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poleshift.org | 1,638,174,634,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00054.warc.gz | 66,441,068 | 4,965 | From: "Uhomann" To: Subject: Re: Homann's paper on Precession Date: Sat, 20 May 2000 18:48:12 -0600 MIME-Version: 1.0 X-Priority: 3
Dear Mr. Robbins,
I owe you a belated but still sincere thank you for posting my paper 'The Precession -Time Paradox' on your website.
Only a few days ago I noticed my article to be on the web by using the Infoseek search engine. As you can tell I am not a good "Internet-browser".
By the way, thanks to Mr. Greene of COSMOLOGYREVIEW my paper has also been posted on his website at www.cosmolgyreview.com , along with some other letters and articles written by my son (see Reader's Forum Section).
In case you are interested in this issue, I have also enclosed the latest information on the subject Precession.
With best regards,
Karl-Heinz Homann
Experts at the International Astronomical Union (IAU) are in an uproar. Worldwide, thousands of students and teachers have been confronted with the most controversial astronomical problem in the history of science. In 1955 the IAU substituted the tropical year of 31,556,925.97474 seconds for the sidereal year as the fundamental unit of time. But in authoritative textbooks it is asserted that the time interval of the sidereal year or Earth's complete period of revolution measured with respect to inertial space is about 31,558,149.5 seconds. The IAU refuses to confirm this assertion. Experts have recognized the fact that such a sidereal year does NOT exist in reality. The IAU is accused of willfully misleading the scientific community. www.cosmologyreview.com/uhomann_letter.html
Introduction:
The following paper examines the simple mathematical and physical relationship between earth's complete rotation period, its complete revolution period and precession. It questions the assumption that the sidereal year is supposedly about 1223 seconds longer than the tropical year. This time difference is the most crucial scientific argument in order to prove whether earth's precession is a physical fact or NOT. The solution to this unique problem must be based on the laws of logic and mathematics, as well as on the results of practical observations.
Definitions:
Tropical-sidereal year: 31,556,925.97474 seconds
Mean sidereal day: 86164.0905382 seconds
Mean solar day: 86400 seconds
Sidereal day: 86164.09966 seconds
Assertions:
1. No matter what the motion of the earth in its orbit or the orientation of its axis in space is, the complete 360° rotation period of the earth on its axis is 86164.0905382 seconds.
2. The orientation of the earth's polar axis in space (e.g. with respect to the sun) has no influence on earth's period of revolution.
3. The axis of the earth slowly changes its orientation in space - either (a) directly due to a precession of earth's polar axis or (b) indirectly as a result of our entire solar system moving around a point x in space. Please note, that if the axis of the earth remains aligned to point x, then earth's period of rotation measured with respect to point x has to be about 86164.0905382 seconds.
4. The time interval of a complete rotation of the earth measured with respect to inertial space is about 9.12 ms more; i.e. an actual time difference of about 9.12 ms exists between a sidereal day and the mean sidereal day.
5. The civil calendar is aligned to the time interval of the tropical-sidereal year, which is determined by measuring earth's daily rotation period with respect to inertial space.
6. A time span of about 31,558,150 seconds relates to an orbit period of 360° plus 50.26". Such a time interval has no relevance to astronomy.
7. The equation 365.24219878 × 86400s = 31,556,925.97s = 366.24219878 × 86164.09054s describes a 360° revolution period of the earth.
8. The equation 365.256361 × 86400s = 31,558,149.59s = 366.256361 × 86164.09966s describes a different 360° revolution period, whereby the solar day has also 86400 s but the sidereal day has 86164.09966 s.
9. The precise time interval of a sidereal year - i.e. a complete 360° revolution around an inner fixed point of reference (sun) with respect to an outer fixed point of reference (inertial space) - can ONLY be determined by:
a. Measuring the daily rotations of the earth with respect to the outer point of reference (e.g. fixed stars), in order to determine earth's mean sidereal rotation period.
b. Applying the physical relationship that the earth describes after a complete 360° period of revolution EXACTLY ONE rotation more relative to the outer point of reference than relative to the sun.
c. Considering the physical fact that only the actual 360° rotation period of the earth on its axis can be used for the determination of earth's complete 360° period of revolution.
Experts claim that:
A precession of the earth's polar axis must occur and the rate of precession is about 50.26" or 3.35 seconds per tropical year.
Due to a precession of the earth a time difference of about 1223 seconds occurs between a sidereal year and a tropical year (i.e. a difference of about 3.35 seconds per day), since the vernal point of reference retrogrades around the sun.
The time interval of a tropical year relates to an orbit period of 360° minus 50.26".
The time interval of about 31,558,150 seconds is earth's true 360° orbit period, measured with respect to inertial space.
The sidereal year of about 31,558,150 s is no longer a fundamental time in astronomy.
Conclusion:
Conforming to mathematical and physical laws precession relates to periods of rotation and NOT to periods of revolution. Assuming that the precession period (the period of time for the earth's polar axis to describe a complete circle in space) is approx. 8.142 × 1011 seconds, then the total number of earth's rotations that could in fact be measured during such a period of time is:
8.142 × 1011 s ÷ 86164.09054 s = 9,449,412 rotations (w.r.t. the moving equinox or point x)
8.142 × 1011 s ÷ 86164.09966 s = 9,449,411 rotations (w.r.t. the inertial space)
In order to simplify the problem let us assume that the earth does NOT revolve around the sun. There are two possibilities:
1. Precession occurs
This means, 0 revolution periods or 0 sidereal years. However, due to precession a time interval of exactly one tropical year or about 365 days w.r.t the sun is seemingly created (shifting of earth's seasons w.r.t the sun). But in reality only a time difference of exactly one rotation (day) can be measured w.r.t. the moving point of reference, which has a retrograde angular velocity of about 0.1368" per day. The assumption that a measurable time difference equal to a period of one year or more than 365 days occurs, supposedly due to the moving point of reference retrograding around the sun, is therefore false.
Note: Since this time-discrepancy does not exist in reality, a so-called sidereal year of about 31,558,150 s was declared to be earth's true orbit time, which is about 1223 seconds longer than the actual tropical-sidereal year of 31,556,925.97474 seconds.
2. Precession does not occur and our entire solar system revolves around a point x
This means, 0 revolutions, 0 sidereal years and also 0 tropical years. However, due to the revolution period of our solar system around point x, a time difference of exactly one rotation (day) can be measured w.r.t. to inertial space. Consequently, the vernal point (and point x) moves with respect to the fixed stars by about 9.12 ms per day, and NOT by about 3.35 seconds per day or 1223 seconds per year around the sun.
If precession were to occur as claimed, not a single fixed star can be measured that has a mean meridian transition time of about 86164.09054 s (equal to the vernal-equinox transition period). But the transit periods of Sirius, as described in the paper www.cosmologyreview.com/beel_dog.html, prove otherwise.
Dr. Myles Standish, an expert on Planetary Ephemerides at NASA's Jet Propulsion Laboratory, made the following comment on March 3, 2000 after reading my letter www.cosmologyreview.com/uhomann_standish.html:
"… I believe that your timings (of the transits of Sirius) are accurate, but I think that you are misinterpreting what they are measuring. Again, they are being taken from a precessing earth and are, therefore, subject to precession."
If the measurement of Sirius is taken from a precessing earth, can anyone explain why the mean transit time of Sirius is identical to the mean transit time of the vernal equinox, considering the indisputable fact that Sirius does not retrograde around our sun by about 1223 seconds per year?
As you will undoubtedly agree, physical relationships are established or disproved by whether they work in practice and not by a vote of majority. The assumptions regarding the theory of earth's precession are inconsistent with practical observations and mathematical results. | 2,107 | 8,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | longest | en | 0.902892 |
https://smart2write.com/blog/eco-2-unit-2-as-assignment/ | 1,675,830,552,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500671.13/warc/CC-MAIN-20230208024856-20230208054856-00861.warc.gz | 536,613,841 | 17,855 | ## Eco 2- Unit 2 AS: Assignment
Paper details:
Begin working on your assignment for Unit 2 and compiling your data on market demand. This assignment below will not be due until the end of Unit 2.
Answer the following questions based on the table. You will need to submit this as a Word document in proper APA formatting. You will need to provide support for any outside resources that you find.
Table 4-5 below contains information about the corn market. An agricultural price floor is a price that the government guarantees farmers will receive for a particular crop. Suppose the federal government sets a price floor for corn at \$12 per bushel.
Price per Bushel
(dollars)
Quantity Demanded (bushels)
Quantity Supplied (bushels)
\$2
40,000
0
4
34,000
4,000
6
28,000
8,000
8
24,000
16,000
10
20,000
20,000
12
18,000
28,000
14
12,000
36,000
16
6,000
40,000
What is the amount of shortage or surplus in the corn market as result of the price floor?
If the government agrees to purchase any surplus output at \$12, how much will it cost the government?
If the government buys all of the farmers’ output at the floor price, how many bushels of corn will it have to purchase and how much will it cost the government?
Suppose the government buys up all of the farmers’ output at the floor price and then sells the output to consumers at whatever price it can get. Under this scheme, what is the price at which the government will be able to sell off all of the output it had purchased from farmers? What is the revenue received from the government’s sale?
In this problem we have considered two government schemes:
A price floor is established and the government purchases any excess output and
The government buys all the farmers’ output at the floor price and resells at whatever price it can get. Which scheme will taxpayers prefer?
Consider again the two schemes. Which scheme will the farmers prefer?
Consider again the two schemes. Which scheme will corn buyers prefer?
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Place your order with us and get a high quality, unique and plagiarism free paper that will guarantee you amazing results!! | 494 | 2,136 | {"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} | 2.640625 | 3 | CC-MAIN-2023-06 | latest | en | 0.929151 |
https://luanvan68.com/170%E2%80%B3-in-ft-convert-170-inches-to-feet-online-inch-calculator/ | 1,725,861,673,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00189.warc.gz | 341,877,977 | 50,598 | # 170″ in ft. Convert 170 inches to feet. Online inch calculator.
If you want to know how many feet are in 170 inches, you can obtain the answer in several ways. Here, you will find all the methods for converting and calculating inches to feet and vice versa. On this page, we consider all the detailed variants for converting 170 feet to inches and provide comprehensive examples, conversion tables, and charts to assist you in understanding how to convert inches to feet and vice versa.
• Convert 170 inches using the online converter on this website;.
• Calculate 170 inches using the InchPro calculator from our software collection for offline unit conversion.
• Perform mathematical computations and conversions for 170 inches specified in this article.
• We will explore conversion between inches and feet, and we will look in more detail at individual topics below. We have options to convert from “ft” to “in” and from “in” to “ft”. When using the abbreviation “ft”, it indicates feet, and when using the abbreviation “in”, it indicates inches.
We can easily convert values between inches and feet. By clicking on the arrow icons between input fields, you can swap the fields and perform other calculations. If you enter a value in any field, you will get the result in the opposite field. The final result of the conversion, “Feet 14.1666666667 = Inches 170,” is displayed under the green string of input fields. You can see a more detailed result of the calculation, which uses a coefficient of 0.0833333333333, under each field. If you want to convert any inch value, you only need to enter the required value in the left input field (or the right result field will be automatically filled). The online inch converter has an adaptive shape for different devices, so it looks the same on both input fields and monitors. The online converter has a very simple interface that will help us convert inches quickly. If we consider using the online converter on the web page, we can convert 170 inches into feet.
#### Let’s look at the result of calculate for the current value 170 inches
The conversion of 170 inches to feet equals 14.1666666667 feet.
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If you came to this page already to see the result of the value of “ft” equaling 14.1666666667, briefly write it in the box on the right (or in the field on the left) and see the work of the online calculator.
## Convert 170 inches in ft by conversion tables
Please find these 4 tables at the bottom of this page, which include the Astronomical chart conversion, International chart conversion, US survey chart conversion, and Metric measure of units basic conversion. We have made it possible to compute all possible values for units of measure in the lower tables, which is an interesting possibility. This is only a part of the features of the service page, but we have briefly reviewed how to use the unit Converter on this page.
• All conversions of 170 inches in the Metric System Units.
• All translations of 170 inches in the US Survey Units.
• All conversions of 170 inches into the International Units.
• All conversions of 170 inches in the Astronomical Units.
• All transformations in the page calculator are performed synchronously, and the data in the conversion tables are dynamically calculated. The conversion results in the lower tables will be recalculated for a value of 170 (inches), so you will see all the conversion results in those tables. To calculate all possible results for the main units, you can use our conversion tables on other pages of the website by entering the value of 170 inches into the calculator and deleting it again. Without using our conversion tables, you can also search and compute a huge list of computed values for all types of units in the lower tables. However, now you will only get the result of 14.1666666667 feet for the example of 170 inches, no matter which field (feet or inches) you enter in any web calculator test number.
### How many feet are in 170 inches?
Let’s do a simple calculation. We can calculate values using simple multiplication by knowing that 1 inch is equal to 0.0833333333333 feet (ft). It is important to remember this value, as it determines the basic value of the multiplier for calculating other lengths, sizes, and transformations for these units (feet and inches). This coefficient answers the question of how many feet are equivalent to a given number of inches. We can convert from one length unit (foot and inch) to another using a conversion factor, which is equal to 0.0833333333333. To answer this question, let’s start with a brief definition of foot and inch.
170 inches multiplied by 0.0833333333333 equals 14.1666666667 feet.
Therefore, it can be observed that by multiplying with the factor, we obtain the subsequent correlation:
Maybe you are interested 681
170 inches is equivalent to approximately 14.1666666667 feet.
### How much is 170 inches in feet?
You can write a summary in which all possible results have the same meaning. We have already seen how to change feet to inches and how to convert these two values.
The length of 14.1666666667 feet is equal to 170 inches, and when converted to inches, it is also equal to 170 inches. Additionally, the length of 14.1666666667 feet can be expressed as 170 inches or seventy hundred one feet.
To obtain a comprehensive evaluation of analogous data, kindly refer to the subsequent pages.
• Convert 169 inches to feet.
• ### How to convert 170 inches into feet? All rules and methods.
There are several methods to convert 170 inches into feet.
• Computation utilizing the equation;
• Calculation using the ratios;.
• Computation utilizing the web-based converter of the present webpage;
• Computation utilizing the offline calculator “InchPro Decimal”.
• #### Calculating 170 inches to ft formula for lengths and values.
In the calculations for inches and feet, we will utilize the formula provided below that will rapidly obtain the desired outcome.
Y (in) × 0.0833333333333 = X (ft) Y – measurement in inches X – outcome in feet.
To transform a set of values from feet and inches into inches, we can use the following examples as a result. Remember that when converting inches to feet, you just need to multiply the number of inches by a factor of 0.0833333333333 (which is equal to 1 inch).
14.5 feet is equal to 0.0833333333333 multiplied by 174.0 inches. 14.4166666667 inches is equal to 0.0833333333333 multiplied by 173.0 inches. 14.3333333333 inches is equal to 0.0833333333333 multiplied by 172.0 inches. 14.25 inches is equal to 0.0833333333333 multiplied by 171.0 inches. 14.1666666667 inches is equal to 0.0833333333333 multiplied by 170.0 inches.
We multiplied all the variants in the range of inches, from 170.0″ to 174.0″, with the same ratio of 0.0833333333333 and obtained the correct results in the calculations.
#### The calculation using mathematical proportions to convert 170 inches into feet
To calculate the values in feet and inches, we need to use the proportional form of “172.0″, 171.0″, and 170.0″. Let’s see some examples next. We can calculate the value in feet for any length in inches according to the rules of arithmetic. However, to calculate the proportions, you need to know the reference value of 1 inch in feet.
To convert X from feet to inches, you can multiply X by 170.0 and then divide it by 1. This will give you X in inches. Then, you can convert X back to feet by multiplying it by 0.0833333333333 and dividing it by 1.
All ratios are based on the reference value of 1 inch = 0.0833333333333 feet.
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#### Calculation of values using inch online calculator on the page
You can utilize our online universal converter on the current web page to convert and obtain any directions in feet and inches, as well as distances and dimensions, quickly and for free.
How can you use the online inch calculator to read more details about the manual for this calculator? To quickly get the result in feet for any value in inches, simply enter any number into the inches field, such as 174.0, 173.0, 172.0, or 171.0, for example. Currently, the inches field contains the number 170 (in), but you can change it to any other value.
Please see the results in the table next to it. We use the ratio of 0.0833333333333 to help us obtain the desired values in the computation of feet in the results. You can now check the calculator on this site to ensure that it works quickly and accurately before leaving. After the calculation, you should see the values in the right margin written in inches, and we will set up a table in the left margin to display the values. Additionally, you can find the web calculator at the top of this page, which we will also use. As an example, let’s try to calculate the values in feet and convert them into inches, taking 11 as the value.
#### Convert 170 inches with the use of calculator “InchPro Decimal”
Please see the screenshots for more information. In the “Software” menu of this website or app, you can find a link to Decimal InchPro, which describes how to convert distances, sizes, lengths, widths, and heights between different units of measurement, such as inches. This application also has an offline mode, allowing you to easily convert 170 inches into other units. If you download and install the software on your computer, you can access all of these calculations and use the included conversion tables. This calculator not only converts distances and lengths into inches, but also allows you to convert any value into other units. We will briefly describe the possibility of using our calculator to convert 170 inches.
### Visual charts conversion of 170 inches.
Additionally, the proportion of the lengths of the segments remains consistent across screens of varying resolutions, including small mobile devices and large monitors. In practical terms, this picture allows for a clear visual representation of the ratio between these quantities. The correlation between inches and feet is often difficult for many individuals to grasp.
#### The graphical representation of scales for comparing values.
The graph shows the relative values in inches, represented by different lengths and colors of rectangular segments, as well as a visual reference with feet as the measurement unit.
Your browser does not have the capability to display the canvas element.
The charts depicting the correlation between inches and feet are represented in the following shades:
• Green is the initial measurement or distance in inches.
• The scale in inches is represented by the color blue.
• The color yellow represents the measurement in feet.
• The diagram depicts the correlation between inches and feet for equivalent lengths and magnitudes, with the scale potentially fluctuating based on the numerical value displayed on the page. Please refer to the charts illustrating the blue and yellow hues.
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## Square Root of 361
In this article, we will also look at different methods for calculating the square root of 361 using a computer or calculator. We’ll answer some common questions… | 2,706 | 12,420 | {"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} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.815217 |
http://newtoyslist.com/rubik/rubix-building-solutions-rubiks-cube-building.html | 1,566,314,577,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315544.11/warc/CC-MAIN-20190820133527-20190820155527-00035.warc.gz | 134,524,733 | 3,034 | Hi, your aluminum cube is beautiful. My 6 yr old son has asked me to help him make a cube. So I went to the Home Depot today and looked around the plumbing and hardware sections. I only found some PVC pipe attachments but nothing with 6 knobs. I also found some nuts and bolts but I don't know which kind to get. Could you give me some idea of what to buy? I don't have very many tools at home but I do have a drill if I needed to drill through the PVC pipe and copper pipe cutters if I needed to use metal piping. Please help.
The project uses the Pi to directly solve the Rubik’s cube. The BrickPi3 takes the unsolved Rubik’s cube and the Raspberry Pi takes a picture of each side of the Rubik’s cube with the Raspberry Pi Camera. The Pi creates a text map of the color squares that shows where they are located on the cube. When it has fully mapped the cube, the Pi uses the “kociemba” python library to map out the moves needed to solve the Rubik’s cube. This information is taken by the Pi and BrickPi3 to solve the Rubik’s cube using the LEGO motors. The result: a solved Rubik’s cube. Rubix Building Solutions
Rubik’s Build It, Solve It is similar to the conventional Rubik’s block, but with a twist. This block includes each the tools, bits and directions kids need to be able to construct a Rubik’s Cube of the own. After this block was assembled together, there’s an education booklet (it’s’s 10-pages in duration) which will direct you through the procedure for solving the Rubik’s Cube (eventually). Here, you’ll find everything from identifying the areas of the block to solving fundamental puzzles. With this toy, kids will be provided a slow and continuous introduction about the best way best to use the block and progress to harder struggles. | 399 | 1,759 | {"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 | 3 | CC-MAIN-2019-35 | latest | en | 0.938931 |
https://dsp.stackexchange.com/questions/51032/selecting-gradient-orientation-for-generalized-hough-transform-while-scaling-and | 1,713,403,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00186.warc.gz | 196,382,632 | 44,680 | # Selecting gradient orientation for Generalized Hough Transform while scaling and rotation is present
I am using this link to study Generalized Hough Transform. I will explain it briefly. In order to detect a shape, an R-table is created for the shape with $\phi$ indicating gradient orientation for points on the edge of the shape. Then $(r, \beta)$ is used to indicate the vector from each of these points to the centroid in the shape. While detecting the shape, we will first find all the edge points and their gradient orientation $(\phi)$. Based on this $\phi$, $(r, \beta)$ values are used to calculate centroid points. Based on the votes on centroid points (and R table), the shape can be detected.
My question is related to rotation of the shape which is explained towards the end. If the shape is rotated by $\theta$, then the gradient orientation $(\phi)$ for a given edge point changes. So, shouldn't we do either one of the following:
1) Rotate all the $\phi$ values in R table by $\theta$?
OR
2) Rotate gradient vector by $(-\theta)$ and then calculate $\phi$ for the edge point?
• Can I please ask what are you trying to achieve? Also, what is perceived as "the problem" here?
– A_A
Sep 18, 2018 at 22:47
• I am trying to find the right method to get the correct $\phi$ from the R-table when rotation is present. As the shape rotates, the tangent at a point on the shape also rotates. So, does the gradient vector. This means $\phi$ changes. I am trying to see what is the best way to get the right $\phi$ and corresponding entries from the R-table.
– skr
Sep 19, 2018 at 0:55
– A_A
Sep 21, 2018 at 14:44
• @A_A The answer you provided gives an explanation of the algorithm itself, but it is still not clear to me how the correct phi is picked. Let me review this answer again in sometime.
– skr
Sep 21, 2018 at 15:04
• No problem. The "correct" $\phi$ is selected by rounding to the nearest angle pointed by $\angle G$. The gradient indexes $\phi$.
– A_A
Sep 21, 2018 at 17:07
I was having the same issue as you had: most resources I found online seem to miss this issue about the rotating gradient orientation, as did A_A in their answer (which to be fair is very clear in the other aspects of the GHT).
I found that the original paper by Ballard addresses this very thing in section 4.5 (the emphasis is mine):
We denote a particular $$R$$-table for a shape $$S$$ by $$R(\phi)$$. $$R$$ can be viewed as a multiply-vector-valued function. It is easy to see that simple transformations to this table will allow it to detect scaled or rotated instances of the same shape.
For example if the shape is scaled by $$s$$ and this transformation is denoted by $$T_s$$, then $$T_s[R(\phi)] = sR(\phi)$$ i.e.,all the vectors are scaled by $$s$$. Also, if the object is rotated by $$\theta$$ and this transformation is denoted by $$T_\theta$$, then $$T_\theta[R(\phi)] = Rot\{R[(\phi-\theta)\ (mod\ 2\pi)],\ \theta\}$$ i.e., all the indices are incremented by $$-\theta$$ modulo $$2\pi$$, the appropriate vectors $$r$$ are found, and then they are rotated by $$\theta$$.
So I am pretty sure that your option (1) is a correct way of going about this: when varying $$\theta$$ you index an offset-by-$$\theta$$ entry on the $$R$$-table.
A disclaimer is in order: I am no expert on the matter. In any case, it seems pretty clear to me from the paper, but perhaps I'm missing some invariance which lets us ignore the offset. Anyway, hope this helps.
Edit: Indeed, I just finished implementing the GHT and it won't work properly without the $$\theta$$-offset.
This first image shows the result without applying the offset before entering the table (the red crosses on the left indicate detected centroids corresponding with the object on the right). As you can see, the rotated object is not detected (the false positive on the lower-left side disappears if I use a stricter threshold (i.e, minimum amount of votes)).
The following results from offsetting by $$-\theta$$ before retrieving the $$R$$-table entries (and is also resistant to a higher threshold, meaning it's not just a translation of the false positive):
• Well, having read this, I am trying to understand what is the difference between what you have written and "$S$ modulates your $r$ (the radius of the template) and $\theta$ offsets your $\beta$ (the bearing of your template)." (?)
– A_A
Jan 18, 2019 at 0:04
If the shape is rotated by $\theta$, then the gradient orientation ($\phi$) for a given edge point changes. So, shouldn't we do either one of the following:
1. Rotate all the ϕ values in R table by θ?
OR
1. Rotate gradient vector by ($-\theta$) and then calculate $\phi$ for the edge point?
The generalised Hough transform does #1 implicitly by the $S, \theta$ parameters when accumulating.
Note here that usually, you don't really know that the extracted contour has been rotated by some angle (let's call it $\alpha$). You usually are trying to find that angle.
During the preparation stage, you construct the "template", $\phi$. This quantises the circumference of a circle that encloses the shape to be recognised into $k$ levels ($k$ distinct angles). Then, for each $k^{th}$ level, you look at the contours that make up your template image at the $\phi_k$ bearing and note down their $(r, \beta)$. The $(r,\beta)$ is nothing more than the polar coordinates of an element of your template image's contour that is oriented at angle $\phi_k$.
If you wanted to recognise a simple circle, then you would only have one column of $(r, \beta)$. If that circle enclosed another, smaller circle (imagine for example a road sign), then at some orientation $\phi_k$ you would have more than one "edges" and therefore more than one columns of $(r, \beta)$ data.
Now, during the recognition phase, you use the $\angle G$ to lookup an entry in $\phi$. Once you have found that entry, your take that entry's $(r, \beta)$ and use them to find the "right Hough counter cell" $H(x_c,y_c)$ and increment it.
The point here is that the $(r, \beta)$ are calculated with respect to the centroid and it is that centroid $x_c,y_c$ that accumulates quickly in $H$. When the shape is matching, it means that your test image contains a lot of $\angle G$ entries that have resulted in the $H(x_c,y_c)$ counter getting a very high value.
The next step is to be able to recognise the template $\phi$ irrespectively of scale $S$ and orientation $\theta$.
Notice here that if you were to take the Hough transform of the template with its centroid translated at $(0,0)$, then different rotations $\theta_1, \theta_2, \theta_3, \ldots$ are basically cyclic lateral shifts in the Hough space (where the $x$ dimension now corresponds to bearing $\theta$) and different (uniform) scalings $S_1, S_2, S_3, \ldots$ correspond to different "arc amplitudes" (in the $y$ direction that corresponds to distance from origin).
So, the template doesn't really change, all that you do is consider different scalings and rotations at the same time. From another point of view it is as if you were running $N$ different Hough transforms where you scale / rotate the original image by $(S,\theta)_n$. You would produce one $H$ for every $(S, \theta)_n$ such as $\left(1,0\right), \left(1, \frac{\pi}{4}\right), \left(1, \frac{\pi}{2}\right), \left(1, \frac{3 \pi}{4}\right), \left(1, \pi\right), \left(0.5,0\right), \left(0.5, \frac{\pi}{4} \right), \ldots, \left(0.25, 0 \right), \ldots$. In this case here we are looking for 4 orientations at scale 1, 0.5, 0.25 and so on.
But, rotating and scaling the original image can be costly because there are definitely more pixels in an image than elements in the $k$ level quantised periphery of the curve you are trying to find.
For this reason, instead of rotating the original image and running parallel Hough transforms, you simply scale and rotate $\phi$ but you do it on the fly. Instead of modifying the $\phi$ you just introduce $S, \theta$. $S$ modulates your $r$ (the radius of the template) and $\theta$ offsets your $\beta$ (the bearing of your template).
In this way, you produce $N$ matrices $H(x,y)$, one for each $(S, \theta)_n$ you would like to consider. Again, the $x_c, y_c$ counter that crosses a certain threshold marks the location of the recognised contour but in addition now you can also infer at which scale and rotation by fixing the $(S, \theta)$ pair.
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# Hardest Area Questions: "Probability and Combinations"
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Hardest Area Questions: "Probability and Combinations" [#permalink] 20 Sep 2010, 00:04
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BOOKMARKED
Hardest Area Questions: "Probability and Combinations Questions With Solutions".
This is the third part of the tough questions with solutions, this time PS Probability and Combinations Questions.
DS questions at: 700-gmat-data-sufficiency-questions-with-explanations-100617.html
PS questions at: tough-problem-solving-questions-with-solutions-100858.html
Hope it helps.
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Probability and Combinations.doc [316 KiB]
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 15 Aug 2011, 05:13
1 basic question though..are these questions relevant wrt the GMAT type questions or are these random questions that have interested you..
my question is primarily for Inequalities,700+ DS questions and P&C questions..
Thanks,
LC
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 11 Oct 2011, 01:30
Awesome collection!! Thanks
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 17 Oct 2011, 03:33
The Question on Denise : Q. No.2
In my Opinion has a simpler solution than the one stated in the document.
We shall look at it in the following way :
P = {1- (Probability of getting that One right combination in the 75 trials) }
P = 1 -{ C(75,1) / C(4000,1) }
P = 157/160
Option C.
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 07 Nov 2011, 08:37
Thanks. Great questions.
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 14 Mar 2012, 15:36
Thank you for this resource! Very helpful.
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 23 Apr 2012, 17:24
thank you... this is one of my weaker points on the GMAT. This will come in handy. Thanks again.
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 04 Feb 2013, 01:46
Expert's post
Bunuel wrote:
Hardest Area Questions: "Probability and Combinations Questions With Solutions".
This is the third part of the tough questions with solutions, this time PS Probability and Combinations Questions.
DS questions at: 700-gmat-data-sufficiency-questions-with-explanations-100617.html
PS questions at: tough-problem-solving-questions-with-solutions-100858.html
Hope it helps.
Hi Bunuel,
This is an awesome collection man followed by great explanations...Kudos !
Could you please let me know whether there are any further CONSOLIDATED COLLECTION of questions in you sig.(although apparently I don't find any other consolidated one...) or anywhere in the forum (which have been reviewed by you/bb/walker or any other experts) on 'Probability and Combinations' especially for 700+ category ?
It would of great help if you kindly let me know whether any discussion on the followings are available in the forum or not (as referred in the GMAT Club Math book):
1.The Official Guide,Quantitative 2th Edition: PS #79; PS #160
2.The Official Guide, 11th Edition: DT #4; DT #7; PS #10; PS #64; PS #173; PS #217; PS #231; DS #82; DS #114;
As I don't have the OG11 and Quant OG 2,so please share the links,if any.
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 04 Feb 2013, 03:48
Expert's post
debayan222 wrote:
Bunuel wrote:
Hardest Area Questions: "Probability and Combinations Questions With Solutions".
This is the third part of the tough questions with solutions, this time PS Probability and Combinations Questions.
DS questions at: 700-gmat-data-sufficiency-questions-with-explanations-100617.html
PS questions at: tough-problem-solving-questions-with-solutions-100858.html
Hope it helps.
Hi Bunuel,
This is an awesome collection man followed by great explanations...Kudos !
Could you please let me know whether there are any further CONSOLIDATED COLLECTION of questions in you sig.(although apparently I don't find any other consolidated one...) or anywhere in the forum (which have been reviewed by you/bb/walker or any other experts) on 'Probability and Combinations' especially for 700+ category ?
It would of great help if you kindly let me know whether any discussion on the followings are available in the forum or not (as referred in the GMAT Club Math book):
1.The Official Guide,Quantitative 2th Edition: PS #79; PS #160
2.The Official Guide, 11th Edition: DT #4; DT #7; PS #10; PS #64; PS #173; PS #217; PS #231; DS #82; DS #114;
As I don't have the OG11 and Quant OG 2,so please share the links,if any.
DS Question Directory by Topic and Difficulty: ds-question-directory-by-topic-difficulty-128728.html
PS Question Directory by Topic and Difficulty: gmat-ps-question-directory-by-topic-difficulty-127957.html
All OG13 questions with solutions: the-official-guide-quantitative-question-directory-143450.html
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 04 Feb 2013, 04:21
Expert's post
Bunuel wrote:
debayan222 wrote:
Bunuel wrote:
Hardest Area Questions: "Probability and Combinations Questions With Solutions".
This is the third part of the tough questions with solutions, this time PS Probability and Combinations Questions.
DS questions at: 700-gmat-data-sufficiency-questions-with-explanations-100617.html
PS questions at: tough-problem-solving-questions-with-solutions-100858.html
Hope it helps.
Hi Bunuel,
This is an awesome collection man followed by great explanations...Kudos !
Could you please let me know whether there are any further CONSOLIDATED COLLECTION of questions in you sig.(although apparently I don't find any other consolidated one...) or anywhere in the forum (which have been reviewed by you/bb/walker or any other experts) on 'Probability and Combinations' especially for 700+ category ?
It would of great help if you kindly let me know whether any discussion on the followings are available in the forum or not (as referred in the GMAT Club Math book):
1.The Official Guide,Quantitative 2th Edition: PS #79; PS #160
2.The Official Guide, 11th Edition: DT #4; DT #7; PS #10; PS #64; PS #173; PS #217; PS #231; DS #82; DS #114;
As I don't have the OG11 and Quant OG 2,so please share the links,if any.
DS Question Directory by Topic and Difficulty: ds-question-directory-by-topic-difficulty-128728.html
PS Question Directory by Topic and Difficulty: gmat-ps-question-directory-by-topic-difficulty-127957.html
All OG13 questions with solutions: the-official-guide-quantitative-question-directory-143450.html
Thanks bunuel..
Can you give any idea on the above highlighted part ?
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 04 Feb 2013, 04:26
Expert's post
I think that all these questions were already discussed on the forum, so you could try to search.
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 08 Feb 2013, 09:59
Expert's post
Bunuel wrote:
I think that all these questions were already discussed on the forum, so you could try to search.
Bunuel,
Actually I don't have either of OG Quantitative 2th Edition or OG 11th Edition..So it's bit tough for me to search this individual question..Not only these but also all the questions(from these two books) referred by you/bb in GMAT Club Math book.
So,can you please let me know if there is any particular thread in the forum that contains the link to the discussions of ALL questions from OG Quantitative 2th Edition or OG 11th Edition..?
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 24 Feb 2013, 09:00
Hello Bunuel, Appreciate your help so far. Do you have a similar list consolidated for Weighted Avg problem types?
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 24 Feb 2013, 10:08
Expert's post
debayan222 wrote:
Bunuel wrote:
I think that all these questions were already discussed on the forum, so you could try to search.
Bunuel,
Actually I don't have either of OG Quantitative 2th Edition or OG 11th Edition..So it's bit tough for me to search this individual question..Not only these but also all the questions(from these two books) referred by you/bb in GMAT Club Math book.
So,can you please let me know if there is any particular thread in the forum that contains the link to the discussions of ALL questions from OG Quantitative 2th Edition or OG 11th Edition..?
bb/Bunuel- any information on this guys?
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 21 Mar 2013, 12:38
Thanks Bunuel! great job!
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 06 Apr 2013, 05:03
Hi Bunuel, Please tag this forum with quantitative and Problem Solving. This will make the search on your valuable inputs easier.
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 09 May 2013, 01:41
Bunuel wrote:
prashantbacchewar wrote:
Thanks Bunuel.
Is there any document which will explain me Probability and Combinatrics.?
Check Probability and Combinatorics chapters of Math Book:
math-probability-87244.html
math-combinatorics-87345.html
Hope it helps.
Hi banuel,
Is there any documents where concepts and approach methods of inequalities explained ? I am looking for it for a long time but haven't found one.
Regards
kabilan.K
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 09 May 2013, 01:43
1
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Expert's post
kabilank87 wrote:
Bunuel wrote:
prashantbacchewar wrote:
Thanks Bunuel.
Is there any document which will explain me Probability and Combinatrics.?
Check Probability and Combinatorics chapters of Math Book:
math-probability-87244.html
math-combinatorics-87345.html
Hope it helps.
Hi banuel,
Is there any documents where concepts and approach methods of inequalities explained ? I am looking for it for a long time but haven't found one.
Regards
kabilan.K
Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 10 Jul 2013, 00:17
Expert's post
Bumping for review*.
*New project from GMAT Club!!! Check HERE
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 21 Nov 2013, 01:33
Bunuel wrote:
sandeep800 wrote:
thanx for that bunuel!!wish u could have posted answer instead if links!!we need to connect to net evry time we wanaa see a solution..anyways good doc for P& C
Every question has a detailed solution with answers. So that's not true.
All questions have links to the discussions on Gmat Club forum in case of any questions.
Do You have such collection for Modulus Question?
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 21 Nov 2013, 02:18
Expert's post
honchos wrote:
Bunuel wrote:
sandeep800 wrote:
thanx for that bunuel!!wish u could have posted answer instead if links!!we need to connect to net evry time we wanaa see a solution..anyways good doc for P& C
Every question has a detailed solution with answers. So that's not true.
All questions have links to the discussions on Gmat Club forum in case of any questions.
Do You have such collection for Modulus Question?
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Re: Hardest Area Questions: "Probability and Combinations" [#permalink] 21 Nov 2013, 02:18
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So this is my first time posting anything like this, but here goes!
Spoiler
Goals for this year:
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Other fun stuff:
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Anyways that’s all from me for now, enjoy
(edited 1 year ago)
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Study chemistry 🧪
By practicing ofc
I like this. Straight and to the point! Good luck with your goals and studies.
Original post by 0ptics
I like this. Straight and to the point! Good luck with your goals and studies.
Why thank you! Good luck with everything too 🥳
Ah I haven’t posted in a month so we are back at it!
Since it’s 9:45pm rn, I am going to post my todo list that I did today and then we will start tracking my progress from tomorrow
Today’s todo list:
- Chapter 11 revision [maths]
- Practice NMR spectroscopy [chemistry] - I have no clue what I’m doing but after watching allergy chemistry, I think i got the hang of it
- Tutor [job] - literally love this job!
So that is all I did today but tbf i did do a lot in the weekend so im going to let myself have a chill day today anyways !!
(edited 1 year ago)
Right so today we did quite a bit.
Had biology&chemistry and then maths after school
Very much worn out
Today’s todo list is very boring but needs to be done because I’ve got an integration exam tomorrow and I’ve revised one section from the chapter
ToDo:
- Reverse chain rule
- Integration by substitution
- Integration by parts
- Partial fractions
- Finding areas
- Trapezium rule
- Differential equations
Wish me luck
(edited 1 year ago)
Original post by teebslmao
Right so today we did quite a bit.
Had biology&chemistry and then maths after school
Very much worn out
Today’s todo list is very boring but needs to be done because I’ve got an integration exam tomorrow and I’ve revised one section from the chapter
ToDo:
- Reverse chain rule
- Integration by substitution
- Integration by parts
- Partial fractions
- Finding areas
- Trapezium rule
- Differential equations
Wish me luck
Good luck!
Original post by teebslmao
Right so today we did quite a bit.
Had biology&chemistry and then maths after school
Very much worn out
Today’s todo list is very boring but needs to be done because I’ve got an integration exam tomorrow and I’ve revised one section from the chapter
ToDo:
- Reverse chain rule
- Integration by substitution
- Integration by parts
- Partial fractions
- Finding areas
- Trapezium rule
- Differential equations
Wish me luck
Good luckkkk 😁
Original post by TriplexA
Good luck!
Thank you! Will need as much luck as I can get
Original post by starinthenight
Good luckkkk 😁
Thank you very much
Original post by teebslmao
Thank you! Will need as much luck as I can get
You need self belief - you can do this!
Original post by TriplexA
You need self belief - you can do this!
Loving the support!! Maths is one of my weaker subjects so i got to be more confident!
Original post by teebslmao
Right so today we did quite a bit.
Had biology&chemistry and then maths after school
Very much worn out
Today’s todo list is very boring but needs to be done because I’ve got an integration exam tomorrow and I’ve revised one section from the chapter
ToDo:
- Reverse chain rule
- Integration by substitution
- Integration by parts
- Partial fractions
- Finding areas
- Trapezium rule
- Differential equations
Wish me luck
ugh im literally on a break from doing 11D rn 😃😃
Original post by bbdiya
ugh im literally on a break from doing 11D rn 😃😃
No way - same 😝 genuinely struggling though but we move 😭
How are you finding integration so far? I would say its one of the most unnecessary subjects I’ve come across 😪
Original post by teebslmao
Right so today we did quite a bit.
Had biology&chemistry and then maths after school
Very much worn out
Today’s todo list is very boring but needs to be done because I’ve got an integration exam tomorrow and I’ve revised one section from the chapter
ToDo:
- Reverse chain rule
- Integration by substitution
- Integration by parts
- Partial fractions
- Finding areas
- Trapezium rule
- Differential equations
Wish me luck
Ah, chain rule is complete. I’m still a bit iffy on it so im not 100% ready for tomorrow but I hope that once I do some questions from the mixed exercise, I will begin to get the hang of it
Original post by teebslmao
Ah, chain rule is complete. I’m still a bit iffy on it so im not 100% ready for tomorrow but I hope that once I do some questions from the mixed exercise, I will begin to get the hang of it
You’ve got this girl! Just a few months more of a level torture then you will be free from it 😋
Original post by Immaculatevibes
You’ve got this girl! Just a few months more of a level torture then you will be free from it 😋
Thank you bb
It doesn’t sound like that much but it will drag on and on and on..
How you holding up med-wise?
Original post by teebslmao
Right so today we did quite a bit.
Had biology&chemistry and then maths after school
Very much worn out
Today’s todo list is very boring but needs to be done because I’ve got an integration exam tomorrow and I’ve revised one section from the chapter
ToDo:
- Reverse chain rule
- Integration by substitution
- Integration by parts
- Partial fractions
- Finding areas
- Trapezium rule
- Differential equations
Wish me luck
don't even talk to me about integration by substitution. i'm not convinced it works tbh. i remember you from the plym a100 forum. what are your uni plans atm?
Original post by z-u-l-u
don't even talk to me about integration by substitution. i'm not convinced it works tbh. i remember you from the plym a100 forum. what are your uni plans atm?
Heyy, i think i remember you too from Plymouth
Well, I’ve got anglia Ruskin left to hear back from but I’ve got no hope of an offer because it didn’t go that great 💀
I am planning a gap year because at this point, i think it will be best for me. However, the grades must come first ! So a-levels have been my main focus so far
How is your uni application going so far? Hopefully its going well 😁
Original post by teebslmao
No way - same 😝 genuinely struggling though but we move 😭
How are you finding integration so far? I would say its one of the most unnecessary subjects I’ve come across 😪
it’s so hard😩😩 literally was saying in class today when will i ever use this **** in my life again ????? honestly just wanna get the next few months over with😭 | 1,568 | 6,373 | {"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-2024-10 | latest | en | 0.920133 |
http://www.jiskha.com/display.cgi?id=1344820053 | 1,462,438,501,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860126377.4/warc/CC-MAIN-20160428161526-00026-ip-10-239-7-51.ec2.internal.warc.gz | 612,400,821 | 3,845 | Thursday
May 5, 2016
# Homework Help: physics
Posted by Ben on Sunday, August 12, 2012 at 9:07pm.
A 1.5-kg playground ball is moving with a velocity of 3.0 m/s directed 30
°
below thehorizontal just before it strikes a horizontal surface. The ball leaves this surface 0.50 slater with a velocity of 2.0 m/s directed 60
°
above the horizontal. What is themagnitude of the average resultant force on the ball?
• physics - drwls, Sunday, August 12, 2012 at 9:55pm
Divide the magnitude of the momentum changr by 0.50 s. You will need to compute both the horizontal and vertical components of the momentum change. | 174 | 612 | {"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} | 3.09375 | 3 | CC-MAIN-2016-18 | longest | en | 0.901051 |
https://math.stackexchange.com/questions/2160069/solving-2d-heat-equation-with-separation-of-variables | 1,719,231,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00453.warc.gz | 349,160,223 | 36,730 | # Solving 2D heat equation with separation of variables
Solve:
$$\begin{cases} u_t = Ku_{xx}\\ u(0,y,t) = u(\pi,y,t) = 0\\ u_y(x,0,t) = u_y(x,\pi,t) = 0\\ u(x,y,0) = 1 \end{cases}$$
for $0 < x < \pi$, $0 < y < \pi$, $t>0$:
This was the problem given to me, but I don't believe it has a nontrivial solution (correct me if I'm wrong). Instead, I think the problem was meant to say: $$u_t = K(u_{xx} + u_{yy})$$ In this case,
Let $u(x,y,t) = f(x)g(y)h(t)$: $$\frac{h'(t)}{h(t)} = \frac{f"(x)}{f(x)} + \frac{g"(y)}{g(y)}$$ Let: $$\frac{f"(x)}{f(x)} = -n^2$$ $$\frac{g"(y)}{g(y)} = -m^2$$ $$\frac{h'(t)}{h(t)} = -(m^2 + n^2)$$ This leads to the seaparted solutions: $$f(x) = A\cos(nx) + B\sin(mx)$$ $$g(y) = C\cos(my) + D\sin(my)$$ $$h(t) = Ee^{-(n^2 + m^2)t}$$
$$u(o,y,t) = u(\pi,y,t) = 0 \space\text{ implies } \space A = 0$$ $$u_y(x,0,t) = u_y(x,\pi,t) = 0 \space\text{ implies } \space D = 0$$ Thus, $$u(x,y,t) = B\sin(nx)\cos(my)e^{-(n^2 + m^2)t}$$ To satisfy the initial value, we can exploit the superposition principle:
$$u(x,y,t) = \sum_{m=0}^\infty \sum_{n=0}^\infty B_{nm}\sin(nx)\cos(my)e^{-(n^2 + m^2)t}$$ $$u(x,y,0) = \sum_{m=0}^\infty \sum_{n=0}^\infty B_{nm}\sin(nx)\cos(my) = 1$$
This was as far as I was able to get. I'm unsure how to satisfy the initial condition given this double sum. Some help would be appreciated!
You can find $$m$$ and $$n$$ using boundary conditions. Instead of calling your constant $$n$$ or $$m$$, call them $$k$$ or $$\lambda$$. $$m$$ and $$n$$ are used frequently for natural numbers.
$$u(x=\pi) = 0 \Rightarrow B\sin(\lambda\pi) = 0$$
And we want a non trivial solution, so $$B\ne 0$$. Therefore $$\sin(\lambda \pi)=0$$
$$\lambda \pi = \pi n \Rightarrow \lambda = n$$
I didn't see you use the BVs so I'm not sure if you did. that step.
Now all that's left is to find the coefficient $$B_{nm}$$ using the orthogonality properties of your eigenfunctions.
$$B_{nm} = \frac{4}{\pi^2}\int_0^\pi\int_0^\pi f(x,y)\sin(n'x)\sin(m'y)dxdy$$
where $$f(x)=1$$ in this case.
You don't even have to memorize the integral above to find the coefficient in the future.
You can simply multiply both sides by $$\sin(n'x)\sin(m'y)$$ and integrate on the domain. Using properties of Kronecker delta, only when $$m' = m$$ and $$n'=n$$ will get something that isn't zero. The left hand side will simply always be:
$$\\B_{nm}\int_0^\pi \sin^2(nx)dx\int_0^\pi \sin^2(my)dy =B_{nm}\frac{\pi}{2}\frac{\pi}{2} = B_{nm}\frac{\pi^2}{4}$$
Now we have:
$$B_{nm}\frac{\pi^2}{4} = \int_0^\pi\int_0^\pi 1\sin(nx)\sin(my)dxdy$$
• I got fooled by my own choice of notation in assuming that $n$ and $m$ were integers, but still, it makes no difference for my final result. I'm unsure how to use the orthogonality condition in 2D to obtain $B_{nm}$ Commented Feb 24, 2017 at 23:40
• multiply both sides by sin(nx)sin(my) and integrate Commented Feb 24, 2017 at 23:43
• wouldn't you want to multiply by $\sin(nx)\cos(my)$ instead? Otherwise multiplying through by $\sin(nx)\sin(my)$ and integrating would result in 0, as $\cos(my)$ and $\sin(my)$ are orthogonal for all $n,m$ Commented Feb 25, 2017 at 0:06 | 1,156 | 3,131 | {"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-26 | latest | en | 0.820999 |
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# Solve this
Asked by venkat 25th December 2017, 8:56 PM
Since Q is projected after 1 second, P will reach the maximum height and meet Q when it is returning back.
Let maximum height reached by P is H.
H = (u×u) /(2×g), where u is initial projection velocity 30 m/s
H = (30×30) / (2*10) = 45 m.
after reaching the maximumheight let P travels back for t seconds and meet Q.
Let the distance travelled by P in that t seconds is h and it is given by
h = (1/2)g×t2 = (1/2)×10×t2 = 5×t2 .........................(1)
Then distance travelled by Q when it meets P is 45-h and the time taken is (2+t)s. This distance is given by
45-h = 30×(2+t) - (1/2)×10×(2+t)2
after substituting for h from eqn.(1), we will get
45 - 5×t2 = 30×(2+t)-5×(2+t)2
solving the above eqn for t, we get t = 0.5 s.
Hence P travelled 3.5 s before meetin Q
Answered by Expert 27th December 2017, 3:13 PM
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You have rated this answer 10/10 | 432 | 1,199 | {"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-2022-33 | latest | en | 0.910619 |
https://en.wikipedia.org/wiki/Talk:Tangent_vector | 1,490,618,742,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189472.3/warc/CC-MAIN-20170322212949-00440-ip-10-233-31-227.ec2.internal.warc.gz | 791,809,541 | 8,990 | # Talk:Tangent vector
## First major revision
Please provide suggestions for improvement! I think I may have made the first major revision to this article. I reorganized the old material into the intro and added a definition and properties as well as reference. I may add more later. Gray's definition of a tangent vector is actually prior to his definition of a directional derivative. I may add this material which talks about a vector part and a point of application as comprising a tangent vector which is just these two pair of points in Euclidean space. I don't like this definition however. I need a little more mathematical sophistication, but I intend to cover tangent vectors as elements of a tangent space [in] a differentiable manifold. I do not think that this material is too technical. A tangent is part of elementary calculus, but a tangent vector is part of vector calculus. Therefore, our readers will have some very small bit of mathematical sophistication (like me).Stewart.M.Nash (talk) 02:34, 15 May 2015 (UTC)
## Motivation sample with error
${\displaystyle \mathbf {T} (0)={\frac {\mathbf {r} ^{\prime }(0)}{|\mathbf {r} ^{\prime }(0)|}}=\left.{\frac {(2t,2e^{2t},\sin {t})}{\sqrt {4t^{2}+e^{2t}+\sin ^{2}{t}}}}\right|_{t=0}=(0,1,0)\,.}$
Here the bottom part is not equal to the vector length. AFAIK it must be
${\displaystyle \mathbf {T} (0)={\frac {\mathbf {r} ^{\prime }(0)}{|\mathbf {r} ^{\prime }(0)|}}=\left.{\frac {(2t,2e^{2t},\sin {t})}{\sqrt {4t^{2}+4e^{4t}+\sin ^{2}{t}}}}\right|_{t=0}=(0,1,0)\,.}$ Arkadi kagan (talk) 17:34, 12 December 2015 (UTC)
## Expression in denominator of example
The example only makes sense if the denominator evaluates to the value 2 at t=0; e.g., (4t2 + 4e4t + sin2t )0.5 = 2. Arkadi is correct.
Krakengreen (talk) 08:37, 23 December 2015 (UTC) | 545 | 1,816 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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} | 3.9375 | 4 | CC-MAIN-2017-13 | latest | en | 0.939828 |
https://bridgitmendlermusic.com/how-do-you-interpret-a-regression-graph/ | 1,721,564,977,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00417.warc.gz | 122,042,060 | 9,395 | ## How do you interpret a regression graph?
Interpreting the slope of a regression line The slope is interpreted in algebra as rise over run. If, for example, the slope is 2, you can write this as 2/1 and say that as you move along the line, as the value of the X variable increases by 1, the value of the Y variable increases by 2.
## What does the regression analysis tell you?
Regression analysis is a reliable method of identifying which variables have impact on a topic of interest. The process of performing a regression allows you to confidently determine which factors matter most, which factors can be ignored, and how these factors influence each other.
How do you interpret regression intercepts?
The intercept (often labeled the constant) is the expected mean value of Y when all X=0. Start with a regression equation with one predictor, X. If X sometimes equals 0, the intercept is simply the expected mean value of Y at that value.
### What does R Squared mean in EViews?
The R-squared ( ) statistic measures the success of the regression in predicting the values of the dependent variable within the sample. The statistic will equal one if the regression fits perfectly, and zero if it fits no better than the simple mean of the dependent variable.
### What is the main purpose of regression analysis?
Typically, a regression analysis is done for one of two purposes: In order to predict the value of the dependent variable for individuals for whom some information concerning the explanatory variables is available, or in order to estimate the effect of some explanatory variable on the dependent variable.
What are the objectives of regression analysis?
Objective of Regression analysis is to explain variability in dependent variable by means of one or more of independent or control variables.
#### How do you interpret multiple regression intercepts?
Intercept: the intercept in a multiple regression model is the mean for the response when all of the explanatory variables take on the value 0. In this problem, this means that the dummy variable I = 0 (code = 1, which was the queen bumblebees) and log(duration) = 0, or duration is 1 second.
#### What are two regression lines?
In regression analysis, there are usually two regression lines to show the average relationship between X and Y variables. It means that if there are two variables X and Y, then one line represents regression of Y upon x and the other shows the regression of x upon Y (Fig. 35.2).
Which is the first variable in the regression in EViews?
In EViews you specify a regression with the ls command followed by a list of variables. (“LS” is the name for the EViews command to estimate an ordinary Least Squares regression.) The first variable is the dependent variable, the variable we’d like to explain pce in this case.
## Where do I find the regression command in EViews?
EVIEWS commands or menu commands are printed in the courier type. The View button on the top left of the Equation window gives you a list of options in order to visualise your regression results and to perform a variety of tests.
## What does the keyword C mean in EViews?
“C” is a special keyword telling EViews to estimate the equation with an intercept. (You have simply told EViews to regress the dependent variable, pce, on the explanatory variable, income and a constant). Therefore, whether you use the menu or type a command, EViews churns out the regression results shown below:
What are the features of a regression output?
For instance, in undertaking an ordinary least squares (OLS) estimation using any of these applications, the regression output will give the ANOVA (analysis of variance) table, F-statistic, R-squared, prob-values, coefficient, standard error, t-statistic, sum of squared residuals and so on. These are some common features of a regression output.
How do you interpret a regression graph? Interpreting the slope of a regression line The slope is interpreted in algebra as rise over run. If, for example, the slope is 2, you can write this as 2/1 and say that as you move along the line, as the value of the X variable increases by… | 865 | 4,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.895407 |
http://perplexus.info/show.php?pid=2758&cid=20171 | 1,544,559,496,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823702.46/warc/CC-MAIN-20181211194359-20181211215859-00046.warc.gz | 225,997,126 | 5,170 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
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Who will win? (Posted on 2005-01-10)
There are 2n cards labeled 1, 2, ..., 2n respectively, and the cards are distributed randomly between two players so that each has n cards. Each player takes turns to place one card, and you win if you put down a card so that the current sum of all the played cards is divisible by 2n+1.
For example, if n=10, and the previously placed cards are 5, 8, 9, 19, then if player A now places 1, he wins since 5+8+9+19+1 = 42 is divisible by 2*10+1=21.
Assuming both players want to win, what strategy should one adopt in order to win? Following the strategy, is there a consistent winner of this game?
See The Solution Submitted by Bon Rating: 2.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(5): Solution | Comment 8 of 9 |
(In reply to re(4): Solution by David Shin)
Yes, I see now I had a bug in my program that was following that local strategy (or I intended to make follow that strategy). The bug caused 2 to lose sometimes. Now that it follows that strategy perfectly, 2 never loses. The commented out IF i <> j THEN and its corresponding END IF, were still present, so if the winning card was the one that produced the need to play that card again (which the opponent could not do), that move was not recognized as a winner. It just looked at the already played cards and the other cards in its hand to see if they were needed to win by the opponent.
n2 = 8
DIM deck(n2)
DO
FOR i = 1 TO n2
deck(i) = i
NEXT
FOR i = 1 TO n2
j = INT(RND(1) * n2 + 1)
IF i <> j THEN SWAP deck(i), deck(j)
NEXT
FOR i = 1 TO n2 STEP 2
PRINT deck(i);
NEXT
PRINT
FOR i = 2 TO n2 STEP 2
PRINT deck(i);
NEXT
PRINT
REDIM down(n2)
winner = -1
played = 0
pl = 0
DO
t = 0
FOR i = 1 TO played
t = t + down(i)
IF t > n2 THEN t = t - (n2 + 1)
NEXT
toPlay = 0
FOR i = 1 + pl TO n2 STEP 2
IF deck(i) > 0 AND t + deck(i) = (n2 + 1) THEN
winner = pl
toPlay = i
EXIT FOR
END IF
NEXT
IF toPlay = 0 THEN
FOR i = 1 + pl TO n2 STEP 2
IF deck(i) THEN
FOR j = 1 TO played
IF down(j) THEN
t2 = t + deck(i) + down(j)
DO
IF t2 > n2 THEN t2 = t2 - (n2 + 1)
LOOP UNTIL t2 < (n2 + 1)
IF t2 = 0 THEN
toPlay = i
END IF
END IF
NEXT
END IF
NEXT
END IF
IF toPlay = 0 THEN
FOR i = 1 + pl TO n2 STEP 2
IF deck(i) THEN
FOR j = 1 + pl TO n2 STEP 2
IF deck(j) THEN
' IF i <> j THEN
t2 = t + deck(i) + deck(j)
DO
IF t2 > n2 THEN t2 = t2 - (n2 + 1)
LOOP UNTIL t2 < (n2 + 1)
IF t2 = 0 THEN
IF deck(j) < deck(i) THEN toPlay = j: ELSE toPlay = i
END IF
' END IF
END IF
NEXT
END IF
NEXT
END IF
IF toPlay = 0 THEN
FOR i = 1 + pl TO n2 STEP 2
IF deck(i) THEN toPlay = i
NEXT
END IF
played = played + 1
COLOR 13 + pl
PRINT deck(toPlay);
down(played) = deck(toPlay)
deck(toPlay) = 0
pl = 1 - pl
LOOP UNTIL winner > -1
wins(winner) = wins(winner) + 1
COLOR 7
PRINT
PRINT wins(0), wins(1)
LOOP
Posted by Charlie on 2005-01-10 20:12:13
Search: Search body:
Forums (0) | 1,046 | 3,005 | {"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-2018-51 | longest | en | 0.959463 |
https://completesuccess.in/index.php/2018/06/08/analytical-reasoning-set-107/ | 1,656,363,705,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00679.warc.gz | 215,735,028 | 24,801 | # Analytical Reasoning Set 107
Directions (1-5) Study the following information carefully to answer the given questions.
Ten candidates namely viz. P, Q, R, S, T, U, V, W, X and Y attend the interview but not necessarily in the same order having interview on five different days starting from Monday to Friday of the same week. Each day, two candidates have interview at two different time slots, i.e 9.00 AM and 2 P.M.
X has an interview on Tuesday at 9.00 A.M. The number of people who have interview between V and S is same as the number of people who have interview between R and W. Q has his interview immediately before X. X does not have interview on any of the days before V. The one who has interview at 9.00 A.M has his interview immediately before Y. W does not has interview at 2 P.M S has his interview immediately after the day of one who has interview on Monday. U does not have interview at 2 P.M. S does not has interview on any one of the days after T. Only three people have interview between V and T. Neither T nor V have interview on Friday. Only two people have interview between U and Y. U does not have interview on any of the days after W.
Explanation
Q1. Which of the following is correctly matched?
(a) S – Monday
(b) R – Tuesday
(c) P – Friday
(d) V – Tuesday
(e) T – Wednesday
Q2. Who among the following have interview on Friday?
(a) X, T
(b) W, Y
(c) Q, P
(d) S, T
(e) T, R
Q3. Four among the following form a group in a certain way. Which of the following does not belong to Group?
(a) V
(b) X
(c) U
(d) P
(e) T
Q4. Who among the following person has interview at 2 P.M?
(a) R
(b) V
(c) X
(d) U
(e) T
Q5. How many persons have seminar at 2’0 clock between V and R?
(a) 5
(b) 6
(c) 2
(d) 4
(e) None of these. | 536 | 1,742 | {"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} | 3.578125 | 4 | CC-MAIN-2022-27 | latest | en | 0.869313 |
https://mathoverflow.net/questions/251497/taking-away-the-almost-sure | 1,571,241,337,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00191.warc.gz | 585,004,029 | 22,337 | # Taking away the “almost sure” [closed]
Given an arbitrary sequence of random variables (or say measurable functions on a finite-measure space) $\xi_n$, one can show by a truncation and Borel-Cantelli argument that there always exists a sequence $c_n>0$ such that $$\sum_{n=1}^\infty c_n \xi_n \quad \text{converges almost surely.}$$
Can one give an example to show that the "almost surely" in the statement CANNOT be strengthened to "pointwise"?
## closed as off-topic by R W, Franz Lemmermeyer, Ryan Budney, András Bátkai, WolfgangOct 6 '16 at 7:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – R W, Franz Lemmermeyer, Ryan Budney, Wolfgang
If this question can be reworded to fit the rules in the help center, please edit the question.
• RVs in general are not defined pointwise EVERYWHERE, so this question does not make a lot of sense. – Asaf Oct 5 '16 at 20:21
• To make sense of the question, you might say: find a sequence of real-valued Borel functions $\xi_n$ on, say, $[0,1]$, such that there is no sequence $c_n > 0$ for which $\sum_{n=1}^\infty c_n \xi_n$ converges pointwise. – Robert Israel Oct 5 '16 at 20:24
Let $\Omega=(0,1)$ and $\xi_n:\omega\in\Omega\mapsto$ the $n$-th term of the continued fraction expansion of $\omega$. Given a sequence $c_n$, there is another sequence $m_n\in\mathbb{N}$ such that $\sum_{n=1}^\infty c_nm_n$ diverges. Let $x=[m_1,m_2,\dots,m_n,\dots]$. Then $\sum_{n=1}^\infty c_n\xi_n(x)$ diverges. | 488 | 1,653 | {"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} | 3.09375 | 3 | CC-MAIN-2019-43 | latest | en | 0.846275 |
https://holooly.com/solutions-v2-1/in-the-evening-a-contractor-uses-the-wire-rope-on-a-large-crane-to-lift-a-1000-lb-air-compressor-10-ft-above-the-ground-which-will-help-prevent-mischief-occurring-overnight-the-wire-rope-has-a-diame/ | 1,680,200,355,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00222.warc.gz | 346,016,009 | 20,732 | ## Q. 3.4
In the evening a contractor uses the wire rope on a large crane to lift a 1000-lb air compressor 10 ft above the ground, which will help prevent mischief occurring overnight. The wire rope has a diameter of d = 1/4 in.
and is composed of IPS wires in a 6 × 19 Seale IWRC construction; see Table 3.1.The wire rope modulus is $E_r$ = 12,000 ksi, and its compactness factor is CF = 0.40.The rope runs 10 ft between the crane winch at A and a pulley (sheave) at B, and it then runs L = 40 ft down to the attachment C at the compressor on the ground. The hook at the end of the wire rope, position C, has an attachment efficiency of 90%. (a) Calculate the elongation of the wire rope between the winch at A and the compressor at C, just as the compressor is lifted from the ground.
This is the length of wire rope which must be wound onto the winch drum before the compressor will move. (b) Calculate the factor of safety for the wire rope in this operation, assuming that failure occurs at the recommended maximum load of Table 3.1.
TABLE 3 . 1 Weight per Length and Recommended Allowable Loads (in tons of 2000 lb) Are Shown for 6 × 19 Seale IWRC (independent wire rope core) Construction Wire Ropes Using IPS (improved plough steel) Wires.
Diameter (in.) Weight, lb/ft. Allowable load, tons 1/4 0.18 2.94 1/2 0.46 11.5 1 0.58 44.9 2 7.39 172
## Verified Solution
(a) It is necessary to first calculate the metallic area of the wire rope.
From Eq. 3.21, the metallic area of the rope is
$A_m$ = CF × d² = 0.40 × (0.25 in.)² = 0.025 in.²
The flexibility coefficient, from Eq. 3.19, is
$f_r = \frac{L}{A_mE_r} =\frac{(10 ft + 40 ft)} {(0.025 in^2)(12,000 ksi)} = 0.167 \frac{in.} {kip}$
Note that the entire length of the wire rope between the winch and the compressor (10 ft + 40 ft) elongates under the loading. You can verify this using free-body diagrams. The elongation of the wire rope is then
e = $f_rP = \left(0.167\frac{in.}{kip}\right)$ (1 kip) = 0.167 in. Ans. (a)
(b) Calculation of the expected failure load, $P_F$, must account for the efficiency of the hook end attachment at C. Using the maximum load suggested for this wire rope in Table 3.1 of 5.88 kips, the allowable load after adjusting for the attachment efficiency is
$P_F$ = 0.90 × 5.88 kips = 5.29 kips
The greatest load on the wire rope occurs at B and is the 1000 lbs. (1 kip) of the compressor plus the weight of the wire rope. The total weight of wire rope hanging from the sheave at B to the attachment at C is,
W = (0.18 lb/ft)(40 ft) = 7.2 lb
This is, of course, a negligibly small load compared to that of the compressor.
Thus, we may reasonably neglect the weight of the wire rope for this relatively short length. The factor of safety is then
FS = $\frac{5.29 kips }{1 kip}$ = 5.29 Ans. (b) | 839 | 2,808 | {"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": 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": 7, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-14 | latest | en | 0.866172 |
http://www.ub.edu/rfa/R/regression_with_categorical_dependent_variables.html | 1,718,829,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00505.warc.gz | 53,401,836 | 7,084 | Regression with Categorical Dependent Variables
Montserrat Guillén
This page presents regression models where the dependent variable is categorical, whereas covariates can either be categorical or continuous, using data from the book Predictive Modeling Applications in Actuarial Science. A methodological overview can be found in:
• Frees, E.W. (2010). Regression modeling with actuarial and financial applications. Cambridge University Press. New York.
• Greene, W.H. (2011). Econometric analysis. 7th edition. Prentice Hall. New York.
DATA DESCRIPTION
Name Content description FullCoverage.csv 4000 policy holders of motor insurance VehOwned.csv 2067 customers of an insurance firm who are offered unordered options VehChoicePrice.csv This example corresponds to a similar situation to previous example
LOGISTIC REGRESSION MODEL
In the logistic regression model the dependent variable is binary. This model is the most popular for binary dependent variables. It is highly recommended to start from this model setting before more sophisticated categorical modeling is carried out. Dependent variable yi can only take two possible outcomes. We assume yi follows a Bernoulli distribution with probability πi. The probability of the 'event' response πi depends on a set of individual characteristics xi, i = 1, . . ., n, where n is the number of observations.
- Specification
Show/Hide
The logistic regression model specifies that:
$$Pr(y_1=1|\mathbf{x}_i)=\pi_i=\frac{1}{1+exp(-\mathbf{x}_i^\prime \mathbf{\beta})}=\frac{exp(\mathbf{x}_i^\prime \bf{\beta})}{1+exp(\mathbf{x}_i^\prime \bf{\beta})}$$
and the inverse of this relationship, called the link function in generalized linear models, expresses x'i β as a function of πi as:
$$\mathbf{x}_i^\prime \beta=\ln \left(\frac{\pi_i}{1-\pi_i}\right)= logit(\pi_i).$$
- Parameter estimation
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- Script
- Results
MODELS FOR ORDINAL CATEGORICAL DEPENDENT VARIABLES
In ordinal categorical dependent variable models the responses have a natural ordering. This is quite common in insurance, an example is to model possible claiming outcomes as ordered categorical responses.
- Specification
Show/Hide
Let us assume that an ordinal categorical variable has J possible choices. The most straightforward model in this case is the cumulative logit model, also known as ordered logit. Let us denote by yi the choice of individual i for a categorical ordered response variable. Let us assume that πij is the probability that i choses j, j=1,...,J. So, πi1i2+ ... + πiJ = 1. Response probabilities depend on the individual predictors, again, we assume they depend on x'i β. It is important to bear in mind that the ordered logit model concentrates on the cumulative probabilities Pr(yi ≤ j | xi ). Then,
$$logit(Pr(y_i\le j|\mathbf{x}_i ))=\alpha_j+\mathbf{x}_i^\prime \beta.$$ Note that, $$logit(Pr(y_i\le j| \mathbf{x}_i))=\ln \left(\frac{Pr(y_i\le j|\mathbf{x}_i )} {1-Pr(y_i\le j|\mathbf{x}_i )}\right).$$
- Script
- Results
MODELS FOR NOMINAL CATEGORICAL DEPENDENT VARIABLES
Let us start with the generalized logit model. This model is often called the multinomial logit model, which we will present later and which is a bit more general. However, the generalized logit model is so widely used that this is the reason why it is often called the multinomial logit model. It is used to predict the probabilities of the different possible outcomes of a categorically distributed dependent variable, given a set of independent variables that measure individual risk factors.
- Specification
Show/Hide
Let us denote by yi the choice of individual i for a nominal categorical response variable. Let us assume that πij is the probability that i choses j, j=1,...,J and i=1,...,n. So, πi1+ πi2 + ... + πiJ = 1. The probabilities depend on the individual predictors, and we assume these choice probabilities depend on x'i βj.
We assume that the J-th alternative is the baseline choice. Then, the generalized logit regression model is specified as:
\begin{eqnarray} Pr(y_1=j|\mathbf{x}_i)&=&\frac{exp(\mathbf{x}_i^\prime \beta_j)}{1+\sum_{k=1}^{J-1}exp(\mathbf{x}_i^\prime \beta_k)}, \,\, j=1,...,J-1 \nonumber\\ Pr(y_1=J|\mathbf{x}_i)&=&\frac{1}{1+\sum_{k=1}^{J-1}exp(\mathbf{x}_i^\prime \beta_k)}.\nonumber \end{eqnarray}
So there are J-1 vectors of parameters to be estimated, namely β1, β2, ... , βJ-1. We set vector βJ to zero for identification purposes.
- Script
- Results
MULTINOMIAL LOGISTIC REGRESSION MODEL
In the multinomial logistic regression model individual characteristics can be different for different choices. This model is also known as the conditional logit model due to the fact that individual characteristics depend on the chosen alternative.
- Specification
Show/Hide
The multinomial logistic regression model specification is,
\begin{eqnarray} Pr(y_1=j|\mathbf{x}_{ij})=\frac{exp(\mathbf{x}_{ij}^\prime \beta)}{\sum_{k=1}^{J}exp(\mathbf{x}_{ik}^\prime \beta)}, \,\, j=1,...,J. \end{eqnarray}
There is only one vector of unknown parameters β, but we have J vectors of known characteristics xi1, xi2, ..., xiJ.
- Script
- Results
REFERENCES
[1] Allison, P. D. (1999). Logistic regression using the SAS system: theory and application. Cary, NC: SAS Institute.
[2] Cameron, A. C. and Trivedi, P. K. (2005) Microeconometrics: methods and applications. Cambridge University Press. New York.
[3] Frees, E. W. (2010). Regression modeling with actuarial and financial applications. Cambridge University Press. New York.
[4] Greene, W. H. (2011). Econometric analysis. 7th edition. Prentice Hall. New York.
[5] Hilbe, J. M. (2009). Logistic regression models. CRC Press, Chapman & Hall. Boca Raton, FL.
[6] Hosmer, D. W. and Lemeshow, S. (2000). Applied logistic regression. John Wiley & Sons, New York, 2nd edition.
[7] Long, J. S. (1997). Regression models of categorical and limited dependent variables. Sage, Thousand Oaks, CA.
• Universitat de Barcelona - Last Updated: 05-23-2014 | 1,557 | 5,989 | {"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": 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": 5, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.724167 |
https://cloud.tencent.com/developer/article/1010165 | 1,553,538,700,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204086.87/warc/CC-MAIN-20190325174034-20190325200034-00084.warc.gz | 449,215,217 | 18,315 | # LWC 57：721. Accounts Merge
## LWC 57：721. Accounts Merge
Problem:
Given a list accounts, each element accounts[i] is a list of strings, where the first element accounts[i][0] is a name, and the rest of the elements are emails representing emails of the account. Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email that is common to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name. After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails in sorted order. The accounts themselves can be returned in any order.
Example 1:
Input: accounts = [[“John”, “johnsmith@mail.com”, “john00@mail.com”], [“John”, “johnnybravo@mail.com”], [“John”, “johnsmith@mail.com”, “john_newyork@mail.com”], [“Mary”, “mary@mail.com”]] Output: [[“John”, ‘john00@mail.com’, ‘john_newyork@mail.com’, ‘johnsmith@mail.com’], [“John”, “johnnybravo@mail.com”], [“Mary”, “mary@mail.com”]] Explanation: The first and third John’s are the same person as they have the common email “johnsmith@mail.com”. The second John and Mary are different people as none of their email addresses are used by other accounts. We could return these lists in any order, for example the answer [[‘Mary’, ‘mary@mail.com’], [‘John’, ‘johnnybravo@mail.com’], [‘John’, ‘john00@mail.com’, ‘john_newyork@mail.com’, ‘johnsmith@mail.com’]] would still be accepted.
Note:
The length of accounts will be in the range [1, 1000
The length of accounts[i] will be in the range [1, 10].
The length of accounts[i][j] will be in the range [1, 30].
• 同名， 但没有邮箱重复，说明不是同一个人，直接加入数据库。
• 同名，且邮箱有重复，说明是同一个人，则把所有关联的邮箱进行合并即可。
``` public List<List<String>> accountsMerge(List<List<String>> accounts) {
Map<String, List<Set<String>>> mem = new HashMap<>();
int n = accounts.size();
for (int i = 0; i < n; ++i) {
List<String> account = accounts.get(i);
String name = account.get(0);
if (mem.containsKey(name)) {
List<Set<String>> mm = mem.get(name);
int idx = -1;
Set<String> hh = new HashSet<>();
for (int k = 1; k < account.size(); ++k){
}
Set<Integer> mer = new HashSet<>();
for (int k = 1; k < account.size(); ++k) {
String email = account.get(k);
for (int j = 0; j < mm.size(); ++j) {
if (mm.get(j).contains(email)) {
idx = j;
}
}
}
if (mer.size() == 0) { // 同名，属于两个人
mem.computeIfAbsent(name, k -> new ArrayList<>()).add(hh);
}
else { // 属于同一个人
// 邮箱合并
List<Integer> merge = new ArrayList<>();
List<Set<String>> removes = new ArrayList<>();
for (int j = 1; j < merge.size(); ++j) {
}
// 删除之前的邮箱
for (Set<String> re : removes) {
mm.remove(re);
}
}
}
else {
// 数据库中不存在该用户，直接加入数据库
Set<String> mails = new HashSet<>();
for (int j = 1; j < account.size(); ++j) {
}
mem.computeIfAbsent(name, k -> new ArrayList<>()).add(mails);
}
}
List<List<String>> ans = new ArrayList<>();
for (String key : mem.keySet()) {
List<Set<String>> val = mem.get(key);
for (int i = 0; i < val.size(); ++i) {
List<String> vv = new ArrayList<>();
Collections.sort(vv);
List<String> tmp = new ArrayList<>();
}
}
return ans;
} ```
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1546 | 1,312 | 3,942 | {"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} | 2.765625 | 3 | CC-MAIN-2019-13 | latest | en | 0.838003 |
https://www.vedantu.com/maths/reducing-equations-to-simpler-form | 1,619,109,275,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594341.91/warc/CC-MAIN-20210422160833-20210422190833-00503.warc.gz | 1,154,977,212 | 90,786 | # Reducing Equations To Simpler Form
View Notes
## How to Reduce Equations to Simpler Form
Reducing equations is the process of simplifying them. Since not all equations are presented in a linear form, it is vital to reduce them in simpler forms that are easy to understand. Performing it includes various mathematical approaches, and they vary as per the need of a particular equation.
The primary aim of this process is to ease the calculation process. Simplifying such complicated equations into their linear form means effortless calculation, and lesser mistakes. In this chapter of reducing method class 10 students will come across various examples of such equations, and the process to abridge them to perform a hassle-free calculation.
## What is a Linear Equation?
Before moving on to the simplification process, one needs to learn more about linear equations. These equations are known as the first order. Also, these are known as equations for a straight line.
A linear equation is an algebraic expression where every term is a product of single variables and constants, or they remain constant themselves. It carries the first-order power of variables. Such equations are typically represented as Xa+Y=0, where X and Y are constants and X is not equal to 0.
Reducing method class 10 teaches students to learn converting different complicated equations into linear or simpler form.
## Types of Linear Equation
Here are different calculation methods of the same:
• Some of them have one variable on the left-hand side, like 4x + 5 = 30.
• Some may have one variable, but on both sides, like 4x + 5 = 20 + 6x.
Apart from these, some linear equations may have non-linear forms. It requires reducing equations to linear form to make the calculation process simple. An example here is (2y + 5)/(4y + 2) = 1/4. Such equations are not easy to solve; it requires simplification.
## What is Reducing Equation?
Reducing Equations is the process that converts non-linear ones into linear ones. Since every equation is not always available in a simple and straightforward format, it is essential to break them down to make solving easy.
Solving these equations requires usage of some mathematical applications such as cross multiplication, division, etc. on both sides. It helps to convert complicated equations to their linear forms. Following this conversion, it becomes easy to find the value of the variables.
### Tactics of Simplification
Among many tactics to simplify non-linear equations, cross multiplication is one of the prominent ones. In this method, students can multiply the numerator of one fraction with another’s denominator, and vice-versa.
Cross multiplication of equation reducible to linear form example includes the following:
(a - 2)/(a + 8) = ⅔
Now, cross multiplying this equation will result in, 3(a - 2) = 2(a + 8).
Following this cross multiplication, one needs to implement another mathematical operation to move a step closer to solve this equation. It is known as opening the brackets. Additionally, another law used here is called distributive law. Under this, students need to multiply any value within the brackets with the one outside of it.
Now, on using this law on the above mentioned equation, one will get:
3a - 6 = 2a + 16
After implementing distributive law, one needs to arrange the variables on one side and constants on one side. While performing this step, students need to remember that, when they move any value from the RHS to the LHS, it will shift from its negative value to a positive one, and vice-versa. Implementing that in this equation results in:
3a - 2a = 16 + 6
a = 22
A point to note here is that, if students do addition, or subtract, or even perform multiplication with the same value on either side, they will get the value of a variable without changing the final equation.
Now, this is a relatively simple example of the concept of reducing method class 10. There are more complex examples as well, where students need to employ more mathematical applications like LCM to find the desired result.
Point to Note: Equations are a condition of a particular variable.
Reducing method class 10 is an essential chapter of mathematics, and helps students get a clear idea of solving equations. Since it is a vital chapter for the upcoming board exams as well as for higher studies, one must learn it in detail, and thoroughly.
Along with the traditional textbooks, and practice sets, online platforms like Vedantu can be a big help for students. The availability of exam notes, mock question papers, study material coupled with live online classes, and doubt clearing sessions let individuals better their exam preparations.
1. What is LCM?
Ans. LCM stands for least common multiple, and it is the smallest multiple, which are common between two numbers. It represents a value that can evenly divide any two numbers is derived from. It is also known as the least common divisor. For instance, the LCM of 4 and 6 is 24, and this 24 is divisible by both 4, 6.
LCM has different applications based on an equation that you are solving. You can use it to add or subtract any two fractions when their denominators are different. During any mathematical operations such as addition, subtraction, mainly with fractions, LCM is used to make the denominators similar. It simplifies the entire process and helps with the calculation.
2. What is HCF?
Ans. HCF stands for the highest common factor, and it represents the greatest number that divides two numbers. It is also termed as Greatest Common Divisor and Greatest Common Measure. Now, for example, the HCF of 60 and 75 is 15, as it is the highest number that can divide 60 and 75 properly. To find HCF, two methods are being employed chiefly, prime factorisation and division. Similar to the LCM, HCF is another vital mathematical application that is often used while calculating fractions to make equations simple and concise.
3. What are the Methods of Solving an Equation?
Ans. There are three methods that one can use to solve any equation. These are substitution, elimination and augmented matrices. Among these three, substitution and elimination are widely used to solve equations, as they allow it to happen in two simple steps. On the other hand, augmented matrices involve a long process, and it has a wide variety of applications compared to the other two. A notable advantage of augmented matrices is that it can solve a system of 3-4 equations. The other two methods are difficult to employ or will make the solution process lengthy. | 1,399 | 6,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2021-17 | longest | en | 0.953824 |
https://discussions.unity.com/t/how-do-i-rotate-my-camera-smoothly/75789 | 1,708,600,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00691.warc.gz | 212,073,139 | 5,963 | # How do I rotate my camera smoothly?
I have this code for rotating camera… now I wanna rotate it smoothly so when I drop my mouse , camera not to stop immediatly.
``````using System.Collections;
public class rotation : MonoBehaviour {
public float horizontalSpeed = 0.001F;
public float verticalSpeed = 0.001F;
// Update is called once per frame
void Update () {
if (Input.GetMouseButton(0)){
float h = horizontalSpeed * Input.GetAxis("Mouse X");
float v = verticalSpeed * (-Input.GetAxis("Mouse Y"));
h/=10;
v/=10;
transform.RotateAroundLocal(Vector3.forward,h);
transform.RotateAroundLocal(Vector3.forward,v);
}
}
}
``````
I don’t understand the goal from this code. First, RotateAroundLocal() is not documented. So you want to fix or change that. Most undocumented rotation functions take radians. You should change to using Transform.Rotate(). Second both line 14 and line 15 use ‘Vector3.forward’ as the axis of rotation. I’m assuming since you calculate both ‘h’ and ‘v’ that you want to rotate on two axes, but I’m not sure how you want that done. When you rotate on two local axes, your object motion will diverge from your mouse motion making it difficult for users, so you need to figure out how you want it to rotate.
Here is a script for you to play with. It takes your code and add some momentum so that the object continues to spin after the mouse is no longer moved. It rotates on two axes, so you can get an idea of the problems with rotating on two local axes. Put the script on a cube and play before adding it to a camera:
``````using System.Collections;
using UnityEngine;
public class Bug27 : MonoBehaviour {
public float horizontalSpeed = 1.1F;
public float verticalSpeed = 1.1F;
public float drag = 0.95f;
Vector3 v3Rotation = Vector3.zero;
private float h = 0.0f;
private float v = 0.0f;
void Update () {
if (Input.GetMouseButton(0)){
v3Rotation.z += horizontalSpeed * Input.GetAxis("Mouse X");
v3Rotation.x += verticalSpeed * Input.GetAxis("Mouse Y");
}
v3Rotation *= drag;
transform.Rotate (v3Rotation);
}
}
``````
Note the drag solution here is very simple and not frame rate independent. | 512 | 2,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-10 | latest | en | 0.779462 |
http://forums.wolfram.com/mathgroup/archive/2009/Mar/msg00716.html | 1,596,637,493,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735958.84/warc/CC-MAIN-20200805124104-20200805154104-00261.warc.gz | 45,477,185 | 7,541 | Re: How to map a list on function
• To: mathgroup at smc.vnet.net
• Subject: [mg97715] Re: How to map a list on function
• From: Vince <blueschi at gmail.com>
• Date: Thu, 19 Mar 2009 02:13:04 -0500 (EST)
• References: <gpg18t\$ceh\$1@smc.vnet.net>
```On Mar 14, 6:35 am, buts <mange... at yahoo.com> wrote:
> Hello,
>
> Could anyone explain me how to do the following:
> I have a long list of integer numbers in groups of four:
>
> list= {{10,3,5,7},{4,6,8,9},{0,8,3,6}, ...... }
>
> or its Flatten version.
>
> How to write a fast function g[list] which does this:
>
> g[list] =x^10*y^3*z^5*u^7 + x^4*y^6*z^8*u^9 + y^8*z^3*u^6+ ...
>
> It should be done many times (say, 10^4-6) on lists with length > 1000, so taking parts is not a good idea.
> How to use Map or similar ?
> Thanks.
Use Listability. Try:
Plus @@ Times @@ Power[{x, y, z, u}, Transpose@{{10, 3, 5, 7}, {4, 6,
8, 9}, {0, 8, 3, 6}}]
Some timing for other list lengths---1 000 000 gives ~ 18 seconds, 10
000 gives ~ 0.1 seconds, and 1000 gives ~ 0 seconds.
Vince Virgilio
```
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• Next by thread: importing fixed width data | 450 | 1,249 | {"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 | 0.721235 |
https://oeis.org/A210277 | 1,723,428,138,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00481.warc.gz | 333,090,992 | 4,476 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A210277 a(n) = (3*n)!/3^n. 5
1, 2, 80, 13440, 5913600, 5381376000, 8782405632000, 23361198981120000, 94566133475573760000, 553211880832106496000000, 4492080472356704747520000000, 49017582114356362204938240000000, 699971072593008852286518067200000000 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..100 D. Bevan, D. Levin, P. Nugent, J. Pantone and L. Pudwell, Pattern avoidance in forests of binary shrubs, arXiv preprint arXiv:1510:08036 [math.CO], 2015-2016. FORMULA E.g.f.: 1/(1-x^3/3). a(n) = Product_{i=1..n} (2*binomial(3i,3)). - James Mahoney, Apr 04 2012 From Amiram Eldar, Jan 18 2021: (Start) Sum_{n>=0} 1/a(n) = exp(3^(1/3))/3 + (2/3)*exp(-3^(1/3)/2)*cos(3^(5/6)/2). Sum_{n>=0} (-1)^n/a(n) = exp(-3^(1/3))/3 + (2/3)*exp(3^(1/3)/2)*cos(3^(5/6)/2). (End) MATHEMATICA Table[(3 n)!/3^n, {n, 0, 15}] (* Vincenzo Librandi, Feb 15 2013 *) PROG (Magma)[Factorial(3*n)/3^n: n in [0..15]]; // Vincenzo Librandi, Feb 15 2013 CROSSREFS Cf. A210278, A000680, A067630, A084939, A084940, A084941, A084942, A084943, A084944, A087127, A001147, A132101. Sequence in context: A123828 A260659 A351854 * A008563 A059487 A156932 Adjacent sequences: A210274 A210275 A210276 * A210278 A210279 A210280 KEYWORD nonn,easy AUTHOR Mohammad K. Azarian, Mar 20 2012 STATUS approved
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Last modified August 11 21:40 EDT 2024. Contains 375073 sequences. (Running on oeis4.) | 700 | 1,819 | {"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} | 3.28125 | 3 | CC-MAIN-2024-33 | latest | en | 0.514962 |
http://www.quhasa.com/96023--Two-horses-start-trotting-towards-each-other--one-from-A-to-B-and-another-from-B-to-A- | 1,597,100,067,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738699.68/warc/CC-MAIN-20200810205824-20200810235824-00312.warc.gz | 164,803,173 | 15,669 | Two horses start trotting towards each other, one from A to B and another from B to A. They cross each other after one hour and the first horse reaches B, 5/6 hour before the secon What is the speed of the slower horse?
Two horses start trotting towards each other, one from A to B and another from B to A. They cross each other after one hour and the first horse reaches B, 5/6 hour before the secon What is the speed of the slower horse?
1)30 km/h
2)15 km/h
3)25 km/h
4)20 km/h
• : 76
• : 0
20 km/h | 145 | 509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-34 | latest | en | 0.883808 |
http://blog.sigfpe.com/2007/02/comonads-and-reading-from-future.html | 1,469,653,865,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827079.61/warc/CC-MAIN-20160723071027-00208-ip-10-185-27-174.ec2.internal.warc.gz | 28,306,755 | 15,024 | # A Neighborhood of Infinity
## Saturday, February 03, 2007
### Comonads and reading from the future
I'm too busy reading Bernie Pope and Russell O'Connor's articles in the latest Monad Reader to think about much else right now, so I'm only going to make a brief post. I want to mention something pointed out by Nick Frisby over here. The loeb function is in fact closely related to cfix, the dual of mfix for monads, defined by Dave Menendez. (Nick Frisby's statement isn't quite correct but it's interest value outweighs its truthiness.) In fact, I can borrow Russell's language to say that whereas mfix sends data back in time, cfix and loeb read data from the future!
Think of it like this: when you use cobind with a comonad you provide some function that distills down a comonadic data structure into a single value and then cobind applies this distillation repeatedly at every point in the comonadic structure to give you another comonadic structure. My canonical example is the cellular automaton where you give a rule that defines how to compute each individual cell as a function of the whole grid, and cobind applies this throughout an entire input grid to give you back an entire output grid. The functions cfix and loeb allow you to give rules for each cell where you're allowed to make reference to the entire output grid. If you think of the output grid as something in the future that you haven't yet computed, then cfix and loeb allow you to read from the future.
Here's an example of cfix it in use (where I make [] a comonad by interpreting it as a kind of one-sided zipper):
> instance Show (x -> a)> instance Eq (x -> a)> instance (Num a,Eq a) => Num (x -> a) where> fromInteger = const . fromInteger> f + g = \x -> f x + g x> f * g = \x -> f x * g x> negate = (negate .)> abs = (abs .)> signum = (signum .)> class Comonad w where> coreturn :: w a -> a> cobind :: (w a -> b) -> w a -> w b> instance Comonad [] where> coreturn (x:xs) = x> cobind f [] = []> cobind f (x:xs) = f (x:xs) : cobind f xs> cfix :: Comonad d => d (d a -> a) -> a> cfix d = coreturn d (cobind cfix d)> ouroboros = [2*head.tail,1+head.tail,17]> test = cfix ouroboros
In the expression test we can use head and tail to refer to the sublist to the right of each element from within the list itself. For example, in the subexpression 2*head.tail the head.tail is being applied to cfix [1+head.tail,17].
Note how in this example (1) loeb gives you back an entire list, not just one element and (2) cfix respects the comonadic structure in the sense that you can only refer to elements to the right of the current one whereas loeb lets you refer to the entire output list. I think this helps to make clear the differences and similarities between cfix and loeb.
That wasn't too brief. But I did cheat by copying and pasting code from an old post. Anyway, back to tinkering with my implementation of dropWhile...
#### 8 comments:
Kea said...
Your blog continues to amaze me, and I just wanted to say thanks. Most of the CompSci jargon is gobbledygook to me, but I really appreciate the effort you're taking to talk about monads and the like.
sigfpe said...
kea,
If you think some of the jargon looks like gobbledygook, you should check out the published papers that some of this stuff comes from :-) It's been hard work trying to make inroads into this stuff over the last two years.
kotsu said...
That's a nice contrast you've developed.
My first post was inspired mostly by type signatures. Now that I've looked at your actual definition of loeb :), I readily see the difference you're highlighting. As you pointed out, loeb does not need the comonadic structure. This highlights the big difference: comonadic structure (and thus cfix) caters to relative references [1] whereas loeb allows for absolute references.
In your spreadsheet example, you use !! to get the nth cell in the list. A comonadic approach would allow a cell to refer to "the cell to my left" without needing to know its own cell number. Both features are supported in various ways by spreadsheets: whenever I do a fill-down in Excel, every now and then I'm surprised that the relative references worked how I had intended.
Just a couple of notes: to get cfix to "return a whole list" you can use cobind cfix. Also, to have access to the "entire output list", you can use a more robust comonad (like Uustalu and Vene's LVS) with an absolute access function a la !!.
Thanks for the discussion, that was a nice contrast to appreciate.
[1] - This is why zippers induce comonads; I can talk about neighbors in various directions relative to wherever I'm currently at in the structure.
kotsu said...
Forgot to sign! Nick Frisby here again :)
sigfpe said...
(I guess kotsu=Nick Frisby.)
Of course! cfix and loeb deal with relative and absolute references. I hadn't thought of it that way. It's truly bizarre how fiddling about semi-randomly with types leads to new ways to view concepts that are already well known and considered natural.
I'll have a look at Uustalu and Vene's LVS. Their recent papers have been really interesting.
geophf said...
I've been reading your explorations of comonads -- your explanations using examples of cellular automata and image processing have been very helpful for me to understand them a bit better. In parallel, I've been reading about Haskell attribute grammars. Do comonads address a class of problems well that attribute grammars do not?
One of the issues with monads, even with transformers, is that they "stack" poorly (the result depends on order of composition and each monad's internal implementation detail); do comonads compose better than monads?
sigfpe said...
geophf,
By coincidence, I've been thinking about stacking comonads for the last couple of days. Not got very far yet. A web search reveals almost no hits on "comonad transformer".
geophf said...
re: your thinking about composing comonads -- excellent, as I've observed that you thinking about classes of problems results in solutions on your blog.
As to attribute grammars for Haskell, have you looked at them? Applied them? Do you see them being useful in these problem domains, or do they address an entirely different class of problems? | 1,519 | 6,281 | {"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} | 2.875 | 3 | CC-MAIN-2016-30 | longest | en | 0.898652 |
https://quizizz.com/admin/quiz/581a1c682bd55be011d9db43 | 1,601,582,607,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131986.91/warc/CC-MAIN-20201001174918-20201001204918-00372.warc.gz | 554,736,522 | 8,569 | Speed, Acceleration, and Velocity
3 years ago
mrsbrockmyre
Save
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Homework
Solo Practice
Practice
• Question 1
30 seconds
Q. A child rides a bike around the block.
Speed
Velocity
Acceleration
• Question 2
45 seconds
Q. Describe the speed of the rocket in the speed vs. time graph.
Constant speed
Decreasing speed
Increasing speed
Not moving
• Question 3
30 seconds
Q. A dog runs on the sidewalk 9 mi/hr.
Speed
Acceleration
Velocity
• Question 4
30 seconds
Q. Describe the speed of the object from 4-6 seconds using the distance vs. time graph.
Increasing speed
Decreasing speed
Not moving
Constant speed
• Question 5
30 seconds
Q. While going to school, a student walks an average of 5 km/hr west for 12 blocks.
Speed
Acceleration
Velocity
• Question 6
30 seconds
Q. Describe the speed of the object from
0-2 seconds using the distance vs. time graph.
Constant
Speeding up
Slowing down
Not moving
• Question 7
30 seconds
Q. While on a trip, a family travels east from San Antonio to Houston going 70 mph.
Speed
Acceleration
Velocity
• Question 8
45 seconds
Q. Describe the speed of the object from
2-5 seconds using the speed vs. time graph.
Increasing speed
Not moving
Decreasing speed
Constant speed
• Question 9
30 seconds
Q. A bicyclist takes a curved, scenic ride through the park going 4.5 m/s.
Speed
Acceleration
Velocity
• Question 10
45 seconds
Q. Describe the speed of the object from
5-6 seconds using the speed vs. time graph
Speeding up
Slowing down
Not moving
Constant speed
• Question 11
30 seconds
Q. Some motorcycles can go from 0 mi/hr to 60 mi/hr in under 4 seconds. | 440 | 1,614 | {"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} | 2.734375 | 3 | CC-MAIN-2020-40 | latest | en | 0.789199 |
https://bsh-online.nl/how_much_is_1_ton_of_boiler_.html | 1,627,605,238,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153899.14/warc/CC-MAIN-20210729234313-20210730024313-00199.warc.gz | 158,926,740 | 5,294 | ## how much is 1 ton of boiler
### how many kw equal 1.5 ton steam boiler – oil fired boiler
Jul 17, 2020 · Normally, 1 ton of water needs to absorb 600,000 kcal of heat value to turn into steam, so 1 ton of gas The boiler output is 600,000 kcal, according to the formula: Hourly gas consumption of 1 ton gas boiler = 600,000 kcal ÷ 98% ÷ 8600 kcal per cubic meter = 71.19m3
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### how much is 4 ton steam boiler – Coal Fired Boiler
how much gas required to generate 1 ton steam - Zozen Boiler. energy required to produce 1 ton of steam – oil fired. Kefid · How much natural gas is required to produce 1 ton of steam? 1 ton is about 1000 kg …
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### How Much Gas Required to Generate 1 Ton Steam
How Much Diesel Oil Is Required to Produce 1 Ton of Steam
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### how much gas need for 1ton boiler
how many kw equal 1.5 ton steam boiler – oil fired boiler for sale
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### How much water is in a one-ton steam boiler? - Quora
1 ton = 10000 kg. water density approximately 1000 kg/m3. so. it has 10m3. or you can consider the kgs of steam =liters of water. so you have 10000 kg of steam = 10000 liters of water. 1.5K views. Hafiz Bilal., former CEO at AJM Boiler Manufacturer.
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### How Much Diesel Oil Is Required to Produce 1 Ton of Steam
How Much Gas Required to Generate 1 Ton Steam
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### how much natural gas comsumed by boiler to produce 1 ton
How much water is in a one-ton steam boiler? - Quora
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### How Much Steam Produced by 1 Ton Standard Coal at a Boiler
According to 0.8MPa (G) saturated steam, the theoretical energy consumption of ton of steam (calculated as 0 ℃ water) is 662,766 kcal, according to this calculation, 1 ton of steam consumption of standard coal is 94.7 kg of standard coal / ton of steam, steam boiler efficiency 70% of the total coal consumption is 135kg standard coal / ton of steam, almost 1 ton coal can produce about 7 tons of …
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### energy required to produce 1 ton of steam – oil fired
the cost of generation and the cost of consumption. If the plant has only one steam this much steam, of generating steam from the boiler(s): 1. how much steam produce 1 tonn water boiler; create 1 ton of steam. … How much natural gas is Boiler Fuel Consumption. We have one 1 Ton capacity
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### Calculate How To
Jul 14, 2020 · Calculation formula of oil consumption of diesel-fired steam boiler: fuel oil consumption of fuel-fired steam boiler per hour = 3600*heat power/heat value of diesel/boiler efficiency. If it is burning diesel, the fuel consumption per hour is about 132kg. A 1 ton diesel steam boiler produces 1 ton of steam per hour, so the fuel consumption of a 1 ton diesel boiler is about 70 liters.
Get a Quote | 738 | 2,763 | {"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-2021-31 | latest | en | 0.901635 |
https://scicomp.stackexchange.com/questions/2482/intersection-of-hyperplanes?noredirect=1 | 1,713,679,446,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00281.warc.gz | 470,909,186 | 39,566 | Intersection of hyperplanes
A very basic question but i couldn't find another post about it:
Given $p$ non parallel hyper-plane in $\mathbb{R}^p$:
$\left(\begin{array}{cccc} c_{11} & a_{11} & .... & a_{1p} \\ ... & ... & ... & .... \\ c_{p1} & a_{p1} & .... & a_{pp} \end{array} \right)$
$||a_{i.}||=1$
What is the best (from a numerical standpoint of view) way to get the $p$ vector of coordinates $x$ of their intersection?
If the matrix A is non-singular, the intersection of the hyperplanes is simply the the solution of the linear system of equations $Ax=b$, where
$A= \left(\begin{array}{cccc} a_{11} & .... & a_{1p} \\ ... & ... & .... \\ a_{p1} & .... & a_{pp} \end{array} \right)$
$b= -\left(\begin{array}{cccc} c_{11} \\ ... \\ c_{p1} \end{array} \right)$ | 257 | 771 | {"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} | 3.34375 | 3 | CC-MAIN-2024-18 | latest | en | 0.681003 |
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## Re: DAE daspk solver problem
From: Lazy_Tom Subject: Re: DAE daspk solver problem Date: Sat, 29 Oct 2016 10:57:50 -0700 (PDT)
```Sebastian Schöps wrote
>
> Lazy_Tom wrote
>> the output was :
>>
>> warning: Option "RelTol" not set, new value 1.0e-006 is used
>> warning: called from
>> radau5_test_1_V04 at line 19 column 6
>> warning: Option "AbsTol" not set, new value 1.0e-006 is used
>> warning: Option "NormControl" will be ignored by this solver
>> warning: Option "InitialStep" not set, new value 1.0e-006 is used
>> warning: Option "MaxStep" not set, new value 1.4e+000 is used
>> warning: Option "Mass" only supports constant mass matrices M() and not
>> M(t,y)
> Did you see the last line? The solver does not like function handles for
> the mass matrix. However, you can just use the matrix itself:
>
> vmass = diag([1, 1, 1, 1, 0]);
> vopt = odeset ('Mass', vmass, 'NormControl', 'on');
>
> I will try to look to your example in more detail next week.
>
> Sebastian
The reasono why I didn't use the matrix and opted for the ready-made example
is that the matrix didnt work.
In any case, I send you the code and the error, hope it helps
function [vyd] = frobertson (vt, vy, varargin)
vyd(1,1) = vy(2);
vyd(2,1) = -vy(1) * vy(5);
vyd(3,1) = vy(4);
vyd(4,1) = -vy(3) * vy(5) - 10;
vyd(5,1) = vy(1)^2 + vy(3)^2 - 1;
endfunction
vmass = diag([1, 1, 1, 1, 0]);
vopt = odeset ('Mass', vmass, 'NormControl', 'on');
vsol = ode5r (@frobertson, [0, 17], [1, 0, 0, 0, 0], vopt);
plot (vsol.x, vsol.y);
output:
warning: Option "RelTol" not set, new value 1.0e-006 is used
warning: called from
radau5_test_1_V05 at line 18 column 6
warning: Option "AbsTol" not set, new value 1.0e-006 is used
warning: Option "NormControl" will be ignored by this solver
warning: Option "InitialStep" not set, new value 1.0e-006 is used
warning: Option "MaxStep" not set, new value 1.4e+000 is used
warning: Option "NewtonTol" not set, default value is used
warning: Option "MaxNewtonIterations" not set, default value 7 is used
EXIT OF RADAU5 AT X= 0.2428E-02
STEP SIZE T0O SMALL, H= 2.1389523049476520E-018
error: missing implementation
error: called from radau5_test_1_V05 at line 18 column 6
Marco
--
View this message in context:
http://octave.1599824.n4.nabble.com/DAE-daspk-solver-problem-tp4680377p4680421.html
Sent from the Octave - General mailing list archive at Nabble.com.
``` | 857 | 2,434 | {"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-2019-18 | latest | en | 0.659708 |
http://www.mathworks.com/help/ident/ref/damp.html?requestedDomain=www.mathworks.com&nocookie=true | 1,513,095,605,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948517350.12/warc/CC-MAIN-20171212153808-20171212173808-00518.warc.gz | 417,365,198 | 15,389 | # Documentation
### This is machine translation
Translated by
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Note: This page has been translated by MathWorks. Please click here
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# damp
Natural frequency and damping ratio
## Syntax
```damp(sys) [Wn,zeta] = damp(sys) [Wn,zeta,P] = damp(sys) ```
## Description
`damp(sys)` displays a table of the damping ratio (also called damping factor), natural frequency, and time constant of the poles of the linear model `sys`. For a discrete-time model, the table also includes the magnitude of each pole. Frequencies are expressed in units of the reciprocal of the `TimeUnit` property of `sys`. Time constants are expressed in the same units as the `TimeUnit` property of `sys`.
This command requires a Control System Toolbox™ license.
```[Wn,zeta] = damp(sys)``` returns the natural frequencies, `Wn`, and damping ratios,`zeta`, of the poles of `sys`.
```[Wn,zeta,P] = damp(sys)``` returns the poles of `sys`.
## Input Arguments
`sys` Any linear dynamic system model.
## Output Arguments
`Wn` Vector containing the natural frequencies of each pole of `sys`, in order of increasing frequency. Frequencies are expressed in units of the reciprocal of the `TimeUnit` property of `sys`. If `sys` is a discrete-time model with specified sample time, `Wn` contains the natural frequencies of the equivalent continuous-time poles (see Algorithms). If `sys` has an unspecified sample time `(Ts = -1)`, then the software uses ```Ts = 1``` and calculates `Wn` accordingly. `zeta` Vector containing the damping ratios of each pole of `sys`, in the same order as `Wn`. If `sys` is a discrete-time model with specified sample time, `zeta` contains the damping ratios of the equivalent continuous-time poles (see Algorithms). If `sys` has an unspecified sample time `(Ts = -1)`, then the software uses ```Ts = 1``` and calculates `zeta` accordingly. `P` Vector containing the poles of `sys`, in order of increasing natural frequency. `P` is the same as the output of `pole(sys)`, except for the order.
## Examples
collapse all
Create the following continuous-time transfer function:
`H = tf([2 5 1],[1 2 3]);`
Display the natural frequencies, damping ratios, time constants, and poles of H.
`damp(H)`
``` Pole Damping Frequency Time Constant (rad/seconds) (seconds) -1.00e+00 + 1.41e+00i 5.77e-01 1.73e+00 1.00e+00 -1.00e+00 - 1.41e+00i 5.77e-01 1.73e+00 1.00e+00 ```
Obtain vectors containing the natural frequencies and damping ratios of the poles.
`[Wn,zeta] = damp(H);`
Calculate the associated time constants.
`tau = 1./(zeta.*Wn);`
Create a discrete-time transfer function.
`H = tf([5 3 1],[1 6 4 4],0.01);`
Display information about the poles of H.
`damp(H)`
``` Pole Magnitude Damping Frequency Time Constant (rad/seconds) (seconds) -3.02e-01 + 8.06e-01i 8.61e-01 7.74e-02 1.93e+02 6.68e-02 -3.02e-01 - 8.06e-01i 8.61e-01 7.74e-02 1.93e+02 6.68e-02 -5.40e+00 5.40e+00 -4.73e-01 3.57e+02 -5.93e-03 ```
The `Magnitude` column displays the discrete-time pole magnitudes. The `Damping`, `Frequency`, and `Time Constant` columns display values calculated using the equivalent continuous-time poles.
Obtain vectors containing the natural frequencies and damping ratios of the poles.
`[Wn,zeta] = damp(H);`
Calculate the associated time constants.
`tau = 1./(zeta.*Wn);`
## Algorithms
The natural frequency, time constant, and damping ratio of the system poles are defined in the following table:
Continuous TimeDiscrete Time with Sample Time Ts
Pole Location
`$s$`
`$z$`
Equivalent Continuous-Time Pole
`$s=\frac{ln\left(z\right)}{{T}_{s}}$`
Natural Frequency
`${\omega }_{n}=|s|$`
`${\omega }_{n}=|s|=|\frac{ln\left(z\right)}{{T}_{s}}|$`
Damping Ratio
`$\zeta =-cos\left(\angle s\right)$`
`$\begin{array}{lll}\zeta \hfill & =-cos\left(\angle s\right)\hfill & =-cos\left(\angle ln\left(z\right)\right)\hfill \end{array}$`
Time Constant
`$\tau =\frac{1}{{\omega }_{n}\zeta }$`
`$\tau =\frac{1}{{\omega }_{n}\zeta }$`
## See Also
#### Introduced before R2006a
Was this topic helpful?
Get trial now | 1,234 | 4,254 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "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} | 2.765625 | 3 | CC-MAIN-2017-51 | latest | en | 0.75003 |
https://math.stackexchange.com/questions/2167197/suppose-gch-then-aleph-function-and-beth-function-are-isomorphic | 1,696,279,292,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00537.warc.gz | 413,902,242 | 34,713 | # Suppose GCH, then Aleph function and Beth function are isomorphic?
GHC claims that:
2^($\aleph_0$) = $\aleph_1 \$
2^($\aleph_1$) = $\aleph_2 \$ and so on.
Since the aleph function defined as:
$\aleph_0$ = $\omega$
2^($\aleph_\alpha$) = $\aleph_\alpha^+$ and so on.
Since the beth function defined as:
$\beth_0$=$\aleph_0$
2^($\beth_0$) = $\beth_1 \$
2^($\beth_1$) = $\beth_2 \$ and so on.
If we assume GCH then aleph numbers and beth numbers would not be same?
• Yes. ${{{{}}}}$ Mar 1, 2017 at 17:17
## 1 Answer
You have a mistake in the definition of the $\aleph$ cardinals. $2^{\aleph_\alpha}$ is not $\aleph_\alpha^+$, at least not in general. This is $\sf GCH$, and not the definition of the $\aleph$ numbers.
The $\aleph$ numbers are defined as follows: - $\aleph_0=\omega$; - $\aleph_{\alpha+1}$ is the least ordinal whose cardinality is larger than $\aleph_\alpha$ (such ordinal exists due to Hartogs' theorem); and - If $\alpha$ is a limit ordinal, then $\aleph_\alpha=\sup\{\aleph_\beta\mid\beta<\alpha\}$.
The $\beth$ numbers, on the other hand, are defined using power sets for successor steps.
Since $\sf GCH$ is often formulated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$, we can now prove by transfinite induction that $\aleph_\alpha=\beth_\alpha$ for all $\alpha$.
• You have of course right. In the script of my lecture there are several definitions and theorems before the aleph sequences. I did not state them all. On the 40th page of the script (math.uni-bonn.de/ag/logik/teaching/2016WS/Aktuelles_Skript.pdf) you can see that it is stated similar to your statement with the previous Definition. (I mean the Def. 105 and then Def. 107) I see if we assume GCH, then the answer is obvious but I wanted to be sure that I do not skip any important step. Mar 2, 2017 at 11:11 | 567 | 1,797 | {"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} | 3.40625 | 3 | CC-MAIN-2023-40 | latest | en | 0.843494 |
https://www.cfd-online.com/Forums/fluent/29205-hvn-tell-me-detail-relative-pressure.html | 1,512,999,014,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513512.31/warc/CC-MAIN-20171211125234-20171211145234-00160.warc.gz | 737,637,770 | 14,885 | # HVN tell me the detail of relative pressure
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December 31, 2001, 03:29 HVN tell me the detail of relative pressure #1 chong chee nan Guest Posts: n/a please tell me how to get relative pressure, i mean the solution or the formula calculation. and the outlet pressure shown in a straight line that mean all the point and the outlet have same value of pressure. can u give me your mail address.
January 2, 2002, 05:55 Re: HVN tell me the detail of relative pressure #2 Jin-Wook LEE Guest Posts: n/a For incompressible flow simulation, the relative pressure can be considered as the difference from the given(or assumed) value at the given point. Given point might be arbitrary. That is, if you know the pressure at the outlet, you can add the pressure difference between the outlet(where you know the pressure) and the each point, and you can obtain the real pressure. Example : - You know outlet pressure, say 101325 Pa - Outlet pressure of CFD result, say -100Pa - CFD result for the pressure at the certain point, say -50 Pa. Don't stick at the value of -100 or -50, You can interprete the result as, presuure at the outlet is 101325Pa(which you have information) and the pressure at the certain point is 101325-(-50)=101375 Pa. Sincerely, Jinwook
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7.5kviews
The crank pin circle radius of a horizontal engine is $300 mm.$ the mass of the reciprocating parts is $250 kg.$ When the crank has traveled $60^\circ$ from IDC,
2
741views
The difference between the driving and back pressure is $0.35 N/mm^2$. The connecting rod length between centers is $1.2 m$ and the cylinder bore is $0.5 m.$ If the engine runs at $250 rpm$ and if the effect of piston rod diameter is neglected, Calculate:
1] Pressure on slide bars.
2] Thrust in the connecting rod.
3] Tangential force on the crank pin.
4] Turning moment on the crankshaft.
Given: $r = 300mm$
$M_R = 250 kg$
$\theta = 60°$
$\triangle P \ = \ P \ = \ 0.35 \ N/mm^2$
$L = 1.2m$
$D = 0.5m$
$N = 250\ rpm$
$\therefore \omega = \frac{2 \pi (250)}{60} = 26.18 \ rad/s.$
$\rightarrow$ $n = \frac{L}{r} = \frac{1.2}{0.3} = 4$
$\rightarrow$ Net load on piston = $F_L = 0.35 \times \frac{\pi }{4} (500)^2$
$F_L = 68.72 \times 10^3 N$
$\rightarrow$ Inertia force = $F_I = M_R. \omega^2. r. [cos \ \theta + \frac{cos \ 2 \theta}{n}]$
$= (250) (26.18)^2 (0.3) [ cos \ 60 + \frac{cos \ (2 \times 60)}{4}]$
$F_I = 19.28 \times 10^3 \ N$
$\rightarrow$ Piston Effort = $F_P \ = \ F_L \ – \ F_I$
$=(68.72 – 19.28) \times 10^3$
$F_P = 49.44 \times 10^3 N$
$\rightarrow$ $sin \ \phi = \frac{sin \ 60}{4} = 0.2165$
$\phi \ = \ 12.50°$
$\rightarrow$ Pressure on slide bars:
$F_N \ = \ F_Q . \ sin \ \phi$
$= (\frac{F_p}{cos \ \phi}) sin \ \phi$
$= f_p \ tan \ \phi$
$= (49.44 \times 10^3) tan \ (12.50)$
$F_N = 10.96 \times 10^3 \ N$
$\rightarrow$ Thrust in the connecting rod:
$F_Q = \frac{F_p}{cos \ \phi} = \frac{49.44 \times 10^3}{cos \ (12.5)}$
$F_Q = 50.64 \times 10^3 \ N$
$\rightarrow$ Tangential force on crank pin:
$F_T = F_Q . sin \ (\theta \ + \ \phi)$
$= (50.64 \times 10^3) sin \ (60 + 12.5)$
$F_T = 48.29 \times 10^3 \ N$
$\rightarrow$ Turning moment on crankshaft:
$T \ = \ F_T . \ r$
$= (48.29 \times 10^3) \ (0.3)$
$= 14.48 \times 10^3 \ N-m$ | 792 | 1,981 | {"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} | 4.15625 | 4 | CC-MAIN-2024-33 | latest | en | 0.56918 |
http://www.haskell.org/pipermail/haskell-cafe/2009-November/068850.html | 1,411,223,466,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657133417.25/warc/CC-MAIN-20140914011213-00237-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 568,016,794 | 2,167 | # [Haskell-cafe] Area from [(x,y)] using foldl
michael rice nowgate at yahoo.com
Sun Nov 8 16:30:18 EST 2009
```That's certainly better than mine, but I'm lost again, with the following. What seemed like a simple improvement doesn't compile.
Michael
===============
This works.
area :: [(Double,Double)] -> Double
area ps = abs \$ (/2) \$ area' (last ps) ps
where area' _ [] = 0
area' (x0,y0) ((x,y):ps) = (x0-x)*(y0+y) + area' (x,y) ps
*Main> let p = [(0.0,0.0),(1.0,0.0),(1.0,1.0),(0.0,1.0),(0.0,0.0)]
*Main> area (last p) p
1.0
*Main>
===============
This doesn't.
area :: [(Double,Double)] -> Double
area p = abs \$ (/2) \$ area' (last p):p
where area' [] = 0
area' ((x0,y0),(x,y):ps) = ((x0-x)*(y0+y)) + area' (x,y):ps
Subject: Re: [Haskell-cafe] Area from [(x,y)] using foldl
To: "michael rice" <nowgate at yahoo.com>
Date: Sunday, November 8, 2009, 3:52 PM
On Sun, Nov 8, 2009 at 9:04 PM, michael rice <nowgate at yahoo.com> wrote:
Of course! Back to the drawing board.
If I understand the problem correctly, I'm not convinced that foldl is the right approach (nevermind that foldl is almost never what you want, foldl' and foldr being the correct choice almost always). My proposition would be the following :
> area ps = abs . (/2) . sum \$ zipWith (\(x,y) (x',y') -> (x - x') * (y + y')) ps (tail \$ cycle ps)
I think it express the algorithm more clearly.
--
Jedaï
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https://number.academy/621793 | 1,680,127,356,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00372.warc.gz | 478,238,280 | 12,448 | # Number 621793
Number 621,793 spell 🔊, write in words: six hundred and twenty-one thousand, seven hundred and ninety-three . Ordinal number 621793rd is said 🔊 and write: six hundred and twenty-one thousand, seven hundred and ninety-third. Color #621793. The meaning of the number 621793 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 621793. What is 621793 in computer science, numerology, codes and images, writing and naming in other languages
## What is 621,793 in other units
The decimal (Arabic) number 621793 converted to a Roman number is (D)(C)(X)(X)MDCCXCIII. Roman and decimal number conversions.
#### Weight conversion
621793 kilograms (kg) = 1370804.8 pounds (lbs)
621793 pounds (lbs) = 282043.5 kilograms (kg)
#### Length conversion
621793 kilometers (km) equals to 386365 miles (mi).
621793 miles (mi) equals to 1000680 kilometers (km).
621793 meters (m) equals to 2039979 feet (ft).
621793 feet (ft) equals 189525 meters (m).
621793 centimeters (cm) equals to 244800.4 inches (in).
621793 inches (in) equals to 1579354.2 centimeters (cm).
#### Temperature conversion
621793° Fahrenheit (°F) equals to 345422.8° Celsius (°C)
621793° Celsius (°C) equals to 1119259.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
621793 seconds equals to 1 week, 4 hours, 43 minutes, 13 seconds
621793 minutes equals to 1 year, 3 months, 1 week, 4 days, 19 hours, 13 minutes
### Codes and images of the number 621793
Number 621793 morse code: -.... ..--- .---- --... ----. ...--
Sign language for number 621793:
Number 621793 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 621793
### Multiplications
#### Multiplication table of 621793
621793 multiplied by two equals 1243586 (621793 x 2 = 1243586).
621793 multiplied by three equals 1865379 (621793 x 3 = 1865379).
621793 multiplied by four equals 2487172 (621793 x 4 = 2487172).
621793 multiplied by five equals 3108965 (621793 x 5 = 3108965).
621793 multiplied by six equals 3730758 (621793 x 6 = 3730758).
621793 multiplied by seven equals 4352551 (621793 x 7 = 4352551).
621793 multiplied by eight equals 4974344 (621793 x 8 = 4974344).
621793 multiplied by nine equals 5596137 (621793 x 9 = 5596137).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 621793
Half of 621793 is 310896,5 (621793 / 2 = 310896,5 = 310896 1/2).
One third of 621793 is 207264,3333 (621793 / 3 = 207264,3333 = 207264 1/3).
One quarter of 621793 is 155448,25 (621793 / 4 = 155448,25 = 155448 1/4).
One fifth of 621793 is 124358,6 (621793 / 5 = 124358,6 = 124358 3/5).
One sixth of 621793 is 103632,1667 (621793 / 6 = 103632,1667 = 103632 1/6).
One seventh of 621793 is 88827,5714 (621793 / 7 = 88827,5714 = 88827 4/7).
One eighth of 621793 is 77724,125 (621793 / 8 = 77724,125 = 77724 1/8).
One ninth of 621793 is 69088,1111 (621793 / 9 = 69088,1111 = 69088 1/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
621793
#### Is Prime?
The number 621793 is not a prime number. The closest prime numbers are 621779, 621799.
#### Factorization and factors (dividers)
The prime factors of 621793 are 709 * 877
The factors of 621793 are 1 , 709 , 877 , 621793
Total factors 4.
Sum of factors 623380 (1587).
#### Powers
The second power of 6217932 is 386.626.534.849.
The third power of 6217933 is 240.401.672.983.364.256.
#### Roots
The square root √621793 is 788,538522.
The cube root of 3621793 is 85,352309.
#### Logarithms
The natural logarithm of No. ln 621793 = loge 621793 = 13,340363.
The logarithm to base 10 of No. log10 621793 = 5,793646.
The Napierian logarithm of No. log1/e 621793 = -13,340363.
### Trigonometric functions
The cosine of 621793 is -0,903566.
The sine of 621793 is 0,42845.
The tangent of 621793 is -0,474177.
### Properties of the number 621793
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 621793 in Computer Science
Code typeCode value
PIN 621793 It's recommended that you use 621793 as your password or PIN.
621793 Number of bytes607.2KB
CSS Color
#621793 hexadecimal to red, green and blue (RGB) (98, 23, 147)
Unix timeUnix time 621793 is equal to Thursday Jan. 8, 1970, 4:43:13 a.m. GMT
IPv4, IPv6Number 621793 internet address in dotted format v4 0.9.124.225, v6 ::9:7ce1
621793 Decimal = 10010111110011100001 Binary
621793 Decimal = 1011120221101 Ternary
621793 Decimal = 2276341 Octal
621793 Decimal = 97CE1 Hexadecimal (0x97ce1 hex)
621793 BASE64NjIxNzkz
621793 MD5ac933df4719357c74689779e35fe052f
621793 SHA1068269ac816266246a3fd0440ba9e36a5a6ac822
621793 SHA2245698685e713e4c338c65383c3d11e1a6baf144d8b19164c16db5e1e8
621793 SHA38469a854d7f43860007b167fa26b0127fa6e2ed563247e6c2ebf23d7419f37cc4e7d1f18529611118691d68e5f095841d3
More SHA codes related to the number 621793 ...
If you know something interesting about the 621793 number that you did not find on this page, do not hesitate to write us here.
## Numerology 621793
### Character frequency in the number 621793
Character (importance) frequency for numerology.
Character: Frequency: 6 1 2 1 1 1 7 1 9 1 3 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 621793, the numbers 6+2+1+7+9+3 = 2+8 = 1+0 = 1 are added and the meaning of the number 1 is sought.
## № 621,793 in other languages
How to say or write the number six hundred and twenty-one thousand, seven hundred and ninety-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 621.793) seiscientos veintiuno mil setecientos noventa y tres German: 🔊 (Nummer 621.793) sechshunderteinundzwanzigtausendsiebenhundertdreiundneunzig French: 🔊 (nombre 621 793) six cent vingt et un mille sept cent quatre-vingt-treize Portuguese: 🔊 (número 621 793) seiscentos e vinte e um mil, setecentos e noventa e três Hindi: 🔊 (संख्या 621 793) छः लाख, इक्कीस हज़ार, सात सौ, तिरानवे Chinese: 🔊 (数 621 793) 六十二万一千七百九十三 Arabian: 🔊 (عدد 621,793) ستمائة و واحد و عشرون ألفاً و سبعمائةثلاثة و تسعون Czech: 🔊 (číslo 621 793) šestset dvacet jedna tisíc sedmset devadesát tři Korean: 🔊 (번호 621,793) 육십이만 천칠백구십삼 Danish: 🔊 (nummer 621 793) sekshundrede og enogtyvetusindsyvhundrede og treoghalvfems Dutch: 🔊 (nummer 621 793) zeshonderdeenentwintigduizendzevenhonderddrieënnegentig Japanese: 🔊 (数 621,793) 六十二万千七百九十三 Indonesian: 🔊 (jumlah 621.793) enam ratus dua puluh satu ribu tujuh ratus sembilan puluh tiga Italian: 🔊 (numero 621 793) seicentoventiunomilasettecentonovantatré Norwegian: 🔊 (nummer 621 793) seks hundre og tjue-en tusen, syv hundre og nitti-tre Polish: 🔊 (liczba 621 793) sześćset dwadzieścia jeden tysięcy siedemset dziewięćdzisiąt trzy Russian: 🔊 (номер 621 793) шестьсот двадцать одна тысяча семьсот девяносто три Turkish: 🔊 (numara 621,793) altıyüzyirmibinyediyüzdoksanüç Thai: 🔊 (จำนวน 621 793) หกแสนสองหมื่นหนึ่งพันเจ็ดร้อยเก้าสิบสาม Ukrainian: 🔊 (номер 621 793) шiстсот двадцять одна тисяча сiмсот дев'яносто три Vietnamese: 🔊 (con số 621.793) sáu trăm hai mươi mốt nghìn bảy trăm chín mươi ba Other languages ...
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## Comment
If you know something interesting about the number 621793 or any other natural number (positive integer), please write to us here or on Facebook. | 2,622 | 7,601 | {"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-2023-14 | latest | en | 0.67846 |
https://discuss.boardinfinity.com/t/bfgs-optimization-algorithm/5862 | 1,721,320,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00357.warc.gz | 178,855,920 | 6,725 | # BFGS Optimization Algorithm
BFGS is a second-order optimization algorithm.
It is an acronym, named for the four co-discovers of the algorithm: Broyden, Fletcher, Goldfarb, and Shanno.
It is a local search algorithm, intended for convex optimization problems with a single optima.
The BFGS algorithm is perhaps best understood as belonging to a group of algorithms that are an extension to Newton’s Method optimization algorithm, referred to as Quasi-Newton Methods.
Newton’s method is a second-order optimization algorithm that makes use of the Hessian matrix.
A limitation of Newton’s method is that it requires the calculation of the inverse of the Hessian matrix. This is a computationally expensive operation and may not be stable depending on the properties of the objective function.
Quasi-Newton methods are second-order optimization algorithms that approximate the inverse of the Hessian matrix using the gradient, meaning that the Hessian and its inverse do not need to be available or calculated precisely for each step of the algorithm. The main difference between different Quasi-Newton optimization algorithms is the specific way in which the approximation of the inverse Hessian is calculated.
The BFGS algorithm is one specific way for updating the calculation of the inverse Hessian, instead of recalculating it every iteration. It, or its extensions, may be one of the most popular Quasi-Newton or even second-order optimization algorithms used for numerical optimization. A benefit of using the Hessian, when available, is that it can be used to determine both the direction and the step size to move in order to change the input parameters to minimize (or maximize) the objective function.
Quasi-Newton methods like BFGS approximate the inverse Hessian, which can then be used to determine the direction to move, but we no longer have the step size.
The BFGS algorithm addresses this by using a line search in the chosen direction to determine how far to move in that direction.
For the derivation and calculations used by the BFGS algorithm, I recommend the resources in the further reading section at the end of this tutorial.
The size of the Hessian and its inverse is proportional to the number of input parameters to the objective function. As such, the size of the matrix can become very large for hundreds, thousand, or millions of parameters.
Limited memory BFGS (or L-BFGS) is an extension to the BFGS algorithm that addresses the cost of having a large number of parameters. It does this by not requiring that the entire approximation of the inverse matrix be stored, by assuming a simplification of the inverse Hessian in the previous iteration of the algorithm (used in the approximation). | 545 | 2,735 | {"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} | 3.1875 | 3 | CC-MAIN-2024-30 | latest | en | 0.905138 |
https://couponsanddiscouts.com/present-value-discount-rate-2019/ | 1,656,884,120,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00690.warc.gz | 233,758,285 | 8,305 | # Present Value Discount Rate 2019
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Filter Time:
### Discount Rate Formula: Calculating Discount Rate …
Aug 16, 2019 · As stated above, net present value (NPV) and discounted cash flow (DCF) are methods of valuation used to assess the quality of an investment opportunity, and both of them use discount rate as a key element. Net Present …
https://www.profitwell.com/recur/all/discount-rate-formula
### Appendix C | Present Discounted Value – UH Microeconomics 2019
Table C2 Computing the Present Discounted Value of a Bond; Stream of Payments (for the 8% interest rate) Present Value (for the 8% interest rate) Stream of Payments (for the 11% interest rate) Present Value (for the 11% interest rate) \$240 payment after one year. \$240/(1 + 0.08)1 = \$222.20. \$240 payment after one year. \$240/(1 + 0.11)1 = \$216.20
http://pressbooks.oer.hawaii.edu/microeconomics2019/chapter/appendix-c-present-discounted-value/
### Federal Reserve Board - The Discount Window and …
The discount rate is the interest rate charged to commercial banks and other depository institutions on loans they receive from their regional Federal Reserve Bank's lending facility—the discount window. The Federal Reserve Banks offer three types of credit to depository institutions: primary credit, secondary credit, and seasonal credit, each with ...
https://www.federalreserve.gov/monetarypolicy/discountrate.htm
### Understanding Discount Rate, Present Value and Net Present Value
Multiplication is a form of repeated addition. In this next equation to solve for the present value, he’ll divide to subtract the discount rate. Division similarly is a form of repeated subtraction. Here’s the one-year formula: (Future Value) divided by (1+the discount rate) \$110 / (1 + .05) \$110 / 1.05 = \$104.76 (the present value)
https://costarmanager.com/the-internets-most-simple-explanations-of-discount-rate-present-value-and-net-present-value/
### Appendix C: Present Discounted Value – Principles of …
Stream of Payments (for the 8% interest rate) Present Value (for the 8% interest rate) Stream of Payments (for the 11% interest rate) Present Value (for the 11% interest rate) \$240 payment after one year: \$240/(1 + 0.08) 1 = \$222.20: \$240 payment after one year: \$240/(1 + 0.11) 1 = \$216.20: \$3,240 payment after second year: 2: \$3,240 payment after second year: 2: Total
https://opentextbc.ca/principlesofeconomics/back-matter/appendix-c-present-discounted-value/
### Federal Discount Rate | Federal Reserve Rates - Bankrate
Jun 28, 2022 · Prime rate, federal funds rate, COFI Updated: 06/28/2022; This week Month ago Year ago; Federal Discount Rate: 1.75: 1.00: 0.25
https://www.bankrate.com/rates/interest-rates/federal-discount-rate/
### US Discount Rate - YCharts
Jun 24, 2022 · Basic Info. US Discount Rate is at 1.75%, compared to 1.75% the previous market day and 0.25% last year. This is lower than the long term average of 1.89%. Report.
https://ycharts.com/indicators/us_discount_rate
### Present value of 1 table — AccountingTools
Feb 14, 2022 · A present value of 1 table states the present value discount rates that are used for various combinations of interest rates and time periods. A discount rate selected from this table is then multiplied by a cash sum to be received at a future date, to arrive at its present value. The interest rate selected in the table can be based on the current amount the investor is obtaining …
https://www.accountingtools.com/articles/present-value-of-1-table
Feb 06, 2022 · In order to calculate the net present value of the investment, an analyst uses a 5% hurdle rate and calculates a value of \$578.64. This compares to a non-discounted total cash flow of \$700. Essentially, an investor is saying …
https://corporatefinanceinstitute.com/resources/knowledge/finance/discount-rate/
### Discount Rates For Social Security Or Pension Decisions
Jul 19, 2017 · Which means the cash flows are \$40,000/year for the next 4 years, with a total final payment of \$1,140,000 at the end of the 5 th year (which represents the last profit distribution, plus the \$1.1M terminal value). At a 5% discount rate, the present value of these future cash flows is \$1,035,058, which makes the stock an appealing investment at “just” a \$1,000,000 price.
https://www.kitces.com/blog/net-present-value-discount-rate-formula-retirement-plan-pension-lump-sum-or-social-security-breakeven/
### How to calculate Discount Rate with Examples - EDUCBA
Discount Rate is calculated using the formula given below. Discount Rate = T * [ (Future Cash Flow / Present Value) 1/t*n – 1] Discount Rate = 2 * [ (\$10,000 / \$7,600) 1/2*4 – 1] Discount Rate = 6.98%. Therefore, the effective discount rate for David in this case is 6.98%.
https://www.educba.com/discount-rate-formula/
### Present Value Calculator
Present Value. Present Value, or PV, is defined as the value in the present of a sum of money, in contrast to a different value it will have in the future due to it being invested and compound at a certain rate. Net Present Value. A popular concept in finance is the idea of net present value, more commonly known as NPV.
https://www.calculator.net/present-value-calculator.html
### Discount Rate Impact Dec 2019 Actuarial Valuations | Kapadia Global
Jan 10, 2020 · This means that, if there was a liability of Rs. 100 as at 31st Mar 2019, and the discount rate falls from 7.35% to 6.95% in Dec 2019, the liability will increase by approximately 4.00%, all other things being same.
### What Is the Discount Rate? - The Motley Fool
Jun 30, 2022 · While originally considered "transitory," the Fed changed its tune in late 2021 and began taking action to curb inflation in early 2022. The Fed raised the discount rate to 0.5% in early 2022, and ...
https://www.fool.com/investing/how-to-invest/stocks/discount-rate/
### Present Value of Future Savings Calculator - Pigly!
Nov 13, 2020 · Because each of the increments are a few years apart, we need to adjust the formula. There are two values to consider in this situation. To find the present value of the total, we must identify the present values of both increments. We then add these to our starting present value. Partial Present Value = \$250 Discount rate = 2% annual; 0.17% monthly
https://pigly.com/savings/present-value.php | 1,613 | 6,346 | {"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} | 3.703125 | 4 | CC-MAIN-2022-27 | latest | en | 0.865913 |
https://secondsminutes.com/374-6-minutes-in-seconds | 1,675,864,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00069.warc.gz | 522,182,037 | 7,060 | # 374.6 minutes in seconds
## Result
374.6 minutes equals 22476 seconds
You can also convert 374.6 minutes to minutes and seconds.
## Conversion formula
Multiply the amount of minutes by the conversion factor to get the result in seconds:
374.6 min × 60 = 22476 s
## How to convert 374.6 minutes to seconds?
The conversion factor from minutes to seconds is 60, which means that 1 minutes is equal to 60 seconds:
1 min = 60 s
To convert 374.6 minutes into seconds we have to multiply 374.6 by the conversion factor in order to get the amount from minutes to seconds. We can also form a proportion to calculate the result:
1 min → 60 s
374.6 min → T(s)
Solve the above proportion to obtain the time T in seconds:
T(s) = 374.6 min × 60 s
T(s) = 22476 s
The final result is:
374.6 min → 22476 s
We conclude that 374.6 minutes is equivalent to 22476 seconds:
374.6 minutes = 22476 seconds
## Result approximation:
For practical purposes we can round our final result to an approximate numerical value. In this case three hundred seventy-four point six minutes is approximately twenty-two thousand four hundred seventy-six seconds:
374.6 minutes ≅ 22476 seconds
## Conversion table
For quick reference purposes, below is the minutes to seconds conversion table:
minutes (min) seconds (s)
375.6 minutes 22536 seconds
376.6 minutes 22596 seconds
377.6 minutes 22656 seconds
378.6 minutes 22716 seconds
379.6 minutes 22776 seconds
380.6 minutes 22836 seconds
381.6 minutes 22896 seconds
382.6 minutes 22956 seconds
383.6 minutes 23016 seconds
384.6 minutes 23076 seconds
## Units definitions
The units involved in this conversion are minutes and seconds. This is how they are defined:
### Minutes
The minute is a unit of time or of angle. As a unit of time, the minute (symbol: min) is equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). As a unit of angle, the minute of arc is equal to 1⁄60 of a degree, or 60 seconds (of arc). Although not an SI unit for either time or angle, the minute is accepted for use with SI units for both. The SI symbols for minute or minutes are min for time measurement, and the prime symbol after a number, e.g. 5′, for angle measurement. The prime is also sometimes used informally to denote minutes of time. In contrast to the hour, the minute (and the second) does not have a clear historical background. What is traceable only is that it started being recorded in the Middle Ages due to the ability of construction of "precision" timepieces (mechanical and water clocks). However, no consistent records of the origin for the division as 1⁄60 part of the hour (and the second 1⁄60 of the minute) have ever been found, despite many speculations.
### Seconds
The second (symbol: s) (abbreviated s or sec) is the base unit of time in the International System of Units (SI). It is qualitatively defined as the second division of the hour by sixty, the first division by sixty being the minute. The SI definition of second is "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom". Seconds may be measured using a mechanical, electrical or an atomic clock. SI prefixes are combined with the word second to denote subdivisions of the second, e.g., the millisecond (one thousandth of a second), the microsecond (one millionth of a second), and the nanosecond (one billionth of a second). Though SI prefixes may also be used to form multiples of the second such as kilosecond (one thousand seconds), such units are rarely used in practice. The more common larger non-SI units of time are not formed by powers of ten; instead, the second is multiplied by 60 to form a minute, which is multiplied by 60 to form an hour, which is multiplied by 24 to form a day. The second is also the base unit of time in other systems of measurement: the centimetre–gram–second, metre–kilogram–second, metre–tonne–second, and foot–pound–second systems of units. | 1,048 | 4,292 | {"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} | 4.375 | 4 | CC-MAIN-2023-06 | latest | en | 0.827432 |
http://home.earthlink.net/~enigmaep/annihilation/enterprises2.html | 1,519,048,788,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812665.41/warc/CC-MAIN-20180219131951-20180219151951-00321.warc.gz | 166,990,658 | 4,688 | Purveyor of All Your Weapons of Mass Destruction Needs. Extensive resources and information concerning all aspects of Nuclear Weapons. Photo gallery, Links,...
Optimal: 1024x768, 16m colors, N4.x+
Purveyor of All Your Weapons of Mass Destruction Needs
Section II
HOW MANY ARE THERE?
A Lot More Than You May Think
A Matter of Scale To fully appreciate the dynamics of a nuclear explosion, one must understand the scale of the events involved, both incredibly large and infinitesimally small. These can be summed up in four key concepts: time, distance, temperature, and pressure. TIME A microsecond is one-millionth of one second A nanosecond is one-billionth of one second Both of these increments are exceedingly small, yet the difference between the two is amazingly huge. To illustrate, let us consider instead the difference between one million and one billion: one million seconds is about eleven and a half days one billion seconds is over thirty one and a half years Many complex nuclear reactions involving astronomical numbers of atomic nuclei occur within an exploding nuclear weapon in a few hundred microseconds. DISTANCE Although most fission and fusion stages in a nuclear warhead are located within inches of each other, this distance is enormous when compared to the spaces between particles in an atom: If a hydrogen atom (the simplest atom, with 1 proton in the nucleus orbited by 1 electron) were enlarged so that its nucleus were 1 inch in diameter, its electron would orbit at a distance of 1400 feet. Thus, atoms are mostly empty space. The apparent solidity of matter is an illusion, due in part to the enormous number in a small space. Atoms are so small, that if a drop of water were enlarged to the size of the earth, the individual atoms would be smaller than oranges. TEMPERATURE The temperatures inside a nuclear explosion are well over 100 million degrees. The energy at these temperatures is very powerful: it is sufficient to strip a hydrogen atom of its electron, give the nucleus enough velocity to overcome the repulsive force of another hydrogen nucleus, smash them together, forming a new (helium) nucleus and release energy. This type of collision can only occur at extremely high temperatures, and is called a thermonuclear reaction. PRESSURE Along with high temperatures, a thermonuclear reaction requires extremely high pressures. 1 bar equals 98.69% of 1 atmosphere, or 14.5 lbs. per square inch (psi) at sea level Typical daily fluctuations in air pressure are measured in millibars (1/1000 bar), as seen on weather maps. In an exploding thermonuclear weapon, pressures are as high as a trillion atmospheres--about 8,000,000,000 tons psi. The result of such enormous pressures on a substance is an increase in its density. The more closely compacted the atoms of a fissile or fusionable material, the more likely they will undergo a collision (reaction), and also a higher frequency of such collisions. Also, the denser the material, the better (and longer) it can withstand outwardly directed forces which seek to dissemble it, and end the nuclear reactions occuring within it.
HOW MANY IS ENOUGH...
...To End Civilization As We Know It?
A Lot Fewer Than You May Think...
HOW SAFE ARE YOU?
Find Out With the Nuclear Blast Mapper
The Hiroshima Panorama Project
The Aoi T-Bridge in Hiroshima, Japan, August 1945; Ground Zero of the first atomic bomb used in combat. The only buildings that remain are Western type concrete structures. Note the depression in the roof of the building in the foreground.
A History of the
US Nuclear Weapons Complex
Last Update: 29Dec02 | 777 | 3,646 | {"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-2018-09 | latest | en | 0.93931 |
http://www.talkstats.com/threads/lottery-question.67236/?p=194818 | 1,516,424,412,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889325.32/warc/CC-MAIN-20180120043530-20180120063530-00183.warc.gz | 582,687,671 | 8,476 | # lottery question
#### tdx
##### New Member
Connecticut has a lottery game called Cash5. There are 35 numbers to choose from.
You have to select 5 numbers.
5 numbers are then drawn from an airball machine.
What are the odds of the 5 numbers you have selected being the same as the 5 numbers drawn from the airball machine?
The lottery says the odds are 1 in 324,632 as shown in the link below on page 5
( hope the link works.....payoff is \$ 100,000 if you guess all 5 numbers ) | 124 | 485 | {"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-2018-05 | latest | en | 0.931591 |
http://azfoo.net/gdt/babs/numbers/n/number250.html | 1,568,717,710,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573070.36/warc/CC-MAIN-20190917101137-20190917123137-00222.warc.gz | 18,642,218 | 2,000 | About the Number 250 (two hundred fifty)
250 was the smallest number that didn't have a nBAB on 7 April 2016. MathBabbler is trying to ensure that the numbers 0-300 have nBABs by the end of the Spring 2016 Semester. With the posting of this nBAB for 250, 252 became the smallest number without a nBAB on 7 April 2016.
```MathBabbler Number Analyst (MBNA) output:
=========================================
250 is a natural, whole, integer
250 is even
250 proper divisors are: 1,2,5,10,25,50,125,
250 has 7 proper divisors
250 is deficient (sum of divisors is 218; ratio: 0.872)
250 is unhappy
250 is not a Squarefree Number: 25,
250 is composite (not prime)
250 has the prime factors: 2*5*5*5 (sum=17)
250 is a hoax number
250 is a 5-smooth number
250 in octal is 0372
250 in binary is 11111010 (is evil)
250 nearest square numbers: -25...6 (225...256 [16])
sqrt(250) = 15.8114
ln(250) = 5.52146
log(250) = 2.39794
250 reciprocal is .00400000000000000000000000000000
250! is 3.23286e+492
250 is 79.5775 Pi years
250 is 12 score and 10 years
250 written as a Roman numeral is CCL
250 is a multiple of 2 & it contains a 2 (A121022)
250 is a multiple of 5 & it contains a 5 (A121025)
250 is a multiple of 25 & it contains a 25
250 is a multiple of 50 & it contains a 50
```
Creator: Gerald Thurman [gthurman@gmail.com]
Created: 07 April 2016 | 436 | 1,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-39 | latest | en | 0.890106 |
http://www.numbersaplenty.com/211100 | 1,597,443,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740343.48/warc/CC-MAIN-20200814215931-20200815005931-00402.warc.gz | 160,662,352 | 3,498 | Search a number
211100 = 22522111
BaseRepresentation
bin110011100010011100
3101201120112
4303202130
523223400
64305152
71536311
oct634234
9351515
10211100
1113466a
12a21b8
1375116
1456d08
1542835
hex3389c
211100 has 18 divisors (see below), whose sum is σ = 458304. Its totient is φ = 84400.
The previous prime is 211097. The next prime is 211129. The reversal of 211100 is 1112.
Adding to 211100 its reverse (1112), we get a palindrome (212212).
It is a happy number.
211100 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a Harshad number since it is a multiple of its sum of digits (5).
It is a plaindrome in base 11 and base 16.
It is a nialpdrome in base 10.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 956 + ... + 1155.
2211100 is an apocalyptic number.
211100 is a gapful number since it is divisible by the number (20) formed by its first and last digit.
It is an amenable number.
211100 is an abundant number, since it is smaller than the sum of its proper divisors (247204).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
211100 is a wasteful number, since it uses less digits than its factorization.
211100 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 2125 (or 2118 counting only the distinct ones).
The product of its (nonzero) digits is 2, while the sum is 5.
The square root of 211100 is about 459.4562003064. The cubic root of 211100 is about 59.5428216094.
The spelling of 211100 in words is "two hundred eleven thousand, one hundred", and thus it is an iban number. | 517 | 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} | 3.375 | 3 | CC-MAIN-2020-34 | latest | en | 0.898857 |
https://www.slideserve.com/may/content-session-4 | 1,575,735,283,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540500637.40/warc/CC-MAIN-20191207160050-20191207184050-00328.warc.gz | 864,901,547 | 12,772 | Content Session 4
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# Content Session 4 - PowerPoint PPT Presentation
## Content Session 4
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##### Presentation Transcript
1. Content Session 4 July 7, 2009
2. Addition & Subtraction M3N5. Students will understand the meaning of decimal fractions and common fractions in simple cases and apply them in problem-solving situations. e. Understand the concept of addition and subtraction of decimal fractions and common fractions with like denominators. f. Model addition and subtraction of decimal fractions and common fractions with like denominators. g. Use mental math and estimation strategies to add and subtract decimal fractions and common fractions with like denominators.
3. Addition & Subtraction M4N5. Students will further develop their understanding of the meaning of decimals and use them in computations. • Add and subtract both one and two digit decimals. M4N6. Students will further develop their understanding of the meaning of decimal fractions and common fractions and use them in computations. b. Add and subtract fractions and mixed numbers with like denominators. (Denominators should not exceed twelve.)
4. Addition & Subtraction M5N4. Students will continue to develop their understanding of the meaning of common fractions and compute with them. g. Add and subtract common fractions and mixed numbers with unlike denominators.
5. Goals (Grades 3 & 4) • Grade 3 • The meaning of addition and subtraction remains the same even when numbers become decimal numbers or fractions • Decimal numbers and fractions are numbers, just like whole numbers • Grade 4 • Fluency with decimal addition and subtraction • The sum of two fractions may exceed 1 (improper fractions or mixed numbers), and the minuend may also exceed 1.
7. Key Ideas • Unitary perspective of numbers • 0.3 is 3 0.1-units; 3/5 is 3 1/5-units; etc. • Relative size of (decimal) numbers • 0.32 is 32 0.01-units • Addition/Subtraction can be performed only when the two numbers are referring to the same unit
8. How do these relate to 3 + 4? • 30 + 40 • 300 + 400 • 3000 + 4000 • etc. How about these? • 300 + 40 • 30 + 4000
9. If you put a tape that is 0.3 meters long and another tape that is 0.4 meters long together, end to end, how long will it be? • What math sentence will represent this problem? • What does 0.3 meters mean? • What does 0.4 meters mean? • How many 0.1 meter will there be altogether? • What is the answer?
10. If you put a tape that is 3/8 meters long and another tape that is 3/8 meters long together, end to end, how long will it be? • What math sentence will represent this problem? • What does 3/8 meters mean? • What does 4/8 meters mean? • How many 1/8 meter will there be altogether? • What is the answer?
11. If you put a tape that is 0.6 meters long and another tape that is 0.8 meters long together, end to end, how long will it be? What math sentence will represent this problem? What does 0.6 meters mean? What does 0.8 meters mean? How many 0.1 meter will there be altogether? What is the answer?
12. If you put a tape that is 0.07 meters long and another tape that is 0.05 meters long together, end to end, how long will it be? What math sentence will represent this problem? What does 0.07 meters mean? What does 0.05 meters mean? How many 0.01 meter will there be altogether? What is the answer?
13. If you put a tape that is 3.6 meters long and another tape that is 2.2 meters long together, end to end, how long will it be? What math sentence will represent this problem? What does 3.6 meters mean? What does 2.2 meters mean?
14. If you put a tape that is 3.6 meters long and another tape that is 2.2 meters long together, end to end, how long will it be? What math sentence will represent this problem? What does 3.6 meters mean? What does 2.2 meters mean? 3 meters + 2 meters, and 6 0.1-meters + 2 0.1-meters 5 meters and 8 0.1-meters, or 5.8 meters
15. If you put a tape that is 3.6 meters long and another tape that is 2.2 meters long together, end to end, how long will it be? What math sentence will represent this problem? What does 3.6 meters mean? What does 2.2 meters mean? 36 0.1-meters + 22 0.1-meters 58 0.1-meters, or 5.8 meters
16. If you put a tape that is 3.73 meters long and another tape that is 2.2 meters long together, end to end, how long will it be? What math sentence will represent this problem? What does 3.73 meters mean? What does 2.2 meters mean?
17. If you put a tape that is 5/8 meters long and another tape that is 7/8 meters long together, end to end, how long will it be? • What math sentence will represent this problem? • What does 5/8 meters mean? • What does 7/8 meters mean? • How many 1/8 meter will there be altogether? • What is the answer?
18. If you put a tape that is 5/8 meters long and another tape that is 3/4 meters long together, end to end, how long will it be? • What math sentence will represent this problem? • What is different about this problem? • What does 5/8 meters mean? • What does 3/4 meters mean? • What can we do?
19. If you put a tape that is 5/8 meters long and another tape that is 3/4 meters long together, end to end, how long will it be? • Let’s make these fractions refer to the same unit. • We can change 3/4 into 6/8 [or we can change both to 10/16 and 12/16, or some other equivalent fraction pairs] • Now, we know how to add those fractions.
20. Do we need the least common denominator? • No – we just need a common denominator (common unit) in order to add.
21. Multiplying & Dividing Decimals M4N5. • Multiply and divide both one and two digit decimals by whole numbers. M5N3. • Explain the process of multiplication and division, including situations in which the multiplier and divisor are both whole numbers and decimals.
22. Multiplying & Dividing Decimals M4N5. • Multiply and divide both one and two digit decimals by whole numbers. M5N3. • Explain the process of multiplication and division, including situations in which the multiplier and divisor are both whole numbers and decimals.
23. Which problem can we use our whole number multiplication knowledge to solve? • 1m of wire weighs 1.4 lb. How much will 6m of the same wire weigh? • 1m of wire weighs 6 grams. How much will 1.4m of the sa,e wore weigh?
24. What does 1.4 x 6 mean?
25. Double Number Line
26. Decimal Unit Approach • We have 6 groups of 14 0.1 grams. • 14 x 6 = 84; Altogether, we have 84 0.1 grams. • 84 0.1 grams 8.4 grams • 1.4 x 6 = 8.4
27. Which problem can we use our whole number division knowledge to solve? • 4 m of iron pipe weighs 3.6 kg. How much will 1m of the same pipe weigh? • 3.6 m of iron pipe weighs 4 kg. How much will 1m of the same pipe weigh?
28. What does 3.6 ÷ 4 mean?
29. Double Number Line
30. Decimal Unit Approach • Divide 36 0.1-kg to make 4 equal groups. • 36 ÷ 4 = 9; Each group will have 9 0.1-kg. • 9 0.1-kg = 0.9 kg. • 3.6 ÷ 4 = 0.9
31. What if we had 3.7 ÷ 4 mean? • 37 0.1-kg: make 4 equal groups • BUT, 37 ÷ 4 = 9 rem. 1 • Each group will get 9 0.1-kg and there will be 1 0.1-kg left over. • 3.7 ÷ 4 = 0.9 rem. 0.1
32. Dividing on: 3.7 ÷ 4 • Model 3.7 ÷ 4 using base-10 blocks – use a flat as 1. • What will be left over? • Can we trade it? With what?
33. What is 8 ÷ 5? | 2,108 | 7,342 | {"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} | 4.59375 | 5 | CC-MAIN-2019-51 | latest | en | 0.854768 |
www.nicerodds.co.uk | 1,653,053,324,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00303.warc.gz | 1,075,920,706 | 44,725 | # Fractional vs Decimal odds
What are the fractional odd?
Fractional odds are the most traditional form of odds that are also known as UK Odds. It mostly used to indicate the probabilities, as the probability of something happening, and about the amount that the bookmaker thinks the result is worth.
Fractional odds are always displayed as two numbers that are separated by a forward slash i.e.
5/1 – ‘five to one’ you can also read as five to one against
1/1 – ‘evens’
1/2 – ‘one to two’ also read as two to one on
What is the decimal odd?
Decimal odds mostly used in Europe, Australia, New Zealand, and the Canadian continent. It's a little easier to understand and work with. Favorites and underdogs can be seen immediately by looking at the numbers.
The decimal number represents the amount of each win for \$1. For decimal odds, the number represents the total payout of the profit. In other words, your stake is already included in the decimal number that makes calculation easier for total payout.
Total payout can be calculated as
Total Payout = Stake x Decimal Odd Number
Understanding of Fractional & Decimal Betting Odds
Fractional Odds Explained
When you see two numbers that are separated by a slash, i.e. 5/1, it’s a fractional betting odds. There are two numbers matter one at the right side which means the unit stake you are betting in and second is the left hand which means is the potential return per unit stake.
In this case of 5/1 bet, your bet is £1, so if it is successful than you’ll win £4. The total return on this bet would be £5. (This all means that initial £1 stake plus £4 winnings)
The probability of a 5/1 chance winning is 16% - based on his calculation 1 / (5 + 1) = 0.16
It means (potential return / stake = winnings)
Decimal Odds Explained
Decimal odds are alternative in the fractional format to seeing betting odds. Which preference to view odds in is a personal preference for customers. For the decimal odds, we can calculate the winnings with
Winnings = (odds × stake) – stake
Decimal odds are always written with the decimal sign. For example, we can write 9.0, and if we write it in fractional odd than we can write 8/1. When you see decimal odds showed its means these include your unit stake of one.
If you bet £1 at 9.0 and you win then you will get back £9, its means initial £1 stake and £8 winnings. The possibility of a 9.0 chance winning is 10% that’s based on the following calculation
1 / (9 + 1) = 0.10
You can calculate (9.0 chance = 8/1)
1 / (9 + 1) × 100 = 10% winnings
This equation is known as the decimal odd equation
How Do Odds Work in Betting?
For your knowledge, according to well-known betting websites, there is a 2% chance that Dwayne "The Rock" Johnson will win the US presidential election in 2020. "The Rock" odd against winning the election are listed as 50/1 which means there are 50 chances to lose and one chance to win. You can calculate it as 1/51 = 0.0196 which is the probability of winning.
There are three types of odds which are fractional odds, decimal odds, and American odds. These have different ways of resenting the same thing and no specific difference in terms of payouts.
How Fractional Odds Work
Fractional odds are written with a slash (/)” or a “hyphen (-) i.e. 9/1 or 9-1 and announced as “nine-to-one.” These are the most preferred odds across the world and the world’s largest bookmakers use it.
A fractional listing of 9/1 odds means that you can win \$9 against every \$1 you wager its means you can get back your investment with winnings. In other words, you will receive your stake that’s \$1 in addition to the profit which is \$9, which results in total payout \$10. So if you stake \$10 at 9/1, you should get the total payout of \$100, it means \$90 profit with\$10 stakes.
Where total potential will calculated by
Total Payout = [Stake x (Numerator/Denominator)] + Stake
For example, popular sports betting websites showed the fractional odds for the betting on the team to win 2017-18, NBA Championship. They have selected three teams with the lowest odds against winning.
Golden State Warriors: 10/11
Houston Rockets: 9/4
Cleveland Cavaliers: 7/1
How Decimal Odds Work
Decimal odds are written with the decimal sign and a bit easier to understand and work with. Favorites and underdogs can be seen immediately by looking at the numbers, through odds for vs odds against you can find what are your best options.
The decimal odd number represents the amount of each win for \$1 each. It represents the total payout instead of profit. In other words, having the decimal number stakes are included in it and you have no need to add these. It makes the total payout calculation easier.
You can calculate the total payout with the following equation
Total Payout = Stake x Decimal Odd Number
For example, a popular betting websites prices some candidates to win the 2020 U.S. Presidential Election. They make a list of decimal odds for the top three candidates as of March 21, 2018.
Donald Trump: 3.00
Bernie Sanders: 11.00
Elizabeth Warren: 13.00
This number represents only the amount that can be won against every \$1 at stake. So, if anyone bets \$100 on Donald Trump to be re-elected, president. The total payout for this person is \$300, it includes \$100 of initial stake with a net profit of \$200.
Odds Table for Fractional to Decimal Betting
You can convert the fractional betting into its decimal amount. In the following table, the left-hand column is the fractional betting that is from smallest to largest and right side showing fractions grouped by their denominator.
1/1000 = 0.001 1/4 = 0.25 1/100 = 0.01 3/4 = 0.75 1/50 = 0.02 5/4 = 1.25 1/33 = 0.03 7/4 = 1.75 1/25 = 0.04 9/4 = 2.25 1/20 = 0.05 11/4 = 2.75 1/16 = 0.06 13/4 = 3.25 1/14 = 0.07 15/4 = 3.75 1/12 = 0.08 1/11 = 0.09 1/10 = 0.1 1/5 = 0.2 1/9 = 0.11 2/5 = 0.4 1/8 = 0.12 3/5 = 0.6 1/7 = 0.14 4/5 = 0.8 3/20 = 0.15 6/5 = 1.2 1/6 = 0.16 7/5 = 1.4 2/11 = 0.18 8/5 = 1.6 1/5 = 0.2 9/5 = 1.8 1/9 = 0.22 11/5 = 2.2 1/4 = 0.25 12/5 = 2.4 2/7 = 0.28 13/5 = 2.6 3/10 = 0.3 14/5 = 2.8 1/3 = 0.33 7/20 = 0.35 4/10 = 0.4 1/6 = 0.16 9/20 = 0.45 2/6 = 0.32 1/2 or 5/10 = 0.5 3/6 = 0.5 11/20 = 0.55 4/6 = 0.66 6/10 = 0.6 5/6 = 0.83 13/20 = 0.65 = 7/10 = 0.7 = 3/4 = 0.75 1/7 = 0.14 8/10 = 0.8 2/7 = 0.28 17/20 = 0.85 3/7 = 0.42 9/10 = 0.9 4/7 = 0.57 19/20 = 0.95 5/7 = 0.71 1/1 = 1 6/7 = 0.85 21/20 = 1.05 11/10 = 1.1 23/20 = 1.15 1/8 = 0.125 12/10 = 1.2 3/8 = 0.375 5/4 = 1.25 5/8 = 0.625 13/10 = 1.3 7/8 = 0.875 27/20 = 1.35 14/10 = 1.4 29/20 = 1.45 1/9 = 0.11 15/10 = 1.5 2/9 = 0.22 31/20 = 1.55 4/9 = 0.44 16/10 = 1.6 5/9 = 0.55 33/20 = 1.65 7/9 = 0.77 17/10 = 1.7 8/9 = 0.88 7/4 = 1.75 18/10 = 1.8 37/20 = 1.85 1/10 = 0.1 19/10 = 1.9 2/10 = 0.2 39/20 = 1.95 3/10 = 0.3 2/1 = 2 4/10 = 0.4 21/10 = 2.1 5/10 = 0.5 22/10 = 2.2 6/10 = 0.6 9/4 = 2.25 7/10 = 0.7 23/10 = 2.3 8/10 = 0.8 24/10 = 2.4 9/10 = 0.9 5/2 = 2.5 11/10 = 1.1 26/10 = 2.6 12/10 = 1.2 27/10 = 2.7 13/10 = 1.3 11/4 = 2.75 14/10 = 1.4 28/10 = 2.8 15/10 = 1.5 29/10 = 2.9 16/10 = 1.6 3/1 = 3 17/10 = 1.7 31/10 = 3.1 18/10 = 1.8 32/10 = 3.2 19/10 = 1.9 13/4 = 3.25 2/1 = 2 33/10 = 3.3 34/10 = 3.4 7/2 = 3.5 1/11 = 0.09 36/10 = 3.6 2/11 = 0.18 37/10 = 3.7 3/11 = 0.27 15/4 = 3.75 4/11 = 0.36 38/10 = 3.8 6/11 = 0.54 39/10 = 3.9 7/11 = 0.63 4/1 = 4 8/11 = 0.72 9/2 = 4.5 9/11 = 0.81 5/1 = 5 10/11 = 0.9090 11/2 = 5.5 6/1 = 6 13/2 = 6.5 1/12 = 0.08 7/1 = 7 1/14 = 0.07 15/2 = 7.5 1/16 = 0.06 8/1 = 8 1/20 = 0.05 17/2 = 8.5 1/25 = 0.04 9/1 = 9 1/33 = 0.03 19/2 = 9.5 1/50 = 0.02 10/1 = 10 1/100 = 0.01
In this table any number that is over 10 will simply be its number to 1. I.e.15 = 15/1. 125 = 125/1. Etc. | 2,879 | 7,691 | {"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} | 4.09375 | 4 | CC-MAIN-2022-21 | latest | en | 0.958805 |
http://www.docstoc.com/docs/3426144/Algebra-II-Assignment-Guide | 1,397,983,439,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00132-ip-10-147-4-33.ec2.internal.warc.gz | 397,434,751 | 14,856 | # Algebra II Assignment Guide
Document Sample
``` Algebra II Assignment Guide
Fall Semester
Chapter 0 Problem Solving- 7 Days
Days Topic Assignment Other Notes
1 0.1 Pictures Graphs 2-6 even, 7-11, 13, 16 Project: Design a
and Diagrams Computer Icon and
Presentation-Due Day 4
2 0.2 Symbolic 1-7, 10, 12-14
Representation
3 0.3 Organizing 1-7, 9, 11-14
Information
4 Review Day 1 Review Exercises 1-9 Presentation of Computer
Icon Project
5 Review Day 2 10-15
6 Test Chapter 0
Notes:
Chapter 1 Patterns and Recursion-11 Days
Day Topic Assignment Other Notes
1 1.1 Patterns and 1-6, 8, 9, 11, 14-17
Recursion
2 1.2 Modeling Growth 1-7, 10, 13, 16-20 Quiz on Day 3
and Decay
3 1.3 First Look at Limits 1-8, 11, 13-15
4 1.4 Graphing Sequences 1-6, 9, 13-15
5 GSP Basics/ Catch-up None
6 GSP Exploration Questions 1-3
7 1.5 Loans and Investment 1-7, 9, 10, 11, 12-15 Project: Pyramid
Plan Due Day 2 of
Review
Quiz on Day 8
8 Exploration Refining the Questions 1-3
Growth Model
9 Review Day 1 Review Exercises 1-5
10 Review Day 2 Review Exercises 6-10 Project
Presentations Due
11 Test
Notes:
Chapter 2 Describing Data- 7 Days
Day Topic Assignment Other Notes
1 2.1 Measures of Central 1-9, 13, 15-19
Tendency and Box Plots
2 2.2 Measures of Spread 1-4, 6-8, 10, 11, 15, 16-18 Quiz on Day 3
3 2.3 Histograms and 1-6 Quiz on Day 4
Percentile Ranks
4 2.3 Histograms and 7-9, 11-13
Percentile Ranks
5 Exploration: Census Questions 1, 2
Microdata Review Exercises 1-8
6 Review Day TAL 1-4 AWYL Write Test
Items
7 Test Email-AWYL Begin a
Portfolio
Notes:
Chapter 3 Linear Models and Systems-16 Days
Day Topic Assignment Other Notes
1 3.1 Linear Equations and 1-4, 6, 7, 8, 10, 12
Arithmetic Sequences
2 3.2 Revisiting Slope 1-4, 5, 6
Sec. 3.1: 13-16
3 3.2 Revisiting Slope 7-10, 12, 13-17
4 3.3 Fitting a Line to Data 1-8, 10, 11-14 Quiz on Day 5
Project: Talking
Trash Due Day 11
5 3.4 The Median-Median 1-6, 7, 12, 13, 14
Line
6 3.4 The Median-Median 9, 10, 11, 15, 16
Line
7 3.5 Residuals 1-6, 12, 13
8 3.5 Residuals 7-11, 14, 15
9 Exploration: Residual Questions 1-5 Quiz on Day 10
Plots and Least Squares
10 3.6 Linear Systems 1-7, 9, 10, 12, 13, 14
11 3.7 Substitution and 1-5, 8, 10, 12 Project from Day 4
Elimination Due
Quiz on Day 12
12 Review Day 1-10
13 Chapter Test
14 Chapters 1-3 Review Mixed Review 11-15
15 Chapters 1-3 Review Mixed Review 16-20
16 Test Chapters 1-3 Email AWYL-Write in
your journal by the end of
study hall
Notes:
Chapter 4 Functions, Relations, and Transformations
Day Topic Assignment Other Notes
1 4.1 Interpreting Graphs 1-4, 6, 7, 10, 11
2 4.2 Function Notation 1-5, 7-10, 12, 16, 17 Project: Step
Functions due Day 8
Quiz on Day 3
3 4.3 Lines in Motion 1-6, 9, 10-13
4 4.4 Translations and the 1-11
5 4.5 Reflections and the Square 1-11 Quiz on Day 6
Root Family
6 Exploration: Rotation as a Questions 1 and 2
Composition of Transformations Sec 4.5: 12-14
7 4.6 Stretches and Shrinks and the 1-6, 9, 10, 12, 13
Absolute Value Family
8 4.7 Transformations and the 1-7, 9-13 Project Due
Circle Family
9 4.8 Composition of Functions 1-6, 13, 14
10 4.8 Composition of Functions 8, 9, 10, 12, 15, 16
11 Review 1-9
12 Test
Note:
```
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views: 20 posted: 1/10/2009 language: English pages: 5 | 1,406 | 4,379 | {"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} | 3.171875 | 3 | CC-MAIN-2014-15 | longest | en | 0.67189 |
https://numbermatics.com/n/39339715091044/ | 1,653,770,457,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00660.warc.gz | 485,229,580 | 6,684 | # 39339715091044
## 39,339,715,091,044 is an even composite number composed of three prime numbers multiplied together.
What does the number 39339715091044 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 27 divisors.
39339715091044 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of twenty-seven divisors.
## Prime factorization of 39339715091044:
### 22 × 672 × 468072
(2 × 2 × 67 × 67 × 46807 × 46807)
See below for interesting mathematical facts about the number 39339715091044 from the Numbermatics database.
### Names of 39339715091044
• Cardinal: 39339715091044 can be written as Thirty-nine trillion, three hundred thirty-nine billion, seven hundred fifteen million, ninety-one thousand and forty-four.
### Scientific notation
• Scientific notation: 3.9339715091044 × 1013
### Factors of 39339715091044
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 6
• Sum of prime factors: 46876
### Divisors of 39339715091044
• Number of divisors d(n): 27
• Complete list of divisors:
• Sum of all divisors σ(n): 69888860676243
• Sum of proper divisors (its aliquot sum) s(n): 30549145585199
• 39339715091044 is a deficient number, because the sum of its proper divisors (30549145585199) is less than itself. Its deficiency is 8790569505845
### Bases of 39339715091044
• Binary: 10001111000111011111011011111001011010011001002
• Base-36: DY0EYN6G4
### Squares and roots of 39339715091044
• 39339715091044 squared (393397150910442) is 1547613183444515033209009936
• 39339715091044 cubed (393397150910443) is 60882661707850834393180412997889340613184
• 39339715091044 is a perfect square number. Its square root is 6272138
• The cube root of 39339715091044 is 34010.2953513663
### Scales and comparisons
How big is 39339715091044?
• 39,339,715,091,044 seconds is equal to 1,250,881 years, 13 weeks, 1 day, 12 hours, 24 minutes, 4 seconds.
• To count from 1 to 39,339,715,091,044 would take you about three million, one hundred twenty-seven thousand, two hundred three years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 39339715091044 cubic inches would be around 2834.2 feet tall.
### Recreational maths with 39339715091044
• 39339715091044 backwards is 44019051793393
• The number of decimal digits it has is: 14
• The sum of 39339715091044's digits is 58
• More coming soon!
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The information we have on file for 39339715091044 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 39339715091044, math, Factors of 39339715091044, curriculum, school, college, exams, university, Prime factorization of 39339715091044, STEM, science, technology, engineering, physics, economics, calculator, thirty-nine trillion, three hundred thirty-nine billion, seven hundred fifteen million, ninety-one thousand and forty-four.
Oh no. Javascript is switched off in your browser.
Some bits of this website may not work unless you switch it on. | 1,152 | 4,241 | {"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} | 3.1875 | 3 | CC-MAIN-2022-21 | latest | en | 0.782604 |
https://www.cs.ox.ac.uk/people/nathanael.fijalkow/Talk/ALGA_Marseille/ | 1,526,928,375,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864466.23/warc/CC-MAIN-20180521181133-20180521201133-00617.warc.gz | 738,392,600 | 3,128 | ## The star-height problem
• $(a + b^*)^* + a b^*$ has star-height $2$
• $a^* b^* + (ba)^*$ has star-height $1$
Star-height problem: given a regular language $L$, what is the minimal star-height of a regular expression denoting $L$?
## Cost automata
Theorem (Hashiguchi): the star-height problem can be reduced to the boundedness problem of cost automata
Induces $f : A^* \to \mathbb{N} \cup \{ \infty \}$ $$f(w) = \text{min} \left\{ \text{max} \text{ counter value in } \rho \mid \rho \text{ accepting run} \right\}$$
## Reduction
A string expression of height $h$ and width $m$ is $$\bigcup w_1\ f_1^*\ w_2\ f_2^*\ \ldots\ w_i\ f_i^*$$ where:
• $i \le m$
• $|w_\ell| \le m$
• $f_\ell$ are string expressions of height $h-1$ and width $m$
Proposition (Kirsten 2005): For every regular language $L$ and $h \in \mathbb{N}$ one can compute a cost automaton such that for every $w \in A^*$, the following are equivalent:
• there is a string expression $e$ of height $h$ and width $m$ such that $w \in e \subseteq L$
• the automaton has a run on $w$ with value at most $m$
### Boundedness of cost automata
$$\exists N \in \mathbb{N},\ \forall w \in A^*,\ f(w) \le N$$ Remark: generalises universality
Proposition (Bojańczyk 2015): A cost automaton is bounded if, and only if, Eve wins the Gale-Stewart game where:
• Eve outputs sets of transitions.
Eve wins if there exists $N$ such that:
• at every point there exists an accepting run,
• in all runs the value of the counters are bounded by $N$.
## Bottom line
### To solve the star-height problem, it suffices to solve boundedness games
Corollary: the star-height problem is decidable.
## An example
$$f(a^{n_1} b a^{n_2} b \cdots a^{n_k} b) = \text{min}\ n_\ell$$ $n = 2$
### Constructing history-deterministic automata
Idea (F., Colcombet 2016): composition of:
• A history-deterministic automaton $\mathcal{B}$ that inputs letters and outputs a run tree. To prove that $\mathcal{B}$ is history-deterministic we need a positionality result for cost games.
• A deterministic B-automaton $\mathcal{C}$ that inputs a run tree and checks whether all paths have small values. To obtain a deterministic B-automaton $\mathcal{C}$ we rely on Safra's construction used as a black-box. | 692 | 2,244 | {"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} | 3.203125 | 3 | CC-MAIN-2018-22 | latest | en | 0.671472 |
https://communities.sas.com/t5/New-SAS-User/WHAT-DO-THESE-PLOTS-SAY-ABOUT-THE-REGRESSION-OUTCOMES/td-p/562473 | 1,723,700,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00086.warc.gz | 137,181,791 | 36,910 | Pyrite | Level 9
## WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
I have run a regression, and in the output the following charts are generated. I cannot relate these charts with the parameter estimates. Could someone give some idea?
For example, does the chart imply that the predicted value of Abs_Jump is 0 by all the regressors except Abs_Surp_Mex14? How could this be if I have the parameter estimate for Abs_Surp_Mex8 which is 0.469 with p-value 0.01? Please enlighten me.
8 REPLIES 8
Diamond | Level 26
## Re: WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
Can we see the code you used? Can we see the rest of the output from this regression?
--
Paige Miller
Pyrite | Level 9
## Re: WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
``````proc qlim data= MXN_Jump_News;
model Abs_Jump= DW_Mon DW_Tue DW_Wed DW_Thu Sin1 Sin2 Sin3 Sin4 Cos1 Cos2 Cos3 Cos4 Abs_Surp_US4 Abs_Surp_US7 Abs_Surp_US8 Abs_Surp_US9 Abs_Surp_US11 Abs_Surp_US16 Abs_Surp_US20 Abs_Surp_US22 PS_US30 Abs_Surp_GER12 Abs_Surp_MEX5 Abs_Surp_MEX6 Abs_Surp_MEX8 Abs_Surp_MEX10 Abs_Surp_MEX14 PS_MEX16 SP4 SP16 SP21 SP23; nloptions maxiter= 500;
endogenous Abs_Jump ~ censored (lb= 0);
run;``````
DW_Mon to DW_Thu are dummies. SP4 SP16 SP21 SP23 are also dummies. Sin1-Sin4 and Cos1-Cos4 are sine and cosine values. The dependent variable has only positive values. The other independent variables take both positive and negative values.
Super User
## Re: WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
When I see code such as (reformatted for legibility on the forum)
```proc qlim data= MXN_Jump_News;
model Abs_Jump= DW_Mon DW_Tue DW_Wed DW_Thu
Sin1 Sin2 Sin3 Sin4 Cos1 Cos2 Cos3 Cos4
Abs_Surp_US4 Abs_Surp_US7 Abs_Surp_US8 Abs_Surp_US9
Abs_Surp_US11 Abs_Surp_US16 Abs_Surp_US20 Abs_Surp_US22 PS_US30
Abs_Surp_GER12
Abs_Surp_MEX5 Abs_Surp_MEX6 Abs_Surp_MEX8 Abs_Surp_MEX10 Abs_Surp_MEX14 PS_MEX16
SP4 SP16 SP21 SP23; nloptions maxiter= 500;
endogenous Abs_Jump ~ censored (lb= 0);
run;```
I wonder if when the values for the variables with US in the name are populated whether the ones with GER or MEX have values other than 0 and vice versa. Maybe just too many variables with 0 for values.
It may not hurt to describe, if not list some data, why the Abs_Surp variables have such different suffixes.
Super User
## Re: WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
Why is your title in upper case?
Random guess based on not a whole lot, you have a rare event or data that's very weirdly distributed in some manner. What does the distribution of your response variable look like?
Pyrite | Level 9
## Re: WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
Sorry for the upper case.
Pyrite | Level 9
## Re: WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
Would you please look at the code and the full result above? Much thanks.
Diamond | Level 26
## Re: WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
As a general rule, when you get such un-intuitive results from a model fit, the causes are most likely one of the following
1. Extreme outlier(s)
2. Overfitting
3. Extreme multi-collinearity between the X-variables
4. Humongous amounts of noise in the y-variables — possibly because of an extreme outlier(s)
Now, I have never used PROC QLIM and so I can't really say any more about your results, and there may be other reasons why this is happening specifically to PROC QLIM. Nevertheless, the above 4 possibilities are things you need to investigate yourself, since you have the data.
--
Paige Miller
Super User
## Re: WHAT DO THESE PLOTS SAY ABOUT THE REGRESSION OUTCOMES?
@d6k5d3 wrote:
Would you please look at the code and the full result above? Much thanks.
My previous question is still unanswered, which will likely help answer the other questions. What does the distribution of your response variable look like before modelling? If you take a close look at your output it seems like there isn't even an upper bound which seems a bit weird to me, which may mean your data is massively skewed as I suspected.
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• 4 in conversation | 1,180 | 4,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.601795 |
https://askworksheet.com/coordinate-grid-super-teacher-worksheets-answer-key/ | 1,669,573,515,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00541.warc.gz | 146,477,563 | 21,316 | # Coordinate Grid Super Teacher Worksheets Answer Key
Showing top 8 worksheets in the category coordinates answer sheet. Super teacher worksheets coordinate grid.
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Coordinate Grid Ordered Pairs Coordinate Plane Worksheets Super Teacher Worksheets Coordinate Grid | 777 | 4,603 | {"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} | 2.828125 | 3 | CC-MAIN-2022-49 | latest | en | 0.747072 |
https://ohbug.com/uva/11783/ | 1,726,708,902,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00229.warc.gz | 403,759,977 | 2,253 | Nails
The World Of Nails (WON) is a famous hardware store specialized in selling hundreds of different kinds of nails. Some of them very old, rare and expensive. However, the store also offers some standard hand tools and wood. Rado, the store’s manager, has two kids who usually go to the store after school and spend the afternoon playing with the tools. During all this time, the kids play with hammers, nails, screws, wood, etc. (quite dangerous isn’t it?). One day, the kids were playing in the wood deposit. By accident, they hit a huge pile of wood sticks which then collapsed. A lot of wood sticks fell to the floor, but did not hit any of the kids. The kids did not get scare and so they continued playing. They took a bag of very expensive nails and a hammer and decided to hammer the nails on the wood sticks, just for fun. If a wood stick were laying in top of another, they joined these two pieces together by hammering exactly one nail where they crossed, see Figure 1. No more than two stick are crossed in the same point such as shown on third configuration in figure below. If a wood stick was laying on the floor all alone, with no other stick on top of it, they hammered the stick on to the floor using two nails, one for each end of the stick. Figure 1: Different configurations After a couple of hours, Rado discovered the disaster in the deposit. When he noticed what kind of nails were the kids using, he was shocked. A lot of money wasted on a little game. Rado wants you to help him calculate how many nails did they kids use, to calculate how much money he has lost. Input The input consists in several test cases. Each case corresponds to a set of wooden sticks. The first line of a test case contains one integer (0 ≤ N ≤ 1000), indicating the number of sticks. Next, N lines follow, each one corresponding to a stick. Each line contains four integer values (0 ≤ X1, Y1, X2, Y2 ≤ 1000) separated of each other by exactly one blank space. The first two integers correspond to the Cartesian coordinates of one end of the wood stick. The later two integers correspond to the Cartesian coordinate of the other end of the wood stick. Assume no stick is completely superposed by another. The end of input is indicated by a test case with N = 0. Output For each case specified in the input, the output is a single line containing the number of nails used by the kids.
2/2 Sample Input 3 0030 1133 1222 3 0030 1100 1022 3 1122 0050 0105 0 Sample Output 3 2 6 | 582 | 2,480 | {"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} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.968172 |
https://puzzling.stackexchange.com/questions/110229/up-like-a-rocket-ship | 1,627,956,912,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00485.warc.gz | 478,881,477 | 37,997 | # Up like a rocket ship
This is a Palisade (or 6 Cells) puzzle. The rules are:
• Thicken some of the pentagons' sides so the thick borders outline regions of six pentagons apiece.
• Any pentagon with a number in it indicates how many of its sides are thickened. Note that if the pentagon is along the border of the entire diagram, then that side counts as one of the thickened sides.
• Thickened sides can be used only as parts of regions' outlines: no thickened side of a pentagon can have both of its sides in the same region.
• Besides the outer border of the diagram, two sides are thickened to get you started.
• This was nice. I solved it using similar logic to Deusovi, a few minutes at a time over a few sessions. Thanks! – Harfatum May 27 at 23:29
To start:
The 0 makes a 6-cell region by itself. This creates a wall by the nearby 1.
Then, make sure not to block off any cells where they can't be part of 6-cell regions:
Next,
the 4 clue does some work - if it went up to join the 2, it would stop the 2 from making a region.
And we can do something similar with the 4 clue down below. (I also notice that the 2 in the bottom left is already satisfied.)
We can complete the bottom-right corner:
The 4 clue going up and right would block off the region under it. And if that region doesn't take both cells next to it, it blocks off the one it doesn't take.
The 2 near the bottom middle cannot join with the other 2, because the region would be too large.
Now we've finished off a clue:
We've both of that 2's walls, so it claims the other three cells next to it; this makes the bottom-left region complete.
The chokepoint on the left blocks off six cells.
Putting a border on the left side of the 3 would cause it to gain a fourth wall. Putting a border on the right side of the 3 would either block the 2 region or cause it to merge with the 3 and be too big. So the 3's remaining border must be on the bottom.
This completes the 1 region, and the rest of the puzzle falls into place. The solved puzzle:
• Good explanation. I'd appreciate your feedback on the puzzle. – msh210 May 24 at 7:25
• @msh210 It was pretty simple overall, but I had fun getting used to the logic of the genre. I liked how the ending required more complicated logic, too - overall it felt like there was a nice gradual increase in difficulty. – Deusovi May 24 at 19:37
• Thanks! [15 characters] – msh210 May 24 at 20:37 | 612 | 2,421 | {"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} | 4.21875 | 4 | CC-MAIN-2021-31 | longest | en | 0.950126 |
https://reibot.org/2011/10/01/sd/?replytocom=1523 | 1,669,707,080,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00334.warc.gz | 506,094,723 | 27,161 | # Standard Deviation and Moving Average
Recently my neighbor paid me to build a key less entry system for his dorm room. I decided to go the economical route and use a button/potentiometer that sits outside the door and an Arduino on the inside that controls a servo connected to the lock. For my room, I thought it would be interesting to use a Ping))) ultrasonic distance sensor instead of the potentiometer and lose the button.
The Ping))) sensor kept taking readings while my hand was moving. In order to fix this I decided use a Moving Average filter, then calculate the Standard Deviation of the values currently included in the M.A. filter. When my hand is still, the Standard Deviation will become very small.
Example code:
M.A._and_S.D.(pdf)- “storeValue(variable);” is how to enter data into the array, then call M.A. and S.D.
Not much of a circuit required! Arduino's regulator also powers Ping))). Servo has it's own 5v regulator... needs capacitors
pingDoorLocker(pdf) – As you can see, this program blew up a little…
I tried to make the M.A. and S.D. code very easy to follow. Some things could have been combined in the S.D. and Variance method, but to the beginner what I wrote above is probably easier to understand since it follows the equations. As for the pingDoorLocker – I threw that code together very quickly.
Follow up notes: That was probably the worst way to do this project… I thought of a few ways how to write the program that would chop the code WAY down, but this is an example about using M.A. and S.D.! Pretty bad use of a M.A. filter if you ask me!
His door unlocker.
Since I did put a few hours into building my neighbor’s door opener, here’s an image of it! He didn’t want numbers on the potentiometer dial, so I made the LED flash the number that is currently being entered.
Code for his door opener (pdf) – leave a comment if you want schematics/code on rapid share since pdf loses tabs.
Electrical Engineer who loves to bike!
### 6 Responses to Standard Deviation and Moving Average
1. i like your ping door locker!:D
2. aashishasanand says:
Well! It is pretty good actually. It can be upgraded as per requirements. Can you please give the schematics and components used in this circuit?
Nice work!!
3. Hello there! Do you know if they make any plugins to safeguard against
hackers? I’m kinda paranoid about losing everything
I’ve worked hard on. Any tips?
• Moser says:
Don’t put your micro controller or anything outside. Think of if there’s any way it could be taken over from what’s accessible outside.
Check your code and think about how the system would react if it got rebooted or powered off.
Other than that, just write good code. If you only have 3 digits in your password then there’s only 999 combinations (still more than a normal hacker would want to try…)
4. Cameron says: | 650 | 2,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.949061 |
https://de.mathworks.com/matlabcentral/profile/authors/12883640-steven-yeh | 1,571,260,663,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669546.24/warc/CC-MAIN-20191016190431-20191016213931-00196.warc.gz | 460,141,031 | 17,798 | Community Profile
# Steven Yeh
7 total contributions since 2018
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How to detect the change point
You pretty much have it laid out: diff_data=diff(data); diff_data=[0;diff_data]; percentageChange = abs(diff_data...
mehr als ein Jahr ago | 2
| accepted
How do I efficiently replace data in data sets where I want to find a value in the data that is in one array and replace it with the data in the same position in another array.
You can use a single loop with find function: a = 1:10; b = 11:20; c = 1:2:10; for i = 1:length(c) inde...
mehr als ein Jahr ago | 0
| accepted
xlswrite text with multiple empty cells inbetween
There are many ways to do this, one of them is to create padding cells: a = cell(1,20); Header1 = {'All',a{:}, 'Develope...
mehr als ein Jahr ago | 1
| accepted
How can I animate the drawing of a figure 8 shape in a polar plot to confirm the path it takes?
Try this: start = 0; end_ = 2*pi; disc = pi/16; theta4 = 0:0.01:2*pi; rho4 = sqrt(4*cos(2*theta4).*(sec(theta4)...
mehr als ein Jahr ago | 0
| accepted
Divide y values of two graph?
You could create a new time series, and use linear interpolation to find corresponding values. For example: ax = linspac...
mehr als ein Jahr ago | 1 | 364 | 1,267 | {"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-2019-43 | latest | en | 0.519129 |
https://vickreymathcircle.wordpress.com/2015/04/06/my-pretend-birthday-party-3to/ | 1,524,255,907,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944682.35/warc/CC-MAIN-20180420194306-20180420214306-00353.warc.gz | 741,828,051 | 19,901 | # My Pretend Birthday Party 3
## The Activities
1. Topic: Charts. Book: What Do You Like? Math Concept Reader
2. Topic: Charts. My birthday was last week! The kids helped me plan a pretend birthday party by voting on different items using various charts and graphs.
1. What drink?
2. What snack?
3. Where should the party be?
4. What game?
5. What kind of cake?
3. Topic: Search, Number Ordering. The evil, made-up Easter Goblin loves to steal eggs. She thinks of a number between 1 and 40, and if you don’t guess it in time, she gets an egg. I started by letting the kids have 10 guesses to find the number. For example: Is it 30? No, it’s higher than 30. Then I used a number line to help them keep track of which numbers it could be.
The dastardly Easter Goblin vs the Math Circle kids.
The number line showing that the number must be between 20 and 25.
## How did it go?
I led the younger circle this week, and all 5 kids attended. The kids were very engaged in all the activities, and did a good job waiting their turns.
#### What do you like?
This is a simple Math Concept reader that describes a group of kids planning a neighborhood block party. It’s pretty dry, but the kids were interested in the graphs and charts and voting.
#### My Pretend Party!
I’ve planned a pretend party each of the last two years with the older kids. This is the younger kids first chance. Before circle I had prepared 5 different types of charts for voting. During circle, the kids helped me fill in the chart, suggesting what types of cake, etc we should vote on. The kids really enjoyed this part…everyone was constantly raising their hands, ready to contribute.
Next we passed each chart around to each kid so they could vote on each topic. We had a chart where you write you name under the option you choose, make a tick mark next to your two favorites, fill in a box in a bar chart, and write your name + write an X in the box you choose.
At the end of circle, I handed one chart to each kid, and had all the parents and the kids from the older circle vote. Each kid had to explain what their chart’s topic was, and how to vote.
In the end, my pretend party was a Pump It Up, with chocolate cake and crepes and fruit. We’ll drink water, and play soccer at my party. Sounds fun!
The kids were a bit sad when they learned they would not get to actually go to this party, but they sang me Happy Birthday and we shared a pretend cake.
#### The Easter Goblin Guessing Game
I told the kids about the made up Easter Goblin who steals eggs and keeps them in her purse. My daughter and I had made some nice construction paper eggs before circle. The eggs were taped on the wall between the Easter Goblin, and a picture I drew of the 5 Math Circle kids.
Each round the Easter Goblin (me) wrote down a secret number between 1 and 40. The kids got 10 guesses to figure out my number. If they did, then Math Circle got an egg. Otherwise the Easter Goblin got one.
The first couple rounds I had a number line out on the table, but didn’t really help them figure out the possible range for the answer. We had guesses like: 20? no, lower. 30? no, lower. Also, one of the kids loved to guess huge numbers like 100 million or infinity.
After this, I started tracking the range of possible numbers using two glass beads on the numberline. This helped the kids a lot, and I only gave them 6 guesses.
The kids did understand that they should only guess numbers in the range, but they still need practice with number recognition. They are not sure if 27 is twenty-seven or seventy-two. When we do this again, I’ll make sure all the kids are seeing the number line rightside-up.
The kids loved beating the Easter Goblin, and they won 5 of the 6 eggs (one time I gave them extra guesses). | 897 | 3,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.953913 |
https://www.printablemultiplication.com/worksheets/Fun-Multiplication-Coloring-Worksheets | 1,607,165,616,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747774.97/warc/CC-MAIN-20201205104937-20201205134937-00494.warc.gz | 829,171,907 | 76,629 | ## Multiplication Chart X30
…of arithmetic to find out. Review basics of multiplication. Also, evaluate the fundamentals how to use a multiplication table. Multiplication Chart X30 Free Printable Download”] We will review a multiplication
## Multiplication Chart 80×80
…making use of the Multiplication Table. Incoming search terms: 4s multiplication chart 4s multiplication practice long multiplication worksheet and answers probability multiplication rule worksheet what is a mu;ltiplication grid 2×2…
## Multiplication Chart Coloring Activity
…for addition. At this moment, children use a organization grasp of addition. For that reason, multiplication is the up coming plausible form of arithmetic to learn. Assess fundamentals of multiplication….
## Multiplication Chart 90×90
…basics using a multiplication table. Let us overview a multiplication instance. Using a Multiplication Table, increase 4 times 3 and get a response 12: 4 x 3 = 12. The…
## Multiplication Chart Sbac
…of arithmetic to find out. Assess basics of multiplication. Also, evaluate the fundamentals utilizing a multiplication table. Allow us to evaluation a multiplication instance. Employing a Multiplication Table, flourish four…
## Multiplication Flash Cards Virtual
…evaluate the basic principles how to use a multiplication table. In 2020 | Multiplication Flashcards, Flashcards, Distance Learning”] We will assessment a multiplication instance. Utilizing a Multiplication Table, flourish 4…
## Multiplication Chart Multiplication Chart
…point, kids have got a firm understand of addition. As a result, multiplication may be the after that reasonable method of arithmetic to learn. Assess fundamentals of multiplication. Also, assess…
## Multiplication Flash Cards X2
multiplication will be the next reasonable type of arithmetic to discover. Assess basics of multiplication. Also, review the fundamentals how to use a multiplication table. Allow us to overview a…
## Printable Blank Multiplication Table 0-10
…utilizing a multiplication table. Allow us to evaluation a multiplication case in point. Utilizing a Multiplication Table, increase four times about three and acquire a response 12: 4 x 3…
## Multiplication Chart Of 6
…the basic principles how to use a multiplication table. We will evaluation a multiplication example. Utilizing a Multiplication Table, increase several times about three and have a response twelve: 4… | 491 | 2,429 | {"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} | 3.953125 | 4 | CC-MAIN-2020-50 | longest | en | 0.725473 |
https://zbmath.org/?q=an:0063.06802 | 1,618,542,639,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00222.warc.gz | 1,051,998,932 | 10,060 | # zbMATH — the first resource for mathematics
Infinite soluble groups. (Russian. English summary) Zbl 0063.06802
From the author’s summary: In the present paper the following definition of a soluble group is given: A (finite or infinite) group $$\Gamma$$ is called soluble if there exist soluble sets in all the quotient groups $$\Phi/\Psi$$, where $$\Phi$$ is an arbitrary subgroup of $$\Gamma$$ (in particular, $$\Gamma$$ itself) and $$\Psi$$ is an arbitrary normal subgroup of $$\Phi$$ (in particular, the unit of the group).
The importance of the “local” properties of groups is further established. A group is said to be locally soluble if all its subgroup generated by finite systems of elements are soluble in the above sense. We have the fundamental
Theorem 3. A locally soluble group is soluble.
We further have: Theorem 4. If a group $$\Gamma$$ is soluble, then for every normal subgroup $$\Theta$$ a soluble set of $$\Gamma$$ can be chosen, which contains $$\Theta$$ as an element.
Theorem 5. If a group possesses a well-ordered ascending soluble set, then the group is soluble.
A group is called locally finite if every subgroup generated by a finite set of elements is finite. For locally finite groups the property to be soluble is equivalent to the existence of a soluble set. The theory of such groups has been given by S. Chernikov (S. Tchernikov) [Mat. Sb., N. Ser. 13(55), 317–333 (1943; Zbl 0063.07318)].
We introduce the notion of weakly soluble set. The difference between a soluble set mentioned above and weakly soluble set of a group $$\Gamma$$ is that the elements of the latter are subgroups, not necessarily normal in $$\Gamma$$; the second condition is replaced by the following one: whenever there are two immediately neighbouring subgroups belonging to the weakly soluble set, the smaller subgroup is normal in the larger one and the corresponding quotient group is abelian.
If a locally finite group possesses at least one weakly soluble set, then the group is soluble (Chernikov, loc. cit.).
Theorem 6. If a normal subgroup $$\Theta$$ of a group $$\Gamma$$ and the quotient group $$\Gamma/\Theta$$ are locally finite, then $$\Gamma$$ itself is locally finite.
Several classes of groups that had been defined before, such as nilpotent groups, locally finite $$p$$-groups, periodical special groups [O. Schmidt (O. Yu. Shmidt), Rec. Math. Moscou, n. Ser. 8, 363–375 (1940; Zbl 0024.25403)], proved to be particular cases of soluble groups.
It is our theory that gives simple proofs of Chernikov’s theorems on locally soluble groups with the minimality condition [S. Chernikov , loc. cit.]; these theorems are thereby generalized.
Some subclasses can be distinguished in the class of soluble groups in the sense of ours that now will be called A-soluble. By a B-soluble group we shall mean a soluble group, which possesses at least one well-ordered ascending soluble set, by a C-soluble group – that possessing at least one well-ordered descending soluble set. The property B is, evidently, a generalization of the property of finite soluble group that they possess abelian normal subgroups different from the unit, while C is a generalization of another property of these groups – to be different from their commutants.
The notions of solubility introduced by R. Baer [Trans. Am. Math. Soc. 47, 393–434 (1940; Zbl 0023.30002)] are various particular cases of our B-solubility.
Finally, we give examples of groups that are B-soluble, but not C-soluble, and conversely. The first example is, at the same time, an example of infinite $$p$$-group, which coincides with its commutant, while the second one shows an infinite $$p$$-group without abelian normal subgroups.
##### MSC:
20F16 Solvable groups, supersolvable groups
Full Text: | 916 | 3,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-17 | latest | en | 0.882808 |
https://metric-calculator.com/convert-troy-pound-to-gram.htm | 1,701,590,008,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00077.warc.gz | 439,470,468 | 7,572 | # Troy Pounds to Grams Converter
Select conversion type:
Rounding options:
Convert Grams to Troy Pounds (g to lb t) ▶
## Conversion Table
troy pounds to grams lb t g 1 lb t 373.2417 g 2 lb t 746.4834 g 3 lb t 1119.7252 g 4 lb t 1492.9669 g 5 lb t 1866.2086 g 6 lb t 2239.4503 g 7 lb t 2612.6921 g 8 lb t 2985.9338 g 9 lb t 3359.1755 g 10 lb t 3732.4172 g 11 lb t 4105.6589 g 12 lb t 4478.9007 g 13 lb t 4852.1424 g 14 lb t 5225.3841 g 15 lb t 5598.6258 g 16 lb t 5971.8675 g 17 lb t 6345.1093 g 18 lb t 6718.351 g 19 lb t 7091.5927 g 20 lb t 7464.8344 g
## How to convert
1 troy pound (lb t) = 373.2417216 gram (g). Troy Pound (lb t) is a unit of Weight used in Standard system. Gram (g) is a unit of Weight used in Metric system.
## Troy Pounds: A Unit of Weight
Troy pounds are a unit of weight that are used for measuring precious metals, such as gold, silver and platinum. Troy pounds are derived from the French word troye, which was the name of a market town in France where English merchants traded in the Middle Ages. The symbol for troy pound is lb t.
## Definition of the Troy Pound
The troy pound is defined as 12 troy ounces, which are each 20 pennyweights, which are each 24 grains. The troy grain is equal to the avoirdupois grain, which is one seven-thousandth of an avoirdupois pound. The troy ounce is heavier than the avoirdupois ounce, but the troy pound is lighter than the avoirdupois pound.
The troy pound is equal to about 0.8229 avoirdupois pounds or 0.3732 kilograms. The troy ounce is equal to about 1.0971 avoirdupois ounces or 31.1035 grams. The troy grain is equal to about 0.0648 milligrams.
## How to Convert Troy Pounds
Troy pounds can be converted to other units of weight by using conversion factors or formulas. Here are some examples of how to convert troy pounds to other units of weight in the US customary system and the SI system:
• To convert troy pounds to avoirdupois pounds, multiply by 0.8229. For example, 10 lb t = 10 x 0.8229 = 8.229 lb.
• To convert troy pounds to avoirdupois ounces, multiply by 13.1657. For example, 5 lb t = 5 x 13.1657 = 65.8285 oz.
• To convert troy pounds to tons (short), divide by 1216.05. For example, 20 lb t = 20 / 1216.05 = 0.0164 ton.
• To convert troy pounds to kilograms, multiply by 0.3732. For example, 15 lb t = 15 x 0.3732 = 5.598 kg.
• To convert troy pounds to grams, multiply by 373.2417. For example, 25 lb t = 25 x 373.2417 = 9331.0425 g.
• To convert troy pounds to milligrams, multiply by 373241.7216. For example, 30 lb t = 30 x 373241.7216 = 11197251.648 mg.
## Where Troy Pounds are Used
Troy pounds are used in different countries and regions for different applications and purposes. Here are some examples of where troy pounds are used:
• In most countries that use the SI system, troy pounds are not used for measuring weight, but only for measuring precious metals, such as gold, silver and platinum.
• In the United States, troy pounds are used for measuring precious metals, especially for gold and silver coins and bullion.
• In Canada, Australia and New Zealand, troy pounds are used for measuring precious metals, especially for gold and silver coins and bullion.
• In the United Kingdom, troy pounds are used for measuring precious metals, especially for gold and silver coins and bullion.
• In India, troy pounds are used for measuring precious metals, especially for gold and silver jewelry.
## History of Troy Pounds
Troy pounds have a long history that dates back to ancient times. Here are some highlights of the history of troy pounds:
• The troy pound was originally based on the weight of a Roman libra, which was a unit of mass used in ancient Rome that was equivalent to about 328 grams.
• The troy pound was used in different systems of measurement, such as the Roman system, the Byzantine system, the Arabic system and the English system. It varied from about 350 grams to about 400 grams depending on the region and the time period.
• The troy pound was standardized by royal statutes and international agreements in different periods of history. For example, in 1527 an act of Henry VIII fixed the troy pound at exactly 5760 grains; in 1828 an act of Congress adopted the troy pound as the official unit of weight for coinage in the United States; in 1959 an international agreement defined the international avoirdupois ounce as exactly 28.349523125 grams, which made the troy ounce exactly 31.1034768 grams.
## Example Conversions of Troy Pounds to Other Units
Here are some examples of conversions of troy pounds to other units of weight:
• 1 lb t = 0.8229 lb
• 1 lb t = 13.1657 oz
• 1 lb t = 0.000411 ton
• 1 lb t = 0.3732 kg
• 1 lb t = 373.2417 g
• 1 lb t = 373241.7216 mg
• 1 lb t = 12 oz t
• 1 lb t = 240 dwt
• 1 lb t = 5760 gr
• 1 lb t = 0.3671 stone
## Grams: A Unit of Weight
Grams are a unit of weight that are used in the International System of Units (SI), also known as the metric system. Grams are derived from the French word gramme, which was taken from the Late Latin term gramma, meaning a small weight. The symbol for gram is g.
## Definition of the Gram
The gram is defined as one one-thousandth of a kilogram, which is the base unit of mass in the SI. The kilogram is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10 −34 when expressed in the unit J s, which is equal to kg m 2 s −1, where the meter and the second are defined in terms of c and ∆ν Cs. The Planck constant is a fundamental physical constant that relates the energy of a photon to its frequency.
The gram is a very small unit of weight, equivalent to about 0.035 ounces or 0.002 pounds. It is commonly used for measuring non-liquid ingredients in cooking, such as flour, sugar, salt and spices. It is also used for measuring small masses, such as jewelry, coins, medicines and chemicals.
## How to Convert Grams
Grams can be converted to other units of weight by using conversion factors or formulas. Here are some examples of how to convert grams to other units of weight in the US customary system and the SI system:
• To convert grams to ounces, divide by 28.349523125. For example, 100 g = 100 / 28.349523125 = 3.527 oz.
• To convert grams to pounds, divide by 453.59237. For example, 500 g = 500 / 453.59237 = 1.102 lb.
• To convert grams to tons (short), divide by 907184.74. For example, 1000 g = 1000 / 907184.74 = 0.0011 ton.
• To convert grams to kilograms, divide by 1000. For example, 200 g = 200 / 1000 = 0.2 kg.
• To convert grams to milligrams, multiply by 1000. For example, 50 g = 50 x 1000 = 50000 mg.
• To convert grams to micrograms, multiply by 1000000. For example, 10 g = 10 x 1000000 = 10000000 µg.
## Where Grams are Used
Grams are used in different countries and regions for different applications and purposes. Here are some examples of where grams are used:
• In most countries that use the SI system, grams are used for measuring food portions and ingredients, postal items, drugs and medicines, precious metals and gems and scientific experiments.
• In the United States, grams are sometimes used for measuring food portions and ingredients, especially for nutritional information labels, drugs and medicines, precious metals and gems and scientific experiments.
• In Canada, grams are used for measuring food portions and ingredients, postal items, drugs and medicines, precious metals and gems and scientific experiments.
• In Australia and New Zealand, grams are used for measuring food portions and ingredients, postal items, drugs and medicines, precious metals and gems and scientific experiments.
• In China, grams are used for measuring food portions and ingredients, postal items, drugs and medicines, precious metals and gems and scientific experiments.
## History of Grams
Grams have a long history that dates back to ancient times. Here are some highlights of the history of grams:
• The gram was originally defined in 1795 as the absolute weight of a volume of pure water equal to the cube of the hundredth part of a meter at the temperature of melting ice.
• The gram was later changed to be the weight of a volume of pure water equal to the cube of the hundredth part of a meter at the temperature of maximum density of water (4 °C).
• The gram was part of the centimeter–gram–second system of units (CGS) that was developed in the 19th century as an alternative to the meter–kilogram–second system of units (MKS).
• The gram was replaced by the kilogram as the base unit of mass in the SI system that was adopted in 1960 as an international standard for measurements.
• The gram was redefined in terms of the kilogram and the Planck constant in 2019 as part of a major revision of the SI system.
## Example Conversions of Grams to Other Units
Here are some examples of conversions of grams to other units of weight:
• 1 g = 0.035274 oz
• 1 g = 0.002205 lb
• 1 g = 0.0000011 ton
• 1 g = 0.001 kg
• 1 g = 1000 mg
• 1 g = 1000000 µg
• 1 g = 0.032151 troy oz
• 1 g = 0.032151 apothecaries’ oz
• 1 g = 0.035274 Spanish oz
• 1 g = 0.033814 French oz
Grams also can be marked as grammes (alternative British English spelling in UK).
Español Russian Français | 2,546 | 9,194 | {"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} | 2.96875 | 3 | CC-MAIN-2023-50 | longest | en | 0.80966 |
https://www.valleyislelighting.net/observation-brain-teaser-do-you-have-a-sharp-eye-find-the-clock-in-less-than-20-seconds/ | 1,696,005,699,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00012.warc.gz | 1,123,759,375 | 67,510 | Home » Observation brain teaser: Do you have a sharp eye? Find the clock in less than 20 seconds!
Observation brain teaser: Do you have a sharp eye? Find the clock in less than 20 seconds!
Come on and try to find the clock among the fruits in less than 20 seconds! It’s a fun visual brain teaser that will test your observation skills. See if you can beat the clock!
The game of the day is an exciting and visually challenging task. How fast can you find a clock among fruits in less than 20 seconds?
It may sound like an easy task, but it will be harder than you think! The visual test can be found below and not on the image at the top of the article which is there for illustration purposes only.
So, put your skills to the test and see how well you do!
Test your observation skills and find the clock in 20 seconds or less
Put your observation skills to the test and see if you can find the clock hidden among the fruits in this image. You’ll have only 20 seconds to spot it, so be quick!
See also Spot the difference: Can you bridge the gap and find the 4 differences? Only geniuses will succeed!
In order to make it a fair game, remember that you must look at the image below and not at the one at the top of the article, which is purely for illustration purposes.
So, are you up for the challenge? Give it a shot and see how fast you can spot it!
Finding a clock among the fruits within 20 seconds is a challenging task. It requires you to focus on the details and observe keenly.
To complete this task, start by scanning the image from top to bottom and side to side. Make sure that you look for all the possible hidden shapes, angles, shadows, and colors.
Besides, try to look for any visual clues that may hint at the presence of a clock – such as a long thin object that looks like an arm or circles of different sizes.
See also Spot the 3 differences in 25 seconds: Only 15% manage to do so in that time!
If you use these techniques, you will be able to find the clock quickly.
Have you found the clock among the fruits? Let’s see if you’ve managed to solve the puzzle on the next page! We can figure it out together and have some fun along the way.
Can you hit the high note in this ‘spot the difference’ game? Only true virtuosos can discern all 3!
Written by : Jessica H. Duran
I'm a content writer who loves learning new things and discovering new ideas. I'm an avid cook, always trying out new recipes and pushing myself in the kitchen. I love spending time in my garden, growing new plants and flowers. I'm also a big fan of puzzles, brain-teasers, and logical challenges. I'm constantly looking for new ways to exercise my mind and keep my brain sharp. I'm also a lifestyle enthusiast. I'm always on the lookout for new experiences and unique ways to make my life more meaningful. This drives me to create content that is both informative and entertaining. I strive to provide readers with content that is both educational and enjoyable. | 662 | 2,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-40 | latest | en | 0.93096 |
http://stackoverflow.com/questions/13626172/create-sum-of-cell-matrices | 1,406,896,067,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510274979.56/warc/CC-MAIN-20140728011754-00203-ip-10-146-231-18.ec2.internal.warc.gz | 267,684,848 | 16,014 | # create sum of cell matrices
I want to have the sum of all matrices stored in a cell array. Currently I do this
``````StackSummImage = zeros(size(imageList{1}));
for k = 1:stackSize
StackSummImage = StackSummImage + imageList{k};
end
``````
But I would rather want to write it in a single line if possible.
-
If each of the N images is K x K, you can use `cat` to concatenate all the images into a K x K x N array, and then sum it along the third dimension:
``````>> imageList = {[1 2; 3 4], [5 6; 7 8], [9 10; 11 12]};
>> stackSummImage = sum(cat(3,imageList{:}),3)
ans =
15 18
21 24
``````
Edit: You mentioned in the comments that you can't create a single array, because of memory limits. Below is the memory usage profile when I first allocate a 1500x1500x1500 array of doubles (which takes around 30 GB) and deallocate it, followed by allocating a cell array of 1500 arrays, each of which is a 1500x1500 double array. As you can see, the total memory usage is the same in both cases.
-
I can not create such an array. I save the data in a cell array because the whole data is 4-10 GB. – Matthias Pospiech Nov 29 '12 at 13:13
@MatthiasPospiech Why do you think in a cell array it takes less space? – angainor Nov 29 '12 at 13:23
Because cells are pointers to arrays and a single 3D array is a single block in memory. I can load all data with cells, but could not if I tried to allocate a single 3D array. – Matthias Pospiech Nov 29 '12 at 13:43
@MatthiasPospiech Do you mean that you have enough memory to store that data in a cell array, but not enough memory to create a copy? – Chris Taylor Nov 29 '12 at 13:43
This line should do:
``````StackSummImage = sum([imageList{:}])
``````
-
This will result in a vector, not an array of the same dimensions as the original images. – Chris Taylor Nov 29 '12 at 12:54 | 546 | 1,837 | {"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-2014-23 | latest | en | 0.863386 |
http://poj.org/problem?id=1897 | 1,529,411,049,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267862929.10/warc/CC-MAIN-20180619115101-20180619135101-00168.warc.gz | 255,171,785 | 4,569 | Online JudgeProblem SetAuthorsOnline ContestsUser
Web Board
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Submit Problem
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Prob.ID:
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Language:
Data Mining
Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 382 Accepted: 105
Description
Dr. Tuple is working on the new data-mining application for Advanced Commercial Merchandise Inc. One of the subroutines for this application works with two arrays P and Q containing N records of data each (records are numbered from 0 to N - 1). Array P contains hash-like structure with keys. Array P is used to locate records for processing and the data for the corresponding record is later retrieved from the array Q.
All records in array P have a size of SP bytes and records in array Q have size of SQ bytes. Dr. Tuple needs to implement this subroutine with the highest possible performance because it is a hot-spot of the whole data-mining application. However, SP and SQ are only known at run-time of application which complicates or makes impossible to make certain well-known compile-time optimizations.
The straightforward way to find byte-offset of i-th record in array P is to use the following formula:
Pofs(i) = SP * i, (1)
and the following formula for array Q:
Qofs(i) = SQ * i. (2)
However, multiplication computes much slower than addition or subtraction in modern processors. Dr. Tuple avoids usage of multiplication while scanning array P by keeping computed byte-offset Pofs(i) of i-th record instead of its index i in all other data-structures of data-mining application. He uses the following simple formulae when he needs to compute byte-offset of the record that precedes or follows i-th record in array P:
Pofs(i + 1) = Pofs(i) + SP
Pofs(i - 1) = Pofs(i) - SP
Whenever a record from array P is located by either scanning of the array or by taking Pofs(i) from other data structures, Dr. Tuple needs to retrieve information from the corresponding record in array Q. To access record in array Q its byte-offset Qofs(i) needs to be computed. One can immediately derive formula to compute Qofs(i) with known Pofs(i) from formulae (1) and (2):
Qofs(i) = Pofs(i) / SP * SQ (3)
Unfortunately, this formula not only contains multiplication, but also contains division. Even though only integer division is required here, it is still an order of magnitude slower than multiplication on modern processors. If coded this way, its computation is going to consume the most of CPU time in data-mining application for ACM Inc.
After some research Dr. Tuple has discovered that he can replace formula (3) with the following fast formula:
Qofs'(i) = (Pofs(i) + Pofs(i) << A) >> B (4)
where A and B are non-negative integer numbers, "<< A" is left shift by A bits (equivalent to integer multiplication by 2A ), " >> B" is right shift by B bits (equivalent to integer division by 2B ).
This formula is an order of magnitude faster than (3) to compute, but it generally cannot always produce the same result as (3) regardless of the choice for values of A and B. It still can be used if one is willing to sacrifice some extra memory.
Conventional layout of array Q in memory (using formula (2)) requires N * SQ bytes to store the entire array. Dr. Tuple has found that one can always choose such K that if he allocates K bytes of memory for the array Q (where K >= N * SQ ) and carefully selects values for A and B, the fast formula (4) will give non-overlapping storage locations for each of the N records of array Q.
Your task is to write a program that finds minimal possible amount of memory K that needs to be allocated for array Q when formula (4) is used. Corresponding values for A and B are also to be found. If multiple pairs of values for A and B give the same minimal amount of memory K, then the pair where A is minimal have to be found, and if there is still several possibilities, the one where B is minimal. You shall assume that integer registers that will be used to compute formula (4) are wide enough so that overflow will never occur.
Input
The input consists of three integer numbers N , SP, and SQ separated by spaces (1 <= N <= 220, 1 <= SP <= 210, 1 <= SQ <= 210).
Output
Write to the output file a single line with three integer numbers K, A, and B separated by spaces.
Sample Input
`1024 7 1`
Sample Output
```1119 2 5
```
Source
[Submit] [Go Back] [Status] [Discuss] | 1,048 | 4,473 | {"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-2018-26 | longest | en | 0.885298 |
https://www.varsitytutors.com/high_school_math-help/understanding-angles-in-the-unit-circle | 1,669,530,073,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00606.warc.gz | 1,126,709,120 | 57,978 | # High School Math : Understanding Angles in the Unit Circle
## Example Questions
### Example Question #52 : The Unit Circle And Radians
In the unit circle, what is the angle in radians that corresponds to the point (0, -1)?
Explanation:
On the unit circle, (0,-1) is the point that falls between the third and fourth quadrant. This corresponds to .
### Example Question #51 : The Unit Circle And Radians
What is the reference angle for ? | 104 | 446 | {"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} | 2.921875 | 3 | CC-MAIN-2022-49 | longest | en | 0.77806 |
https://secondsminutes.com/377625-minutes-in-seconds | 1,680,071,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00320.warc.gz | 567,410,799 | 7,038 | # 377625 minutes in seconds
## Result
377625 minutes equals 22657500 seconds
## Conversion formula
Multiply the amount of minutes by the conversion factor to get the result in seconds:
377625 min × 60 = 22657500 s
## How to convert 377625 minutes to seconds?
The conversion factor from minutes to seconds is 60, which means that 1 minutes is equal to 60 seconds:
1 min = 60 s
To convert 377625 minutes into seconds we have to multiply 377625 by the conversion factor in order to get the amount from minutes to seconds. We can also form a proportion to calculate the result:
1 min → 60 s
377625 min → T(s)
Solve the above proportion to obtain the time T in seconds:
T(s) = 377625 min × 60 s
T(s) = 22657500 s
The final result is:
377625 min → 22657500 s
We conclude that 377625 minutes is equivalent to 22657500 seconds:
377625 minutes = 22657500 seconds
## Result approximation:
For practical purposes we can round our final result to an approximate numerical value. In this case three hundred seventy-seven thousand six hundred twenty-five minutes is approximately twenty-two million six hundred fifty-seven thousand five hundred seconds:
377625 minutes ≅ 22657500 seconds
## Conversion table
For quick reference purposes, below is the minutes to seconds conversion table:
minutes (min) seconds (s)
377626 minutes 22657560 seconds
377627 minutes 22657620 seconds
377628 minutes 22657680 seconds
377629 minutes 22657740 seconds
377630 minutes 22657800 seconds
377631 minutes 22657860 seconds
377632 minutes 22657920 seconds
377633 minutes 22657980 seconds
377634 minutes 22658040 seconds
377635 minutes 22658100 seconds
## Units definitions
The units involved in this conversion are minutes and seconds. This is how they are defined:
### Minutes
The minute is a unit of time or of angle. As a unit of time, the minute (symbol: min) is equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). As a unit of angle, the minute of arc is equal to 1⁄60 of a degree, or 60 seconds (of arc). Although not an SI unit for either time or angle, the minute is accepted for use with SI units for both. The SI symbols for minute or minutes are min for time measurement, and the prime symbol after a number, e.g. 5′, for angle measurement. The prime is also sometimes used informally to denote minutes of time. In contrast to the hour, the minute (and the second) does not have a clear historical background. What is traceable only is that it started being recorded in the Middle Ages due to the ability of construction of "precision" timepieces (mechanical and water clocks). However, no consistent records of the origin for the division as 1⁄60 part of the hour (and the second 1⁄60 of the minute) have ever been found, despite many speculations.
### Seconds
The second (symbol: s) (abbreviated s or sec) is the base unit of time in the International System of Units (SI). It is qualitatively defined as the second division of the hour by sixty, the first division by sixty being the minute. The SI definition of second is "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom". Seconds may be measured using a mechanical, electrical or an atomic clock. SI prefixes are combined with the word second to denote subdivisions of the second, e.g., the millisecond (one thousandth of a second), the microsecond (one millionth of a second), and the nanosecond (one billionth of a second). Though SI prefixes may also be used to form multiples of the second such as kilosecond (one thousand seconds), such units are rarely used in practice. The more common larger non-SI units of time are not formed by powers of ten; instead, the second is multiplied by 60 to form a minute, which is multiplied by 60 to form an hour, which is multiplied by 24 to form a day. The second is also the base unit of time in other systems of measurement: the centimetre–gram–second, metre–kilogram–second, metre–tonne–second, and foot–pound–second systems of units. | 1,035 | 4,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-14 | latest | en | 0.792588 |
https://juicystudio.com/services/readability.php?url=https://probablyasd.com | 1,656,426,448,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103556871.29/warc/CC-MAIN-20220628142305-20220628172305-00245.warc.gz | 374,720,542 | 4,977 | ## Contents
Gunning Fog, Flesch Reading Ease, and Flesch-Kincaid are reading level algorithms that can be helpful in determining how readable your content is. Reading level algorithms only provide a rough guide, as they tend to reward short sentences made up of short words. Whilst they're rough guides, they can give a useful indication as to whether you've pitched your content at the right level for your intended audience.
## Interpreting the Results
This service analyses the readability of all rendered content. Unfortunately, this will include navigation items, and other short items of content that do not make up the part of the page that is intended to be the subject of the readability test. These items are likely to skew the results. The difference will be minimal in situations where the copy content is much larger than the navigation items, but documents with little content but lots of navigation items will return results that aren't correct.
Philip Chalmers of Benefit from IT provided the following typical Fog Index scores, to help ascertain the readability of documents.
Typical Fog Index Scores
Fog Index Resources
6 TV guides, The Bible, Mark Twain
8 - 10 Most popular novels
10 Time, Newsweek
11 Wall Street Journal
14 The Times, The Guardian
Over 20 Only government sites can get away with this, because you can't ignore them.
Over 30 The government is covering something up
The following table contains the readability results for https://probablyasd.com.
Summary Value
Total sentences 135
Total words 1045
Average words per Sentence 7.74
Words with 1 Syllable 794
Words with 2 Syllables 96
Words with 3 Syllables 45
Words with 4 or more Syllables 110
Percentage of word with three or more syllables 14.83%
Average Syllables per Word 1.49
Gunning Fog Index 9.03
## Gunning-Fog Index
The following is the algorithm to determine the Gunning-Fog index.
• Calculate the average number of words you use per sentence.
• Calculate the percentage of difficult words in the sample (words with three or more syllables).
• Add the totals together, and multiply the sum by 0.4.
• Algorithm: (average_words_sentence + number_words_three_syllables_plus) * 0.4
The result is your Gunning-Fog index, which is a rough measure of how many years of schooling it would take someone to understand the content. The lower the number, the more understandable the content will be to your visitors. Results over seventeen are reported as seventeen, where seventeen is considered post-graduate level.
The following is the algorithm to determine the Flesch Reading Ease.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of syllables per word multiplied by 84.6 and subtract it from the average number of words multiplied by 1.015.
• Subtract the result from 206.835.
• Algorithm: 206.835 - (1.015 * average_words_sentence) - (84.6 * average_syllables_word)
The result is an index number that rates the text on a 100-point scale. The higher the score, the easier it is to understand the document. Authors are encouraged to aim for a score of approximately 60 to 70.
The following is the algorithm to determine the Flesch-Kincaid grade level.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of words by 0.39 and add it to the average number of syllables per word multiplied by 11.8.
• Subtract 15.50 from the result.
• Algorithm: (0.39 * average_words_sentence) + (11.8 * average_syllables_word) - 15.9
The result is the Flesch-Kincaid grade level. Like the Gunning-Fog index, it is a rough measure of how many years of schooling it would take someone to understand the content. Negative results are reported as zero, and numbers over twelve are reported as twelve. | 880 | 3,874 | {"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-2022-27 | latest | en | 0.8847 |
https://wenshengchen.com/2019/08/29/747-largest-numer-at-least-twice-of-others | 1,719,282,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00537.warc.gz | 547,160,246 | 4,899 | In a given integer array nums, there is always exactly one largest element.
Find whether the largest element in the array is at least twice as much as every other number in the array.
If it is, return the index of the largest element, otherwise return -1.
Example 1:
Input: nums = [3, 6, 1, 0]
Output: 1
Explanation: 6 is the largest integer, and for every other number in the array x, 6 is more than twice as big as x. The index of value 6 is 1, so we return 1.
Example 2:
Input: nums = [1, 2, 3, 4]
Output: -1
Explanation: 4 isn’t at least as big as twice the value of 3, so we return -1.
Note:
1. nums will have a length in the range [1, 50].
2. Every nums[i] will be an integer in the range [0, 99].
## Solution
from typing import List
class Solution:
def dominantIndex(self, nums: List[int]) -> int:
largestNumIndex = -1
largestNum = 0
secondLargestNum = 0
for index, num in enumerate(nums):
if num >= largestNum:
secondLargestNum = largestNum
largestNum = num
largestNumIndex = index
elif num > secondLargestNum:
secondLargestNum = num
if largestNum >= secondLargestNum * 2:
return largestNumIndex
return -1
## Test Cases
test = Solution() | 338 | 1,166 | {"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} | 3.6875 | 4 | CC-MAIN-2024-26 | latest | en | 0.698703 |
https://hackage-origin.haskell.org/package/kan-extensions-5.2.2/docs/Data-Functor-Day-Curried.html | 1,722,669,530,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00196.warc.gz | 227,059,410 | 3,926 | kan-extensions-5.2.2: Kan extensions, Kan lifts, the Yoneda lemma, and (co)density (co)monads
Copyright 2013-2016 Edward Kmett and Dan Doel BSD Edward Kmett experimental rank N types Safe-Inferred Haskell2010
Data.Functor.Day.Curried
Description
Day f -| Curried f
Day f ~ Compose f when f preserves colimits / is a left adjoint. (Due in part to the strength of all functors in Hask.)
So by the uniqueness of adjoints, when f is a left adjoint, Curried f ~ Rift f
Synopsis
# Right Kan lifts
newtype Curried g h a Source #
Constructors
Curried FieldsrunCurried :: forall r. g (a -> r) -> h r
#### Instances
Instances details
Functor g => Functor (Curried g h) Source # Instance detailsDefined in Data.Functor.Day.Curried Methodsfmap :: (a -> b) -> Curried g h a -> Curried g h b #(<\$) :: a -> Curried g h b -> Curried g h a # (Functor g, g ~ h) => Applicative (Curried g h) Source # Instance detailsDefined in Data.Functor.Day.Curried Methodspure :: a -> Curried g h a #(<*>) :: Curried g h (a -> b) -> Curried g h a -> Curried g h b #liftA2 :: (a -> b -> c) -> Curried g h a -> Curried g h b -> Curried g h c #(*>) :: Curried g h a -> Curried g h b -> Curried g h b #(<*) :: Curried g h a -> Curried g h b -> Curried g h a #
toCurried :: (forall x. Day g k x -> h x) -> k a -> Curried g h a Source #
The universal property of Curried
fromCurried :: Functor f => (forall a. k a -> Curried f h a) -> Day f k b -> h b Source #
toCurried . fromCurried ≡ id
fromCurried . toCurried ≡ id
applied :: Functor f => Day f (Curried f g) a -> g a Source #
This is the counit of the Day f -| Curried f adjunction
unapplied :: g a -> Curried f (Day f g) a Source #
This is the unit of the Day f -| Curried f adjunction
adjointToCurried :: Adjunction f u => u a -> Curried f Identity a Source #
Curried f Identity a is isomorphic to the right adjoint to f if one exists.
adjointToCurried . curriedToAdjoint ≡ id
curriedToAdjoint . adjointToCurried ≡ id
curriedToAdjoint :: Adjunction f u => Curried f Identity a -> u a Source #
Curried f Identity a is isomorphic to the right adjoint to f if one exists.
composedAdjointToCurried :: (Functor h, Adjunction f u) => u (h a) -> Curried f h a Source #
Curried f h a is isomorphic to the post-composition of the right adjoint of f onto h if such a right adjoint exists.
curriedToComposedAdjoint :: Adjunction f u => Curried f h a -> u (h a) Source #
Curried f h a is isomorphic to the post-composition of the right adjoint of f onto h if such a right adjoint exists.
curriedToComposedAdjoint . composedAdjointToCurried ≡ id
composedAdjointToCurried . curriedToComposedAdjoint ≡ id
liftCurried :: Applicative f => f a -> Curried f f a Source #
The natural isomorphism between f and Curried f f. lowerCurried . liftCurried ≡ id liftCurried . lowerCurried ≡ id
lowerCurried (liftCurried x) -- definition
lowerCurried (Curried (<*> x)) -- definition
(<*> x) (pure id) -- beta reduction
pure id <*> x -- Applicative identity law
x
lowerCurried :: Applicative f => Curried f g a -> g a Source #
Lower Curried by applying pure id to the continuation.
See liftCurried.
rap :: Functor f => Curried f g (a -> b) -> Curried g h a -> Curried f h b Source #
Indexed applicative composition of right Kan lifts. | 988 | 3,300 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.689758 |
http://upuzzledme.com/userProfilePage?puzzlePage=10&ansPage=1&pageConn=&user=admin | 1,531,730,295,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589237.16/warc/CC-MAIN-20180716080356-20180716100356-00425.warc.gz | 369,308,098 | 15,466 | "Life is puzzle!!!"
Pune ,India
A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest. What percent of his apples does the vendor throw?
A. 17 B. 23 C. 77 D. None of these
A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest. What percent of his apples does the vendor throw?
A. 17 B. 23 C. 77 D. None of these
There are 9 coins, out of which one is odd, i.e. its weight is either less or more than that of the other 8 coins. How many iterations of weighing using a pan balance are required to find the odd coin and to find whether it is heavier or lighter?
A. 2 B. 3 C. 4 D. 5 E. It is not possible to find out the odd weight coin
P and Q can complete a work in 20 days and 12 days respectively. P alone started the work and Q joined him after 4 days till the completion of the work. How long did the work last? A. 5 days B. 10 days C. 14 days D. 22 days
A ship develops a leak 12 km from the shore. Despite the leak, the ship is able to move towards the shore at a speed of 8 km/hr. However, the ship can stay afloat only for 20 minutes. If a rescue vessel were to leave from the shore towards the ship and it takes 4 minutes to evacuate the crew and passengers of the ship, what should be the minimum speed of the rescue vessel in order to be able to successfully rescue the people aboard the ship?
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?
A. 43 units B. 8 units C. 12 units D. Data inadequate E. None of these
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 173 m B. 200 m C. 273 m D. 300 m
Look at this series: 7, 10, 8, 11, 9, 12, ... What number should come next?
A. 7 B. 10 C. 12 D. 13
Look at this series: 2, 1, (1/2), (1/4), ... What number should come next?
A. (1/3) B. (1/8) C. (2/8) D. (1/16)
If A + B means A is the mother of B; A - B means A is the brother B; A % B means A is the father of B and A x B means A is the sister of B, which of the following shows that P is the maternal uncle of Q? A.Q - N + M x P
B. P + S x N - Q
C. P - M + N x Q
D. Q - S % P
What is the next number in the sequence?
61 57 50 61 43 36 61 ? ? | 791 | 2,733 | {"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} | 3.578125 | 4 | CC-MAIN-2018-30 | latest | en | 0.963327 |
https://briangordon.github.io/2014/06/sqrts-and-fixed-points.html | 1,511,150,681,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805911.18/warc/CC-MAIN-20171120032907-20171120052907-00597.warc.gz | 576,476,538 | 6,862 | # Square roots and fixed points
math programming 02 June 2014
There’s an old comment on HN that annoys me. In it, a developer whines bitterly and eloquently about his experience being stumped by a common interview question: “write an algorithm for computing the square root of a number.”
I tried a couple more times to answer this question that is completely unrelated to the job I’m interviewing for. Finally, I get so fed up with this moron that my internal ticker clicks over and I realize, “Even if I get this job, I’m going to be dealing with these kinds of nazel-gazing engineers every single day. Not an environment I want to be in.”
“Completely unrelated?” The problem is easily solved with binary search; is it inconceivable that a frontend developer would ever use binary search? For that matter, is this interview for a contract to build a single front-end, or for a full-time developer position which may continue indefinitely? Because if it’s the latter, general problem solving questions are completely fair game. The company is going to need to make an investment of institutional knowledge in a new developer. If the developer can pivot to a new role as needed then the company may be able to avoid hiring someone new in the future and save the cost of bringing another employee up to speed. There is therefore legitimate value added by being able to tackle problems outside of your comfort zone. Petulantly hanging up on the interviewer and deciding he’s not worth your time for asking such a question is childish.
Anyway, other than the obvious solution of binary search, how might we solve the problem of calculating $\sqrt{a}$? Let’s first introduce the concept of a fixed point. A fixed point is a value which is unchanged by the function - that is, $f(p) = p$. For example, a fixed point of the sine function is 0 because $\sin(0)=0$. A fixed point of the cosine function is located around 0.739085133 because $\cos(0.739085133)\approx0.739085133$. In fact, if we plot the cosine function on top of $f(p) = p$ then we can see that they intersect at exactly that point:
One interesting fact which might be of use is that $\sqrt{a}$ is a fixed point of the function $f(x)=\frac{a}{x}$:
We can therefore graph $f(x)=\frac{a}{x}$ to find its fixed point $\sqrt{a}$. Here’s a plot for various values of a:
By subtracting x from the function it becomes a root finding problem:
This we can solve with Newton’s method:
Of course, we could have just used Newton’s method from the beginning. If we’re trying to calculate $\sqrt{a}$ then the value we’re looking for is a zero of the function $f(x)=x^{2}-a$. In this case we have:
These are different functions but they have the same fixed point. This is a plot of $x_{n+1} - x_n$ for the two different functions (with a=7) showing that they have the same zeros:
Newton’s method corresponds closely to a concept called fixed point iteration. In fixed point iteration, we repeatedly evaluate a function and feed its output back into its input. Eventually we hope it will converge on a fixed point. We mentioned the cosine function earlier; this is one such function which always converges on a particular fixed point. Enter any value into a scientific calculator and repeatedly press the cosine button:
-1.0000, 0.5403, 0.8575, 0.6542, 0.7934, 0.7013, 0.7639, 0.7221, 0.7504, 0.7314, 0.7442, 0.7356
It will eventually converge to a value near 0.7391:
Another function which converges to a fixed point is the expression given by Newton’s method applied to the function $f(x)=x^2-a$, which we derived before:
If we rearrange the right side we obtain:
This is called the Babylonian method, or Heron’s method, and it was actually discovered long before Newton’s method. The idea was that if $x_n$ is an overestimate to $\sqrt{a}$ then $\frac{a}{x_n}$ is an underestimate, and you can average them to get a better approximation. Repeated iteration improves this approximation until it converges on $\sqrt{x}$, which we know is a fixed point.
Remember that before we said $\sqrt{a}$ is a fixed point of the function $f(x)=\frac{a}{x_n}$. Unfortunately if you iterate that function, you will not approach $\sqrt{a}$. Fixed point iteration doesn’t always work and this is one such case. The math behind being able to tell whether an arbitrary function will converge to a fixed point under fixed point iteration is complicated.
Now it would be very simple to wrap the Babylonian method in a loop and perform a couple steps of fixed point iteration to get a decent sqrt(a). But since we’re finding a fixed point, this seems like a nice time to break out something called a fixed point combinator. The best-known fixed point combinator is the Y combinator. You’ve probably heard of it due to the eponymous startup incubator founded by Lisp greybeard Paul Graham, of the famous Paul Graham essays.
This is the lambda calculus definition of the Y combinator, due to Haskell Curry:
The reason y is called a fixed-point combinator is because of what happens when you apply it to a function and reduce. By following the lambda calculus reduction rules you can find that the Y combinator satisfies the equation $y\ f = f\ (y\ f)$. This matches the form $something = f\ (something)$ - the definition of a fixed point. So, $y\ f$ is a fixed point of $f$! Therefore all we have to do is apply the Y combinator to obtain a fixed point of $f$, right?
Well, not really. $f$ takes and returns a function so the fixed point of $f$ isn’t a number at all, it’s the function $f'$ which $f$ maps to $f'$. In other words, all the Y combinator is doing is facilitating fixed-point iteration:
It appears that this expansion will continue forever and never terminate. But we can build a termination condition into the function $f$ so that it stops expanding. Let’s see how that would work with the sqrt example. Our $f$ could look like this in JavaScript:
function step (callback) {
return function (originalValue, approxSqrt) {
// Babylonian method formula to improve our approximation of the square root
// of originalValue.
var improvedApproxSqrt = (approxSqrt + (originalValue / approxSqrt)) / 2;
// How far off the mark we are.
var discrepancy = Math.abs(originalValue -
(improvedApproxSqrt * improvedApproxSqrt));
// Termination condition
if(discrepancy < 0.00001) {
return improvedApproxSqrt;
}
return callback(originalValue, improvedApproxSqrt);
};
}
This looks a lot like recursion, except that we’ve never referred to an free variable name. This is called anonymous recursion, and it’s useful in systems (notably lambda calculus) where functions cannot refer to themselves by name.
Now we need the magic which will repeatedly invoke step and feed its output back into its input: the Y combinator. How would it look in JavaScript? Here’s the Y combinator in lambda calculus again:
This corresponds pretty directly to a JavaScript function:
function Y(f) {
return (function (x) {
return f(x(x));
})(function (x) {
return f(x(x));
});
}
Unfortunately, when we try to use this version of the Y combinator, we get a stack overflow. If you trace out the execution you’ll see that x(x) must be evaluated in order to get a final return value for Y, and this causes infinite recursion. The reason this works at all in lambda calculus is that lambda calculus is call by name so $f (x\ x)$ is evaluated by expanding the definition of $x\ x$ and passing that function to f. JavaScript, on the other hand, is call by value, so the x function is actually evaluated with x as an argument.
If we η-reduce the Y combinator then we obtain an alternate fixed-point combinator, called the Z combinator, which contains an extra layer of indirection and prevents runaway recursion:
Here’s the JavaScript code corresponding to the Z combinator:
function Z(f) {
return (function (x) {
return f(function (v) { return x(x)(v); });
})(function (x) {
return f(function (v) { return x(x)(v); });
});
}
We still have a bit of a problem: our step function takes two variables, while Z only calls it with one. So let’s modify Z:
function Z(f) {
return (function (x) {
return f(function (v1, v2) { return x(x)(v1, v2); });
})(function (x) {
return f(function (v1, v2) { return x(x)(v1, v2); });
});
}
Now we can see it working:
> var sqrt = Z(step);
> sqrt(2, 1); // sqrt of 2, with a starting estimate of 1
1.4142156862745097
It’s rather awkward to always have to provide a starting estimate, so we can add a wrapper which always guesses 1 to start:
> function sqrt (num) { return Z(step)(num, 1); }
> sqrt(2)
1.4142156862745097 | 2,060 | 8,578 | {"found_math": true, "script_math_tex": 34, "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} | 3.84375 | 4 | CC-MAIN-2017-47 | longest | en | 0.949331 |
http://www.nap.edu/openbook.php?record_id=9822&page=137 | 1,369,157,986,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700264179/warc/CC-MAIN-20130516103104-00054-ip-10-60-113-184.ec2.internal.warc.gz | 617,198,025 | 10,458 | Questions? Call 800-624-6242
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## Adding It Up: Helping Children Learn Mathematics (2001) Center for Education (CFE)
### Citation Manager
. "4 The Strands of Mathematical Proficiency." Adding It Up: Helping Children Learn Mathematics. Washington, DC: The National Academies Press, 2001.
Page 137
The following HTML text is provided to enhance online readability. Many aspects of typography translate only awkwardly to HTML. Please use the page image as the authoritative form to ensure accuracy.
Adding + It Up: Helping Children Learn Mathematics
The same is true for rational numbers. Only 35% of 13-year-olds correctly ordered three fractions, all in reduced form,56 and only 35%, asked for a number between .03 and .04, chose the correct response.57 These findings suggest that students may be calculating with numbers that they do not really understand.
#### Procedural Fluency
An overall picture of procedural fluency is provided by the NAEP long-term trend mathematics assessment,58 which indicates that U.S. students’ performance has remained quite steady over the past 25 years (see Box 4–4). A closer look reveals that the picture of procedural fluency is one of high levels of proficiency in the easiest contexts. Questions in which students are asked to add or subtract two- and three-digit whole numbers presented numerically in the standard format are answered correctly by about 90% of 13-year-olds, with almost as good performance among 9-year-olds.59 Performance is slightly lower among 13-year-olds for division.60
Box 4–4 NAEP Scale Scores, Long-Term Trend Assessment, 1973–1999 SOURCE: Campbell, Hombo, and Mazzeo, 2000, p. 9. These scale scores include all content areas: number, geometry, algebra, and so on.
Page 137
Front Matter (R1-R20) Executive Summary (1-14) 1 Looking at Mathematics and Learning (15-30) 2 The State of School Mathematics in the United States (31-70) 3 Number: What Is There to Know? (71-114) 4 The Strands of Mathematical Proficiency (115-156) 5 The Mathematical Knowledge Children Bring to School (157-180) 6 Developing Proficiency with Whole Numbers (181-230) 7 Developing Proficiency with Other Numbers (231-254) 8 Developing Mathematical Proficiency Beyond Number (255-312) 9 Teaching for Mathematical Proficiency (313-368) 10 Developing Proficiency in Teaching Mathematics (369-406) 11 Conclusions and Recommendations (407-432) Biographical Sketches (433-440) Index (441-454) | 586 | 2,483 | {"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-2013-20 | longest | en | 0.896447 |
https://www.rdocumentation.org/packages/stats/versions/3.2.2/topics/varimax | 1,611,812,055,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00759.warc.gz | 945,645,278 | 6,098 | # varimax
0th
Percentile
Keywords
multivariate
##### Usage
varimax(x, normalize = TRUE, eps = 1e-5)
promax(x, m = 4)
##### Arguments
x
A loadings matrix, with $p$ rows and $k < p$ columns
m
The power used the target for promax. Values of 2 to 4 are recommended.
normalize
logical. Should Kaiser normalization be performed? If so the rows of x are re-scaled to unit length before rotation, and scaled back afterwards.
eps
The tolerance for stopping: the relative change in the sum of singular values.
##### Details
These seek a ‘rotation’ of the factors x %*% T that aims to clarify the structure of the loadings matrix. The matrix T is a rotation (possibly with reflection) for varimax, but a general linear transformation for promax, with the variance of the factors being preserved.
##### Value
A list with components
The ‘rotated’ loadings matrix, x %*% rotmat, of class "loadings".
rotmat
The ‘rotation’ matrix.
##### References
Hendrickson, A. E. and White, P. O. (1964) Promax: a quick method for rotation to orthogonal oblique structure. British Journal of Statistical Psychology, 17, 65--70.
Horst, P. (1965) Factor Analysis of Data Matrices. Holt, Rinehart and Winston. Chapter 10.
Kaiser, H. F. (1958) The varimax criterion for analytic rotation in factor analysis. Psychometrika 23, 187--200.
Lawley, D. N. and Maxwell, A. E. (1971) Factor Analysis as a Statistical Method. Second edition. Butterworths.
factanal, Harman74.cor.
library(stats) ## varimax with normalize = TRUE is the default fa <- factanal( ~., 2, data = swiss) varimax(loadings(fa), normalize = FALSE) promax(loadings(fa)) | 439 | 1,612 | {"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} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.634564 |
http://clay6.com/qa/15691/mole-fraction-of-i-2-in-c-6h-6-is-0-2-the-molality-of-i-2-in-c-6h-6-is | 1,481,422,751,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00242-ip-10-31-129-80.ec2.internal.warc.gz | 55,229,035 | 27,370 | Browse Questions
# Mole fraction of $I_2$ in $C_6H_6$ is 0.2.The molality of $I_2$ in $C_6H_6$ is
$\begin{array}{1 1}(a)\;3.2&(b)\;6.4\\(c)\;1.6&(d)\;2.3\end{array}$
$\large\frac{n}{n+N}=$$0.2 \large\frac{N}{n+N}$$=0.8$
Or $\large\frac{n}{N}=\frac{1}{4}$
$\large\frac{nI_2}{WC_6H_6}$$\times MC_6H_6=\large\frac{1}{4} \thereforeMolality =\large\frac{nI_2}{wC_6H_6}$$\times 1000$
$\qquad\qquad=\large\frac{1}{4}\times\frac{1000}{MC_6H_6}$
Molality =$\large\frac{1}{4}\times \frac{1000}{78}$
$\qquad\;\;\;\;=3.2$
Hence (a) is the correct answer. | 279 | 545 | {"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} | 3.625 | 4 | CC-MAIN-2016-50 | longest | en | 0.286353 |
https://brilliant.org/problems/another-problem-using-herons/ | 1,627,969,035,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00534.warc.gz | 155,378,852 | 8,588 | # Another Problem using Heron's
Geometry Level 3
Let $\triangle ABC$ have a circle inscribed inside of it such that the circle is tangent at points $X$, $Y$, and $Z$ along sides $AB$, $AC$, and $BC$, respectively. Let $AX=3$, $BZ=4$, and $CY=5$. Find the area of $\triangle ABC$.
If your answer can be expressed as $a\sqrt{b}$ where $a$ and $b$ are positive integers with $b$ squarefree, find $a+b$.
× | 125 | 405 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "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} | 3.15625 | 3 | CC-MAIN-2021-31 | latest | en | 0.832265 |
https://dict.youdao.com/w/eng/power-factor/ | 1,601,386,998,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401643509.96/warc/CC-MAIN-20200929123413-20200929153413-00014.warc.gz | 379,870,764 | 10,689 | go top
#### 您要找的是不是:
power factor 功率因数；功率系数
## power factor 英
• [电] 功率因数；功率系数
### 网络释义专业释义英英释义
功率因数
AT89C52单片机对电力系统中低压无功功率进行监测和控制（毕业设计）_论文 关键词： 功率因数 补偿器 电容器 AT89C52 [gap=1361]Keywords： AT89C52 Power-factor Compensator Capacitor
power-factor compensation 功率因数的补偿 ; 功率因数补偿 ; 相位的补偿 ; 功率系数的补偿
power-factor compensator 功率因数补偿器
power-factor measurement 功率因数测量 ; 功率因素测量
power-factor meter 功率因数计英语
power-limit factor 功率极限因数
power-factor regulating 功率因数调整
更多收起网络短语
• 功率因数 - 引用次数：67
A high precision method of power-factor measurement is researched,and a reactive power compensation control system to low voltage network with capacitor bank is realized by sing-chip computer 8031.
提出一种精度较高的功率因数测量方法。
参考来源 - 电网功率因数的测量及无功自动补偿控制 Power
• 功率因数 - 引用次数：4
According to the principle of "compensation graded,and keeping balance on the spot" to deal with the reactive power,we could improve the power-factor on the user side and encourage the clients to compensate the reactive power on the spot themselves.
按照无功功率“分级补偿,就地平衡”的原则,改善用户侧功率因数,提高用户无功功率的自给能力及水平,激励其进行就地补偿,是降低线损、提高电压质量、保证电网安全稳定运行的关键。
参考来源 - 功率因数调整电费模型与激励约束机制研究
·2,447,543篇论文数据，部分数据来源于NoteExpress
#### Power factor
• In competitive practical shooting, power factor refers to a ranking system for the momentum of pistol cartridges. The power factor is used in competitions sponsored by IPSC, USPSA, The Bianchi Cup, Steel Challenge, and IDPA.
### 柯林斯英汉双解大词典
#### power factor
• 1.
N (in an electrical circuit) the ratio of the power dissipated to the product of the input volts times amps (电路中)功率因数
### 双语例句权威例句
• This power-factor monitoring and compensation system was made up of electrical parameters detection system and compensation controlling system.
功率因数监测补偿系统参数检测系统补偿控制系统两部分组成。
• Among the three-phase power-factor correction circuits, the full decoupling circuit is especially suitable for high power fields because of its high input power and good electrical performance.
三相功率因数校正电路中，解耦型电路由于输入功率大，电气性能较好优势特别适用于大功率场合。
• A 300w experimental prototype of power supply with power-factor-correction function based on dual-loop control strategy is presented in this paper.
基于电流电压的双环控制策略，研制了台输出功率300w功率因数校正电源样机
\$firstVoiceSent
- 来自原声例句 | 804 | 2,213 | {"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-2020-40 | latest | en | 0.440236 |
http://www.thecodingforums.com/threads/c99-does-const-int-x-5-make-x-a-constant-expression.515017/ | 1,477,058,279,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718278.43/warc/CC-MAIN-20161020183838-00346-ip-10-171-6-4.ec2.internal.warc.gz | 733,639,816 | 13,051 | # (C99) Does "const int x=5;" make x a "constant expression"?
Discussion in 'C Programming' started by jaime, Jun 16, 2007.
1. ### jaimeGuest
Hi again all.
Given the line:
const int x=5;
Can I then use "x" as a constant expression? (By "constant expression", I
mean "constant expression" as defined in the C99 standard)
I've been searching google for 2 days now trying to answer this myself,
and I'm just getting more and more confused (some things I read make me
think "yes", while some things I read make me think "no").
I have many questions I'd like to ask on this topic, but rather than bore
you all rigid with the results of all of my research (points for and
against), I thought I'd just try this short question first.
Also, could anyone answering please give me an idea of how I can infer the
9899:TC2).
So, baffled once again, I humbly seek wise words from the council of
elders...
Ta, Jaime
jaime, Jun 16, 2007
2. ### CBFalconerGuest
jaime wrote:
>
> Given the line:
> const int x=5;
> Can I then use "x" as a constant expression? (By "constant
> expression", I mean "constant expression" as defined in the C99
> standard)
No. It is a constant object, not expression. I.E. it cannot be
altered.
--
<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>
cbfalconer at maineline dot net
--
Posted via a free Usenet account from http://www.teranews.com
CBFalconer, Jun 16, 2007
3. ### Joe WrightGuest
jaime wrote:
> Hi again all.
>
> Given the line:
> const int x=5;
> Can I then use "x" as a constant expression? (By "constant expression", I
> mean "constant expression" as defined in the C99 standard)
>
> I've been searching google for 2 days now trying to answer this myself,
> and I'm just getting more and more confused (some things I read make me
> think "yes", while some things I read make me think "no").
>
> I have many questions I'd like to ask on this topic, but rather than bore
> you all rigid with the results of all of my research (points for and
> against), I thought I'd just try this short question first.
>
> Also, could anyone answering please give me an idea of how I can infer the
> answer by reading the "Standard" (which I _think_ is currently ISO/IEC
> 9899:TC2).
>
> So, baffled once again, I humbly seek wise words from the council of
> elders...
>
> Ta, Jaime
No. const != constant in C. In your example..
const int x = 5;
...x is const and 5 is constant. You can't use x where you need a constant.
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
Joe Wright, Jun 16, 2007
4. ### Ben BacarisseGuest
jaime <> writes:
> Hi again all.
>
> Given the line:
> const int x=5;
> Can I then use "x" as a constant expression? (By "constant expression", I
> mean "constant expression" as defined in the C99 standard)
Short: no.
Medium: not if you want portable code.
Long: read all of section 6.6 of the standard. The trouble is it says
what you *can* have. The value of a variable, const, or otherwise is
not one of these.
Section 6.6 does permit implementations to allow other forms, so you
might be able to do it, but your code will not be portable. Such am
implementation would be violating the spirit in which const was
introduced (to signify a read-only, run-time object).
--
Ben.
Ben Bacarisse, Jun 16, 2007
5. ### Keith ThompsonGuest
jaime <> writes:
> Hi again all.
>
> Given the line:
> const int x=5;
> Can I then use "x" as a constant expression? (By "constant expression", I
> mean "constant expression" as defined in the C99 standard)
>
> I've been searching google for 2 days now trying to answer this myself,
> and I'm just getting more and more confused (some things I read make me
> think "yes", while some things I read make me think "no").
[...]
No, "const" in C really means "read-only", not "constant".
expression in C++</OT>.
--
Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Keith Thompson, Jun 16, 2007 | 1,156 | 4,237 | {"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-2016-44 | latest | en | 0.918377 |
http://www.differencebetween.com/difference-between-atomic-radius-and-vs-ionic-radius/ | 1,387,552,903,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345772708/warc/CC-MAIN-20131218054932-00049-ip-10-33-133-15.ec2.internal.warc.gz | 363,493,574 | 17,777 | • A A
We can define a radius for a circle or a ball. In that case, we say that the radius is the distance between the centre of the circle to a point in its circumference. Atoms and ions are also regarded as having a structure similar to a ball. Therefore, we can define a radius for them too. As in the general definition, for atoms and ions we say that the radius is the distance between the centre and the boundary. | 94 | 419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2013-48 | longest | en | 0.964169 |
http://forums.wolfram.com/mathgroup/archive/2010/Sep/msg00370.html | 1,601,389,127,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401643509.96/warc/CC-MAIN-20200929123413-20200929153413-00779.warc.gz | 55,962,976 | 8,087 | Map Correlation[] across sublists of matrices
• To: mathgroup at smc.vnet.net
• Subject: [mg112490] Map Correlation[] across sublists of matrices
• From: Garapata <warsaw95826 at mypacks.net>
• Date: Fri, 17 Sep 2010 06:41:39 -0400 (EDT)
```Hi everyone,
I have a matrix of times series data with the following dimensions:
Dimensions[timeSeriesMatrix]
{281, 5}
I want to partition the data into sliding windows of varying lengths
then find all the correlations of each window as it slides across the
data. To make this a bit clearer=E2=80=A6
I set up the following list to hold the window lengths:
window = {2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233}
(Nothing important about the Fibonacci ratio, just using it to look at
different chunks of data. Eventually I'll do it continuously.)
I then partition the data like this:
partitionedData = Partition[timeSeriesMatrix, #, 1] & /@ windows;
This gives me a list of sublists of matrices with the following
dimensions:
Dimensions[#] & /@ partitionedData
{{280, 2, 5}, {279, 3, 5}, {277, 5, 5}, {274, 8, 5}, {269, 13, 5},
{261, 21, 5}, {248, 34, 5}, {227, 55, 5}, {193, 89, 5}, {138, 144, 5},
{49, 233, 5}}
To explain a bit, the first sublist in partitionedData (dimensions
{280, 2, 5}) has 280 matrices each with a 2 data point "window" for 5
time series. The last matrix has 49 matrices each with 233 data
points for the 5 time series.
Now I need to map Correlation[] across each matrix in each sublist.
The following gets me part of the way there, but I don't think I've
thought about the problem in the right way yet:
Correlation[partitionedData[[#, 1]]] & /@ Range[Length[windows]];
=E2=80=A8Dimensions[%]
{11, 5, 5}
This above only gives me the first matrix in each sublist. I need
them all.
I think I've missed something simple. Any suggestions much
appreciated.
G
P.S., Once I have this I hope to wrap a manipulate around it with
controls to let me display and examine how the correlations change as
the different windows slide across the original time series data. So
if anyone has some terrific short cut to do it, let me know. | 629 | 2,108 | {"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-2020-40 | longest | en | 0.854975 |
https://blog.intzone.com/the-polya-kipling-method/ | 1,718,656,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00184.warc.gz | 123,316,830 | 23,155 | # The Pólya-Kipling Method
Had a chat recently with a colleague and the topic of Software Development Lifecycle (SDLC) came up. Whenever I talk about concepts or technical stuff with other people, I typically try to explain it in a way that is easy to understand for a non-technical person. It is a way of challenging myself cos if you can’t explain something simply, it means you do not understand it well enough. One of my favourite approaches is Concrete, Pictorial, Abstract – e.g. if explaining fractions to a kid, use a real pizza first (Concrete), followed by diagrams (Pictorial) and finally concepts (Abstract).
There’s a lot of good articles on SDLC, far more comprehensive than what I can muster, so I’m not going to try to come up with a new SDLC concept. Instead, I’m going to talk about a method I came up with, and try to apply it to SDLC. I’m sure others would have thought of something similar before, just that I’m too lazy to Google for related articles lol. I call this method the Pólya-Kipling Method.
Pólya – Understand, Plan, Do, Check
The first part of the method is named after George Pólya. He wrote a book published in 1945 titled How to Solve It which talks about a system of thinking which can help you solve any problem. This system is his famous Four Principles of Problem Solving – Understand, Plan, Do Check. One of my Math lecturers used to drum this mantra into our heads and it is still ringing in mine after more than a decade 😛
Kipling – 5W1H
The second part of the method is named after Rudyard Kipling. He wrote a book published in 1902 titled Just So Stories for Little Children. In it was a tale, “The Elephant’s Child”, which contains this poem:
I keep six honest serving-men
(They taught me all I knew);
Their names are What and Why and When
And How and Where and Who.
These are the Five Ws and One H (5W1H) frequently taught to students of English in primary schools – Who, What, When, Where, Why and How. While he may not have been the first one to come up with them, he certainly made them memorable 🙂
The Pólya-Kipling Method
This method is designed to be easy to remember and simple to apply. It has 4 main stages, which corresponds to Pólya’s Four Principles – Understand, Plan, Do, Check. In each stage, the Five Ws and One H are used as guiding questions. And now, let’s try to apply it to SDLC. The context is that of a new feature request for an existing software product. The software is split into several components (backend API, frontend UI, Android, iOS, etc.), each undertaken by a team of developers and a team lead.
Understand
Who: Who is involved? Client, Business Development, Project Manager.
What: What are the requirements?
When: When is this due?
Where: Where in the software are changes needed?
Why: Why is this feature needed?
How: How to measure results?
Plan
Who: Who is involved? Project Manager, Team Lead for each affected component.
What: What are the changes for each component?
When: When are the tasks for each component due? When to do testing and integration?
Where: Where in each component are changes needed?
Why: Why are the changes needed?
How: How will all the components integrate? Architecture, API, workflow.
Do
Who: Who is involved? Developers.
What: What are the tasks and tests for each component and developer?
When: When does each task need to be completed?
Where: Where in the code needs to be changed? Where to commit the code? This can also refer to which stage the task is at – research, in progress, review, QA, done.
Why: Why should a certain design or approach be adopted?
How: How best to code? How to commit code? Tests, best practices, PSR for PHP code, Git flow methodology for committing code, etc.
Check
Who: Who is involved? Team Lead for each component, Project Manager, Business Development, Client.
What: What to check?
When: When does each involved party come in to check?
Where: Where to check? Code, API request/response, UI.
Why: Why did this changed code fail?
How: How to check? Unit tests and linting checks by automated CI, code review by Team Lead, integration testing by Project Manager, user acceptance testing (UAT) by Business Development and Client.
Conclusion
An idea is still just an idea – how well it is implemented in real life is another matter. Obviously, this short exercise does not cover all the possible questions and answers, but hopefully it gives a glimpse on how this method may be used. Try it out today and have fun – adhuc! | 1,033 | 4,486 | {"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-2024-26 | latest | en | 0.955472 |
https://www.developerhelps.com/swap-two-numbers-using-bitwise-operator/ | 1,701,321,692,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00613.warc.gz | 822,601,358 | 17,490 | # How to Swap Two Numbers in JAVA?
In this tutorial we will learn how to swap two numbers in Java
swap two numbers in Java programming means swapping the values of two variables, which can be done using a temporary variable or without using a temporary variable. Swapping values is useful in programming, such as sorting or reordering elements in an array.
Example:
There are two variables m & n.
Value of x is β4β
Value of y is β5β.
Before Swapping: x value = 4; y value = 5
After Swapping: x value = 5; y value = 4
## How to Swap Two Numbers Using the third variable in JAVA
`````` import java.util.Scanner;
import java.util.NoSuchElementException;
public class MyClass {
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
try {
int m, n, temp;
Scanner s = new Scanner(System.in);
System.out.print("Enter the first number:");
m = s.nextInt();
System.out.print("Enter the second number:");
n = s.nextInt();
temp = m;
m = n;
n = temp;
System.out.println("After Swapping");
System.out.println("First number:"+m);
System.out.println("Second number:"+n);
}
catch (NoSuchElementException e) {
System.out.println("Type something in the Stdin box above....");
}
}
}``````
### Output :
``````Enter the first number: 25
Enter the second number: 37
After Swapping
First number: 37
Second number: 25
``````
### Complexities :
Time Complexity: O(1)
The time complexity of this program is O(1) since the number of operations performed is constant and does not depend on the size of the input.
Space Complexity: O(1)
The space complexity of this program is O(1) since only a constant amount of space is used to store the variables and the Scanner object.
## How to Swap two Numbers without using a third variable in JAVA
``````import java.util.Scanner;
public class Swap_Integers
{
public static void main(String args[])
{
int m, n;
Scanner s = new Scanner(System.in);
System.out.print("Enter the first number:");
m = s.nextInt();
System.out.print("Enter the second number:");
n = s.nextInt();
m = m + n;
n = m - n;
m = m - n;
System.out.println("After Swapping");
System.out.println("First number:"+m);
System.out.println("Second number:"+n);
}
}
``````
### Output :
``````Enter the first number: 4
Enter the second number: 5
After Swapping
First number: 5
Second number:4
``````
### Complexities :
Time Complexity: O(1)
The time complexity of the above program is constant time or O(1), as the number of operations performed does not change with the size of the input.
Space Complexity: O(1)
The space complexity of the program is O(1), as it only uses a constant amount of memory to store the input variables and the Scanner object.
## How to Swap Two Numbers Using Bitwise Operator in JAVA
Swap two numbers using bitwise operator There are a number of methods to swap two numbers using Java. Today we will focus on swapping those numbers using a bitwise operator in Java. In computers, arithmetic calculations such as addition, division, subtraction, multiplication etc are done at a bit level. To perform these calculations, we have bitwise operators such as OR(|), AND(&), XOR(^), complement(~), shift right (>>), shift left(<<) etc. Bitwise is an XOR operator that compares bits of two numbers. If the numbers are equal, it returns the output as 1. If the numbers are not equal, it returns the output as 0.
For example, if we take the binary output of say number 1 which is: 00000101 and binary output of number 5 is: 00001010. As we compare the outputs of these 2 binary numbers, we can conclude that they are not the same on comparison. Hence, the output for the above numbers will be 0. The other methods for swapping two numbers are by using the temp variable, using an array, using the third variable with the arithmetic operator, using temp with multiplication and division etc. Swapping is a process in which the values of two integer numbers are exchanged with each other.
Below is a program to understand how we can swap two numbers using Java. For this, we will use the nextInt() method of the Scanner class.
### Swap two numbers using bitwise operator
``````import java.util.Scanner;
public class DeveloperHelps
{
public static void main(String args[])
{
int num1, num2;
Scanner scanner = new Scanner(System.in);
System.out.print("Enter first number:");
num1 = scanner.nextInt();
System.out.print("Enter second number:");
num2 = scanner.nextInt();
num1 = num1 ^ num2;
num2 = num1 ^ num2;
num1 = num1 ^ num2;
scanner.close();
System.out.println("First number after swapping is:"+num1);
System.out.println("Second number after swapping is:"+num2);
}
}``````
The output of the above program for swapping two number using bitwise operator in Java will be:
``````Enter first number: 160
Enter second number: 260
First number after swapping is: 260
Second number after swapping is: 160``````
Here we have used nextInt() method of the Scanner class to swap two numbers in Java. We use scanner class to break down the input into small token values. To know more about the scanner class, click here. | 1,214 | 5,065 | {"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-2023-50 | longest | en | 0.644452 |
http://gmatclub.com/forum/revenue-80225.html?fl=similar | 1,484,652,297,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279657.18/warc/CC-MAIN-20170116095119-00120-ip-10-171-10-70.ec2.internal.warc.gz | 115,983,430 | 42,327 | revenue : PS Archive
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. The annual revenue of a computer company is 5% of the total world revenue on the industry, y. If the annual growth prediction of this company is 10%, what is the revenue of this company at the end of future 3 years?
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09 Jul 2009, 09:13
Is there any information at what rate revenue of the industry would grow?
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Looks like you are missing a major chunk of question here.
Re: revenue [#permalink] 22 Jul 2009, 00:38
Display posts from previous: Sort by | 484 | 1,760 | {"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-2017-04 | latest | en | 0.886758 |
https://www.vidyakul.com/course/relations-functions | 1,575,936,685,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525598.55/warc/CC-MAIN-20191209225803-20191210013803-00493.warc.gz | 891,473,377 | 71,204 | ## Relations & Functions
4.8 (185 Users)
Class 12 , Maths
Board: CBSE & State Boards , Language: Hinglish
Mode- Online
Videos- 7
Validity- Till 31 Mar 2020
Language- Hinglish
No. of Views- Unlimited
## WHAT WILL I LEARN?
Relations & Functions comes under the unit Relations & Functions which accounts for 10 marks in board exams. This chapter covers the following: Types of Relations, Types of Functions, Composition of Functions and Invertible Function, and Binary Operations. The entire chapter contains many theories and each theory has exercises which help the student understand each topic better. This chapter is not only useful for board exams but also for various competitive exams.
## Curriculum
Relations and functions
• Types of Relation Demo
• NCERT Exercise on Relations
• Types of Functions
• NCERT Exercise on Functions
• Composition of Functions
• Binary Operations
• NCERT Exercise on Binary Operations
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## 8999 6300
PCM & Biology | Class 11th, 12th & Competitive exams | 35+Years Experience | Pradeep Kshetrapal is a philanthropist and educator. His engaging physics lectures helps learner to grasp puzzling concepts of physics in an interesting manner. The video lectures and lecture notes are exclusively created by him to help learner.
## 1 Free
Mode- Online
Videos- 7
Validity- Till 31 Mar 2020
Language- Hinglish
No. of Views- Unlimited | 562 | 2,032 | {"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} | 2.703125 | 3 | CC-MAIN-2019-51 | longest | en | 0.826063 |
http://hackage.haskell.org/package/base-4.3.1.0/docs/src/Data-Function.html | 1,484,916,310,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280834.29/warc/CC-MAIN-20170116095120-00371-ip-10-171-10-70.ec2.internal.warc.gz | 122,573,599 | 2,389 | ```-----------------------------------------------------------------------------
-- |
-- Module : Data.Function
-- Copyright : Nils Anders Danielsson 2006
-- License : BSD-style (see the LICENSE file in the distribution)
--
-- Maintainer : libraries@haskell.org
-- Stability : experimental
-- Portability : portable
--
-- Simple combinators working solely on and with functions.
module Data.Function
( -- * "Prelude" re-exports
id, const, (.), flip, (\$)
-- * Other combinators
, fix
, on
) where
import Prelude
infixl 0 `on`
-- | @'fix' f@ is the least fixed point of the function @f@,
-- i.e. the least defined @x@ such that @f x = x@.
fix :: (a -> a) -> a
fix f = let x = f x in x
-- | @(*) \`on\` f = \\x y -> f x * f y@.
--
-- Typical usage: @'Data.List.sortBy' ('compare' \`on\` 'fst')@.
--
-- Algebraic properties:
--
-- * @(*) \`on\` 'id' = (*)@ (if @(*) ∉ {⊥, 'const' ⊥}@)
--
-- * @((*) \`on\` f) \`on\` g = (*) \`on\` (f . g)@
--
-- * @'flip' on f . 'flip' on g = 'flip' on (g . f)@
-- Proofs (so that I don't have to edit the test-suite):
-- (*) `on` id
-- =
-- \x y -> id x * id y
-- =
-- \x y -> x * y
-- = { If (*) /= _|_ or const _|_. }
-- (*)
-- (*) `on` f `on` g
-- =
-- ((*) `on` f) `on` g
-- =
-- \x y -> ((*) `on` f) (g x) (g y)
-- =
-- \x y -> (\x y -> f x * f y) (g x) (g y)
-- =
-- \x y -> f (g x) * f (g y)
-- =
-- \x y -> (f . g) x * (f . g) y
-- =
-- (*) `on` (f . g)
-- =
-- (*) `on` f . g
-- flip on f . flip on g
-- =
-- (\h (*) -> (*) `on` h) f . (\h (*) -> (*) `on` h) g
-- =
-- (\(*) -> (*) `on` f) . (\(*) -> (*) `on` g)
-- =
-- \(*) -> (*) `on` g `on` f
-- = { See above. }
-- \(*) -> (*) `on` g . f
-- =
-- (\h (*) -> (*) `on` h) (g . f)
-- =
-- flip on (g . f)
on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
(.*.) `on` f = \x y -> f x .*. f y
``` | 723 | 1,871 | {"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-2017-04 | longest | en | 0.278165 |
https://calculator.academy/hp-to-amps-calculator/ | 1,695,792,271,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00011.warc.gz | 181,914,025 | 57,533 | Enter the horsepower, voltage, efficiency, and power factor into the calculator to convert hp to amps. This calculator can also evaluate the hp volts, efficiency, or power factor given the other variables.
## HP to Amps Formula
The following formula is used to calculate the total horsepower from amps.
A = HP * 746 / (V* n * PF)
• Where A is the amps
• HP is the horsepower
• V is the volts
• n is the efficiency
• PF is the power factor
To calculate amps from HP, multiply the hp by 746, then divide the result by the product of the volts, efficiency, and power factor.
## HP to Amps Definition
HP to amps is the conversion from the standard unit of mechanical power to electrical power of amps.
## HP to Amps Example
How to calculate HP to Amps?
1. First, determine the total horsepower.
Measure the horsepower of the engine.
2. Next, determine the volts.
Measure the volts.
3. Next, determine the efficiency.
Calculate the efficiency.
4. Next, determine the power factor.
Calculate the power factor.
5. Finally, calculate hp to amps.
Calculate hp to amps using the formula above.
## FAQ
What is horsepower?
A horsepower is a standard unit for electrical power that is equal to exactly 746 watts.
How is Amps calculated from HP?
If you are converting directly from watts, you simply need to divide the watts by 746 to get horsepower. If you are converting from Amps, you must include more information about the motor as described in the formula A = HP * 746 / (V* n * PF). | 339 | 1,498 | {"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} | 3.625 | 4 | CC-MAIN-2023-40 | latest | en | 0.895973 |
https://book.thedatascienceinterviewproject.com/sql/windows-functions | 1,695,758,907,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510219.5/warc/CC-MAIN-20230926175325-20230926205325-00318.warc.gz | 162,527,195 | 142,027 | # Windows Functions
Since solving any reasonable SQL problem requires a combination of all the topics covered here, hence it becomes difficult to seggregate problems based on one topic alone. So for SQL we are creating a dedicated Problems section. Theoretical and Basic questions will still be under their dedicated sections.
Window (also, windowing or windowed) functions perform a calculation over a set of rows. I like to think of “looking through the window” at the rows that are being returned and having one last chance to perform a calculation. The window is defined by the OVER clause which determines if the rows are partitioned into smaller sets and if they are ordered. They allow you to add your favourite aggregate function to a non-aggregate query. Similar to Transform is pandas group by clause.
GROUP vs WINDOW
## Common Windows Functions
• Ranking functions
• ROW_NUMBER: is used to add unique row numbers to a partition or to the entire result set. It has the ability to turn non-unique rows into unique rows
• RANK: it will give numbers same as row_number just that same data will get same rank
• DENSE_RANK: doesnot skip and rank number
• NTILE: It assigns bucket numbers to the rows instead of row numbers or ranks
Ranking functions, check playground to work with this
• Offset functions
• LAG: the function allows you to pull columns or expressions from a row before the current row
• LEAD: the function allows you to pull columns or expressions from a row after the current row
• FIRST_VALUE: the functions allows you to return values from the first row of the partition
• LAST_VALUE: the functions allows you to return values from the last row of the partition
• Statistical functionsPERCENT_RANK: returns the percentage of rows that rank lower than the current row, its formula is
$\frac{\text{Rank} -1}{\text{Row count} -1}$
• CUME_DIST: cumulative distribution, returns the exact rank, its formula is
$\frac{\text{Rank}}{\text{Row count}}$
• PERCENTILE_DISC & PERCENTILE_CONT: these two work in the opposite way. Given a percent rank, find the value at that rank. They differ in that PERCENTILE_DISC will return a value that exists in the set while PERCENTILE_CONT will calculate an exact value if none of the values in the set falls precisely at that rank. You can use PERCENTILE_CONT to calculate a median by supplying 0.5 as the percent rank. For example, which temperature ranks at 50% in St. Louis?
One problem is you cannot add window functions to the WHERE clause. But certain Data Bases like TeraData, Snowflake, Databricks, Big Query, etc. have support for something called QUALIFY. Let's take an example to understand the difference:
In traditional SQL you will write the below SQL:
select * from (SELECT Product, Region, Revenue, ROW_NUMBER() OVER (PARTITION BY Region ORDER BY Revenue DESC) as rn FROM Sales ) where rn=1
Here's how you can use the QUALIFY keyword to achieve this:
SELECT Product, Region, Revenue FROM Sales QUALIFY ROW_NUMBER() OVER (PARTITION BY Region ORDER BY Revenue DESC) = 1;
Remember you CAN include window function and GROUP BY in the same statement, in such case the WINDOW function is evaluated after GROUP BY. As an example, below we can get both the user2 count at a user1 level along with total number of users in the table in the same query:
select user1
, 100*(CAST(count(distinct(user2)) as FLOAT)/CAST (count(*) over () as FLOAT)) as popularity_percent | 765 | 3,429 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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 | 3 | CC-MAIN-2023-40 | longest | en | 0.865921 |
http://temposchlucker.blogspot.com/2008/09/influence-of-king.html | 1,531,841,878,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589752.56/warc/CC-MAIN-20180717144908-20180717164908-00087.warc.gz | 362,866,074 | 14,255 | ## Thursday, September 25, 2008
### The influence of the King
Any pawngame without kings can be evaluated to the end. There are mainly two parameters for a pawn:
• What is the distance to the promotion square
• Are there impediments on the road to the promotion square
The weapons of the pawn are:
• Restraint
• Zugzwang
• Sacrifice
What happens when a king enters the arena? What is the influence of the King?
When there are two kings on an empty board the result is a draw. This means that the influence of the king is derived from how he effects the parameters of the pawns:
• Attack: can he free the road to promotion of his own pawn?
• Defence: can he impede the road to promotion of the enemy pawn?
This gives the king clear goals to head for. The power of the king is that he can make multipurpose moves. He can head for two (or more) goals at the same time.
The most famous example of all is of course Reti's position:
Diagram 1.
White to move and draw.
When there would be no kings on the board, this would be a clear win for white. It is very important to take note of this advantage for white (which I hadn't realized before).
When evaluating the kings you must first have a clear picture of the potential targets of both the kings and the pawns.
Black:
• Attack: de black pawn can promote on its own.
• Defence: the black king is already in the square of the white pawn and he must keep it that way.
White:
The white king has two targets.
• For attack: the key square d7, where the total path to promotion is protected.
• For defence: the square of the h5 pawn. Notice how big that target area is.
Multi-purpose moves.
Multiple goals add a whole new element to the battle. White has 4 moves at his disposal:
• c7 (serves 1 goal)
• Kg8 (1 goal)
• Kh7 (1 goal)
• Kg7 (2 goals)
Black has 5 moves at his disposal:
• Ka7 (0 goals)
• Ka5 (0 goals)
• Kb5 (0 goals)
• Kb6 (1 goal)
• h4 (1 goal)
The goals of black and white are mutual complementary by nature. Every attack (pawn promotion) has a defense attached which is mandatory. The one who makes a move which serves more goals than his opponent, comes one step closer to the realization of one of them.
Let's see what happens if we make a quantum leap and extrapolate this conclusion to the whole game.
In the startposition both players have 16 goals each. 8 attacking goals (pawn promotion) and 8 defensive goals (prevent the opponent's pawn promotion). If the game is not decided by accidents, like winning material or mating (winning) the king, one of the pawns will have to do the job. The pieces are circling around the pawns either assisting them or impeding them. Every multi-purpose move that you make (in relation to the 16 goals) while the opponent makes single-purpose moves, will bring you closer to the realization of one of the goals.
The value of each piece movement must be seen against this background. Thus fullfilling the words of Philidor which always seemed incomprehendsible to me until now.
There is more to say about the king and pawn, but I must digest this unsuspected conclusion first. | 727 | 3,080 | {"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-2018-30 | latest | en | 0.951918 |
https://socratic.org/questions/if-a-projectile-is-shot-at-an-angle-of-pi-8-and-at-a-velocity-of-14-m-s-when-wil | 1,720,825,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00799.warc.gz | 432,516,858 | 6,015 | # If a projectile is shot at an angle of (pi)/8 and at a velocity of 14 m/s, when will it reach its maximum height?
May 15, 2016
$t = 0.55 \text{ } s$
#### Explanation:
$\text{it can be calculated using t=} \frac{{v}_{i} \cdot \sin \alpha}{g}$
${v}_{i} = 14 \text{ } \frac{m}{s}$
$\alpha = \frac{\pi}{8}$
$g = 9.81 \frac{N}{k g}$
$t = \frac{14 \cdot 0.383}{9.81}$
$t = 0.55 \text{ } s$ | 160 | 394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.727321 |
http://math.ubooks.pub/Books/ON/M1/1703/C12S4M003.html | 1,519,345,842,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814300.52/warc/CC-MAIN-20180222235935-20180223015935-00414.warc.gz | 210,498,171 | 7,312 | 12-4. The Most General Applications of Bernoulli’s Equation
Learning Objectives
• Calculate using Torricelli’s theorem.
• Calculate power in fluid flow.
Torricelli’s Theorem
Figure 1 shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance $h$ from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is
$P1+12ρv12+ρgh1=P2+12ρv22+ρgh2.$
Both ${P}_{1}$ and ${P}_{2}$ equal atmospheric pressure (${P}_{1}$ is atmospheric pressure because it is the pressure at the top of the reservoir. ${P}_{2}$ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving
$12ρv12+ρgh1=12ρv22+ρgh2.$
Solving this equation for ${v}_{2}^{2}$, noting that the density $\rho$ cancels (because the fluid is incompressible), yields
$v22=v12+2g(h1−h2).$
We let $h={h}_{1}-{h}_{2}$; the equation then becomes
$v22=v12+2gh$
where $h$ is the height dropped by the water. This is simply a kinematic equation for any object falling a distance $h$ with negligible resistance. In fluids, this last equation is called Torricelli’s theorem. Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.
Figure 1: (a) Water gushes from the base of the Studen Kladenetz dam in Bulgaria. (credit: Kiril Kapustin; http://www.ImagesFromBulgaria.com) (b) In the absence of significant resistance, water flows from the reservoir with the same speed it would have if it fell the distance $h$ without friction. This is an example of Torricelli’s theorem.
Figure 2: Pressure in the nozzle of this fire hose is less than at ground level for two reasons: the water has to go uphill to get to the nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes, by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.
All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change. (See Figure 2.)
Example 1: Calculating Pressure: A Fire Hose Nozzle
Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$. The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle?
Strategy
Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant.
Solution
Bernoulli’s equation states
$P 1 + 1 2 ρv 1 2 +ρ gh1 =P2+ 12 ρv22 +ρ gh2 ,$
where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds ${v}_{1}$ and ${v}_{2}$. Since $Q={A}_{1}{v}_{1}$ , we get
$v1=QA1=40.0×10−3m3/sπ(3.20×10−2m)2=12.4m/s.$
Similarly, we find
$v2=56.6 m/s.$
(This rather large speed is helpful in reaching the fire.) Now, taking ${h}_{1}$ to be zero, we solve Bernoulli’s equation for ${P}_{2}$:
$P2=P1+12ρ v 1 2 − v 2 2 −ρgh2.$
Substituting known values yields
$P2=1.62×106N/m2+12(1000kg/m3)(12.4m/s)2−(56.6m/s)2−(1000kg/m3)(9.80m/s2)(10.0m)=0.$
Discussion
This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure equals atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.
Power in Fluid Flow
Power is the rate at which work is done or energy in any form is used or supplied. To see the relationship of power to fluid flow, consider Bernoulli’s equation:
$P+12ρv2+ρgh=constant.$
All three terms have units of energy per unit volume, as discussed in the previous section. Now, considering units, if we multiply energy per unit volume by flow rate (volume per unit time), we get units of power. That is, $\left(E/V\right)\left(V/t\right)=E/t$. This means that if we multiply Bernoulli’s equation by flow rate $Q$, we get power. In equation form, this is
$P+12ρv2+ρghQ=power.$
Each term has a clear physical meaning. For example, $\text{PQ}$ is the power supplied to a fluid, perhaps by a pump, to give it its pressure $P$. Similarly, $\frac{1}{2}{\mathrm{\rho v}}^{2}Q$ is the power supplied to a fluid to give it its kinetic energy. And $\rho \text{ghQ}$ is the power going to gravitational potential energy.
Making Connections: Power:
Power is defined as the rate of energy transferred, or $E/t$. Fluid flow involves several types of power. Each type of power is identified with a specific type of energy being expended or changed in form.
Example 2: Calculating Power in a Moving Fluid
Suppose the fire hose in the previous example is fed by a pump that receives water through a hose with a 6.40-cm diameter coming from a hydrant with a pressure of $0\text{.}\text{700}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$. What power does the pump supply to the water?
Strategy
Here we must consider energy forms as well as how they relate to fluid flow. Since the input and output hoses have the same diameters and are at the same height, the pump does not change the speed of the water nor its height, and so the water’s kinetic energy and gravitational potential energy are unchanged. That means the pump only supplies power to increase water pressure by $0\text{.}\text{92}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ (from $0.700×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ to $1.62×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$).
Solution
As discussed above, the power associated with pressure is
$power = PQ = 0.920×106N/m2 40.0×10−3m3/s. = 3.68×104W=36.8kW$
Discussion
Such a substantial amount of power requires a large pump, such as is found on some fire trucks. (This kilowatt value converts to about 50 hp.) The pump in this example increases only the water’s pressure. If a pump—such as the heart—directly increases velocity and height as well as pressure, we would have to calculate all three terms to find the power it supplies.
Summary
• Power in fluid flow is given by the equation $\left({P}_{1}+\frac{1}{2}{\mathrm{\rho v}}^{2}+\rho \text{gh}\right)Q=\text{power}\text{,}$ where the first term is power associated with pressure, the second is power associated with velocity, and the third is power associated with height.
Conceptual Questions
Exercise 1
Based on Bernoulli’s equation, what are three forms of energy in a fluid? (Note that these forms are conservative, unlike heat transfer and other dissipative forms not included in Bernoulli’s equation.)
Exercise 2
Water that has emerged from a hose into the atmosphere has a gauge pressure of zero. Why? When you put your hand in front of the emerging stream you feel a force, yet the water’s gauge pressure is zero. Explain where the force comes from in terms of energy.
Exercise 3
The old rubber boot shown in Figure 3> has two leaks. To what maximum height can the water squirt from Leak 1? How does the velocity of water emerging from Leak 2 differ from that of leak 1? Explain your responses in terms of energy.
Figure 3: Water emerges from two leaks in an old boot.
Exercise 4
Water pressure inside a hose nozzle can be less than atmospheric pressure due to the Bernoulli effect. Explain in terms of energy how the water can emerge from the nozzle against the opposing atmospheric pressure.
Problems & Exercises
Exercise 1
Hoover Dam on the Colorado River is the highest dam in the United States at 221 m, with an output of 1300 MW. The dam generates electricity with water taken from a depth of 150 m and an average flow rate of $\text{650}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s}$. (a) Calculate the power in this flow. (b) What is the ratio of this power to the facility’s average of 680 MW?
Show/Hide Solution
Solution
(a) $\text{9.56}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{W}$
(b) 1.4
Exercise 2
A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) (a) At takeoff, an aircraft travels at 60.0 m/s, so that the air speed relative to the bottom of the wing is 60.0 m/s. Given the sea level density of air to be $1\text{.}\text{29}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$, how fast must it move over the upper surface to create the ideal lift? (b) How fast must air move over the upper surface at a cruising speed of 245 m/s and at an altitude where air density is one-fourth that at sea level? (Note that this is not all of the aircraft’s lift—some comes from the body of the plane, some from engine thrust, and so on. Furthermore, Bernoulli’s principle gives an approximate answer because flow over the wing creates turbulence.)
Exercise 3
The left ventricle of a resting adult’s heart pumps blood at a flow rate of $\text{83}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{3}\text{/s}$, increasing its pressure by 110 mm Hg, its speed from zero to 30.0 cm/s, and its height by 5.00 cm. (All numbers are averaged over the entire heartbeat.) Calculate the total power output of the left ventricle. Note that most of the power is used to increase blood pressure.
Show/Hide Solution
Solution
1.26 W
Exercise 4
A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.750 L/s, with an output pressure of $3.00×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$. (a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point? (b) The hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem. | 2,882 | 10,656 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 46, "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} | 4.59375 | 5 | CC-MAIN-2018-09 | latest | en | 0.907212 |
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By michaellin250, history, 3 weeks ago, ,
Problem
Editorial
I don't understand the editorial when it says:
"dp0[i][j] = The higher i digits has already been decided, and there are j non-zero digits, and it has already been determined that it is less than N
dp1[i][j] = The higher i digits has already been decided, and there are j non-zero digits, and it has not yet been determined that it is less than N"
What does this mean?
• +1
» 3 weeks ago, # | +1 I do not even get the first sentence: "It is enough to consider the integers of N digits, by filling the higher digits with 0 if necessary.". What is "N digits" here?
• » » 3 weeks ago, # ^ | +1 In the problem, the number N is the number we're reading in. I'm guessing that's what they mean by N in this case? I don't know.
» 3 weeks ago, # | +1 The editorial doesn't make sense to me, but since k = 3 and sz(n) = 100, I wrote a very straightforward brute force (solve it differently for each k): https://atcoder.jp/contests/abc154/submissions/15264239
• » » 3 weeks ago, # ^ | +1 Thank you so much!! Can you explain what you're doing in K == 2 compared to k == 3?
• » » » 3 weeks ago, # ^ | 0 I'm just going over every single possible number with exactly k nonzero digits and then checking if it is bad, i.e., if it is > N. If it is not bad then the answer gets incremented by one.
• » » » » 3 weeks ago, # ^ | 0 Is pref a prefix sum?
» 3 weeks ago, # | 0 I think what they're talking about is ensuring that the number is bounded under $N$ whilst building it. Since the number is built from left to right, when we append a digit to the right, we need to know which digits we're allowed to append. If $N = 54321$ and our current number is $5432$, then "it has not yet been determined that it is less than $N$", so we are only allowed to append a $0$ or $1$. If we append something bigger like a $2$, then the number becomes $54322 > N$, which is not allowed.On the other hand, if our current number is $5431$, then no matter what digit we append next, our number will still be less than $N$. Even the max digit of $9$ gives us $54319 < N$. Thus, "it has already been determined that it is less than $N$".$dp0$ and $dp1$ thus represent the number of ways to build numbers fulfilling these conditions.
• » » 3 weeks ago, # ^ | 0 Thanks!! What is dp0 and dp1 for index i and index j representing then?
• » » » 3 weeks ago, # ^ | 0 $i$ would be having the first $i$ digits of the number already set, and $j$ would be the number of non-zero digits used so far. When transitioning from $dp0[i][j]$ or $dp1[i][j]$ to another state, we iterate over all valid digits as mentioned above and move to the corresponding state. For example, if $N = 54321$ and we are currently at $dp1[4][j]$, then choosing $0$ as the next digit transitions to $dp0[5][j]$, choosing $1$ transitions to $dp1[5][j+1]$, and choosing any other digit is not permitted, since the 5th digit of $N$ is $1$ and we have not yet determined our number to be less than $N$.When the editorial says "the higher $i$ digits", I think that's just poor translation. I'd assume higher is akin to first or leftmost. | 889 | 3,175 | {"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} | 3.671875 | 4 | CC-MAIN-2020-34 | latest | en | 0.946692 |
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# Note-taking during Verbal?
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19 Jul 2011, 18:47
Do you take notes during the Verbal section? I am starting to do so in my prep books, but I may be starting a bad habit because I won't have the advantage of marking up text during the CAT. Furthermore, the practice of taking notes is also time-consuming. Have you developed your own personal GMAT note-taking shorthand? Looking for tips. Thanks!
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19 Jul 2011, 20:49
While some people swear by it, note-taking consumed too much time for me to actually use it during a CAT. I also didn't want to become reliant on this method, since the test wouldn't be in paper format for me. Having to adjust to this on test day felt too risky.
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Re: Note-taking during Verbal? [#permalink] 19 Jul 2011, 20:49
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# Note-taking during Verbal?
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https://lxxb.cstam.org.cn/cn/article/doi/10.6052/0459-1879-21-612 | 1,713,126,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00867.warc.gz | 361,541,197 | 64,898 | EI、Scopus 收录
引用本文: 施力维, 马强, 舒进辉. 条形荷载下梯度非均匀非饱和土的动力响应分析. 力学学报, 2022, 54(7): 2008-2018.
Shi Liwei, Ma Qiang, Shu Jinhui. Dynamic responses of graded non-homogeneous unsaturated soils under a strip load. Chinese Journal of Theoretical and Applied Mechanics, 2022, 54(7): 2008-2018.
Citation: Shi Liwei, Ma Qiang, Shu Jinhui. Dynamic responses of graded non-homogeneous unsaturated soils under a strip load. Chinese Journal of Theoretical and Applied Mechanics, 2022, 54(7): 2008-2018.
## DYNAMIC RESPONSES OF GRADED NON-HOMOGENEOUS UNSATURATED SOILS UNDER A STRIP LOAD
• 摘要: 基于多孔介质混合物理论, 建立了梯度非均匀非饱和土地基模型, 研究了条形荷载作用下梯度非均匀非饱和土地基的动力响应问题. 通过傅里叶积分变换和Helmholtz矢量分解原理, 获得频域内非饱和土地基动力响应问题的通解, 结合回传射线矩阵法和边界条件, 求解获得了非均匀非饱和土层中位移、应力以及孔隙压力的计算列式. 假设沿深度方向梯度非均匀非饱和土的物理力学性质按幂函数连续变化, 通过数值傅里叶逆变换得到了非均匀非饱和土地基中的应力、位移以及孔隙压力等物理量的数值解, 分析讨论了土体非均匀性对非饱和土介质动力响应的影响规律. 结果表明: 土体非均匀性显著改变了非饱和土中竖向位移、正应力和孔隙压力在其深度方向上的振动模态, 其中孔隙气压在其深度方向的振动频率随着梯度因子的增加而不断增大, 波峰值不断靠近地表处附近; 竖向位移随着梯度因子的增大不断减小; 正应力和孔隙水压随着梯度因子的增大先增大后减小, 并且土体非均匀程度越高, 正应力与孔隙水压的幅值越大.
Abstract: Based on the three-phase porous media mixed theory, a graded non-homogeneous unsaturated foundation model is established and the dynamic response of graded non-homogeneous unsaturated soils subjected to a strip load is addressed. The general solutions of dynamic response for unsaturated foundation in frequency domain are derived by using the Fourier transform and Helmholtz vector decomposition. Then the calculation formula of displacement, stress, and pore pressure of graded non-homogeneous unsaturated soil is derived by combining with the reverberation-ray matrix method (RRMM), boundary conditions and general solutions of dynamic response for unsaturated foundation in frequency domain. Assuming that the continuous variation of physical and mechanical properties of unsaturated soils along the thickness-coordinate by exponential law distribution, the numerical solutions of displacement, stress, and pore pressure then obtained by using numerical inverse Fourier transformation, and the influence of soil heterogeneity on the dynamic response of unsaturated soil is discussed. The results show that the non-homogeneous of unsaturated soil has a considerable effect on the dynamic response of unsaturated soil, which significantly changes the vibration modes of vertical displacement, normal stress and pore pressure in the depth direction. The vibration frequency of pore air pressure in the depth direction increases with the increase of gradient factor, and the peak wave is constantly near the surface. The vertical displacement decreases with the increase of gradient factor, but the normal stress and pore water pressure first increase and then decrease with the increase of gradient factor, and the higher nonhomogeneity of soil is, the greater amplitude of normal stress and pore water pressure is.
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http://www.ics.uci.edu/~eppstein/PADS/StrongConnectivity.py | 1,519,261,549,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-09/segments/1518891813832.23/warc/CC-MAIN-20180222002257-20180222022257-00048.warc.gz | 470,301,119 | 2,320 | """StrongConnectivity.py DFS-based algorithm for computing strongly connected components. If G is a graph, then - StronglyConnectedComponents(G) returns a list of its components, each represented as a subgraph of G - Condensation(G) returns a directed acyclic graph, the vertices of which are strongly connected components of G. Each vertex of the condensation is represented as a frozenset of the vertices of G within a single strongly connected component. D. Eppstein, July 2005. """ import unittest import DFS class StronglyConnectedComponents(DFS.Searcher): """ Generate the strongly connected components of G. G should be represented in such a way that "for v in G" loops through the vertices, and "G[v]" produces a list of the neighbors of v; for instance, G may be a dictionary mapping each vertex to its neighbor set. The result of StronglyConnectedComponents(G) is a sequence of subgraphs of G. """ def __init__(self,G): """Search for strongly connected components of graph G.""" # set up data structures for DFS self._components = [] self._dfsnumber = {} self._activelen = {} self._active = [] self._low = {} self._biglow = len(G) self._graph = G # perform the Depth First Search DFS.Searcher.__init__(self,G) # clean up now-useless data structures del self._dfsnumber, self._activelen, self._active, self._low def __iter__(self): """Return iterator for sequence of strongly connected components.""" return iter(self._components) def __len__(self): """How many components are there?""" return len(self._components) def _component(self,vertices): """Make a new SCC.""" vertices = set(vertices) induced = {v:{w for w in self._graph[v] if w in vertices} for v in vertices} self._components.append(induced) def preorder(self,parent,child): """Handle first visit to vertex in DFS search for components.""" if parent == child: self._active = [] self._activelen[child] = len(self._active) self._active.append(child) self._low[child] = self._dfsnumber[child] = len(self._dfsnumber) def backedge(self,source,destination): """Handle non-tree edge in DFS search for components.""" self._low[source] = min(self._low[source],self._low[destination]) def postorder(self,parent,child): """Handle last visit to vertex in DFS search for components.""" if self._low[child] == self._dfsnumber[child]: self._component(self._active[self._activelen[child]:]) for v in self._components[-1]: self._low[v] = self._biglow del self._active[self._activelen[child]:] else: self._low[parent] = min(self._low[parent],self._low[child]) def Condensation(G): """Return a DAG with vertices equal to sets of vertices in SCCs of G.""" components = {} GtoC = {} for C in StronglyConnectedComponents(G): C = frozenset(C) for v in C: GtoC[v] = C components[C] = set() for v in G: for w in G[v]: if GtoC[v] != GtoC[w]: components[GtoC[v]].add(GtoC[w]) return components # If run as "python StrongConnectivity.py", run tests on various small graphs # and check that the correct results are obtained. class StrongConnectivityTest(unittest.TestCase): G1 = { 0:[1], 1:[2,3], 2:[4,5], 3:[4,5], 4:[6], 5:[], 6:[] } C1 = [[0],[1],[2],[3],[4],[5],[6]] # Work around http://bugs.python.org/issue11796 by using a loop # instead of a dict/set comprehension in a class variable initializer # should be: # Con1 = {frozenset([v]):{frozenset([w]) for w in G1[v]} for v in G1} Con1 = {} for v in G1: Con1[frozenset([v])] = {frozenset([w]) for w in G1[v]} G2 = { 0:[1], 1:[2,3,4], 2:[0,3], 3:[4], 4:[3] } C2 = [[0,1,2],[3,4]] f012 = frozenset([0,1,2]) f34 = frozenset([3,4]) Con2 = {f012:{f34}, f34:set()} knownpairs = [(G1,C1),(G2,C2)] def testStronglyConnectedComponents(self): """Check known graph/component pairs.""" for (graph,expectedoutput) in self.knownpairs: output = [list(C) for C in StronglyConnectedComponents(graph)] for component in output: component.sort() output.sort() self.assertEqual(output,expectedoutput) def testSubgraph(self): """Check that each SCC is an induced subgraph.""" for (graph,expectedoutput) in self.knownpairs: components = StronglyConnectedComponents(graph) for C in components: for v in C: for w in graph: self.assertEqual(w in graph[v] and w in C, w in C[v]) def testCondensation(self): """Check that the condensations are what we expect.""" self.assertEqual(Condensation(self.G1),self.Con1) self.assertEqual(Condensation(self.G2),self.Con2) if __name__ == "__main__": unittest.main() | 1,178 | 4,375 | {"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-2018-09 | latest | en | 0.768674 |
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Take a look at the video of this trick. Can you perform it yourself? Why is this maths and not magic?
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Find the sum of all three-digit numbers each of whose digits is odd.
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What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Roll over the Dice
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Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded?
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Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
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Got It game for an adult and child. How can you play so that you know you will always win?
### Next-door Numbers
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Florence, Ethan and Alma have each added together two 'next-door' numbers. What is the same about their answers?
### Round the Four Dice
##### Age 7 to 11 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand. | 2,032 | 8,886 | {"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-2019-39 | latest | en | 0.922622 |
https://slideplayer.com/slide/220208/ | 1,545,200,744,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831334.97/warc/CC-MAIN-20181219045716-20181219071716-00461.warc.gz | 744,172,190 | 26,125 | # Solving Inequalities by Adding or Subtracting
## Presentation on theme: "Solving Inequalities by Adding or Subtracting"— Presentation transcript:
Solving Inequalities by Adding or Subtracting
3-2 Solving Inequalities by Adding or Subtracting Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz
Warm Up Graph each inequality. Write an inequality for each situation.
1. The temperature must be at least –10°F. 2. The temperature must be no more than 90°F. –10 10 x ≥ –10 x ≤ 90 –90 90 Solve each equation. 3. x – 4 = 10 14 4. 15 = x + 1.1 13.9
Objectives Solve one-step inequalities by using addition.
Solve one-step inequalities by using subtraction.
Solving one-step inequalities is much like solving one-step equations
Solving one-step inequalities is much like solving one-step equations. To solve an inequality, you need to isolate the variable using the properties of inequality and inverse operations.
Helpful Hint Use an inverse operation to “undo” the operation in an inequality. If the inequality contains addition, use subtraction to undo the addition.
Example 1A: Using Addition and Subtraction to Solve
Inequalities Solve the inequality and graph the solutions. x + 12 < 20 x + 12 < 20 Since 12 is added to x, subtract 12 from both sides to undo the addition. –12 –12 x + 0 < 8 x < 8 Draw an empty circle at 8. –10 –8 –6 –4 –2 2 4 6 8 10 Shade all numbers less than 8 and draw an arrow pointing to the left.
Example 1B: Using Addition and Subtraction to Solve
Inequalities Solve the inequality and graph the solutions. d – 5 > –7 d + 0 > –2 d > –2 d – 5 > –7 Since 5 is subtracted from d, add 5 to both sides to undo the subtraction. Draw an empty circle at –2. –10 –8 –6 –4 –2 2 4 6 8 10 Shade all numbers greater than –2 and draw an arrow pointing to the right.
Example 1C: Using Addition and Subtraction to Solve
Inequalities Solve the inequality and graph the solutions. 0.9 ≥ n – 0.3 0.9 ≥ n – 0.3 Since 0.3 is subtracted from n, add 0.3 to both sides to undo the subtraction. 1.2 ≥ n – 0 1.2 ≥ n 1.2 Draw a solid circle at 1.2. 1 2 Shade all numbers less than 1.2 and draw an arrow pointing to the left.
Solve each inequality and graph the solutions.
Check It Out! Example 1 Solve each inequality and graph the solutions. a. s + 1 ≤ 10 Since 1 is added to s, subtract 1 from both sides to undo the addition. s + 1 ≤ 10 –1 –1 9 –10 –8 –6 –4 –2 2 4 6 8 10 s + 0 ≤ 9 s ≤ 9 b > –3 + t Since –3 is added to t, add 3 to both sides to undo the addition. > –3 + t +3 > 0 + t –10 –8 –6 –4 –2 2 4 6 8 10 t <
Solve the inequality and graph the solutions.
Check It Out! Example 1c Solve the inequality and graph the solutions. q – 3.5 < 7.5 Since 3.5 is subtracted from q, add 3.5 to both sides to undo the subtraction. q – 3.5 < 7.5 q – 0 < 11 q < 11 –7 –5 –3 –1 1 3 5 7 9 11 13
Since there can be an infinite number of solutions to an inequality, it is not possible to check all the solutions. You can check the endpoint and the direction of the inequality symbol. The solutions of x + 9 < 15 are given by x < 6.
Example 2: Problem-Solving Application
Sami has a gift card. She has already used \$14 of the of the total value, which was \$30. Write, solve, and graph an inequality to show how much more she can spend. Understand the problem 1 The answer will be an inequality and a graph that show all the possible amounts of money that Sami can spend. List important information: • Sami can spend up to, or at most \$30. • Sami has already spent \$14.
Example 2 Continued 2 Make a Plan Write an inequality. Let g represent the remaining amount of money Sami can spend. Amount remaining plus \$30. is at most amount used g + 14 30 g + 14 ≤ 30
Draw a solid circle at 0 and16.
Example 2 Continued Solve 3 g + 14 ≤ 30 Since 14 is added to g, subtract 14 from both sides to undo the addition. – 14 – 14 g + 0 ≤ 16 g ≤ 16 Draw a solid circle at 0 and16. 2 4 6 8 10 12 14 16 18 Shade all numbers greater than 0 and less than 16.
Example 2 Continued Look Back 4 Check Check a number less than 16.
g ≤ 30 ≤ 30 20 ≤ 30 Check the endpoint, 16. g + 14 = 30 30 30 Sami can spend from \$0 to \$16.
Check It Out! Example 2 The Recommended Daily Allowance (RDA) of iron for a female in Sarah’s age group (14-18 years) is 15 mg per day. Sarah has consumed 11 mg of iron today. Write and solve an inequality to show how many more milligrams of iron Sarah can consume without exceeding RDA.
Check It Out! Example 2 Continued
Understand the problem 1 The answer will be an inequality and a graph that show all the possible amounts of iron that Sami can consume to reach the RDA. List important information: • The RDA of iron for Sarah is 15 mg. • So far today she has consumed 11 mg.
Check It Out! Example 2 Continued
Make a Plan Write an inequality. Let x represent the amount of iron Sarah needs to consume. Amount taken plus 15 mg is at most amount needed 11 + x 15 11 + x 15
Check It Out! Example 2 Continued
Solve 3 11 + x 15 Since 11 is added to x, subtract 11 from both sides to undo the addition. – –11 x 4 Draw a solid circle at 4. 1 2 3 4 5 6 7 8 9 10 Shade all numbers less than 4. x 4. Sarah can consume 4 mg or less of iron without exceeding the RDA.
Check It Out! Example 2 Continued
Look Back 4 Check Check a number less than 4. 15 15 14 15 Check the endpoint, 4. 11 + x = 15 Sarah can consume 4 mg or less of iron without exceeding the RDA.
Example 3: Application Mrs. Lawrence wants to buy an antique bracelet at an auction. She is willing to bid no more than \$550. So far, the highest bid is \$475. Write and solve an inequality to determine the amount Mrs. Lawrence can add to the bid. Check your answer. Let x represent the amount Mrs. Lawrence can add to the bid. \$475 plus amount can add is at most \$550. x + 475 550 475 + x ≤ 550
Example 3 Continued 475 + x ≤ 550 – – 475 x ≤ 75 0 + x ≤ 75 Since 475 is added to x, subtract 475 from both sides to undo the addition. Check the endpoint, 75. Check a number less than 75. x = 550 x ≤ 550 ≤ 550 525 ≤ 550 Mrs. Lawrence is willing to add \$75 or less to the bid.
Let p represent the number of additional pounds Josh needs to lift.
Check It Out! Example 3 What if…? Josh wants to try to break the school bench press record of 282 pounds. He currently can bench press 250 pounds. Write and solve an inequality to determine how many more pounds Josh needs to lift to break the school record. Check your answer. Let p represent the number of additional pounds Josh needs to lift. 250 pounds plus additional pounds is greater than 282 pounds. 250 + p > 282
Check It Out! Example 3 Continued
– –250 p > 32 Since 250 is added to p, subtract 250 from both sides to undo the addition. Check Check the endpoint, 32. Check a number greater than 32. p = 282 p > 282 > 282 283 > 282 Josh must lift more than 32 additional pounds to reach his goal.
Lesson Quiz: Part I Solve each inequality and graph the solutions. 1. 13 < x + 7 x > 6 2. –6 + h ≥ 15 h ≥ 21 y ≤ –2.1 y ≤ –8.8
Lesson Quiz: Part II 4. A certain restaurant has room for 120 customers. On one night, there are 72 customers dining. Write and solve an inequality to show how many more people can eat at the restaurant. x + 72 ≤ 120; x ≤ 48, where x is a natural number
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# Topic: Primitive element (field theory)
Primitive element (field theory) - Wikipedia, the free encyclopedia In mathematics, a primitive element for an extension of fields L/K is an element ζ of L such that If the extension L/K admits a primitive element, then L is either a finite extension of K, in case ζ is an algebraic element of L over K, or L is isomorphic to the field of rational functions over K in one indeterminate, if ζ is a transcendental element of L over K. The primitive element theorem of field theory answers the question of which finite field extensions have primitive elements. en.wikipedia.org /wiki/Primitive_element_(field_theory) (468 words)
PlanetMath: primitive element theorem (Site not responding. Last check: 2007-10-22) Note that this implies that every finite separable extension is not only finitely generated, it is generated by a single element. Cross-references: Galois theory, Galois group, minimal polynomials, splitting field, characteristic, corollary, theorem, infinite, algebraic closure, algebraic, clear, roots, polynomial, isomorphic, generate, transcendental, indeterminate, generated by, finitely generated, separable extension, degree, finite, extension, fields This is version 12 of primitive element theorem, born on 2001-10-15, modified 2005-01-22. planetmath.org /encyclopedia/PrimitiveElementTheorem.html (247 words)
PlanetMath: proof of primitive element theorem (Site not responding. Last check: 2007-10-22) So if we can show that any field extension generated by two elements is also generated by one element, we will be done: simply apply the result to the last two elements "proof of primitive element theorem" is owned by archibal. This is version 1 of proof of primitive element theorem, born on 2004-03-20. planetmath.org /encyclopedia/ProofOfPrimitiveElementTheorem2.html (186 words)
YourArt.com >> Encyclopedia >> primitive (Site not responding. Last check: 2007-10-22) The Amish are a particularly obvious example of this, and the primitive baptist church is another. Indigenous peoples and their beliefs and practices are sometimes described as "primitive", a usage that is seen as unhelpful and inaccurate by the vast majority of contemporary anthropologists and similar professionals. Primitive communism postulates a pre-agrarian form of communism. www.yourart.com /research/encyclopedia.cgi?subject=/primitive (169 words)
Finite Fields K[x] is a principal ideal domain (an ideal is a subring that is closed under multiplication, a principal ideal is an ideal generated by a single element, and a principal ideal domain is a commutative ring with unity all of whose ideals are principal). We denote the group of all automorphisms of a field L by G(L) and the subgroup of G(L) that fixes all elements of the subfield K of L by G(L/K). Now the primitive elements are to be found among the roots of the irreducible polynomials (they cannot be elements of the prime field). www-math.cudenver.edu /~wcherowi/courses/finflds.html (3085 words)
Primitive Polynomial Computation Theory and Algorithm The field is exactly the set of all polynomials of degree 0 to n-1 with the two field operations being addition and multiplication of polynomials modulo g(x) and with modulo p integer arithmetic on the polynomial coefficients. f(x) is a primitive polynomial iff the field element x generates the cyclic group of non-zero field elements of the finite field. A primitive polynomial is the minimal polynomial of a generator, and its roots are conjugates of the generator. www.seanerikoconnor.freeservers.com /Mathematics/AbstractAlgebra/PrimitivePolynomials/theory.html (3266 words)
[No title] (Site not responding. Last check: 2007-10-22) This is called finding a primitive element theta of K. If alpha_1,..., alpha_n are the roots of f, we find a primitive element by induction, finding one for Q(theta_i, alpha_{i+1}) where theta_i is a primitive element of Q(alpha_1,..., alpha_i). Note that theta_{i-1} is the final theta you want.) The "theorem on the primitive element" says that a primitive element for Q(theta_i, alpha_{i+1}) is of the form theta_i + k * alpha_{i+1} for almost all rational integers k. A maximal-degree factor of the resultant should be the minimal polynomial of a primitive element of Q(theta_i, alpha_{i+1}), as long as k is generic enough. www.math.niu.edu /~rusin/known-math/95/splitting.fld (619 words)
Number Theory Glossary Fields with a finite number of elements are called Galois fields. A Galois Field is a field with finite number of elements. A primitive element in a group is an element whose powers exhaust the entire group. www.math.umbc.edu /~campbell/NumbThy/Class/Glossary.html (827 words)
Primitive element - Wikipedia, the free encyclopedia in number theory, a primitive root modulo n in field theory, an element that generates a given field extension, see primitive element (field theory) in a Hopf algebra, a particular kind of element on which the comultiplication takes a certain simple form. en.wikipedia.org /wiki/Primitive_element (123 words)
The Graduate Research of Johnny Crypto A primitive element of a Galois field of size p^n is an element whose powers are all different. By definition of a field extension, the field that is being extended is a subfield of the larger field. We want to find primitive elements to use for the coefficients of the polynomials that are used to generate the S-Boxes. www.artfags.org /~jnycrpto/graduate.htm (495 words)
Open Questions: Algebraic Number Theory Galois theory is a way to "map" extensions of fields to groups and their subgroups in such a way that most of the interesting details about the extension are reflected in details about the groups, and vice versa. Class field theory has a reputation for being a very difficult subject, and it is. There is a fair bit of abstract conceptual machinery involved even to explain many of the results, and (of course) much more to prove the results. As far as ramification is concerned, the class field K is the maximal abelian extension of k which is unramified except for primes that divide the conductor F. This is an alternative definition of K as the class field of k, so this too is a direct generalization of Hilbert's results. www.openquestions.com /oq-ma018.htm (19624 words)
Galois Theory Glossary An element a of a field L is algebraic of degree n over a subfield K if there is an irreducible polynomial f(t) in K[t] of degree n such that f(a) = 0. An extension of fields L/K (this notation does not denote any sort of quotient) is a ring homomorphism K --> L. Such a homomorphism has to be injective, so that K is isomorphic to a subfield of L. It is often convenient to identify K with this subfield. An element a in a field L is a separably algebraic element over a subfield K if there is a polynomial f(t) in K[t] such that f(a) = 0 and f'(a) is nonzero. www.wra1th.plus.com /Galois/gloss.html (892 words)
Primitive Element Theorem (Site not responding. Last check: 2007-10-22) PRIMITIVE ELEMENTS IN FINITE FIELDS WITH ARBITRARY TRACE... Primitive element (field theory) article - Primitive element (field theory) math... Algorithms for primitive elements of free Lie algebras and Lie superalgebras... www.scienceoxygen.com /math/290.html (102 words)
UNDERSTANDING CONFLICT. VIOLENCE, AND WAR: FOUNDATIONS It has led to the development of a field theory of behavior which, I believe, integrates a variety of theoretical and philosophical approaches to war and violence and serves as the phenomenological framework for analyzing whether war is inevitable and what might be done about it. The mathematical structure of the field is more precise but loses much of the rich meaning of the field interpretation and narrows communication to the small number familiar with the particular mathematics (linear algebra) involved. Dynamic fields entail a continuous extension of energy or potentiality throughout a space or region which is the spring or seat of forces, powers, or influences. www.hawaii.edu /powerkills/NOTE10.HTM (6974 words)
Quantum Field Theory Quantum field theory can be developed by adopting the path integral formulism as mentioned earlier, or by combining the field equation with canonical quantization as shown in the next sub-topic. A renormalizable theory is one in which the details of a deeper scale are not needed to describe the physics at the present scale, save for a few experimentally measurable parameters (see more in the section about "Renormalizable Theories"). In the Standard model the scalar field is identified as the Higgs field responsiable for the mass of fermions and gauge bosons. universe-review.ca /R15-12-QFT.htm (11980 words)
Jo on the web. Galois Fields GF(q) is a field with q elements, also called a finite field because there are a finite number (q) of elements. A Primitive Element of GF(q) is an element ‘a‘such that every field element except zero can be expressed as a power of a. The primitive element of such a field would itself be such a polynomial. nislab.bu.edu /nislab/projects/bchsync/BCH.html (615 words)
Document title goes here. We'll show how they can be derived from Galois field theory and derive their sharply peaked autocorrelation properties. When the Galois field elements are written out as a list of polynomial coefficients, the PN sequence is the last column. We don't need to generate the field and do time consuming finite field arithmetic to find a PN sequence, but can use a recursive method instead. www.seanerikoconnor.freeservers.com /CommunicationTheory/PseudoNoiseSequences/MechanicalAutocorrelatorBasedOnPseudoNoiseSequences.html (600 words)
list of theorems - Article and Reference from OnPedia.com In some fields, theorem can be considered as a courtesy title, given to major results, although with a content that would not satisfy a mathematician. No attempt is made here to comment on that aspect of usage: this is a list of results known as theorems. Haboush's theorem (algebraic groups, representation theory, invariant theory) www.onpedia.com /encyclopedia/list-of-theorems (172 words)
The Creative Process in Nature It is a tenet of General Systems theory that certain fundamental processes in nature are isomorphic; that if we can understand one of these processes, we can use this knowledge to help us comprehend others of its kind. The scientific view of this natural creative process is represented in a 4x3 model of the unified field theory and the creation of matter in the Big Bang, while the occult or mythological creation story, with a similar 4x3 pattern, as found in the astrological tradition, is mapped as an overlay. Finally, a numbering system is developed for the elements of counting and a notation system for the operations, relationships, and quantitative concepts of mathematics. www.people.cornell.edu /pages/jag8/shaman.html (8441 words)
12F: Field extensions Once upon a time, mathematicians (and others) would spend time on a subject call the "Theory of equations", which was just chock-full of algorithms and the theory of polynomials and their roots. Nowadays, this is the subject of ring theory or numerical analysis, but we chose to keep much of that material here since it often involves a consideration of the splitting fields of that polynomial. An example from Galois theory: calculating the fixed field K(X)^G, for a certain small G. Example of the fixed field under a subgroup of the Galois group (= Sym(3)). www.math.niu.edu /~rusin/known-math/index/12FXX.html (965 words)
Field Theory Prove that the Galois group of a finite extension of finite fields is cyclic. Describe the construction of an algebraic closure of a field. Show that any finite subgroup of the multiplicative group of a field is cyclic. www.math.dartmouth.edu /graduate-students/syllabi/sample-questions/algebra/node5.html (140 words)
Sophie Huczynska (Site not responding. Last check: 2007-10-22) Over the years, number theory has expanded to the study of more complicated structures than the set of natural numbers; it is now a vast subject area, with strong links to other branches of mathematics. elements of a finite field which are simultaneously additive and multiplicative generators of the field. The result which establishes the existence of such elements is the Primitive Normal Basis Theorem: it states that, for every prime power q and positive integer n, there exists a primitive normal basis of GF(q^n) over GF(q). homepages.inf.ed.ac.uk /shuczyns (802 words)
ABSTRACT ALGEBRA ON LINE: Ideal Theory of Commutative Rings Let D be an integral domain with quotient field F, and let I be an ideal of D that is invertible when considered as a fractional ideal. Let D be an integral domain with quotient field Q, and let F be a finite extension field of Q. If D* is the set of all elements of F that are integral over D, then D* is a Dedekind domain. Let F be a field, and first consider the ring F[x] of polynomials in one indeterminate. www.math.niu.edu /~beachy/aaol/commutative.html (2296 words)
DR Donald Mills (Site not responding. Last check: 2007-10-22) My primary interest is in the theory of finite fields and the applications of this theory to communications issues. While I was at West Point, I did work regarding the existence of primitive elements in cubic extensions of finite fields where the elements take on the form aC+b, C being a defining element of the cubic extension and a and b belonging to the underlying field. The paper corresponding to this is entitled "Primitive roots in cubic extensions of finite fields" (joint with G. McNay), and has recently appeared in the proceedings for the Sixth International Conference on Finite Fields and Applications. www.math.siu.edu /mills/default1.htm (2056 words)
http://www.math.wisc.edu/graduate/guide-qe.htm (Site not responding. Last check: 2007-10-22) Galois extensions and the fundamental theorem of Galois theory. Theory of Fourier Series; Orthogonal functions; Sturm-Liouville theory and connections with Fourier series; Special Fourier bases (Bessel functions, Legendre polynomials); Fourier transforms (Fourier and Fourier sine and Fourier cosine); Laplace transform and solution of initial-boundary value problems for equations; Evaluation of integrals via complex variables techniques. The advanced Model Theory, Recursion Theory, and Set Theory sections correspond, roughly, to the contents of 776, 773, and 771, respectively. www.math.wisc.edu /graduate/guide-qe.htm (1829 words)
SYLLABUS FOR M.A./M.SC. PART II (Site not responding. Last check: 2007-10-22) FIELD THEORY: Construction of fields, Algebraic extensions, Transcendence basis transcendence degree, Degree of algebraic extensions, Group action and transitive groups, Frobenius kernel and primitive group, Jordan's theorem on sharp multiple transitivity affine and projective geometry, Iwasawa theorem, Witt's theorem and Mathiem groups, Sharp 3-transitive groups, Zassenhaus groups. Arithmetic in Quadratic Number fields: Integers, Units Primes and irreducible elements, Failure of unique factorization, (Informal) definition of Ideal class group, Pell's equation and relation to continued fractions. math.mu.ac.in /syllabus/partII (3320 words)
Lecture Summary MP473, Number Theory IIIH/IVH The algebraic numbers form a field, the algebraic integers form a ring. The field polynomial is a power of the minimal polynomial, properties of norm and trace, splitting fields and normal extensions, K-isomorphisms and K-automorphisms, there are [L : K] K-isomorphims of L, a normal extension L has [L : under the isomorphisms of K, discriminant of a field basis, non-vanishing of the discriminant, effect of change of basis on the discriminant, discriminant in terms of conjugates of basis elements, resultant R(f(x),g(x)) of two polynomials, R(f(x),g(x))=0 if and only if f(x) and g(x) have a non-trivial factor in common, Disc(f(x)) the discriminant of f(x). www.numbertheory.org /courses/MP473/lectures.html (849 words)
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https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-15-differentiation-in-several-variables-15-2-limits-and-continuity-in-several-variables-exercises-page-771/2 | 1,685,483,079,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646144.69/warc/CC-MAIN-20230530194919-20230530224919-00052.warc.gz | 836,169,356 | 12,522 | ## Calculus (3rd Edition)
$$2$$
Given $$\lim _{(x, y) \rightarrow\left(\frac{4}{9}, \frac{2}{9}\right)} \frac{x}{y}$$ Since $\frac{x}{y}$ is a rational function and is continuous at $\left(\frac{4}{9}, \frac{2}{9}\right)$, then by using substitution, we get \begin{align*} \lim _{(x, y) \rightarrow\left(\frac{4}{9}, \frac{2}{9}\right)} \frac{x}{y}&= \frac{4/9}{2/9}\\ &=\frac{4}{2}\\ &=2 \end{align*} | 159 | 402 | {"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} | 4.03125 | 4 | CC-MAIN-2023-23 | latest | en | 0.527099 |
https://brilliant.org/problems/its-nice-to-see-less-numbers-p/ | 1,544,828,121,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826354.54/warc/CC-MAIN-20181214210553-20181214232553-00236.warc.gz | 547,987,481 | 8,989 | It's nice to see less numbers
Let the sum of first $$n$$ natural numbers be $$N$$. Suppose two numbers $$k$$ , $$k+1$$ are removed from the sum $$N$$ , the new sum becomes 1224. Find the value of $$k-20$$.
× | 64 | 209 | {"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} | 2.640625 | 3 | CC-MAIN-2018-51 | latest | en | 0.866805 |
https://cs.stackexchange.com/questions/37595/why-is-the-complexity-of-this-nested-for-loop-not-on2 | 1,709,565,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00157.warc.gz | 186,595,740 | 38,975 | # Why is the complexity of this nested for loop not $O(n^2)$?
I have the following pseudo-code:
mystery(n):
if n <= 50 :
for i = 1 ... n :
for j = 1 ... n :
print i*j
else :
mystery(n-1)
For the following nested for loop:
for i = 1 ... n :
for j = 1 ... n :
For every i in n, j iterates through n as many as i times. So, why is it that the complexity is not $O(n^2)$?
Because $n$ is bounded by $50$, which is simply a constant. Observe that when $n$ is larger than $50$, you simply have a recursive call. It's only when $n \leq 50$ that you go into the first portion of the if statement and execute the for-loops. Therefore, both nested loops take $\Theta(1)$ time. Hence the total time complexity of the mystery function is linear.
Note that $O(n^2)$ is technically correct, though not tight since $O(1) \subset O(n^2)$. The exact running time is $\Theta(1)$ as pointed above. | 256 | 885 | {"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} | 3.40625 | 3 | CC-MAIN-2024-10 | latest | en | 0.936318 |
https://physics.stackexchange.com/questions/741710/absorption-spectrum-dependence-on-concentration-pressure | 1,708,736,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00778.warc.gz | 476,477,473 | 39,878 | # Absorption spectrum dependence on concentration/pressure
I am having difficulties understanding the relation between absorption spectrum and gas concentration. The online resources I found, including many questions here, do not clarify my doubts.
Take for example this picture from wikipedia. It lacks any information about what concentration/partial pressure of each species results in the given absorption spectrum:
I don't understand how the absorption spectrum in the picture above relates to the concentration of the species in the atmosphere. For example, assuming that the concentration used for that spectrum is the average earth atmospheric concentration (i.e. 340ppm of CO2), I wonder what happens when the concentration halves. Do we get only 50% absorbance in the 4.3 microns band for CO2 or do we get a bit more than that because CO2 will still be fairly opaque at this concentration/wavelength?
The Beer-Lambert Law says that the transmittance of a sample is given by $$T = e^{-n \sigma \ell}$$, where $$\ell$$ is the path length, $$\sigma$$ is the absorption (or scattering) cross-section of the scattering species, and $$n$$ is the concentration of the species (molecules per volume.) If you halve the concentration ($$n' = n/2$$), then we can see from the above formula that the new transmittance is $$T' = e^{-n' \sigma \ell} = (e^{-n \sigma \ell})^{1/2} =\sqrt{T}$$.
In the case of the graphs above, it's important to note that what you're calling "100%" absorption is really something like 99.999999% absorption and 0.000001% transmission. There's never going to be perfect absorption in this sort of situation, because the scattering/absorption process is probabilistic. In general, the above argument says that if a fraction of $$1 - T$$ of the photons are absorbed before you halve the concentration, then $$1 - \sqrt{T}$$ will be absorbed afterwards. Exactly how big of a difference this makes depends on exactly how close you are to 100%. If you were scattering 99% of the light before, then you'll be scattering only 90% after. If it was 99.999999% $$(1 - 10^{-8})$$ before, then it will be 99.99% after ($$1 - 10^{-4}$$).
The graph below shows the new absorbance $$A'$$ (vertical axis) as a function of the old absorbance $$A$$ (horizontal axis). The fact that the slope of the graph becomes infinite as $$A \to 1$$ reflects the idea that it really matters how close to 100% absorbance you are. On the other hand, at very low absorbance (near $$A = 0$$), you can see that halving the concentration simply halves the absorbance.
• That is extremely clear, thank you! From wikipedia it seemed that the beer-lambert law was not exponential, maybe because it is assuming a logarithmic scale for A? Dec 21, 2022 at 21:09 | 658 | 2,751 | {"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": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-10 | longest | en | 0.897107 |
https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=90266§ion=1 | 1,627,302,354,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152129.33/warc/CC-MAIN-20210726120442-20210726150442-00268.warc.gz | 907,621,442 | 18,232 | Everyday maths 2 (Wales)
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# 1 Units of measure
A unit of measure is simply what you measure something in and you will already be familiar with using centimetres, metres, kilograms, grams, litres and millilitres for measurement. When measuring something, you need to choose the appropriate unit of measure for the item you are measuring. You would not, for example, measure the length of a room in millimetres. Have a go at the short activity below and choose the most appropriate unit for each example.
## Activity 1: Choosing the unit
Using the following two lists, match each numbered item with the correct letter.
1. Millilitres (ml)
2. Metres (m)
3. Kilograms (kg)
4. Grams (g)
5. Centimetres (cm)
6. Litres (l)
• a.Length of a garden fence
• b.Weight of an egg
• c.Amount of water in a paddling pool
• d.Amount of liquid in a glass
• e.Length of a computer screen
• f.Weight of a dog
The correct answers are:
• 1 = d
• 2 = a
• 3 = f
• 4 = b
• 5 = e
• 6 = c
Hopefully you found that activity fairly straightforward. Next you’ll take a closer look at some units of measure and how to convert between them.
FSM_2_CYMRU
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Request an Open University prospectus371 | 531 | 2,301 | {"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} | 3.546875 | 4 | CC-MAIN-2021-31 | longest | en | 0.893134 |
https://basicexceltutorial.com/how-to-calculate-the-95-percentile/ | 1,726,417,237,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00520.warc.gz | 106,766,298 | 14,837 | # How to calculate the 95 percentile
The handling of data requires numerous manipulations to be done to particular data. You need to find the mean, mode, median, standard deviation, maximum, minimum, and also the percentile of the data in question.
These manipulations can be referred to as mathematical functions and are way too much essential in any data that involves numeric values. These mathematical manipulations once are done make data valid and accurate to use.
When we calculate or get the percentile of any sort of data, means that the value in question if for example is the 90th percentile, the values in that data set are lower than or equal to 90 and the remaining 10 is greater or higher than that amount.
Any of the ninety-nine points that divide an ordered distribution into one hundred parts and each part containing one percent of the population is what is referred to as the percentile. The percentile helps divide the data into smaller units all with value and also gives one the data range.
Below are some of the steps to achieve the 95 percentile of a given data set.
Step 1
The first step is to enter data into an empty excel sheet, open an excel workbook, and record names in one column and marks in the second column as in the case below.
Your excel sheet should look like the one above.
Step 2
In this step we are now going to calculate the 95 percentile, we do this with the help of the PERCENTILE function, one of the functions provided by Excel. We incorporate this function in a formula, the general formula, therefore, will be; =PERCENTILE (B2: B10, 0.95). Since our data set has ten cells, we are going to find the percentile from cell B2 to cell B10. We are using 0.95 because the whole data set is 100% and 95 divided by 100 is 0.95.
From the above scenario, the 95 percentile of this data set is 97.6, this is to means that the majority of the values are lower than or equal to 97.6. The highlighted values are all lower than the percentile. | 444 | 1,989 | {"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} | 4.5625 | 5 | CC-MAIN-2024-38 | latest | en | 0.909753 |
https://tex.stackexchange.com/questions/448774/how-to-get-this-simple-histogram | 1,719,104,454,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00133.warc.gz | 488,664,485 | 35,321 | # How to get this simple histogram?
I want to create this:
But I've come up with this one:
I've searched the internet first but the histograms they were plotting were so much complicated.
\begin{tikzpicture}
\begin{axis}[
symbolic x coords={$0$, $1$, $2$},
xtick=data]
($0$, 0.25
($1$, 0.50)
($2$, 0.25)
};
\end{axis}
\end{tikzpicture}
You easily can draw it like normal rectangles using tikz and fill with north east lines from the patterns library.
\documentclass[border=5pt,tikz]{standalone}
\usetikzlibrary{patterns}
\begin{document}
\begin{tikzpicture}[yscale=4]
\draw (-1.5,0)--(3,0) (-1,0)--(-1,.75);
\draw[pattern=north east lines] (-.5,0) rectangle (.5,.25) (.5,0) rectangle (1.5,.5) (1.5,0) rectangle (2.5,.25);
\foreach \x in {0,1,2}
\draw (\x,0) -- (\x,-1pt) node[below]{\footnotesize\x};
\foreach \y in {0.25,0.50}
\draw (-1,\y) -- ({-1cm-4pt},\y) node[left]{\footnotesize\y};
\end{tikzpicture}
\end{document}
• The filling part wasn't actually necessary for me. So, it's much appreciated, thanks a lot! Commented Sep 1, 2018 at 1:39
• You're welcome, good luck! Commented Sep 1, 2018 at 2:59 | 393 | 1,116 | {"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} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.761345 |
http://www.jiskha.com/display.cgi?id=1350397339 | 1,495,860,093,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608773.42/warc/CC-MAIN-20170527040600-20170527060600-00093.warc.gz | 661,093,395 | 3,871 | # Physics
posted by on .
The ACE towing company tows a disabled 10,500 N automobile off the road at a constant speed. If the tow line makes an angle of 10.0° with the vertical as shown, what is the tension in the line supporting the car?
• Physics - ,
T•cosα=m•g
T=mg/cosα =
=10500/cos10°= 10660 N
• Physics - ,
10,448 N | 100 | 326 | {"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-2017-22 | latest | en | 0.88868 |
http://www.everything2.com/title/Why+beverage+cans+have+concave+bottoms | 1,408,510,245,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500800168.29/warc/CC-MAIN-20140820021320-00278-ip-10-180-136-8.ec2.internal.warc.gz | 352,836,351 | 14,764 | Actually, this node has so far entirely missed the reason that the bottoms of aluminum beverage cans are curved. It's simple:
Curved walls are stronger
This is true not only for cans, but for all sorts of structures and containers, it's a fundamental engineering principle. Here are some other examples of this principle in action:
• Automobiles. Notice that the hood of your car has few (or no) flat surfaces, it's all curves. Same is true for most of the rest of the surfaces (windshield is another good example - side windows don't have to be as strong, and they have to open, so they aren't curved).
• Corrugated boxes, roofs and so on. The corrugation makes for strength.
• Other containers, such as propane tanks, scuba tanks, all sorts of things. If they have to stack, then the bottom is either curved inward (which reduces the volume the container can hold), or a false bottom is placed around the real bottom to make them stackable.
Why is it this way? Because the strength of a cross-section goes by its cross-sectional moment of inertia. If this doesn't make sense to you, think about this:
Hold a flat sheet of paper in your hand. It's floppy and has no structural strength to speak of. This is because it has almost no moment of inertia around the bending axis (ie, all of its mass is on the bending axis). Now, roll the paper into a tube. Suddenly, the same sheet of paper has considerable stiffness. By rolling it into a tube, you increased its moment of inertia around the bending axis (by moving its mass away from the axis and out to the edge of the tube). This is also why I-beams have that "I" shape, but you're probably tired of this by now.
Keep an eye on the world around you. You will see this principle in action over and over again.
Nifty, eh?
The bottom of the soda can is concave. (The bottom of plastic bottles is convex with a false bottom to make it stand up.)
The sides of the can are (gasp!) not flat either! They are also round--convex in this case. Milk cartons have flat sides. Soda cans do not. Also note that plastic milk jugs are frequently not flat either, but have some kind of indentation in them to increase strength.
The bottom of your propane tank pops from concave to convex immediately prior to catastrophic failure, just because the force to pop the bottom is less than the force to burst the tank, and so that will usually occur first. I don't think it has anything to do with a built in warning. If there was such a thing, it would be in the form of a relief valve, like what is on most liquid air tanks.
Plastic soda bottles probably have a convex bottom because the pressure needed to collapse a plastic concave bottom is so low that the fizz in the soda itself would do it immediately. Jhonbus's explanation below of the bottom sounds right to me.
Beverage cans (usually made to contain carbonated drinks) are designed to hold their shape while pressurised, in fact non-carbonated drinks are pressurised with nitrogen to give their cans strength while being shipped and stored.
Cans must be stackable; that is it must be possible to place one on top of the other in a stable manner. If the bottom of aluminium cans were not made in this concave shape, and were made flat instead, they would tend to bulge outward slightly from the pressure, thus making stacking of the cans impossible. Such methods as adding plastic sleeves to the bottoms of cans is not really feasible, as the cost per can must be kept down to an absolute minimum when such a large volume of product is produced. (This is somewhere round about the trillion can per year mark).
Here in England, and indeed all the other European countries I have visited, our plastic beverage bottles are not designed with a concave bottom and a false base, but are moulded into a pentagonal curvy shape which gives it both strength under pressure, and the ability to stand up on a flat surface. (I can't really describe the shape, but I'm sure there are pictures of it somewhere on the www.)
Previous writeups in this node have suggested a wide variety of potential answers to the “concave beverage can bottom” question, ranging from the fanciful to the mundane to the practical:
• Hovercraft” Theory: One theory says that the concave bottom allows the beverage companies (and for purposes of this writeup, we’ll stick to beer and soda distributors) to blow air under the cans, allowing them to glide along the assembly line like a puck in an air hockey game. This idea really doesn’t hold water, though. Not only doesn’t it have any references to back it up, but it’s pretty clear that the rollers along the average assembly line would allow the cans to move at least as fast as if they were “floating,” making any extravagant “hovercraft” model pretty much superfluous. < /li>
• Capitalist Conspiracy” Theory: This idea says that the concave bottom is really a deep, capitalist conspiracy to cheat “Joe Six Pack” out of a few extra ounces of tasty beverage per can. The problem with this idea is that the product – beer, soda, whatever -- is sold by the volume of the product, not the volume of the can. Twelve fluid ounces, or 355 ml, to be exact, making any reduction in can volume completely irrelevant. So this theory likewise falls short. < /li>
• “Depressurization” Theory: Under this approach, the concave bottom of the can is meant to handle rapid changes in air pressure when the beverage is transported in an unpressurized aircraft. Aside from the fact that most beer and soda is generally manufactured and canned near the market in which it is sold, the simple fact is that the theory doesn’t work. If you take any can of pressurized beverage (beer, soda, carbonated water) on an unpressurized plane up to 35,000 feet, you’ll have an explosion on your hands. I guarantee it. < /li>
• ”Stacking the Cans” Theory: Here, the theory is that the concave bottom is needed to allow the cans to be stacked. Otherwise, the theory goes, the cans would all bulge out from the internal pressure, making it impossible for distributors and grocers to stack the product. That’s all well and good, in theory, but the simple fact is that we don’t see this kind of bulging in real life, whether it’s on the concave bottom or the flat top of beverage cans. So if the flat tops don’t “bulge,” it seems pretty clear that the bottoms could be flat, too, without “bulging.”
• By the way, if you ever do see bulging like this, DON’T CONSUME THE PRODUCT. It’s a sign that something may have gone wrong in the canning process, and that the product may be toxic. Of course, it may just be a sign that the deliveryman dropped the can, in which case it will probably explode in your face. Either way, opening it is probably a bad idea.< /li>
• ”Structural Strength” Theory: Of all the “concave bottom” theories put forth in this node, this one comes closest to being scientifically accurate. But while it’s not inherently wrong –- concave bottoms are structurally stronger –- it doesn’t satisfactorily explain why distributors use them in their product. I mean, steel is stronger than aluminum, so if strength were the only concern, wouldn’t distributors use steel in their cans, instead of aluminum? The fact that they don’t tells us that something besides mere “structural strength” is at work here.
• In fact, that last question gets you to the real answer, now, doesn’t it? Manufacturers want to use the strongest material and shape that they can, while minimizing weight in the process. Why do they want to do this? Money, that’s why.
Companies that manufacture beer and soda containers in the United States produce over 300 million aluminum beverage cans per day. That’s 100 billion cans per year. With that level of production -– more than one can per person in the United States per day -– even a small reduction in unit cost will result in huge savings for the industry. And a reduction in weight translates directly into a reduction in cost -- a one percent reduction in a can’s weight means a savings for the industry of as much as \$40 million.
So there’s big money to be made. In fact, the amount of money at stake is enough to suggest that the shape and materials in your average aluminum can have probably been structurally optimized. The average beverage consumer, in the U.S. and elsewhere, might be surprised to learn that manufacturers of aluminum cans exercise the same amount of attention and precision as engineers designing aircraft wings for the newest experimental planes. It’s that important.
The Optimal Can Shape
All right. Suppose I had twelve ounces of a pressurized liquid. The average pressure in a modern-day beverage can is 90 psi, so let’s assume that’s what I’ve got. Here’s the question -- ignoring all other considerations, what would be the optimal shape for a container holding these twelve ounces of liquid?
The answer is a sphere. Mathematically speaking, a sphere has the smallest possible surface area for any given volume. At least in three dimensions. The reason for this is a question of calculus that goes beyond the scope of this writeup. But just picture the lowly soap bubble. The soap film naturally gravitates to the most efficient –- lowest energy –- state of being, a sphere.
Needless to say, when it comes to manufacturing a container, the smallest surface area means the least amount of material, hence the lowest cost. So why don’t we see spherical beer cans? Well, to some extent, we do. Some manufacturers even make spherical kegs, which are, after all, nothing more than big beer cans. But spheres are hard to distribute. They roll, they don’t stack well, and they’d probably be pretty difficult to drink from.
So what’s the next best thing?
Cylindrical Cans
Well, let’s start with the sphere, a shape we know to be optimal. Now, cut a two-dimensional slice in the sphere. What do you have? A circle. And while a sphere is an optimal shape in three dimensions, a circle is optimal in two. It is the shape that requires the least circumference for a given area to be enclosed.
But beer and soda –- and everything else, for that matter -– don’t exist in two dimensions, they exist in three. So can we take a circle –- the best container shape in two dimensions -– and expand it out to three? The answer, of course, is yes. Just take the circle, and extend it along a third axis as far as necessary to enclose the required volume.
In other words, a cylinder. Which is nothing more than a can. Not only that, it’s a can holding a pressurized liquid, which engineers refer to as a pressure vessel. The can relies on the pressure from within to retain its structural integrity, just as a fire hose relies on the water pressure to maintain its shape.
You’re Tops In My Book
Now that we have a cylinder, what do we do with the top and bottom? I mean, if we’re not careful, we’re going to wind up with something no more useful to drink from than a fire hose. Looking first at the top, it’s clear that it needs to be flat –- not for any engineering reason, but for customer convenience. It’s hard to get your mouth around a spherical top, after all. Insert juvenile joke here.
But while the amount of aluminum necessary to withstand the lateral 90 psi pressure was fairly minimal for the can’s cylinder –- the circular shape was, after all, structurally optimal –- the flat top needs to be significantly stronger. Just imagine trying to hold back the water at the end of a fire hose.
To increase the lid’s strength, manufacturers reduce the amount of manganese in the aluminum alloy, while increasing the amount of magnesium. The typical aluminum alloy in the sides of a beverage can incorporates by weight 1 percent magnesium, 1 percent manganese, 0.4 percent iron, 0.2 percent silicon, and 0.15 percent copper, in addition to aluminum. It is ironed to tolerances within 0.0001 inch, and is made slightly thicker at the top and the bottom for added integrity. In addition to withstanding the 90 psi lateral pressure, the typical aluminum can is able to support up to 250 pounds on its lateral axis. In other words, you can stand on it.
Now, with the top lid, the magnesium content can reach a full 2 percent of the alloy’s weight, with the manganese content being reduced to a trace. While this shift in alloy content makes the lid stronger, it also makes it significantly heavier. To reduce the added weight, manufacturers make the diameter of the lid less than the body of the rest of the can, resulting in the characteristic taper you see in cans today. Even with this weight reduction measure, however, the lid of an average aluminum beverage can often makes up 25 percent or more of the total weight of the can.
Bottoms Up
OK, how about the other end of the can? We still need something to withstand the increased longitudinal pressures of the liquid, but we’re no longer constrained by the consumer’s need to drink out of it. That is, unless the consumer is a frat boy shotgunning a Bud.
The first question is what material are we going to use? Do we use the stronger magnesium-rich material from the top, or do we stick with the alloy we’ve used for the sides? Well, we’d rather not add the extra weight from the top lid all over again, so what we really want to do is to come up with a shape that will be strong enough to hold the liquid, even with the lighter alloy.
What’s the strongest shape we can think of? Fortunately, the answer has been around for thousands of years. Ever seen a Roman Arch? The guiding principle behind this little engineering feat was that it allowed the Romans to cut down on material when they were building stuff –- bridges, aqueducts, you know –- by using a shape that diverted the gravitational pressure from above down to the base, without the need for material to fill the base.
So think the Arc de Triomphe. Think the Roman aqueducts leading into the fabled city. Think pretty much any bridge –- from antiquity to the present –- that spans more than twenty feet. The arch –- a simple yet utilitarian design –- buys you a lot of architectural bang for the buck.
So how does this apply to the bottom of your average Diet Pepsi can? Well, imagine an arch in three dimensions – an arch “in the round” if you will. That’s exactly what you see at the bottom of every soda and beer can made today. The curved shape dissipates the pressure from the beverage around the rim at the bottom, while still allowing the manufacturer to use the lighter, manganese-rich alloy.
Money, Money, Money
So in the end, it all comes down to money. The concave bottom on beverage cans isn’t for “floating,” or for “stacking,” or for “pressure stabilization.” And while the shape is structurally stronger, that’s not the real reason it’s used. At the end of the day, the concave shape allows manufacturers to use the cheapest, lightest materials –- today’s beverage cans now weigh less than 0.48 ounces, compared to 0.66 ounces in the 1960’s –- as a means to reduce costs.
Welcome to Capitalism.
Bibliography
• The Aluminum Beverage Can, William F. Hosford and John L. Duncan, Scientific American, September 1994. More than you would ever want to know, from two writers who have devoted 30 years of their lives to the lowly aluminum can.
• History of the Beverage Can, Museum of Beverage Containers and Advertising (not a joke) (http://www.gono.com/v-tours/sodacone/scone3.htm)
• The Evolution of Useful Things (http:www.eiu.edu/scienced/3290/science/discrepant/popcan.html)
• Soap Bubble (http://wn.wikipedia.org/wiki/Soap_bubble)
• Arch (http://wn.wikipedia.org/wiki/Arch)
• Two years as an undergraduate physics major, combined with years of being the guinea pig test reader for my father's books such as The Scientist Goes to the Seashore, The Scientist Goes to the Mountains, and The Scientist Goes to the City
• Additional experience with several of my father's lesser-known works, such as The Scientist Abandons His Family, The Scientist Refuses to See His First Grandson, and The Scientist Finds Out That His Wife Cheated On Him With A Fireman. I understand that The Scientist Burns in Hell will be a posthumous work. | 3,556 | 16,128 | {"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-2014-35 | longest | en | 0.96163 |
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