\ 74 & 66 & 90 & 58 \\ \end{bmatrix}$
For each level of the DWT using the Haar wavelet, we will perform the transform on the rows first and then the columns. For the row pass, we transform each row as follows:
• Take row i = [ i1, i2, i3, i4], and let i_t = [a1, a2, d1, d2] represent the transformed row
• a1 = (i1 + i2)/2
• a2 = (i3 + i4)/2
• d1 = (i1 - i2)/2
• d2 = (i3 - i4)/2
After the row transforms, the images looks as follows: $\begin{bmatrix} 75 & 105 & 25 & -45 \\ 40 & 35 & -20 & 5 \\ 70 & 76 & -20 & -6 \\ 70 & 74 & 4 & 16 \\ \end{bmatrix}$
Now we apply the same method to the columns in the exact same way.
## Proposed Method
The proposed method uses subbands from the second level FWT and discards the first level subbands. The authors postulate that this method is more 'organic' in capturing the data compression and will create less artifacts that may affect the image classification.
### Forward Propagation
FWT can be expressed by [Insert equation] and [Insert Equation] where [] is the approximation function, [] is the detail function, W, W, are approximation and detail coefficients, h and h are time reversed scaling and wavelet vectors, (n) represents the sample in the vector, and j denotes the resolution level. To apply to images, FWT is first applied on the rows and then the columns. If a low (L) and high(H) sub-band is extracted from the rows and similarly for the columns than at each level there is 4 sub-bands (LH, HL, HH, and LL) where LL will further be decomposed into the level 2 decomposition.
Using the level 2 decomposition sub-bands, the Inverse Fast Wavelet Transform (IFWT) is used to obtain the resulting sub-sampled image, which is sub-sampled by a factor of two. The Equation for IFWT is [] where the parameters are the same as previously explained. Figure 4 displays the algorithm for the forward propagation.
### Back Propagation
This is simply the reverse of the forward propagation. The FWT of the image is upsampled to be used as the level 2 decomposition. Then IFWT is performed to obtain the original image which is upsampled by a factor of two using wavelet methods. Figure 5 displays the algorithm.
## Results
The authors tested on MNIST, CIFAR-10, SHVN, and KDEF and the paper provides comprehensive results for each. Stochastic gradient descent was used and the Haar wavelet is used due to its even, square subbands. The network for all datasets except MNIST is loosedly based on (Zeiler & Fergus, 2013). The authors keep the network consistent, but change the pooling method for each dataset. They also experiment with dropout and Batch Normalization to examine the effects of regularization on their method. All pooling methods compared use a 2x2 window. The overall results teach us that the pooling method should be chosen specific to the type of data we have. In some cases wavelet pooling may perform the best, and in other cases, other methods may perform better, if the data is more suited for those types of pooling.
### MNIST
Figure 6 shows the network and Table 1 shows the accuracy. It can be seen that wavelet pooling achieves the best accuracy from all pooling methods compared.
### CIFAR-10
Figure 7 shows the network and Tables 2 and 3 shows the accuracy without and with dropout. Average pooling achieves the best accuracy but wavelet pooling is still competitive.
## Computational Complexity
The authors explain that their paper is a proof of concept and is not meant to implement wavelet pooling in the most efficient way. The table below displays a comparison of the number of mathematical operations for each method according to the dataset. It can be seen that wavelet pooling is significantly worse. The authors explain that through good implementation and coding practices, the method can prove to be viable.
## Criticism
### Positive
• Wavelet Pooling achieves competitive performance with standard go to pooling methods
• Leads to comparison of discrete transformation techniques for pooling (DCT, DFT)
### Negative
• Only 2x2 pooling window used for comparison
• Highly computationally extensive
• Not as simple as other pooling methods
• Only one wavelet used (HAAR wavelet) | 2,081 | 9,177 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-21 | latest | en | 0.905921 |
https://codereview.stackexchange.com/questions/272779/first-hangman-game-need-honest-opinion | 1,660,468,580,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00649.warc.gz | 189,817,874 | 70,475 | First Hangman Game. Need Honest Opinion
Doing Angela Yu's 100 days of code. Have not seen the videos on this section yet. This is just my attempt.
here is the link to run the code. Hangman game
"""Hangman Game"""
import random
import os
#word list
word_list = ["aardvark", "baboon", "camel"]
HANGMANPICS = ['''
+---+
| |
|
|
|
|
=========''', '''
+---+
| |
O |
|
|
|
=========''', '''
+---+
| |
O |
| |
|
|
=========''', '''
+---+
| |
O |
/| |
|
|
=========''', '''
+---+
| |
O |
/|\ |
|
|
=========''', '''
+---+
| |
O |
/|\ |
/ |
|
=========''', '''
+---+
| |
O |
/|\ |
/ \ |
|
=========''']
#Randomly choose a word from the word_list and assign it to a variable called chosen_word.
chosen_word = random.choice(word_list)
#creating a list for the guesses
blanks = list("_" * len(chosen_word))
lives = 6 #total no of lives
h = 0 #hangman array index
#Ask the user to guess a letter and assign their answer to a variable called guess. Make guess lowercase.
while True:
guess = input("Guess a letter? ").lower()
if guess.isalpha() and len(guess) == 1:
return guess
# to replace the blanks
def replace_blank(guess):
"""checks if the guess is part of array or not """
global lives
global h
if guess not in chosen_word:
lives -= 1
h += 1
else:
for i in range(len(chosen_word)):
if chosen_word[i] == guess:
blanks[i] = chosen_word[i]
hangman = HANGMANPICS[h]
os.system('clear')
if guess not in chosen_word:
print(f"You guessed {guess}. That's not in the word. You lose a life.\n")
print(blanks)
print(hangman)
return blanks
#Check if the letter the user guessed (guess) is one of the letters in the chosen_word.
def testing():
while lives > 0:
blanks = replace_blank(guess)
word_f = ''.join(blanks)
if word_f == chosen_word:
print("\n ****You win**** \n ")
break
elif lives == 0:
print("You lose")
def main():
testing()
if __name__ == '__main__':
main()
$$$$
Currently, you have print statements mixed throughout your game logic. I would suggest separating your code into the part that decides what to do with the latest input, and the part that prints a summary of what happened. To do this, you can replace your prints with properties on a game state object. By extracting a GameState object, you can also replace globals with immutable variables, which can make your code easier to think through.
Here is an example GameState object. When updating the game state by a guess, it returns a new GameState that's an update of the old game state.
class GameState:
def __init__(self, lives, h, blanks, chosen_word):
self.lives = lives
self.h = h
self.blanks = blanks
self.chosen_word = chosen_word
return GameState(self.lives - 1, self.h + 1, self.blanks.copy(), self.chosen_word)
def next_state_for_good_guess(self, i, character):
new_blanks = self.blanks.copy()
new_blanks[i] = character
return GameState(self.lives, self.h, new_blanks, self.chosen_word)
def get_guessed_word(self):
return ''.join(self.blanks)
def get_hangman_pic(self):
return HANGMANPICS[self.h]
def did_win(self):
return self.get_guessed_word() == self.chosen_word
def did_lose(self):
return self.lives <= 0
def is_game_over(self):
return self.did_lose() or self.did_win()
Here's how your code changes if you're using this game state object:
def get_next_state(guess, state):
if guess not in state.chosen_word:
else:
for i in range(len(state.chosen_word)):
if state.chosen_word[i] == guess:
return state.next_state_for_good_guess(i, guess)
def print_state(guess, state):
os.system('clear')
if guess not in state.chosen_word:
print(f"You guessed {guess}. That's not in the word. You lose a life.\n")
print(state.blanks)
print(state.get_hangman_pic())
if state.did_win():
print("\n ****You win**** \n ")
elif state.did_lose():
print("You lose")
def testing():
chosen_word = random.choice(word_list)
starting_lives = 6 # total no of lives
starting_blanks = list("_" * len(chosen_word))
state = GameState(starting_lives, 0, starting_blanks, chosen_word)
while not state.is_game_over():
state = get_next_state(guess, state)
print_state(guess, state)
By splitting up the responsibilities like this you can think about fewer things at once. When you're thinking through the game rules, you don't have to think about how to print. When you're thinking about printing, you don't have to think about how the state changes -- only what the current state is. This is especially important as the program grows larger than something that can be entirely fit inside your head.
Regarding immutability, I can see at least two benefits:
1. Like single responsibility, this reduces the number of things you think about at once. There is exactly one place where the state can change, so if you want to understand state change you look there. Whereas, with multiple global variables, the state can be changed anywhere in the program so to understand state change you have to look at the whole program
2. You get a lot of powerful options for free once things are immutable. With this version, you could check the consequences of a guess without taking the guess. If you wanted to provide three hints to players, you could implement the hint by calling get_next_state and reporting whether the guess was good or not, without storing the result in state and printing the update. I'm not suggesting you go design you code for features that you don't need; I'm illustrating how separating the state change logic from the state storage logic opens up cool possibilities that we don't have when the state change logic is writing directly to our state variables.
If a function just calls another function, delete one of them. You don't need both main() and testing(), since the former does nothing other than call the latter.
Global variables are almost never needed. Don't organize programs around the need to modify global variables. Instead, organize your programs around functions that rely only on their local variables (or methods that follow similar principles). If some functions need access to the same information, pass that data back and forth using function arguments and return values.
Don't put logic in top-level code. At the top level of a program, you can perform imports or define constants, functions, or classes. Put all other code inside of functions or methods. If we apply all of the preceding ideas to your current code, we end up with the following structure:
def main():
lives = 6
chosen_word = random.choice(WORD_LIST)
...
while lives > 0:
blanks, lives = replace_blank(guess, chosen_word, blanks, lives)
...
...
def replace_blank(guess, chosen_word, blanks, lives):
...
return (blanks, lives)
Try to design functions so that they perform a single task. This reorganization of the code highlights some awkwardness in the implementation. The main() function initializes lives and blanks but then delegates responsibility for updating them to the replace_blank() function. Meanwhile, that function has various other responsibilities, like printing messages to the user and clearing the terminal.
Missing features. Many hangman games tell the user what letters have already been guessed and I've never seen one that penalizes a player for guessing an already-guessed letter. I mention this missing behavior because it points the way toward a less awkward code design. The first step is to select a data structure to hold prior guesses. A set is a natural fit. In fact, if we know the initial number of lives, the secret word, and the prior guesses, all other information about game state can be derived (remaining lives, whether the game is finished, how the secret word should be displayed, etc).
Try to separate logic/algorithm from interactivity and printing. Code that interacts with users is a hassle to test and debug. Ideally, you want most of your program to operate purely in the data realm: for functions that means receiving data and returning data; for methods it might mean receiving data and then returning data or modifying internal data. However, many programs require interactivity and related printing. Programs like this are often good fits for a design that used a main function to handle interactions and an immutable object that knows hold to hold the current state and then how to compute the next state. Let's start with a sketch of the interactive layer. This is where we perform all printing and collecting of user input. Notice especially that this part of the program contains very little algorithmic logic.
import random
import os
WORD_LIST = [...]
HANGMANPICS = [...]
def main():
# Setup.
gs = GameState(random.choice(WORD_LIST))
# Play until done.
while not gs.is_finished:
# Printing.
os.system('clear')
print(gs.graphic)
print('Secret word:', gs.display_word)
print('Prior guesses:', gs.display_guesses)
if gs.message:
print(gs.message)
# Derive the next state based on user input.
# Wrap up.
print(gs.outcome)
while True:
guess = input("Guess a letter? ").lower()
if guess.isalpha() and len(guess) == 1:
return guess
if __name__ == '__main__':
main()
The rest of the code just requires implementing the needed data-retrieval behaviors in GameState. As noted, the initial state will hold an empty set for prior guesses. A message attribute will be used to store information needed by our interactive layer when printing game status.
from dataclasses import dataclass, field
@dataclass(frozen = True)
class GameState:
word : str
prior_guesses : set = field(default_factory = set)
message : str = ''
INITIAL_LIVES = len(HANGMANPICS) - 1
BLANK = '_'
Inside the GameState class, we need to implement various read-only properties to get information derived from our core pieces of data (the word and the prior guesses):
class GameState:
@property
def lives(self):
return self.INITIAL_LIVES - sum(
char not in self.word
for char in self.prior_guesses
)
@property
def display_word(self):
return ''.join(
char if char in self.prior_guesses else self.BLANK
for char in self.word
)
@property
def display_guesses(self):
return ' '.join(sorted(self.prior_guesses))
@property
def is_finished(self):
return self.lives <= 0 or self.word == self.display_word
@property
def graphic(self):
return HANGMANPICS[self.INITIAL_LIVES - self.lives]
@property
def outcome(self):
return (
None if not self.is_finished else
'You won' if self.display_word == self.word else
'You lost'
)
Finally, the class needs a method to generate a new GameState instance based on the current state and the user's guess:
class GameState:
def next_state(self, guess):
w = self.word
pg = self.prior_guesses
message = (
f'Already guessed: {guess}' if guess in pg else
f'Correct: {guess}' if guess in w else
f'Not in word: {guess}. You lose a life'
)
return GameState(w, pg | {guess}, message)
With a design like this, the vast majority of your functionality could be tested and debugged purely by working with the data-oriented GameState class.
• Why are you declaring basically all of the GameState methods as properties? This doesn't make much sense to me.
• @ades Because the alternatives are worse. Consider lives as an example. (1) Option 1: store it as an attribute. Bad idea, because lives is purely derived from word and prior guesses. If you store lives separately, you have to manage correlated data, which is error prone. (2) Option 2: provide lives via a regular method. That works alright, but it does force the user to type gs.lives() rather than gs.lives.
• I think having is_finished as a property makes sense, but things like outcome and display_word should not be properties. I think you can take two routes - either follow google's style guide and define properties by how trivial the operation is, or try to deduce according to behaviour. And in either of those cases I don't think all those methods should be properties.
• Properties for me are for attributes/members. Displaying a word is a process, which is confusing to mix with the operations that you use for attributes. I'd wonder what was going on if I read a call that was GameState.display_word - I could see how GameState.current_word` would be a property. That would be the behavioural deduction. If we instead use Google's "classifier", you could also argue that the operation that you perform isn't trivial enough; you have a conditional and you transform type. This one is less clear ("what is trivial"), but I think you get the point. | 2,855 | 12,382 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-33 | latest | en | 0.782517 |
https://discourse.mcneel.com/t/merging-a-lot-of-solids/112839 | 1,660,132,668,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00203.warc.gz | 219,328,583 | 6,642 | # Merging a lot of solids
Hello everybody!
So after an intense period of learning, I managed to design a trippy ashtray! I created it in Grasshopper, baked it and that did a few “manual” adaptations in Rhino. Each one of these is a separate Solid. How could I close this all up in one single solid that is “filled” and not hollowed out? I tried to do a “Solid Union” on these cubes but I get a “Boolean Union is set empty” error.
I imagine that maybe there could be gaps between the cubes so that’s why it is not working? But also I imagine I first need to create a base in the shape of the bottom row.
Impossible to see details. Could be a problem with non-manifold edges.
I think the first thing I would try is scale every cube by a tiny factor, let’s say 1.001 and try to merge everything then
I think to make that a solid with breps(or even meshes) will be a huge hustle.
Why do you ant to solidify it? For 3d-print?
1 Like
Yeah the plan is to 3D print it!
Is this step even necessary for 3D printing?
Try looking into Marching cubes.
You would lose the ‘voxel’ look though. I’m not sure if that’s something you aimed for, or just a limitation of how you modeled it.
some printers dont care if its joined(booled)together.
general thoughts:
-you may think of the resolution of the printer
-the wall thickness(some parts look thin, cause the cubes arent touching much)
-the material(it may become hot when putting the cigarette)
Thanks for the inputs! I am gonna try to print it without modifications to see what comes out. Fingers crossed!
And yeah, the plan is to silicon mold it afterward and then cast it in concrete, so the cigarette heat won’t be a problem
Hi, if you want it only for 3d-print, MeshMixer from Autodesk can be another option you should take a look.
This processes meshes containing errors (yours is not errors though) through voxels and convert it to a single solid.
------tutorial--------
Import an .stl, Edit->Make Solid (try raising all the options to the highest if you are not sure about the options)
Thank you! I will give it a look | 493 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | latest | en | 0.931974 |
https://brainly.in/question/128032 | 1,484,630,027,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279468.17/warc/CC-MAIN-20170116095119-00486-ip-10-171-10-70.ec2.internal.warc.gz | 825,316,755 | 10,352 | # The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. NCERT Class X Mathematics - Mathematics Chapter _SOME APPLICATIONS OF TRIGONOMETRY
1
by shivaKalita199
2015-06-12T09:20:33+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Let the distance between the foot of the tower and the foot of the building be d.
height of the building = H
height of the tower = h = 50 m
Tan 30⁰ = 1/√3 = H/d
Tan 60⁰ = √3 = h/d
=> (1/√3)/(√3) = H / h
=> H = h/3 => 50 m / 3 = 16.667 meters
click on thanks button above pls | 274 | 958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-04 | latest | en | 0.910707 |
https://es.mathworks.com/company/newsletters/articles/run-matlab-image-processing-algorithms-on-raspberry-pi-and-nvidia-jetson.html | 1,653,726,336,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663013003.96/warc/CC-MAIN-20220528062047-20220528092047-00525.warc.gz | 303,516,058 | 18,027 | # Run MATLAB Image Processing Algorithms on Raspberry Pi and NVIDIA Jetson
By Jim Brock and Murat Belge, MathWorks
Thanks to low-cost hardware platforms such as Raspberry Pi™, it is now easier than ever to prototype image processing algorithms on hardware. Most image processing algorithms are computationally intensive, and it can be challenging to run them on an embedded platform with acceptable frame rates. While Raspberry Pi is sufficient for running simple image processing algorithms, large images and complex algorithms are best run on more powerful hardware such as NVIDIA® Jetson.
Using a chroma key effect as an example, this article describes a simple workflow for deploying a MATLAB® image processing algorithm to embedded hardware. We'll generate C code from the algorithm with MATLAB Coder™, and then use the Run on Hardware utility to prototype the algorithm on a Raspberry Pi board. Finally, we'll move the algorithm to an NVIDIA Jetson Tx1 platform to achieve real-time performance.
## The Chroma Keying Algorithm
Widely used in TV weather reports, movie production, and photo editing applications, chroma keying is a video processing technique in which a foreground subject is shot against a solid color background, such as a green screen, that is later replaced by a different scene (Figure 1).
Figure 1. Before-and-after example of chroma keying.
The chroma keying algorithm compares each pixel in the image with a reference color representing the solid background color. If the color of the pixel is close enough to the reference color, the pixel is replaced with the corresponding pixel from a pre-selected scene image. Mathematically, the chroma keying algorithm can be formulated as the following:
$P_{final}(j,k)=m(j,k)*P_{original}(j,k)+(1-m(j,k))*P_{scene}(j,k)$
Where $$P_{final}(j,k)$$ represents the final pixel value at location $$(j,k)$$ after chroma keying, $$P_{original}(j,k)$$ is the pixel value corresponding to the original image, $$P_{scene}(j,k)$$ is the pixel value representing the scene that replaces the solid background color, and $$m(j,k)∈[0,1]$$ is a mask value. The mask value $$m(j,k)$$ should be 1 for foreground pixels and 0 for background pixels. A mask value between 0 and 1 provides a smooth transition from background to foreground.
The mask value at each pixel is usually computed in the YCbCr color space instead of the usual RGB color space. The Y component of the YCbCr image represents the luminance component and determines how light or dark the image is. Cb and Cr components represent the chroma components that can be used to measure similarity to a reference color. Measuring color similarity using only the Cb and Cr components of the image makes the algorithm robust to variations in luminance values in light and dark areas of the solid background color.
To measure the similarity of a pixel color to a reference color, we use the squared Euclidian distance in chroma space:
$d^2 (j,k)=(Cb(j,k)-Cb_{ref} (j,k))^2+(Cr(j,k)-Cr_{ref} (j,k))^2$
Finally, we compute the mask value at location $$(j,k)$$ in the image using the following formula:
$m(j,k)=\left\{\begin{matrix} 1 & if d(j,k)>t_2 \\ 0 & if d(j,k)<t_1 \\ \frac{d^2 (j,k)-t_1^2}{t_2^2-t_1^2} & if t_1<d(j,k)<t_2 \end{matrix}\right.$
Where $$t_{1}$$ and $$t_{2}$$ with $$t_{2} > t_{1}$$ represent threshold values to be determined.
## MATLAB Implementation
Here is the MATLAB implementation of the chroma keying algorithm.
function Pfinal = chromaKey(P, Pscene, refColorYCbCr, t1, t2)
Cbref = double(refColorYCbCr(1,1,2));
Crref = double(refColorYCbCr(1,1,3));
PYCbCr = rgb2ycbcr(P);
Cb = double(PYCbCr(:,:,2));
Cr = double(PYCbCr(:,:,3));
d = (Cb - Cbref).^2 + (Cr - Crref).^2;
t1 = t1^2;
t2 = t2^2;
m = zeros([size(d,1) size(d,2)]);
for j = 1:size(m,1)
for k = 1:size(m,2)
if d(j,k) > t2
m(j,k) = 1;
elseif d(j,k) > t1
m(j,k) = (d(j,k) - t1) / (t2 - t1);
end
end
end
m = repmat(imgaussfilt(m,0.8), [1 1 3]);
Pfinal = uint8(double(P).*m + double(Pscene).*(1-m));
end
In MATLAB, images are represented as [N, M, 3] arrays of type uint8. This means that we'll need to convert the image data type to 'double' before performing mathematical operations. To avoid abrupt transitions from background to foreground, we apply a Gaussian filter to the computed mask.
## Determining Reference Color and Thresholds
A chroma keying algorithm requires a reference color and thresholds. Using the camera interface in MATLAB Support Package for Raspberry Pi, we capture images of the actual scene. We can then empirically determine the approximate reference color for the background and the approximate threshold values.
r = raspi;
cam = cameraboard;
for k = 1:10
img = snapshot(cam);
end
The img = snapshot(cam); command plots the image captured from Raspberry Pi camera in MATLAB. We use the Data Cursor tool in the MATLAB plot to specify the background color (Figure 2).
Figure 2. The MATLAB Data Cursor tool, used to determine background color values.
To determine the thresholds, we run the algorithm in a loop and adjust the threshold values:
refColorRGB = zeros([1,1,3],'uint8');
refColorRGB(1,1,:) = uint8([93 177 21]);
refColorYCbCr = rgb2ycbcr(refColorRGB);
t1 = 28;
t2 = 29;
scene = data.bg;
% Main loop
for k = 1:1000
img = snapshot(cam);
imgFinal = chromaKey(img, scene, refColorYCbCr, t1, t2);
figure(1),image(img);
figure(2),image(imgFinal);
drawnow;
end
When we run the code we get an image shown against the background we selected (Figure 3).
Figure 3. Left: original image. Right: image obtained after running the chroma keying algorithm.
## Deploying the Chroma Keying Algorithm to Raspberry Pi
Before deploying the code, we need to write a loop around the chroma keying algorithm to capture images from a camera and display them on a monitor attached to Raspberry Pi:
function chromaKeyApp()
%Chroma keying example for Raspberry Pi hardware.
%#codegen
% Copyright 2017 The MathWorks, Inc.
w = matlab.raspi.webcam(0,[1280,720]);
d = matlab.raspi.SDLVideoDisplay;
refColorYCbCr = zeros([1,1,3],'uint8');
refColorYCbCr(1,1,:) = uint8([0 76 98]);
scene = imrotate(data.bg,90);
% Main loop
for k = 1:60
img = snapshot(w);
img = chromaKey(img, scene, refColorYCbCr, 28, 29);
displayImage(d,img);
end
release(w);
release(d);
end
matlab.raspi.webcam and matlab.raspi.SDLVideoDisplay are System objects™ in the Run on Hardware utility that facilitate use of camera and Raspberry Pi display in a deployment workflow. To compile and run the code, we execute the following command:
runOnHardware(r,'chromaKeyApp')
The function runOnHardware creates a MATLAB Coder configuration for Raspberry Pi hardware, generates code for the chromaKeyApp.m script, and deploys it. In order to run the algorithm at a reasonable frame rate, the image size can be reduced to 640x480 or 320x240.
## Generating GPU Code
The algorithm is working on the Raspberry Pi, but it is not achieving the real-time performance we're looking for. To accelerate the algorithm, we will use GPU Coder™ to deploy it to the NVIDIA Jetson platform. We need to generate GPU code to take advantage of the inherent parallelism in the algorithm. First, we write a wrapper main function that uses OpenCV to access a USB camera connected to the NVIDIA Jetson. This function will marshal video frames from the camera to our chromaKey algorithm and then display the output on the screen.
When generating GPU code, we first create a GPU Coder configuration object, set the GPU parameters to target the NVIDIA Jetson platform, and include our custom main function. We will not compile the code on the MATLAB host computer, because we are generating code specifically for the NVIDIA Jetson board. We will create a script to set up the GPU Coder configuration, input example data, and generate source code for our application.
% Create GPU Coder configuration for Jetson TX2
cfg = coder.gpuConfig('exe');
cfg.GpuConfig.MallocMode = 'Unified';
cfg.GpuConfig.ComputeCapability = '6.2';
cfg.GenCodeOnly = 1;
cfg.CustomSource = 'main_webcam.cu';
% Create sample inputs
refColorRGB = [70 130 85]; % RGB light Green
tmpColor = zeros([1,1,3],('uint8');
tmpColor(1,1,:) = uint8(refColorRGB);
refColor = rgb2ycbcr(tmpColor);
threshold1 = 14;
threshold2 = 20;
% Generate CUDA code for chromaKey
codegen -config cfg -args {fg,bg,refColor,threshold1,threshold2} chromaKey
We then run the script in MATLAB to generate CUDA code for the chromaKey algorithm.
## Deploying a Green Screen Algorithm to NVIDIA Jetson
To deploy the generated code to the NVIDIA Jetson, we need to package all the required files into the codegen directory , with the following MATLAB commands.
% Prepare files for transfer to NVIDIA Jetson TX2
copyfile('Scenery.jpg','codegen/exe/chromaKey/');
copyfile('main_webcam.cu','codegen/exe/chromaKey/');
copyfile(fullfile(matlabroot,'extern','include','tmwtypes.h'),'codegen/exe/chromaKey/');
copyfile('buildAndRun.sh','codegen/exe/chromaKey/');
The next step is to copy the entire generated codegen folder from the host machine to the NVIDIA Jetson board. After the files have been transferred, we sign in to the NVIDIA Jetson directly to build and run the application.
Once logged in to the NVIDIA Jetson, we run the jetson_clocks.sh script provided by NVIDIA to maximize the performance of the board, change to the codegen directory containing the generated source code we just transferred, and execute the compile command shown below.
Once the executable (chromaKey) has been built, the application is run with a USB-connected webcam on the NVIDIA Jetson board with the following command. The frames-per-second rate will be displayed on the output.
$> sudo ./jetson_clocks.sh$> cd codegen/exe/chromaKey
$> nvcc -o chromaKey *.cu -rdc=true -arch sm_62 -O3 pkg-config --cflags --libs opencv -lcudart$> ./chromaKey 1
Figure 4 shows the output from the NVIDIA Jetson board's USB camera before and after the green screen effect.
Figure 4. Before-and-after example of applying the green screen effect.
## Comparing Raspberry Pi and NVIDIA Jetson Performance
The greater parallel processing power of the GPU on the NVIDIA Jetson significantly improves the algorithm's performance. The Raspberry Pi achieved approximately 1 frame per second, while the NVIDIA Jetson achieved more than 20 frames per second for an image size of 1280x720—we gained a more than twentyfold speedup without making any modifications or optimizations to our algorithm. We could improve performance even more by optimizing the MATLAB algorithm for more efficient GPU code generation.
## Summary
In this example we saw how to rapidly generate code for a MATLAB algorithm and deploy it to embedded hardware like the Raspberry Pi. We quickly determined that our algorithm was working correctly and needed to be parallelized. Using MATLAB and GPU Coder, we generated a highly parallel implementation of the algorithm and deployed it to an NVIDIA Jetson board, achieving a significant performance improvement.
Published 2018 | 2,730 | 11,029 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-21 | latest | en | 0.861003 |
https://www.jiskha.com/questions/1002685/for-a-monthly-customer-charge-of-5-69-plus-8-48-cents-per-kilowatt-hour-for-up-to-1000 | 1,576,248,016,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540555616.2/warc/CC-MAIN-20191213122716-20191213150716-00167.warc.gz | 742,509,842 | 5,296 | pre cal
for a monthly customer charge of \$5.69 plus 8.48 cents per kilowatt-hour for up to 1000 kilowatt-hours. (a) Write a linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month, 0 ≤ x ≤ 1000.
1. 👍 0
2. 👎 0
3. 👁 212
1. C = 5.69 + .0848x
1. 👍 0
2. 👎 0
posted by Steve
Similar Questions
1. pre cal
Electricity Rates in Florida. Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of \$5.69 plus 8.48 cents per kilowatt-hour for up to 1000 kilowatt-hours. (a) Write a linear
asked by kisha on January 15, 2014
2. math
Power companies typically bill customers based on the number of kilowatt-hours used during a single billing period. A kilowatt is a measure of how much power (energy) a customer is using, while a kilowatt-hour is one kilowatt of
asked by HELLO on March 9, 2013
3. Physics
At 7.65 cents per kilowatt-hour, what does it cost to operate a 10.0 hp motor for 6.00 hr? At 7.65 cents per kilowatt-hour, what does it cost to leave a 75.0 W light burning 24.0 hours a day?
asked by Ana on March 6, 2011
4. Algebra
A phone company offers two monthly charge plans. In Plan A, the customer pays a monthly fee of \$13 and then an additional 6 cents per minute of use. In Plan B, there is no monthly fee, but the customer pays 8 cents per minute of
asked by Gilbert on September 25, 2015
5. Chemistry
Assume electricity cost 15 cents kw per hour....using the energy requirements.what is the per gram cost of energy to produce Na + AL? Useful info: 2NaCL --> 2Na + CL2 around 600 degree C; 14 KJ per g Na. Al- 2Al2O3-->4Al+3O2 1000
asked by Liliana on November 14, 2006
6. Physics
One kilowatt-hour (kWh) is the amount of work or energy generated when one kilowatt of power is supplied for a time of one hour. A kilowatt-hour is the unit of energy used by power companies when figuring your electric bill.
asked by Candy on February 29, 2012
7. algebra
Express each situation as an algebraic expression: 1. An electric company charges \$10 a month plus \$.15 per kilowatt-hour of electricity used. What is the monthly bill in dollars if K kilowatt-hours are used?
asked by Anonymous on October 26, 2010
one kilwatt/hour (kWh) is the amount of work or energy generated when one kilowatt of power is supplied fr a time of one hour. a kilowatt/hour is the unit of energy used by power companies when figuring your electric bill.
asked by taylor on February 24, 2012
9. programming logic
The No Interest Credit Company provides zero-interest loans to customers. (It makes a profi t by selling advertising space in its monthly statements and selling its customer lists.) Design an application that gets customer account
asked by carl on December 5, 2013
10. statistics
6. According to the records of Enersource, an electric company serving the Mississauga area, the mean electricity consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electricity
asked by david on June 23, 2013
More Similar Questions | 858 | 3,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-51 | latest | en | 0.898797 |
http://forums.wolfram.com/mathgroup/archive/2007/Dec/msg00629.html | 1,582,163,660,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144498.68/warc/CC-MAIN-20200220005045-20200220035045-00294.warc.gz | 63,605,049 | 7,744 | Re: Re: Problem with Syntax in Mathematica Summation
• To: mathgroup at smc.vnet.net
• Subject: [mg84425] Re: [mg84419] Re: Problem with Syntax in Mathematica Summation
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Sun, 23 Dec 2007 04:37:09 -0500 (EST)
```Needs["HypothesisTesting`"];
The function NormalPValue returns a rule rather than a value
NormalPValue[x]
OneSidedPValue ->
(1/2)*(1 - Erf[Abs[x]/Sqrt[2]])
Consequently, in an expression it would be used in a form similar to
(OneSidedPValue /. NormalPValue[x])
(1/2)*(1 - Erf[Abs[x]/Sqrt[2]])
(TwoSidedPValue /. NormalPValue[x, TwoSided -> True])
1 - Erf[Abs[x]/Sqrt[2]]
Bob Hanlon
---- magma <maderri2 at gmail.com> wrote:
> There were a couple of syntax errors:
> in the summation with ebitda and in kinv0
> Below is the corrected code.
>
> wacc = ((rf + \[Beta] ((1 + tax) (d/(d + e))) (rf - rm)) (e/(d +
>
> e))) ((rf + (1 - (NormalPValue[(Log[(d/(d + e))]/((d/(d +
> e)) - 1)) (1/
> stde)])) r) (d/(d + e)))
> value = \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(t = 1\), \(\[Infinity]\)]
> \*FractionBox[\((
> \*SubscriptBox[\(ebitda\), \(t\)]\ \((1 - tax)\))\), \(\((1 + wacc)\)^
> t\)]\) + \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(t = 1\), \(\[Infinity]\)]
> \*FractionBox[\(\((
> \*SubscriptBox[\(dep\), \(t\)])\)\ tax\), \(\((1 + wacc)\)^
> t\)]\) - \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(t = 1\), \(n\)]
> \*FractionBox[\(\((
> \*SubscriptBox[\(d\), \(0\)] -
> \*UnderoverscriptBox[\(\[Sum]\), \(t = 1\), \(n\)]
> \*SubscriptBox[\(d\), \((t - 1)\)])\) \((rf + 1 - \((NormalPValue[\((
> \*FractionBox[\(Log[\((d/\((d + e)\))\)]\), \(\((d/\((d + e)\))\) -
> 1\)])\) \((
> \*FractionBox[\(1\), \(stde\)])\)])\))\)
> r \((1 - tax)\)\), \(\((1 + wacc)\)^t\)]\) - \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(t = 1\), \(n\)]
> \*FractionBox[\(
> \*SubscriptBox[\(d\), \(0\)]/n\), \(\((1 + wacc)\)^t\)]\) - Subscript[
> kinv, 0]
>
```
• Prev by Date: how fill PolarPlot?
• Next by Date: Re: Mathematica 6 Documentation - how does one add to it?
• Previous by thread: Re: Problem with Syntax in Mathematica Summation
• Next by thread: sum over i+j+k==n | 851 | 2,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-10 | latest | en | 0.493755 |
http://www.kaizou.org/2023/05/machine-learning-quantization-introduction.html | 1,701,357,520,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00778.warc.gz | 80,167,099 | 11,273 | # A brief introduction to Machine Learning models quantization
25 May 2023 by David Corvoysier
Even before the development of Large Language Models (LLM), the increasing memory and computing requirements of Deep Neural Networks (DNN) has been a concern.
Functionally, DNN are graphs of arithmetic operations: the inputs are fed at the stem and the chain of operations produces the outputs at the head.
From an implementation perspective, the operations are performed on floating point numbers, which are a digital representation of decimal numbers composed of a mantissa and an exponent:
$x = mantissa . 2^{exponent}$
The 32-bit floating point representation if the most common, as it allows to represent numbers in a range that is sufficient for most operations. The float32 mantissa is composed of 24-bit (including sign), and the exponent is 8-bit.
Each operation performed at an operating node in the inference device requires its inputs to be transferred from either a static memory location or the previous processing nodes.
The cost of these transfers adds-up with the cost of the operations themselves.
The DNN terminology for operation data is “weights” for static inputs and “activations” for dynamic inputs/outputs.
Note: the outputs of an operation are designated as “activations” even if it is not actually an activation.
The process of representating the n-bit weights and activations of a DNN into a smaller number of bits is called quantization1.
It is typically used in DNN to “quantize” float32 weights and activations into 8-bit integer.
This brings several benefits:
• reducing the weights to 8-bit requires 4 times less memory on the device to store them,
• reducing the activations to 8-bits reduces the amount of data exchanged between nodes, which impacts latency,
• using 8-bit instead of 32-bit inputs for an operation improves vectorization (multiple data processed at the same time for a single operation),
• all standard integer arithmetic operations but the division are faster than their floating point counterpart,
• GPU devices may include specific mechanisms to process 8-bit inputs (like NVIDIAS 8-bit Tensor cores).
## A mathematical formulation of linear quantization
The most widespread type of quantization is the linear or affine quantization scheme first introduced in tensorflow lite2.
The representation of a linearly quantized number is composed of:
• an integer mantissa,
• a float scale,
• an integer zero-point.
$x = (mantissa - zeropoint).scale$
The scale is used to project back the integer numbers into a float representation.
The zero point corresponds to the value that zero takes in the target representation.
If we compare that formula with the floating point representation one can see immediately that each floating point number can be represented exactly with the same mantissa, a scale corresponding to the exponent and a null zero-point.
Of course this representation would be very inefficient because it would require two integer and a float to represent each number.
## Applicability of quantization to Machine-Learning
When quantizing Machine-Learning models, one can take advantage of the fact that the training produces weights and activations stay within reasonably stable ranges for a given operation.
This comes from several empirical techniques used to improve convergence:
• weights initialization3,
• weights and/or activation regularization4,
• explicit normalization layers5.
This means that the weights and activations tensors for a specific operation can be represented using the same scale and zero-point, thus leading to a very compact representation.
Note: this is why quantization is often categorized as a form of compression, although unlike most compression techniques, it produces numbers that can be directly used for arithmetic operations.
There are various subtypes of quantization.
The first two subtypes are related to the dimensions of the scale and zero-point:
• per-tensor quantization uses a single scalar value for scale and zero-point for a whole tensor of weights or activations,
• per-axis quantization uses a vector of scales and zero-points whose length corresponds to a single axis of the tensor (typically the channels or embeddings axis).
The second subtypes are related to the symmetry of the resulting quantized numbers:
• symmetric quantization assumes that the quantization range is symmetric, which leads to a zero-point equal to zero and a signed integer representation of the values,
• asymmetric quantization does not assume anything, and zero-point is typically non-null.
Weights are typically quantized symmetrically per-axis.
Activations are typically quantized asymmetrically, most of the time per-tensor.
## Quantizing a float tensor
The first step to quantize a float tensor is to choose the quantization range, i.e. the minimum and maximum float values one wants to represent: $[Min, Max]$.
Since the weights are constant tensors, they are typically quantized using the mimimum and maximum values of the tensor, globally or along the channel axis.
Evaluating the quantization range of the activations is more difficult as they are dependent of the inputs of the previous operation. Their range is therefore evaluated globally inside a model, as explained in the next paragraph.
For a target bit width of n for the mantissa, one evaluates the scale as:
$scale = \frac{Max - Min}{2^n - 1}$
The zero-point is then deduced from the scale to make sure that $Min$ is mapped to the lowest integer value and $Max$ to the highest integer value.
This leads to the following formulas for signed/unsigned representations:
• unsigned: $zeropoint = round(\frac{Min}{scale})$
• signed: $zeropoint = round(\frac{Min}{scale}) - 2^{n - 1}$
The quantization of a float tensor is then:
$mantissa = saturate(round(\frac{x}{scale}) + zeropoint)$
Again, the saturation depends of the signed of the target representation:
• unsigned: $[0, 2n - 1]$,
• signed: $[-2^{n-1}, 2^{n-1} - 1]$.
Note that the zero-point always has the same signedness as the mantissa.
## Quantizing a Machine Learning Model
As mentioned before, a Machine Learning model uses two types of tensors: weights and activations.
The static weights need to be quantized only once, each weight tensor producing three new static tensors for the mantissa, scale and zeropoint.
Since weights can contain positive and negative values, they are typically quantized into int8.
.----------.
| Weights |
| float32 |
| constant |
+----+-----+
/ | \
v v v
.----------. .----------. .------------.
| Weights | | scale | | zero-point |
| int8 | | float32 | | int8 |
| constant | | constant | | constant |
'----------' '----------' '------------'
The dynamic activations on the other hand need to be quantized on-the-fly by inserting the quantization operations in the graph:
• evaluate the quantization range,
• quantize.
The evaluation of the quantization range is costly because is requires a full-scan of the activations tensor, which is a bottleneck for parallel processing.
For that reason, the activations quantization ranges are often evaluated before the inference on a selected number of samples: this is called the calibration of the quantized model.
Note: the operations that clip their outputs like the bounded ReLU are an exception and don’t require an explicit calibration, since the exact range of their outputs is known in advance.
After calibration, each activation float variable is mapped to an integer variable and two static tensors.
.-----------.
| Activations |
| float32 |
| variable |
/'-----+-----'\
/ | \
v v v
.-----------. .----------. .------------.
| Activations | | scale | | zero-point |
| (u)int8 | | float32 | | (u)int8 |
| variable | | constant | | constant |
'-----------' '----------' '------------'
Note: the activations can be quantized into either int8 or uint8. It is simpler to quantize them to uint8 if they correspond to the output of a ReLU operation, since zero-point will be in that case 0.
Conceptually, the resulting graph is a clone of the original graph where all compatible operations are replaced by a version that operates on tuples of (mantissa, scale, zero-point).
Separating the constant and variable tensors, this leads to the following graphs:
.---------. .--------. .----------. .------------.
| Inputs | | Inputs | | scale | | zero-point |
| float32 | | (u)int8 | | float32 | | (u)int8 |
| variable | | variable | | constant | | constant |
'----+----' '----+---' '-----+----' '------+-----'
| . '------------+-------------'
.----------. v |\ .----------. |
| Weights | .------. +--' \ | Weights | |
| float32 +->| Matmul | +--. / | int8 +-. |
| constant | '---+--' |/ | constant | | | .------------.
'----------' | ' '----------' | | | scale |
v | | .-+ float32 |
.---------. .----------. | v | | constant |
| Outputs | | scale | | .-------. | '------------'
| float32 | | float32 +-+->| QMatMul |<-+
| variable | | constant | | '---+---' | .------------.
'---------' '----------' | | | | zero-point |
| | '-+ (u)int8 |
.----------. | | | constant |
|zero-point| | | '------------'
| int8 +-' |
| constant | |
'----------' |
v
.--------.
| Outputs |
| (u)int8 |
| variable |
'--------'
## Quantized linear operations
Most basic Machine Learning operations can be performed using integer arithmetics, which makes them compatible with linearly quantized inputs.
This does not mean however that one can just replace all floating point operations by an equivalent integer operation: the scale and zeropoint of all weights and activations must be taken into account to produce an equivalent graph.
Also, there are two important restrictions with respect to the inputs quantization:
• additions between the integer mantissa of inputs can only be performed if they are in the same scale,
• operations that combine the integer mantissa of inputs channels can only be performed if the channels are in the same scale, i.e if the inputs are quantized per-tensor.
Note: in another post I explain how it is possible to add two inputs quantized with different scales by adding an explicit alignment operation beforehand.
From an implementation perspective, operations accepting linearly quantized inputs are very specific to each device.
In the next paragraph, I will detail a possible implementation of a quantized matrix multiplication.
## Wrap-up example: a quantized matrix multiplication
Let’s consider a simple matrix multiplication of an $X(I, J)$ input by a $W(J, K)$ set of weights:
$Y = X.W$
Since the matrix multiplication multiplies all inputs along the dimension of length $J$ and adds them, $X$ cannot be quantized per-axis, because it will lead to the addition of quantized numbers that are not in the same scale.
There is no such restriction on $W$, since the filters along $K$ are all applied independently.
After quantization of the weights per-axis and calibration of the inputs per-tensor, we obtain:
$X \approx X_s * (X_q - X_{zp})$, with $X_s()$, $X_q(I, J)$, $X_{zp}()$
$W \approx W_s * (W_q - W_{zp})$, with $W_s(K)$, $W_q(J, K)$, $W_{zp}(K)$
We can also approximate the outputs per-axis, assuming that the next operation does not require per-tensor inputs.
$Y \approx Y_s * (Y_q - Y_{zp})$, with $Y_s(K)$, $Y_q(I, K)$, $Y_{zp}(K)$
The operation is summarized on the graph below (note that the intermediate integer output Y_q can be implicit):
.-----. .-----. .------.
| X_q | | X_s | | X_zp |
'--+--' '--+--' '--+---'
'--------+-------'
.-----. |
| W_q +-. |
'-----' | | .-----.
| v .-+ Y_s |
.-----. | .---------. | '-----'
| W_s +-+->| QMatMul |<--+
'-----' | '----+----' | .-----.
| | '-+ Y_zp|
.-----. | | '-----'
|W_zp +-' |(Y_q)
'-----' |
v
.-.
| Y |
'-'
Going through the graph step by step:
• evaluate the matrix multiplication of the quantized inputs to produce float outputs
$O = X_s * (X_q - X_{zp}) . W_s * (W_q - W_{zp})$
• quantize the float outputs to obtain 8-bit integer outputs
$Y_q = saturate(round(\frac{O}{Y_s}) + Y_{zp})$
• convert back the 8-bit integer outputs to float outputs
$Y \approx Y_s * (Yq - Y_{zp})$
Since $X_s$ is a scalar, and $W_s$ has the same dimension as the outputs last dimension, the first operation can also be written:
$O = (X_s * W_s) * (X_q - X_{zp}) . (W_q - W_{zp})$
This means that the matrix multiplication can be operated equivalently on integer values, and the result is a quantized integer number with a scale corresponding to the product of the inputs and weights scale and a null zero-point.
The quantized sequence of operations is then to:
• evaluate the matrix multiplication of the 8-bit integer inputs to produce n-bit integer outputs
$O_q = (X_q - X_{zp}) . (W_q - W_{zp})$
• convert the n-bit integer outputs to float outputs
$O = (X_s * W_s) * O_q$
• quantize the float outputs to obtain 8-bit integer outputs
$Y_q = saturate(round(\frac{O}{Y_s}) + Y_{zp})$
• convert back the 8-bit integer outputs to float outputs
$Y \approx Y_s * (Yq - Y_{zp})$
The question that should immediately arise at this stage is why we need another quantization operation after the matrix multiplication, since we already have a quantized output ?
The reason is simply the bitwidth of the outputs: we need an explicit quantization to make sure that the results of the integer matrix multiplication fit in 8-bit.
Note: when the operation is followed by a bias addition, the biases are typically quantized to 32-bit with a scale precisely equal to $X_s * W_s$ so that they can be added directly to the outputs before quantizing.
Going one step further and replacing $O$, since $Y_s$ has the same shape as $X_s * W_s$, we can omit the third step and write directly:
• evaluate the matrix multiplication of the integer inputs to produce n-bit integer outputs
$O_q = (X_q - X_{zp}) . (W_q - W_{zp})$
• quantize the n-bit integer outputs to obtain 8-bit integer outputs
$Y_q = saturate(round(\frac{X_s * W_s}{Y_s} * O_q) + Y_{zp})$
• convert back the 8-bit integer outputs to float outputs
$Y \approx Y_s * (Yq - Y_{zp})$
This reveals that we can directly ‘downscale’ the integer outputs of the operation with a folded scale $F_s = \frac{Y_s}{X_s * W_s}$.
The downscaling operation can be implemented as a float division and a round.
Note: I will detail in another post an implementation using only integer arithmetic.
The simplified graph can be summarized below:
.-----. .------.
| X_q | | X_zp |
'--+--' '--+---'
'----+----'
.-----. |
| W_q +-. v
'-----' | .----------.
+->|IntMatMul |
.-----. | '----+-----'
|W_zp +-' | .-----.
'-----' v .-+ F_s |
.---------. | '-----'
|Downscale|<-+
'----+----' | .-----.
v '-+ Y_zp|
.-. '-----'
| Y |
'-'
This can be further simplified by removing the zero-points if we assume a symmetric quantization.
.-----.
| X_q |
'--+--'
|
v
.-----. .----------.
| W_q +->|IntMatMul |
'-----' '----+-----'
|
v
.---------. .-----.
|Downscale|<-+ Y_s |
'----+----' '-----'
v
.-.
| Y |
'-'
Note: the quantized matrix multiplication can be implemented in very different ways on devices that do not have efficient implementations of the integer Matrix Multiplication.
## References
1. Yunchao Gong, Liu Liu, Ming Yang, Lubomir Bourdev, “Compressing Deep Convolutional Networks using Vector Quantization” arxiv, 2014.
2. Benoit Jacob, Skirmantas Kligys, Bo Chen, Menglong Zhu, Matthew Tang, Andrew Howard, Hartwig Adam, Dmitry Kalenichenko, “Quantization and Training of Neural Networks for Efficient Integer-Arithmetic-Only Inference” arxiv, 2017.
3. Stone Yun, Alexander Wong, “Where Should We Begin? A Low-Level Exploration of Weight Initialization Impact on Quantized Behaviour of Deep Neural Networks”, arxiv, 2020.
4. Arash Ahmadian, Saurabh Dash, Hongyu Chen, Bharat Venkitesh, Stephen Gou, Phil Blunsom, Ahmet Üstün, Sara Hooker, “Intriguing Properties of Quantization at Scale”, arxiv, 2023.
5. Elaina Teresa Chai, “Analysis of quantization and normalization effects in deep neural networks”, stanford, 2021. | 4,041 | 16,860 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-50 | latest | en | 0.926354 |
https://coursescholars.com/stat-19/ | 1,675,234,569,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499911.86/warc/CC-MAIN-20230201045500-20230201075500-00594.warc.gz | 194,612,710 | 13,989 | # STAT
This course is concerned with the basic understanding of probability and it’s applications. The goal is to learn and you get back what you put in. Briefly state the question and use the Equation Editor to present all equations clearly. Answer any 2 questions.
For probability questions, define events as:
A: drawing a 10♣
B: drawing a 9♥
P(A&B) = P(A)P(B|A) = ??
Special Exercises (1to 5)
1. What is the probability of drawing at random the 4♥ and the 9♣ from a deck of cards on two successive draws? Use Insert/Insert symbol to get the heart, club, diamond, spade symbol by selecting the drop down menu to the left of Paragraph. Also, click on the drop down menu next to the Black Square icon in the menu of the Editor to change colors of the characters.
2. 18 people were poled and 10 said they have shopped at Wal-Mart and 8 said they have shopped at Target. 4 said they shopped at both. What is the probability that the group shopped at Wal-Mart or Target?.
3. Define independent and mutually exclusive events mathematically by using P(A) & P(B) for events A & B.
4. What is the probability of randomly drawing a 10 or a ♦ from a deck of cards?
5. What are the total possible outcomes of rolling a pair of dice? Note that the pair 3,4 is not the same as the pair 4,3. What is the probability of rolling a 7 with a pair of dice? P(7) = No. of successful outcomes/Total No. of outcomes
In questions 1,2,4, begin by clearly defining events A & B as shown below.
For example, A: 4♥ and B: 9♣. State P(A)=??, P(B)=?? and then apply the appropriate equations , such as
P(A&B) = P(A)P(B|A) or P(AorB) = P(A)+P(B)-P(A&B)
Substitute values and compute the answer. I urge you to read the attachments in the Videos-Topics In Stat 230/Probability Rules.
I prefer to use P(A&B) or P(AorB) instead of P(A∩B) or P(A∪B). You will see the latter on the final so be sure you know what the “∩” and “U” symbols mean.
COMPLEMENTARY EVENTS & Try It 3.19 in Illowsky
Complement_&_Rule.swf
The complement of an event A is the event A does not occur. The complement of event A is denoted by Ac and either event A or Ac must occur. Therefore,
P(A) + P(Ac) = 1.0
or,
P(A) = 1.0 – P(Ac)
For example, consider tossing a fair coin 10 times. Define event A as;
A: Observe at least one head
Ac: TTTTTTTTTT (observe no heads, 10 tails are tossed)
P(Ac) = (12)10=11024
P(A) = 1.0 − P(Ac) =1−11024=.999
We are fairly sure of observing at least one head in ten tosses.
Dependent vs Independent Events
Independent Events: Consider tossing two die. What is the probability of throwing a “One” on die #1 and then a “Six” on die # 2? Answer is 1/6 times 1/6 = 1/36. These events are independent of each other and P(A&B) = P(A)P(B).
Dependent events: What is the probability of drawing the Queen of Spades and the King of Spades if no card is repaced after the first draw. We apply the rule that P(A&B) = P(A)P(B|A) = P(B)P(A|B). Since A is the queen, P(A) = 1/52 and to then draw the kink, P(B|A) = 1/51, the answer is 1/52 times 1/51. Recall that P(B|A) means the probability that event B occurs given that event A has already occurred. Event B is dependent upon event A occurring.
Don't use plagiarized sources. Get Your Custom Essay on
STAT
Just from \$13/Page
## Calculate the price of your order
550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
\$26
The price is based on these factors:
Number of pages
Urgency
Basic features
• Free title page and bibliography
• Unlimited revisions
• Plagiarism-free guarantee
• Money-back guarantee
On-demand options
• Writer’s samples
• Part-by-part delivery
• Overnight delivery
• Copies of used sources
Paper format
• 275 words per page
• 12 pt Arial/Times New Roman
• Double line spacing
• Any citation style (APA, MLA, Chicago/Turabian, Harvard)
## Our guarantees
Delivering a high-quality product at a reasonable price is not enough anymore.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.
### Money-back guarantee
You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.
### Zero-plagiarism guarantee
Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.
### Free-revision policy
Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 1,285 | 4,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-06 | latest | en | 0.910698 |
https://www.convert-measurement-units.com/convert+Ounce+fluid+US+to+Decaliter.php | 1,675,410,043,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00771.warc.gz | 725,570,592 | 13,366 | Convert Ounce fluid US to dal (Ounce fluid US to Decaliter)
Ounce fluid US into Decaliter
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Ounce+fluid+US+to+Decaliter.php
How many Decaliter make 1 Ounce fluid US?
1 Ounce fluid US = 0.002 957 352 956 25 Decaliter [dal] - Measurement calculator that can be used to convert Ounce fluid US to Decaliter, among others.
Convert Ounce fluid US to Decaliter (Ounce fluid US to dal):
1. Choose the right category from the selection list, in this case 'Volume'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Ounce fluid US'.
4. Finally choose the unit you want the value to be converted to, in this case 'Decaliter [dal]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '238 Ounce fluid US'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Volume'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '16 Ounce fluid US to dal' or '27 Ounce fluid US into dal' or '79 Ounce fluid US -> Decaliter' or '43 Ounce fluid US = dal' or '40 Ounce fluid US to Decaliter' or '6 Ounce fluid US into Decaliter'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(53 * 43) Ounce fluid US'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '238 Ounce fluid US + 714 Decaliter' or '84mm x 82cm x 30dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 4.627 278 970 237 4×1025. For this form of presentation, the number will be segmented into an exponent, here 25, and the actual number, here 4.627 278 970 237 4. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 4.627 278 970 237 4E+25. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 46 272 789 702 374 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 950 | 3,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-06 | latest | en | 0.836326 |
http://gmatclub.com/forum/ab-and-cd-are-two-chords-of-a-circle-intersecting-at-the-24027.html?fl=similar | 1,481,351,468,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542939.6/warc/CC-MAIN-20161202170902-00030-ip-10-31-129-80.ec2.internal.warc.gz | 111,525,972 | 41,946 | AB and CD are two chords of a circle intersecting at the : Quant Question Archive [LOCKED]
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 09 Dec 2016, 22:31
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
AB and CD are two chords of a circle intersecting at the
Author Message
Intern
Joined: 26 Nov 2005
Posts: 18
Followers: 0
Kudos [?]: 19 [0], given: 0
AB and CD are two chords of a circle intersecting at the [#permalink]
Show Tags
02 Dec 2005, 11:50
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
AB and CD are two chords of a circle intersecting at the point E outside the circle so that
B and D are closer to E. If BE = 3 inches, AB = 5 inches, and ED = 4 inches, what is the
length of CD in inches?
(A) 2
(B) 5
(C) 6
(D) 7
(E) 8
Senior Manager
Joined: 03 Nov 2005
Posts: 395
Location: Chicago, IL
Followers: 3
Kudos [?]: 48 [0], given: 17
Show Tags
02 Dec 2005, 16:13
AE/CE=ED/EB let CE be x and AE=AB+BE=8
8/x=4/3 24=4x; x=CE=6
CE=ED+DC; DC=CE-ED=6-4=2
_________________
Hard work is the main determinant of success
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 287 [0], given: 0
Show Tags
02 Dec 2005, 20:09
rlevochkin wrote:
AE/CE=ED/EB let CE be x and AE=AB+BE=8
8/x=4/3 24=4x; x=CE=6
CE=ED+DC; DC=CE-ED=6-4=2
great, you remember that similar-triangle rule of a circle!! ..i'm going to draw and prove how it works
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 287 [0], given: 0
Show Tags
02 Dec 2005, 20:25
if you're interested in
btw, hadzamir, i find your geo problems very interesting. Can you let me know the source or post more here?!! Thanks
Attachments
similar.doc [25.5 KiB]
Intern
Joined: 26 Nov 2005
Posts: 18
Followers: 0
Kudos [?]: 19 [0], given: 0
Show Tags
03 Dec 2005, 03:32
rlevochkin wrote:
AE/CE=ED/EB let CE be x and AE=AB+BE=8
8/x=4/3 24=4x; x=CE=6
CE=ED+DC; DC=CE-ED=6-4=2
thanks
Intern
Joined: 26 Nov 2005
Posts: 18
Followers: 0
Kudos [?]: 19 [0], given: 0
Show Tags
03 Dec 2005, 04:23
laxieqv wrote:
if you're interested in
btw, hadzamir, i find your geo problems very interesting. Can you let me know the source or post more here?!! Thanks
one thing in your prove that i don't get:
why do you assume that ab and cd intersect in point O which is the center of the circle, or in other word - why do you assume that angle DOA is 180 degrees (why AD is a straight line?) same with line CB of course.
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 287 [0], given: 0
Show Tags
03 Dec 2005, 04:32
laxieqv wrote:
if you're interested in
btw, hadzamir, i find your geo problems very interesting. Can you let me know the source or post more here?!! Thanks
one thing in your prove that i don't get:
why do you assume that ab and cd intersect in point O which is the center of the circle, or in other word - why do you assume that angle DOA is 180 degrees (why AD is a straight line?) same with line CB of course.
Did i say so?! ..You should look more carefully at my illustration
The first part i was trying to prove that every angle form by two ends of a cord and a certain point in the circle is always = 1/2 angle formed by that cord's ends and the circle's centre ( This is a basic principle of geometry)
Then, i apply this rule to get DAB = DCB , COZ they're both =1/2 DOC ( YOU have to imagine this centre on your own, i don't put it in my illustration)
Again, there's nothing related to O in the second figure.
Intern
Joined: 26 Nov 2005
Posts: 18
Followers: 0
Kudos [?]: 19 [0], given: 0
Show Tags
03 Dec 2005, 04:58
laxieqv wrote:
laxieqv wrote:
if you're interested in
btw, hadzamir, i find your geo problems very interesting. Can you let me know the source or post more here?!! Thanks
one thing in your prove that i don't get:
why do you assume that ab and cd intersect in point O which is the center of the circle, or in other word - why do you assume that angle DOA is 180 degrees (why AD is a straight line?) same with line CB of course.
Did i say so?! ..You should look more carefully at my illustration
The first part i was trying to prove that every angle form by two ends of a cord and a certain point in the circle is always = 1/2 angle formed by that cord's ends and the circle's centre ( This is a basic principle of geometry)
Then, i apply this rule to get DAB = DCB , COZ they're both =1/2 DOC ( YOU have to imagine this centre on your own, i don't put it in my illustration)
Again, there's nothing related to O in the second figure.
you are right, now i get it. THANKS!!!
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 287 [0], given: 0
Show Tags
03 Dec 2005, 05:07
your geo problems are very interesting!! Can you please tell me the source or post more of them?! Thank you.
Intern
Joined: 26 Nov 2005
Posts: 18
Followers: 0
Kudos [?]: 19 [0], given: 0
Show Tags
03 Dec 2005, 06:51
laxieqv wrote:
your geo problems are very interesting!! Can you please tell me the source or post more of them?! Thank you.
they are not from a book. i got them from a friend of mine (the paper called "hard questions"), but the real hard questions i have published here and i will continue to publish if i'll encounter with more.
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 287 [0], given: 0
Show Tags
03 Dec 2005, 07:14
thank you ..and i'm sorry if my attitude was not friendly just now..i'm now much stressful due to my upcoming finals
VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6
Kudos [?]: 325 [0], given: 0
Show Tags
03 Dec 2005, 09:37
i got CD=20/3 ? BD is parallel to AC, hence EB/EA=ED/EC ! so 3/8=4/(4+x) => x=20/3 ! what i am missing ?
_________________
If your mind can conceive it and your heart can believe it, have faith that you can achieve it.
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 287 [0], given: 0
Show Tags
03 Dec 2005, 09:56
christoph wrote:
i got CD=20/3 ? BD is parallel to AC, hence EB/EA=ED/EC ! so 3/8=4/(4+x) => x=20/3 ! what i am missing ?
you're missing from the bold part, buddy ...the two lines can't always be parallel
Display posts from previous: Sort by | 2,118 | 6,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2016-50 | latest | en | 0.85853 |
http://easy-ciphers.com/sauciest | 1,558,762,130,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257889.72/warc/CC-MAIN-20190525044705-20190525070705-00151.warc.gz | 63,195,806 | 25,365 | Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: sauciest
cipher variations: tbvdjftu ucwekguv vdxflhvw weygmiwx xfzhnjxy ygaiokyz zhbjplza aickqmab bjdlrnbc ckemsocd dlfntpde emgouqef fnhpvrfg goiqwsgh hpjrxthi iqksyuij jrltzvjk ksmuawkl ltnvbxlm muowcymn nvpxdzno owqyeaop pxrzfbpq qysagcqr rztbhdrs
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: sauciest Cipher: hzfxrvhg
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: sauciest Cipher: BAAAB AAAAA BAABB AAABA ABAAA AABAA BAAAB BAABA
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: sauciest
cipher variations:
tbvdjftu
dbjhzndg
nbxlpvns
xblpfdxe
hbztvlhq
rbnxltrc
lbpfrjla
vbdjhrvm
fbrnxzfy
pbfrnhpk
zbtvdpzw
jbhztxji
ucwekguv
eckiaoeh
ocymqwot
ycmqgeyf
icauwmir
scoymusd
mcqgskmb
wcekiswn
gcsoyagz
qcgsoiql
acuweqax
kciauykj
vdxflhvw
fdljbpfi
pdznrxpu
zdnrhfzg
jdbvxnjs
tdpznvte
ndrhtlnc
xdfljtxo
hdtpzbha
rdhtpjrm
bdvxfrby
ldjbvzlk
weygmiwx
gemkcqgj
qeaosyqv
aeosigah
kecwyokt
ueqaowuf
oesiumod
yegmkuyp
ieuqacib
seiuqksn
cewygscz
mekcwaml
xfzhnjxy
hfnldrhk
rfbptzrw
bfptjhbi
lfdxzplu
vfrbpxvg
pftjvnpe
zfhnlvzq
jfvrbdjc
tfjvrlto
dfxzhtda
nfldxbnm
ygaiokyz
igomesil
sgcquasx
cgqukicj
mgeyaqmv
wgscqywh
qgukwoqf
agiomwar
kgwscekd
ugkwsmup
egyaiueb
ogmeycon
zhbjplza
jhpnftjm
thdrvbty
dhrvljdk
nhfzbrnw
xhtdrzxi
rhvlxprg
bhjpnxbs
lhxtdfle
vhlxtnvq
fhzbjvfc
phnfzdpo
aickqmab
kiqogukn
uieswcuz
eiswmkel
oigacsox
yiuesayj
siwmyqsh
cikqoyct
miyuegmf
wimyuowr
giackwgd
qiogaeqp
bjdlrnbc
ljrphvlo
vjftxdva
fjtxnlfm
pjhbdtpy
zjvftbzk
tjxnzrti
djlrpzdu
njzvfhng
xjnzvpxs
hjbdlxhe
rjphbfrq
ckemsocd
mksqiwmp
wkguyewb
gkuyomgn
qkiceuqz
akwgucal
ukyoasuj
ekmsqaev
okawgioh
ykoawqyt
ikcemyif
skqicgsr
dlfntpde
nltrjxnq
xlhvzfxc
hlvzpnho
rljdfvra
blxhvdbm
vlzpbtvk
flntrbfw
plbxhjpi
zlpbxrzu
jldfnzjg
tlrjdhts
emgouqef
omuskyor
ymiwagyd
imwaqoip
smkegwsb
cmyiwecn
wmaqcuwl
gmouscgx
qmcyikqj
amqcysav
kmegoakh
umskeiut
fnhpvrfg
pnvtlzps
znjxbhze
jnxbrpjq
tnlfhxtc
dnzjxfdo
xnbrdvxm
hnpvtdhy
rndzjlrk
bnrdztbw
lnfhpbli
vntlfjvu
goiqwsgh
qowumaqt
aokyciaf
koycsqkr
uomgiyud
eoakygep
yocsewyn
ioqwueiz
soeakmsl
coseaucx
mogiqcmj
woumgkwv
hpjrxthi
rpxvnbru
bplzdjbg
lpzdtrls
vpnhjzve
fpblzhfq
zpdtfxzo
jprxvfja
tpfblntm
dptfbvdy
nphjrdnk
xpvnhlxw
iqksyuij
sqywocsv
cqmaekch
mqaeusmt
wqoikawf
gqcmaigr
aqeugyap
kqsywgkb
uqgcmoun
equgcwez
oqikseol
yqwoimyx
jrltzvjk
trzxpdtw
drnbfldi
nrbfvtnu
xrpjlbxg
hrdnbjhs
brfvhzbq
lrtzxhlc
vrhdnpvo
frvhdxfa
prjltfpm
zrxpjnzy
ksmuawkl
usayqeux
esocgmej
oscgwuov
ysqkmcyh
iseockit
csgwiacr
msuayimd
wsieoqwp
gswieygb
qskmugqn
asyqkoaz
ltnvbxlm
vtbzrfvy
ftpdhnfk
ptdhxvpw
ztrlndzi
jtfpdlju
dthxjbds
ntvbzjne
xtjfprxq
htxjfzhc
rtlnvhro
btzrlpba
muowcymn
wucasgwz
guqeiogl
queiywqx
ausmoeaj
kugqemkv
euiykcet
ouwcakof
yukgqsyr
iuykgaid
sumowisp
cuasmqcb
nvpxdzno
xvdbthxa
hvrfjphm
rvfjzxry
bvtnpfbk
lvhrfnlw
fvjzldfu
pvxdblpg
zvlhrtzs
jvzlhbje
tvnpxjtq
dvbtnrdc
owqyeaop
ywecuiyb
iwsgkqin
swgkaysz
cwuoqgcl
mwisgomx
gwkamegv
qwyecmqh
awmisuat
kwamickf
uwoqykur
ewcuosed
pxrzfbpq
zxfdvjzc
jxthlrjo
txhlbzta
dxvprhdm
nxjthpny
hxlbnfhw
rxzfdnri
bxnjtvbu
lxbnjdlg
vxprzlvs
fxdvptfe
qysagcqr
kyuimskp
uyimcaub
eywqsien
oykuiqoz
iymcogix
syageosj
cyokuwcv
mycokemh
wyqsamwt
gyewqugf
rztbhdrs
bzhfxlbe
lzvjntlq
vzjndbvc
fzxrtjfo
pzlvjrpa
jzndphjy
tzbhfptk
dzplvxdw
nzdplfni
xzrtbnxu
hzfxrvhg
sauciest
caigymcf
mawkoumr
wakoecwd
gaysukgp
qamwksqb
kaoeqikz
uacigqul
eaqmwyex
oaeqmgoj
yasucoyv
iagyswih
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: sauciest Cipher: fnhpvrfg
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: sauciest Cipher: 3411543142513444
Extended Methods:
Method #1
Plaintext: sauciest
method variations: xfzhokxy clentpcd hqksyuhi nvpxdzno
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```s a u c i e s t
3 1 5 3 4 5 3 4
4 1 4 1 2 1 4 4 ```
They are then read out in rows:
3153453441412144
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: sauciest Cipher: cpysddbt
Method #3
Plaintext: sauciest
method variations: dvoqwlto voqwltod oqwltodv qwltodvo wltodvoq ltodvoqw todvoqwl odvoqwlt
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: sauciest
first 5040 cipher variations(40320 total)
sauciest
sauciets
sauciset
sauciste
saucitse
saucites
sauceist
sauceits
saucesit
saucesti
saucetsi
saucetis
saucseit
saucseti
saucsiet
saucsite
saucstie
saucstei
sauctesi
saucteis
sauctsei
sauctsie
sauctise
saucties
sauicest
sauicets
sauicset
sauicste
sauictse
sauictes
sauiecst
sauiects
sauiesct
sauiestc
sauietsc
sauietcs
sauisect
sauisetc
sauiscet
sauiscte
sauistce
sauistec
sauitesc
sauitecs
sauitsec
sauitsce
sauitcse
sauitces
saueicst
saueicts
saueisct
saueistc
saueitsc
saueitcs
sauecist
sauecits
sauecsit
sauecsti
sauectsi
sauectis
sauescit
sauescti
sauesict
sauesitc
sauestic
sauestci
sauetcsi
sauetcis
sauetsci
sauetsic
sauetisc
sauetics
sausiect
sausietc
sausicet
sausicte
sausitce
sausitec
sauseict
sauseitc
sausecit
sausecti
sausetci
sausetic
sausceit
sausceti
sausciet
sauscite
sausctie
sausctei
sausteci
sausteic
saustcei
saustcie
saustice
saustiec
sautiesc
sautiecs
sautisec
sautisce
sauticse
sautices
sauteisc
sauteics
sautesic
sautesci
sautecsi
sautecis
sautseic
sautseci
sautsiec
sautsice
sautscie
sautscei
sautcesi
sautceis
sautcsei
sautcsie
sautcise
sautcies
sacuiest
sacuiets
sacuiset
sacuiste
sacuitse
sacuites
sacueist
sacueits
sacuesit
sacuesti
sacuetsi
sacuetis
sacuseit
sacuseti
sacusiet
sacusite
sacustie
sacustei
sacutesi
sacuteis
sacutsei
sacutsie
sacutise
sacuties
saciuest
saciuets
saciuset
saciuste
saciutse
saciutes
sacieust
sacieuts
saciesut
saciestu
sacietsu
sacietus
saciseut
sacisetu
sacisuet
sacisute
sacistue
sacisteu
sacitesu
saciteus
sacitseu
sacitsue
sacituse
sacitues
saceiust
saceiuts
saceisut
saceistu
saceitsu
saceitus
saceuist
saceuits
saceusit
saceusti
saceutsi
saceutis
sacesuit
sacesuti
sacesiut
sacesitu
sacestiu
sacestui
sacetusi
sacetuis
sacetsui
sacetsiu
sacetisu
sacetius
sacsieut
sacsietu
sacsiuet
sacsiute
sacsitue
sacsiteu
sacseiut
sacseitu
sacseuit
sacseuti
sacsetui
sacsetiu
sacsueit
sacsueti
sacsuiet
sacsuite
sacsutie
sacsutei
sacsteui
sacsteiu
sacstuei
sacstuie
sacstiue
sacstieu
sactiesu
sactieus
sactiseu
sactisue
sactiuse
sactiues
sacteisu
sacteius
sactesiu
sactesui
sacteusi
sacteuis
sactseiu
sactseui
sactsieu
sactsiue
sactsuie
sactsuei
sactuesi
sactueis
sactusei
sactusie
sactuise
sactuies
saicuest
saicuets
saicuset
saicuste
saicutse
saicutes
saiceust
saiceuts
saicesut
saicestu
saicetsu
saicetus
saicseut
saicsetu
saicsuet
saicsute
saicstue
saicsteu
saictesu
saicteus
saictseu
saictsue
saictuse
saictues
saiucest
saiucets
saiucset
saiucste
saiuctse
saiuctes
saiuecst
saiuects
saiuesct
saiuestc
saiuetsc
saiuetcs
saiusect
saiusetc
saiuscet
saiuscte
saiustce
saiustec
saiutesc
saiutecs
saiutsec
saiutsce
saiutcse
saiutces
saieucst
saieucts
saieusct
saieustc
saieutsc
saieutcs
saiecust
saiecuts
saiecsut
saiecstu
saiectsu
saiectus
saiescut
saiesctu
saiesuct
saiesutc
saiestuc
saiestcu
saietcsu
saietcus
saietscu
saietsuc
saietusc
saietucs
saisuect
saisuetc
saisucet
saisucte
saisutce
saisutec
saiseuct
saiseutc
saisecut
saisectu
saisetcu
saisetuc
saisceut
saiscetu
saiscuet
saiscute
saisctue
saiscteu
saistecu
saisteuc
saistceu
saistcue
saistuce
saistuec
saituesc
saituecs
saitusec
saitusce
saitucse
saituces
saiteusc
saiteucs
saitesuc
saitescu
saitecsu
saitecus
saitseuc
saitsecu
saitsuec
saitsuce
saitscue
saitsceu
saitcesu
saitceus
saitcseu
saitcsue
saitcuse
saitcues
saeciust
saeciuts
saecisut
saecistu
saecitsu
saecitus
saecuist
saecuits
saecusit
saecusti
saecutsi
saecutis
saecsuit
saecsuti
saecsiut
saecsitu
saecstiu
saecstui
saectusi
saectuis
saectsui
saectsiu
saectisu
saectius
saeicust
saeicuts
saeicsut
saeicstu
saeictsu
saeictus
saeiucst
saeiucts
saeiusct
saeiustc
saeiutsc
saeiutcs
saeisuct
saeisutc
saeiscut
saeisctu
saeistcu
saeistuc
saeitusc
saeitucs
saeitsuc
saeitscu
saeitcsu
saeitcus
saeuicst
saeuicts
saeuisct
saeuistc
saeuitsc
saeuitcs
saeucist
saeucits
saeucsit
saeucsti
saeuctsi
saeuctis
saeuscit
saeuscti
saeusict
saeusitc
saeustic
saeustci
saeutcsi
saeutcis
saeutsci
saeutsic
saeutisc
saeutics
saesiuct
saesiutc
saesicut
saesictu
saesitcu
saesituc
saesuict
saesuitc
saesucit
saesucti
saesutci
saesutic
saescuit
saescuti
saesciut
saescitu
saesctiu
saesctui
saestuci
saestuic
saestcui
saestciu
saesticu
saestiuc
saetiusc
saetiucs
saetisuc
saetiscu
saeticsu
saeticus
saetuisc
saetuics
saetusic
saetusci
saetucsi
saetucis
saetsuic
saetsuci
saetsiuc
saetsicu
saetsciu
saetscui
saetcusi
saetcuis
saetcsui
saetcsiu
saetcisu
saetcius
sascieut
sascietu
sasciuet
sasciute
sascitue
sasciteu
sasceiut
sasceitu
sasceuit
sasceuti
sascetui
sascetiu
sascueit
sascueti
sascuiet
sascuite
sascutie
sascutei
sascteui
sascteiu
sasctuei
sasctuie
sasctiue
sasctieu
sasiceut
sasicetu
sasicuet
sasicute
sasictue
sasicteu
sasiecut
sasiectu
sasieuct
sasieutc
sasietuc
sasietcu
sasiuect
sasiuetc
sasiucet
sasiucte
sasiutce
sasiutec
sasiteuc
sasitecu
sasituec
sasituce
sasitcue
sasitceu
saseicut
saseictu
saseiuct
saseiutc
saseituc
saseitcu
saseciut
sasecitu
sasecuit
sasecuti
sasectui
sasectiu
saseucit
saseucti
saseuict
saseuitc
saseutic
saseutci
sasetcui
sasetciu
sasetuci
sasetuic
sasetiuc
saseticu
sasuiect
sasuietc
sasuicet
sasuicte
sasuitce
sasuitec
sasueict
sasueitc
sasuecit
sasuecti
sasuetci
sasuetic
sasuceit
sasuceti
sasuciet
sasucite
sasuctie
sasuctei
sasuteci
sasuteic
sasutcei
sasutcie
sasutice
sasutiec
sastieuc
sastiecu
sastiuec
sastiuce
sasticue
sasticeu
sasteiuc
sasteicu
sasteuic
sasteuci
sastecui
sasteciu
sastueic
sastueci
sastuiec
sastuice
sastucie
sastucei
sastceui
sastceiu
sastcuei
sastcuie
sastciue
sastcieu
satciesu
satcieus
satciseu
satcisue
satciuse
satciues
satceisu
satceius
satcesiu
satcesui
satceusi
satceuis
satcseiu
satcseui
satcsieu
satcsiue
satcsuie
satcsuei
satcuesi
satcueis
satcusei
satcusie
satcuise
satcuies
saticesu
saticeus
saticseu
saticsue
saticuse
saticues
satiecsu
satiecus
satiescu
satiesuc
satieusc
satieucs
satisecu
satiseuc
satisceu
satiscue
satisuce
satisuec
satiuesc
satiuecs
satiusec
satiusce
satiucse
satiuces
sateicsu
sateicus
sateiscu
sateisuc
sateiusc
sateiucs
satecisu
satecius
satecsiu
satecsui
satecusi
satecuis
satesciu
satescui
satesicu
satesiuc
satesuic
satesuci
sateucsi
sateucis
sateusci
sateusic
sateuisc
sateuics
satsiecu
satsieuc
satsiceu
satsicue
satsiuce
satsiuec
satseicu
satseiuc
satseciu
satsecui
satseuci
satseuic
satsceiu
satsceui
satscieu
satsciue
satscuie
satscuei
satsueci
satsueic
satsucei
satsucie
satsuice
satsuiec
satuiesc
satuiecs
satuisec
satuisce
satuicse
satuices
satueisc
satueics
satuesic
satuesci
satuecsi
satuecis
satuseic
satuseci
satusiec
satusice
satuscie
satuscei
satucesi
satuceis
satucsei
satucsie
satucise
satucies
suaciest
suaciets
suaciset
suaciste
suacitse
suacites
suaceist
suaceits
suacesit
suacesti
suacetsi
suacetis
suacseit
suacseti
suacsiet
suacsite
suacstie
suacstei
suactesi
suacteis
suactsei
suactsie
suactise
suacties
suaicest
suaicets
suaicset
suaicste
suaictse
suaictes
suaiecst
suaiects
suaiesct
suaiestc
suaietsc
suaietcs
suaisect
suaisetc
suaiscet
suaiscte
suaistce
suaistec
suaitesc
suaitecs
suaitsec
suaitsce
suaitcse
suaitces
suaeicst
suaeicts
suaeisct
suaeistc
suaeitsc
suaeitcs
suaecist
suaecits
suaecsit
suaecsti
suaectsi
suaectis
suaescit
suaescti
suaesict
suaesitc
suaestic
suaestci
suaetcsi
suaetcis
suaetsci
suaetsic
suaetisc
suaetics
suasiect
suasietc
suasicet
suasicte
suasitce
suasitec
suaseict
suaseitc
suasecit
suasecti
suasetci
suasetic
suasceit
suasceti
suasciet
suascite
suasctie
suasctei
suasteci
suasteic
suastcei
suastcie
suastice
suastiec
suatiesc
suatiecs
suatisec
suatisce
suaticse
suatices
suateisc
suateics
suatesic
suatesci
suatecsi
suatecis
suatseic
suatseci
suatsiec
suatsice
suatscie
suatscei
suatcesi
suatceis
suatcsei
suatcsie
suatcise
suatcies
sucaiest
sucaiets
sucaiset
sucaiste
sucaitse
sucaites
sucaeist
sucaeits
sucaesit
sucaesti
sucaetsi
sucaetis
sucaseit
sucaseti
sucasiet
sucasite
sucastie
sucastei
sucatesi
sucateis
sucatsei
sucatsie
sucatise
sucaties
suciaest
suciaets
suciaset
suciaste
suciatse
suciates
sucieast
sucieats
suciesat
suciesta
sucietsa
sucietas
suciseat
suciseta
sucisaet
sucisate
sucistae
sucistea
sucitesa
suciteas
sucitsea
sucitsae
sucitase
sucitaes
suceiast
suceiats
suceisat
suceista
suceitsa
suceitas
suceaist
suceaits
suceasit
suceasti
suceatsi
suceatis
sucesait
sucesati
sucesiat
sucesita
sucestia
sucestai
sucetasi
sucetais
sucetsai
sucetsia
sucetisa
sucetias
sucsieat
sucsieta
sucsiaet
sucsiate
sucsitae
sucsitea
sucseiat
sucseita
sucseait
sucseati
sucsetai
sucsetia
sucsaeit
sucsaeti
sucsaiet
sucsaite
sucsatie
sucsatei
sucsteai
sucsteia
sucstaei
sucstaie
sucstiae
sucstiea
suctiesa
suctieas
suctisea
suctisae
suctiase
suctiaes
sucteisa
sucteias
suctesia
suctesai
sucteasi
sucteais
suctseia
suctseai
suctsiea
suctsiae
suctsaie
suctsaei
suctaesi
suctaeis
suctasei
suctasie
suctaise
suctaies
suicaest
suicaets
suicaset
suicaste
suicatse
suicates
suiceast
suiceats
suicesat
suicesta
suicetsa
suicetas
suicseat
suicseta
suicsaet
suicsate
suicstae
suicstea
suictesa
suicteas
suictsea
suictsae
suictase
suictaes
suiacest
suiacets
suiacset
suiacste
suiactse
suiactes
suiaecst
suiaects
suiaesct
suiaestc
suiaetsc
suiaetcs
suiasect
suiasetc
suiascet
suiascte
suiastce
suiastec
suiatesc
suiatecs
suiatsec
suiatsce
suiatcse
suiatces
suieacst
suieacts
suieasct
suieastc
suieatsc
suieatcs
suiecast
suiecats
suiecsat
suiecsta
suiectsa
suiectas
suiescat
suiescta
suiesact
suiesatc
suiestac
suiestca
suietcsa
suietcas
suietsca
suietsac
suietasc
suietacs
suisaect
suisaetc
suisacet
suisacte
suisatce
suisatec
suiseact
suiseatc
suisecat
suisecta
suisetca
suisetac
suisceat
suisceta
suiscaet
suiscate
suisctae
suisctea
suisteca
suisteac
suistcea
suistcae
suistace
suistaec
suitaesc
suitaecs
suitasec
suitasce
suitacse
suitaces
suiteasc
suiteacs
suitesac
suitesca
suitecsa
suitecas
suitseac
suitseca
suitsaec
suitsace
suitscae
suitscea
suitcesa
suitceas
suitcsea
suitcsae
suitcase
suitcaes
sueciast
sueciats
suecisat
suecista
suecitsa
suecitas
suecaist
suecaits
suecasit
suecasti
suecatsi
suecatis
suecsait
suecsati
suecsiat
suecsita
suecstia
suecstai
suectasi
suectais
suectsai
suectsia
suectisa
suectias
sueicast
sueicats
sueicsat
sueicsta
sueictsa
sueictas
sueiacst
sueiacts
sueiasct
sueiastc
sueiatsc
sueiatcs
sueisact
sueisatc
sueiscat
sueiscta
sueistca
sueistac
sueitasc
sueitacs
sueitsac
sueitsca
sueitcsa
sueitcas
sueaicst
sueaicts
sueaisct
sueaistc
sueaitsc
sueaitcs
sueacist
sueacits
sueacsit
sueacsti
sueactsi
sueactis
sueascit
sueascti
sueasict
sueasitc
sueastic
sueastci
sueatcsi
sueatcis
sueatsci
sueatsic
sueatisc
sueatics
suesiact
suesiatc
suesicat
suesicta
suesitca
suesitac
suesaict
suesaitc
suesacit
suesacti
suesatci
suesatic
suescait
suescati
suesciat
suescita
suesctia
suesctai
suestaci
suestaic
suestcai
suestcia
suestica
suestiac
suetiasc
suetiacs
suetisac
suetisca
sueticsa
sueticas
suetaisc
suetaics
suetasic
suetasci
suetacsi
suetacis
suetsaic
suetsaci
suetsiac
suetsica
suetscia
suetscai
suetcasi
suetcais
suetcsai
suetcsia
suetcisa
suetcias
suscieat
suscieta
susciaet
susciate
suscitae
suscitea
susceiat
susceita
susceait
susceati
suscetai
suscetia
suscaeit
suscaeti
suscaiet
suscaite
suscatie
suscatei
suscteai
suscteia
susctaei
susctaie
susctiae
susctiea
susiceat
susiceta
susicaet
susicate
susictae
susictea
susiecat
susiecta
susieact
susieatc
susietac
susietca
susiaect
susiaetc
susiacet
susiacte
susiatce
susiatec
susiteac
susiteca
susitaec
susitace
susitcae
susitcea
suseicat
suseicta
suseiact
suseiatc
suseitac
suseitca
suseciat
susecita
susecait
susecati
susectai
susectia
suseacit
suseacti
suseaict
suseaitc
suseatic
suseatci
susetcai
susetcia
susetaci
susetaic
susetiac
susetica
susaiect
susaietc
susaicet
susaicte
susaitce
susaitec
susaeict
susaeitc
susaecit
susaecti
susaetci
susaetic
susaceit
susaceti
susaciet
susacite
susactie
susactei
susateci
susateic
susatcei
susatcie
susatice
susatiec
sustieac
sustieca
sustiaec
sustiace
susticae
susticea
susteiac
susteica
susteaic
susteaci
sustecai
sustecia
sustaeic
sustaeci
sustaiec
sustaice
sustacie
sustacei
sustceai
sustceia
sustcaei
sustcaie
sustciae
sustciea
sutciesa
sutcieas
sutcisea
sutcisae
sutciase
sutciaes
sutceisa
sutceias
sutcesia
sutcesai
sutceasi
sutceais
sutcseia
sutcseai
sutcsiea
sutcsiae
sutcsaie
sutcsaei
sutcaesi
sutcaeis
sutcasei
sutcasie
sutcaise
sutcaies
suticesa
suticeas
suticsea
suticsae
suticase
suticaes
sutiecsa
sutiecas
sutiesca
sutiesac
sutieasc
sutieacs
sutiseca
sutiseac
sutiscea
sutiscae
sutisace
sutisaec
sutiaesc
sutiaecs
sutiasec
sutiasce
sutiacse
sutiaces
suteicsa
suteicas
suteisca
suteisac
suteiasc
suteiacs
sutecisa
sutecias
sutecsia
sutecsai
sutecasi
sutecais
sutescia
sutescai
sutesica
sutesiac
sutesaic
sutesaci
suteacsi
suteacis
suteasci
suteasic
suteaisc
suteaics
sutsieca
sutsieac
sutsicea
sutsicae
sutsiace
sutsiaec
sutseica
sutseiac
sutsecia
sutsecai
sutseaci
sutseaic
sutsceia
sutsceai
sutsciea
sutsciae
sutscaie
sutscaei
sutsaeci
sutsaeic
sutsacei
sutsacie
sutsaice
sutsaiec
sutaiesc
sutaiecs
sutaisec
sutaisce
sutaicse
sutaices
sutaeisc
sutaeics
sutaesic
sutaesci
sutaecsi
sutaecis
sutaseic
sutaseci
sutasiec
sutasice
sutascie
sutascei
sutacesi
sutaceis
sutacsei
sutacsie
sutacise
sutacies
scuaiest
scuaiets
scuaiset
scuaiste
scuaitse
scuaites
scuaeist
scuaeits
scuaesit
scuaesti
scuaetsi
scuaetis
scuaseit
scuaseti
scuasiet
scuasite
scuastie
scuastei
scuatesi
scuateis
scuatsei
scuatsie
scuatise
scuaties
scuiaest
scuiaets
scuiaset
scuiaste
scuiatse
scuiates
scuieast
scuieats
scuiesat
scuiesta
scuietsa
scuietas
scuiseat
scuiseta
scuisaet
scuisate
scuistae
scuistea
scuitesa
scuiteas
scuitsea
scuitsae
scuitase
scuitaes
scueiast
scueiats
scueisat
scueista
scueitsa
scueitas
scueaist
scueaits
scueasit
scueasti
scueatsi
scueatis
scuesait
scuesati
scuesiat
scuesita
scuestia
scuestai
scuetasi
scuetais
scuetsai
scuetsia
scuetisa
scuetias
scusieat
scusieta
scusiaet
scusiate
scusitae
scusitea
scuseiat
scuseita
scuseait
scuseati
scusetai
scusetia
scusaeit
scusaeti
scusaiet
scusaite
scusatie
scusatei
scusteai
scusteia
scustaei
scustaie
scustiae
scustiea
scutiesa
scutieas
scutisea
scutisae
scutiase
scutiaes
scuteisa
scuteias
scutesia
scutesai
scuteasi
scuteais
scutseia
scutseai
scutsiea
scutsiae
scutsaie
scutsaei
scutaesi
scutaeis
scutasei
scutasie
scutaise
scutaies
scauiest
scauiets
scauiset
scauiste
scauitse
scauites
scaueist
scaueits
scauesit
scauesti
scauetsi
scauetis
scauseit
scauseti
scausiet
scausite
scaustie
scaustei
scautesi
scauteis
scautsei
scautsie
scautise
scauties
scaiuest
scaiuets
scaiuset
scaiuste
scaiutse
scaiutes
scaieust
scaieuts
scaiesut
scaiestu
scaietsu
scaietus
scaiseut
scaisetu
scaisuet
scaisute
scaistue
scaisteu
scaitesu
scaiteus
scaitseu
scaitsue
scaituse
scaitues
scaeiust
scaeiuts
scaeisut
scaeistu
scaeitsu
scaeitus
scaeuist
scaeuits
scaeusit
scaeusti
scaeutsi
scaeutis
scaesuit
scaesuti
scaesiut
scaesitu
scaestiu
scaestui
scaetusi
scaetuis
scaetsui
scaetsiu
scaetisu
scaetius
scasieut
scasietu
scasiuet
scasiute
scasitue
scasiteu
scaseiut
scaseitu
scaseuit
scaseuti
scasetui
scasetiu
scasueit
scasueti
scasuiet
scasuite
scasutie
scasutei
scasteui
scasteiu
scastuei
scastuie
scastiue
scastieu
scatiesu
scatieus
scatiseu
scatisue
scatiuse
scatiues
scateisu
scateius
scatesiu
scatesui
scateusi
scateuis
scatseiu
scatseui
scatsieu
scatsiue
scatsuie
scatsuei
scatuesi
scatueis
scatusei
scatusie
scatuise
scatuies
sciauest
sciauets
sciauset
sciauste
sciautse
sciautes
sciaeust
sciaeuts
sciaesut
sciaestu
sciaetsu
sciaetus
sciaseut
sciasetu
sciasuet
sciasute
sciastue
sciasteu
sciatesu
sciateus
sciatseu
sciatsue
sciatuse
sciatues
sciuaest
sciuaets
sciuaset
sciuaste
sciuatse
sciuates
sciueast
sciueats
sciuesat
sciuesta
sciuetsa
sciuetas
sciuseat
sciuseta
sciusaet
sciusate
sciustae
sciustea
sciutesa
sciuteas
sciutsea
sciutsae
sciutase
sciutaes
scieuast
scieuats
scieusat
scieusta
scieutsa
scieutas
scieaust
scieauts
scieasut
scieastu
scieatsu
scieatus
sciesaut
sciesatu
sciesuat
sciesuta
sciestua
sciestau
scietasu
scietaus
scietsau
scietsua
scietusa
scietuas
scisueat
scisueta
scisuaet
scisuate
scisutae
scisutea
sciseuat
sciseuta
sciseaut
sciseatu
scisetau
scisetua
scisaeut
scisaetu
scisauet
scisaute
scisatue
scisateu
scisteau
scisteua
scistaeu
scistaue
scistuae
scistuea
scituesa
scitueas
scitusea
scitusae
scituase
scituaes
sciteusa
sciteuas
scitesua
scitesau
sciteasu
sciteaus
scitseua
scitseau
scitsuea
scitsuae
scitsaue
scitsaeu
scitaesu
scitaeus
scitaseu
scitasue
scitause
scitaues
sceaiust
sceaiuts
sceaisut
sceaistu
sceaitsu
sceaitus
sceauist
sceauits
sceausit
sceausti
sceautsi
sceautis
sceasuit
sceasuti
sceasiut
sceasitu
sceastiu
sceastui
sceatusi
sceatuis
sceatsui
sceatsiu
sceatisu
sceatius
sceiaust
sceiauts
sceiasut
sceiastu
sceiatsu
sceiatus
sceiuast
sceiuats
sceiusat
sceiusta
sceiutsa
sceiutas
sceisuat
sceisuta
sceisaut
sceisatu
sceistau
sceistua
sceitusa
sceituas
sceitsua
sceitsau
sceitasu
sceitaus
sceuiast
sceuiats
sceuisat
sceuista
sceuitsa
sceuitas
sceuaist
sceuaits
sceuasit
sceuasti
sceuatsi
sceuatis
sceusait
sceusati
sceusiat
sceusita
sceustia
sceustai
sceutasi
sceutais
sceutsai
sceutsia
sceutisa
sceutias
scesiuat
scesiuta
scesiaut
scesiatu
scesitau
scesitua
scesuiat
scesuita
scesuait
scesuati
scesutai
scesutia
scesauit
scesauti
scesaiut
scesaitu
scesatiu
scesatui
scestuai
scestuia
scestaui
scestaiu
scestiau
scestiua
scetiusa
scetiuas
scetisua
scetisau
scetiasu
scetiaus
scetuisa
scetuias
scetusia
scetusai
scetuasi
scetuais
scetsuia
scetsuai
scetsiua
scetsiau
scetsaiu
scetsaui
scetausi
scetauis
scetasui
scetasiu
scetaisu
scetaius
scsaieut
scsaietu
scsaiuet
scsaiute
scsaitue
scsaiteu
scsaeiut
scsaeitu
scsaeuit
scsaeuti
scsaetui
scsaetiu
scsaueit
scsaueti
scsauiet
scsauite
scsautie
scsautei
scsateui
scsateiu
scsatuei
scsatuie
scsatiue
scsatieu
scsiaeut
scsiaetu
scsiauet
scsiaute
scsiatue
scsiateu
scsieaut
scsieatu
scsieuat
scsieuta
scsietua
scsietau
scsiueat
scsiueta
scsiuaet
scsiuate
scsiutae
scsiutea
scsiteua
scsiteau
scsituea
scsituae
scsitaue
scsitaeu
scseiaut
scseiatu
scseiuat
scseiuta
scseitua
scseitau
scseaiut
scseaitu
scseauit
scseauti
scseatui
scseatiu
scseuait
scseuati
scseuiat
scseuita
scseutia
scseutai
scsetaui
scsetaiu
scsetuai
scsetuia
scsetiua
scsetiau
scsuieat
scsuieta
scsuiaet
scsuiate
scsuitae
scsuitea
scsueiat
scsueita
scsueait
scsueati
scsuetai
scsuetia
scsuaeit
scsuaeti
scsuaiet
scsuaite
scsuatie
scsuatei
scsuteai
scsuteia
scsutaei
scsutaie
scsutiae
scsutiea
scstieua
scstieau
scstiuea
scstiuae
scstiaue
scstiaeu
scsteiua
scsteiau
scsteuia
scsteuai
scsteaui
scsteaiu
scstueia
scstueai
scstuiea
scstuiae
scstuaie
scstuaei
scstaeui
scstaeiu
scstauei
scstauie
scstaiue
scstaieu
sctaiesu
sctaieus
sctaiseu
sctaisue
sctaiuse
sctaiues
sctaeisu
sctaeius
sctaesiu
sctaesui
sctaeusi
sctaeuis
sctaseiu
sctaseui
sctasieu
sctasiue
sctasuie
sctasuei
sctauesi
sctaueis
sctausei
sctausie
sctauise
sctauies
sctiaesu
sctiaeus
sctiaseu
sctiasue
sctiause
sctiaues
sctieasu
sctieaus
sctiesau
sctiesua
sctieusa
sctieuas
sctiseau
sctiseua
sctisaeu
sctisaue
sctisuae
sctisuea
sctiuesa
sctiueas
sctiusea
sctiusae
sctiuase
sctiuaes
scteiasu
scteiaus
scteisau
scteisua
scteiusa
scteiuas
scteaisu
scteaius
scteasiu
scteasui
scteausi
scteauis
sctesaiu
sctesaui
sctesiau
sctesiua
sctesuia
sctesuai
scteuasi
scteuais
scteusai
scteusia
scteuisa
scteuias
sctsieau
sctsieua
sctsiaeu
sctsiaue
sctsiuae
sctsiuea
sctseiau
sctseiua
sctseaiu
sctseaui
sctseuai
sctseuia
sctsaeiu
sctsaeui
sctsaieu
sctsaiue
sctsauie
sctsauei
sctsueai
sctsueia
sctsuaei
sctsuaie
sctsuiae
sctsuiea
sctuiesa
sctuieas
sctuisea
sctuisae
sctuiase
sctuiaes
sctueisa
sctueias
sctuesia
sctuesai
sctueasi
sctueais
sctuseia
sctuseai
sctusiea
sctusiae
sctusaie
sctusaei
sctuaesi
sctuaeis
sctuasei
sctuasie
sctuaise
sctuaies
siucaest
siucaets
siucaset
siucaste
siucatse
siucates
siuceast
siuceats
siucesat
siucesta
siucetsa
siucetas
siucseat
siucseta
siucsaet
siucsate
siucstae
siucstea
siuctesa
siucteas
siuctsea
siuctsae
siuctase
siuctaes
siuacest
siuacets
siuacset
siuacste
siuactse
siuactes
siuaecst
siuaects
siuaesct
siuaestc
siuaetsc
siuaetcs
siuasect
siuasetc
siuascet
siuascte
siuastce
siuastec
siuatesc
siuatecs
siuatsec
siuatsce
siuatcse
siuatces
siueacst
siueacts
siueasct
siueastc
siueatsc
siueatcs
siuecast
siuecats
siuecsat
siuecsta
siuectsa
siuectas
siuescat
siuescta
siuesact
siuesatc
siuestac
siuestca
siuetcsa
siuetcas
siuetsca
siuetsac
siuetasc
siuetacs
siusaect
siusaetc
siusacet
siusacte
siusatce
siusatec
siuseact
siuseatc
siusecat
siusecta
siusetca
siusetac
siusceat
siusceta
siuscaet
siuscate
siusctae
siusctea
siusteca
siusteac
siustcea
siustcae
siustace
siustaec
siutaesc
siutaecs
siutasec
siutasce
siutacse
siutaces
siuteasc
siuteacs
siutesac
siutesca
siutecsa
siutecas
siutseac
siutseca
siutsaec
siutsace
siutscae
siutscea
siutcesa
siutceas
siutcsea
siutcsae
siutcase
siutcaes
sicuaest
sicuaets
sicuaset
sicuaste
sicuatse
sicuates
sicueast
sicueats
sicuesat
sicuesta
sicuetsa
sicuetas
sicuseat
sicuseta
sicusaet
sicusate
sicustae
sicustea
sicutesa
sicuteas
sicutsea
sicutsae
sicutase
sicutaes
sicauest
sicauets
sicauset
sicauste
sicautse
sicautes
sicaeust
sicaeuts
sicaesut
sicaestu
sicaetsu
sicaetus
sicaseut
sicasetu
sicasuet
sicasute
sicastue
sicasteu
sicatesu
sicateus
sicatseu
sicatsue
sicatuse
sicatues
siceaust
siceauts
siceasut
siceastu
siceatsu
siceatus
siceuast
siceuats
siceusat
siceusta
siceutsa
siceutas
sicesuat
sicesuta
sicesaut
sicesatu
sicestau
sicestua
sicetusa
sicetuas
sicetsua
sicetsau
sicetasu
sicetaus
sicsaeut
sicsaetu
sicsauet
sicsaute
sicsatue
sicsateu
sicseaut
sicseatu
sicseuat
sicseuta
sicsetua
sicsetau
sicsueat
sicsueta
sicsuaet
sicsuate
sicsutae
sicsutea
sicsteua
sicsteau
sicstuea
sicstuae
sicstaue
sicstaeu
sictaesu
sictaeus
sictaseu
sictasue
sictause
sictaues
sicteasu
sicteaus
sictesau
sictesua
sicteusa
sicteuas
sictseau
sictseua
sictsaeu
sictsaue
sictsuae
sictsuea
sictuesa
sictueas
sictusea
sictusae
sictuase
sictuaes
siacuest
siacuets
siacuset
siacuste
siacutse
siacutes
siaceust
siaceuts
siacesut
siacestu
siacetsu
siacetus
siacseut
siacsetu
siacsuet
siacsute
siacstue
siacsteu
siactesu
siacteus
siactseu
siactsue
siactuse
siactues
siaucest
siaucets
siaucset
siaucste
siauctse
siauctes
siauecst
siauects
siauesct
siauestc
siauetsc
siauetcs
siausect
siausetc
siauscet
siauscte
siaustce
siaustec
siautesc
siautecs
siautsec
siautsce
siautcse
siautces
siaeucst
siaeucts
siaeusct
siaeustc
siaeutsc
siaeutcs
siaecust
siaecuts
siaecsut
siaecstu
siaectsu
siaectus
siaescut
siaesctu
siaesuct
siaesutc
siaestuc
siaestcu
siaetcsu
siaetcus
siaetscu
siaetsuc
siaetusc
siaetucs
siasuect
siasuetc
siasucet
siasucte
siasutce
siasutec
siaseuct
siaseutc
siasecut
siasectu
siasetcu
siasetuc
siasceut
siascetu
siascuet
siascute
siasctue
siascteu
siastecu
siasteuc
siastceu
siastcue
siastuce
siastuec
siatuesc
siatuecs
siatusec
siatusce
siatucse
siatuces
siateusc
siateucs
siatesuc
siatescu
siatecsu
siatecus
siatseuc
siatsecu
siatsuec
siatsuce
siatscue
siatsceu
siatcesu
siatceus
siatcseu
siatcsue
siatcuse
siatcues
siecaust
siecauts
siecasut
siecastu
siecatsu
siecatus
siecuast
siecuats
siecusat
siecusta
siecutsa
siecutas
siecsuat
siecsuta
siecsaut
siecsatu
siecstau
siecstua
siectusa
siectuas
siectsua
siectsau
siectasu
siectaus
sieacust
sieacuts
sieacsut
sieacstu
sieactsu
sieactus
sieaucst
sieaucts
sieausct
sieaustc
sieautsc
sieautcs
sieasuct
sieasutc
sieascut
sieasctu
sieastcu
sieastuc
sieatusc
sieatucs
sieatsuc
sieatscu
sieatcsu
sieatcus
sieuacst
sieuacts
sieuasct
sieuastc
sieuatsc
sieuatcs
sieucast
sieucats
sieucsat
sieucsta
sieuctsa
sieuctas
sieuscat
sieuscta
sieusact
sieusatc
sieustac
sieustca
sieutcsa
sieutcas
sieutsca
sieutsac
sieutasc
sieutacs
siesauct
siesautc
siesacut
siesactu
siesatcu
siesatuc
siesuact
siesuatc
siesucat
siesucta
siesutca
siesutac
siescuat
siescuta
siescaut
siescatu
siesctau
siesctua
siestuca
siestuac
siestcua
siestcau
siestacu
siestauc
sietausc
sietaucs
sietasuc
sietascu
sietacsu
sietacus
sietuasc
sietuacs
sietusac
sietusca
sietucsa
sietucas
sietsuac
sietsuca
sietsauc
sietsacu
sietscau
sietscua
sietcusa
sietcuas
sietcsua
sietcsau
sietcasu
sietcaus
siscaeut
siscaetu
siscauet
siscaute
siscatue
siscateu
sisceaut
sisceatu
sisceuat
sisceuta
siscetua
siscetau
siscueat
siscueta
siscuaet
siscuate
siscutae
siscutea
siscteua
siscteau
sisctuea
sisctuae
sisctaue
sisctaeu
sisaceut
sisacetu
sisacuet
sisacute
sisactue
sisacteu
sisaecut
sisaectu
sisaeuct
sisaeutc
sisaetuc
sisaetcu
sisauect
sisauetc
sisaucet
sisaucte
sisautce
sisautec
sisateuc
sisatecu
sisatuec
sisatuce
sisatcue
sisatceu
siseacut
siseactu
siseauct
siseautc
siseatuc
siseatcu
sisecaut
sisecatu
sisecuat
sisecuta
sisectua
sisectau
siseucat
siseucta
siseuact
siseuatc
siseutac
siseutca
sisetcua
sisetcau
sisetuca
sisetuac
sisetauc
sisetacu
sisuaect
sisuaetc
sisuacet
sisuacte
sisuatce
sisuatec
sisueact
sisueatc
sisuecat
sisuecta
sisuetca
sisuetac
sisuceat
sisuceta
sisucaet
sisucate
sisuctae
sisuctea
sisuteca
sisuteac
sisutcea
sisutcae
sisutace
sisutaec
sistaeuc
sistaecu
sistauec
sistauce
sistacue
sistaceu
sisteauc
sisteacu
sisteuac
sisteuca
sistecua
sistecau
sistueac
sistueca
sistuaec
sistuace
sistucae
sistucea
sistceua
sistceau
sistcuea
sistcuae
sistcaue
sistcaeu
sitcaesu
sitcaeus
sitcaseu
sitcasue
sitcause
sitcaues
sitceasu
sitceaus
sitcesau
sitcesua
sitceusa
sitceuas
sitcseau
sitcseua
sitcsaeu
sitcsaue
sitcsuae
sitcsuea
sitcuesa
sitcueas
sitcusea
sitcusae
sitcuase
sitcuaes
sitacesu
sitaceus
sitacseu
sitacsue
sitacuse
sitacues
sitaecsu
sitaecus
sitaescu
sitaesuc
sitaeusc
sitaeucs
sitasecu
sitaseuc
sitasceu
sitascue
sitasuce
sitasuec
sitauesc
sitauecs
sitausec
sitausce
sitaucse
sitauces
siteacsu
siteacus
siteascu
siteasuc
siteausc
siteaucs
sitecasu
sitecaus
sitecsau
sitecsua
sitecusa
sitecuas
sitescau
sitescua
sitesacu
sitesauc
sitesuac
sitesuca
siteucsa
siteucas
siteusca
siteusac
siteuasc
siteuacs
sitsaecu
sitsaeuc
sitsaceu
sitsacue
sitsauce
sitsauec
sitseacu
sitseauc
sitsecau
sitsecua
sitseuca
sitseuac
sitsceau
sitsceua
sitscaeu
sitscaue
sitscuae
sitscuea
sitsueca
sitsueac
sitsucea
sitsucae
sitsuace
sitsuaec
situaesc
situaecs
situasec
situasce
situacse
situaces
situeasc
situeacs
situesac
situesca
situecsa
situecas
situseac
situseca
situsaec
situsace
situscae
situscea
situcesa
situceas
situcsea
situcsae
situcase
situcaes
seuciast
seuciats
seucisat
seucista
seucitsa
seucitas
seucaist
seucaits
seucasit
seucasti
seucatsi
seucatis
seucsait
seucsati
seucsiat
seucsita
seucstia
seucstai
seuctasi
seuctais
seuctsai
seuctsia
seuctisa
seuctias
seuicast
seuicats
seuicsat
seuicsta
seuictsa
seuictas
seuiacst
seuiacts
seuiasct
seuiastc
seuiatsc
seuiatcs
seuisact
seuisatc
seuiscat
seuiscta
seuistca
seuistac
seuitasc
seuitacs
seuitsac
seuitsca
seuitcsa
seuitcas
seuaicst
seuaicts
seuaisct
seuaistc
seuaitsc
seuaitcs
seuacist
seuacits
seuacsit
seuacsti
seuactsi
seuactis
seuascit
seuascti
seuasict
seuasitc
seuastic
seuastci
seuatcsi
seuatcis
seuatsci
seuatsic
seuatisc
seuatics
seusiact
seusiatc
seusicat
seusicta
seusitca
seusitac
seusaict
seusaitc
seusacit
seusacti
seusatci
seusatic
seuscait
seuscati
seusciat
seuscita
seusctia
seusctai
seustaci
seustaic
seustcai
seustcia
seustica
seustiac
seutiasc
seutiacs
seutisac
seutisca
seuticsa
seuticas
seutaisc
seutaics
seutasic
seutasci
seutacsi
seutacis
seutsaic
seutsaci
seutsiac
seutsica
seutscia
seutscai
seutcasi
seutcais
seutcsai
seutcsia
seutcisa
seutcias
secuiast
secuiats
secuisat
secuista
secuitsa
secuitas
secuaist
secuaits
secuasit
secuasti
secuatsi
secuatis
secusait
secusati
secusiat
secusita
secustia
secustai
secutasi
secutais
secutsai
secutsia
secutisa
secutias
seciuast
seciuats
seciusat
seciusta
seciutsa
seciutas
seciaust
seciauts
seciasut
seciastu
seciatsu
seciatus
secisaut
secisatu
secisuat
secisuta
secistua
secistau
secitasu
secitaus
secitsau
secitsua
secitusa
secituas
secaiust
secaiuts
secaisut
secaistu
secaitsu
secaitus
secauist
secauits
secausit
secausti
secautsi
secautis
secasuit
secasuti
secasiut
secasitu
secastiu
secastui
secatusi
secatuis
secatsui
secatsiu
secatisu
secatius
secsiaut
secsiatu
secsiuat
secsiuta
secsitua
secsitau
secsaiut
secsaitu
secsauit
secsauti
secsatui
secsatiu
secsuait
secsuati
secsuiat
secsuita
secsutia
secsutai
secstaui
secstaiu
secstuai
secstuia
secstiua
secstiau
sectiasu
sectiaus
sectisau
sectisua
sectiusa
sectiuas
sectaisu
sectaius
sectasiu
sectasui
sectausi
sectauis
sectsaiu
sectsaui
sectsiau
sectsiua
sectsuia
sectsuai
sectuasi
sectuais
sectusai
sectusia
sectuisa
sectuias
seicuast
seicuats
seicusat
seicusta
seicutsa
seicutas
seicaust
seicauts
seicasut
seicastu
seicatsu
seicatus
seicsaut
seicsatu
seicsuat
seicsuta
seicstua
seicstau
seictasu
seictaus
seictsau
seictsua
seictusa
seictuas
seiucast
seiucats
seiucsat
seiucsta
seiuctsa
seiuctas
seiuacst
seiuacts
seiuasct
seiuastc
seiuatsc
seiuatcs
seiusact
seiusatc
seiuscat
seiuscta
seiustca
seiustac
seiutasc
seiutacs
seiutsac
seiutsca
seiutcsa
seiutcas
seiaucst
seiaucts
seiausct
seiaustc
seiautsc
seiautcs
seiacust
seiacuts
seiacsut
seiacstu
seiactsu
seiactus
seiascut
seiasctu
seiasuct
seiasutc
seiastuc
seiastcu
seiatcsu
seiatcus
seiatscu
seiatsuc
seiatusc
seiatucs
seisuact
seisuatc
seisucat
seisucta
seisutca
seisutac
seisauct
seisautc
seisacut
seisactu
seisatcu
seisatuc
seiscaut
seiscatu
seiscuat
seiscuta
seisctua
seisctau
seistacu
seistauc
seistcau
seistcua
seistuca
seistuac
seituasc
seituacs
seitusac
seitusca
seitucsa
seitucas
seitausc
seitaucs
seitasuc
seitascu
seitacsu
seitacus
seitsauc
seitsacu
seitsuac
seitsuca
seitscua
seitscau
seitcasu
seitcaus
seitcsau
seitcsua
seitcusa
seitcuas
seaciust
seaciuts
seacisut
seacistu
seacitsu
seacitus
seacuist
seacuits
seacusit
seacusti
seacutsi
seacutis
seacsuit
seacsuti
seacsiut
seacsitu
seacstiu
seacstui
seactusi
seactuis
seactsui
seactsiu
seactisu
seactius
seaicust
seaicuts
seaicsut
seaicstu
seaictsu
seaictus
seaiucst
seaiucts
seaiusct
seaiustc
seaiutsc
seaiutcs
seaisuct
seaisutc
seaiscut
seaisctu
seaistcu
seaistuc
seaitusc
seaitucs
seaitsuc
seaitscu
seaitcsu
seaitcus
seauicst
seauicts
seauisct
seauistc
seauitsc
seauitcs
seaucist
seaucits
seaucsit
seaucsti
seauctsi
seauctis
seauscit
seauscti
seausict
seausitc
seaustic
seaustci
seautcsi
seautcis
seautsci
seautsic
seautisc
seautics
seasiuct
seasiutc
seasicut
seasictu
seasitcu
seasituc
seasuict
seasuitc
seasucit
seasucti
seasutci
seasutic
seascuit
seascuti
seasciut
seascitu
seasctiu
seasctui
seastuci
seastuic
seastcui
seastciu
seasticu
seastiuc
seatiusc
seatiucs
seatisuc
seatiscu
seaticsu
seaticus
seatuisc
seatuics
seatusic
seatusci
seatucsi
seatucis
seatsuic
seatsuci
seatsiuc
seatsicu
seatsciu
seatscui
seatcusi
seatcuis
seatcsui
seatcsiu
seatcisu
seatcius
sesciaut
sesciatu
sesciuat
sesciuta
sescitua
sescitau
sescaiut
sescaitu
sescauit
sescauti
sescatui
sescatiu
sescuait
sescuati
sescuiat
sescuita
sescutia
sescutai
sesctaui
sesctaiu
sesctuai
sesctuia
sesctiua
sesctiau
sesicaut
sesicatu
sesicuat
sesicuta
sesictua
sesictau
sesiacut
sesiactu
sesiauct
sesiautc
sesiatuc
sesiatcu
sesiuact
sesiuatc
sesiucat
sesiucta
sesiutca
sesiutac
sesitauc
sesitacu
sesituac
sesituca
sesitcua
sesitcau
sesaicut
sesaictu
sesaiuct
sesaiutc
sesaituc
sesaitcu
sesaciut
sesacitu
sesacuit
sesacuti
sesactui
sesactiu
sesaucit
sesaucti
sesauict
sesauitc
sesautic
sesautci
sesatcui
sesatciu
sesatuci
sesatuic
sesatiuc
sesaticu
sesuiact
sesuiatc
sesuicat
sesuicta
sesuitca
sesuitac
sesuaict
sesuaitc
sesuacit
sesuacti
sesuatci
sesuatic
sesucait
sesucati
sesuciat
sesucita
sesuctia
sesuctai
sesutaci
sesutaic
sesutcai
sesutcia
sesutica
sesutiac
sestiauc
sestiacu
sestiuac
sestiuca
sesticua
sesticau
sestaiuc
sestaicu
sestauic
sestauci
sestacui
sestaciu
sestuaic
sestuaci
sestuiac
sestuica
sestucia
sestucai
sestcaui
sestcaiu
sestcuai
sestcuia
sestciua
sestciau
setciasu
setciaus
setcisau
setcisua
setciusa
setciuas
setcaisu
setcaius
setcasiu
setcasui
setcausi
setcauis
setcsaiu
setcsaui
setcsiau
setcsiua
setcsuia
setcsuai
setcuasi
setcuais
setcusai
setcusia
setcuisa
setcuias
seticasu
seticaus
seticsau
seticsua
seticusa
seticuas
setiacsu
setiacus
setiascu
setiasuc
setiausc
setiaucs
setisacu
setisauc
setiscau
setiscua
setisuca
setisuac
setiuasc
setiuacs
setiusac
setiusca
setiucsa
setiucas
setaicsu
setaicus
setaiscu
setaisuc
setaiusc
setaiucs
setacisu
setacius
setacsiu
setacsui
setacusi
setacuis
setasciu
setascui
setasicu
setasiuc
setasuic
setasuci
setaucsi
setaucis
setausci
setausic
setauisc
setauics
setsiacu
setsiauc
setsicau
setsicua
setsiuca
setsiuac
setsaicu
setsaiuc
setsaciu
setsacui
setsauci
setsauic
setscaiu
setscaui
setsciau
setsciua
setscuia
setscuai
setsuaci
setsuaic
setsucai
setsucia
setsuica
setsuiac
setuiasc
setuiacs
setuisac
setuisca
setuicsa
setuicas
setuaisc
setuaics
setuasic
setuasci
setuacsi
setuacis
setusaic
setusaci
setusiac
setusica
setuscia
setuscai
setucasi
setucais
setucsai
setucsia
setucisa
setucias
ssucieat
ssucieta
ssuciaet
ssuciate
ssucitae
ssucitea
ssuceiat
ssuceita
ssuceait
ssuceati
ssucetai
ssucetia
ssucaeit
ssucaeti
ssucaiet
ssucaite
ssucatie
ssucatei
ssucteai
ssucteia
ssuctaei
ssuctaie
ssuctiae
ssuctiea
ssuiceat
ssuiceta
ssuicaet
ssuicate
ssuictae
ssuictea
ssuiecat
ssuiecta
ssuieact
ssuieatc
ssuietac
ssuietca
ssuiaect
ssuiaetc
ssuiacet
ssuiacte
ssuiatce
ssuiatec
ssuiteac
ssuiteca
ssuitaec
ssuitace
ssuitcae
ssuitcea
ssueicat
ssueicta
ssueiact
ssueiatc
ssueitac
ssueitca
ssueciat
ssuecita
ssuecait
ssuecati
ssuectai
ssuectia
ssueacit
ssueacti
ssueaict
ssueaitc
ssueatic
ssueatci
ssuetcai
ssuetcia
ssuetaci
ssuetaic
ssuetiac
ssuetica
ssuaiect
ssuaietc
ssuaicet
ssuaicte
ssuaitce
ssuaitec
ssuaeict
ssuaeitc
ssuaecit
ssuaecti
ssuaetci
ssuaetic
ssuaceit
ssuaceti
ssuaciet
ssuacite
ssuactie
ssuactei
ssuateci
ssuateic
ssuatcei
ssuatcie
ssuatice
ssuatiec
ssutieac
ssutieca
ssutiaec
ssutiace
ssuticae
ssuticea
ssuteiac
ssuteica
ssuteaic
ssuteaci
ssutecai
ssutecia
ssutaeic
ssutaeci
ssutaiec
ssutaice
ssutacie
ssutacei
ssutceai
ssutceia
ssutcaei
ssutcaie
ssutciae
ssutciea
sscuieat
sscuieta
sscuiaet
sscuiate
sscuitae
sscuitea
sscueiat
sscueita
sscueait
sscueati
sscuetai
sscuetia
sscuaeit
sscuaeti
sscuaiet
sscuaite
sscuatie
sscuatei
sscuteai
sscuteia
sscutaei
sscutaie
sscutiae
sscutiea
ssciueat
ssciueta
ssciuaet
ssciuate
ssciutae
ssciutea
sscieuat
sscieuta
sscieaut
sscieatu
sscietau
sscietua
ssciaeut
ssciaetu
ssciauet
ssciaute
ssciatue
ssciateu
ssciteau
ssciteua
sscitaeu
sscitaue
sscituae
sscituea
ssceiuat
ssceiuta
ssceiaut
ssceiatu
ssceitau
ssceitua
ssceuiat
ssceuita
ssceuait
ssceuati
ssceutai
ssceutia
ssceauit
ssceauti
ssceaiut
ssceaitu
ssceatiu
ssceatui
sscetuai
sscetuia
sscetaui
sscetaiu
sscetiau
sscetiua
sscaieut
sscaietu
sscaiuet
sscaiute
sscaitue
sscaiteu
sscaeiut
sscaeitu
sscaeuit
sscaeuti
sscaetui
sscaetiu
sscaueit
sscaueti
sscauiet
sscauite
sscautie
sscautei
sscateui
sscateiu
sscatuei
sscatuie
sscatiue
sscatieu
ssctieau
ssctieua
ssctiaeu
ssctiaue
ssctiuae
ssctiuea
sscteiau
sscteiua
sscteaiu
sscteaui
sscteuai
sscteuia
ssctaeiu
ssctaeui
ssctaieu
ssctaiue
ssctauie
ssctauei
ssctueai
ssctueia
ssctuaei
ssctuaie
ssctuiae
ssctuiea
ssicueat
ssicueta
ssicuaet
ssicuate
ssicutae
ssicutea
ssiceuat
ssiceuta
ssiceaut
ssiceatu
ssicetau
ssicetua
ssicaeut
ssicaetu
ssicauet
ssicaute
ssicatue
ssicateu
ssicteau
ssicteua
ssictaeu
ssictaue
ssictuae
ssictuea
ssiuceat
ssiuceta
ssiucaet
ssiucate
ssiuctae
ssiuctea
ssiuecat
ssiuecta
ssiueact
ssiueatc
ssiuetac
ssiuetca
ssiuaect
ssiuaetc
ssiuacet
ssiuacte
ssiuatce
ssiuatec
ssiuteac
ssiuteca
ssiutaec
ssiutace
ssiutcae
ssiutcea
ssieucat
ssieucta
ssieuact
ssieuatc
ssieutac
ssieutca
ssiecuat
ssiecuta
ssiecaut
ssiecatu
ssiectau
ssiectua
ssieacut
ssieactu
ssieauct
ssieautc
ssieatuc
ssieatcu
ssietcau
ssietcua
ssietacu
ssietauc
ssietuac
ssietuca
ssiauect
ssiauetc
ssiaucet
ssiaucte
ssiautce
ssiautec
ssiaeuct
ssiaeutc
ssiaecut
ssiaectu
ssiaetcu
ssiaetuc
ssiaceut
ssiacetu
ssiacuet
ssiacute
ssiactue
ssiacteu
ssiatecu
ssiateuc
ssiatceu
ssiatcue
ssiatuce
ssiatuec
ssitueac
ssitueca
ssituaec
ssituace
ssitucae
ssitucea
ssiteuac
ssiteuca
ssiteauc
ssiteacu
ssitecau
ssitecua
ssitaeuc
ssitaecu
ssitauec
ssitauce
ssitacue
ssitaceu
ssitceau
ssitceua
ssitcaeu
ssitcaue
ssitcuae
ssitcuea
sseciuat
sseciuta
sseciaut
sseciatu
ssecitau
ssecitua
ssecuiat
ssecuita
ssecuait
ssecuati
ssecutai
ssecutia
ssecauit
ssecauti
ssecaiut
ssecaitu
ssecatiu
ssecatui
ssectuai
ssectuia
ssectaui
ssectaiu
ssectiau
ssectiua
sseicuat
sseicuta
sseicaut
sseicatu
sseictau
sseictua
sseiucat
sseiucta
sseiuact
sseiuatc
sseiutac
sseiutca
sseiauct
sseiautc
sseiacut
sseiactu
sseiatcu
sseiatuc
sseituac
sseituca
sseitauc
sseitacu
sseitcau
sseitcua
sseuicat
sseuicta
sseuiact
sseuiatc
sseuitac
sseuitca
sseuciat
sseucita
sseucait
sseucati
sseuctai
sseuctia
sseuacit
sseuacti
sseuaict
sseuaitc
sseuatic
sseuatci
sseutcai
sseutcia
sseutaci
sseutaic
sseutiac
sseutica
sseaiuct
sseaiutc
sseaicut
sseaictu
sseaitcu
sseaituc
sseauict
sseauitc
sseaucit
sseaucti
sseautci
sseautic
sseacuit
sseacuti
sseaciut
sseacitu
sseactiu
sseactui
sseatuci
sseatuic
sseatcui
sseatciu
sseaticu
sseatiuc
ssetiuac
ssetiuca
ssetiauc
ssetiacu
sseticau
sseticua
ssetuiac
ssetuica
ssetuaic
ssetuaci
ssetucai
ssetucia
ssetauic
ssetauci
ssetaiuc
ssetaicu
ssetaciu
ssetacui
ssetcuai
ssetcuia
ssetcaui
ssetcaiu
ssetciau
ssetciua
ssacieut
ssacietu
ssaciuet
ssaciute
ssacitue
ssaciteu
ssaceiut
ssaceitu
ssaceuit
ssaceuti
ssacetui
ssacetiu
ssacueit
ssacueti
ssacuiet
ssacuite
ssacutie
ssacutei
ssacteui
ssacteiu
ssactuei
ssactuie
ssactiue
ssactieu
ssaiceut
ssaicetu
ssaicuet
ssaicute
ssaictue
ssaicteu
ssaiecut
ssaiectu
ssaieuct
ssaieutc
ssaietuc
ssaietcu
ssaiuect
ssaiuetc
ssaiucet
ssaiucte
ssaiutce
ssaiutec
ssaiteuc
ssaitecu
ssaituec
ssaituce
ssaitcue
ssaitceu
ssaeicut
ssaeictu
ssaeiuct
ssaeiutc
ssaeituc
ssaeitcu
ssaeciut
ssaecitu
ssaecuit
ssaecuti
ssaectui
ssaectiu
ssaeucit
ssaeucti
ssaeuict
ssaeuitc
ssaeutic
ssaeutci
ssaetcui
ssaetciu
ssaetuci
ssaetuic
ssaetiuc
ssaeticu
ssauiect
ssauietc
ssauicet
ssauicte
ssauitce
ssauitec
ssaueict
ssaueitc
ssauecit
ssauecti
ssauetci
ssauetic
ssauceit
ssauceti
ssauciet
ssaucite
ssauctie
ssauctei
ssauteci
ssauteic
ssautcei
ssautcie
ssautice
ssautiec
ssatieuc
ssatiecu
ssatiuec
ssatiuce
ssaticue
ssaticeu
ssateiuc
ssateicu
ssateuic
ssateuci
ssatecui
ssateciu
ssatueic
ssatueci
ssatuiec
ssatuice
ssatucie
ssatucei
ssatceui
ssatceiu
ssatcuei
ssatcuie
ssatciue
ssatcieu
sstcieau
sstcieua
sstciaeu
sstciaue
sstciuae
sstciuea
sstceiau
sstceiua
sstceaiu
sstceaui
sstceuai
sstceuia
sstcaeiu
sstcaeui
sstcaieu
sstcaiue
sstcauie
sstcauei
sstcueai
sstcueia
sstcuaei
sstcuaie
sstcuiae
sstcuiea
ssticeau
ssticeua
ssticaeu
ssticaue
ssticuae
ssticuea
sstiecau
sstiecua
sstieacu
sstieauc
sstieuac
sstieuca
sstiaecu
sstiaeuc
sstiaceu
sstiacue
sstiauce
sstiauec
sstiueac
sstiueca
sstiuaec
sstiuace
sstiucae
sstiucea
ssteicau
ssteicua
ssteiacu
ssteiauc
ssteiuac
ssteiuca
ssteciau
ssteciua
sstecaiu
sstecaui
sstecuai
sstecuia
ssteaciu
ssteacui
ssteaicu
ssteaiuc
ssteauic
ssteauci
ssteucai
ssteucia
ssteuaci
ssteuaic
ssteuiac
ssteuica
sstaiecu
sstaieuc
sstaiceu
sstaicue
sstaiuce
sstaiuec
sstaeicu
sstaeiuc
sstaeciu
sstaecui
sstaeuci
sstaeuic
sstaceiu
sstaceui
sstacieu
sstaciue
sstacuie
sstacuei
sstaueci
sstaueic
sstaucei
sstaucie
sstauice
sstauiec
sstuieac
sstuieca
sstuiaec
sstuiace
sstuicae
sstuicea
sstueiac
sstueica
sstueaic
sstueaci
sstuecai
sstuecia
sstuaeic
sstuaeci
sstuaiec
sstuaice
sstuacie
sstuacei
sstuceai
sstuceia
sstucaei
sstucaie
sstuciae
sstuciea
stuciesa
stucieas
stucisea
stucisae
stuciase
stuciaes
stuceisa
stuceias
stucesia
stucesai
stuceasi
stuceais
stucseia
stucseai
stucsiea
stucsiae
stucsaie
stucsaei
stucaesi
stucaeis
stucasei
stucasie
stucaise
stucaies
stuicesa
stuiceas
stuicsea
stuicsae
stuicase
stuicaes
stuiecsa
stuiecas
stuiesca
stuiesac
stuieasc
stuieacs
stuiseca
stuiseac
stuiscea
stuiscae
stuisace
stuisaec
stuiaesc
stuiaecs
stuiasec
stuiasce
stuiacse
stuiaces
stueicsa
stueicas
stueisca
stueisac
stueiasc
stueiacs
stuecisa
stuecias
stuecsia
stuecsai
stuecasi
stuecais
stuescia
stuescai
stuesica
stuesiac
stuesaic
stuesaci
stueacsi
stueacis
stueasci
stueasic
stueaisc
stueaics
stusieca
stusieac
stusicea
stusicae
stusiace
stusiaec
stuseica
stuseiac
stusecia
stusecai
stuseaci
stuseaic
stusceia
stusceai
stusciea
stusciae
stuscaie
stuscaei
stusaeci
stusaeic
stusacei
stusacie
stusaice
stusaiec
stuaiesc
stuaiecs
stuaisec
stuaisce
stuaicse
stuaices
stuaeisc
stuaeics
stuaesic
stuaesci
stuaecsi
stuaecis
stuaseic
stuaseci
stuasiec
stuasice
stuascie
stuascei
stuacesi
stuaceis
stuacsei
stuacsie
stuacise
stuacies
stcuiesa
stcuieas
stcuisea
stcuisae
stcuiase
stcuiaes
stcueisa
stcueias
stcuesia
stcuesai
stcueasi
stcueais
stcuseia
stcuseai
stcusiea
stcusiae
stcusaie
stcusaei
stcuaesi
stcuaeis
stcuasei
stcuasie
stcuaise
stcuaies
stciuesa
stciueas
stciusea
stciusae
stciuase
stciuaes
stcieusa
stcieuas
stciesua
stciesau
stcieasu
stcieaus
stciseua
stciseau
stcisuea
stcisuae
stcisaue
stcisaeu
stciaesu
stciaeus
stciaseu
stciasue
stciause
stciaues
stceiusa
stceiuas
stceisua
stceisau
stceiasu
stceiaus
stceuisa
stceuias
stceusia
stceusai
stceuasi
stceuais
stcesuia
stcesuai
stcesiua
stcesiau
stcesaiu
stcesaui
stceausi
stceauis
stceasui
stceasiu
stceaisu
stceaius
stcsieua
stcsieau
stcsiuea
stcsiuae
stcsiaue
stcsiaeu
stcseiua
stcseiau
stcseuia
stcseuai
stcseaui
stcseaiu
stcsueia
stcsueai
stcsuiea
stcsuiae
stcsuaie
stcsuaei
stcsaeui
stcsaeiu
stcsauei
stcsauie
stcsaiue
stcsaieu
stcaiesu
stcaieus
stcaiseu
stcaisue
stcaiuse
stcaiues
stcaeisu
stcaeius
stcaesiu
stcaesui
stcaeusi
stcaeuis
stcaseiu
stcaseui
stcasieu
stcasiue
stcasuie
stcasuei
stcauesi
stcaueis
stcausei
stcausie
stcauise
stcauies
sticuesa
sticueas
sticusea
sticusae
sticuase
sticuaes
sticeusa
sticeuas
sticesua
sticesau
sticeasu
sticeaus
sticseua
sticseau
sticsuea
sticsuae
sticsaue
sticsaeu
sticaesu
sticaeus
sticaseu
sticasue
sticause
sticaues
stiucesa
stiuceas
stiucsea
stiucsae
stiucase
stiucaes
stiuecsa
stiuecas
stiuesca
stiuesac
stiueasc
stiueacs
stiuseca
stiuseac
stiuscea
stiuscae
stiusace
stiusaec
stiuaesc
stiuaecs
stiuasec
stiuasce
stiuacse
stiuaces
stieucsa
stieucas
stieusca
stieusac
stieuasc
stieuacs
stiecusa
stiecuas
stiecsua
stiecsau
stiecasu
stiecaus
stiescua
stiescau
stiesuca
stiesuac
stiesauc
stiesacu
stieacsu
stieacus
stieascu
stieasuc
stieausc
stieaucs
stisueca
stisueac
stisucea
stisucae
stisuace
stisuaec
stiseuca
stiseuac
stisecua
stisecau
stiseacu
stiseauc
stisceua
stisceau
stiscuea
stiscuae
stiscaue
stiscaeu
stisaecu
stisaeuc
stisaceu
stisacue
stisauce
stisauec
stiauesc
stiauecs
stiausec
stiausce
stiaucse
stiauces
stiaeusc
stiaeucs
stiaesuc
stiaescu
stiaecsu
stiaecus
stiaseuc
stiasecu
stiasuec
stiasuce
stiascue
stiasceu
stiacesu
stiaceus
stiacseu
stiacsue
stiacuse
stiacues
steciusa
steciuas
stecisua
stecisau
steciasu
steciaus
stecuisa
stecuias
stecusia
stecusai
stecuasi
stecuais
stecsuia
stecsuai
stecsiua
stecsiau
stecsaiu
stecsaui
stecausi
stecauis
stecasui
stecasiu
stecaisu
stecaius
steicusa
steicuas
steicsua
steicsau
steicasu
steicaus
steiucsa
steiucas
steiusca
steiusac
steiuasc
steiuacs
steisuca
steisuac
steiscua
steiscau
steisacu
steisauc
steiausc
steiaucs
steiasuc
steiascu
steiacsu
steiacus
steuicsa
steuicas
steuisca
steuisac
steuiasc
steuiacs
steucisa
steucias
steucsia
steucsai
steucasi
steucais
steuscia
steuscai
steusica
steusiac
steusaic
steusaci
steuacsi
steuacis
steuasci
steuasic
steuaisc
steuaics
stesiuca
stesiuac
stesicua
stesicau
stesiacu
stesiauc
stesuica
stesuiac
stesucia
stesucai
stesuaci
stesuaic
stescuia
stescuai
stesciua
stesciau
stescaiu
stescaui
stesauci
stesauic
stesacui
stesaciu
stesaicu
stesaiuc
steaiusc
steaiucs
steaisuc
steaiscu
steaicsu
steaicus
steauisc
steauics
steausic
steausci
steaucsi
steaucis
steasuic
steasuci
steasiuc
steasicu
steasciu
steascui
steacusi
steacuis
steacsui
steacsiu
steacisu
steacius
stscieua
stscieau
stsciuea
stsciuae
stsciaue
stsciaeu
stsceiua
stsceiau
stsceuia
stsceuai
stsceaui
stsceaiu
stscueia
stscueai
stscuiea
stscuiae
stscuaie
stscuaei
stscaeui
stscaeiu
stscauei
stscauie
stscaiue
stscaieu
stsiceua
stsiceau
stsicuea
stsicuae
stsicaue
stsicaeu
stsiecua
stsiecau
stsieuca
stsieuac
stsieauc
stsieacu
stsiueca
stsiueac
stsiucea
stsiucae
stsiuace
stsiuaec
stsiaeuc
stsiaecu
stsiauec
stsiauce
stsiacue
stsiaceu
stseicua
stseicau
stseiuca
stseiuac
stseiauc
stseiacu
stseciua
stseciau
stsecuia
stsecuai
stsecaui
stsecaiu
stseucia
stseucai
stseuica
stseuiac
stseuaic
stseuaci
stseacui
stseaciu
stseauci
stseauic
stseaiuc
stseaicu
stsuieca
stsuieac
stsuicea
stsuicae
stsuiace
stsuiaec
stsueica
stsueiac
stsuecia
stsuecai
stsueaci
stsueaic
stsuceia
stsuceai
stsuciea
stsuciae
stsucaie
stsucaei
stsuaeci
stsuaeic
stsuacei
stsuacie
stsuaice
stsuaiec
stsaieuc
stsaiecu
stsaiuec
stsaiuce
stsaicue
stsaiceu
stsaeiuc
stsaeicu
stsaeuic
stsaeuci
stsaecui
stsaeciu
stsaueic
stsaueci
stsauiec
stsauice
stsaucie
stsaucei
stsaceui
stsaceiu
stsacuei
stsacuie
stsaciue
stsacieu
staciesu
stacieus
staciseu
stacisue
staciuse
staciues
staceisu
staceius
stacesiu
stacesui
staceusi
staceuis
stacseiu
stacseui
stacsieu
stacsiue
stacsuie
stacsuei
stacuesi
stacueis
stacusei
stacusie
stacuise
stacuies
staicesu
staiceus
staicseu
staicsue
staicuse
staicues
staiecsu
staiecus
staiescu
staiesuc
staieusc
staieucs
staisecu
staiseuc
staisceu
staiscue
staisuce
staisuec
staiuesc
staiuecs
staiusec
staiusce
staiucse
staiuces
staeicsu
staeicus
staeiscu
staeisuc
staeiusc
staeiucs
staecisu
staecius
staecsiu
staecsui
staecusi
staecuis
staesciu
staescui
staesicu
staesiuc
staesuic
staesuci
staeucsi
staeucis
staeusci
staeusic
staeuisc
staeuics
stasiecu
stasieuc
stasiceu
stasicue
stasiuce
stasiuec
staseicu
staseiuc
staseciu
stasecui
staseuci
staseuic
stasceiu
stasceui
stascieu
stasciue
stascuie
stascuei
stasueci
stasueic
stasucei
stasucie
stasuice
stasuiec
stauiesc
stauiecs
stauisec
stauisce
stauicse
stauices
staueisc
staueics
stauesic
stauesci
stauecsi
stauecis
stauseic
stauseci
stausiec
stausice
stauscie
stauscei
staucesi
stauceis
staucsei
staucsie
staucise
staucies
Read more ...[1] , [2] , [3] | 26,567 | 54,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-22 | latest | en | 0.834718 |
http://forums.wolfram.com/mathgroup/archive/2012/Apr/msg00274.html | 1,590,741,555,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347402457.55/warc/CC-MAIN-20200529054758-20200529084758-00103.warc.gz | 50,275,132 | 7,761 | Re: troublesome integral
• To: mathgroup at smc.vnet.net
• Subject: [mg126011] Re: troublesome integral
• From: "Stephen Luttrell" <steve at _removemefirst_stephenluttrell.com>
• Date: Fri, 13 Apr 2012 04:44:42 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <jm64ib\$6la\$1@smc.vnet.net>
```You can rewrite the integral as
Integrate[
Cos[\[Alpha] + \[Beta]] Exp[I z Cos[\[Beta]]], {\[Beta], 0, 2 \[Pi]},
Assumptions -> z \[Element] Reals]
which can be rearranged as (the Sin[\[Alpha]] Sin[\[Beta]] part integrates
to zero)
Integrate[
Cos[\[Alpha]] Cos[\[Beta]] Exp[I z Cos[\[Beta]]], {\[Beta], 0,
2 \[Pi]}, Assumptions -> z \[Element] Reals]
which evaluates to
2 I \[Pi] BesselJ[1, z] Cos[\[Alpha]]
Over the years I too have encountered this problem with evaluating integral
representations of Bessel functions, and the only solution I have found is
to give mathematica a helping hand, as above.
--
Stephen Luttrell
West Malvern, UK
"peter lindsay" <pl0 at me.com> wrote in message
news:jm64ib\$6la\$1 at smc.vnet.net...
>
> wolfram who are escalating it to the developers. Possibly someone here has
>
> Integrate[Cos[\[Beta]] Exp[I z Cos[\[Beta]-\[Alpha]]],{\[Beta],0,2
> \[Pi]},Assumptions->z\[Element]Reals]
>
> doesn't seem to run.
>
>
> 2 I \[Pi] BesselJ[1,z] Cos[\[Alpha]] [ I think ]
>
> thanks
>
>
> Peter Lindsay
>
>
```
• Prev by Date: Re: Replace a vertical line in ListPlot
• Next by Date: Re: What characters are allowed in mathematica variable names? i.e. how
• Previous by thread: Re: troublesome integral
• Next by thread: Re: troublesome integral | 521 | 1,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-24 | latest | en | 0.758819 |
https://www.r-bloggers.com/2020/12/abs-and-relu-are-not-mercer-kernels/ | 1,726,236,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00897.warc.gz | 880,054,278 | 20,995 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
I am sharing some rough notes (in R and Python) here on how while `dot(a, b)` fulfills “Mercer’s condition” (by definition!, and I’ll just informally call these beasts a “Mercer Kernel”), the seemingly harmless variations `abs(dot(a, b))` `relu(dot(a, b))` are not Mercer Kernels (`relu(x) = max(0, x) = (abs(x) + x)/2`). It turns out they fail the required positive-semi-definiteness checks.
It is kind of a tricky point, though very close to the definitions. Here I’ll just try to state what is true, without confusing it with the derivations why it is true.
If `a` and `b` are in `R1`, then these two forms are Mercer Kernels! This is because in this case `abs(dot(a, b)) = dot(abs(a), abs(b))` plus the usual rules for building new kernels.
And if we only check up-to 3 by 3 systems for positive semi-definiteness we also get deceived into thinking `abs` and `relu` are Mercer Kernels by variations of Sylvester’s criterion.
At this point one is sufficiently confused/frustrated that it is worth re-checking that the dot-product itself (which is the prototype for a Mercer Kernel) is in fact a Mercer Kernel under the check-definition.
The most common reason one cares, is the positive-semi-definiteness is used to establish convexity in support vector machines, which is turn is used to establish the associated optimization problem is convex and “easy.”
If you are interested in kernelized machine learning and (like everyone) need to see the so-called “obvious” steps checked, I invite you to check out these very rough notes. Or, at least be innoculated to know `abs(dot(a, b))` `relu(dot(a, b))` are not Mercer Kernels in general, even if you can’t immediately regurgitate why. | 440 | 1,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-38 | latest | en | 0.907567 |
http://www.numberempire.com/742637 | 1,498,472,449,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320707.69/warc/CC-MAIN-20170626101322-20170626121322-00460.warc.gz | 590,936,169 | 4,738 | Home | Menu | Get Involved | Math forum
Number properties
# Number 742637
seven hundred forty two thousand six hundred thirty seven
### Properties of the number 742637
Factorization 7 * 277 * 383 Divisors 1, 7, 277, 383, 1939, 2681, 106091, 742637 Count of divisors 8 Sum of divisors 854016 Is prime? NO Previous prime 742619 Next prime 742657 742637th prime 11261291 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 10110101010011101101 Octal 2652355 Duodecimal 2b9925 Hexadecimal b54ed Square 551509713769 Square root 861.76388877697 Natural logarithm 13.517962644423 Decimal logarithm 5.8707765830473 Sine 0.81095908865663 Cosine -0.58510285978211 Tangent -1.386011151883
Number 742637 is pronounced seven hundred forty two thousand six hundred thirty seven. Number 742637 is a composite number. Factors of 742637 are 7 * 277 * 383. Number 742637 has 8 divisors: 1, 7, 277, 383, 1939, 2681, 106091, 742637. Sum of the divisors is 854016. Number 742637 is not a Fibonacci number. It is not a Bell number. Number 742637 is not a Catalan number. Number 742637 is not a regular number (Hamming number). It is a not factorial of any number. Number 742637 is a deficient number and therefore is not a perfect number. Binary numeral for number 742637 is 10110101010011101101. Octal numeral is 2652355. Duodecimal value is 2b9925. Hexadecimal representation is b54ed. Square of the number 742637 is 551509713769. Square root of the number 742637 is 861.76388877697. Natural logarithm of 742637 is 13.517962644423 Decimal logarithm of the number 742637 is 5.8707765830473 Sine of 742637 is 0.81095908865663. Cosine of the number 742637 is -0.58510285978211. Tangent of the number 742637 is -1.386011151883
742636 742638 | 575 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-26 | longest | en | 0.646639 |
https://www.excelforum.com/excel-tips/843509-dynamic-cascading-drop-downs-using-index-and-match.html?s=896d3e5a56805f0d58fa3791ec00945d | 1,506,345,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818691830.33/warc/CC-MAIN-20170925130433-20170925150433-00200.warc.gz | 753,719,720 | 15,792 | Dynamic Cascading Drop-Downs Using INDEX & MATCH
1. Dynamic Cascading Drop-Downs Using INDEX & MATCH
I dug this up after reading Jbeaucaire's thread (link here). This is just another way to solve that problem using Index and Match. I like learning numerous ways to do things like this, if you do too, then please read on.
Create the top Categories as a header row above the associated subcategories (fig1).
TOP CATEGORY (FIRST-DROP-DOWN)
Create a named range for the top Categories (header row). Using Index, we can create a dynamic range (expands / contracts when you add / remove columns). The Refers-To box for Categories reads
=Lists!\$A\$1:INDEX(Lists!\$1:\$1,COUNTA(Lists!\$1:\$1))
"Categories" is used in the data validation list source, for the first drop down (fig3 & 4).
SUB-CATEGORIES (DEPENDENT-DROP-DOWN)
The second drop down menu, is defined by the users choice in the top category. A second named range, Sub_Categories provides the dynamic range needed for that data validation list (fig5)
The data validation for this cell is similar to the top category drop down, it is simply the name "sub_categories" (sorry, only 5 figures per posting).
In this version of the solution, the named formulas do the work. As shown in fig2, Sub_Categories makes use of the index function to define a subset of a column. The Refers To field for Sub_Categories reads:
=INDEX(Lists!\$2:\$2,colChoice):INDEX(Lists!\$1:\$1048576,COUNTA(INDEX(Lists!\$1:\$1048576,,colChoice)),colChoice)
colChoice is a third named formula, added just to make sub_categories more readable. colChoice is a standard use of the Match function used to "match" the users top category selection with the corresponding column on the "Lists" worksheet (also shown in fig2).
colChoice = Match(Selection!\$C\$2,Lists!\$1:\$1,0)
If you are unfamiliar with the INDEX function, this may look a little confusing. The main idea is, the first instance of the Index function (to the left of the Colon operator), defines a single cell reference corresponding to the top of the selected choices (note, this the choices start in row 2). In this example, it will place return A2, B2, or C2.
The second and third instances are a bit trickier, but they function to return a reference to the bottom cell (last) of the appropriate column. They will evaluate to either A7, B6, or C10 in this example. The entire formula will return one of the following (again, in this example) A2:A7, B2:B6, or C2:C10. Note, if you add or remove choices from any of the list, this function will expand/contract to reflect the change!
For a very good explanation of INDEX used to create named formulas, read up here.
My example is attached to the next post
2. Re: Dynamic Cascading Drop-Downs Using INDEX & MATCH
attachment for described solution.
3. Re: Dynamic Cascading Drop-Downs Using INDEX & MATCH
Lists!\$1:\$1048576
why not:
Lists!\$A:\$A
Regards, TMS
4. Re: Dynamic Cascading Drop-Downs Using INDEX & MATCH
Well, let me break this down the second index third INDEX function:
First off, I'm using index with three arguments (Table, Row, Column) to return a reference to a cell. The reference will bottom-out my second drop down menu. The overall goal is to return a reference to the bottom-most cell of the column matching the user's choice (from the first drop down).
=INDEX(Lists!\$2:\$2,colChoice):INDEX(Lists!\$1:\$1048576,COUNTA(INDEX(Lists!\$1:\$1048576,,colChoice)),colChoice)
Table argument(Lists!\$1:\$1048576)
The table I'm interested in sits in the Lists worksheet, and I don't know how many rows and columns will be used. To allow for and unknown number of columns, I'll just assume the table starts in the upper left cell and pass in the entire sheet. That's what the Lists!\$1:\$1048576 does for us, it passes in all rows (entire-rows).
The second use of the index function highlighted in blue, returns an entire column (the one matching the users selection "colChoice"). By leaving the row argument blank (INDEX(Lists!\$1:\$1048576,,colChoice)) - we tell index to return the entire column.
In words: INDEX(Lists!\$1:\$1048576,,colChoice) means "return to me the entire column from the worksheet named list, whose column number is colChoice". By including that entire column in the COUNTA() function, we return the count of non-blank cells in the colChoice column (Since the column starts in the first row, this count is the row number of the bottom-most cells we are looking for -- assuming we have no blanks in the drop-down menu. A safe assumption if we build the list in the first place with that in mind).
SO, this blue highlighted section returns the row number we need.
Putting this together, we have:
=INDEX(Lists!\$2:\$2,colChoice):INDEX(from-entire-Lists-worksheet, Bottom-most-row#-with-text-in-colChoice-column, colChoice-column#)
And since we have indexed the entire sheet, we will get a reference like cells(bottom-row,choosen-column).address on the right side of the colon operator to complete our drop-down list
5. Re: Dynamic Cascading Drop-Downs Using INDEX & MATCH
This is great and I want to be able to use this system.
But!
How do I expand it so as when I am on the next line down I can make an independent selection.
In the part "colChoice = Match(Selection!\$C\$2,Lists!\$1:\$1,0)" I need this to work in the next row down 5000 rows
e.g.
colChoice = Match(Selection!\$C\$3,Lists!\$1:\$1,0)
colChoice = Match(Selection!\$C\$4,Lists!\$1:\$1,0)
colChoice = Match(Selection!\$C\$5,Lists!\$1:\$1,0)
How can I edit this without having an independent colChoice in name manager for each row?
Users Browsing this Thread
There are currently 1 users browsing this thread. (0 members and 1 guests)
Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
Search Engine Friendly URLs by vBSEO 3.6.0 RC 1 | 1,495 | 5,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-39 | latest | en | 0.780057 |
https://blenderartists.org/t/what-happens-when-you-shift-d-duplicate-a-cube/590647 | 1,571,387,424,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986679439.48/warc/CC-MAIN-20191018081630-20191018105130-00499.warc.gz | 392,107,336 | 5,494 | # What happens when you (Shift D) duplicate a cube?
What characteristics is the second cube given?
By this I mean:
Let’s say you shift a and create 2 cubes.
[ ] [ ]
Now you create a third cube and duplicate it shift D
[ ] [ ]
Now you have 4 cubes
[ ] [ ] - [ ] [ ]
What’s the difference between these 2 pair of cubes? Everything that you can tell me?
The main difference in your very simple example may be, where the cubes are placed by creation, a new ‘shift a’ cube is placed at the position of the 3d-cursor a ‘shift d’ cube is placed, where the orignal cube is.
But I don’t think this example really shows the use of the ‘shift d’ operation.
You could have modified the cube into a lionhead and if you duplicate it, you get 2 lion heads.
If you gave it a material, they would have had, the very same material, not a copy, see in the outliner.
You could have selted more then one cube and duplicated it.
A more interesting question would have been, what is different with ‘alt d’ (duplicate linked), where you get an object with the same mesh as the original object, but a different position, as you can see in the outliner.
First just to clarify, an ‘object’ primarily represents a location, rotation, and scale in 3D space (think of an empty and what you can do in object mode). A mesh is a collection of vertices connected to describe, a surface and you can edit it in blender’s edit mode. A mesh is linked to an object so that it can appear in the 3D Viewport with the location, rotation, and scale of that object. The mesh itself may have things applied to it such as a material and modifiers (look at the outliner heirarchy).
Every object that is duplicated with shift-d creates a new object with a new, unique copy of the mesh. By default the same things that were applied to the mesh you copied are applied to your newly duplicated mesh, but you could change that. The mesh is independent and you could apply a different material and modifiers if you wanted.
If you alt-d you create a new object linked with the same mesh. Because it uses the same mesh data, whatever form the mesh takes and whatever is applied to the mesh will occur in all instances of it. You haven’t created a new mesh, just linked a pre-existing mesh asset to a new object (location, rotation, scale).
I don’t believe you can have an object with multiple meshes, although theoretically it would be nice if you wanted to change things like the materials and modifiers of specific parts of a mesh while keeping the same object.
The current ui can obscure these concepts so hopefully clarifying these things to users is in the cards (for example having a properties panel that is specific to what you click and not showing everything on an equal level). | 614 | 2,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-43 | latest | en | 0.945896 |
https://byjus.com/questions/an-angle-is-one-fifth-of-its-supplement-find-the-measure-of-the-angle/?replytocom=235221 | 1,632,806,859,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00106.warc.gz | 196,142,261 | 37,642 | # An angle is one-fifth of its supplement, find the measure of the angle.
Let one angle = x°
It’s supplementary angle = ( 180 – x )°
According to the problem given ,
x = (180 – x ) / 5
5x = 180 – x
5x + x = 180
6x = 180
x = 180/6
x = 30°
Therefore ,
First angle = x = 30°
It’s supplement = 180 – x
= 180 – 30
= 150°
4.5 (21)
(41)
(9)
#### Choose An Option That Best Describes Your Problem
Thank you. Your Feedback will Help us Serve you better. | 158 | 462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-39 | latest | en | 0.856247 |
https://studydaddy.com/question/analysis-an-australian-manufacturing-company-is-keen-to-develop-new-products-and-1 | 1,563,746,970,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527204.71/warc/CC-MAIN-20190721205413-20190721231413-00523.warc.gz | 561,169,642 | 8,894 | Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# ANALYSIS An Australian manufacturing company is keen to develop new products and develop new product line of shoes so that the company can expand into Asian markets more than before. The data have bee
ANALYSIS
An Australian manufacturing company is keen to develop new products and develop new product line of shoes so that the company can expand into Asian markets more than before. The data have been collected for 99 products from the market. The data are in an excel file named “SHOES”. The file includes: PRICE: Price of the shoes in dollars.
GENDER: 1 for Female product and 2 for Male product.
COUNTRY: the country in which the shoes is produced: 1 made in Thailand, 2 made in Singapore and 3 made in China.
COST: Production cost of the shoes in dollars.
QUESTIONS Part 1:
1. Calculate the Mean, Median, Mode, Standard Deviation and Coefficient of Variation, for prices for men and women shoes, separately. Compare the figures and explain that what conclusions you can draw from these analyses? Draw a box and Whisker plot for men and female shoes prices and comment on the shape of the graph.
2. Calculate the Mean, Median, Mode, Standard Deviation and Coefficient of Variation for prices for the three countries – Thailand, Singapore and China, separately. Compare the figures and explain what conclusions you can draw from these analyses? Draws a Box and Whisker plot for each country and comment on the shape of the graph.
Part 2: Answer the following hypothesis questions:
1. Determine if average prices for female shoes is less than average prices for male shoes. Compare the result with part 1 question 1. Does the result confirm your previous findings? (Follow the hypothesis testing steps, 0.05 level of significance, assuming “none equal variances” of populations).
2. Determine if there is a difference in shoes prices in different countries. Compare the result with part 1 question 2. Does the result confirm you previous findings? (Follow the hypothesis testing steps, 0.05 level of significance).
3. Using a scatter graph of price (Y variable on vertical axis) and production cost (X variable, horizontal axis), comment on the relationship between price and cost.
4. Using regression analysis comment on the relationship between price and production cost. (comment about the estimated coefficient (b1), the coefficient of determination, model validity and residuals plot). Does the result confirm your previous findings in question 3?
Files: SHOES.xlsx | 543 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-30 | longest | en | 0.938136 |
https://www.taylorfrancis.com/books/9780080521626/chapters/10.4324/9780080521626-22 | 1,597,110,610,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738723.55/warc/CC-MAIN-20200810235513-20200811025513-00215.warc.gz | 836,740,949 | 32,690 | chapter 2
9 Pages
## Using ratio analysis
A ratio is a relationship between one number and another. If one workteam has twelve people and another has eight people, the ratio of their numbers could be expressed in any of the following ways:
as 12:8 or as 12
8 or as 3:2 or as
2 or as 1.5:1 or as 1.5 to 1
Financial indicators are usually presented in one of the last two forms in the list above. | 109 | 401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-34 | latest | en | 0.985086 |
https://www.geeksforgeeks.org/class-11-rd-sharma-solutions-chapter-33-probability-exercise-33-3-set-1/?ref=rp | 1,660,896,685,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573630.12/warc/CC-MAIN-20220819070211-20220819100211-00671.warc.gz | 663,611,623 | 27,432 | # Class 11 RD Sharma Solutions – Chapter 33 Probability – Exercise 33.3 | Set 1
• Last Updated : 08 May, 2021
### Question 1. Which of the following cannot be valid assignment of probability for elementary events of outcomes of sample space S={w1, w2, w3, w4, w5, w6, w7} :
Elementary events are:
Solution:
We know that the value of probability of an event must be between 0 and 1 and that the sum of all probabilities of a event must be equal to one. Using this rationale,
1. This is valid since each probability of event w1 lies between 0 and 1 and the sum of all probabilities of event w1 is 1, or P{w1}= 1.
2. This is valid since each probability of event w2 lies between 0 and 1 and the sum of all probabilities of event w2 is 1, or P{w2}=2.
3. This is not valid as the sum of all probabilities is 2.8, which is more than one. P{wi} = 2.8
4. This is not valid since the probability of w7 is 15/14= 1.07 > 1.
Hence, the valid events are only 1 and 2.
### (i) a prime number
Solution:
Since a die is thrown, the number of outcomes in sample space must be 6. Thus, n{S}= 6.
Let E be the event of getting a prime number.
∴ E = {2,3,5} or n{E} = 3
We know, Probability of an event = Number of outcomes in that event / number of outcomes in sample space
∴ P{E} = n{E}/ n{S} = 3/6
∴ P{E}= 1/2
### (ii) 2 or 4
Solution:
Let N be the event of getting a 2 or 4.
Hence, N = (2,4} or n{N} = 2
Probability of event N or P{N} = n{N}/n{S}
or P{N} = 2/6
∴ P{N} = 1/3
### (iii) a multiple of 2 or 3
Solution:
Let R be the event of getting a multiple of 2 or 3 when a die is thrown.
∴ R = {2,3,4,6}
Thus, n{R} = 4
∴ P{R} = n{R}/ n{S} = 4/6
∴ P{R} = 2/3
### (i) 8 as the sum
Solution:
Since a pair of die have been thrown, the total number of outcomes in the sample space S become
6×6 = 62= 36.
Let R be the event of getting 8 as the sum when a pair of dice is thrown together.
∴ R = {(2,6), (3,5), (4,9), (5,3), (6,2)}
∴ n{R} = 5
Hence probability of R = n{R}/ n{S}
∴ P{R} = 5/36
### (ii) a doublet
Solution:
Let D be the event that a doublet appear on the dice when a pair of dice is thrown together.
∴ D = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
Thus, n{D} = 6
Hence probability of D = n{D}/ n{S}
P{D} = 6/36 = 1/6
∴ P{D} = 1/6
### (iii) a doublet of prime numbers
Solution:
Let F be the event of getting a doublet of prime numbers when a pair of dice is thrown together.
∴ F = {(2,2), (3,3), (5,5)}
Thus, n{F} = 3
Hence probability of F = n{F}/ n{S}
P{F}= 3/36 = 1/12
∴ P{F} = 1/12
### (iv) a doublet of odd numbers
Solution:
Let Q be the event of getting a doublet of odd numbers when a pair of dice is thrown.
∴ Q = {(1,1), (3,3), (5,5)}
Thus, n{Q} = 3
Probability of Q = n{Q}/ n{S}
P{Q} = 3/36 = 1/12
∴ P{Q} = 1/12
### (v) a sum greater than 9
Solution:
Let K be the event of getting a sum greater than 9 when a pair of dice is thrown.
∴ K = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
Thus, n{K} = 6
Probability of K = n{K}/n{S}
P{K}= 6/36 = 1/6
∴ P{K} = 1/6
### (vi) an even number on first
Solution:
Let W be the event of getting an even number on first dice when a pair of dice is thrown. If the first number has to be even, it means that any other number can appear on the second dice, be it even or odd. Using this rationale:
∴ W ={(2,1), (2,2), (2,3), (2,4),(2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Thus, n{W} = 18
Probability of W = n{W}/ n{S}
= 18/36 = 1/2
∴ P{W} = 1/2
### (vii) an even number on one and a multiple of 3 on the other
Solution:
Let V be the event of that we get an even number on one and a multiple of 3 on the other die when a pair of dice is thrown simultaneously. In this case, it is not specified that the even number shall have to be on the face of first or the second dice.
So long as an even number appears on any of the two dices, it is to be taken as an outcome of the given event. Same goes for multiple of 3 as well. Using this rationale:
∴ V = {(2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4)}
Thus, n{V} = 11
Probability of V = n{V}/ n{S}
∴ P{V} = 11/36
### (viii) neither 9 nor 11 as the sum of the numbers on the faces
Solution:
Let H be the event of getting neither 9 nor 11 as the sum of the numbers on the faces of two dice when they are thrown together.
Therefore, H’ shall be the event that either 9 or 11 appear as sum of the two numbers on the faces of the two dices.
H’ = {(3,6), (4,5), (5,4), (5,6), (6,3), (6,5)}
n{H’} = 6
P{H’} = n{H’}/ n{S} = 6/36 = 1/6
Thus, P{H} = 1−P{H’} = 1−1/6 = 5/6
### (ix) a sum less than 6
Solution:
Let E be the event of obtaining a sum less then 6 when a pair of die is thrown together.
∴ E = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}
∴ n{E} =10
Probability of E = P{E} = n{E}/ n{S} = 10/36 = 5/18
∴ P(E) = 5/18
### (x) a sum less than 7
Solution:
Let C be the event that a sum less than 7 appears on the faces of the two dice when thrown together.
∴ C = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3).(4,1),(4,2),(5,1)}
Thus, n{C} = 15
Probability of C = P{C} = n{C}/ n{S}
= 15/36 = 5/12
∴ P(C) = 5/12
### (xi) a sum more than 7
Solution:
Let X be an event of getting a sum more than 7 from the numbers on the faces of the two dice when thrown together.
∴ X = {(2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3),(5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)}
Thus n{X} = 15
Probability of X = P{X} = n{X}/ n{S} = 15/36 = 5/12
∴ P(X) = 5/12
### (xii) neither a doublet nor a total of 10
Solution:
Let A be the event of getting neither a doublet nor a total of 10 when a pair of dice is thrown simultaneously.
∴ A’ is the event of getting either a doublet or a total of 10 on the faces of the two dices.
∴ A’ = {(1,1), (2,2), (3,3), (4,6), (5,5), (6,4), (6,6)}
Thus, n{A’} = 8
Probability of A’ = n{A’}/ n{S} = 8/36 = 2/9
Probability of A = P{A} = 1−P{A’}
= 1−2/9
∴ P(A’} = 7/9
### (xiii) odd number on the first and 6 on the second
Solution:
Let B be the event of getting an odd number on the first and 6 on the second face of the two dices.
∴ B = {(1,6), (3,6), (5,6)}
Thus n{B} = 3
Probability of B = P{B} = n{B}/ n{S} = 3/36 = 1/12
∴ P(B} = 1/12
### (xiv) a number greater than 4 on each side
Solution:
Let J be the event of getting a number greater than 4 on each side of dice when both dices are thrown simultaneously.
∴ J = {(5,5), (5,6), (6,5), (6,6)}
Thus n{J} = 4
Probability of J = P{J} = n{J}/ n{S} = 4/36 = 1/9
∴ P(J} = 1/9
### (xv) a total of 9 or 11
Solution:
Let I be the event of getting a total of 9 or 11 when two dice are thrown.
∴ I = {(3,6),(4,5),(5,4),(5,6),(6,3),(6,5)}
Thus n{I} = 6
Probability of I = P{I} = n{I}/ n{S} = 6/36 = 1/6
∴ P(I} = 1/6
### (xvi) a total greater than 8
Solution:
Let Z be the event of getting a total greater than 8.
∴ Z = {(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}
∴ n{Z} = 10
Probability of Z = n{Z}/ n{S} = 10/36 = 5/18
∴ P(Z} = 5/18
### Question 4. In a single throw of three dice, find the probability of getting a total of 17 or 18.
Solution:
Since three dice have been thrown simultaneously, the total number of outcomes in that sample space become 6×6×6= 63 = 216.
Let E be the event of getting a total of 17 or 18. Thus E = {(6,6,5), (6,5,6), (5,6,6), (6,6,6)}
Hence, n{E} = 4
Probability of event E = P{E} = n{E}/ n{S} = 4/216 = 1/54
∴ P{E} = 1/54
### Question 5. Three coins are tossed together. Find the probability of getting:
Solution:
Since 2 coins have been tossed together, the total outcomes of the sample space are n{S} =
2×2×2 = 23 = 8.
Let A be the event of getting exactly 2 heads.
∴ A = {HHT, HTH, THH} or, n{A} = 3
Probability of event A = P{A} = n{A}/n{S} = 3/8
∴ P{A} = 3/8
### (ii) at least two heads
Solution:
Let B be the event of getting at least 2 heads when three coins are thrown together.
∴ B = {HHH, HHT, THH, HTH} or, n{B} = 4
Probability of event B = P{B} = n{B}/ n{S} = 4/8 = 1/2
∴ P{B} = 1/2
### (iii) at least one head and one tail
Solution:
Let C be the event of getting at least one head and one tail .
C= {(HHT, THT, HTT, TTH, HTH, THH} or, n{C}= 6
P{C}= n{C}/ N(S) = 6/8 = 3/4
∴ P{C}= 3/4
### Question 6. What is the probability that an ordinary year has 53 Sundays?
Solution:
We know that an ordinary year has 52 weeks and 1 day.
52 weeks in a year obviously implies 52 Sundays.
But in the question, we have to find the probability of having 53 Sundays which means that we have to fine the probability of the last 1 day of an ordinary year to be a Sunday.
Total number of days in a week = Number of outcomes of sample space = n(S) = 7
S= {MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY, SATURDAY, SUNDAY}
Hence, the probability of that one day being a Sunday = 1/7
Thus, the probability of having 53 Sundays in an ordinary year is 1/7.
### Question 7. What is the probability that a leap year has 53 Sundays and 53 Mondays?
Solution:
We know that in a leap year we have 52 weeks and 2 days (366 days)
The sample space for the last 2 days shall be
S= {(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday)} or, n{S}=7
Let E denote the event of getting a Sunday and a Monday as the last 2 days in a leap year.
∴ E= {Sunday, Monday} or, n{E} = 1
Probability of E= P{E}= n{E}/ n{S} = 1/7
Thus, the probability that a leap year has 53 Sundays and 53 Mondays is 1/7.
### (a) all three balls are white
Solution:
Since there are 8+5= 13 balls in the bag and we are required to draw any three balls at random.
Total number of outcomes shall be= n{S}= 13C3 = 286.
Let A be the event that all the three balls drawn are white.
Hence the number of outcomes in event A= n{A} = 5C3= 10.
Probability of event A= n{A}/n{S} = 10/286= 5/143
Hence the probability of drawing all three white balls is 5/143.
### (b) all three balls are red
Solution:
Let R denote the event that all three balls drawn are red.
Since total number of red balls is 8, the number of ways of drawing 3 red balls out of 8 = 8C3 = 56 = n{R}.
We know, n{S}= 286
∴ Probability of R= n{R}/n{S}= 56/286 = 28/143
Hence, the probability of drawing all three red balls is 28/143.
### (c) one ball is red and two balls are white
Solution:
Let E be the event that one ball drawn is red and the other two balls are white.
Number of outcomes of E= 8C1 × 5C2 = 8 ×10 = 80.
∴ Probability of E= n{E}/n{S} = 80/286 = 40/143
Hence the probability of getting one red and other two balls as white is 40/143.
### Question 9. In a single throw of three dice, find the probability of getting same numbers on all the three dice.
Solution:
Since three dice have been rolled together, the total number of outcomes in the sample space= n{S}= 63= 6×6×6= 216.
Let E be the event of getting the same numbers on all the three dice.
∴ E= {(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)}
Number of outcomes in E= n{E} = 6
Probability of E= n{E}/n{S}= 6/216 = 1/36
Hence the probability of getting the same numbers on all the three dice is 1/36.
### Question 10. Two unbiased dice are thrown together. Find the probability that the total of numbers on the dice is greater than 10.
Solution:
Since two dice have been rolled together, the total number of outcomes in the sample space= n{S}= 62= 6×6= 36.
Let E be the event of getting a total of numbers on the dice greater than 10.
∴E= {(5,6), (6,5), (6,6)}
Number of outcomes in E= n{E}= 3
Since n{S}= 52, Probability of E= n{E}/n{S}= 3/36 = 1/12
Hence the probability of getting a total of numbers on the dice greater than 10 is 1/12.
### (i) a black king
Solution:
Since a card is drawn from a pack of 52 cards, so number of elementary events in the sample space = n {S} = 52C1 = 52.
Let E be the event of drawing a black king. Since there are two black kings, one of spade and other of club,
∴n {E} = 2C1 = 2
Since n{S}= 52, Probability of event E = n{E}/n{S} = 2/52 = 1/26.
Hence the probability of drawing a black king is 1/26.
### (ii) either a black card or a king
Solution:
Let A be the event of drawing a black card or a king. We know that there are 26 black cards and 4 kings out of which 2 kings are black.
Hence, total number of outcomes of event A= n{E} = 26C1 + 4C12C1= 28
In this case, 2 needs to be subtracted from total because there are two black kings which are already counted in black cards and to avoid double counting.
Since n{S}= 52, Probability of A= n{A}/n{S} = 28/52 = 7/13.
Hence, the probability of getting either a black card or a king is 7/13.
### (iii) black and a king
Solution:
Let W be the event of drawing a black card and a king. We know there are two black kings, one of spade and other of club.
Hence number of outcomes of W= n{W} = 2C1 = 2.
Since n{S}=52, Probability of W= n{W}/n{S} = 2/52 = 1/26.
Hence the probability of drawing a black card and a king is 1/26.
### (iv) a jack, queen, or a king
Solution:
Let B be the event of drawing a jack, queen or king. We know there are 4 kings, 4 queens and 4 jacks in a deck of cards.
Hence total number of outcomes of event B = n{B} = 4C1 + 4C1 + 4C1 = 12.
Since n{S}= 52, Probability of B = P{B} = n{B}/n{S} = 12/52 = 3/13
Hence the probability of drawing a jack, queen or king is 3/13.
### (v) neither a heart nor a king
Solution:
Let L be the event of drawing neither a heart nor a king and let L′ as the event that either a heart or king appears.
Since there are 13 hearts and 4 kings total number of outcomes = n (L′) = 6C1 + 4C1 – 1=16 ,deducting the king of hearts.
Probability of L’ = P{L′} = n{L’}/n{S} = 16/52 = 4/13
So, P{L} = 1 – P{L’} = 1 – 4/13 = 9/13
Hence the probability of drawing neither a heart nor a king is 9/13.
### (vi) spade or an ace
Solution:
Let D be the event of drawing a spade or king. We know there are 13 spades and 4 kings, but 1 king is already included in the 4 kings.
Number of outcomes in D = n{D} = 13C1 + 4C1 – 1=16
Since N{S} = 52, Probability of event D = P{D} = n{D}/n{S} = 16/52 = 4/13
Hence the probability of drawing either a spade or an ace is 4/13.
### (vii) neither an ace nor a king
Solution:
Let K be the event of drawing neither an ace nor a king and K′ as the event that either an ace or king appears.
Number of outcomes in K’ = n{K’} = 4C1 + 4C1 = 8
Since n{S}= 52, Probability of K’ = P{K’} = n{K’}/n{S} = 8/52= 2/13
So, P{K} = 1 – P{K’} = 1 – 2/13= 11/13
Hence the probability of drawing neither an ace nor a king is 11/13.
### (viii) a diamond card
Solution:
Let X be the event of drawing a diamond card. We know there are 13 diamond cards.
Hence number of outcomes of event X = n{X} = 13C1 = 13.
Since n{S} = 52, Probability of event X = P{X} = n{X}/n{S} = 13/52 = 1/4
Hence the probability of drawing a diamond card is 1/4.
### (ix) not a diamond card
Solution:
Let X be the event of drawing not a diamond card and X′ as the event that diamond card appears. We know there are 13 diamond cards.
Hence number of outcomes of event X = n{X} = 13C1 = 13.
Since n{S} = 52, Probability of event X = P{X} = n{X}/n{S} = 13/52 = 1/4
So, P{X’} = 1 – P{X’} = 1 – 1/4 = 3/4
Hence the probability of not drawing a diamond card is 3/4.
### (x) a black card
Solution:
Let E be the event of drawing a black card. We know there are 26 black cards (spades and clubs).
Hence total number of outcomes of event E = n{E} = 26C1 = 26
Since n{S} = 52, Probability of event E = P{E} = n{E}/n{S} = 26/52= 1/2
Hence the probability of drawing a black card is 1/2.
### (xi) not an ace
Solution:
Let E be the event of drawing not an ace and E′ as the event that ace card appears. We know that there are 4 ace cards.
Number of outcomes in E’ = n{E’} = 4C1 = 4.
Since n{S} = 52, Probability of event E’ = P{E’} = n{E’}/n{S} = 4/52 = 1/13
So, P (E) = 1 – P (E′) = 1 – 1/13 =12/13
Hence the probability of not drawing an ace is 12/13.
### (xii) not a black card
Solution:
Let E be the event of not drawing a black card. We know there are 26 cards other than black cards (red cards of hearts and diamonds)
Hence number of outcomes of drawing a red card = n{E} = 26C1 = 26 .
Since n{S} = 52, Probability of event E = P{E} = n{E}/n{S} = 26/52 = 1/2.
Hence the probability of not drawing a black card is 1/2.
### Question 12. In shuffling a pack of 52 playing cards, four are accidentally dropped. Find the chance that the missing cards should be one from each suit.
Solution:
A pack of 52 cards from which 4 are dropped. We now have to find the probability that the missing cards should be one from each suit.
We know that, from well shuffled pack of cards, 4 cards missed out total possible outcomes = n{S} = 52C4 = 270725
Let E be the event that four missing cards are from each suite
Hence total number of outcomes of event E = n{E} = 13C1 × 13C1 × 13C1 × 13C1 = 134
Probability of event E =P{E} = n{E}/n{S} = 134/270725 = 2197/20825
Hence the probability that the missing cards should be one from each suit is 2197/20825.
### Question 13. From a deck of 52 cards, four cards are drawn simultaneously. Find the chance that they will be the four honors of the same suit.
Solution:
We have to find the probability that all the face cards drawn from a pack of 52 cards are of same suits.
Total possible outcomes in sample space =n{S} = 52C4
Let E be the event that all the cards drawn are face cards of same suit.
Hence number of outcomes of event E = n (E)= 4 × 4C4 = 4 × 1 = 4.
Probability of event E =P{E} = n{E}/n{S}= 4/270725
Hence the probability that the face cards drawn from a pack of 52 cards are of same suits is 4/270725.
### Question 14. Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?
Solution:
We have to find the probability that the ticket has a number which is a multiple of 3 or 7.
Total possible outcomes in sample space =n{S} = 20C1 = 20.
Let E be the event of getting ticket which has number that is multiple of 3 or 7.
Hence outcomes of event E are {3,6,9,12,15,18,7,14}. Number of outcomes of event E are n{E} = 8.
Probability of event E =P{E} = n{E}/n{S}= 8/20 = 4/5
Therefore, the probability that the ticket has a number which is a multiple of 3 or 7 is 4/5.
### Question 15. A bag contains 6 red, 4 white, and 8 blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue.
Solution:
We have to find the probability that the one is red, one is white and one is blue.
Since three balls are drawn so, total number of outcomes for drawing 3 balls is n{S} = 18C3 = 816
Let E be the event that one red, one white and one blue ball is drawn.
n{E} = 6C1 × 4C1 × 8C1 = 192
Probability of event E =P{E} = n{E}/n{S} = 192 / 816 = 4/17
Therefore, the probability that one is red, one is white and one is blue is 4/17.
My Personal Notes arrow_drop_up | 6,837 | 19,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-33 | latest | en | 0.9175 |
https://us.metamath.org/mpeuni/df-bj-sethom.html | 1,720,875,823,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00886.warc.gz | 506,320,706 | 3,883 | Mathbox for BJ < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > df-bj-sethom Structured version Visualization version GIF version
Definition df-bj-sethom 34533
Description: Define the set of functions (morphisms of sets) between two sets. Same as df-map 8395 with arguments swapped. TODO: prove the same staple lemmas as for ↑m. Remark: one may define Set⟶ = (𝑥 ∈ dom Struct , 𝑦 ∈ dom Struct ↦ {𝑓 ∣ 𝑓:(Base‘𝑥)⟶(Base‘𝑦)}) so that for morphisms between other structures, one could write ... = {𝑓 ∈ (𝑥 Set⟶ 𝑦) ∣ ...}. (Contributed by BJ, 11-Apr-2020.)
Assertion
Ref Expression
df-bj-sethom Set⟶ = (𝑥 ∈ V, 𝑦 ∈ V ↦ {𝑓𝑓:𝑥𝑦})
Distinct variable group: 𝑥,𝑓,𝑦
Detailed syntax breakdown of Definition df-bj-sethom
StepHypRef Expression
1 csethom 34532 . 2 class Set
2 vx . . 3 setvar 𝑥
3 vy . . 3 setvar 𝑦
4 cvv 3444 . . 3 class V
52cv 1537 . . . . 5 class 𝑥
63cv 1537 . . . . 5 class 𝑦
7 vf . . . . . 6 setvar 𝑓
87cv 1537 . . . . 5 class 𝑓
95, 6, 8wf 6324 . . . 4 wff 𝑓:𝑥𝑦
109, 7cab 2779 . . 3 class {𝑓𝑓:𝑥𝑦}
112, 3, 4, 4, 10cmpo 7141 . 2 class (𝑥 ∈ V, 𝑦 ∈ V ↦ {𝑓𝑓:𝑥𝑦})
121, 11wceq 1538 1 wff Set⟶ = (𝑥 ∈ V, 𝑦 ∈ V ↦ {𝑓𝑓:𝑥𝑦})
Colors of variables: wff setvar class This definition is referenced by: (None)
Copyright terms: Public domain W3C validator | 582 | 1,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.518865 |
https://analystnotes.com/cfa_question.php?p=AQHEROUPT | 1,679,686,032,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945288.47/warc/CC-MAIN-20230324180032-20230324210032-00513.warc.gz | 125,459,444 | 6,170 | ### CFA Practice Question
A firm has a Euro 100 million loan that requires it to pay LIBOR + 75 basis points. LIBOR is currently 4.3%. The firm is apprehensive of rising interest rates, and therefore enters into a 3 year swap agreement that requires it to pay fixed 5.2% and receive LIBOR every 90 days with the notional amount Euro 100 million. Fixed payments are calculated assuming 360 days in a year, whereas floating payments (both loan and swap) are calculated using 365 days in a year. LIBOR on the first settlement date is 3.95%. The net interest payment (both swap and loan) on the first settlement date is:
A. 1,558,904
B. 2,545,205
C. 1,484,932
Explanation: Fixed payments is Euro 100 million * 5.2% * (90/360) = Euro 1,300,000
Floating payment is Euro 100 million * (4.3% + 0.75% - 4.3%) * (90/365) = Euro 184,932
Total payments = Fixed + Floating = 1,484,932
It is important to remember that the payments on the first settlement date are made on the LIBOR today, rather than LIBOR that prevails at that time.
User Comment
moll It actually does not matter which LIBOR to use in this question, as the two rates are canceled out. However if we were required to calculate the floating payments alone then it would matter.
jpducros Under the loan, the firm has to pay 4,3+0,75 but receives 4,3 from the floating leg of the swap. Here both are done in the same calculation. | 375 | 1,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-14 | latest | en | 0.960873 |
https://hackage.haskell.org/package/strict-base-types-0.6.1/docs/Data-Tuple-Strict.html | 1,558,657,152,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00494.warc.gz | 493,807,821 | 6,743 | strict-base-types-0.6.1: Strict variants of the types provided in base.
Copyright (c) 2006-2007 Roman Leshchinskiy(c) 2013 Simon Meier BSD-style (see the file LICENSE) Simon Meier experimental GHC None Haskell98
Data.Tuple.Strict
Contents
Description
The strict variant of the standard Haskell pairs and the corresponding variants of the functions from Data.Tuple.
Synopsis
# Documentation
data Pair a b :: * -> * -> * #
The type of strict pairs.
Constructors
!a :!: !b infixl 2
Instances
fst :: Pair a b -> a #
Extract the first component of a strict pair.
snd :: Pair a b -> b #
Extract the second component of a strict pair.
curry :: (Pair a b -> c) -> a -> b -> c #
Curry a function on strict pairs.
uncurry :: (a -> b -> c) -> Pair a b -> c #
Convert a curried function to a function on strict pairs.
swap :: Pair a b -> Pair b a Source #
Analagous to swap from Data.Tuple
zip :: [a] -> [b] -> [Pair a b] Source #
Zip for strict pairs (defined with zipWith).
unzip :: [Pair a b] -> ([a], [b]) Source #
Unzip for stict pairs into a (lazy) pair of lists.
# Orphan instances
Source # Methodsbitraverse :: Applicative f => (a -> f c) -> (b -> f d) -> Pair a b -> f (Pair c d) # Source # Methodsbifold :: Monoid m => Pair m m -> m #bifoldMap :: Monoid m => (a -> m) -> (b -> m) -> Pair a b -> m #bifoldr :: (a -> c -> c) -> (b -> c -> c) -> c -> Pair a b -> c #bifoldl :: (c -> a -> c) -> (c -> b -> c) -> c -> Pair a b -> c # Source # Methodsbimap :: (a -> b) -> (c -> d) -> Pair a c -> Pair b d #first :: (a -> b) -> Pair a c -> Pair b c #second :: (b -> c) -> Pair a b -> Pair a c # Source # Methodsswapped :: (Profunctor p, Functor f) => p (Pair b a) (f (Pair d c)) -> p (Pair a b) (f (Pair c d)) # Functor (Pair e) Source # Methodsfmap :: (a -> b) -> Pair e a -> Pair e b #(<\$) :: a -> Pair e b -> Pair e a # Foldable (Pair e) Source # Methodsfold :: Monoid m => Pair e m -> m #foldMap :: Monoid m => (a -> m) -> Pair e a -> m #foldr :: (a -> b -> b) -> b -> Pair e a -> b #foldr' :: (a -> b -> b) -> b -> Pair e a -> b #foldl :: (b -> a -> b) -> b -> Pair e a -> b #foldl' :: (b -> a -> b) -> b -> Pair e a -> b #foldr1 :: (a -> a -> a) -> Pair e a -> a #foldl1 :: (a -> a -> a) -> Pair e a -> a #toList :: Pair e a -> [a] #null :: Pair e a -> Bool #length :: Pair e a -> Int #elem :: Eq a => a -> Pair e a -> Bool #maximum :: Ord a => Pair e a -> a #minimum :: Ord a => Pair e a -> a #sum :: Num a => Pair e a -> a #product :: Num a => Pair e a -> a # Source # Methodstraverse :: Applicative f => (a -> f b) -> Pair e a -> f (Pair e b) #sequenceA :: Applicative f => Pair e (f a) -> f (Pair e a) #mapM :: Monad m => (a -> m b) -> Pair e a -> m (Pair e b) #sequence :: Monad m => Pair e (m a) -> m (Pair e a) # (Data a, Data b) => Data (Pair a b) Source # Methodsgfoldl :: (forall d c. Data d => c (d -> c) -> d -> c c) -> (forall g. g -> c g) -> Pair a b -> c (Pair a b) #gunfold :: (forall c r. Data c => c (c -> r) -> c r) -> (forall r. r -> c r) -> Constr -> c (Pair a b) #toConstr :: Pair a b -> Constr #dataTypeOf :: Pair a b -> DataType #dataCast1 :: Typeable (* -> *) t => (forall d. Data d => c (t d)) -> Maybe (c (Pair a b)) #dataCast2 :: Typeable (* -> * -> *) t => (forall d e. (Data d, Data e) => c (t d e)) -> Maybe (c (Pair a b)) #gmapT :: (forall c. Data c => c -> c) -> Pair a b -> Pair a b #gmapQl :: (r -> r' -> r) -> r -> (forall d. Data d => d -> r') -> Pair a b -> r #gmapQr :: (r' -> r -> r) -> r -> (forall d. Data d => d -> r') -> Pair a b -> r #gmapQ :: (forall d. Data d => d -> u) -> Pair a b -> [u] #gmapQi :: Int -> (forall d. Data d => d -> u) -> Pair a b -> u #gmapM :: Monad m => (forall d. Data d => d -> m d) -> Pair a b -> m (Pair a b) #gmapMp :: MonadPlus m => (forall d. Data d => d -> m d) -> Pair a b -> m (Pair a b) #gmapMo :: MonadPlus m => (forall d. Data d => d -> m d) -> Pair a b -> m (Pair a b) # Generic (Pair a b) Source # Associated Typestype Rep (Pair a b) :: * -> * # Methodsfrom :: Pair a b -> Rep (Pair a b) x #to :: Rep (Pair a b) x -> Pair a b # (Semigroup a, Semigroup b) => Semigroup (Pair a b) Source # Methods(<>) :: Pair a b -> Pair a b -> Pair a b #sconcat :: NonEmpty (Pair a b) -> Pair a b #stimes :: Integral b => b -> Pair a b -> Pair a b # (Monoid a, Monoid b) => Monoid (Pair a b) Source # Methodsmempty :: Pair a b #mappend :: Pair a b -> Pair a b -> Pair a b #mconcat :: [Pair a b] -> Pair a b # (Arbitrary a, Arbitrary b) => Arbitrary (Pair a b) Source # Methodsarbitrary :: Gen (Pair a b) #shrink :: Pair a b -> [Pair a b] # (Hashable a, Hashable b) => Hashable (Pair a b) Source # MethodshashWithSalt :: Int -> Pair a b -> Int #hash :: Pair a b -> Int # (ToJSON a, ToJSON b) => ToJSON (Pair a b) Source # MethodstoJSON :: Pair a b -> Value #toEncoding :: Pair a b -> Encoding #toJSONList :: [Pair a b] -> Value #toEncodingList :: [Pair a b] -> Encoding # (FromJSON a, FromJSON b) => FromJSON (Pair a b) Source # MethodsparseJSON :: Value -> Parser (Pair a b) #parseJSONList :: Value -> Parser [Pair a b] # (Binary a, Binary b) => Binary (Pair a b) Source # Methodsput :: Pair a b -> Put #get :: Get (Pair a b) #putList :: [Pair a b] -> Put # (NFData a, NFData b) => NFData (Pair a b) Source # Methodsrnf :: Pair a b -> () # Strict (a, b) (Pair a b) Source # Methodsstrict :: Iso' (a, b) (Pair a b) # ((~) * a a', (~) * b b') => Each (Pair a a') (Pair b b') a b Source # Methodseach :: Traversal (Pair a a') (Pair b b') a b # Field1 (Pair a b) (Pair a' b) a a' Source # Methods_1 :: Lens (Pair a b) (Pair a' b) a a' # Field2 (Pair a b) (Pair a b') b b' Source # Methods_2 :: Lens (Pair a b) (Pair a b') b b' # | 1,926 | 5,618 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-22 | latest | en | 0.708028 |
https://mathematica.stackexchange.com/questions/tagged/plotting?sort=active&page=3 | 1,566,416,467,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316150.53/warc/CC-MAIN-20190821174152-20190821200152-00007.warc.gz | 531,915,304 | 38,671 | Questions tagged [plotting]
Questions on creating visualizations from functions or data using high-level constructors such as Plot, ListPlot, Histogram, etc.
10,226 questions
Filter by
Sorted by
Tagged with
96 views
Unexpected results from ListContourPlot
Bug introduced in 8 or earlier and persisting through 12.0 I want to use ListContourPlot and I have lists such that, ...
18 views
Getting error message RegionPlot3D::boolf
I want to visualize the bounded region for a linear programming problem. I used the folowwing, however it gives me an error : ...
6k views
ListPointPlot3D with connected points?
With the command ListPlot[data,Joined -> True] I can visually connect adjacent data points in the array with lines. However, this option appears to be missing ...
54 views
plot each column separately [duplicate]
I have an 800*161 matrix and I am trying to plot each column with respect to the 1st column.so I tried the following code ...
179 views
How to put the parent node higher than the children nodes in a drawing for a graph?
I have a tree as below. Graph[{7 -> 2, 2 -> 3, 3 -> 4, 3 -> 5}, VertexLabels -> "Name"] However, I want the drawing would put parent nodes at a ...
49 views
Graph lines with points at the ends
I want to graph ten lines (of a single color) with points at the ends. The points are "cells" and the lines are "veins". I have tried it as follows. ...
75 views
Opacity or related issue with DensityPlot3D
I'm trying to make use of the function DensityPlot3D with the following specific example ...
201 views
Graphing a solid between two structures using spherical coordinates
How can I graph the solid that lies above the cone $z=\sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=z$? I tried ...
75 views
How to plot a Histogram and compare with a particular probability density function PDF
I would like to know how to plot a make a comparison between a histogram and a set of random data, I have the following probability density function (PDF) which is given for this problem. I can ...
28 views
How do I add a plot label in this case?
I am fitting these data: ...
315 views
ListPlot draws nothing
Consider the following simplified example: ...
142 views
To plot branch cut of logarithm
I like to see the branch cut of the function: $$1 - z \ln[(1+z)/z].$$ If I plot it in the complex plane: ...
173 views
Show two plots together: a two dimensional curve tangent to the maxima of a three dimensional plot
I have a list containing 3 columns and 6552 rows which can be found here. The plot of data is shown below: We have a cross cut ...
106 views
37 views
ListDensityPlot and interpolation order
I am working on a project that takes arrays of data and visualizes them using ListDensityPlot. I noticed, when I imported an array from a CSV file, the last column ...
651 views
Plotting in polar coordinates, simple implicit curves
How can I plot $\theta=\cos r$ in polar plane? Of course I know that it is different from $r=\cos \theta.$
171 views
What is the best way to use errors in mathematica M12?
Since ErrorListPlot has been superseded by new functionality in ListPlot for Mathematica 12.0, what is the best way to handle ...
801 views
Polar contour plot in Mathematica?
I am following a text on fluid mechanics with MAPLE examples. I want to do the following ContourPlot in Mathematica in Polar coordinates: (r^2-\frac{a^3}{r}) \...
351 views
How to show decimal points on the axes frames in mathematica?
I have a set of data g1 and when plot using n1, i got the plot shown in figure below. I need to change the x-axis label from (1,2,3,4,5...) to (1.0,2.0,3.0,4.0,5.0) and y-axis label from (50000,...
55 views
Plot Using User-defined Coordinates
Suppose instead of using Cartesian coordinate $\vec{x}=(1, 0)$ and $\vec{y}=(0, 1)$, I want to define new coordiates $\vec{a} = (1,0)$ and $\vec{b} = (\frac{1}{2},\frac{\sqrt{3}}{2} )$. And when I ...
60 views
Vector field Stream?
The following graph shows the vector plot. How can I plot the solid lines which represents how the vectors are spatially varying and both on the same graph (vector plot and stream field)? ...
23 views
Plotting an NDEigensystem solution over a helicoid
I am trying to plot the solutions of a particular Hamiltonian eigensystem over a helicoid. In other words, my solutions are a function of the radial variables (r,$\phi$). The Helicoid is given by the ...
58 views
How to plot special contour lines for a “Ridge System” of complex function?
Mathematica makes it very easy to to plot the contour lines for a function of two real variables using ContourPlot. It also makes it very easy to plot the ...
128 views
...
183 views
How to plot a function for various values of 2 constants
I have a function like this: ...
103 views
Does ListPlot Joined Dotted crash front end reproducibly for large number of points?
The problem The following code crashes Mathematica 12 for Windows 64 front-end reproducibly in my computer (Win7 Pro, i7-4770 3.4GHz 16Gb RAM). ...
41 views
How can I Plot lines in different colors from a data in Graphics 3D?
I want to plot all these different line in different colors to visualize them better. But all I get is the black color for all of them. ...
56 views
Callout mystery
I have this function: de[a_, x_] = -1 + (E^-x x^a)/((-a + x) Gamma[a, x]); When I do ...
928 views
Plotting vector field in polar coordinates
Is there an easy way(other than painfully decomposing into Cartesian components) to plot a vector field in polar coordinates? Basically I have a vector function $\vec V (r,\phi,z)$. How do I plot it ?
93 views
116 views
Can you dimension the frame in a plot with absolute unit e.g. cm? [duplicate]
Is it possible to set the height and width of a plot in absolute units for in $\rm{cm}$? Ideally I'd specifically like to control the frame size for when the plot option ...
43 views
Restrictions as string to expression in region plots
I'm trying to build a function that generate a string of restrictons to pass on to a region plot with ToExpression, but I think it fails to pass the plot variables. ...
58 views
How to get separate ListPlots into a single ListPlot with legends and labels
I am currently getting 2 seperate plots for the stress variation with t as shown in the picture below. I would like to get them in a single plot with legend to distinguish them. Could someone please ...
247 views
How can I improve the speed of computing region intersection?
I have a polygon defined by a list of nodes (x,y). I want to cut the polygon by a horizontal line at position y = a and get the new polygon above the position y = a. I am using the RegionIntersect ...
96 views
How can I combine two contour plots into one and give it a bar legend?
I have these contour plots: ...
31 views
Mathematica Scalar Field on 3D Space
Fellow Mathematicai, is there any way to create a visual plot of a Scalar Field on 3D space in Mathematica? With Gratitude, - SDH
30 views
how to change axis range [closed]
I wanted to plot a function with different form of axis range
85 views
Why call to StreamPlot terminates kernel each time?
FYI, reported to WRI [CASE:4288967] By mistake, I put x range starting from negative to make StreamPlot for expression with ...
57 views
why this call to StreamPlot returns unevaluated? (no error message, no beep)
I am really baffled by this. Call to StreamPlot does not return empty plot, nor an error message, nor a beep. It just return unevaluated. I do not think I've seen ...
41 views
Combine a List of Plots: Using MapThread with Show [closed]
I have a list of graphics p1 = Plot[x Sin[x] - 1, {x, -2.5, 2.5}] p2 = Plot[Evaluate[D[x Sin[x] - 1, x]], {x, -2.5, 2.5},PlotStyle -> Red] ListNew = {p1, p2}; ...
121 views
Rescaling and Image Padding Issues with GraphicsRow
I am trying to get a row of plots (which include legends and labels) using GraphicsRow, with vertical heights of the plots equal. I manually set the size and image padding of these plots so that the ... | 2,007 | 8,038 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-35 | latest | en | 0.927201 |
https://egvideos.com/video/new-mexico/grade-1/math/1.g.3/partitioning-into-equal-shares | 1,669,622,590,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00145.warc.gz | 272,335,520 | 10,045 | # New Mexico - Grade 1 - Math - Geometry - Partitioning Into Equal Shares - 1.G.3
### Description
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
• State - New Mexico
• Standard ID - 1.G.3
• Subjects - Math Common Core
• Math
• Geometry
## More New Mexico Topics
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral. | 589 | 2,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-49 | longest | en | 0.904328 |
http://planetmath.org/1121thealgebraicstructureofdedekindreals | 1,521,511,687,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647251.74/warc/CC-MAIN-20180320013620-20180320033620-00013.warc.gz | 218,019,070 | 9,979 | # 11.2.1 The algebraic structure of Dedekind reals
The construction of the algebraic and order-theoretic structure of Dedekind reals proceeds as usual in intuitionistic logic. Rather than dwelling on details we point out the differences between the classical and intuitionistic setup. Writing $L_{x}$ and $U_{x}$ for the lower and upper cut of a real number $x:\mathbb{R}_{\mathsf{d}}$, we define addition as
$\displaystyle L_{x+y}(q)$ $\displaystyle:\!\!\equiv\exists(r,s:\mathbb{Q}).\,L_{x}(r)\land L_{y}(s)\land q% =r+s,$ $\displaystyle U_{x+y}(q)$ $\displaystyle:\!\!\equiv\exists(r,s:\mathbb{Q}).\,U_{x}(r)\land U_{y}(s)\land q% =r+s,$
$\displaystyle L_{-x}(q)$ $\displaystyle:\!\!\equiv\exists(r:\mathbb{Q}).\,U_{x}(r)\land q=-r,$ $\displaystyle U_{-x}(q)$ $\displaystyle:\!\!\equiv\exists(r:\mathbb{Q}).\,L_{x}(r)\land q=-r.$
With these operations $(\mathbb{R}_{\mathsf{d}},0,{+},{-})$ is an abelian group. Multiplication is a bit more cumbersome:
$\displaystyle L_{x\cdot y}(q)$ \displaystyle:\!\!\equiv\begin{aligned} \displaystyle\exists(a,b,c,d:\mathbb{Q% }).&\displaystyle L_{x}(a)\land U_{x}(b)\land L_{y}(c)\land U_{y}(d)\land\\ &\displaystyle\qquad q<\min(a\cdot c,a\cdot d,b\cdot c,b\cdot d),\end{aligned} $\displaystyle U_{x\cdot y}(q)$ \displaystyle:\!\!\equiv\begin{aligned} \displaystyle\exists(a,b,c,d:\mathbb{Q% }).&\displaystyle L_{x}(a)\land U_{x}(b)\land L_{y}(c)\land U_{y}(d)\land\\ &\displaystyle\qquad\max(a\cdot c,a\cdot d,b\cdot c,b\cdot d)
These formulas are related to multiplication of intervals in interval arithmetic, where intervals $[a,b]$ and $[c,d]$ with rational endpoints multiply to the interval
$[a,b]\cdot[c,d]=[\min(ac,ad,bc,bd),\max(ac,ad,bc,bd)].$
For instance, the formula for the lower cut can be read as saying that $q when there are intervals $[a,b]$ and $[c,d]$ containing $x$ and $y$, respectively, such that $q$ is to the left of $[a,b]\cdot[c,d]$. It is generally useful to think of an interval $[a,b]$ such that $L_{x}(a)$ and $U_{x}(b)$ as an approximation ofΒ $x$, see \autorefex:RD-interval-arithmetic.
We now have a commutative ring with unit $(\mathbb{R}_{\mathsf{d}},0,1,{+},{-},{\cdot})$. To treat multiplicative inverses, we must first introduce order. Define $\leq$ and $<$ as
$\displaystyle(x\leq y)$ $\displaystyle\ :\!\!\equiv\ \forall(q:\mathbb{Q}).\,L_{x}(q)\Rightarrow L_{y}(% q),$ $\displaystyle(x $\displaystyle\ :\!\!\equiv\ \exists(q:\mathbb{Q}).\,U_{x}(q)\land L_{y}(q).$
###### Lemma 11.2.1.
For all $x:\mathbb{R}_{\mathsf{d}}$ and $q:\mathbb{Q}$, $L_{x}(q)\Leftrightarrow(q and $U_{x}(q)\Leftrightarrow(x.
###### Proof.
If $L_{x}(q)$ then by roundedness there merely is $r>q$ such that $L_{x}(r)$, and since $U_{q}(r)$ it follows that $q. Conversely, if $q then there is $r:\mathbb{Q}$ such that $U_{q}(r)$ and $L_{x}(r)$, hence $L_{x}(q)$ because $L_{x}$ is a lower set. The other half of the proof is symmetric. β
The relation $\leq$ is a partial order, and $<$ is transitive and irreflexive. Linearity
$(x
is valid if we assume excluded middle, but without it we get weak linearity
$(x (11.2.2)
At first sight it might not be clear whatΒ (11.2.2) has to do with linear order. But if we take $x\equiv u-\epsilon$ and $y\equiv u+\epsilon$ for $\epsilon>0$, then we get
$(u-\epsilon
This is linearity βup to a small numerical errorβ, i.e., since it is unreasonable to expect that we can actually compute with infinite precision, we should not be surprised that we can decideΒ $<$ only up to whatever finite precision we have computed.
To see thatΒ (11.2.2) holds, suppose $x. Then there merely exists $q:\mathbb{Q}$ such that $U_{x}(q)$ and $L_{y}(q)$. By roundedness there merely exist $r,s:\mathbb{Q}$ such that $r, $U_{x}(r)$ and $L_{y}(s)$. Then, by locatedness $L_{z}(r)$ or $U_{z}(s)$. In the first case we get $x and in the second $z.
Classically, multiplicative inverses exist for all numbers which are different from zero. However, without excluded middle, a stronger condition is required. Say that $x,y:\mathbb{R}_{\mathsf{d}}$ are apart from each other, written $x\mathrel{\#}y$, when $(x:
$(x\mathrel{\#}y):\!\!\equiv(x
If $x\mathrel{\#}y$, then $\lnot(x=y)$. The converse is true if we assume excluded middle, but is not provable constructively. Indeed, if $\lnot(x=y)$ implies $x\mathrel{\#}y$, then a little bit of excluded middle follows; see \autorefex:reals-apart-neq-MP.
###### Theorem 11.2.3.
A real is invertible if, and only if, it is apart from $0$.
###### Remark 11.2.4.
We observe that a real is invertible if, and only if, it is merely invertible. Indeed, the same is true in any ring, since a ring is a set, and multiplicative inverses are unique if they exist. See the discussion following \autorefcor:UC.
###### Proof.
Suppose $x\cdot y=1$. Then there merely exist $a,b,c,d:\mathbb{Q}$ such that $a, $c and $0<\min(ac,ad,bc,bd)$. From $0 and $0 it follows that $a$, $b$, and $c$ are either all positive or all negative. Hence either $0 or $x, so that $x\mathrel{\#}0$.
Conversely, if $x\mathrel{\#}0$ then
$\displaystyle L_{x^{-1}}(q)$ $\displaystyle:\!\!\equiv\exists(r:\mathbb{Q}).\,U_{x}(r)\land((0 $\displaystyle U_{x^{-1}}(q)$ $\displaystyle:\!\!\equiv\exists(r:\mathbb{Q}).\,L_{x}(r)\land((01)% \lor(r<0\land 1>qr))$
defines the desired inverse. Indeed, $L_{x^{-1}}$ and $U_{x^{-1}}$ are inhabited because $x\mathrel{\#}0$. β
The archimedean principle can be stated in several ways. We find it most illuminating in the form which says that $\mathbb{Q}$ is dense in $\mathbb{R}_{\mathsf{d}}$.
###### Theorem 11.2.5 (Archimedean principle for $\mathbb{R}_{\mathsf{d}}$).
For all $x,y:\mathbb{R}_{\mathsf{d}}$ if $x then there merely exists $q:\mathbb{Q}$ such that $x.
###### Proof.
By definition of $<$. β
Before tackling completeness of Dedekind reals, let us state precisely what algebraic structure they possess. In the following definition we are not aiming at a minimal axiomatization, but rather at a useful amount of structure and properties.
###### Definition 11.2.6.
An ordered field is a set $F$ together with constants $0$, $1$, operations $+$, $-$, $\cdot$, $\min$, $\max$, and mere relations $\leq$, $<$, $\mathrel{\#}$ such that:
1. 1.
$(F,0,1,{+},{-},{\cdot})$ is a commutative ring with unit;
2. 2.
$x:F$ is invertible if, and only if, $x\mathrel{\#}0$;
3. 3.
$(F,{\leq},{\min},{\max})$ is a lattice;
4. 4.
the strict order $<$ is transitive, irreflexive, and weakly linear ($x);
5. 5.
apartness $\mathrel{\#}$ is irreflexive, symmetric and cotransitive ($x\mathrel{\#}y\Rightarrow x\mathrel{\#}z\lor y\mathrel{\#}z$);
6. 6.
for all $x,y,z:F$:
$\displaystyle x\leq y$ $\displaystyle\Leftrightarrow\lnot(y $\displaystyle x $\displaystyle\Rightarrow x $\displaystyle x\mathrel{\#}y$ $\displaystyle\Leftrightarrow(x $\displaystyle x\leq y $\displaystyle\Rightarrow x $\displaystyle x\leq y$ $\displaystyle\Leftrightarrow x+z\leq y+z,$ $\displaystyle x\leq y\land 0\leq z$ $\displaystyle\Rightarrow xz\leq yz,$ $\displaystyle x $\displaystyle\Leftrightarrow x+z $\displaystyle 0 $\displaystyle\Leftrightarrow xz $\displaystyle 0 $\displaystyle\Rightarrow 0 $\displaystyle 0$ $\displaystyle<1.$
Every such field has a canonical embedding $\mathbb{Q}\to F$. An ordered field is when for all $x,y:F$, if $x then there merely exists $q:\mathbb{Q}$ such that $x.
###### Theorem 11.2.7.
The Dedekind reals form an ordered archimedean field.
###### Proof.
We omit the proof in the hope that what we have demonstrated so far makes the theorem plausible. β
Title 11.2.1 The algebraic structure of Dedekind reals \metatable | 2,518 | 7,623 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 156, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-13 | latest | en | 0.622282 |
https://1hotelsouthbeachforsale.com/qa/question-what-is-isohyetal-line.html | 1,627,231,209,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00321.warc.gz | 93,829,069 | 8,201 | # Question: What Is Isohyetal Line?
## How do you solve rainfall intensity?
Rainfall intensities can be accurately measured by means of a continuously recording autographic rain gauge.
It is also possible to time the length of individual rainstorms and to calculate the average intensities by dividing the measured rainfall depths by the corresponding duration of the storms..
## What is another name of Isolines?
The Basics of Isolines and Contour Lines Isolines, also referred to as contour lines, can be used to represent elevation on a map by connecting points of equal elevation, for instance.
## What is Isohypse?
Isohypse (aka height contour) A line of equal geopotential height. Geopotential assumes the earth is perfectly flat and a perfect sphere. The geopotential height is the distance above the Earth’s surface if it was a perfect and flat sphere. Isohypse are shown on a constant pressure surface.
## What are Isohyets and isobars?
Isobars are used to show the distribution of air pressure . Some common isolines encountered in physical geography are: isotherm: a line connecting points of equal temperature. isohyet: a line that connects points of equal precipitation. … isotach: a line of equal wind speed.
## What is Isohyetal method?
The isohyetal method is used to estimate the mean precipitation across an area by drawing lines of equal precipitation. The method uses topographic and other data to yield reliable estimates. Isohyets are contours of equal precipitation analogous to contour lines on a topographic map.
## What do you mean by contour line?
Contour line, a line on a map representing an imaginary line on the land surface, all points of which are at the same elevation above a datum plane, usually mean sea level. map: contour lines. The diagram illustrates how contour lines show relief by joining points of equal elevation.
## What is the space between contour lines called?
Contour lines–usually curved, parallel brown lines–show the shape and elevation of the land by connecting points of equal elevation. … The space between the contour lines is called the contour interval and represents a specific (set) distance.
## Can contour lines cross?
Contour lines can never cross one another. Each line represents a separate elevation, and you can’t have two different elevations at the same point.
## What are 3 types of contour lines?
There are 3 kinds of contour lines you’ll see on a map: intermediate, index, and supplementary.Index lines are the thickest contour lines and are usually labeled with a number at one point along the line. … Intermediate lines are the thinner, more common, lines between the index lines.More items…
## How is annual rainfall calculated?
(i) Annual rainfall = Sum of rainfall in all twelve months. Therefore, annual rainfall is 129.2 cm. The total annual rainfall = Sum of rainfall in all twelve months. Therefore, annual rainfall is 128.7 cm.
## What are the 5 Rules of contour lines?
Rule 1 – every point of a contour line has the same elevation. Rule 2 – contour lines separate uphill from downhill. Rule 3 – contour lines do not touch or cross each other except at a cliff. Rule 4 – every 5th contour line is darker in color.
## Why do we use contour lines?
The purpose of contour lines is to represent the tridimensional shape of the terrestrial surface on a bidimensional map. Contour lines are the intersection of an horizontal plane parallel to the reference level and the topographical surface to describe. Consequently: Contour lines are always closed curves.
## How do we use contour lines?
Contour lines are lines drawn on a map connecting points of equal elevation, meaning if you physically followed a contour line, elevation would remain constant. Contour lines show elevation and the shape of the terrain. They’re useful because they illustrate the shape of the land surface — its topography — on the map.
## What are the features of contour lines?
Other characteristics of contour lines are: – Horizontal distance between contour lines is inversely proportional to slope. – Uniform slopes have uniformly spaced lines. – Along plane surfaces, contour lines are straight and parallel. – Contour lines are perpendicular to lines of steepest slopes.
## What is Isoheight?
An isoheight or isohypse is a line of constant geopotential height on a constant pressure surface chart. Isohypse and isoheight are simply known as lines showing equal pressure on a map. | 923 | 4,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-31 | latest | en | 0.904583 |
https://plus.maths.org/content/taxonomy/term/272 | 1,529,756,702,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864958.91/warc/CC-MAIN-20180623113131-20180623133131-00032.warc.gz | 656,934,032 | 7,408 | ## Fibonacci number
A quick introduction to a famous number.
Bees do it, rabbits do it, and luckily, we humans can do it too: explore the famous Fibonacci sequence!
Does the famous Fibonacci sequence always appear in sunflower seed heads?
Anything involving bunny rabbits has to be good.
Patterns and structures lie at the heart of mathematics, some even say they are mathematics. But how do they help us do mathematics?
The Fibonacci sequence – 0, 1, 1, 2, 3, 5, 8, 13, ... – is one of the most famous pieces of mathematics. We see how these numbers appear in multiplying rabbits and bees, in the turns of sea shells and sunflower seeds, and how it all stemmed from a simple example in one of the most important books in Western mathematics.
Dan Brown in his book, The Da Vinci Code, talks about the "divine proportion" as having a "fundamental role in nature". Brown's ideas are not completely without foundation, as the proportion crops up in the mathematics used to describe the formation of natural structures like snail's shells and plants, and even in Alan Turing's work on animal coats. But Dan Brown does not talk about mathematics, he talks about a number. What is so special about this number?
Marcus du Sautoy begins a two part exploration of the greatest unsolved problem of mathematics: The Riemann Hypothesis. In the first part, we find out how the German mathematician Gauss, aged only 15, discovered the dice that Nature used to chose the primes.
It was Euclid who first defined the Golden Ratio, and ever since people have been fascinated by its extraordinary properties. Find out if beauty is in the eye of the beholder, and how the Golden Ratio crosses from mathematics to the arts.
During World Mathematical Year 2000 a sequence of posters were displayed month by month in the trains of the London Underground aiming to stimulate, fascinate - even infuriate passengers! Keith Moffatt tells us about three of the posters from the series. | 420 | 1,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-26 | latest | en | 0.937841 |
https://prezi.com/sxioopdtjip6/scientific-notation/ | 1,547,856,144,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660877.4/warc/CC-MAIN-20190118233719-20190119015719-00541.warc.gz | 604,820,511 | 23,616 | ### Present Remotely
Send the link below via email or IM
Present to your audience
• Invited audience members will follow you as you navigate and present
• People invited to a presentation do not need a Prezi account
• This link expires 10 minutes after you close the presentation
• A maximum of 30 users can follow your presentation
Do you really want to delete this prezi?
Neither you, nor the coeditors you shared it with will be able to recover it again.
# Scientific Notation
What you'll learn, to write and evaluate numbers in scientific notation to calutate.
by
## sierra robinson
on 1 March 2013
Report abuse
#### Transcript of Scientific Notation
Objectives: You will learn to evaluate numbers in scientific notation and to learn how to calculate with
scientific notation. Scientific Notation Standard Notation You can change expressions from scientific notation to
standard notation by simplifying the product of the two factors Exploring Scientific Notation 5 California Content Standards Read, write, and compare rational numbers in scientific notation (positive and negative powers of ten), compare rational numbers in general. Writing in Scientific Notation Write in scientific notation.
73,000,000 move seven places
to the left. Scientific Notation provides a way to write numbers
using powers of ten. You write a number in scientific
notation as the product of two factors. 73,000,000= 7.3x10 7 Write each number in standard notation
9.5 x 10
9.500 write zeros while moving the
decimal point. 3 Rewrite in standard notation = 9,500 4 x 10 5 x 10 3 = 5 x 10 2 = ________________ 5 x 10 - 4 = 0.0005 5 x 10 - 3 = 0.005 5,000 = 50,000 500 5 x 10 - 2 = 0.05 7. Look for and make use of structure 1. In what was does this problem connect to other mathematical concepts? 2. What patterns do you find in negative and positive exponents of scientific notation? 6. Attend to precision What would be a more efficient strategy? 1. 2. What mathematical terms apply in this situation? Standards for Mathematical Practice Standards for Mathematical practice Standards for Mathematical Practice Standards for Mathematical Practice 5. Use appropriate tools strategically. 3. Construct viable arguments and critique the reasoning or others. 1. What is the same and what is different about the negative and postive way of scientific notation? 2. How do you test whether your approach worked? 1. What mathematical tools could we use to visualize and represent the situation? 2. What estimate did you make for the solution? By: Gina Leletzopoulos and Tiara Luke
Full transcript | 601 | 2,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-04 | latest | en | 0.860395 |
https://primes.utm.edu/curios/page.php?curio_id=7549 | 1,652,894,517,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00229.warc.gz | 506,800,890 | 3,657 | # 377
This number is a composite.
Single Curio View: (Seek other curios for this number)
377 = 2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2, i.e., the sum of squares of the first six primes. [Mizuki] | 76 | 194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-21 | longest | en | 0.67078 |
https://www.jiskha.com/display.cgi?id=1367087277 | 1,503,325,583,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108709.89/warc/CC-MAIN-20170821133645-20170821153645-00125.warc.gz | 907,228,946 | 3,625 | # trig
posted by .
solve the point given below is on the terminal side of an angle0. find the exact value of the six trigonometric functions of 0. (10,-10)
• trig -
the other long side is 10sqrt2
sinTheta=-10/10sqrt2=-.707
tantheta=-10/10=-1
you do the rest, if necessary, I can check them.
## Similar Questions
1. ### trig
Find the values of the six trigonometric functions of an angle 0 in standard position whose terminal side passes through point P. P(4, -3)
2. ### trig
If (-2,5) is a point on the terminal side of an angle theta, find the exact value of the six trigonometric functions of theta. Then find the actual measure of theta.
3. ### trig
Find the values of the six trigonometric functions of the angle in standard position with the terminal side passing through the point P(-4, 3).
4. ### trig
find the exact values of THE SIX TRIGONOMETRIC FUNCTIONS OF THE ANGLE THETA WHICH HAS A POINT ON THE TERMINAL SIDE OF (-1,4)
5. ### pre calc
find the exact values of the six trigonometric functions of theta if the terminal side of theta in standard position contains the given point: 1. (8, 4) 2. (4, 4√3) 3. (0, -4) 4. (6, 2)
6. ### precalc
Find the exact values of the six trigonometric functions of theta if the terminal side of theta in standard position contains the given point: 1. (8, 4) 2. (4, 4√3) 3. (0, -4) 4. (6, 2)
7. ### pre calc
find the exact values of the six trigonometric functions of theta if the terminal side of theta in standard position contains the given point: 1. (8, 4) 2. (4, 4√3) 3. (0, -4) 4. (6, 2)
8. ### Math- Trigonometry
The point, in on the terminal side of an angle in standard position find the exact value of the six trigonometric functions of the angle a. (8,15) b.(-9,-40)
9. ### trig
(a) sketch an angle A in standard position whose terminal ray passses through the point (1,-4) (b) Find the exact value of the six trigonometric functions of A, (c) Let B be the reference angle of A. Find a point on the terminal rsy …
10. ### Algebra II
The terminal side of θ in standard position contains each point. Find the exact values of the six trigonometric functions of θ. Given point: (7, -6)
More Similar Questions | 624 | 2,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-34 | latest | en | 0.626291 |
https://crayolateachers.ca/lesson/tantalizing-tessellations-irregular-polygons-colour/ | 1,548,295,039,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584445118.99/warc/CC-MAIN-20190124014810-20190124040810-00212.warc.gz | 475,941,295 | 12,680 | # TANTALIZING TESSELLATIONS – Irregular Polygons, Colour
Students use translation to create an irregular shape that will tessellate and use the shape to create a design filled with blended colour.
80 Minutes
Mathematics
Visual Arts
#### Vocabulary
colour irregular polygon plane polygon regular polygon tessellation transformation translation vertex
#### Materials
Crayola Marker and Watercolour Paper – 22.9 cm x 30.5 cm cm (9” X 12”) Clear tape Scissors Heavy Weight Paper 7.6 cm x 7.6 cm (3" x 3") 1 piece per student Fine Line Marker Crayons Twistable Crayons
## Steps
### Step One
1. Start with a square of heavy weight paper 7.5 cm x 7.5 cm.
2. Cut out a section of the square.
3. Start at one vertex and end at the vertex beside it.
### Step Two
1. Translate the shape by sliding it to the opposite side of the square.
2. Tape the shape to the square being sure to fit it right up against the edge of the square.
### Step Three
1. Cut out a new shape from the straight side of the square.
### Step Four
1. Slide it across to the other side and tape it in place.
### Step Five
1. Trace your new shape in at least 3 different positions on a piece of drawing paper.
2. See what they remind you of.
### Step Six
1. Choose the idea you like the best and use it to tessellate the shape across your paper.
2. Draw the shape and details with marker.
### Step Seven
1. Use twistable crayons to fill in the design.
2. Blend at least 2 colours together in each space.
### Step Eight
1. Decide how you want to display your finished design.
## Learning Goals
Students will be able to:
1. Use translation to create an irregular shape that will tessellate;
2. Create a design filled with blended colour;
3. Add details to the shape to make it interesting;
4. Explain why the tessellation works; and
5. Demonstrate technical accomplishment and creativity.
## Extensions
1. Have students create a Mathematics of Tessellation book that includes all or some of the following shapes used in tessellations with explanations of why the shapes tessellate:
- repeated use of ONE regular polygon
- repeated use of a unit of shape made up of TWO OR MORE different regular polygons
- irregular shapes created by transformation and rotation
2. Have students select an art image by M. C. Escher and explain how it works.
3. Have students use the work of M. C. Escher as inspiration for their own tessellating artwork.
## Prepare
1. Prior to this lesson have students explore tessellations involving repeated use of one polygon and shapes made up of 2 or more different polygons.
Pineapple
Honeycomb
Fish Scales
Snake
Den Haag
Princessehof
VoorHout
Alhambra
4. Gather books and pictures of tessellations, for example, Toads and Tessellations: A Math Adventure, by Sharon Morrisette; An Optical Artist: Exploring Patterns and Symmetry, by Greg Roza; Tessellations: The History and Making of Symmetrical Designs, by Pam Stephens; Introduction to Tessellations, by Dale Seymour and Jill Britton
5. Create a sample irregular shape that will tessellate.
## Introduction
1. View the images of tessellations in nature and discuss them. Review and/or introduce the term tessellation and have students describe the characteristics of a tessellation, e.g.,
- patterns of repeated shapes that cover a flat surface with no gaps and no overlaps
2. Discuss what kinds of regular shapes can tessellate and why
- regular polygons (triangles, squares or hexagons)
3. Discuss how to use translation to create an irregular shape that will tessellate.
- a translation is when the shape is slid across the paper and drawn again in another place
4. - regular polygons can only tessellate if the sum of the interior angles is 360 degrees
5. View and discuss the images of Escher's art.
6. Introduce the challenge.
## Activities
### The Challenge
1. Use translation to create an irregular shape that will tessellate.
2. Create a design filled with blended colour.
3. Add details to the shape to make it interesting.
4. Explain why the tessellation works.
5. Demonstrate technical accomplishment and creativity.
### The Process
1. Make sure everyone understands the challenge.
2. Establish success criteria with your students, for example,
I know I am successful when:
- I used translation to create the shape
- the repeated shape covers the paper without any gaps or overlaps
- the use of blended colours is effective
- I added details to the shape to make it interesting
- the paper is in good condition
- I can explain why the shape tessellates
3. Guide students through the steps outlined in this lesson plan.
4. Observe students as they work.
5. Provide individual assistance and encouragement.
## Sharing
1. Once all the designs are complete display them for a group discussion.
Look closely at the tessellations.
- Choose one that interests you for some reason.
- Share thoughts about the work.
2. During the discussion include references to:
blended colours - How does the use of blended colours contribute to the effectiveness of the overall design?
- design - How does the use of detail contribute to the effectiveness of the overall design?
- technique – How and why does the tessellation work?
## Assessment
1. Observe students as they work – thoughtful focus, discriminating, seeking more information, elaborating, experimenting
2. Observe students as they discuss the art works – active listening, insightful contributions, supporting ideas with evidence found in the artwork and from personal experience. | 1,285 | 5,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-04 | longest | en | 0.81697 |
http://www.perlmonks.org/index.pl?node_id=7366 | 1,511,186,045,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00332.warc.gz | 472,591,747 | 8,587 | There's more than one way to do things PerlMonks
### Cartesian Cross-Products
by japhy (Canon)
on Apr 12, 2000 at 10:06 UTC Need Help??
``` 1: =pod What is a Cartesian Cross Product?
2:
3: I think this is just too damn cool to pass up. If you don't
4: know what a Cartesian (Cross-) Product is, it's basically:
5:
6: A = (1,2,3)
7: B = (4,5)
8: CCP(A,B) =>
9: (1,4)
10: (1,5)
11: (2,4)
12: (2,5)
13: (3,4)
14: (3,5)
15:
16: Yay, that's all well and good. Here's how to implement the
17: Cartesian Product generator in Perl:
18:
19: =pod Explanation of algorithm used
20:
21: Given a list of sets, say ([a,b], [c,d,e], [f,g]), we first determine how
22: many sets can be created. Mathematically, this is determined as follows:
23:
24: For a list of sets, { a[1], a[2], ..., a[n] }, to determine how many sets
25: can be created by choosing an element from a[1] as the first element of a
26: set, an element of a[2] for the second element, and so on, picking an
27: element of a[n] as the n-th element, we create a list { s[1], s[2], ...,
28: s[n] }, where each element s[i] is the number of element in a[i]. We can
29: pick any of the s[i] elements from a[i] for the specified element in the
30: set to be created, so the number of sets to be created is
31:
32: n
33: -----
34: | | s[p] .
35: | |
36: p=1
37:
38: That is, the product of the sizes of all the sets.
39:
40: Now that we know how many sets we'll be creating, we start to populate these
41: sets. We modify the same index of each set per loop; that is, we modify
42: a[0][0], a[1][0], a[2][0], ..., a[n][0], before we modify any index in a[1].
43:
44: I utilize a "repetition value", which starts at 1, and is multiplied by the
45: size of the previous set (s[i-1]) when the population of a specific index of
46: the new sets is complete. The repetition value indicates how many times the
47: specific element will be inserted in a row on a pass over an index. The
48: starting value of 1 means that each element in a[0] will be inserted once, and
49: then the next element will be entered, and after all elements have been
50: exhausted, we go back to inserting a[0].
51:
52: After we've exhausted a[0], we multiply the repetition value by s[0], and we
53: move on to a[1]. For each value here, we fill in the next index in the new
54: sets, but we do this R times in succession, where R is the repetition value.
55:
56: We continue through until the new sets are completed.
57:
58: =cut
59:
60: sub cartesian {
61: my \$len = 1;
62: my (@ret,\$rep,\$i,\$j,\$p,\$k);
63:
64: for (@_) { \$len *= @\$_ }
65:
66: for (\$rep = 1, \$i = 0; \$i < @_; \$rep *= @{ \$_[\$i] }, \$i++) {
67: for (\$j = 0, \$p = 0; \$j < \$len; \$j += \$rep, \$p++) {
68: for (\$k = 0; \$k < \$rep; \$k++) {
69: print STDERR << "DEBUGGING" if 0; # set to true to see debug output
70: repetition value: \$rep
71: modifying set[@{[ \$j + \$k]}], index[\$i]
72: value is element @{[ \$p % @{ \$_[\$i] } ]} ('\$_[\$i][\$p % @{ \$_[\$i] }]') of original set[\$i]
73:
74: DEBUGGING
75: \$ret[\$j + \$k][\$i] = \$_[\$i][\$p % @{ \$_[\$i] }]
76: }
77: }
78: }
79:
80: return @ret;
81: }
82:
83: # uncomment to see a test run
84: # print map "@\$_\n", cartesian( [1,2] , [3,4,5] , [6,7] );```
Replies are listed 'Best First'.
Re: Cartesian Cross-Products
by Anonymous Monk on Feb 12, 2005 at 02:01 UTC
Add "\$#ret = \$len -1" at line 64 for better performance. For a 100*65*20 set, benchmark results are: Benchmark: timing 10 iterations of new, old... new: 16 wallclock secs (15.59 usr + 0.09 sys = 15.68 CPU) @ 0.64/s (n=10) old: 21 wallclock secs (20.16 usr + 0.10 sys = 20.26 CPU) @ 0.49/s (n=10)
People who learned C should "learn" Perl... This is the "Perl" way to do it.
```perl -e '
use strict;
my @a=qw{1 2 3};
my @b=qw{4 5};
my @c=();
foreach my \$a (@a) {
foreach my \$b (@b) {
push @c, [\$a, \$b];
}
}
print join("\n", map {join ",", @\$_} @c), "\n"
'
---
1,4
1,5
2,4
2,5
3,4
3,5
Mike (mrdvt92)
It is indeed the Perl way to do it if you know in advance how many arrays you'd like to get the product of.
If you don't, Japhy's algorithm is the way to go - even if it looks C-ish with a lot of indices. I suspect that it may be faster as well, using the array pre-allocation optimisation pointed out earlier.
"The Perl Way"
```sub cartesian {
my @C = map { [ \$_ ] } @{ shift @_ };
foreach (@_) {
my @A = @\$_;
@C = map { my \$n = \$_; map { [ \$n, @\$_ ] } @C } @A;
}
return @C;
}
Create A New User
Node Status?
node history
Node Type: perlcraft [id://7366]
Approved by root
help
Chatterbox?
and all is quiet...
How do I use this? | Other CB clients
Other Users?
Others making s'mores by the fire in the courtyard of the Monastery: (12)
As of 2017-11-20 13:52 GMT
Sections?
Information?
Find Nodes?
Leftovers?
Voting Booth?
In order to be able to say "I know Perl", you must have:
Results (286 votes). Check out past polls.
Notices? | 1,718 | 5,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-47 | longest | en | 0.891155 |
https://www.windows2universe.org/kids_space/black.html&edu=elem | 1,643,287,079,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305260.61/warc/CC-MAIN-20220127103059-20220127133059-00581.warc.gz | 1,088,929,161 | 11,676 | How much (in meters) is a black hole's diameter? Where do you go if you were to be sucked down a black hole? What would happen to you?
The diameter of a black hole depends on the mass of the original object that becomes a black hole. So the more massive the object, the bigger the diameter of a black hole formed from that object. If earth became a black hole it would have a diameter of about 0.017 meters, about the size of a marble! Think of how small your bedroom would seem then! Our sun would have a diameter of about 6000 meters, just a little less than 4 miles. No one really knows what would happen if you were to get sucked into a black hole. The problem is that once you are inside the black hole there is no way to talk to the people outside the black hole. So, we don't know what goes on inside a black hole. A possibility is that you would slowly be stretched out like a string of spaghetti until you snapped! So stay far away from the edges of black holes!
Submitted by John (MI, USA)
(October 13, 1997)
What are the retrograde motions of planets in the sky?
It depends on which type of motion you are asking about. If you take a birds-eye view from the top of the solar system all the planets orbit around the Sun in a counter-clockwise (or direct) direction....more
How do Astronauts Live in Space?
Have you ever wondered how astronauts live in space? Did you know they do a lot of the same things we do here on Earth? Astronauts eat, exercise and sleep just like we do. However, their food isn't always...more
How far is the Earth from the Sun, the Moon and all the other planets? How far are all of the planets from the Sun? Do you know of a software that tracks the planets in real-time?
There is a really neat internet program called Solar System Live that shows where all of the planets and the Sun are. If you go to that page, you'll see an image similar to the one on the left. Below the...more
Is it really true that man never really walked on the Moon?
The picture of the American Flag (the one put there by the Apollo astronauts) is waving (or straight out) in the wind. How could that be possible if there is no atmosphere on the Moon? Was it some sort...more
How many planets orbit the sun?
I was wondering if there is a new planet? Are there planets (a tenth planet?) after Pluto belonging to our solar system? What are the names of the new planets discovered in the solar system? Are there...more
Did the Big Bang create any black holes? Wouldn't the universe evolve differently if some of the original energy had disappeared in black holes?
When an object has a really high energy, it can form a black hole. This is called a primordial black hole. Primordial black holes were formed near the beginning of the universe. Primordial black holes...more
Windows to the Universe, a project of the National Earth Science Teachers Association, is sponsored in part is sponsored in part through grants from federal agencies (NASA and NOAA), and partnerships with affiliated organizations, including the American Geophysical Union, the Howard Hughes Medical Institute, the Earth System Information Partnership, the American Meteorological Society, the National Center for Science Education, and TERC. The American Geophysical Union and the American Geosciences Institute are Windows to the Universe Founding Partners. NESTA welcomes new Institutional Affiliates in support of our ongoing programs, as well as collaborations on new projects. Contact NESTA for more information. | 757 | 3,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-05 | latest | en | 0.967032 |
https://sfepy.org/doc-devel/_sources/examples/navier_stokes-stokes_slip_bc.rst.txt | 1,670,287,035,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00335.warc.gz | 533,357,362 | 1,947 | .. _navier_stokes-stokes_slip_bc: navier_stokes/stokes_slip_bc.py =============================== **Description** Incompressible Stokes flow with Navier (slip) boundary conditions, flow driven by a moving wall and a small diffusion for stabilization. This example demonstrates the use of no-penetration and edge direction boundary conditions together with Navier or slip boundary conditions. Alternatively the no-penetration boundary conditions can be applied in a weak sense using the penalty term dw_non_penetration_p. Find :math:\ul{u}, :math:p such that: .. math:: \int_{\Omega} \nu\ \nabla \ul{v} : \nabla \ul{u} - \int_{\Omega} p\ \nabla \cdot \ul{v} + \int_{\Gamma_1} \beta \ul{v} \cdot (\ul{u} - \ul{u}_d) + \int_{\Gamma_2} \beta \ul{v} \cdot \ul{u} = 0 \;, \quad \forall \ul{v} \;, \int_{\Omega} \mu \nabla q \cdot \nabla p + \int_{\Omega} q\ \nabla \cdot \ul{u} = 0 \;, \quad \forall q \;, where :math:\nu is the fluid viscosity, :math:\beta is the slip coefficient, :math:\mu is the (small) numerical diffusion coefficient, :math:\Gamma_1 is the top wall that moves with the given driving velocity :math:\ul{u}_d and :math:\Gamma_2 are the remaining walls. The Navier conditions are in effect on both :math:\Gamma_1, :math:\Gamma_2 and are expressed by the corresponding integrals in the equations above. The no-penetration boundary conditions are applied on :math:\Gamma_1, :math:\Gamma_2, except the vertices of the block edges, where the edge direction boundary conditions are applied. The penalty term formulation is given by the following equations. Find :math:\ul{u}, :math:p such that: .. math:: \int_{\Omega} \nu\ \nabla \ul{v} : \nabla \ul{u} - \int_{\Omega} p\ \nabla \cdot \ul{v} + \int_{\Gamma_1} \beta \ul{v} \cdot (\ul{u} - \ul{u}_d) + \int_{\Gamma_2} \beta \ul{v} \cdot \ul{u} + \int_{\Gamma_1 \cup \Gamma_2} \epsilon (\ul{n} \cdot \ul{v}) (\ul{n} \cdot \ul{u}) = 0 \;, \quad \forall \ul{v} \;, \int_{\Omega} \mu \nabla q \cdot \nabla p + \int_{\Omega} q\ \nabla \cdot \ul{u} = 0 \;, \quad \forall q \;, where :math:\epsilon is the penalty coefficient (sufficiently large). The no-penetration boundary conditions are applied on :math:\Gamma_1, :math:\Gamma_2. Optionally, Dirichlet boundary conditions can be applied on the inlet in the both cases, see below. For large meshes use the 'ls_i' linear solver - PETSc + petsc4py are needed in that case. Several parameters can be set using the --define option of sfepy-run, see :func:define() and the examples below. Examples -------- Specify the inlet velocity and a finer mesh:: sfepy-run sfepy/examples/navier_stokes/stokes_slip_bc -d shape="(11,31,31),u_inlet=0.5" sfepy-view -f p:p0 u:o.4:p1 u:g:f0.2:p1 -- user_block.vtk Use the penalty term formulation and einsum-based terms with the default (numpy) backend:: sfepy-run sfepy/examples/navier_stokes/stokes_slip_bc -d "mode=penalty,term_mode=einsum" sfepy-view -f p:p0 u:o.4:p1 u:g:f0.2:p1 -- user_block.vtk Change backend to opt_einsum (needs to be installed) and use the quadratic velocity approximation order:: sfepy-run sfepy/examples/navier_stokes/stokes_slip_bc -d "u_order=2,mode=penalty,term_mode=einsum,backend=opt_einsum,optimize=auto" sfepy-view -f p:p0 u:o.4:p1 u:g:f0.2:p1 -- user_block.vtk Note the pressure field distribution improvement w.r.t. the previous examples. IfPETSc + petsc4py are installed, try using the iterative solver to speed up the solution:: sfepy-run sfepy/examples/navier_stokes/stokes_slip_bc -d "u_order=2,ls=ls_i,mode=penalty,term_mode=einsum,backend=opt_einsum,optimize=auto" sfepy-view -f p:p0 u:o.4:p1 u:g:f0.2:p1 -- user_block.vtk .. image:: /../doc/images/gallery/navier_stokes-stokes_slip_bc.png :download:source code .. literalinclude:: /../sfepy/examples/navier_stokes/stokes_slip_bc.py | 1,179 | 3,762 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-49 | latest | en | 0.673529 |
https://managementassignmenthelp.bookmyessay.com/blog/looking-for-guidance-on-factoring-calculator/ | 1,702,305,786,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00376.warc.gz | 425,265,132 | 52,945 | Wants to learn about a Factoring calculator? If this is true so BookMyEssay is here for your help with the Factoring calculator. We’ll guide you about the Factoring calculator. Our experts will provide you with a service of assignment in every subject. If you know how the factor calculator works, click the “Try it” button to see the solution here with the steps. The Process of Factorization of dividing a number into simpler expressions whose product is equal to the main expression. Here you can get the Do My Homework for me service.
The Factoring calculator can convert any algebraic equations to a product of simple number/prime numbers. A factoring calculator can use in any binomial, trinomial, mononomial, rational, or irrational equation. This product of calculator gives you all the steps using different steps like grouping, quadratic root formula, factoring the difference of 2 squares, and so on. For more details read this blog till the end.
## Meaning of Factoring
Sometimes the values of polynomials with rational coefficients can be written as the product of low-degree polynomials they also have rational coefficients in that polynomial. So in this case, the polynomial is called a “factor over the rationals”. The incorporation of Factoring is a very useful technique to solve rational roots and square roots of the values of integral.
All the polynomial with rational coefficient values always have the same number of roots as their powers in the complex plane; Therefore, those root is not rational numbers. In such cases, the polynomial will not factor into the linear polynomials.
All the rational numbers are quotients of the polynomial factors. Like polynomials, rational numbers play a very crucial role in mathematics and physics. Like rational numbers, rational functions are usually expressed in “lowest number”. For a given numerator and denominator pair, this involves finding their greatest common denominator polynomial and removing it from both the numerator and the denominator.
### Why do you need a Factoring calculator?
If you want to use a Factoring calculator so just click on the below solution and solve it. For calculating you can also enter an equation factor on the input field above and click on the “Factor” button to see the solution. With the help of a Factoring calculator, you can easily solve your expression without any worry. The factoring calculator gives you an accurate result of your solution. It will also save you time if you use a Factoring calculator.
### Let’s check how the Factoring calculator works Free
The process of Factoring is breaking down a complex equation into simple terms. The factorized expression must be rewritten as the product of its insoluble number or equation. However, an expression is multiplying only if it can be written as the product number of two or more factors under multiplication number.
Example – x2 +4x-2
(x-4) (x-2) solution | 590 | 2,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-50 | latest | en | 0.889364 |
http://max8us.net/?gmat/forums/post125924.html | 1,534,763,329,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216333.66/warc/CC-MAIN-20180820101554-20180820121554-00425.warc.gz | 255,805,095 | 15,747 | A company plans to assign identification numbers
Math questions from mba.com and GMAT Prep software
nanu.nantaki
Students
Posts: 7
Joined: Wed Aug 05, 2009 12:55 pm
A company plans to assign identification numbers
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000
ankitmisri
Course Students
Posts: 4
Joined: Sat Jul 04, 2009 10:51 pm
Re: A company plans to assign identification numbers
1. Each number is to consist of four different digits from 0 to 9
Hence no number can repeated
2. the first digit cannot be 0
hence you can only choose 1 of the 9 numbers(1-9) for the 1st place i.e. is you have 9 options for 1st digit
For the 2nd digit again you can again choose 1 out 9 numbers (now u have 0 available) in 9c1 ways =9
For the 3rd digit only 8 numbers are available coz of "1."...u can choose that in 8c1 ways = 8
For the 4th place only 7 numbers are available coz of "1."...u can choose that in 7c1 ways = 7
Since all four selected individually, you have to multiply them 9*9*8*7 = 4536
andersonbae
Forum Guests
Posts: 3
Joined: Wed Sep 09, 2009 12:22 am
Re: A company plans to assign identification numbers
Here is an easy solution , if you don't know how to solve this problem.
Don't be embarrassed.
There are three numbers.
0, 1, 2
We can make the identification numbers which does not start "0" as the following
102
120
201
210
The total number of the identification numbers we can make is 4 with 3 different numbers.
We can apply a "SLOT" solution here.
--- --- ---
(3-1) => First Slot : because "O" should not start as a first number
2 => Second Slot
1 => Third Slot
2*2*1 = 4
We apply this way to the original problem.
(10-1) - because "0" should not start as a first number.
(10-1) * 9 * 8 * 7 = 4536
Does it make sense? If I am wrong, please let me know.
Course Students
Posts: 2
Joined: Sat Nov 15, 2008 6:25 am
Re: A company plans to assign identification numbers
I like this method:
First, find out the number of ways you can create the ID numbers WITHOUT any restrictions, then subtract the number of ways we create ID numbers WITH the restriction.
That will give us the number of ways that we can create only valid ID numbers
0->9 inclusive is really 10 digits
The number of ways we can arrange 10 digits in four "slots" is 10 * 9 * 8 * 7 = 5040
So there are 5040 unrestricted numbers
We can find the number of restricted ID numbers by assuming the first digit is already 0. That leaves nine digits left for three "slots":
9 * 8 * 7 = 504
unrestricted ID numbers - restricted ID numbers = Possible ID numbers
5040 - 504 = 4536
RonPurewal
Students
Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am
Re: A company plans to assign identification numbers
ankitmisri wrote:1. Each number is to consist of four different digits from 0 to 9
Hence no number can repeated
2. the first digit cannot be 0
hence you can only choose 1 of the 9 numbers(1-9) for the 1st place i.e. is you have 9 options for 1st digit
For the 2nd digit again you can again choose 1 out 9 numbers (now u have 0 available) in 9c1 ways =9
For the 3rd digit only 8 numbers are available coz of "1."...u can choose that in 8c1 ways = 8
For the 4th place only 7 numbers are available coz of "1."...u can choose that in 7c1 ways = 7
Since all four selected individually, you have to multiply them 9*9*8*7 = 4536
yep. this is the most direct method, and thus the best.
if you have a problem in which ORDER MATTERS, then you should be able to solve the problem by multiplying numbers together.
you shouldn't have to bother with factorials, etc., unless you have a problem on which order DOESN'T matter.
ajaym8
Students
Posts: 24
Joined: Tue Apr 12, 2016 7:32 pm
Re: A company plans to assign identification numbers
Hi Ron,
What is wrong if I do it as indicated below ?
Asking because this method came to my mind first in the mock and I could not get to the correct answer.
I have 4 places to fill.
_ _ _ _
I start with the last digit(one's place), I have 10 numbers to chose from. Similarly for 3rd digit I have 9 numbers to chose from, 8 numbers for the 2nd place.
Now for the first place I cannot include 0, so I have 6 numbers to chose from(7 numbers if 0 were allowed) for the first place (the thousand's place).
I get 6*8*9*10. which gives 4320.
RonPurewal wrote:
ankitmisri wrote:1. Each number is to consist of four different digits from 0 to 9
Hence no number can repeated
2. the first digit cannot be 0
hence you can only choose 1 of the 9 numbers(1-9) for the 1st place i.e. is you have 9 options for 1st digit
For the 2nd digit again you can again choose 1 out 9 numbers (now u have 0 available) in 9c1 ways =9
For the 3rd digit only 8 numbers are available coz of "1."...u can choose that in 8c1 ways = 8
For the 4th place only 7 numbers are available coz of "1."...u can choose that in 7c1 ways = 7
Since all four selected individually, you have to multiply them 9*9*8*7 = 4536
yep. this is the most direct method, and thus the best.
if you have a problem in which ORDER MATTERS, then you should be able to solve the problem by multiplying numbers together.
you shouldn't have to bother with factorials, etc., unless you have a problem on which order DOESN'T matter.
RonPurewal
Students
Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am
Re: A company plans to assign identification numbers
the problem with your approach is that there might be 6 possibilities for the thousands place, but there also might be 7 possibilities.
specifically,
• if one of the last three digits -- the ones you've already chosen -- is 0, then there are still 7 digits available for the thousands place (first slot).
• if none of the last three digits is 0, then, there are only 6 digits available for the first slot.
therefore you can't multiply the possibilities in this order -- unless you want to break them into 2 separate cases. but that would be super annoying, because then you'd have to think about exactly how many different ways you could pick the last three digits with 0 among them, and how many different ways you could pick the last three digits without 0.
RonPurewal
Students
Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am
Re: A company plans to assign identification numbers
more generally, you should ALWAYS pick the items that have RESTRICTIONS ••FIRST•• in problems like this.
otherwise, you get the problem you have here -- if you leave the restricted places for last, then you can't calculate the numbers of possibilities for them (because that depends on the previous choices). | 1,881 | 6,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-34 | latest | en | 0.907549 |
https://arnoldzwicky.org/2018/05/08/four-things-are-two-binary-oppositions/?replytocom=794314 | 1,659,896,287,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00784.warc.gz | 142,331,033 | 16,531 | ## Four things are two binary oppositions
It is really unusual to have a K-12 lesson that allows discussion of dice and playing cards. This may be the first time I’ve encountered it. And I am finding it really disconcerting that the writer keeps dividing suits by color instead of by major/minor.
This was at first completely baffling to me: major/minor?
It turned out to be significant that Bitty is a bridge player, and I am not. Major/minor is a thing in bridge.
In bridge, I have learned, the suits of playing cards come in an order. In ascending order, with minor (MIN) followed by major (MAJ), these divided by a vertical bar:
clubs (B) diamonds (R) | hearts (R) spades (B)
In a chart, to be read like text:
The top row is MIN, the bottom MAJ, and the colors leap out. Four things, two binary oppositions.
But for four things, there are always three possible binary oppositions:
A B C D: AB vs. CD (MIN vs. MAJ); AC vs. BD; AD vs. BC (B vs. R)
So what’s AC vs. BD, the left column vs. the right column? pointed-top (PT) (diamonds, spades) vs. rounded-top (RD) (clubs, hearts).
We are categorizing creatures, and also pattern-seeking creatures, so given four things we’ll break them down into pairs (two-member categories) and see some kind of significance, some kind of “meaning”, in the pairs; we pretty much can’t not do this.
This kind of decomposition into explicit binary oppositions is a hallmark of structuralist thinking. A little essay in NOAD‘s entry:
noun structuralism: a method of interpretation and analysis of aspects of human cognition, behavior, culture, and experience that focuses on relationships of contrast between elements in a conceptual system that reflect patterns underlying a superficial diversity. … Originating in the structural linguistics of Ferdinand de Saussure, and extended into anthropology by Claude Lévi-Strauss, structuralism was adapted to a wide range of social and cultural studies, especially in the 1960s, by writers such as Roland Barthes, Louis Althusser, and Jacques Lacan.
Personal note: my intellectual influences are quite varied, but they include Roman Jakobson, a major figure of structuralist thinking, especially in linguistics. Binarist thinking was as mother’s milk to me.
I waggisly incorporated this thinking into a piece of homoerotic magic realist fiction, told in the voice of Sundance (Sundance is a angel when he flies; yes, he can fly), called Sonny (originally a genuine country boy, clever and sweet, but largely unschooled, also a fool for mansex), called Soní, in a Bolivia of the mind, in a larger world in which both time and death are flexible and variable notions. From my 4/5/11 AZBlogX posting “San Soní”:
…..
I have learned how to do computer searches on titles (there are some truly marvelous things in modern life), and being the sort of person I am I search for SODOM, which nets me a whole pile of stuff I might someday want to take a look at (Sodom’s SongSodomy and the Pirate TraditionThe Haunts of the Sodomites: Where to Fuck and Get Fucked in the Great Cities of the World). Plus the absolute prize-winning title, as far as I’m concerned; The Four First Sisters and the Four Sodomite Martyrs: A Study in Binary Opposition in the Folk Beliefs of Bolivia. By Rosalie Cavanaugh-Charles, who’s at the University of East Anglia. Just published. I am the first user (as they say in libraries) to check this one out.
Chapter 1 tells the legends of the First Mother: how She took the Jaguar as her husband, how the Four First Sisters were born together, all at the same time, that sort of thing. Most of it I already knew, but there was some new stuff; for instance, I didn’t know that all four of the sisters’ names were in some language (or maybe languages) no longer spoken in Bolivia, so that even their usual translations are on the iffy side. Here Cavanaugh-Charles produces the first of her many useful (ok, sometimes baffling; I’m still ferreting my way into academia) tables and charts:
SISTER TRANSLATION CHARACTERISTIC QUALITY Pal’chas ‘mountains’ calm Ixqin ‘sea’ playfulness, good humor Sqonis ‘hot’ or ‘north’ enthusiasm, passion Multhan ‘cold’ or ‘south’ control
When it’s laid out like this you can start to see patterns.
Chapter 2 is about the role the sisters play in Bolivian culture. More familiar stuff, but it’s nice to be reminded.
First, the sisters are used as yardsticks for character. If someone is too serious, humorless, people say that have “not enough Ixqin” in them. If someone is wild, refuses to take things seriously, they have “too much Ixqin”. And someone of good humor is “a perfect Ixqin”. Ok, Piero said I was a perfect Ixqin, and that pleased me a lot.
Second, babies are named for the sisters (but versions of the names that fit Spanish a bit better), in the hope that they’ll turn out to have that sister’s qualities. Max’s mother was named Ixqina; from all accounts, Palchasa would have fit better, but then she was actually named after her father, Ixqino, Bolivia’s first prince, and everyone says the name suited him just perfectly. (Yes, Maximilian I was imported to provide a male ruler for the new nation. Ixqina might have been the most powerful woman in the country, but she was unacceptable as its chief of state. Max the First was, actually, an inspired choice: handsome, charming, enough nobility to be accustomed to command, not enough to want to undo things in a major way.)
Third, and by far most important, each community of women is identified with one of the sisters.
Chapter 3 goes like chapter 1, except that it’s about us faggots. Amazingly, Cavanaugh-Charles talks about fag sex in a direct, un-sneering, un-prissy way; she doesn’t use words off the street, but she’s pretty inventive in getting the picture across with perfectly acceptable words, like “penetrate”. I think I’d like to meet this woman.
But, oh Jesus, it’s still another version of my year with Max, and it’s akilter, askew, discordant with every other version I’ve heard so far, including Consuela’s, which I’d been assuming was the folk take on these events, at least in this century.
In this version, Maximilian III was born the year before I appeared (same date, previous year), and Carlota and the baby sailed happily off for Spain the day of my appearance. A nice, neat, touch.
In this version, Max and I actually ruled together, as equals.
In this version, the couplings that took place in the men’s bedroom of the palace are displaced into sexual performances in the plaza, before every onlooker who happened to come by.
In this version, Simón himself, acting on his own, surprises the unwary four and leads us at gunpoint from the palace. Blood gushes spontaneously from Max’s pectorals. A truly unpleasant picture, to my mind, anyway.
In this version, there’s no shit, and certainly no maggots feeding on our shit. Now there’s a detail I would happily give up.
It goes on and on. I can’t do it any more. My head hurts.
But there’s a table, oh yes a table:
MARTYR EPITHET CHARACTERISTIC QUALITY Maximilian The Christ of Bolivia strength, power Soní The Angel loyalty, good friendship Piero The Teacher learning Francisco The Patient dependability, patience
Chapter 4 is like chapter 2. Yeah, again, about us faggots. It’s all news to me. The idea that we are symbols in modern Bolivian culture is just so bizarre.
But all the stuff from chapter 2 rolls by again, except now it’s about us. And it’s real, meaning that living breathing Bolivians subscribe to these ideas, wow. There are folks who say that some people have “not enough Soní”, meaning they’re fickle, changeable, or “too much Soní”, meaning they’re servile, dependent. Am I a perfect Soní?
Cavanaugh-Charles explains that with another four personages — her word — to identify with, there can be sixteen kinds of communities of women, not just four. Everyone seems to think this is a very good thing.
Apparently there are lots of little Sonís and Sonías running around in Bolivia’s villages these days.
And there’s a Soní cult, with Santa Carlita del Norte (the place where I died with Butch) as its center. I wonder what the rituals are like.
Chapter 5 is completely daunting. It’s all about binary oppositions, with all this stuff about computer logic, somebody named Levi-Strauss (who seems to have thought very deeply about the difference between raw and cooked), and linguistics (here at least I learn a little something, about voiced versus voiceless consonants, rounded versus unrounded vowels — a lot of the terminology strikes me as sexy, though probably that’s an accident).
In chapter 6 we get to what she thinks is the real point of the exercise, breaking down the Four First Sisters and the Four Sodomite Martyrs into binary oppositions. I sort of see the point. I certainly see how Max and I made a couple and how Piero and Francisco did. And I see that I was subordinate to Max (no matter what he said) in the same way that Francisco was subordinate to Piero. But Piero and I made a kind of couple, too, as opposed to Francisco and Manuel, but then Manuel isn’t in this story. So I could group us any which way. Cavanaugh-Charles, though, dismisses the Piero-Soní connection and produces this astonishing chart:
+: TOP/UPWARD -: BOTTOM/DOWNWARD +: SUN/DAY Maximilian Soní -: MOON/NIGHT Piero Francisco
Oh, now I see that this does group me with Piero in a way; he’s my total opposite. Hmmm.
Actually, this comes after she’s done her number on the Four First Sisters. I have to say that it all goes easier for them, you could see it pretty clearly all the way back in chapter 2, what with the sea vs. the mountains and north vs. south and the “hot” sisters vs. the “cool” ones:
+: UP -: DOWN +: DAY/HOT north, hot (Sqonis) sea (Ixqin) -: NIGHT/COOL mountains (Pal’chas) south, cold (Multhan)
Then she goes on to claim that the two charts are really just the same chart. So I’m Ixqin. (What she actually says, omigod, is that “Soní fills the same position in the system of binary oppositions as Ixqin.”)
It gets loonier. She does this analysis over again two more times, with the four elements (fire, water, earth, air) and the four humors (sanguine/passionate, choleric/angry, phlegmatic/stolid, bilious/peevish) and matches these two charts — you really don’t want to see them — with the two from Bolivia. I’m water and angry, in case you’re interested.
The point is that it’s all supposed to be universal, “the same two binary oppositions realized in different material in different domains.”
Actually, she had pretty well sucked me into the whole thing, until she mentioned a guy who’d done this analysis on the Beatles, for Chrissake. That broke the spell for me. I saw that you could jiggle any four things around and make them fit, maybe in more than one way even. Shouldn’t I have been hot, fiery, passionate, like the sun?
But I liked the idea of being connected to Ixqin, to Max’s mother and to his grandfather. And to the sea. Bolivia is a land-locked country; you dream of the far-off sea, you long to walk into waves that sparkle in the sun, you yearn to give your body back to the playful, angry ocean.
…..
The crucial step: “the two charts are really just the same chart”. And, yes, the Beatles: of course, the Beatles.
### 3 Responses to “Four things are two binary oppositions”
1. Robert Coren Says:
I’m not sure, in the case of the “rounded” and “pointed” suits, whether there’s any “meaning” attached to the categories; I think it’s just a convenient way to identify a pair of suits not covered by the other two categories. (Whether it’s actually any more convenient to say that a player had to worry about losing tricks in the “rounded suits” rather than just saying “hearts and clubs” is unclear, and I was briefly puzzled the first time I encountered the usage in a bridge column.)
2. Robert Coren Says:
For another set of four things paired in various ways, see the Wikipedia article on The Quadrivium.
3. The Raw and the Cooked | Arnold Zwicky's Blog Says:
[…] the book figured prominently in my 5/8/18 posting “Four things are two binary […] | 2,893 | 12,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-33 | latest | en | 0.932833 |
https://groups.yahoo.com/neo/groups/wuhu_software_group/conversations/topics/4713?tidx=1 | 1,511,489,636,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807056.67/warc/CC-MAIN-20171124012912-20171124032912-00797.warc.gz | 628,621,641 | 25,553 | ## Weather Question
Expand Messages
• Is there any equipment that us home weather broadcasters can get hold of that can determin cloud height, I have read that it is usaully determined by infrared
Message 1 of 6 , Jan 25, 2007
Is there any equipment that us home weather broadcasters can get hold of
that can determin cloud height, I have read that it is usaully determined by
infrared from the ground. Or is there another way we can do it, besides
just guessing?
thnx
Never-Enuff Technologies
Colver, Pennsylvania
-------------------------------------------
(A)bort, (R)etry, (S)mack the @#\$&*~ thing!
-------------------------------------------
• This is what I use Found it some place on the net: CLOUD HEIGHT The cloud height on this site is an estimate of cumulus clouds using a formula based on
Message 2 of 6 , Jan 26, 2007
This is what I use
Found it some place on the net:
CLOUD HEIGHT
The cloud height on this site is an estimate of cumulus clouds using a formula based on temperature and dew point. Actual measurements of cloud height are made with a Micropulse Lidar (MPL). This device fires a laser into the sky and measures the backscattered signal. Costs for such a device are beyond the scope of weather hobbyists.
You can use the following formulas to calculate the height (H) of a cumulus cloud base given surface temperature and dewpoint.
A) H(meters)=125*(Tc-Tdc)
B) H(feet)=222(Tf-Tdf)
Note:
A) Tc and Tdc in formula are temperature and dewpoint in degrees Celsius.
B) Tf and Tdf in formula are temperature and dewpoint in degrees Fahrenheit.
> -----Original Message-----
> From: wuhu_software_group@yahoogroups.com
> [mailto:wuhu_software_group@yahoogroups.com] On Behalf Of <
> Sent: Thursday, January 25, 2007 9:03 PM
> To: Wuhu_Software_Group
> Subject: [wuhu_software_group] Weather Question
>
> Is there any equipment that us home weather broadcasters can
> get hold of that can determin cloud height, I have read that
> it is usaully determined by infrared from the ground. Or is
> there another way we can do it, besides just guessing?
>
> thnx
>
> Never-Enuff Technologies
> Colver, Pennsylvania
> -------------------------------------------
> (A)bort, (R)etry, (S)mack the @#\$&*~ thing!
> -------------------------------------------
>
>
>
>
> To visit your group on the web, go to:
> http://groups.yahoo.com/group/wuhu_software_group/
>
> Want more or less group messages (Special notices as a
> minimum)?http://groups.yahoo.com/group/wuhu_software_group/join
>
> To unsubscribe from this group, send an email to:
> wuhu_software_group-unsubscribe@yahoogroups.com
>
> http://home.comcast.net/~wuhu_software/
>
>
>
>
• Dickie, Of course their is equipment available, just depends on how much we are willing to spend on our hobby : Check this out, doesn t have a price, but if
Message 3 of 6 , Jan 26, 2007
Dickie,
Of course their is equipment available, just depends on how much we are willing to spend on our hobby :>
Check this out, doesn't have a price, but if you have to ask I guess you can't afford it, isn't that the saying?
http://www.allweatherinc.com/meteorological/8339_ceilometer.html
Or if you don't want to spend any money here is another method:
http://asd-www.larc.nasa.gov/SCOOL/lesson_plans/altlab.html
Now there is also ballons, they can be fun ! Need to know how fast they rise and you can do the calculations, but you could also put them on a long string marked with measurement intervals.
Actually everything I have read is that for us hobby folks is to estimate based on the type of clouds and use the Metar reports from a location close to us.
Steve
• Thnx Steve, I saw the first one when i was looking today, I figured since the price isnt listed, it means we cant afford it...lol Dickie ... From:
Message 4 of 6 , Jan 26, 2007
Thnx Steve,
I saw the first one when i was looking today, I figured since the price isnt listed, it means we cant afford it...lol
Dickie
-----Original Message-----
From: wuhu_software_group@yahoogroups.com [mailto:wuhu_software_group@yahoogroups.com]On Behalf Of steve_03222
Sent: Friday, January 26, 2007 6:37 PM
To: wuhu_software_group@yahoogroups.com
Subject: [wuhu_software_group] Re: Weather Question
Dickie,
Of course their is equipment available, just depends on how much we are willing to spend on our hobby :>
Check this out, doesn't have a price, but if you have to ask I guess you can't afford it, isn't that the saying?
http://www.allweath erinc.com/ meteorological/ 8339_ceilometer. html
Or if you don't want to spend any money here is another method:
http://asd-www. larc.nasa. gov/SCOOL/ lesson_plans/ altlab.html
Now there is also ballons, they can be fun ! Need to know how fast they rise and you can do the calculations, but you could also put them on a long string marked with measurement intervals.
Actually everything I have read is that for us hobby folks is to estimate based on the type of clouds and use the Metar reports from a location close to us.
Steve
• That will work nicely. Thnx Dickie ... From: wuhu_software_group@yahoogroups.com [mailto:wuhu_software_group@yahoogroups.com]On Behalf Of John Liss Sent:
Message 5 of 6 , Jan 27, 2007
That will work nicely.
Thnx
Dickie
-----Original Message-----
From: wuhu_software_group@yahoogroups.com [mailto:wuhu_software_group@yahoogroups.com]On Behalf Of John Liss
Sent: Friday, January 26, 2007 9:44 AM
To: 'wuhu_software_group@yahoogroups.com'
Subject: RE: [wuhu_software_group] Weather Question
This is what I use
Found it some place on the net:
CLOUD HEIGHT
The cloud height on this site is an estimate of cumulus clouds using a formula based on temperature and dew point. Actual measurements of cloud height are made with a Micropulse Lidar (MPL). This device fires a laser into the sky and measures the backscattered signal. Costs for such a device are beyond the scope of weather hobbyists.
You can use the following formulas to calculate the height (H) of a cumulus cloud base given surface temperature and dewpoint.
A) H(meters)=125* (Tc-Tdc)
B) H(feet)=222( Tf-Tdf)
Note:
A) Tc and Tdc in formula are temperature and dewpoint in degrees Celsius.
B) Tf and Tdf in formula are temperature and dewpoint in degrees Fahrenheit.
> -----Original Message-----
> From: wuhu_software_ group@yahoogroup s.com
> [mailto:wuhu_software_ group@yahoogroup s.com] On Behalf Of <
> Sent: Thursday, January 25, 2007 9:03 PM
> To: Wuhu_Software_ Group
> Subject: [wuhu_software_ group] Weather Question
>
> Is there any equipment that us home weather broadcasters can
> get hold of that can determin cloud height, I have read that
> it is usaully determined by infrared from the ground. Or is
> there another way we can do it, besides just guessing?
>
> thnx
>
> Never-Enuff Technologies
> Colver, Pennsylvania
> ------------ --------- --------- --------- ----
> (A)bort, (R)etry, (S)mack the @#\$&*~ thing!
> ------------ --------- --------- --------- ----
>
>
>
>
> To visit your group on the web, go to:
> http://groups. yahoo.com/ group/wuhu_ software_ group/
>
> Want more or less group messages (Special notices as a
> minimum)?http://groups. yahoo.com/ group/wuhu_ software_ group/join
>
> To unsubscribe from this group, send an email to:
> wuhu_software_ group-unsubscrib e@yahoogroups. com
>
> http://home. comcast.net/ ~wuhu_software/
>
>
>
>
• I have been looking for something like this for almost a year. Is there anything that will work as well for stratus, cirrus or some of the other cloud types?
Message 6 of 6 , Jan 27, 2007
I have been looking for something like this for almost a year. Is there anything that will work as well for stratus, cirrus or some of the other cloud types?
Thanks for this one though
Jeff L.
The Wilton Weather Guy
From: wuhu_software_group@yahoogroups.com [mailto: wuhu_software_group@yahoogroups.com ] On Behalf Of < Dickie Bradford >
Sent: Saturday, January 27, 2007 9:26 PM
To: wuhu_software_group@yahoogroups.com
Subject: RE: [wuhu_software_group] Weather Question
That will work nicely.
Thnx
Dickie
-----Original Message-----
From: wuhu_software_ group@yahoogroup s.com [mailto: wuhu_ software_ group@yahoogroup s.com ]On Behalf Of John Liss
Sent: Friday, January 26, 2007 9:44 AM
To: ' wuhu_software_ group@yahoogroup s.com '
Subject: RE: [wuhu_software_ group] Weather Question
This is what I use
Found it some place on the net:
CLOUD HEIGHT
The cloud height on this site is an estimate of cumulus clouds using a formula based on temperature and dew point. Actual measurements of cloud height are made with a Micropulse Lidar (MPL). This device fires a laser into the sky and measures the backscattered signal. Costs for such a device are beyond the scope of weather hobbyists.
You can use the following formulas to calculate the height (H) of a cumulus cloud base given surface temperature and dewpoint.
A) H(meters)=125* (Tc-Tdc)
B) H(feet)=222( Tf-Tdf)
Note:
A) Tc and Tdc in formula are temperature and dewpoint in degrees Celsius.
B) Tf and Tdf in formula are temperature and dewpoint in degrees Fahrenheit.
> -----Original Message-----
> From: wuhu_software_ group@yahoogroup s.com
> [mailto:wuhu_software_ group@yahoogroup s.com] On Behalf Of <
> Sent: Thursday, January 25, 2007 9:03 PM
> To: Wuhu_Software_ Group
> Subject: [wuhu_software_ group] Weather Question
>
> Is there any equipment that us home weather broadcasters can
> get hold of that can determin cloud height, I have read that
> it is usaully determined by infrared from the ground. Or is
> there another way we can do it, besides just guessing?
>
> thnx
>
> Never-Enuff Technologies
> Colver , Pennsylvania
> ------------ --------- --------- --------- ----
> (A)bort, (R)etry, (S)mack the @#\$&*~ thing!
> ------------ --------- --------- --------- ----
>
>
>
>
> To visit your group on the web, go to:
> http://groups. yahoo.com/ group/wuhu_ software_ group/
>
> Want more or less group messages (Special notices as a
> minimum)?http://groups. yahoo.com/ group/wuhu_ software_ group/join
>
> To unsubscribe from this group, send an email to:
> wuhu_software_ group-unsubscrib e@yahoogroups. com
> | 2,732 | 10,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-47 | latest | en | 0.853155 |
https://socratic.org/questions/how-many-grams-of-oxygen-can-be-produced-by-completely-decomposing-10-g-of-mercu | 1,627,228,523,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00056.warc.gz | 535,480,576 | 6,073 | # How many grams of oxygen can be produced by completely decomposing 10 g of mercury(II) oxide?
$2 H g O \left(s\right) + \Delta \rightarrow H g \left(l\right) + {O}_{2} \left(g\right)$
Moles of $H g O$ $=$ $\frac{10 \cdot g}{216.6 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0461$ $m o l$.
There are thus $\frac{0.0461 \cdot \cancel{m o l}}{2} \times 32.00 \cdot g \cdot \cancel{m o {l}^{-} 1}$ $\text{dioxygen gas}$ $=$ ??g | 168 | 419 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-31 | latest | en | 0.505743 |
https://books.google.no/books?id=lQoAAAAAYAAJ&vq=%22can+perform+a+piece+of+work+in+8+days,+A+and+C%22&dq=editions:UOM39015016363460&lr=&hl=no&output=text&source=gbs_navlinks_s | 1,585,879,451,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00062.warc.gz | 367,942,022 | 11,946 | # First Lessons in Algebra: Being an Easy Introduction to that Science ; Designated for the Use of Academies and Common Schools
Russell, Shattuck, & Company, 1835 - 252 sider
### Hva folk mener -Skriv en omtale
Vi har ikke funnet noen omtaler på noen av de vanlige stedene.
### Innhold
INTRODUCTION 13 ADDITION 24 CHAP IV 45 CHAP V 58 FRACTIONS 70 CHAP VII 90 EQUATIONS OF THE FIRST DEGREE 109 EXERCISES IN GENERALIZATION 172
Two Numbers being given to find the Difference of their 174 To prepare Questions whose Answers shall be free from 181 CHAP X 196 Introduction 202 Pure Equations 216 Affected Equations 223 Questions producing Affected Equations 234 MISCELLANEOUS QUESTIONS 241
### Populære avsnitt
Side 246 - Several gentlemen made an excursion, each taking the same sum of money. Each had as many servants attending him as there were gentlemen ; the number of dollars which each had was double the number of all the servants, and the whole sum of money taken out was 3456 dollars.
Side 122 - A man driving his geese to market, was met by another, who said, good morrow, master, with your hundred geese ; says he, I have not a hundred ; but if I had half as many more as I now have, and two geese and a half, I should have a hundred ; how many had he ? /C s"~
Side 141 - A man and his wife usually drank out a cask of beer in 12 days ; but when the man was from home, it lasted the woman 30 days : how many days would the man alone be in drinking it ? Ans.
Side 251 - The fore-wheel of a carriage makes 6 revolutions more than the hind- wheel in going 120 yards; but if the circumference of each wheel be increased one yard, it will make only 4 revolutions more than the hind-wheel in the same distance.
Side 68 - Divide the first term of the dividend by the first term of the divisor, giving the first term of the quotient. Multiply the whole divisor by this term, and subtract the product from the dividend, arranging the remainder in the same order of powers as the dividend and divisor.
Side 237 - A square court yard has a rectangular gravel walk round it. The side of the court wants 2 yards of being 6 times the breadth of the gravel walk ; and the number of square yards in the walk exceeds the number of yards in the periphery of the court by 164. Required the area of the court ? All equations of the second degree may be reduced to one of the following forms.
Side 138 - A gentleman invested -J of his property in a canal. When he sold out, he lost f of the sum invested, receiving only \$1,446. What was the value of his property when he began ? Ans., \$11,568. 29. A gentleman leaves \$315 to be divided among four servants in the following manner : B is to receive as much as A, and |- as much more; C is to receive as much as A and B, and ^ as much more ; D is to receive as much as the other three, and £ as much more. What is the share of each ? price of the chaise, the...
Side 131 - There is a fish whose head weighs 9 lb. his tail weighs as much as his head and half his body, and his body weighs as much as his head and tail both.
Side 166 - If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C in 10 days : how many days would it take each person to perform the same work alone ? Ans.
Side 125 - The distance from A to D is 34 miles. The distance from A to B is to the distance .from C to D as 2 to 3. And ± of the distance from A to B, added to half the distance from C to D, is three times the distance from B to C. What are the respective distances ? Ans. | 895 | 3,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-16 | latest | en | 0.936493 |
https://jp.mathworks.com/matlabcentral/answers/1732750-how-to-compare-two-values-that-are-repeated-in-two-vectors?s_tid=prof_contriblnk | 1,726,663,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00332.warc.gz | 310,554,395 | 29,383 | # how to compare two values that are repeated in two vectors?
5 ビュー (過去 30 日間)
giancarlo maldonado cardenas 2022 年 6 月 2 日
Hello everyone, I'm new to matlab, I wanted to know how I can compare 2 values that are repeated between two vectors, I'll explain myself better, for example I have these two vectors.
A = [10 56 34 11 34 ];
B = [10 56 34 11 ];
Now I need to compare element by element, I mean...
vector B = [(10) 56 34 11]; element 10 is repeated in vector A? No then it is assigned the number 1.
vector B = [10 (56) 34 11 ]; element 56 is repeated in vector A? No then it is assigned the number 1.
vector B = [10 56 (34) 11 ]; element 34 is repeated in vector A? Yes! then it is assigned the number 0.
vector B = [10 56 34 (11) ]; element 34 is repeated in vector A? Yes! then it is assigned the number 1.
the result has to be an array, equal to this.
result= [10 1
56 1
34 0
11 1]
any help will be very helpful.
##### 2 件のコメントなしを表示なしを非表示
Image Analyst 2022 年 6 月 2 日
What if the numbers are not in the same index in the two vectors, like
A = [10 56 34 11 34 ];
B = [11 10 56 34 5 ];
are the 10, 56, and 34 repeated in B? What's your definition of repeated?
And is this your homework (sounds like it)?
giancarlo maldonado cardenas 2022 年 6 月 2 日
Maybe I expressed myself in the wrong way!
I meant for example in the vector B = [10 56 (34) 11 ]; in vector A element 34 is repeated twice? Yes! then it is assigned the number 0.
then all the other values of B are not repeated twice in A so it is assigned the label 1, the only value of B repeated twice in A is the value 34, therefore it is assigned the label 0.
サインインしてコメントする。
### 採用された回答
Voss 2022 年 6 月 2 日
A = [10 56 34 11 34 ];
B = [10 56 34 11 ];
result = [B.' sum(A==B.',2)<=1]
result = 4×2
10 1 56 1 34 0 11 1
Breaking that expression down:
A==B.' % compare each element of A to every element of B
ans = 4×5 logical array
1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0
sum(A==B.',2) % sum each row to get the number of occurrences in A of each element of B
ans = 4×1
1 1 2 1
sum(A==B.',2)<=1 % if the number of occurrences is NOT 2 or more, that's a 1 in the result, otherwise a 0
ans = 4×1 logical array
1 1 0 1
##### 2 件のコメントなしを表示なしを非表示
giancarlo maldonado cardenas 2022 年 6 月 5 日
is there a way to do it with a for loop, and not altogether?
I have to compare each element of vector B and traverse vector A in search of if that element is repeated twice, then an IF condition, if label 0 is repeated, if label 1 is not repeated, but this label can also be 0 with a probability of 10%. this is my code when comparing repeated elements in a single vector.
can u help me?
Voss 2022 年 6 月 5 日
I don't understand what you mean by "if label 0 is repeated, if label 1 is not repeated, but this label can also be 0 with a probability of 10%", but here is how you can perform the task from the original question using a for loop:
A = [10 56 34 11 34 ];
B = [10 56 34 11 ];
result = [B.' zeros(numel(B),1)];
for ii = 1:numel(B)
result(ii,2) = nnz(A==B(ii))<=1;
end
disp(result);
10 1 56 1 34 0 11 1
サインインしてコメントする。
### カテゴリ
Help Center および File ExchangeLTE Toolbox についてさらに検索
R2020a
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 1,094 | 3,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.796034 |
https://akjournals.com/view/journals/10473/129/1-2/article-p1.xml | 1,611,586,506,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703581888.64/warc/CC-MAIN-20210125123120-20210125153120-00126.warc.gz | 202,846,328 | 26,242 | Author:
View More View Less
• 1 Institute of Mathematics A, Graz University of Technology, Steyrergasse 30, 8010 Graz, Austria
Restricted access
## Abstract
Let (nk)k≧1 be a lacunary sequence of positive integers, i.e. a sequence satisfying nk+1/nk > q > 1, k ≧ 1, and let f be a “nice” 1-periodic function with ∝01f(x) dx = 0. Then the probabilistic behavior of the system (f(nkx))k≧1 is very similar to the behavior of sequences of i.i.d. random variables. For example, Erdős and Gál proved in 1955 the following law of the iterated logarithm (LIL) for f(x) = cos 2πx and lacunary
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \usepackage{bbm} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$(n_k )_{k \geqq 1}$$ \end{document}
:
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \usepackage{bbm} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\mathop {\lim \sup }\limits_{N \to \infty } (2N\log \log N)^{1/2} \sum\limits_{k = 1}^N {f(n_k x)} = \left\| f \right\|_2$$ \end{document}
((1))
for almost all x ∈ (0, 1), where ‖f2 = (∝01f(x)2dx)1/2 is the standard deviation of the random variables f(nkx). If (nk)k≧1 has certain number-theoretic properties (e.g. nk+1/nk → ∞), a similar LIL holds for a large class of functions f, and the constant on the right-hand side is always ‖f2. For general lacunary (nk)k≧1 this is not necessarily true: Erdős and Fortet constructed an example of a trigonometric polynomial f and a lacunary sequence (nk)k≧1, such that the lim sup in the LIL (1) is not equal to ‖f2 and not even a constant a.e. In this paper we show that the class of possible functions on the right-hand side of (1) can be very large: we give an example of a trigonometric polynomial f such that for any function g(x) with sufficiently small Fourier coefficients there exists a lacunary sequence (nk)k≧1 such that (1) holds with √‖f22 + g(x) instead of ‖f2 on the right-hand side.
### Monthly Content Usage
Abstract Views Full Text Views PDF Downloads
Aug 2020 0 0 0
Sep 2020 0 0 0
Oct 2020 0 0 0
Nov 2020 0 0 0
Dec 2020 0 0 0
Jan 2021 0 0 0
Feb 2021 0 0 0
## On relatively weakly almost Lindelöf subsets
Author: Yan-Kui Song | 893 | 2,618 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.549084 |
https://math.stackexchange.com/questions/1327171/trigonometry-application-tire-durability | 1,563,373,722,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525312.3/warc/CC-MAIN-20190717141631-20190717163631-00263.warc.gz | 469,637,877 | 34,771 | # trigonometry application : tire durability
Two new cars are each driven at an average speed of 60 mph for an extended highway test drive of 2000 miles. The diameter of the wheels of the two cars are 15 in. and 16 in., respectively. If the cars use tires of equal durability and profile, differing only by the diameter, which car will probably need new tires first.
How do I answer this kind of problem? Shall I use linear velocity? In what way can I provide a better explanation that the smaller tire will need new tires? It's just a hypothesis I made since if the tire is smaller, it will surely wear off than that of the bigger tire because it will rotate faster than that of the bigger one. So the surface of the smaller tire is used more than that of the bigger tire. I don't know how to explain mathematically. Please help.
The number of revolutions of the wheels is given by: $N=\dfrac{L}{\pi d}$ where $d$ is the diameter and $L$ is the path length. No trigonometry is needed andyour intition is correct.
If $N_0$ is the the number of revolutions after wich we must change the wheelers, than, from the velocity you can find the time as $t_0=\dfrac{N_0 \pi d}{v}$. | 286 | 1,175 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-30 | latest | en | 0.95084 |
https://www.sporcle.com/games/Jurve/himym-teds-last-words-to-his-kids/results | 1,544,523,946,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823614.22/warc/CC-MAIN-20181211083052-20181211104552-00408.warc.gz | 1,034,391,535 | 26,000 | ### Television Quiz
Random Television Quiz
# HIMYM: Ted's last words to his kids. Quiz Stats
#### by Jurve Plays Quiz not verified by Sporcle
Challenge
Share
Tweet
Embed
Score 0/202 Timer 07:00
0 Plays Today
% Correct
i
94.9%
i
94.9%
i
94.9%
i
94.9%
i
94.9%
I
94.9%
I
94.9%
i
94.9%
I
94.9%
I
94.9%
I
94.9%
I
94.9%
i
94.9%
a
85.8%
a
85.8%
and
76.6%
And
76.6%
and
76.6%
and
76.6%
and
76.6%
and
76.6%
and
76.6%
You
76.6%
that
75.1%
that
75.1%
that
75.1%
That
75.1%
that
75.1%
that
75.1%
that
75.1%
the
74.6%
the
74.6%
the
74.6%
the
74.6%
the
74.6%
the
74.6%
your
68.5%
me
68.5%
me.
68.5%
me
68.5%
It
67%
it
67%
kids,
67%
it
67%
is
64.5%
was
64%
was
64%
was
64%
met
62.9%
love
62.4%
to
61.4%
to
61.4%
to
61.4%
to
61.4%
she
60.4%
be
58.9%
be
58.9%
her,
57.4%
her,
57.4%
her
57.4%
mom
56.9%
when
56.9%
ever
56.3%
sick.
56.3%
ever
56.3%
got
55.3%
of
53.3%
of
53.3%
% Correct
we
52.8%
every
51.3%
every
51.3%
every
51.3%
every
51.3%
every
51.3%
every
51.3%
long
49.2%
long
49.2%
long
49.2%
as
48.7%
as
48.7%
as
48.7%
as
48.7%
as
48.7%
even
48.2%
Even
48.2%
on
48.2%
on
48.2%
have
47.2%
have
47.2%
this
47.2%
for
47.2%
for
47.2%
47.2%
47.2%
not
46.2%
not
46.2%
can,
45.2%
can
45.2%
can
45.2%
can
45.2%
But
43.7%
in
43.1%
with
42.6%
with
42.6%
with
42.6%
do
41.6%
40.1%
if
40.1%
what
40.1%
at
39.6%
woman
39.1%
times
38.6%
then,
38.6%
times,
38.6%
my
38.1%
knew:
37.6%
fight
37.6%
could
37.6%
could
37.6%
much
37.6%
difficult
37.1%
im
37.1%
difficult
37.1%
there,
37.1%
there
37.1%
moment
36.5%
up,
36.5%
been
36%
morning,
36%
will
35.5%
all
35.5%
second.
35%
through
34.5%
see
34.5%
through
34.5%
christmas
34.5%
% Correct
through
34.5%
or
34.5%
or
34.5%
or
34.5%
or
34.5%
because
34%
train
34%
33.5%
get
33.5%
from
33.5%
never
33%
Sunday
33%
girl
33%
speak.
33%
carried
32.5%
carried
32.5%
carried
32.5%
loving
32%
stupid
32%
God.
32%
God
32%
saw
32%
tap
32%
lesson
31.5%
lesson
31.5%
lesson
31.5%
worst
31.5%
shoulder,
31.5%
might
31%
only
31%
only
31%
thank
31%
Thank
31%
thank.
31%
platform,
31%
stop
30.5%
beautiful
30.5%
hell
29.9%
open
29.4%
mouth
29.4%
gone
28.9%
27.9%
clear.
27.9%
guts
27.9%
walk
27.9%
sleepy
27.4%
stand
27.4%
over
27.4%
5 am.
26.9%
called
24.9%
afternoon,
23.4%
right
22.8%
bump,
22.3%
jealousy
21.8%
pang
17.3%
our
17.3%
way.
17.3%
speed
15.2%
universe
13.7%
came
13.2%
possibly
13.2%
anyone
12.2%
whole
11.2%
boredom
10.2%
else
10.2%
uncertainty
6.6%
### You're not logged in!
Compare scores with friends on all Sporcle quizzes.
OR
### Extras
Created Apr 2, 2014Source
Tags: | 1,277 | 2,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-51 | latest | en | 0.957144 |
https://www.coursehero.com/file/9054023/which-yields-v-12-m-s-The-second-part-involves-either/ | 1,512,975,320,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512208.1/warc/CC-MAIN-20171211052406-20171211072406-00111.warc.gz | 720,985,515 | 22,514 | HW6-phy2048-Spring-2014-solution
# Which yields v 12 m s the second part involves either
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: P46. Our +x direction is east and +y direction is north. The linear momenta for the two m = 2.0 kg parts are then ! ! p1 = mv1 = mv1 " j where v1 = 3.0 m/s, and ! ! p2 = mv2 = m v2 x ˆ + v2y ˆ = mv2 cosθ ˆ + sinθ ˆ i j i j ( ) ( ) where v2 = 5.0 m/s and θ = 30°. The combined linear momentum of both parts is then rrr P = p1 + p2 = mv1 ˆ + mv2 cos θ ˆ + sinθ ˆ = ( mv2 cos θ ) ˆ + ( mv1 + mv2 sin θ ) ˆ j i j i j ( ) = ( 2.0 kg ) ( 5.0 m/s ) ( cos 30° ) ˆ + ( 2.0 kg ) (3.0 m/s + (5.0 m/s ) (sin 30° ) ) ˆ i j ( ) = 8.66 ˆ + 11 ˆ kg ⋅ m/s. i j From conservation of linear momentum we know that this is also the linear momentum of the whol...
View Full Document
{[ snackBarMessage ]}
Ask a homework question - tutors are online | 362 | 984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-51 | latest | en | 0.772973 |
https://www.jiskha.com/questions/107330/to-post-proof-for-the-day-means-to-a-total-the-columns-on-the-journal-and-add-and | 1,600,646,397,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00100.warc.gz | 925,554,064 | 5,428 | # accounting
To post proof for the day means to
A. total the columns on the journal and add and subtract totals as required.
B. match the journal entries against the ledger to detect errors.
C. prepare a bank reconciliation statement.
D. add in any accounts receivable that came in the mail
1. 👍 0
2. 👎 0
3. 👁 179
1. This is a Homework HELP board, not a Homework DO board.
1. 👍 0
2. 👎 0
👩🏫
Ms. Sue
## Similar Questions
1. ### Algebra
How many pivot columns must a 5x7 matrix have if its columns span R^5? Why?
2. ### Language Arts
In "Dust of Snow," the speaker says that something "Has given my heart / A change of mood." What has caused this change of mood? A: Seeing a crow perched in a tree. B: Taking a walk in the woods. C: Thinking about his day.** D:
3. ### discrete math
prove that if n is an integer and 3n+2 is even, then n is even using a)a proof by contraposition b)a proof by contradiction I'll try part b, you'll have to refresh me on what contraposition means here. Here is the claim we start
4. ### accounting (balance sheet, retained earnings...)
Problem P3-5A On September 1, 2002, the account balances of Rijo Equipment Repair Corp. were as follows: Nbr: Debits Nbr: Credits 101 Cash \$4,880 154 Accumulated depreciation \$1,500 112 Accounts receivable 3,520 201 Accounts
1. ### Accounting
how and what do I post from the cash receipts journal to the general journal
2. ### accounting
P4-2A The adjusted trial balance columns of the worksheet for Porter Company are as follows. PORTER COMPANY Worksheet For the Year Ended December 31, 2008 Adjusted Account Trial Balance No. Account Titles Dr. Cr. 101 Cash 18,800
3. ### Physics
A tunnel of length L = 141 m, height H = 6.9 m high, and width 6.0 m (with a flat roof) is to be constructed at distance d = 60 m beneath the ground. The tunnel roof is to be supported entirely by square steel columns, each with a
4. ### Math-calculus
At a certain point on the beach, a post sticks out of the sand, it's top being 76 cm above the beach. The depth of the water at the post varies sinusoidally with time due to the motion of the tides. The depth d is modeled by the
1. ### math solve the problem using the four step plan
At the end of their 3-day vacation, the Palmers traveled a total of 530 miles. On the third day,they drove 75 miles. On the second day,they drove 330 miles. How many miles did they drive the first day? I can only solve this by
2. ### accounting
The cash payments journal used in this problem has only three special amount columns. Under what circumstances would you recommend that additional special amount columns be added to a cash payments journal?
3. ### college
accounting/// which journal would be used if you returned merchandise that had been bought on account , sales journal, purchase journal, cash receipts journal, cash payments journal or general journal. i thought it would be on a
4. ### Lin. Alegra Help
Suppose that you are given the following system of linear equations [variables are x1, x2, x3 and x4] x1 + 3x2 + 9x3 + 2x4 =5 x1 + 3x3 - 4x4 =5 x2 + 2x3 + 3x4 =-1 -2x1 + 3x2 + 5x4 =-1 Let b be the vector such that B = [A b] , and | 862 | 3,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-40 | latest | en | 0.904094 |
https://lkpy.readthedocs.io/projects/lenskit-tf/en/latest/ | 1,656,818,999,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00752.warc.gz | 417,292,687 | 6,451 | # TensorFlow for LensKit
This package provides algorithm implementations, particularly matrix factorization, using TensorFlow. These algorithms serve two purposes:
• Provide classic algorithms ready to use for recommendation or as baselines for new techniques.
• Demonstrate how to connect TensorFlow to LensKit for use in your own experiments.
To install:
pip install lenskit-tf
Or (preferred, once published):
conda install -c conda-forge lenskit-tf
Warning
These implementations are not yet battle-tested — they are here primarily for demonstration purposes at this time.
## Biased MF
These models implement the standard biased matrix factorization model, like lenskit.algorithms.als.BiasedMF, but learn the model parameters using TensorFlow’s gradient descent instead of the alternating least squares algorithm.
### Bias-Based
class lenskit_tf.BiasedMF(features=50, *, bias=True, damping=5, epochs=5, batch_size=10000, reg=0.02, rng_spec=None)
Biased matrix factorization model for explicit feedback, optimized with TensorFlow.
This is a basic TensorFlow implementation of the biased matrix factorization model for rating prediction:
$s(i|u) = b + b_u + b_i + \vec{p}_u \cdot \vec{q_i}$
User and item embedding matrices are regularized with $$L_2$$ regularization, governed by a regularization term $$\lambda$$. Regularizations for the user and item embeddings are then computed as follows:
$\begin{split}\lambda_u = \lambda / |U| \\ \lambda_i = \lambda / |I| \\\end{split}$
This rescaling allows the regularization term to be independent of the number of users and items.
Because the model is very simple, this algorithm works best with large batch sizes.
This implementation uses lenskit.algorithms.bias.Bias for computing the biases, and uses TensorFlow to fit a matrix factorization on the residuals. It then extracts the resulting matrices, and relies on MFPredictor to implement the prediction logic, like lenskit.algorithms.als.BiasedMF. Its code is suitable as an example of how to build a Keras/TensorFlow algorithm implementation for LensKit where TF is only used in the train stage.
A variety of resources informed the design, most notably this one.
Parameters
• features (int) – The number of latent features to learn.
• bias – The bias model to use.
• damping – The bias damping, if bias is True.
• epochs (int) – The number of epochs to train.
• batch_size (int) – The Keras batch size.
• reg (double) – The regularization term $$\lambda$$ used to derive embedding vector regularizations.
• rng_spec – The random number generator initialization.
fit(ratings, **kwargs)
Train a model using the specified ratings (or similar) data.
Parameters
• ratings (pandas.DataFrame) – The ratings data.
• kwargs – Additional training data the algorithm may require. Algorithms should avoid using the same keyword arguments for different purposes, so that they can be more easily hybridized.
Returns
The algorithm object.
predict_for_user(user, items, ratings=None)
Compute predictions for a user and items.
Parameters
• user – the user ID
• items (array-like) – the items to predict
• ratings (pandas.Series) – the user’s ratings (indexed by item id); if provided, they may be used to override or augment the model’s notion of a user’s preferences.
Returns
scores for the items, indexed by item id.
Return type
pandas.Series
### Fully Integrated
class lenskit_tf.IntegratedBiasMF(features=50, *, epochs=5, batch_size=10000, reg=0.02, bias_reg=0.2, rng_spec=None)
Bases: lenskit.algorithms.Predictor
Biased matrix factorization model for explicit feedback, optimizing both bias and embeddings with TensorFlow.
This is a basic TensorFlow implementation of the biased matrix factorization model for rating prediction:
$s(i|u) = b + b_u + b_i + \vec{p}_u \cdot \vec{q_i}$
User and item embedding matrices are regularized with $$L_2$$ regularization, governed by a regularization term $$\lambda$$. Regularizations for the user and item embeddings are then computed as follows:
$\begin{split}\lambda_u = \lambda / |U| \\ \lambda_i = \lambda / |I| \\\end{split}$
This rescaling allows the regularization term to be independent of the number of users and items. The same rescaling applies to the bias regularization.
Because the model is very simple, this algorithm works best with large batch sizes.
This implementation uses TensorFlow to fit the entire model, including user/item biases and residuals, and uses TensorFlow to do the final predictions as well. Its code is suitable as an example of how to build a Keras/TensorFlow algorithm implementation for LensKit where TF used for the entire process.
A variety of resources informed the design, most notably this one and Chin-chi Hsu's example code_.
Parameters
• features (int) – The number of latent features to learn.
• epochs (int) – The number of epochs to train.
• batch_size (int) – The Keras batch size.
• reg (double) – The regularization term for the embedding vectors.
• bias_reg (double) – The regularization term for the bias vectors.
• rng_spec – The random number generator initialization.
model
The Keras model.
fit(ratings, **kwargs)
Train a model using the specified ratings (or similar) data.
Parameters
• ratings (pandas.DataFrame) – The ratings data.
• kwargs – Additional training data the algorithm may require. Algorithms should avoid using the same keyword arguments for different purposes, so that they can be more easily hybridized.
Returns
The algorithm object.
predict_for_user(user, items, ratings=None)
Compute predictions for a user and items.
Parameters
• user – the user ID
• items (array-like) – the items to predict
• ratings (pandas.Series) – the user’s ratings (indexed by item id); if provided, they may be used to override or augment the model’s notion of a user’s preferences.
Returns
scores for the items, indexed by item id.
Return type
pandas.Series
## Bayesian Personalized Rating
class lenskit_tf.BPR(features=50, *, epochs=5, batch_size=10000, reg=0.02, neg_count=1, neg_weight=True, rng_spec=None)
Bases: lenskit.algorithms.Predictor
Bayesian Personalized Ranking with matrix factorization, optimized with TensorFlow.
This is a basic TensorFlow implementation of the BPR algorithm _[BPR].
User and item embedding matrices are regularized with $$L_2$$ regularization, governed by a regularization term $$\lambda$$. Regularizations for the user and item embeddings are then computed as follows:
$\begin{split}\lambda_u = \lambda / |U| \\ \lambda_i = \lambda / |I| \\\end{split}$
This rescaling allows the regularization term to be independent of the number of users and items.
Because the model is relatively simple, optimization works best with large batch sizes.
Parameters
• features (int) – The number of latent features to learn.
• epochs (int) – The number of epochs to train.
• batch_size (int) – The Keras batch size. This is the number of positive examples to sample in each batch. If neg_count is greater than 1, the batch size will be similarly multipled.
• reg (double) – The regularization term for the embedding vectors.
• neg_count (int) – The number of negative examples to sample for each positive one.
• neg_weight (bool) – Whether to weight negative sampling by popularity (True) or not.
• rng_spec – The random number generator initialization.
model
The Keras model.
fit(ratings, **kwargs)
Train a model using the specified ratings (or similar) data.
Parameters
• ratings (pandas.DataFrame) – The ratings data.
• kwargs – Additional training data the algorithm may require. Algorithms should avoid using the same keyword arguments for different purposes, so that they can be more easily hybridized.
Returns
The algorithm object.
predict_for_user(user, items, ratings=None)
Compute predictions for a user and items.
Parameters
• user – the user ID
• items (array-like) – the items to predict
• ratings (pandas.Series) – the user’s ratings (indexed by item id); if provided, they may be used to override or augment the model’s notion of a user’s preferences.
Returns
scores for the items, indexed by item id.
Return type
pandas.Series | 1,825 | 8,169 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-27 | longest | en | 0.795171 |
https://msdn.microsoft.com/en-us/library/system.numerics.complex.asin(v=vs.110).aspx | 1,448,982,526,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398468233.50/warc/CC-MAIN-20151124205428-00344-ip-10-71-132-137.ec2.internal.warc.gz | 836,455,087 | 12,916 | Asin Method
# Complex.Asin Method (Complex)
.NET Framework 4.6 and 4.5
Returns the angle that is the arc sine of the specified complex number.
Namespace: System.Numerics
Assembly: System.Numerics (in System.Numerics.dll)
## Syntax
```public static Complex Asin(
Complex value
)
```
#### Parameters
value
Type: System.Numerics.Complex
A complex number.
#### Return Value
Type: System.Numerics.Complex
The angle which is the arc sine of value.
## Remarks
The Asin method for complex numbers corresponds to the Math.Asin method for real numbers.
The Asin method uses the following formula:
-ImaginaryOne * Log(ImaginaryOne * value + Sqrt(One - value * value))
## Examples
The following example illustrates the Asin method. It shows that passing the value returned by the Asin method to the Sin method returns the original Complex value.
```using System;
using System.Numerics;
public class Example
{
public static void Main()
{
Complex[] values = { new Complex(2.3, 1.4),
new Complex(-2.3, 1.4),
new Complex(-2.3, -1.4),
new Complex(2.3, -1.4) };
foreach (Complex value in values)
Console.WriteLine("Sin(Asin({0})) = {1}",
value, Complex.Sin(Complex.Asin(value)));
}
}
// The example displays the following output:
// Sin(Asin((2.3, 1.4))) = (2.3, 1.4)
// Sin(Asin((-2.3, 1.4))) = (-2.3, 1.4)
// Sin(Asin((-2.3, -1.4))) = (-2.3, -1.4)
// Sin(Asin((2.3, -1.4))) = (2.3, -1.4)
```
## Version Information
Universal Windows Platform
Available since 4.5
.NET Framework
Available since 4.0
Portable Class Library
Supported in: portable .NET platforms
Silverlight
Available since 4.0
Windows Phone
Available since 8.1 | 490 | 1,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2015-48 | longest | en | 0.538516 |
https://manpages.debian.org/bullseye/liblapack-doc/zlangb.3.en.html | 1,638,486,839,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362297.22/warc/CC-MAIN-20211202205828-20211202235828-00576.warc.gz | 429,461,142 | 8,159 | complex16GBauxiliary(3) LAPACK complex16GBauxiliary(3)
# NAME¶
complex16GBauxiliary
# SYNOPSIS¶
## Functions¶
double precision function zlangb (NORM, N, KL, KU, AB, LDAB, WORK)
ZLANGB returns the value of the 1-norm, Frobenius norm, infinity-norm, or the largest absolute value of any element of general band matrix. subroutine zlaqgb (M, N, KL, KU, AB, LDAB, R, C, ROWCND, COLCND, AMAX, EQUED)
ZLAQGB scales a general band matrix, using row and column scaling factors computed by sgbequ.
# Detailed Description¶
This is the group of complex16 auxiliary functions for GB matrices
# Function Documentation¶
## double precision function zlangb (character NORM, integer N, integer KL, integer KU, complex*16, dimension( ldab, * ) AB, integer LDAB, double precision, dimension( * ) WORK)¶
ZLANGB returns the value of the 1-norm, Frobenius norm, infinity-norm, or the largest absolute value of any element of general band matrix.
Purpose:
```
ZLANGB returns the value of the one norm, or the Frobenius norm, or
the infinity norm, or the element of largest absolute value of an
n by n band matrix A, with kl sub-diagonals and ku super-diagonals.```
Returns
ZLANGB
```
ZLANGB = ( max(abs(A(i,j))), NORM = 'M' or 'm'
(
( norm1(A), NORM = '1', 'O' or 'o'
(
( normI(A), NORM = 'I' or 'i'
(
( normF(A), NORM = 'F', 'f', 'E' or 'e'
where norm1 denotes the one norm of a matrix (maximum column sum),
normI denotes the infinity norm of a matrix (maximum row sum) and
normF denotes the Frobenius norm of a matrix (square root of sum of
squares). Note that max(abs(A(i,j))) is not a consistent matrix norm.```
Parameters
NORM
```
NORM is CHARACTER*1
Specifies the value to be returned in ZLANGB as described
above.```
N
```
N is INTEGER
The order of the matrix A. N >= 0. When N = 0, ZLANGB is
set to zero.```
KL
```
KL is INTEGER
The number of sub-diagonals of the matrix A. KL >= 0.```
KU
```
KU is INTEGER
The number of super-diagonals of the matrix A. KU >= 0.```
AB
```
AB is COMPLEX*16 array, dimension (LDAB,N)
The band matrix A, stored in rows 1 to KL+KU+1. The j-th
column of A is stored in the j-th column of the array AB as
follows:
AB(ku+1+i-j,j) = A(i,j) for max(1,j-ku)<=i<=min(n,j+kl).```
LDAB
```
LDAB is INTEGER
The leading dimension of the array AB. LDAB >= KL+KU+1.```
WORK
```
WORK is DOUBLE PRECISION array, dimension (MAX(1,LWORK)),
where LWORK >= N when NORM = 'I'; otherwise, WORK is not
referenced.```
Author
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date
December 2016
## subroutine zlaqgb (integer M, integer N, integer KL, integer KU, complex*16, dimension( ldab, * ) AB, integer LDAB, double precision, dimension( * ) R, double precision, dimension( * ) C, double precision ROWCND, double precision COLCND, double precision AMAX, character EQUED)¶
ZLAQGB scales a general band matrix, using row and column scaling factors computed by sgbequ.
Purpose:
```
ZLAQGB equilibrates a general M by N band matrix A with KL
subdiagonals and KU superdiagonals using the row and scaling factors
in the vectors R and C.```
Parameters
M
```
M is INTEGER
The number of rows of the matrix A. M >= 0.```
N
```
N is INTEGER
The number of columns of the matrix A. N >= 0.```
KL
```
KL is INTEGER
The number of subdiagonals within the band of A. KL >= 0.```
KU
```
KU is INTEGER
The number of superdiagonals within the band of A. KU >= 0.```
AB
```
AB is COMPLEX*16 array, dimension (LDAB,N)
On entry, the matrix A in band storage, in rows 1 to KL+KU+1.
The j-th column of A is stored in the j-th column of the
array AB as follows:
AB(ku+1+i-j,j) = A(i,j) for max(1,j-ku)<=i<=min(m,j+kl)
On exit, the equilibrated matrix, in the same storage format
as A. See EQUED for the form of the equilibrated matrix.```
LDAB
```
LDAB is INTEGER
The leading dimension of the array AB. LDA >= KL+KU+1.```
R
```
R is DOUBLE PRECISION array, dimension (M)
The row scale factors for A.```
C
```
C is DOUBLE PRECISION array, dimension (N)
The column scale factors for A.```
ROWCND
```
ROWCND is DOUBLE PRECISION
Ratio of the smallest R(i) to the largest R(i).```
COLCND
```
COLCND is DOUBLE PRECISION
Ratio of the smallest C(i) to the largest C(i).```
AMAX
```
AMAX is DOUBLE PRECISION
Absolute value of largest matrix entry.```
EQUED
```
EQUED is CHARACTER*1
Specifies the form of equilibration that was done.
= 'N': No equilibration
= 'R': Row equilibration, i.e., A has been premultiplied by
diag(R).
= 'C': Column equilibration, i.e., A has been postmultiplied
by diag(C).
= 'B': Both row and column equilibration, i.e., A has been
replaced by diag(R) * A * diag(C).```
Internal Parameters:
```
THRESH is a threshold value used to decide if row or column scaling
should be done based on the ratio of the row or column scaling
factors. If ROWCND < THRESH, row scaling is done, and if
COLCND < THRESH, column scaling is done.
LARGE and SMALL are threshold values used to decide if row scaling
should be done based on the absolute size of the largest matrix
element. If AMAX > LARGE or AMAX < SMALL, row scaling is done.```
Author
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date
December 2016
# Author¶
Generated automatically by Doxygen for LAPACK from the source code.
Sat Aug 1 2020 Version 3.9.0 | 1,675 | 5,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-49 | latest | en | 0.705877 |
https://chemistry.stackexchange.com/questions/12256/which-ion-of-iron-is-produced-in-a-reaction-between-iron-and-copperii-sulfate | 1,642,625,688,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00516.warc.gz | 213,440,512 | 35,308 | # Which ion of iron is produced in a reaction between iron and copper(II) sulfate?
Using these equations: \begin{aligned} \text{(Fe2):}&&\ce{ Fe + CuSO4 &-> FeSO4 + Cu}\\ \text{(Fe3):}&&\ce{ 2Fe + 3CuSO4 &-> Fe2(SO4)3 + 3Cu} \end{aligned}
Beginning with $$\pu{0.78 g}$$ of iron, the theoretical yields of copper are $$\pu{0.89 g}$$ (with $$\ce{Fe^{2+}}$$) and $$\pu{1.33 g}$$ (with $$\ce{Fe^{3+}}$$), and the actual yield was $$\pu{0.78 g}$$ of copper. I know to find of the ratio of the moles of iron used to the moles of copper formed (88% for $$\ce{Fe^{2+}}$$, 58% for $$\ce{Fe^{3+}}$$), and so the clear choice would be that this reaction formed $$\ce{Fe^{2+}}$$ ions.
But my teacher said this would form $$\ce{Fe^{3+}}$$ ions, but why is that? Shouldn't it have formed $$\ce{Fe^{2+}}$$? The reaction formed $$\pu{0.12 mol}$$ of copper, compared to the $$\pu{0.14 mol}$$ in the theoretical yield, and the other $$\pu{0.02 mol}$$ could have just been lost somewhere along the reaction process.
When "doing" science, usually you try to (in-)validate a model of the workings of the world. In the case of chemists, this world is mostly confined to the scale of molecules.
That said, let's proceed with
# A Theoretical Model
Listed below are some standard reduction potentials I've gathered from the Handbook of Chemistry and Physics, 95th ed. (probably paywalled for you): \begin{align} &&E^0_\text{red} / \mathrm{V}\\ \hline \ce{Fe^{2+} + 2e- & <=> Fe} & -0.447 \\ \ce{Fe^{3+} + 3e- & <=> Fe} & -0.037 \\ \ce{Cu^{2+} + 2e- & <=> Cu} & 0.342 \\ \end{align}
Now we can calculate the potential (and the resulting electromotive force EMF) for each of the reactions: \begin{align} E^0_\text{(Fe2)} = E^{0}_\text{red}(\text{red}) - E^0_\text{red}(\text{ox}) &= 0.342~\mathrm{V} - (-0.447~\mathrm{V}) = 0.789~\mathrm{V} \\ E^0_\text{(Fe3)} &= 0.342~\mathrm{V} - (-0.037~\mathrm{V}) = 0.379~\mathrm{V} \end{align}
It is clear that the reaction (Fe2) releases more energy and thus should be preferred over the other reaction. This means that we expect a yield of 100% Cu and not 150% Cu.
Now that we have gathered some theoretical understanding of the matter, let's proceed to the interesting part, which is
# Confirmation With Experimental Results
You have correctly calculated the theoretical yields for copper to be 0.89 g and 1.33 g, respectively. The corresponding amounts of copper are 0.014 mol and 0.021 mol.
You have measured the amount of copper produced, and it clocks in at 0.78 g. This is already a strong indication that we should side with (Fe2) on this one. But just to really be sure, let's look at the error that we would have if it was (Fe3): $$\varepsilon = \frac{\Delta x}{x} = 41.4\%$$ This is a very high error, especially when compared to the error for (Fe2), which is only 12.1%. Assuming that you did not perform the reaction with pure Fe but a mixture of iron with some iron oxides as contaminants, did not use an analytical scale and performed the experiment in fairly uncontrolled conditions (are you sure the reaction was complete when you aborted?) this error is not as bad as it seems.
With the experimental results in our pocket, we can now move on to the final part:
# The Conclusion
We have shown, using theoretical predictions and experimental confirmation, that the reaction observed was the oxidation of iron to iron(II) under reduction of copper(II) to copper.
By the way, most of the calculations I did with an iPython Notebook, which you can view here. I might have messed something up, you never know.
• Your analysis misses a subtle point. If you consider just the ocidationnof Fe(II) to Fe(III), that has a standard potential of +0.783 V. Therefore Fe(II) is more stable at the potential required to reduce copper, and in fact Fe(III) would oxidize copper metal back to Cu(II) (which then reacts with more elemental iron). This reaction cycle tends to have ugly consequences in pickling of steel strip. Sep 15 '20 at 16:23 | 1,163 | 3,976 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-05 | latest | en | 0.867779 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/190/2/b/a/ | 1,723,285,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00644.warc.gz | 658,210,565 | 58,162 | # Properties
Label 190.2.b.a Level $190$ Weight $2$ Character orbit 190.b Analytic conductor $1.517$ Analytic rank $0$ Dimension $4$ Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [190,2,Mod(39,190)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(190, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([1, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("190.39");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$190 = 2 \cdot 5 \cdot 19$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 190.b (of order $$2$$, degree $$1$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$1.51715763840$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\zeta_{8})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{4} + 1$$ x^4 + 1 Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{8}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q - \zeta_{8}^{2} q^{2} + (\zeta_{8}^{3} - \zeta_{8}^{2} + \zeta_{8}) q^{3} - q^{4} + (\zeta_{8}^{3} + 2 \zeta_{8}) q^{5} + ( - \zeta_{8}^{3} + \zeta_{8} - 1) q^{6} + (\zeta_{8}^{3} + 3 \zeta_{8}^{2} + \zeta_{8}) q^{7} + \zeta_{8}^{2} q^{8} + ( - 2 \zeta_{8}^{3} + 2 \zeta_{8}) q^{9} +O(q^{10})$$ q - z^2 * q^2 + (z^3 - z^2 + z) * q^3 - q^4 + (z^3 + 2*z) * q^5 + (-z^3 + z - 1) * q^6 + (z^3 + 3*z^2 + z) * q^7 + z^2 * q^8 + (-2*z^3 + 2*z) * q^9 $$q - \zeta_{8}^{2} q^{2} + (\zeta_{8}^{3} - \zeta_{8}^{2} + \zeta_{8}) q^{3} - q^{4} + (\zeta_{8}^{3} + 2 \zeta_{8}) q^{5} + ( - \zeta_{8}^{3} + \zeta_{8} - 1) q^{6} + (\zeta_{8}^{3} + 3 \zeta_{8}^{2} + \zeta_{8}) q^{7} + \zeta_{8}^{2} q^{8} + ( - 2 \zeta_{8}^{3} + 2 \zeta_{8}) q^{9} + ( - 2 \zeta_{8}^{3} + \zeta_{8}) q^{10} + (\zeta_{8}^{3} - \zeta_{8}) q^{11} + ( - \zeta_{8}^{3} + \zeta_{8}^{2} - \zeta_{8}) q^{12} + ( - 2 \zeta_{8}^{3} + \cdots - 2 \zeta_{8}) q^{13} + \cdots - 4 q^{99} +O(q^{100})$$ q - z^2 * q^2 + (z^3 - z^2 + z) * q^3 - q^4 + (z^3 + 2*z) * q^5 + (-z^3 + z - 1) * q^6 + (z^3 + 3*z^2 + z) * q^7 + z^2 * q^8 + (-2*z^3 + 2*z) * q^9 + (-2*z^3 + z) * q^10 + (z^3 - z) * q^11 + (-z^3 + z^2 - z) * q^12 + (-2*z^3 - 3*z^2 - 2*z) * q^13 + (-z^3 + z + 3) * q^14 + (-2*z^3 + z^2 + z - 3) * q^15 + q^16 + z^2 * q^17 + (-2*z^3 - 2*z) * q^18 + q^19 + (-z^3 - 2*z) * q^20 + (2*z^3 - 2*z + 1) * q^21 + (z^3 + z) * q^22 + (3*z^3 - 5*z^2 + 3*z) * q^23 + (z^3 - z + 1) * q^24 + (3*z^2 - 4) * q^25 + (2*z^3 - 2*z - 3) * q^26 + (z^3 + z^2 + z) * q^27 + (-z^3 - 3*z^2 - z) * q^28 + (2*z^3 - 2*z + 3) * q^29 + (-z^3 + 3*z^2 - 2*z + 1) * q^30 + (-3*z^3 + 3*z + 2) * q^31 - z^2 * q^32 + (z^3 - 2*z^2 + z) * q^33 + q^34 + (6*z^3 + z^2 - 3*z - 3) * q^35 + (2*z^3 - 2*z) * q^36 + (-6*z^3 - 6*z) * q^37 - z^2 * q^38 + (-z^3 + z + 1) * q^39 + (2*z^3 - z) * q^40 + (3*z^3 - 3*z) * q^41 + (2*z^3 - z^2 + 2*z) * q^42 + (3*z^3 - 6*z^2 + 3*z) * q^43 + (-z^3 + z) * q^44 + (6*z^2 + 2) * q^45 + (-3*z^3 + 3*z - 5) * q^46 + (z^3 - z^2 + z) * q^48 + (6*z^3 - 6*z - 4) * q^49 + (4*z^2 + 3) * q^50 + (z^3 - z + 1) * q^51 + (2*z^3 + 3*z^2 + 2*z) * q^52 + (-6*z^3 + 3*z^2 - 6*z) * q^53 + (-z^3 + z + 1) * q^54 + (-3*z^2 - 1) * q^55 + (z^3 - z - 3) * q^56 + (z^3 - z^2 + z) * q^57 + (2*z^3 - 3*z^2 + 2*z) * q^58 + (7*z^3 - 7*z + 3) * q^59 + (2*z^3 - z^2 - z + 3) * q^60 + (-3*z^3 + 3*z + 10) * q^61 + (-3*z^3 - 2*z^2 - 3*z) * q^62 + (6*z^3 + 4*z^2 + 6*z) * q^63 - q^64 + (-6*z^3 - 2*z^2 + 3*z + 6) * q^65 + (-z^3 + z - 2) * q^66 + (-3*z^3 + 9*z^2 - 3*z) * q^67 - z^2 * q^68 + (-8*z^3 + 8*z - 11) * q^69 + (3*z^3 + 3*z^2 + 6*z + 1) * q^70 + (z^3 - z - 12) * q^71 + (2*z^3 + 2*z) * q^72 + (-6*z^3 - 3*z^2 - 6*z) * q^73 + (6*z^3 - 6*z) * q^74 + (-z^3 + 4*z^2 - 7*z + 3) * q^75 - q^76 + (-3*z^3 - 2*z^2 - 3*z) * q^77 + (-z^3 - z^2 - z) * q^78 + (-6*z^3 + 6*z - 2) * q^79 + (z^3 + 2*z) * q^80 + (-6*z^3 + 6*z - 1) * q^81 + (3*z^3 + 3*z) * q^82 + (-6*z^3 - 6*z^2 - 6*z) * q^83 + (-2*z^3 + 2*z - 1) * q^84 + (2*z^3 - z) * q^85 + (-3*z^3 + 3*z - 6) * q^86 + (5*z^3 - 7*z^2 + 5*z) * q^87 + (-z^3 - z) * q^88 + (5*z^3 - 5*z) * q^89 + (-2*z^2 + 6) * q^90 + (-9*z^3 + 9*z + 13) * q^91 + (-3*z^3 + 5*z^2 - 3*z) * q^92 + (-z^3 + 4*z^2 - z) * q^93 + (z^3 + 2*z) * q^95 + (-z^3 + z - 1) * q^96 + (4*z^3 - 6*z^2 + 4*z) * q^97 + (6*z^3 + 4*z^2 + 6*z) * q^98 - 4 * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q - 4 q^{4} - 4 q^{6}+O(q^{10})$$ 4 * q - 4 * q^4 - 4 * q^6 $$4 q - 4 q^{4} - 4 q^{6} + 12 q^{14} - 12 q^{15} + 4 q^{16} + 4 q^{19} + 4 q^{21} + 4 q^{24} - 16 q^{25} - 12 q^{26} + 12 q^{29} + 4 q^{30} + 8 q^{31} + 4 q^{34} - 12 q^{35} + 4 q^{39} + 8 q^{45} - 20 q^{46} - 16 q^{49} + 12 q^{50} + 4 q^{51} + 4 q^{54} - 4 q^{55} - 12 q^{56} + 12 q^{59} + 12 q^{60} + 40 q^{61} - 4 q^{64} + 24 q^{65} - 8 q^{66} - 44 q^{69} + 4 q^{70} - 48 q^{71} + 12 q^{75} - 4 q^{76} - 8 q^{79} - 4 q^{81} - 4 q^{84} - 24 q^{86} + 24 q^{90} + 52 q^{91} - 4 q^{96} - 16 q^{99}+O(q^{100})$$ 4 * q - 4 * q^4 - 4 * q^6 + 12 * q^14 - 12 * q^15 + 4 * q^16 + 4 * q^19 + 4 * q^21 + 4 * q^24 - 16 * q^25 - 12 * q^26 + 12 * q^29 + 4 * q^30 + 8 * q^31 + 4 * q^34 - 12 * q^35 + 4 * q^39 + 8 * q^45 - 20 * q^46 - 16 * q^49 + 12 * q^50 + 4 * q^51 + 4 * q^54 - 4 * q^55 - 12 * q^56 + 12 * q^59 + 12 * q^60 + 40 * q^61 - 4 * q^64 + 24 * q^65 - 8 * q^66 - 44 * q^69 + 4 * q^70 - 48 * q^71 + 12 * q^75 - 4 * q^76 - 8 * q^79 - 4 * q^81 - 4 * q^84 - 24 * q^86 + 24 * q^90 + 52 * q^91 - 4 * q^96 - 16 * q^99
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/190\mathbb{Z}\right)^\times$$.
$$n$$ $$21$$ $$77$$ $$\chi(n)$$ $$1$$ $$-1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
39.1
−0.707107 − 0.707107i 0.707107 + 0.707107i 0.707107 − 0.707107i −0.707107 + 0.707107i
1.00000i 2.41421i −1.00000 −0.707107 2.12132i −2.41421 1.58579i 1.00000i −2.82843 −2.12132 + 0.707107i
39.2 1.00000i 0.414214i −1.00000 0.707107 + 2.12132i 0.414214 4.41421i 1.00000i 2.82843 2.12132 0.707107i
39.3 1.00000i 0.414214i −1.00000 0.707107 2.12132i 0.414214 4.41421i 1.00000i 2.82843 2.12132 + 0.707107i
39.4 1.00000i 2.41421i −1.00000 −0.707107 + 2.12132i −2.41421 1.58579i 1.00000i −2.82843 −2.12132 0.707107i
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 190.2.b.a 4
3.b odd 2 1 1710.2.d.c 4
4.b odd 2 1 1520.2.d.e 4
5.b even 2 1 inner 190.2.b.a 4
5.c odd 4 1 950.2.a.f 2
5.c odd 4 1 950.2.a.g 2
15.d odd 2 1 1710.2.d.c 4
15.e even 4 1 8550.2.a.bn 2
15.e even 4 1 8550.2.a.cb 2
20.d odd 2 1 1520.2.d.e 4
20.e even 4 1 7600.2.a.v 2
20.e even 4 1 7600.2.a.bg 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
190.2.b.a 4 1.a even 1 1 trivial
190.2.b.a 4 5.b even 2 1 inner
950.2.a.f 2 5.c odd 4 1
950.2.a.g 2 5.c odd 4 1
1520.2.d.e 4 4.b odd 2 1
1520.2.d.e 4 20.d odd 2 1
1710.2.d.c 4 3.b odd 2 1
1710.2.d.c 4 15.d odd 2 1
7600.2.a.v 2 20.e even 4 1
7600.2.a.bg 2 20.e even 4 1
8550.2.a.bn 2 15.e even 4 1
8550.2.a.cb 2 15.e even 4 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{3}^{4} + 6T_{3}^{2} + 1$$ acting on $$S_{2}^{\mathrm{new}}(190, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$(T^{2} + 1)^{2}$$
$3$ $$T^{4} + 6T^{2} + 1$$
$5$ $$T^{4} + 8T^{2} + 25$$
$7$ $$T^{4} + 22T^{2} + 49$$
$11$ $$(T^{2} - 2)^{2}$$
$13$ $$T^{4} + 34T^{2} + 1$$
$17$ $$(T^{2} + 1)^{2}$$
$19$ $$(T - 1)^{4}$$
$23$ $$T^{4} + 86T^{2} + 49$$
$29$ $$(T^{2} - 6 T + 1)^{2}$$
$31$ $$(T^{2} - 4 T - 14)^{2}$$
$37$ $$(T^{2} + 72)^{2}$$
$41$ $$(T^{2} - 18)^{2}$$
$43$ $$T^{4} + 108T^{2} + 324$$
$47$ $$T^{4}$$
$53$ $$T^{4} + 162T^{2} + 3969$$
$59$ $$(T^{2} - 6 T - 89)^{2}$$
$61$ $$(T^{2} - 20 T + 82)^{2}$$
$67$ $$T^{4} + 198T^{2} + 3969$$
$71$ $$(T^{2} + 24 T + 142)^{2}$$
$73$ $$T^{4} + 162T^{2} + 3969$$
$79$ $$(T^{2} + 4 T - 68)^{2}$$
$83$ $$T^{4} + 216T^{2} + 1296$$
$89$ $$(T^{2} - 50)^{2}$$
$97$ $$T^{4} + 136T^{2} + 16$$ | 4,919 | 9,149 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.482385 |
https://math.stackexchange.com/questions/3014033/notation-for-the-derivative-of-a-function-f-or-fx/3014049 | 1,709,315,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00889.warc.gz | 380,953,926 | 39,337 | # Notation for the derivative of a function: $f'$ or $f'(x)\;$?
The derivative of a function is often defined as $$f'$$ and $$f'(x)$$. So which one is it? $$f'(x)$$ is the output of the function $$f'$$, so why do I see people using $$f'$$ and $$f'(x)$$ interchangeably to refer to the derivative of a function?
• By the way, check the first bullet point here $\ddot \smile$ Nov 26, 2018 at 9:53
• $f'$ is the function (therefore a map), $f'(x)$ is the value of the function in the point $x$ (therefore a number). Yes, people mix them up all the times. Nov 26, 2018 at 11:01
• You are correct that $f$ is the function and $f(x)$ is the value of the function when evaluated at a point $x$ in its domain (ditto $f'$ and $f'(x)$). Many elementary texts blur this distinction in an attempt to "dumb down" the material. This causes no end of confusion later on, and you have done well to note the problem. Nov 26, 2018 at 13:20
• @J.Smith In case my answer would be deleted I let here the main reference I've found on that topic Calculus for Dummies
– user
Nov 26, 2018 at 14:36
By definition a function is a triple $$(f,D,C)$$, which is very often denoted by $$f \colon D \to C$$, where $$C,D$$ are two sets and $$f$$ associates to each element of $$D$$ one and only one element of $$C$$.
So when it is clear what $$C$$ and $$D$$ are, or in cases where it is not possible or not necessary to write them down, you just write $$f$$. The expression $$f(x)$$ denotes the element in $$C$$ which $$x \in D$$ is mapped to. So $$f$$ is a function, $$f(x)$$ is an element of $$C$$, two completely different things.
• Nice answer! (+1) Nov 26, 2018 at 8:52
• Thank you @RobertZ. Nov 26, 2018 at 8:59
• I agree with the forst part (+1) but I would be more relaxed for the second part, I think that using f(x) to indicate the function can be tolerated.
– user
Nov 26, 2018 at 9:22
• I think it is a good habit for a beginner to stick to the definitions. After some practice and experience one becomes conscious on where conventions can be relaxed and some abuse of language might be ok. Nov 26, 2018 at 10:06
• The distinction between $f$ and $f(x)$ is important but "let $f(x) =x\sin(1/x)$" has almost become a routine shorthand for "let $f:\mathbb{R} \setminus\{0\}\to\mathbb {R}$ be a function defined by $f(x) =x\sin(1/x)$". Nov 27, 2018 at 5:05
$$f$$ denotes the function and $$f(x)$$ the output of the function when evaluated at $$x$$.
This convention does not differ for the derivative.
• f’ denotes the derivative and f’(x) denotes the output of the derivative, which is the instantaneous rate of change of a function at any point. Correct?
– user618086
Nov 26, 2018 at 10:08
• @J.Smith: you get it.
– user65203
Nov 26, 2018 at 10:13
• @YvesDaoust I think we could be more "relaxed" with that definition and notation, no one will be wound considering the function $f(x)$ :)
– user
Nov 26, 2018 at 10:17
• @gimusi: this would reduce the expressive power (not possible to distinguish the function and the value) and create ambiguities. So, no.
– user65203
Nov 26, 2018 at 10:20
• @YvesDaoust Also f creates ambiguity if we do not specify the domain and the codomain, therefore anytime we refer to a function we should use $f:A\to B$. I don't thing it would be a useful notation. $f(x)$ to indicate the funtion can be used many times without any ambiguity. Of course we ca suggest to do not use that but it is a matter of preferences and not a law.
– user
Nov 26, 2018 at 10:23
The derivative of the function $$f$$ is $$f'$$. People usually make the mistake of saying that it is $$f'(x)$$, just like they talk about, say, the function $$\sin(x)$$, when, in fact, they should be talking about the $$\sin$$ function.
• So when people say 2x is the derivative of x^2, is that incorrect? Also, am I correct in saying that the derivative is the function of the form y=f’(x) whose output represents the instantaneous rate of change at any point of a function or the slope of the tangent line to a point on a curve?
– user618086
Nov 26, 2018 at 8:38
• It is an abuse of language. It would be correct to say that $(\operatorname{Id}^2)'=2\operatorname{Id}$ and, of course, that's what people mean when they say that the derivative of $x^2$ is $2x$. Concerning your second question, the answer is affirmative. Nov 26, 2018 at 8:42
• This is too far away from the original question and I suggest that you post it as another question. But, for me (and, I think, for most users of this forum)$$f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}.$$ Nov 26, 2018 at 8:59
• I have to disagree with José when he says that "of course, that's what people mean when they say that the derivative of $x^2$ is $2x$". I agree that's what people mean when they know what they're talking about and understand the distinction outlined in several answers in this question. The thing is, most people that say this do not understand this distinction, at least I remember a time (pre-university) where I didn't since I didn't even know of a proper definition of function. Nov 26, 2018 at 9:47
• @TaylorRendon Yes, that gives a good idea about the meaning of derivative of a function at a point. Jan 1, 2021 at 23:18
I would read $$f'(x)$$ as "the function $$f'$$ applied to the element $$x$$ of the domain". This gives us a a new element in the range. Meanwhile I read $$f'$$ as a relation, it tells us which elements are mapped to which other elements. The prime just tells us that it is relation to some other function $$f$$ in a very specific way (derivation).
Example: Our 'input' set is $$\{1,2,3 \}$$ our output set is $$\{A,B,C,D\}$$ $$f=\{(1,C),(2,A),(3,D) \}$$ So we now know that $$f(1)=C$$ and $$f(2)=A$$. Notice that the element $$B$$ is not reached and this function is not surjective.
What you should take from this finite example is that a function is a rule that tells us which elements are in a way "paired", while $$f(x)$$ tells us about a specific pair. However sometimes people just represent the function like this by saying:
For arbitrary $$x$$ (so in our example $$1$$,$$2$$ or $$3$$), $$f(x)$$ is given by $$\dots$$ This is indeed another representation of the same idea, but mathematicians ofter prefer the "relation" idea. | 1,799 | 6,210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 47, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.938086 |
https://metanumbers.com/183103000 | 1,601,239,926,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401578485.67/warc/CC-MAIN-20200927183616-20200927213616-00097.warc.gz | 509,773,761 | 8,193 | ## 183103000
183,103,000 (one hundred eighty-three million one hundred three thousand) is an even nine-digits composite number following 183102999 and preceding 183103001. In scientific notation, it is written as 1.83103 × 108. The sum of its digits is 16. It has a total of 9 prime factors and 128 positive divisors. There are 66,211,200 positive integers (up to 183103000) that are relatively prime to 183103000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 9
• Sum of Digits 16
• Digital Root 7
## Name
Short name 183 million 103 thousand one hundred eighty-three million one hundred three thousand
## Notation
Scientific notation 1.83103 × 108 183.103 × 106
## Prime Factorization of 183103000
Prime Factorization 23 × 53 × 19 × 23 × 419
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 9 Total number of prime factors rad(n) 1831030 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 183,103,000 is 23 × 53 × 19 × 23 × 419. Since it has a total of 9 prime factors, 183,103,000 is a composite number.
## Divisors of 183103000
128 divisors
Even divisors 96 32 16 16
Total Divisors Sum of Divisors Aliquot Sum τ(n) 128 Total number of the positive divisors of n σ(n) 4.71744e+08 Sum of all the positive divisors of n s(n) 2.88641e+08 Sum of the proper positive divisors of n A(n) 3.6855e+06 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 13531.6 Returns the nth root of the product of n divisors H(n) 49.682 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 183,103,000 can be divided by 128 positive divisors (out of which 96 are even, and 32 are odd). The sum of these divisors (counting 183,103,000) is 471,744,000, the average is 3,685,500.
## Other Arithmetic Functions (n = 183103000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 66211200 Total number of positive integers not greater than n that are coprime to n λ(n) 188100 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 10180958 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 66,211,200 positive integers (less than 183,103,000) that are coprime with 183,103,000. And there are approximately 10,180,958 prime numbers less than or equal to 183,103,000.
## Divisibility of 183103000
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 3 0 7
The number 183,103,000 is divisible by 2, 4, 5 and 8.
• Arithmetic
• Abundant
• Polite
• Practical
## Base conversion (183103000)
Base System Value
2 Binary 1010111010011110111000011000
3 Ternary 110202112121002121
4 Quaternary 22322132320120
5 Quinary 333333244000
6 Senary 30100311024
8 Octal 1272367030
10 Decimal 183103000
12 Duodecimal 513a2474
20 Vigesimal 2h47ha0
36 Base36 310j6g
## Basic calculations (n = 183103000)
### Multiplication
n×i
n×2 366206000 549309000 732412000 915515000
### Division
ni
n⁄2 9.15515e+07 6.10343e+07 4.57758e+07 3.66206e+07
### Exponentiation
ni
n2 33526708609000000 6138840926433727000000000 1124040190152794714881000000000000 205815130937547170678855743000000000000000
### Nth Root
i√n
2√n 13531.6 567.848 116.325 44.9303
## 183103000 as geometric shapes
### Circle
Diameter 3.66206e+08 1.15047e+09 1.05327e+17
### Sphere
Volume 2.57143e+25 4.21309e+17 1.15047e+09
### Square
Length = n
Perimeter 7.32412e+08 3.35267e+16 2.58947e+08
### Cube
Length = n
Surface area 2.0116e+17 6.13884e+24 3.17144e+08
### Equilateral Triangle
Length = n
Perimeter 5.49309e+08 1.45175e+16 1.58572e+08
### Triangular Pyramid
Length = n
Surface area 5.807e+16 7.23469e+23 1.49503e+08
## Cryptographic Hash Functions
md5 49d36433ae3091ceb4126438c748670d 0078b1f685807e01b5aaeb9ba384e254e70103a5 44513ec8094f7fdec61498030819d4ec8c98f716770f656b23325f0f2d581367 c367ca9c50c814b83c04884e7d5d3ee80369c5eb1abc71a32485d16adc84d5a5ac1032dda748605212320ca17b06eb4947ff6d918ca2aea64f518fd9ea10c00b f176979fc36a17ec8769f21e17056690b3523aea | 1,581 | 4,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-40 | latest | en | 0.798167 |
https://math.stackexchange.com/questions/572196/expectation-value-quantum-mechanics-momentum-operator | 1,566,218,978,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00356.warc.gz | 552,671,163 | 31,362 | # Expectation value quantum mechanics momentum operator
What is the random variable that belongs to the expectation value of momentum in quantum mechanics?
Or in general: Is there any way we can define the expectation values that occur in quantum mechanics via the definition of expectation values by using random variables?
Or more precisely: What are the random variables in quantum mechanics?
For people who are interested in getting some background on what I am asking:
• You get not only the expectation value but the whole probability distribution of any measurable from the wave function, what more are you asking for? – Abdulh Khazzak Gustav ElFakiri Nov 18 '13 at 19:07
• What is the randomn variable that belongs to 'let's say: the expectation value of the momentum?' – user66906 Nov 18 '13 at 19:47
• The measured momentum – Abdulh Khazzak Gustav ElFakiri Nov 18 '13 at 19:51
• could you write down the actual map please? – user66906 Nov 18 '13 at 19:57
• You set up your experiment, then you rpess a button to perform experiment, a computer probably records the momentum and you read off the screen? – Abdulh Khazzak Gustav ElFakiri Nov 18 '13 at 19:58
As the link you posted from another question explain, in the position representation $\psi(x)$, the eigenvectors with eigenvalues $\hbar k$ of the momentum operator $\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}$ are functions of the form $\psi_k(x)=e^{ikx}$. If we want the probability distribution for the measurement of some observable $O$ in a given state, we need to perform the spectral decomposition $O=\int k dP_k$, where $P_k(\cdot)=\psi_k(\psi_k,\cdot)$ is the projector in the eingespace of eigenvalue $k$ (I assumed that the eigenvalue is non-degenerate for simplicity, but the same also holds for the general case). Then the probability distribution will be given by $p(k)=(\psi, P_k \psi)$. So, for the case of the momentum operator, we have the probability distribution $$p(k)=(\psi,\psi_k)(\psi_k,\psi)=|(\psi_k,\psi)|^2=\left|\int e^{-ikx}\psi(x)dx\right|^2=|\tilde{\psi}(k)|^2,$$ where $\tilde{\psi}(k)$ is the Fourier transform of $\psi(x)$. | 561 | 2,136 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-35 | latest | en | 0.800816 |
https://stats.stackexchange.com/questions/56245/goodness-of-fit-for-contingency-tables | 1,719,118,740,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00871.warc.gz | 473,340,967 | 39,584 | # Goodness-of-fit for contingency tables
I have a question on testing mortality table. Suppose I am given a simple mortality table:
age | prob of dying | prob of surviving
---------------------------------------
20 | 0.01 | 0.99
21 | 0.02 | 0.98
22 | 0.03 | 0.97
23 | 0.04 | 0.96
...
I want to test whether the table fit my observed data, i.e., whether the Observed values
age | actual dead | actual survivors
---------------------------------------
20 | 0 | 397
21 | 1 | 189
22 | 0 | 136
23 | 2 | 100
...
fit the expected values
age | actaul dead | actual survivors
---------------------------------------
20 | 3.97 (397*0,01) | 393.03
21 | 3.8 (190*0.02) | 186.2
22 | 4.08 (136*0.03) | 131.92
23 | 4.08 (102*0.04) | 97.92
...
How do I do that? What method should I use? Should I use Chi-Square, although my expected data is quite small? or there is another method?
• Thank you very much for the answer. I alreeady used Chi Square and G test. I am not sure if that is the right apporach, since I already have predetermined probabilities, so my expectations are calculated by the formula $$E(X_{20})=p_{20}*n_{20}$$ and not by the formula used in exact tests for contingcency tables (row*column/total_number) $$E(X_{20})=\frac{n_{20}*d}{n}$$ $p_{20}$ -prob a 20 year old to die $n_{20}$ -total number of 20 year olds $d$- total number of dead \$Could I still use a test of independence, like Fisher's or chi square, or should I consider a different approach? Commented Apr 16, 2013 at 13:04 | 504 | 1,655 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-26 | latest | en | 0.746149 |
https://www.ipl.org/essay/Exponential-Smoothing-Method-FCC8L5YY2R | 1,623,642,909,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00569.warc.gz | 754,031,052 | 13,023 | # Industrial Smoothing Case Study
874 Words4 Pages
Comparing the error estimates among four forecast methods (total demand): Forecasting method MAD MAPEt TS Range Four period Moving average 13,442 13.19 -10.82 to -1.00 Simple exponential smoothing 16,403 19.32 -5.99 to 6.16 Holt’s model 6,475 7.52 -4.01 to 3.84 Winter’s model 5,227 5.56 -4.62 to 5.99 Table 18: Error estimates for HSG steel sheet total demand forecasting It’s obvious from the table 17 that the two simplest methods are useless for the total demand. This shows that the demand includes a observable growing trend, which make the Moving average method always underforecasts the real demand. Holt’s model and Winter’s model in this case both produce the good result but the Winter’s model proves its superiority due to the lower error.…show more content…
The Trend and seasonality-corrected exponential smoothing method (winter’s model) proves its best accuracy in the four methods with the lowest MAD and MAPE. The two partial demands Domestic market demand and Export market demand show the different seasonality. And the Winter’s model is appropriate for both of them. However, the Winter’s model is even more accurate when applying for the total demand which combines these two different seasonal demands. The MAPE 5.56% produced by the Winter’s model in this paper is better than the MAPE 8% produced in “Supply Chain Management” written by Sunil Chopra and Peter Meindl for the Tahoe Salt case. Therefore, in this steel sheet demand case, the company should forecast the total demand instead of forecasting the two market demands separately. 4. Adjusting forecast Winter’s model is chosen to forecast the future demand for the total market with the following forecast results: F17 = 109,494 tons F18 = 139,773 tons F19 = 131,258 tons F20 = 140,829 tons With MAD = 5,227 and MAPE =…show more content…
Which is in the same point with the safety rules of Neal Wagner, Zbigniew Michalewicz, Sven Schellenberg, Constantin Chiriac and Arvind Mohais (2011) (5). With the up-trend demand data like HSG’s steel sheet demand, the safety rule is given by: “If an up-trend is detected, the first three weekly forecasts must be at least as high as the most recent historical demand seen.” Because there is a seasonal factor in this data so the rule will be: every quarter needs to be higher than that of last year. And the forecasts already follow the | 573 | 2,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-25 | latest | en | 0.873594 |
http://www.chegg.com/homework-help/questions-and-answers/calculate-length-copper-wire-diameter-0200-cm-mass-15620-g-density-copper-892-g-cm3-given--q858215 | 1,472,623,419,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983577646.93/warc/CC-MAIN-20160823201937-00227-ip-10-153-172-175.ec2.internal.warc.gz | 367,678,616 | 13,594 | Calculate the length of a copper wire having a diameter of 0.200 cm and a mass of 15.620 g. The density of copper is 8.92 g/cm3. (Given: volume = ??d2 L/4, where ? = 3.14, d = diameter, and L = length) | 69 | 201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-36 | latest | en | 0.781626 |
somedualspace.wordpress.com | 1,623,496,630,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487582767.0/warc/CC-MAIN-20210612103920-20210612133920-00331.warc.gz | 490,626,517 | 16,592 | ## Autoencoders
There are many things, systems, that have Principal Components Analysis as the result of their evolution, their computation; their dynamics. Things like neural networks, for example. So, this time, I decided to play with autoencoders.
An autoencoder is a feed forward neural network that satisfies three properties:
1. It has only one hidden layer
2. If $n_i$ is the dimension of the input layer, $n_o$ the dimension of the output layer and $p$ the dimension of the hidden one; then $n_i = n_o = n$ and $p < n$
3. The output should be as close as possible to the input (In some sense, usually the quadratic error one)
This is the dimensionality reduction setup of the autoencoder, portraying the characteristic funnel architecture shown in figure 1b; it can be seen as a sequence of two affine maps between 3 vector spaces $X$$H$ and $Y$ as in the figure 1a.
## The curvature of curves and its computation
How much does a curve bend? That looks like an important question to ask. Indeed, it is THE question to ask because curvature is everything we need to know about a curve (modulo some annoying groups we will talk about in the future). If you are too shy to ask, you can compute it and that is what this post is about. In order to compute the curvature you need a bunch of things and for each one there is a bunch of ways of doing it, so, let’s talk about some of them.
## On the curve generated by plotting one sine against another
A circle? A line? Actually, it depends. It depends upon parameters like the frequency or the phase; when these change, really interesting things happen. What I mean by “against another” is that the first sine function will be the $x$-coordinate while the other, the $y$-coordinate; then, if you studied engineering you might say “$y$ vs. $x$” but if you studied biology you might be used to say “$x$ vs. $y$” (personal observation). We end up then with a curve, say $\gamma(t) = (x(t), y(t))$, where
$x(t) = A_1\sin(\omega_1 t + \phi_1)$
and
$y(t) = A_2\sin(\omega_2 t + \phi_2)$.
## An asymmetric pdf with infinite support
When was the last time you needed an asymmetric (skewed) probability density function (pdf) with infinite support? Traditional skewed distributions like the gamma family suffer from a semi-infinite support, that is, $\mathrm{supp}(p) = [0, +\infty)$. The support, if you are out of the loop, is the set of values $x$ in the domain of $p$ such that $p(x) > 0$. Why is this inconvenient? well, I will give more details about the specific application later, meanwhile, let’s say that the fact that its derivative is discontinuous at $0$ is problematic; even more, I need my function to be at least in $C^3$, that is, to have at least $3$ continuous derivatives!.
This post constitutes a somewhat dirty solution being as unwilling as I am to review any literature in depth (I might be inventing the wheel again, some wheel, but who cares; this is a blog!). Here we go.
## An example application with JPA and JavaFX
We are going to create an example application with JPA and JavaFX 2 using Netbeans 7.4 as our IDE. It is assumed that you have installed the JDK 7 and have the Java FX Scene Builder installed and configured in your computer. Our example will be scientifically oriented due to the author’s interests (I know, the content of the example is insubstantial.. but just for the fun of imagination). Let’s assume we are asked to build an application to register the data of an experiment in which some parameters need to be collected from a sample of rats in a laboratory. | 868 | 3,570 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 24, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-25 | latest | en | 0.928608 |
https://cboard.cprogramming.com/c-programming/101040-another-simple-array-question.html | 1,493,054,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119642.3/warc/CC-MAIN-20170423031159-00354-ip-10-145-167-34.ec2.internal.warc.gz | 772,842,855 | 14,477 | # Thread: Another Simple Array Question
1. ## Another Simple Array Question
Why does this program not work? Where does it go wrong. I am trying to print out the array.
Code:
``` int a[5], i;
for (i = 0; i < 5; ++i)
printf("%d", a[i]);```
2. But it does work, it just probably doesn't do what you want it to.
It's obvious you're asking homework questions.
3. Sorry, it's a number but it doesn't print the array which is what I'm trying to do.
I'm not a school kid but I am using a text book but it's not homework. Unfortunately I don't have a teacher so I have to rely on boards like these. So please don't make it more difficult than it already is by discouraging people from helping me with misguided comments. I understand the point you are trying to make but I need this.
My text book has the answer but I really need to know why it doesn't work, not simply that the output is garbage (which is the answer in my text.
4. Well look at the code. What do you think a[0] is?
5. start with post incrementing i, also i dunno if you meant to but what about the brackets around the for loop.
what output are you getting compared to what you want?
and what are you giving each element of the array?
why so?
7. wouldn't that start with i[1] not i[0]
8. Originally Posted by guyfromfl
wouldn't that start with i[1] not i[0]
no it wouldn't
the 3rd part of the for statement is alculated after the loop iteration, there is absolutely no difference between pre or post increment (in C). In C++ pre-increment is even more suitable
9. quite interesting... im using my java experience and combining it with c's and will def. check that out. because i know in jscript that would be a different result.
10. Originally Posted by tabstop
Well look at the code. What do you think a[0] is?
I though it would be 0 but I can see that isn't the case.
start with post incrementing i, also i dunno if you meant to but what about the brackets around the for loop.
what output are you getting compared to what you want?
and what are you giving each element of the array?
Will post incrementing have a different effect on this loop and do I really need the brackets in this case?
The output is a huge number which I take to be an overflow error. I thought it would produce a list of five numbers 0 - 4 which are produced by the for loop.
Basically this is a question in the text and it says the answer is: "Garbage", but I can't understand what element I am missing that turns this program into garbage.
11. Code:
``` int a[5], i;
/* i hope you initialized the array before this lol haha :D */
for (i = 0; i < 5; ++i)
printf("%d ", a[i]); /* a space was not being printed. fixed :D */```
12. Code:
`int a[5] ={0};`
this declares an array AND initializes all its members with zeros
13. Originally Posted by guyfromfl
quite interesting... im using my java experience and combining it with c's and will def. check that out. because i know in jscript that would be a different result.
I never used it that much, but I'm pretty sure it's the same in jscript.
14. ++i means you start i as i + what i was, i++ means you start i as what it was then start adding...correct?
++i would mean you start the loop as 1 if i=0?
15. guyfromfl: No!
The i++ part is executed after the condition check, and after a loop if the condition is true. lol haha lol
Popular pages Recent additions | 848 | 3,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-17 | longest | en | 0.978853 |
https://www.enotes.com/homework-help/write-an-equation-line-that-passes-tthrough-points-232371 | 1,516,090,541,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886237.6/warc/CC-MAIN-20180116070444-20180116090444-00754.warc.gz | 885,443,281 | 9,004 | # Write an equation of a line that passes through the points (0,5) and (45,-220)
neela | Student
The equation of the line passing through the two points (x1,y1) and (x2, y2) is given by:
y-y1 = {(y2-y1)/(x2-x1)} (x-x1).
So the equation of the line passing through the given points (0,5) and (45,-225) is:
y- 5 = {(-220-5)/(45-0))(x-0)
y-5 = (-225/45)x
y-5 = -5x.
5x+y -5 = 0.
So the equation of the line is 5x+y-5 = 0.
giorgiana1976 | Student
We'll write the formula of the equation of the line that is passing through 2 given points:
(x2 - x1)/(x - x1) = (y2 - y1)/(y - y1)
We'll identify x1 = 0, x2 = 45, y1 = 5 and y2 = -220. We'll substitute them into the formula above:
(45 - 0)/(x - 0) = (-220 - 5)/(y - 5)
45/x = -225/(y-5)
We'll divide both sides by 45:
1/x = -5/(y-5)
We'll cross multiply and we'll get:
-5x = y - 5
We'll move all terms to the right side and we'll use symmetric property:
5x + y - 5 = 0
The equation of the lines that is passing through the given points is: 5x + y - 5 = 0. | 376 | 1,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-05 | latest | en | 0.924579 |
https://portal.santarosa.edu/srweb/SR_CourseOutlines.aspx?CVID=37000&Semester=20177 | 1,713,091,282,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00079.warc.gz | 434,901,725 | 20,353 | # SRJC Course Outlines
4/14/2024 3:41:22 AM WWTR 112 Course Outline as of Fall 2017 Changed Course CATALOG INFORMATION Discipline and Nbr: WWTR 112 Title: MATH FOR WATER TECH Full Title: Math for Water and Wastewater Technology Last Reviewed:2/13/2023
Units Course Hours per Week Nbr of Weeks Course Hours Total Maximum 3.00 Lecture Scheduled 3.00 17.5 max. Lecture Scheduled 52.50 Minimum 3.00 Lab Scheduled 0 6 min. Lab Scheduled 0 Contact DHR 0 Contact DHR 0 Contact Total 3.00 Contact Total 52.50 Non-contact DHR 0 Non-contact DHR Total 0
Total Out of Class Hours: 105.00 Total Student Learning Hours: 157.50
Title 5 Category: AA Degree Applicable
Repeatability: 00 - Two Repeats if Grade was D, F, NC, or NP
Also Listed As:
Formerly: ENVT 112
Catalog Description:
Untitled document
Practical application of mathematical calculations for hydraulics, chemicals, and solids used in the design, operation, process control and maintenance of water and wastewater distribution systems and treatment facilities.
Prerequisites/Corequisites:
Recommended Preparation:
Limits on Enrollment:
Schedule of Classes Information
Description: Untitled document
Practical application of mathematical calculations for hydraulics, chemicals, and solids used in the design, operation, process control and maintenance of water and wastewater distribution systems and treatment facilities.
Prerequisites:
Recommended:
Limits on Enrollment:
Transfer Credit:
Repeatability:00 - Two Repeats if Grade was D, F, NC, or NP
ARTICULATION, MAJOR, and CERTIFICATION INFORMATION
Associate Degree: Effective: Inactive: Area: CSU GE: Transfer Area Effective: Inactive: IGETC: Transfer Area Effective: Inactive: CSU Transfer: Effective: Inactive: UC Transfer: Effective: Inactive: C-ID:
Certificate/Major Applicable: Both Certificate and Major Applicable
COURSE CONTENT
Student Learning Outcomes:
At the conclusion of this course, the student should be able to:
Untitled document
1. Perform calculations common to the fields of water distribution technology, water treatment
technology, and wastewater treatment technology
2. Use a spreadsheet program such as Excel to perform calculations
Objectives: Untitled document
In order to achieve these learning outcomes, during the course the students will:
1. Convert between common units of time, mass, length, area, volume, and temperature
2. Apply general mathematical concepts to calculations common in the water and wastewater
industry, such as calculation of: areas, volumes, mass, density, specific gravity, ratios,
percentages, median and average values, concentrations, flows, pressures, velocities,
rates, efficiencies, and power usage
3. Complete the mathematical portions of the state certification exams for: Water Distribution
System Operator (Grades 1 or 2), Water Treatment Operator (Grades 1 or 2), and
Wastewater Operator (Grade 1 and 2)
Topics and Scope
Untitled document
I. Units and unit conversion with English and Metric systems
A. Length, areas, and volumes
B. Mass, density, and specific gravity
C. Time and flow rates
D. Temperature
E. Velocity and pressure
F. Price and unit price
G. Power and power cost
H. Dimensional analysis
II. Basic arithmetic
A. Fractions, decimals, and percentages
B. Areas and volumes
C. Ratios and proportions
D. Exponents
III. Intermediate arithmetic
A. Mean, median, and mode
B. Weighted average
C. Efficiency
D. Algebraic variable isolation
IV. Chemistry and related calculations
A. pH
B. Concentrations
C. Dosages
D. Preparation of chemical solutions
V. Pumping calculations
D. Brake and motor horsepower
VI. Collection calculations
A. Average and per capita flow
B. Flow composition and velocity
C. Wet well capacity and pumping rate
D. Grit chamber and sand trap flow velocity and detention times
E. Industrial discharge equivalent population
VII. Primary treatment calculations
B. Biological Oxygen Demand (BOD)
C. Suspended Solids (SS) and Total Dissolved Solids (TDS)
E. Weir overflow rate
VIII. Treatment pond calculations
A. Volume and evaporation rates
IX. Filter calculations
B. Recirculation ratio
X. Activated sludge calculations
B. Mean cell residence time
C. Waste and return pumping rates
D. Aerator air flow
E. Volume and volume index
F. Sludge thickening rate
XI. Solids calculations
A. Centrifuge efficiency
B. Belt press efficiency
C. Volatile solids reduction
D. Volatile acidity/alkalinity ratio
E. Dissolved air flotation
F. Solids mass balance and location
XII. Effluent treatment calculations
A. Concentration-Time (CT)
B. Chlorine dosage, demand, and residual
C. Hypochlorite
D. Dechlorination
E. Ultraviolet (UV) and ozone
F. Log removal and disinfection efficiency
Assignments:
Untitled document
1. In-class exercises (8-10)
2. Problem solving assignments (10-14)
3. Quizzes and/or Midterms (2-10)
4. Final exam | 1,168 | 4,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.82316 |
https://terpconnect.umd.edu/~toh/spectrum/findpeaks.m | 1,542,498,562,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-47/segments/1542039743913.6/warc/CC-MAIN-20181117230600-20181118012600-00350.warc.gz | 739,153,164 | 4,664 | function P=findpeaks(x,y,SlopeThreshold,AmpThreshold,smoothwidth,peakgroup,smoothtype) % function P=findpeaksG(x,y,SlopeThreshold,AmpThreshold,smoothwidth,peakgroup,smoothtype) % Function to locate the positive peaks in a noisy x-y time series data % set. Detects peaks by looking for downward zero-crossings % in the first derivative that exceed SlopeThreshold. % Returns list (P) containing peak number and position, % height, width, and area of each peak. Arguments "slopeThreshold", % "ampThreshold" and "smoothwidth" control peak sensitivity. % Higher values will neglect smaller features. "Smoothwidth" is % the width of the smooth applied before peak detection; larger % values ignore narrow peaks. If smoothwidth=0, no smoothing % is performed. "Peakgroup" is the number points around the top % part of the peak that are taken for measurement. If Peakgroup=0 % the local maximum is takes as the peak height and position. % The argument "smoothtype" determines the smooth algorithm: % If smoothtype=1, rectangular (sliding-average or boxcar) % If smoothtype=2, triangular (2 passes of sliding-average) % If smoothtype=3, pseudo-Gaussian (3 passes of sliding-average) % See http://terpconnect.umd.edu/~toh/spectrum/Smoothing.html and % http://terpconnect.umd.edu/~toh/spectrum/PeakFindingandMeasurement.htm % (c) T.C. O'Haver, 1995, 2014. Version 6, Last revised March, 2016 % Simplified code: Replaced line 91 with call to gaussfit.m function % % Examples: % findpeaksG(0:.01:2,humps(0:.01:2),0,-1,5,5) % x=[0:.01:50];findpeaksG(x,cos(x),0,-1,5,5) % x=[0:.01:5]';findpeaksG(x,x.*sin(x.^2).^2,0,-1,5,5) % x=[-10:.1:10];y=exp(-(x).^2);findpeaksG(x,y,0.005,0.3,3,5,3); % Find, measure, and plot noisy peak with unknown position % x=[-10:.2:10]; % y=exp(-(x+5*randn()).^2)+.1.*randn(size(x)); % P=findpeaksG(x,y,0.003,0.5,7,9,3); % xx=linspace(min(x),max(x)); % yy=P(3).*gaussian(xx,P(2),P(4)); % plot(x,y,'.',xx,yy) % % Related functions: % findvalleys.m, findpeaksL.m, findpeaksb.m, findpeaksb3.m, % findpeaksplot.m, peakstats.m, findpeaksnr.m, findpeaksGSS.m, % findpeaksLSS.m, findpeaksfit.m, findsteps.m, findsquarepulse.m, idpeaks.m % Copyright (c) 2013, 2014 Thomas C. O'Haver % % Permission is hereby granted, free of charge, to any person obtaining a copy % of this software and associated documentation files (the "Software"), to deal % in the Software without restriction, including without limitation the rights % to use, copy, modify, merge, publish, distribute, sublicense, and/or sell % copies of the Software, and to permit persons to whom the Software is % furnished to do so, subject to the following conditions: % % The above copyright notice and this permission notice shall be included in % all copies or substantial portions of the Software. % % THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR % IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, % FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE % AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER % LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, % OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN % THE SOFTWARE. % if nargin~=7;smoothtype=1;end % smoothtype=1 if not specified in argument if smoothtype>3;smoothtype=3;end if smoothtype<1;smoothtype=1;end if smoothwidth<1;smoothwidth=1;end smoothwidth=round(smoothwidth); peakgroup=round(peakgroup); if smoothwidth>1, d=fastsmooth(deriv(y),smoothwidth,smoothtype); else d=deriv(y); end n=round(peakgroup/2+1); P=[0 0 0 0 0]; vectorlength=length(y); peak=1; AmpTest=AmpThreshold; for j=2*round(smoothwidth/2)-1:length(y)-smoothwidth-1, if sign(d(j)) > sign (d(j+1)), % Detects zero-crossing if d(j)-d(j+1) > SlopeThreshold*y(j), % if slope of derivative is larger than SlopeThreshold if or(y(j) > AmpTest, y(j+1) > AmpTest), % if height of peak is larger than AmpThreshold (new version by Anthony Willey) % if y(j) > AmpTest, % if height of peak is larger than AmpThreshold (old version) xx=zeros(size(peakgroup));yy=zeros(size(peakgroup)); for k=1:peakgroup, % Create sub-group of points near peak groupindex=j+k-n+2; if groupindex<1, groupindex=1;end if groupindex>vectorlength, groupindex=vectorlength;end xx(k)=x(groupindex);yy(k)=y(groupindex); end if peakgroup>2, [Height, Position, Width]=gaussfit(xx,yy); PeakX=real(Position); % Compute peak position and height of fitted parabola PeakY=real(Height); MeasuredWidth=real(Width); % if the peak is too narrow for least-squares technique to work % well, just use the max value of y in the sub-group of points near peak. else PeakY=max(yy); pindex=val2ind(yy,PeakY); PeakX=xx(pindex(1)); MeasuredWidth=0; end % Construct matrix P. One row for each peak detected, % containing the peak number, peak position (x-value) and % peak height (y-value). If peak measurement fails and % results in NaN, or if the measured peak height is less % than AmpThreshold, skip this peak if isnan(PeakX) || isnan(PeakY) || PeakY> v=[1 2 3 4 Inf 6 7 Inf 9]; % >> rmnan(v) % ans = % 1 2 3 4 4 6 7 7 9 la=length(a); if isnan(a(1)) || isinf(a(1)),a(1)=0;end for point=1:la, if isnan(a(point)) || isinf(a(point)), a(point)=a(point-1); end end function [Height, Position, Width]=gaussfit(x,y) % Converts y-axis to a log scale, fits a parabola % (quadratic) to the (x,ln(y)) data, then calculates % the position, width, and height of the % Gaussian from the three coefficients of the % quadratic fit. This is accurate only if the data have % no baseline offset (that is, trends to zero far off the % peak) and if there are no zeros or negative values in y. % % Example 1: Simplest Gaussian data set % [Height, Position, Width]=gaussfit([1 2 3],[1 2 1]) % returns Height = 2, Position = 2, Width = 2 % % Example 2: best fit to synthetic noisy Gaussian % x=50:150;y=100.*gaussian(x,100,100)+10.*randn(size(x)); % [Height,Position,Width]=gaussfit(x,y) % returns [Height,Position,Width] clustered around 100,100,100. % % Example 3: plots data set as points and best-fit Gaussian as line % x=[1 2 3 4 5];y=[1 2 2.5 2 1]; % [Height,Position,Width]=gaussfit(x,y); % plot(x,y,'o',linspace(0,8),Height.*gaussian(linspace(0,8),Position,Width)) % Copyright (c) 2012, Thomas C. O'Haver maxy=max(y); for p=1:length(y), if y(p)<(maxy/100),y(p)=maxy/100;end end % for p=1:length(y), z=log(y); coef=polyfit(x,z,2); a=coef(3); b=coef(2); c=coef(1); Height=exp(a-c*(b/(2*c))^2); Position=-b/(2*c); Width=2.35482/(sqrt(2)*sqrt(-c)); | 1,973 | 6,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-47 | latest | en | 0.622485 |
https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=next&oldid=143471 | 1,638,419,257,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00398.warc.gz | 190,146,143 | 13,125 | # Difference between revisions of "2010 AMC 12B Problems/Problem 18"
## Problem
A frog makes $3$ jumps, each exactly $1$ meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than $1$ meter from its starting position?
$\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}$
## Solution 1 (Complex Numbers)
We will let the moves be complex numbers $a$, $b$, and $c$, each of magnitude one. The starts on the origin. It is relatively easy to show that exactly one element in the set $$\{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}$$ has magnitude less than or equal to $1$. (Can you show how?) Hence, the probability is $\boxed{\text{(C)} \frac {1}{4}}$.
## Solution 2 (Simple Calculus)
Yes, we are pulling out calculus...
Represent every jump as a circle of radius 1. The first circle is a circle of radius 1 centered on the origin. WLOG, assume the first jump lands on $(1,0)$. Then, the third circle could be centered be anywhere on the second circle, which is itself centered on $(1,0)$. Let us define $f(\theta)$ as the value of the length of the first circle that lies within the area of the third circle in terms of the angle formed by the two points of intersection and either circle's center (symmetry, you chose!). The intersection of the two circles should form a geometrical lens shape. By sectors, $$f(\theta)=\frac{\theta}{2\pi}*2\pi=\theta$$ As the angle or angle of intersection continuously decreases from $pi$ (when the third circle is on top of the second circle) to $0$ (when the third circle is only touching the first circle at one spot $(1,0)$, I just need to find the average value of this function to find the average arc length where the third jump could land to satisfy the problem. To do this, I can apply average function value with our old buddy calculus, $$\frac{1}{\pi}\int_0^{\pi}\theta\;d\theta=\frac{\pi}{2}$$ The probability that the third jump will land on this arc length is just the arc length divided by the circumference, or $\frac{\frac{\pi}{2}}{2\pi}=\frac{1}{4} \implies \boxed{C}$
~BJHHar
## Solution 3 (Geometric) (Incorrect because probability is not evenly distributed?)(No, correct because actually probability is evenly distributed, except at the very edge and center, but since that has area 0, it is neglegible)
The first hop doesn't matter because no matter where the hops, it lands on the border of the circle you want it to end in. The remaining places that the can jump to form a disk of radius 2 centered at the spot on which the first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.
No matter where we start, we will have the small circle tangent to a point on the big circle. This is just like how $\frac{1}{2}x^2$ and $x^2$ are tangent. The area ratio of the two circles is $$\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}}$$. | 840 | 3,050 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2021-49 | latest | en | 0.854488 |
https://www.knowpia.com/knowpedia/Completeness_(statistics) | 1,638,268,262,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00468.warc.gz | 924,212,516 | 26,351 | BREAKING NEWS
Completeness (statistics)
## Summary
In statistics, completeness is a property of a statistic in relation to a model for a set of observed data. In essence, it ensures that the distributions corresponding to different values of the parameters are distinct.
It is closely related to the idea of identifiability, but in statistical theory it is often found as a condition imposed on a sufficient statistic from which certain optimality results are derived.
## Definition
Consider a random variable X whose probability distribution belongs to a parametric model Pθ parametrized by θ.
Say T is a statistic; that is, the composition of a measurable function with a random sample X1,...,Xn.
The statistic T is said to be complete for the distribution of X if, for every measurable function g,:[1]
${\displaystyle {\text{if }}\operatorname {E} _{\theta }(g(T))=0{\text{ for all }}\theta {\text{ then }}\mathbf {P} _{\theta }(g(T)=0)=1{\text{ for all }}\theta .}$
The statistic T is said to be boundedly complete for the distribution of X if this implication holds for every measurable function g that is also bounded.
### Example 1: Bernoulli model
The Bernoulli model admits a complete statistic.[2] Let X be a random sample of size n such that each Xi has the same Bernoulli distribution with parameter p. Let T be the number of 1s observed in the sample, i.e. ${\displaystyle \textstyle T=\sum _{i=1}^{n}X_{i}}$. T is a statistic of X which has a binomial distribution with parameters (n,p). If the parameter space for p is (0,1), then T is a complete statistic. To see this, note that
${\displaystyle \operatorname {E} _{p}(g(T))=\sum _{t=0}^{n}{g(t){n \choose t}p^{t}(1-p)^{n-t}}=(1-p)^{n}\sum _{t=0}^{n}{g(t){n \choose t}\left({\frac {p}{1-p}}\right)^{t}}.}$
Observe also that neither p nor 1 − p can be 0. Hence ${\displaystyle E_{p}(g(T))=0}$ if and only if:
${\displaystyle \sum _{t=0}^{n}g(t){n \choose t}\left({\frac {p}{1-p}}\right)^{t}=0.}$
On denoting p/(1 − p) by r, one gets:
${\displaystyle \sum _{t=0}^{n}g(t){n \choose t}r^{t}=0.}$
First, observe that the range of r is the positive reals. Also, E(g(T)) is a polynomial in r and, therefore, can only be identical to 0 if all coefficients are 0, that is, g(t) = 0 for all t.
It is important to notice that the result that all coefficients must be 0 was obtained because of the range of r. Had the parameter space been finite and with a number of elements less than or equal to n, it might be possible to solve the linear equations in g(t) obtained by substituting the values of r and get solutions different from 0. For example, if n = 1 and the parameter space is {0.5}, a single observation and a single parameter value, T is not complete. Observe that, with the definition:
${\displaystyle g(t)=2(t-0.5),\,}$
then, E(g(T)) = 0 although g(t) is not 0 for t = 0 nor for t = 1.
## Relation to sufficient statistics
For some parametric families, a complete sufficient statistic does not exist (for example, see Galili and Meilijson 2016 [3]). Also, a minimal sufficient statistic need not exist. (A case in which there is no minimal sufficient statistic was shown by Bahadur in 1957.[citation needed]) Under mild conditions, a minimal sufficient statistic does always exist. In particular, these conditions always hold if the random variables (associated with Pθ ) are all discrete or are all continuous.[citation needed]
## Importance of completeness
The notion of completeness has many applications in statistics, particularly in the following two theorems of mathematical statistics.
### Lehmann–Scheffé theorem
Completeness occurs in the Lehmann–Scheffé theorem,[4] which states that if a statistic that is unbiased, complete and sufficient for some parameter θ, then it is the best mean-unbiased estimator for θ. In other words, this statistic has a smaller expected loss for any convex loss function; in many practical applications with the squared loss-function, it has a smaller mean squared error among any estimators with the same expected value.
Examples exists that when the minimal sufficient statistic is not complete then several alternative statistics exist for unbiased estimation of θ, while some of them have lower variance than others.[5]
### Basu's theorem
Bounded completeness occurs in Basu's theorem,[6] which states that a statistic that is both boundedly complete and sufficient is independent of any ancillary statistic.
Bounded completeness also occurs in Bahadur's theorem. In the case where there exists at least one minimal sufficient statistic, a statistic which is sufficient and boundedly complete, is necessarily minimal sufficient.
## Notes
1. ^ Young, G. A. and Smith, R. L. (2005). Essentials of Statistical Inference. (p. 94). Cambridge University Press.
2. ^ Casella, G. and Berger, R. L. (2001). Statistical Inference. (pp. 285–286). Duxbury Press.
3. ^ Tal Galili & Isaac Meilijson (31 Mar 2016). "An Example of an Improvable Rao–Blackwell Improvement, Inefficient Maximum Likelihood Estimator, and Unbiased Generalized Bayes Estimator". The American Statistician. 70 (1): 108–113. doi:10.1080/00031305.2015.1100683. PMC 4960505. PMID 27499547.CS1 maint: uses authors parameter (link)
4. ^ Casella, George; Berger, Roger L. (2001). Statistical Inference (2nd ed.). Duxbury Press. ISBN 978-0534243128.
5. ^ Tal Galili & Isaac Meilijson (31 Mar 2016). "An Example of an Improvable Rao–Blackwell Improvement, Inefficient Maximum Likelihood Estimator, and Unbiased Generalized Bayes Estimator". The American Statistician. 70 (1): 108–113. doi:10.1080/00031305.2015.1100683. PMC 4960505. PMID 27499547.CS1 maint: uses authors parameter (link)
6. ^ Casella, G. and Berger, R. L. (2001). Statistical Inference. (pp. 287). Duxbury Press.
## References
• Basu, D. (1988). J. K. Ghosh (ed.). Statistical information and likelihood : A collection of critical essays by Dr. D. Basu. Lecture Notes in Statistics. 45. Springer. ISBN 978-0-387-96751-6. MR 0953081.
• Bickel, Peter J.; Doksum, Kjell A. (2001). Mathematical statistics, Volume 1: Basic and selected topics (Second (updated printing 2007) of the Holden-Day 1976 ed.). Pearson Prentice–Hall. ISBN 978-0-13-850363-5. MR 0443141.
• E. L., Lehmann; Romano, Joseph P. (2005). Testing statistical hypotheses. Springer Texts in Statistics (Third ed.). New York: Springer. pp. xiv+784. ISBN 978-0-387-98864-1. MR 2135927. Archived from the original on 2013-02-02.
• Lehmann, E.L.; Scheffé, H. (1950). "Completeness, similar regions, and unbiased estimation. I." Sankhyā: the Indian Journal of Statistics. 10 (4): 305–340. doi:10.1007/978-1-4614-1412-4_23. JSTOR 25048038. MR 0039201.
• Lehmann, E.L.; Scheffé, H. (1955). "Completeness, similar regions, and unbiased estimation. II". Sankhyā: The Indian Journal of Statistics. 15 (3): 219–236. doi:10.1007/978-1-4614-1412-4_24. JSTOR 25048243. MR 0072410. | 1,924 | 6,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-49 | latest | en | 0.814021 |
http://www.masjidtucson.org/publications/books/SP/1990/jan_2/page2.html | 1,369,360,278,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704132298/warc/CC-MAIN-20130516113532-00099-ip-10-60-113-184.ec2.internal.warc.gz | 589,949,110 | 4,322 | January 1990: Page 1, 2, 3, 4 Submitters Perspective Page 2
## Soad Khalifa Rayan, Lisa Spray, Ihsan Ramadan, K. Emami, & Saeed Talari Discover New Proof that Rashad Khalifa is God's Messenger
### Sequential (Conventional) Values
By writing the 28 letters of the Arabic alphabet, then assigning sequential numbers to them, from 1 to 28,we get the sequential or conventional value.
Soad calculated the sequential value of Noah نوح, it is 14+6+8=28. If you look at Sura 71, entitled "Noah," you will note that the number of verses is 28.
She calculated the value of “Abrahamابراهیم ”: 1+2+20+1+5+10+13=52, and 52 is the number of verses in Sura Abraham (No.14).
She then calculated the sequential value of "Muhammad محمد": 13+8+13+4 = 38, and 38 is the number of verses in Sura Muhammad (No.47).
She then calculated the sequential value of “Rashad Khalifa رشاد خلیفه;” 20+21+1+4+24+12+10+17+5=114; all suras. No sura has 114 verses. Her conclusion: this further proves that Rashad Khalifa is God's Messenger of the covenant who is prophesied in 3:81 to prove the Quran's 114 suras.
Soad also noted that the number of suras from Abraham (Sura 14) to Noah (71) is 57, 19x3, with 57 suras outside.
### Keikhusro Emami's Discovery
This old discovery of Mr. K. is relevant to this topic. He added the numbers of the 29 initialed suras: 2 + 3 + 7 + 10 + 11 + 12 + 13 + 14 + 15 + 19 + 20 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 36 + 38 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 50 + 68 = 822. He then added the gematrical values of "Rashad" (505) and "Khalifa" (725). The sum he obtained was 822+505+725 =2052, that is 19x108.
### Lisa Spray's Discovery
A couple of years ago, Lisa gave the 28 letters of the Arabic alphabet sequential values. She then calculated the sum of sequential values of the 14 Quranic Initials that prefix suras 2, 3, 7, 10, 11, 12, 13, 14, 15,19, 20, 26, 27,28, 29, 30,31,32, 36, 38, 40, 41, 42, 43, 44, 45,46, 50, and 68.The letters are A ا, H ه, HH ح, TT ط, Y ی, K ك, L ل, M م, N ن, S س, ‘A ع, SS ص, Q ق, and R ر, The sequential values of these letters are 1, 5, 8, 9, 10, 11, 12, 13, 14, 15,16,18,19, and 20, respectively, and the sum of these numbers is 171, 19x9. Those who are familiar with the Ouran's 19-based miracle appreciate the significance of these findings. This discovery validates the sequential numbering system of the Arabic alphabet.
### SA'EED TALARI'S DISCOVERIES
Saeed was blessed with a huge number of profound discoveries, including what was reported in past issues as part of the "Miracle of Miracles." This time, I wish to report only one of his latest findings:
The parameters of the 3 messengers of Islam, written next to each other, give a multiple of 19. Write down the number of letters in "Ibrahim" (6),then the gematrical value of "Ibrahim" (258), then the number of letters in "Muhammad" (4), then the value of "Muhammad" (92), then the number of letters in "Rashad" (4), then the value of "Rashad" (505). The result is 62584924505, a multiple of 19.
Note: The sum of gematrical values of the 3 messengers, Abraham, Muhammad, & Rashad is 258+92+505=855=19x45.
### Other Discoveries
I have a number of other significant discoveries with which God blessed Brothers Muhammad Azarshahy of San Francisco, Ahmad Yusuf of Nigeria, and Mohamoud Ali Abib of Tucson. I hope I find more space in future issues of the PERSPECTIVE to publish these discoveries. More profound findings were also made by Sister Ihsan Ramadan concerning "Mudathir & Muzzammil."
## AZHAR KHAN'S DISCOVERYVALIDATES THE ARABIC ALPHABET ANDTHE GEMATRICAL VALUES OF ARABIC LETTERS
The Arabic alphabet consists of 28 letters, and each letter has a universally established gematrical value. The 28 letters and their values are shown in the Table to the right. (now below) Brother Azhar Khan wrote down the 28 gematrical values, next to each other, to form one long number as shown below. He then asked his computer to divide this long number by 19, and sure enough, it is a multiple of 19. This proves that Almighty God has created and designed the Arabic language to conform with the Quran's 19-based mathematical code (Please see 30:22).
Arabic: ا ب ج د ه و ز ح ط ی ك ل م ن English: A B G D H W Z HH TT Y K L M N Value: 1 2 3 4 5 6 7 8 9 10 20 30 40 50
Arabic: س ع ف ص ق ر ش ت ث خ ڌ ض ظ غ English: S ‘A F SS Q R Sh T Th Kh Dh DD ZZ Gh Value: 60 70 80 90 100 200 300 400 500 600 700 800 900 1,000
WHY 19!?
Many people still ask this preliminary question. When we tell them that 19 is the gematrical
value of "ONE" (WAHD = 6 + 1 + 8 + 4 = 19), they question the validity of these values. Now we have the ultimate confirmation
## 1234567891020304050607080901002003004005006007008009001000
The gematrical values of all the letters of the Arabic alphabet written next to each other: a multiple of 19. | 1,526 | 4,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2013-20 | latest | en | 0.807076 |
https://www.doubtnut.com/qna/46212 | 1,721,443,748,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00890.warc.gz | 636,930,201 | 31,423 | # If the line, ax+by+c=0 is a normal to the curve xy=2
Video Solution
|
Answer
Step by step video & image solution for If the line, ax+ by + c = 0 is a normal to the curve xy =2 by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One or More
## If the line ax+ by+ c = 0 is a normal to the curve xy = 1, then
Aa>0,b>0
Ba>0,b<0
Ca<0,b>0
Da<0,b<0
• Question 2 - Select One
## If the line ax+by+c=0 is a normal to the curve xy=1, then :
Aa>0,b>0
Ba>0,b<0
Ca>0,b<0
Da<0,b<0
• Question 3 - Select One
## If the line ax+by+c=0 is normal to the curve xy+5=0, then
Aa>0,b>0
Bb>0,a<0
Cb<0,a>0
Dnone of these
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.
Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation
###### Contact Us
• ALLEN CAREER INSTITUTE PRIVATE LIMITED, Plot No. 13 & 14, Dabra Road, Sector 13, Hisar, Haryana 125001
• info@doubtnut.com | 570 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-30 | latest | en | 0.823325 |
https://physics.stackexchange.com/questions/629207/does-the-age-of-the-universe-depend-upon-a-specific-choice-of-referential?noredirect=1 | 1,638,135,014,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00291.warc.gz | 541,348,504 | 31,190 | # Does the age of the universe depend upon a specific choice of referential? [duplicate]
Does the age of the universe depend upon a specific choice of referential or is there a way to define it in a referential-independent way?
• A photon that shot out of the big bang, or if you happened to get caught in a black hole just at the big bang, then you would still see the universe as 0 years old, wouldn't you? Apr 13 at 22:09
• Related, possible duplicate: physics.stackexchange.com/q/104153/123208 Apr 14 at 6:57
• @foolishmuse Why are you asking the OP that question? But anyway, a photon has no rest frame, as discussed in numerous answers on this site. OTOH, there's the Cosmic Neutrino Background, and their proper time is certainly less than the age of the universe measured in the CMB comoving frame. Apr 14 at 8:31
• – mlk
Apr 14 at 12:02
• These are great answers, I would just like to add that the so called cosmological time is the time measured by a comoving clock, a clock which sees the universe isotropic or in other words which is in rest relativ to the microwave background. Since the very beginning such a clock measures the maximum age of the universe. Clocks which aren't comoving would measure a lower age and particles with relativistic speed like neutrinos a much lower age.
– timm
Apr 14 at 16:05
No. The age of the universe does not depend on any referential system. In order to measure time, you need some physical quantity that's changing to measure time against. In the case of cosmology, it's the time perceived by a typical observer based on the expansion parameter $$a$$ --see below. In a manner of speaking, it's the time that follows the galaxies in their mutual expansion. The mathematical basis is the FLRW metric of GR: $$-c^{2}\mathrm{d}\tau^{2}=-c^{2}\mathrm{d}t^{2}+a(t)^{2}d\Sigma^{2}$$ where, $$\mathrm{d}\mathbf{\Sigma}^{2}=\frac{\mathrm{d}r^{2}}{1-kr^{2}}+r^{2}\mathrm{d}\mathbf{\Omega}^{2}$$ and, $$\mathrm{d}\mathbf{\Omega}^{2}=\mathrm{d}\theta^{2}+\sin^{2}\theta\,\mathrm{d}\phi^{2}$$ $$k$$ is the spatial curvature, $$r$$ is the actual distance between two galaxies, and $$a\left(t\right)$$ is called the expansion parameter, and represents the average separation of two standard galaxies measured in this cosmic time. The quantity, $$H=\frac{\dot{a}}{a}$$ is the Hubble parameter, and represents how the scale factor changes with this cosmic time. The age of the universe is essentially the inverse Hubble parameter: $$T=\frac{a}{\dot{a}}$$ And because the universe is large-scale homogeneous and isotropic, the words "typical observer" are justified, and this time does not depend on which observer you choose. Meaning: on which typical galaxy you sit and watch the universe expand. Now, we don't live in a FLRW universe. We live in a De Sitter universe (vacuum dominated), but the argument doesn't change (essentially).
• Is it possible to describe the difference between $t$ and $T$ in layman's terms? Apr 14 at 13:02
• $t$ is the coordinate time: what your "cosmic watch" tells you when you look at it. This "watch" is the metric. But this is misleading, because the rate of expansion has suffered ups and downs during the history of the universe. It was vacuum-dominated during the inflationary era, then radiation-dominated, then matter-dominated, and now back to vacuum-dominated. So to get an accurate $T$ you must keep track of this history. You can picture that on @benrg's diagram. The observers sharing the same $t$ are the circular sections. But for $T$ you need to know what happened before. Apr 15 at 12:02
• Thanks. ${}{}{}$ Apr 15 at 12:03
There is a simple geometric interpretation of the quantity that is called the age of the universe.
Spacetime has a large-scale shape. It looks something like this:
Later times are at the top, earlier times at the bottom.
You shouldn't take this too literally because it isn't a proper embedding (I flipped the sign of the metric to make it Euclidean, and embedded it in Euclidean space), and it's also somewhat misleading in other ways, but it gets one important thing right: spacetime is not maximally symmetric, the way a sphere is, but it does have a kind of "rotational" symmetry, like a pot made on a potter's wheel (though note that it doesn't actually circle back on itself at the distance suggested by this embedding).
On this diagram I drew "vertical" lines in light blue; the brown and yellow worldlines on the left and right are also lines in this category. The age of the universe at any point on the manifold is the distance to the big bang, or end of inflation, measured along one of these lines going through the point in question.
You could object that these lines are a coordinate grid, and indeed they are, but they're a coordinate grid singled out by the inherent shape of the manifold. If the manifold was more symmetric then there would be no natural coordinate system (you can't put polar coordinates on a sphere without choosing some poles). If the manifold was less symmetric then it also wouldn't select a natural coordinate system. But the actual shape of spacetime has the right amount of symmetry to select a preferred time coordinate which can be used to define the age of the universe.
The universe is not actually perfectly symmetric. If you zoom in, there are bumps on the surface, and this means there is some inherent fuzziness in the definition of the age at a point. But it is approximately equal to the distance measured when all of those bumps are smoothed out.
• Did you make the picture yourself? If yes , could you please post the code and specify the software used? Apr 15 at 9:09
• @magma The math of the embedding is described at the Wikimedia Commons page. I used Mathematica's RevolutionPlot3D to do the rendering. Apr 16 at 5:31 | 1,400 | 5,791 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-49 | latest | en | 0.913948 |
https://vzweva.be/ball/Aug-1084.html | 1,623,672,737,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612154.24/warc/CC-MAIN-20210614105241-20210614135241-00575.warc.gz | 525,395,891 | 7,979 | • ### FINENESS MODULUS OF AGGREGATE WHAT, WHY &
01-09-2014· What is Fineness Modulus of Aggregate? Fineness modulus is an empirical factor obtained by adding the cumulative percentages of aggregate retained on each of the standard sieves ranging from 80 mm to 150 micron and dividing this sum by 100. Fineness Modulus of Sand Why to Determine Fineness Modulus? Fineness modulus is generally used to get an idea of how coarse or fine the
### Fineness Modulus of Aggregates: The Inside Scoop Gilson
The fineness modulus (FM) of aggregates can be an intimidating term, but it’s just an empirical value that describes the average size of particles in a sample of aggregate. This factor provides a basis to select estimated proportions for concrete mix design, but its real influence is not always understood.
### What is Fineness Modulus of Coarse Aggregates and Its
Fineness modulus of coarse aggregates represents the average size of the particles in the coarse aggregate by an index number. It is calculated by performing sieve analysis with standard sieves. The cumulative percentage retained on each sieve is added and subtracted by
### What is Fineness Modulus of Sand (Fine Aggregate) and
Therefore, fineness modulus of aggregate = (cumulative % retained) / 100 = (275/100) = 2.75 Fineness modulus of fine aggregate is 2.75. It means the average value of aggregate is in between the 2 nd sieve and 3 rd sieve. It means the average aggregate size is in
### Fineness Of Modulus Of Coarse Aggregates
Fineness Modulus of Coarse Aggregate: Fineness Modulus is defined as an index to the particle size not to the gradation. Fineness Modulus is calculated from the sieve analysis. An empirical factor obtained by adding the total percentage of sample of the aggregate retained on each of a specified series of sieves, and dividing the sum by 100.If the fineness Modulus is constant, the volume of
### (PDF) METHOD OF CALCULATION OF THE FINENESS
The value for tine aggregates commonly ranges from 2.00 to 4.00 and for coarse aggregate from 6.50 to 8.00 when all the material is finer than the 1-1/2-in. 4.2 As noted in the standard definition of fineness modulus (para 1.1) and the example shown in para 4.1, the fineness modulus is based on the percentages of aggregates retained on a specificed series of sieves.
### Fineness Modulus Pavement Interactive
The larger the fineness modulus, the more coarse the aggregate. A typical fineness modulus for fine aggregate used in PCC is between 2.70 and 3.00.
### Solved: Calculate the fineness modulus of aggregate A in
According to the definition of fineness modulus, sieve size 25 mm, 12.5 mm, and 0.075 mm are not included. Sample calculations: Calculate the percentage of individual fraction retained by weight for 19 mm sieve as follows: Here, the percentage of individual fraction retained by weight is and the %passing at 19 mm sieve is . Substitute 92 for .
### Properties of Aggregates Civil Snapshot
13-01-2021· Fineness modulus is an empirical factor obtained by adding the cumulative percentages of aggregate retained on each of the standard sieves ranging from 80 mm to 150 micron and dividing this sum by 100. Fineness modulus is generally used to get an idea of how coarse or fine the aggregate is.
### Fineness Of Modulus Of Coarse Aggregates
Fineness Modulus of Coarse Aggregate: Fineness Modulus is defined as an index to the particle size not to the gradation. Fineness Modulus is calculated from the sieve analysis. An empirical factor obtained by adding the total percentage of sample of the aggregate retained on each of a specified series of sieves, and dividing the sum by 100.If the fineness Modulus is constant, the volume of
### Determination of Fineness Modulus of a Sample of Fine
The object of this experiment is to determine the fineness modulus and hence to plot the particle size distribution curve for the sample of fine aggregate supplied. Theory The fineness modulus of a sample of aggregate is an index number which is roughly proportional to the average size of particle in the aggregate, that is, coarser the aggregate the higher the fineness modulus.
### (PDF) METHOD OF CALCULATION OF THE FINENESS
The value for tine aggregates commonly ranges from 2.00 to 4.00 and for coarse aggregate from 6.50 to 8.00 when all the material is finer than the 1-1/2-in. 4.2 As noted in the standard definition of fineness modulus (para 1.1) and the example shown in para 4.1, the fineness modulus is based on the percentages of aggregates retained on a specificed series of sieves.
### Fineness Modulus Pavement Interactive
For aggregates used in PCC, another common gradation description for fine aggregate is the fineness modulus. It is described in ASTM C 125 and is a single number used to describe a gradation curve. It is defined as: The larger the fineness modulus, the more coarse the aggregate.
### Fineness Modulus Islamic University of Gaza
Fineness modulus of fine aggregate varies from 2.0 to 3.5mm. Fine aggregate having fineness modulus more than 3.2 should not considered as fine aggregate. Various values of fineness modulus for different sands are detailed below. Type of sand Fineness modulus range Fine sand 2.2 2.6 Medium sand 2.6 2.9 Coarse sand 2.9 3.2
When fineness modulus is indicative of the weighted average particle size of the sample to be in two successive sieves from the lowest sieve. For instance, a fineness modulus of 3 represents the third sieve from 150µm. The same fineness modulus can represent a number of different size distributions and grading curves.
### FINENESS MODULUS OF FINE AGGREGATE
FINENESS MODULUS OF FINE AGGREGATE TXDOT DESIGNATION: TEX-402-A CONSTRUCTION DIVISION 1 2 LAST REVIEWED: SEPTEMBER 2014 Test Procedure for FINENESS MODULUS OF FINE AGGREGATE TxDOT Designation: Tex-402-A Effective Date: August 1999 1. SCOPE 1.1 This method determines the fineness modulus of concrete fine aggregate used in
### Properties of Aggregates Civil Snapshot
13-01-2021· Fineness Modulus. Fineness modulus is an empirical factor obtained by adding the cumulative percentages of aggregate retained on each of the standard sieves ranging from 80 mm to 150 micron and dividing this sum by 100. Fineness modulus is generally used to get an idea of how coarse or fine the aggregate is.
### Solved: Calculate the fineness modulus of aggregate A
According to the definition of fineness modulus, sieve size 25 mm, 12.5 mm, and 0.075 mm are not included. Sample calculations: Calculate the percentage of individual fraction retained by weight for 19 mm sieve as follows: Here, the percentage of individual fraction retained by weight is and the %passing at 19 mm sieve is . Substitute 92 for .
### CHAPTER 2 SIEVE ANALYSIS AND FINENESS MODULUS Sampling
SIEVE ANALYSIS AND FINENESS MODULUS Sampling Since the reason for sampling aggregates is to determine the gradation (particle size) of the aggregate, it is necessary that they be sampled correctly. The results of testing will re" ect the condition and characteristics of the
### Fineness modulus Wikipedia
The same value of fineness modulus may therefore be obtained from several different particle size distributions. In general, however, a smaller value indicates a finer aggregate. Fine aggregates range from a FM of 2.00 to 4.00, and coarse aggregates smaller than 38.1 mm range from 6.75 to 8.00.
When fineness modulus is indicative of the weighted average particle size of the sample to be in two successive sieves from the lowest sieve. For instance, a fineness modulus of 3 represents the third sieve from 150µm. The same fineness modulus can represent a number of different size distributions and grading curves.
### Fineness modulus of combined aggregates
The same value of fineness modulus may therefore be obtained from several different particle size distributions. In general, however, a smaller value indicates a finer aggregate. Fine aggregates range from a FM of 2.00 to 4.00, and coarse aggregates smaller than 38.1 mm range from 6.75 to 8.00.
### The Importance of Fineness Modulus| Concrete
01-01-1994· In 1925, Duff Abrams introduced the concept of fineness modulus (FM) for estimating the proportions of fine and course aggregates in concrete mixtures. The premise: "aggregate of the same fineness modulus will require the same quantity of water to produce a mix of the same consistnecy and give a concrete of the same strength."
### What is Fineness Modulus and how to calculate it
How to calculate Fineness Modulus Procure to determine Fineness Modulus. Sieve the aggregate using the appropriate sieves (80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron & 150 micron) Note down the weight of aggregate retained on each sieve.; Calculate the cumulative weight of aggregate retained on each sieve.; Find out the cumulative percentage of
### The fineness modulus of fine aggregate is in the range of
MCQs: The fineness modulus of fine aggregate is in the range of? (A) 2.0 to 3.5 (B) 3.5 to 5.0
### FINENESS MODULUS OF FINE AGGREGATE
FINENESS MODULUS OF FINE AGGREGATE TXDOT DESIGNATION: TEX-402-A CONSTRUCTION DIVISION 1 2 LAST REVIEWED: SEPTEMBER 2014 Test Procedure for FINENESS MODULUS OF FINE AGGREGATE TxDOT Designation: Tex-402-A Effective Date: August 1999 1. SCOPE 1.1 This method determines the fineness modulus of concrete fine aggregate used in
### The fineness modulus of fine aggregate is in the range of
MCQs: The fineness modulus of fine aggregate is in the range of? Civil Engineering Mcqs Design of Concrete Structures Mcqs
### Solved: Calculate the fineness modulus of aggregate A
According to the definition of fineness modulus, sieve size 25 mm, 12.5 mm, and 0.075 mm are not included. Sample calculations: Calculate the percentage of individual fraction retained by weight for 19 mm sieve as follows: Here, the percentage of individual fraction retained by weight is and the %passing at 19 mm sieve is . Substitute 92 for .
### Concrete Aggregates Function, Properties
Empirical factor called the fineness modulus is, often used as an index of the fineness of aggregate, a very important factor when making selection for a source/type of fine aggregates. The fineness modulus is computed from screen analysis data by adding the cumulative percentages of aggregate retained on each of a specified series of sieves, and dividing the sum by 100.
### F.M.:Fineness Modulus of Aggregates LinkedIn
The Fineness modulus (FM) is an empirical figure obtained by adding the total percentage of the sample of an aggregate retained on each of a specified series of sieves, and dividing the sum by 100.
### Fineness Modulus of Fine Aggregate Concrete
Fineness Modulus (FM) is an empirical calculation used to support the optimisation of a concrete mix design. It is used to determine ‘degree of uniformity of the grading’. This can create a relationship between grading and consistence and/or the amount of cement or water to fill the voids.
### THE ROLE OF FINENESS MODULUS Southport Concrete
To determine the fineness modulus, the contractor adds the cumulative percentage-retained values and divides the sum by 100. This process must be repeated for both coarse and fine aggregate sizes. Once the fineness modulus of each size has been determined, they can be combined by means of a special equation to determine the proportions necessary for creating the desired concrete.
### CIVIL ENGINEERING MATERIAL TESTING: FINENESS
Fineness modulus is an empirical factor obtained by adding the cumulative percentages of aggregate retained on each of the standard sieves ranging from 80 mm to 150 micron and dividing this sum by 100.
### HOW TO BLEND AGGREGATE USING FINENESS
10-02-2015· Let “ x ” is the percentage of fine aggregate to be combined with coarse aggregate. x can be calculated using the formula given below. x = [ (F2-F)/ (F-F1)]*100. Where, F1 = Fineness modulus value of fine aggregate. F2 = Fineness modulus value of coarse aggregate. F = Fineness modulus value of combined aggregate.
### The fineness modulus of fine aggregate is in the range of
MCQs: The fineness modulus of fine aggregate is in the range of? (A) 2.0 to 3.5 (B) 3.5 to 5.0
### Solved: Calculate the fineness modulus of aggregate A
According to the definition of fineness modulus, sieve size 25 mm, 12.5 mm, and 0.075 mm are not included. Sample calculations: Calculate the percentage of individual fraction retained by weight for 19 mm sieve as follows: Here, the percentage of individual fraction retained by weight is and the %passing at 19 mm sieve is . Substitute 92 for .
### The fineness modulus of fine aggregate is in the range of
MCQs: The fineness modulus of fine aggregate is in the range of? Civil Engineering Mcqs Design of Concrete Structures Mcqs
### What Is Aggregate? Properties of Aggregates
23-07-2020· Fineness modulus of aggregates is determined by performing a sieve analysis test with the exception that 12.5 mm sieve is omitted. The fineness modulus of aggregates is calculated by using the following formula. Bulk Density. Bulk density of aggregate is
### Concrete Aggregates Function, Properties
Empirical factor called the fineness modulus is, often used as an index of the fineness of aggregate, a very important factor when making selection for a source/type of fine aggregates. The fineness modulus is computed from screen analysis data by adding the cumulative percentages of aggregate retained on each of a specified series of sieves, and dividing the sum by 100. | 3,147 | 13,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-25 | latest | en | 0.895125 |
https://ru.coursera.org/learn/analytics-excel/reviews?authMode=login&page=22 | 1,670,530,257,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00801.warc.gz | 499,926,256 | 105,591 | Вернуться к Mastering Data Analysis in Excel
4.2
звезд
Оценки: 3,868
О курсе
Important: The focus of this course is on math - specifically, data-analysis concepts and methods - not on Excel for its own sake. We use Excel to do our calculations, and all math formulas are given as Excel Spreadsheets, but we do not attempt to cover Excel Macros, Visual Basic, Pivot Tables, or other intermediate-to-advanced Excel functionality. This course will prepare you to design and implement realistic predictive models based on data. In the Final Project (module 6) you will assume the role of a business data analyst for a bank, and develop two different predictive models to determine which applicants for credit cards should be accepted and which rejected. Your first model will focus on minimizing default risk, and your second on maximizing bank profits. The two models should demonstrate to you in a practical, hands-on way the idea that your choice of business metric drives your choice of an optimal model. The second big idea this course seeks to demonstrate is that your data-analysis results cannot and should not aim to eliminate all uncertainty. Your role as a data-analyst is to reduce uncertainty for decision-makers by a financially valuable increment, while quantifying how much uncertainty remains. You will learn to calculate and apply to real-world examples the most important uncertainty measures used in business, including classification error rates, entropy of information, and confidence intervals for linear regression. All the data you need is provided within the course, all assignments are designed to be done in MS Excel, and you will learn enough Excel to complete all assignments. The course will give you enough practice with Excel to become fluent in its most commonly used business functions, and you’ll be ready to learn any other Excel functionality you might need in the future (module 1). The course does not cover Visual Basic or Pivot Tables and you will not need them to complete the assignments. All advanced concepts are demonstrated in individual Excel spreadsheet templates that you can use to answer relevant questions. You will emerge with substantial vocabulary and practical knowledge of how to apply business data analysis methods based on binary classification (module 2), information theory and entropy measures (module 3), and linear regression (module 4 and 5), all using no software tools more complex than Excel....
Лучшие рецензии
TB
16 нояб. 2021 г.
I like and appreciate courses provided through Coursera.This course is very interesting and valuable for those whose jobs do have relevance with data management .God bless Coursera and Duke University
JE
30 окт. 2015 г.
The course deserves a 5-star rating because: (1) content is relevant, (2) the professor is concise and possesses great teaching skills, and (3) the learning modules are applicable to daily problems.
Фильтр по:
526–550 из 921 отзывов о курсе Mastering Data Analysis in Excel
автор: Georgi
28 июня 2017 г.
автор: Gerardo J O
21 нояб. 2015 г.
автор: Bharat M
22 авг. 2020 г.
автор: Boris D
29 апр. 2021 г.
автор: chong d
17 сент. 2019 г.
автор: Tomer W
25 нояб. 2019 г.
автор: chen y
27 нояб. 2015 г.
автор: Paul E H
20 февр. 2016 г.
автор: Maharafa I C
10 нояб. 2015 г.
автор: Kang J N
27 июля 2019 г.
автор: Humberto J
21 авг. 2017 г.
автор: Elvis A
6 февр. 2019 г.
автор: Javier V
3 мая 2018 г.
5 янв. 2016 г.
автор: Smitha
8 нояб. 2015 г.
автор: Shiv P
22 нояб. 2016 г.
автор: Erica M
9 апр. 2016 г.
автор: Łukasz C
21 авг. 2016 г.
автор: Kelly D
6 мар. 2017 г.
автор: Даурен
11 янв. 2018 г.
автор: SATEESHKUMAR I
18 окт. 2016 г.
автор: Ji, C
19 янв. 2018 г.
автор: Colin M
23 сент. 2017 г.
автор: Cliff E
23 мая 2017 г.
автор: Hubert D
31 янв. 2016 г. | 982 | 3,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-49 | latest | en | 0.911922 |
http://www.ni.com/documentation/en/labview-comms/1.0/node-ref/threshold-1d-array/ | 1,597,160,934,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738816.7/warc/CC-MAIN-20200811150134-20200811180134-00486.warc.gz | 169,583,307 | 12,258 | Interpolates points in a 1D array that represents a 2D non-descending graph. This node compares a threshold to the values in an array, starting at a specified index, until it finds a pair of consecutive elements such that the threshold is greater than the value of the first element and less than or equal to the value of the second element.
array of numbers or points
An array of numbers or an array of points where each point is a cluster of x- and y-coordinates.
If this input is an array of numbers, this node assumes the x-coordinates are the same as the indexes of the array itself.
If this input is an array of points, this node uses the second elements in the clusters, or the y-coordinates, to obtain a fractional index that it then uses to interpolate the corresponding x value.
threshold y
The threshold value for the node. If threshold y is less than or equal to the array value at start index, the node returns start index for fractional index or x. If threshold y is greater than every value in the array, the node returns the index of the last value. If the array is empty, the node returns NaN.
start index
The index to start interpolation.
Default: 0—The node returns the result calculated from the entire array rather than a specified section of the array.
fractional index or x
The interpolated result calculated for the array of numbers or points 1D input array.
If array of numbers or points is an array of points where each point is a cluster of x- and y-coordinates, fractional index or x is the interpolated x value corresponding to the interpolated position of threshold y among the y-coordinates, rather than the fractional index of the array. fractional index or x returns the interpolated x value associated with the given y value if you graphed the points. | 376 | 1,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-34 | latest | en | 0.760464 |
https://oeis.org/A289907 | 1,660,206,174,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00357.warc.gz | 404,686,313 | 4,589 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A289907 Initial primes of 5 consecutive primes with consecutive gaps 8,6,4,2. 1
1979, 5399, 11813, 41213, 42443, 44249, 47129, 55799, 57773, 74699, 79613, 84299, 88643, 126473, 143813, 148913, 167099, 176489, 178799, 178889, 209249, 211859, 237143, 266663, 267629, 272249, 272333, 322229, 344153, 348443, 354023, 375083, 391379, 399263, 422069, 449549, 521519, 529673 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS All terms = {23, 29} mod 30. For initial primes of 5 consecutive primes with consecutive gaps 2,4,6,8 see A190814. Number of terms less than 10^k: 0, 0, 0, 2, 13, 65, 317, 1563, 8671, 50643, ..., . - Robert G. Wilson v, Dec 07 2017 LINKS Robert G. Wilson v, Table of n, a(n) for n = 1..10000 (first 3114 terms from Muniru A Asiru) EXAMPLE Prime(299..303) = { 1979, 1987, 1993, 1997, 1999 } and 1979 + 8 = 1987, 1987 + 6 = 1993, 1993 + 4 = 1997, 1997 + 2 = 1999. Also, prime(5852..5856) = { 57773, 57781, 57787, 57791, 57793 } and 5773 + 8 = 57781, 57781 + 6 = 57787, 57787 + 4 = 57791, 57791 + 2 = 57793. MATHEMATICA s = Prepend[Differences@ #, First@ #] & /@ Partition[Prime@ Range[10^5], 5, 1]; Select[s, Drop[#, 1] == Range[8, 2, -2] &][[All, 1]] (* Michael De Vlieger, Jul 14 2017 *) p = {2, 3, 5, 7, 11}; lst = {}; While[ p[[1]] < 530000, If[ Differences@ p == {8, 6, 4, 2}, AppendTo[ lst, p[[1]] ]]; p = Join[Rest@ p, {NextPrime[ p[[-1]]] }]]; lst (* Robert G. Wilson v, Dec 07 2017 *) PROG (GAP) I:=[8, 6, 4, 2];; P:=Filtered([1..1000000], IsPrime);; P1:=List([1..Length(P)-1], i->P[i+1]-P[i]);; Collected(last);; P2:=List([1..Length(P)-Length(I)], i->[P1[i], P1[i+1], P1[i+2], P1[i+3]]);; P3:=List(Positions(P2, I), i->P[i]); (PARI) is(n) = my(q); forstep(i=8, 2, -2, q=nextprime(n+1); if(q-n!=i, return(0)); n=q); return(1) \\ David A. Corneth, Jul 23 2017 CROSSREFS Cf. A078847, A153419, A190814, A190817, A190819, A190838. Sequence in context: A343971 A343972 A225235 * A125490 A237334 A035870 Adjacent sequences: A289904 A289905 A289906 * A289908 A289909 A289910 KEYWORD nonn AUTHOR Muniru A Asiru, Jul 14 2017 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 11 03:41 EDT 2022. Contains 356046 sequences. (Running on oeis4.) | 1,054 | 2,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-33 | latest | en | 0.502197 |
https://www.zbmath.org/?q=ut%3Amarked+metric+graphs+ai%3Avijay.sujith | 1,632,194,448,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00352.warc.gz | 1,114,906,441 | 9,290 | # zbMATH — the first resource for mathematics
The triangular line $$n$$-sigraph of a symmetric $$n$$-sigraph. (English) Zbl 1213.05120
Summary: An $$n$$-tuple $$(a_1,a_2,\dots, a_n)$$ is symmetric, if $$a_k= a_{n-k+1}$$, $$1\leq k\leq n$$. Let $H_n = \{(a_1,a_2,\dots, a_n): a_k\in\{+,-\},\;a_k= a_{n-k+1}, \;1\leq k\leq n\}$ be the set of all symmetric $$n$$-tuples. A symmetric $$n$$-sigraph (symmetric $$n$$-marked graph) is an ordered pair $$S_n= (G,\sigma)$$ $$(S_n= (G,\mu))$$, where $$G= (V, E)$$ is a graph called the underlying graph of $$S_n$$ and $$\sigma: E\to H_n(\mu: V\to H_n)$$ is a function. Analogous to the concept of the triangular line graph of a graph, the triangle line symmetric $$n$$-sigraph of a symmetric $$n$$-sigraph is defined. It is shown that for any symmetric $$n$$-sigraph $$S_n$$, its triangular line symmetric $$n$$-sigraph is $$i$$-balanced. We then give structural characterization of triangular line symmetric $$n$$-sigraphs. Further, we obtain some switching equivalence relationship between triangular line symmetric $$n$$-sigraph and line symmetric $$n$$-sigraph.
##### MSC:
05C22 Signed and weighted graphs | 379 | 1,151 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-39 | latest | en | 0.814271 |
https://www.coin-or.org/GAMSlinks/benchmarks/MINLP/SBB1to5_convex_070902/2-SBB-2.trc.convex_4-SBB-4.trc.convex_sqr.htm | 1,516,340,692,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00000.warc.gz | 888,241,671 | 4,005 | ## Solver Square Comparison: Considers all models.
Date / Time: 08/24/07 21:51:47
Solver comparison utility.
Compares all solver return outcomes (for example optimal, locally optimal, infeasible, unbounded, fail) of one solver with all return outcomes of another solver. Interrupt denotes resource or iteration limit has been reached. Solver SBB-IPOPT is represented on the left (rows) and solver SBB-CONOPT-IPOPT on top (columns). See the solver return definitions for return codes.
Models having trace data only in one trace file are listed in the "no data" column of the other.
Tracefile 1 : 2-SBB-2.trc.convex Tracefile 2 : 4-SBB-4.trc.convex Solvers used : SBB-IPOPT SBB-CONOPT-IPOPT Modeltype(s) MINLP
SBB-CONOPT-IPOPT: optimal SBB-CONOPT-IPOPT: feasible SBB-CONOPT-IPOPT: infeasible SBB-CONOPT-IPOPT: unbounded SBB-CONOPT-IPOPT: fail SBB-CONOPT-IPOPT: no data total SBB-IPOPT SBB-IPOPT: optimal - - - - - - - SBB-IPOPT: feasible - 31 - - - - 31 SBB-IPOPT: infeasible - - - - - - - SBB-IPOPT: unbounded - - - - - - - SBB-IPOPT: fail - - - - 5 - 5 SBB-IPOPT: no data - - - - - - - total SBB-CONOPT-IPOPT - 31 - - 5 - 36
### Solver return definitions:
Outcome Model Status Solver Status optimal 1 or 15 1 locally optimal 2 any feasible 8 or 16 1 or 2 or 3 or 4 or 5 infeasible 4 or 5 or 10 or 19 1 unbounded 3 or 18 1 fail all other all other
### Solver Resource Times
• Models for each solver pair outcome. Listed are the solver resource times TIME(.) in seconds, as well as the ratio RATIO(./.) of resource times for the two solvers if both solved optimally.
• Also listed are the objective values OBJ(.) using both solvers. The better solution found is listed in boldface. A solution is considered better, if the relative objective function difference is greater than 1.00E-05. If both solutions are less than 1e-1, we use the absolute difference.
• Solver resource time ratios for a particular model are listed only if one solver has resource greater than 5.00E-02.
Modelname Time (SBB-IPOPT) Time (SBB-CONOPT-IPOPT) Ratio (SBB-IPOPT/SBB-CONOPT-IPOPT) Obj (SBB-IPOPT) Obj (SBB-CONOPT-IPOPT) alan 0.0800 0.0400 2.000 2.92500000 2.92500000 batch 1.7200 0.7200 2.389 285506.50825285 285506.50815158 du-opt 3.9200 1.7300 2.266 3.56468638 3.56108965 du-opt5 52.3300 20.8200 2.513 8.07365758 8.07365758 ex1223 0.2200 0.0400 5.500 4.57958240 4.57958240 ex1223a 0.0800 0.0100 8.000 4.57958240 4.57958240 ex1223b 0.1700 0.0300 5.667 4.57958240 4.57958240 fac1 1.8800 0.0200 94.000 160912612.35016900 160912612.35016900 fac3 7.2900 0.1900 38.368 31982309.84800030 31982309.84800000 m3 1.9000 1.1700 1.624 37.80000000 37.80000000 m6 261.6500 351.4800 0.744 106.25687691 129.82493644 m7 270.0900 425.1600 0.635 118.75687691 123.96437781 meanvarx 0.2800 0.0100 28.000 14.40406225 14.49698300 nvs03 0.1500 0.0500 3.000 16.00000000 16.00000000 nvs10 0.0600 0.0100 6.000 -310.80000000 -310.80000000 risk2b 32.0900 0.3700 86.730 -55.73616850 -55.73616850 risk2bpb 20.8300 0.4300 48.442 -55.73616850 -55.73616850 st_e14 0.2000 0.0400 5.000 4.57958240 4.57958240 st_miqp1 0.1300 0.0100 13.000 281.00000000 281.00000000 st_miqp2 0.3000 0.0100 30.000 2.00000000 2.00000000 st_miqp4 0.1300 0.0000 +INF.000 -3092.82488276 -4574.00000000 stockcycle 31.3700 98.6000 0.318 119948.68833333 143295.16500000 st_test5 0.7100 0.0400 17.750 -110.00000000 -110.00000000 st_test6 0.7700 0.1100 7.000 471.00000000 471.00000000 st_test8 0.0700 0.0100 7.000 -29604.99999999 -29605.00000000 st_testgr1 0.3000 0.0300 10.000 -12.72810000 -12.72810000 st_testgr3 4.1900 0.0300 139.667 -20.52740000 -20.46880000 st_testph4 0.0700 0.0200 3.500 -80.50000000 -80.50000000 synthes1 0.1200 0.0100 12.000 6.00975891 6.00975891 synthes2 0.4500 0.0800 5.625 73.03531253 73.03531253 synthes3 1.3000 0.1600 8.125 68.00974052 68.00974052
Modelname Time (SBB-IPOPT) Time (SBB-CONOPT-IPOPT) Ratio (SBB-IPOPT/SBB-CONOPT-IPOPT) Status (SBB-IPOPT) Status (SBB-CONOPT-IPOPT) fo7 198.7200 416.1200 --- mstat(14) sstat( 4) mstat(14) sstat( 4) o7 226.7200 437.5700 --- mstat(14) sstat( 4) mstat(14) sstat( 4) tls12 3600.0000 585.2100 --- mstat(14) sstat( 3) mstat(14) sstat( 4) tls6 3600.6100 394.3000 --- mstat(14) sstat( 3) mstat(14) sstat( 4) tls7 2353.8500 297.2600 --- mstat(14) sstat( 4) mstat(14) sstat( 4)
Modelname Time (SBB-IPOPT) Obj (SBB-IPOPT) alan 0.0800 2.92500000 batch 1.7200 285506.50825285 du-opt 3.9200 3.56468638 du-opt5 52.3300 8.07365758 ex1223 0.2200 4.57958240 ex1223a 0.0800 4.57958240 ex1223b 0.1700 4.57958240 fac1 1.8800 160912612.35016900 fac3 7.2900 31982309.84800030 gbd 0.0400 2.20000000 m3 1.9000 37.80000000 m6 261.6500 106.25687691 m7 270.0900 118.75687691 meanvarx 0.2800 14.40406225 nvs03 0.1500 16.00000000 nvs10 0.0600 -310.80000000 risk2b 32.0900 -55.73616850 risk2bpb 20.8300 -55.73616850 st_e14 0.2000 4.57958240 st_e35 4.0000 64868.07685843 st_miqp1 0.1300 281.00000000 st_miqp2 0.3000 2.00000000 st_miqp4 0.1300 -3092.82488276 stockcycle 31.3700 119948.68833333 st_test5 0.7100 -110.00000000 st_test6 0.7700 471.00000000 st_test8 0.0700 -29604.99999999 st_testgr1 0.3000 -12.72810000 st_testgr3 4.1900 -20.52740000 st_testph4 0.0700 -80.50000000 synthes1 0.1200 6.00975891 synthes2 0.4500 73.03531253 synthes3 1.3000 68.00974052
Modelname Time (SBB-IPOPT) Status (SBB-IPOPT) fo7 -- mstat(14) sstat( 4) o7 -- mstat(14) sstat( 4) tls12 -- mstat(14) sstat( 3) tls6 -- mstat(14) sstat( 3) tls7 -- mstat(14) sstat( 4)
Modelname Time (SBB-CONOPT-IPOPT) Obj (SBB-CONOPT-IPOPT) alan 0.0400 2.92500000 batch 0.7200 285506.50815158 du-opt 1.7300 3.56108965 du-opt5 20.8200 8.07365758 ex1223 0.0400 4.57958240 ex1223a 0.0100 4.57958240 ex1223b 0.0300 4.57958240 fac1 0.0200 160912612.35016900 fac3 0.1900 31982309.84800000 m3 1.1700 37.80000000 m6 351.4800 129.82493644 m7 425.1600 123.96437781 meanvarx 0.0100 14.49698300 nvs03 0.0500 16.00000000 nvs10 0.0100 -310.80000000 risk2b 0.3700 -55.73616850 risk2bpb 0.4300 -55.73616850 st_e14 0.0400 4.57958240 st_miqp1 0.0100 281.00000000 st_miqp2 0.0100 2.00000000 st_miqp4 0.0000 -4574.00000000 stockcycle 98.6000 143295.16500000 st_test5 0.0400 -110.00000000 st_test6 0.1100 471.00000000 st_test8 0.0100 -29605.00000000 st_testgr1 0.0300 -12.72810000 st_testgr3 0.0300 -20.46880000 st_testph4 0.0200 -80.50000000 synthes1 0.0100 6.00975891 synthes2 0.0800 73.03531253 synthes3 0.1600 68.00974052 | 2,829 | 6,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-05 | latest | en | 0.769635 |
https://plainmath.net/precalculus/100361-how-can-you-find-the-gcf-of-tw | 1,674,995,613,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00878.warc.gz | 482,899,488 | 22,756 | Raphael Stark
2022-12-26
How can you find the GCF of two polynomials?
juanitacelis1ovy
Expert
The greatest common factor of two polynomials and the greatest common factor of two integers share a similar idea.
The largest integer that is a factor of both the two integers is the greatest common factor of two integers. Finding the common primes and multiplying them together is one method of determining the GCF. Another method is to express the two integers as products of ascending primes.
The greatest common factor of a couple of polynomials is the largest polynomial which is a factor of both of the polynomials. To find the GCF of two polynomials we can factor both of them, identify the common factors and multiply them together.
For example, consider the two polynomials:
$P\left(x\right)=4{x}^{3}+24{x}^{2}+44x+24$
$Q\left(x\right)=6{x}^{3}+42{x}^{2}+84x+48$
To find the GCF of P(x) and Q(x) first factorize them:
$P\left(x\right)=2\cdot 2\cdot \left(x+1\right)\left(x+2\right)\left(x+3\right)$
$Q\left(x\right)=2\cdot 3\cdot \left(x+1\right)\left(x+2\right)\left(x+4\right)$
Picking out the common factors and multiplying them:
$GCF\left(P\left(x\right),Q\left(x\right)\right)=2\cdot \left(x+1\right)\left(x+2\right)=2\left({x}^{2}+3x+2\right)=2{x}^{2}+6x+4$
Do you have a similar question?
Recalculate according to your conditions! | 406 | 1,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 31, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-06 | latest | en | 0.8199 |
https://www.convertunits.com/from/grams+Mustard+Gas/to/moles | 1,607,013,654,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141729522.82/warc/CC-MAIN-20201203155433-20201203185433-00614.warc.gz | 556,857,131 | 13,539 | ››Convert grams Mustard Gas to mole
grams Mustard Gas mol
How many grams Mustard Gas in 1 mol? The answer is 159.07732.
We assume you are converting between grams Mustard Gas and mole.
You can view more details on each measurement unit:
molecular weight of Mustard Gas or mol
The molecular formula for Mustard Gas is C4H8Cl2S.
The SI base unit for amount of substance is the mole.
1 grams Mustard Gas is equal to 0.0062862512393344 mole.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between grams Mustard Gas and mole.
Type in your own numbers in the form to convert the units!
››Convert another chemical substance
Convert grams to moles
››Quick conversion chart of grams Mustard Gas to mol
1 grams Mustard Gas to mol = 0.00629 mol
10 grams Mustard Gas to mol = 0.06286 mol
50 grams Mustard Gas to mol = 0.31431 mol
100 grams Mustard Gas to mol = 0.62863 mol
200 grams Mustard Gas to mol = 1.25725 mol
500 grams Mustard Gas to mol = 3.14313 mol
1000 grams Mustard Gas to mol = 6.28625 mol
››Want other units?
You can do the reverse unit conversion from moles Mustard Gas to grams, or enter other units to convert below:
Enter two units to convert
From: To:
››Details on molecular weight calculations
In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.
Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.
If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.
A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass.
Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.
The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. We use the most common isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass.
Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance.
››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 848 | 3,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-50 | latest | en | 0.867099 |
https://www.physicsforums.com/threads/trigonometric-substitution.185934/ | 1,521,604,342,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647567.36/warc/CC-MAIN-20180321023951-20180321043951-00395.warc.gz | 841,712,665 | 17,432 | # Trigonometric Substitution
1. Sep 20, 2007
### Lanza52
$$\int\sqrt{16-(2x)^{4}}xdx$$
Hint says you may like to use the identity sin(theta)cos(theta)= sin(2theta)/2
However, I think I found a way to use 1-sin^2(theta)=cos^2(theta)
First, (2x)^4 = 16x^4
So make it 16(1-x^2)^2.
Take the 16 out of the root and the integral and you have:
$$4\int\sqrt{1-(x^{2})^{2}}xdx$$
Sub in sin(theta) for x^2. Sub in cos^2(theta) for 1-sin^2(theta). Sub in $$\sqrt{sin(\vartheta)$$ for the x outside the root and $$\frac{1}{2\sqrt{sin(\vartheta)}}$$ for dx.
Cancel out the sqrt{sin} on the top and bottom and drag the 1/2 outside the integral. Anti-derivative of cos is sin and you have 2sin(theta). Sin(theta) = x^2.
And so the final answer is; $$2x^{2}+c$$
Correct?
2. Sep 20, 2007
### Dick
Is the derivative of 2x^2+c the same as the original integrand? I don't think so.
3. Sep 20, 2007
### Lanza52
Hrmm....could you point me in the right direction?
4. Sep 20, 2007
### Dick
Your substitution sin(theta)=x^2 is a great start. So cos(theta)*dtheta=2*x*dx. See where that takes you. I promise that's heading in the right direction.
5. Sep 20, 2007
### Lanza52
Still a little stumped.
Realized I missed the chain rule in 2 steps in my original post...damn rustiness.
Got to $$\frac{-sin\vartheta(cos\vartheta)^{3}}{3}$$
First and foremost, am I correct to this point? Then, I can't figure out the back substitutions here.
6. Sep 20, 2007
### Dick
I get to integral of cos(theta)^2.
7. Sep 20, 2007
### Lanza52
Yea, I posted the anti-derivative of cos(theta)^2. I'm at the point where you need to substitute in x's for the trig functions.
8. Sep 20, 2007
### Dick
That is NOT the antiderivative of cos(theta)^2. To do the antiderivative of cos(theta)^2 use that cos(2*theta)=cos(theta)^2-sin(theta)^2 (from the addition relation for cos) and that sin(theta)^2=1-cos(theta)^2. Solve for cos(theta)^2 in terms of cos(2*theta). Then it's easy to integrate cos(2*theta). After the integration is done you'll find that the hint actually does come in handy for expressing the result in terms of x.
9. Sep 20, 2007
### Dick
Hint: When you think something is the antiderivative of something else then try differentiating it to check. Could save you some time, since you seem to be getting antiderivatives wrong a lot.
10. Sep 20, 2007
### Dick
You could also try checking an integral table or an online site that computes integrals to check yourself. You can't do that on an exam, but it could be handy to avoid wasting time while practicing with homework problems.
11. Sep 20, 2007
### Lanza52
Definitely having problems with the little/old things. That year off killed me. Just have to do all the little things a few times before it comes back to me.
Anyways. Came up with 2cos^2(theta)=cos2theta + 1
So integral cos2theta + 1
Comes to Sintheta * costheta + theta
Don't know what to do with that theta at the end. So I'm guessing I'm wrong.
12. Sep 20, 2007
### Dick
You aren't that wrong this time. theta=arcsin(x^2). And what does sin(theta)*cos(theta) become? Hammer the multiplicative constants down and I think you've got it.
13. Sep 20, 2007
### Lanza52
Just checking to make sure I'm not aiming for the wrong thing;
I'm currently at sin(2theta)/2 + arcsin(x^2)+c
An integration website tells me that I should have $$\sqrt{1-x^{4}}x^{2}$$+ arcsin(x^2)+c
Was the site right?
14. Sep 20, 2007
### Dick
It's time to use the hint for sin(2theta)! You and the site are both right.
Last edited: Sep 20, 2007
15. Sep 20, 2007
### Lanza52
Okay, so I got to the answer that the site gave me. But question, wasn't that 4 brought outside the integral used while converting cos^2theta to cos2theta and when solving for dx?
16. Sep 20, 2007
### Dick
Probably. Sorry, I'm fading a bit and didn't want to check it myself. Just wanted to make sure you did.
17. Sep 20, 2007
### Lanza52
Alright, thanks. =P
I owe ya, but I imagine there isn't much math I could help ya with.
18. Sep 20, 2007
### Dick
You help me by making me feel that my time was not wasted on this internet busyness. The quicker you catch on, the better I feel. You're getting better, right? | 1,280 | 4,205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-13 | longest | en | 0.814822 |
http://www.chegg.com/homework-help/consider-double-slider-two-loop-linkage-illustrated-figure-2-chapter-2-problem-63p-solution-9780201350999-exc | 1,469,450,939,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824226.79/warc/CC-MAIN-20160723071024-00036-ip-10-185-27-174.ec2.internal.warc.gz | 366,146,489 | 17,967 | View more editions
# TEXTBOOK SOLUTIONS FOR Kinematics and Dynamics of Machinery 3rd Edition
• 698 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
Consider the double-slider two-loop linkage illustrated in Figure 2.23, where r1=1, r2=5.2, rF=5, rDE=4, rCD=1.5, and θ1= Π/2 rad. (Distance DE is identified as rDE, etc.) Find θ2, θ3, rD, and rE. Use a numerical method.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 3
Use Newton-Raphson method in mathematical software mathcad to solve the preceding four equations (5), (6), (7) and (8) as below:
Represent form F matrix with equations (5), (6), (7) and (8) as below:
Represent form X matrix with unknowns.
Calculate the G matrix as below:
• Chapter , Problem is solved.
Corresponding Textbook
Kinematics and Dynamics of Machinery | 3rd Edition
9780201350999ISBN-13: 0201350998ISBN: | 292 | 1,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2016-30 | latest | en | 0.808639 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.