url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://www.jiskha.com/display.cgi?id=1382047163 | 1,527,456,612,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00078.warc.gz | 780,225,493 | 3,927 | # math
posted by Taylor
Computer software:\$44
Discount:21%
Sales tax:6%
1. John
Do what you did on the previous problems :P
100/100 - 21/100 = 79/100
100/100 - 6/100 = 94/100
79/100 * 94/100 * 44 = ?
2. John
Whoops, I meant
100/100 - 21/100 = 79/100
100/100 + 6/100 = 106/100
79/100 * 106/100 * 44 = ?
3. Taylor
john is it 36.85
4. John
Yerp
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More Similar Questions | 601 | 2,183 | {"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} | 3.3125 | 3 | CC-MAIN-2018-22 | latest | en | 0.934352 |
https://www.airmilescalculator.com/distance/bdl-to-boi/ | 1,621,089,396,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991370.50/warc/CC-MAIN-20210515131024-20210515161024-00167.warc.gz | 640,873,232 | 77,101 | Distance between Windsor Locks, CT (BDL) and Boise, ID (BOI)
Flight distance from Windsor Locks to Boise (Bradley International Airport – Boise Airport) is 2192 miles / 3528 kilometers / 1905 nautical miles. Estimated flight time is 4 hours 39 minutes.
Driving distance from Windsor Locks (BDL) to Boise (BOI) is 2580 miles / 4152 kilometers and travel time by car is about 42 hours 35 minutes.
Map of flight path and driving directions from Windsor Locks to Boise.
Shortest flight path between Bradley International Airport (BDL) and Boise Airport (BOI).
How far is Boise from Windsor Locks?
There are several ways to calculate distances between Windsor Locks and Boise. Here are two common methods:
Vincenty's formula (applied above)
• 2192.256 miles
• 3528.094 kilometers
• 1905.018 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 2186.450 miles
• 3518.750 kilometers
• 1899.973 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
Airport information
City: Windsor Locks, CT
Country: United States
IATA Code: BDL
ICAO Code: KBDL
Coordinates: 41°56′20″N, 72°40′59″W
B Boise Airport
City: Boise, ID
Country: United States
IATA Code: BOI
ICAO Code: KBOI
Coordinates: 43°33′51″N, 116°13′22″W
Time difference and current local times
The time difference between Windsor Locks and Boise is 2 hours. Boise is 2 hours behind Windsor Locks.
EDT
MDT
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 239 kg (528 pounds).
Frequent Flyer Miles Calculator
Windsor Locks (BDL) → Boise (BOI).
Distance:
2192
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
2192
Round trip? | 500 | 1,900 | {"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} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.839511 |
https://www.coursehero.com/file/6470508/Ch21-2/ | 1,496,008,407,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463611560.41/warc/CC-MAIN-20170528200854-20170528220854-00362.warc.gz | 1,059,048,443 | 23,316 | # Ch21_2 - Balancing Equations Balancing Equations for Redox...
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Unformatted text preview: Balancing Equations Balancing Equations for Redox Reactions for Redox Reactions for Redox 1 2 Balancing Equations Balancing Equations Some redox reactions have equations that must be balanced by special techniques. MnO4- + 5 Fe2+ + 8 H+ 2+ 2+ ---> Fe3+ + H2 O Fe = +2 ---> Mn + 5 Fe Mn = +7 Balancing Equations Balancing Equations Step 4: Multiply each half-reaction by a factor that means the reducing agent supplies as many electrons as the oxidizing agent requires. Reducing agent Cu ---> Cu2+ + 2eOxidizing agent 2 Ag+ + 2 e- ---> 2 Ag Step 5: Add half-reactions to give the overall equation. Cu + 2 Ag+ ---> Cu 2+ + 2Ag The equation is now balanced for both charge and mass. Cu + Ag+ 4 --give--> Cu2+ + Ag Balancing Equations Balancing Equations Balance the following in acid solution— 2+ VO2+ + Zn ---> VO 2+ + Zn2+ VO Zn Step 1: Write the half-reactions Ox Zn ---> Zn 2+ Red VO2+ ---> VO 2+ Step 2: Balance each half-reaction for mass. Ox Zn ---> Zn 2+ Red 2 H+ + VO2+ ---> VO2+ + H2O Add H2O on O-deficient side and add H + on other side for H-balance. Page 1 3 Step 1: Divide the reaction into halfreactions, one for oxidation and the other for reduction. Ox Cu ---> Cu2+ Red Ag+ ---> Ag Step 2: Balance each for mass. Already done in this case. Step 3: Balance each half-reaction for charge by adding electrons. Ox Cu ---> Cu2+ + 2e2eRed Ag+ + e- ---> Ag ---> Consider the reduction of Ag+ ions with copper metal. Mn = +2 Fe = +3 Balancing Equations Balancing Equations 5 Balancing Equations Balancing Equations Step 3: Ox Red Step 4: Ox Red 2e- Balance half-reactions for charge. Zn ---> Zn 2+ + 2ee- + 2 H+ + VO2+ ---> VO2+ + H2 O eMultiply by an appropriate factor. Zn ---> Zn 2+ + 2e2e+ 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2O Step 5: Add half-reactions Zn + 4 H+ + 2 VO2 + ---> Zn 2+ + 2 VO2+ + 2 H2O 6 7 Tips on Balancing Equations • Never add O 2, O atoms, or O2- to balance oxygen. • Never add H2 or H atoms to balance hydrogen. • Be sure to write the correct charges on all the ions. • Check your work at the end to make sure mass and charge are balanced. Page 2 ...
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## This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.
Ask a homework question - tutors are online | 759 | 2,518 | {"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-2017-22 | longest | en | 0.695825 |
https://www.edaboard.com/showthread.php?373061-MIcrostrip-Line-with-COMSOL | 1,532,270,519,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593302.74/warc/CC-MAIN-20180722135607-20180722155607-00386.warc.gz | 892,501,626 | 19,968 | # MIcrostrip Line with COMSOL
1. ## MIcrostrip Line with COMSOL
Hello everyone,
I am simply simulating a quasi-TEM mode of a microstrip line in COMSOL. I got the results as expected, but there is one problem that when I change the length of microstrip, the transmission peak frequency shifts. Can anyone comment on this.
Parameters of line:-
Substrate thickness = 0.9 mm
Substrate relative epsilon = 9.9
Width of metal strip = 0.87
Length of dielectric and ground plate = 60mm (arbitrarly choosen)
Width of dielectric and ground plate= 8 mm (arbitrarly choosen)
I am exciting with the default lumped port of COMSOL at the starting of the transmission line.
•
2. ## Re: MIcrostrip Line with COMSOL
Why you're surprised ?? If the Characteristic Impedance of a Transmission Line is not exactly System's Characteristic Impedance, Input Impedance will draw a small circle around Smith Charts center and will show lower impedance and higher impedance at every Lambda/4.
3. ## Re: MIcrostrip Line with COMSOL
So, should I choose it's length Lambda/4? What about width(not thickness) of dielectric or ground plate? In literature the length and width should be infinite. But for simulations, these have to be finite.
•
4. ## Re: MIcrostrip Line with COMSOL
There is no "correct" length. A frequency dependency is what a transmission line is expected to show. Plot your S11 results in Smith chart and you will see the input impedance curve circle around the line impedance value.
Substrate and ground of 4 * line width should be fine, fields that are further away should not have much impact.
In any case, for such "I modelled something and get unexpected results" questions it is useful to show screenshots of model and/or results.
1 members found this post helpful.
5. ## Re: MIcrostrip Line with COMSOL
Originally Posted by jiten.phy
So, should I choose it's length Lambda/4? What about width(not thickness) of dielectric or ground plate? In literature the length and width should be infinite. But for simulations, these have to be finite.
You talk about unrelated different things.Your phenomena can be illustrated as ;
So, when Input Impedance is decided once, the length can be found OR length is given, Input Impedance can be found.
6. ## Re: MIcrostrip Line with COMSOL
The images that I have attached shows the simulation of microstrip line with physical length 75mm. So as I interpret the S11 parameter plot, can I say that frequencies with lowest S11 value will propagate with minimum reflection. Why are there so many dips? In my simulation the frequency values seems to depend on the physical length of microstrip. Is there a formula from where relating frequency and physical length?
•
7. ## Re: MIcrostrip Line with COMSOL
Regarding the ripple, looks at the posts above and do some reading on this topic. In #5 Bigboss provided the equation relating input impedance, line impedance and load impedance with electrical length (= length measured in wavelength). That is where frequency comes into your results and creates ripple. The same concept is explained in more detail in my old application note on EM simulation of lines, see attachment.
It is obvious that your line impedance is not well matched to the source/load impedance. This can be a correct result from geometry, or an artefact from too coarse meshing in the solver.
Analysis of RFIC Transmission Lines.pdf
--[[ ]]-- | 770 | 3,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-30 | latest | en | 0.890817 |
https://www.sanfoundry.com/c-program-implement-alexander-bogomolnys-unordered-permutation-algorithm-elements/ | 1,718,907,698,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00164.warc.gz | 846,936,176 | 19,804 | # C Program to Implement the Alexander Bogomolny’s UnOrdered Permutation Algorithm for Elements From 1 to N
This C program computes all possible permutations of numbers from 1 to N using Alexander Bogomolyn’s algorithm. This algorithm is implemented in recursive fashion.
Here is the source code of the C program to find permutations of numbers from 1 to N. The C program is successfully compiled and run on a Linux system. The program output is also shown below.
1. `#include <stdio.h>`
2. `void print(const int *v, const int size)`
3. `{`
4. ` int i;`
5. ` if (v != 0) {`
6. ` for ( i = 0; i < size; i++) {`
7. ` printf("%4d", v[i] );`
8. ` }`
9. ` printf("\n");`
10. ` }`
11. `}`
12. `void alexanderbogomolyn(int *Value, int N, int k)`
13. `{`
14. ` static level = -1;`
15. ` int i;`
16. ` level = level+1; Value[k] = level;`
17. ` if (level == N)`
18. ` print(Value, N);`
19. ` else`
20. ` for (i = 0; i < N; i++)`
21. ` if (Value[i] == 0)`
22. ` alexanderbogomolyn(Value, N, i);`
23. ` level = level-1;`
24. ` Value[k] = 0;`
25. `}`
26. `int main()`
27. `{`
28. ` int N = 4;`
29. ` int i;`
30. ` int Value[N];`
31. ` for (i = 0; i < N; i++) {`
32. ` Value[i] = 0;`
33. ` }`
34. ` printf("\nPermutation using Alexander Bogomolyn's algorithm: ");`
35. ` alexanderbogomolyn(Value, N, 0);`
36. ` return 0;`
37. ` `
38. `}`
```\$ gcc permute.c -o permute
\$ ./permute
Permutation using Alexander Bogomolyns algorithm:
1 2 3 4
1 2 4 3
1 3 2 4
1 4 2 3
1 3 4 2
1 4 3 2
2 1 3 4
2 1 4 3
3 1 2 4
4 1 2 3
3 1 4 2
4 1 3 2
2 3 1 4
2 4 1 3
3 2 1 4
4 2 1 3
3 4 1 2
4 3 1 2
2 3 4 1
2 4 3 1
3 2 4 1
4 2 3 1
3 4 2 1
4 3 2 1```
Sanfoundry Global Education & Learning Series – 1000 C Programs. | 809 | 1,923 | {"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-2024-26 | latest | en | 0.365712 |
https://mihaceqycol.carriagehouseautoresto.com/algebra-and-real-life-linear-relationships244112590nh.html | 1,621,100,550,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990551.51/warc/CC-MAIN-20210515161657-20210515191657-00130.warc.gz | 420,363,413 | 3,224 | # Algebra and real life linear relationships
Learning Objectives Apply linear mathematical models to real world problems Key Takeaways Key Points A mathematical model describes a system using mathematical concepts and language.
Example Points: The points are graphed in a scatterplot fashion.
Almost any situation where there is an unknown quantity can be represented by a linear equation, like figuring out income over time, calculating mileage rates, or predicting profit. How fast was his average speed over the course of the run?
Rates Linear equations can be a useful tool for comparing rates of pay. The equation and graph can be used to make predictions.
Outlier Approximated Line: Here is the approximated line given the new outlier point at -1, 6. Linear equations are a tool that make this possible.
## Algebra and real life linear relationships
Notice 4 points are above the line, and 4 points are below the line. At pm he starts to run and leaves his home. Plugging it into the first equation gives us [latex]50 2. Updated March 13, By Jessica Smith Linear equations use one or more variables where one variable is dependent on the other. Key Terms mathematical model: An abstract mathematical representation of a process, device, or concept; it uses a number of variables to represent inputs, outputs, internal states, and sets of equations and inequalities to describe their interaction. The two trains meet at the intersections point [latex] 2. Extrapolation refers to the use of a fitted curve beyond the range of the observed data, and is subject to a greater degree of uncertainty since it may reflect the method used to construct the curve as much as it reflects the observed data. Anything that involves a constant rate of change can be nicely represented with a line with the slope. Real world applications can also be modeled with multiple lines such as if two trains travel toward each other. Linear equations are a tool that make this possible. Our second point is [latex]1. Indeed, so long as you have just two points, if you know the function is linear, you can graph it and begin asking questions! Rates Linear equations can be a useful tool for comparing rates of pay. One difficulty with mathematical models lies in translating the real world application into an accurate mathematical representation. Learning Objectives Apply linear mathematical models to real world problems Key Takeaways Key Points A mathematical model describes a system using mathematical concepts and language.
The simplest and perhaps most common linear regression model is the ordinary least squares approximation.
Rated 9/10 based on 8 review | 510 | 2,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-21 | latest | en | 0.927816 |
https://www.inchcalculator.com/convert/kilowatt-hour-per-hundred-miles-to-mile-per-gallon-equivalent/ | 1,720,840,245,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00097.warc.gz | 675,285,747 | 15,962 | # Kilowatt-Hours per 100 Miles to Miles per Gallon Gasoline Equivalent Converter (KWh/100mi to MPGe)
Enter the electric car efficiency in kilowatt-hours per 100 miles below to get the value converted to miles per gallon gasoline equivalent.
## Result in Miles per Gallon Gasoline Equivalent:
1 kWh/100mi = 3,370.5 MPGe
Do you want to convert MPGe to kWh/100mi?
## How to Convert Kilowatt-Hours per 100 Miles to Miles per Gallon Gasoline Equivalent
Convert kilowatt-hours per 100 miles to miles per gallon gasoline equivalent with this simple formula:
MPGe = 3370.5 ÷ kWh/100mi
Insert the kWh/100mi efficiency measurement in the formula and then solve to find the result.
For example, let's convert 50 kWh/100mi to MPGe:
50 kWh/100mi = ( 3370.5 ÷ 50 ) = 67.41 MPGe
Kilowatt-hours per 100 miles and miles per gallon gasoline equivalent are both units used to measure electric car efficiency. Keep reading to learn more about each unit of measure.
## What Are Kilowatt-Hours per 100 Miles?
Kilowatt-hours per 100 miles are a measure of electric vehicle efficiency measuring the kilowatt-hours of energy required to travel 100 miles.
The kilowatt-hour per 100 miles is a US customary unit of electric car efficiency. Kilowatt-hours per 100 miles can be abbreviated as kWh/100mi; for example, 1 kilowatt-hour per 100 miles can be written as 1 kWh/100mi.
In the expressions of units, the slash, or solidus (/), is used to express a change in one or more units relative to a change in one or more other units.
## What Are Miles per Gallon Gasoline Equivalent?
Miles per gallon gasoline equivalent is a way of measuring the energy needed equivalent to a gallon of gasoline to travel a distance in miles.
Because electric vehicles do not consume gasoline, the EPA set an equivalent amount of energy in kilowatt-hours that is equal to one gallon of gasoline. According to the EPA, one gallon of gasoline is equal to 33.705 kilowatt-hours.[1]
The mile per gallon gasoline equivalent is a US customary unit of electric car efficiency. Miles per gallon gasoline equivalent can be abbreviated as MPGe; for example, 1 mile per gallon gasoline equivalent can be written as 1 MPGe.
## Kilowatt-Hour per 100 Miles to Mile per Gallon Gasoline Equivalent Conversion Table
Table showing various kilowatt-hour per 100 miles measurements converted to miles per gallon gasoline equivalent.
Kilowatt-hours Per 100 Miles Miles Per Gallon Gasoline Equivalent
0.001 kWh/100mi 3,370,500 MPGe
0.002 kWh/100mi 1,685,250 MPGe
0.003 kWh/100mi 1,123,500 MPGe
0.004 kWh/100mi 842,625 MPGe
0.005 kWh/100mi 674,100 MPGe
0.006 kWh/100mi 561,750 MPGe
0.007 kWh/100mi 481,500 MPGe
0.008 kWh/100mi 421,313 MPGe
0.009 kWh/100mi 374,500 MPGe
0.01 kWh/100mi 337,050 MPGe
0.02 kWh/100mi 168,525 MPGe
0.03 kWh/100mi 112,350 MPGe
0.04 kWh/100mi 84,263 MPGe
0.05 kWh/100mi 67,410 MPGe
0.06 kWh/100mi 56,175 MPGe
0.07 kWh/100mi 48,150 MPGe
0.08 kWh/100mi 42,131 MPGe
0.09 kWh/100mi 37,450 MPGe
0.1 kWh/100mi 33,705 MPGe
0.2 kWh/100mi 16,853 MPGe
0.3 kWh/100mi 11,235 MPGe
0.4 kWh/100mi 8,426 MPGe
0.5 kWh/100mi 6,741 MPGe
0.6 kWh/100mi 5,618 MPGe
0.7 kWh/100mi 4,815 MPGe
0.8 kWh/100mi 4,213 MPGe
0.9 kWh/100mi 3,745 MPGe
1 kWh/100mi 3,371 MPGe
## References
1. United States Environmental Protection Agency, Technology, https://www3.epa.gov/otaq/gvg/learn-more-technology.htm | 1,012 | 3,350 | {"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} | 3.390625 | 3 | CC-MAIN-2024-30 | latest | en | 0.787246 |
http://www.sciencedaily.com/releases/2014/07/140716165830.htm | 1,455,050,807,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701157472.18/warc/CC-MAIN-20160205193917-00259-ip-10-236-182-209.ec2.internal.warc.gz | 649,762,922 | 18,347 | Science News
from research organizations
# Fair cake cutting gets its own algorithm
Date:
July 16, 2014
Source:
Springer
Summary:
A mathematician and a political scientist have announced an algorithm by which they show how to optimally share cake between two people efficiently, in equal pieces and in such a way that no one feels robbed.
Share:
FULL STORY
Cutting birthday cake (stock image).
The next time your children quibble about who gets to eat which part of a cake, call in some experts on the art of sharing. Mathematician Julius Barbanel of Union College, and political scientist Steven Brams of New York University, both in the US, published an algorithm in Springer's The Mathematical Intelligencer by which they show how to optimally share cake between two people efficiently, in equal pieces and in such a way that no one feels robbed.
The cut-and-choose method to share divisible goods has been regarded as fair and envy-free since Biblical times, when Abraham divided land equally, and Lot could choose the part he wanted. But being free of envy is not the only consideration when sharing something. What happens when more than two cuts can be made, or when people prefer different, specific sections of whatever is to be divided? Barbanel and Brams believe that with a giveback procedure it is possible to make a perfect division between two people that is efficient, equitable and void of jealousy.
An objective referee (such as a Mom or a computer) is essential to the plan.. The potential cake eaters first tell the referee which parts of the delicacy they value most. In mathematical terms these are called someone's probability density functions, or pdfs. The referee then marks out the cake at all points were the pdfs of the disgruntled would-be cake eaters cross, and assigns portions. If at this point the two parties receive the same size of cake, the task is over. If not, the giveback process starts.
The party who received the larger part of the cake during the first round must give a part of it back to the other person, starting with those parts in which the ratio of their pdfs is the smallest. This goes on until the parties value their portions equally, and have the same volume of cake to eat. This method only works with a finite number of cuts if the players' pdfs are straight-lined, or are so-called piecewise linear sections.
The researchers believe the method can be used to share cake and other divisible goods such as land. In the case of beachfront property being co-owned by two developers, for example, it can help to determine who gets what strips of land to build on based on the pieces of land they value most.
"This allocation is not only equitable but also envy-free and efficient -- that is, perfect," says Barbanel.
"This approach focuses on proving the existence of efficient and envy-free divisions, not on providing algorithms to finding them," emphasizes Brams.
Story Source:
The above post is reprinted from materials provided by Springer. Note: Materials may be edited for content and length.
Journal Reference:
1. J.B. & Brams, S.J. .Two-Person Cake Cutting: The Optimal Number of Cuts. The Mathematical Intelligencer, July 2014 DOI: 10.1007/s00283-013-9442 | 682 | 3,233 | {"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-2016-07 | latest | en | 0.972328 |
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camdota
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A232909 Number of (1+1) X (n+1) 0..2 arrays with every element next to itself plus and minus one within the range 0..2 horizontally or antidiagonally. 1
0, 10, 52, 82, 432, 2372, 8456, 29970, 132972, 547554, 2102096, 8441132, 34612080, 138342906, 551613860, 2224984018, 8958729072, 35906450644, 144189580120, 579895060802, 2329383080028, 9353592833314, 37582177354512, 151003083442364 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 FORMULA Empirical: a(n) = 4*a(n-2) + 28*a(n-3) + 80*a(n-4) + 48*a(n-5) - 67*a(n-6) - 228*a(n-7) - 224*a(n-8). Empirical g.f.: 2*x^2*(5 + 26*x + 21*x^2 - 28*x^3 - 106*x^4 - 104*x^5) / ((1 + x + 7*x^2)*(1 - x - 10*x^2 - 11*x^3 + x^4 + 28*x^5 + 32*x^6)). - Colin Barker, Oct 06 2018 EXAMPLE Some solutions for n=5: ..0..1..0..0..1..0....2..1..2..0..1..0....2..1..0..2..1..0....2..1..0..0..1..2 ..2..0..1..2..1..2....0..1..0..2..1..0....0..1..2..2..1..2....2..1..1..2..1..0 CROSSREFS Row 1 of A232908. Sequence in context: A135242 A041186 A058827 * A028994 A257042 A092966 Adjacent sequences: A232906 A232907 A232908 * A232910 A232911 A232912 KEYWORD nonn AUTHOR R. H. Hardin, Dec 02 2013 STATUS approved
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# Calculation of the CTI
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Calculation of the CTI
Khaled Bekhouche
Biskra/Lancaster University
Calculation of the CTI
Khaled Bekhouche
Biskra/Lancaster University
Published by: Salah Eddine Bekhouche on Jul 19, 2011
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Calculation of the CTI
Meeting 27.02.2009Meeting 27.02.2009
Khaled BekhoucheBiskra/Lancaster University
Calculation method
1-Construction of the distribution of ADC codes of each 10 pixels(Slide 3). The number of frames is 10000.2-Fit the noise and the X-ray by a Gaussian functions (Slide 3).3-For each pixel we calculate the noise centroid and the X-ray centroidby averaging the ADC codes within the two intervals:(X
0noise
-m
noise
0noise
+m
noise
), for the noise(X
0xray
-m
xray
0xray
+m
xray
), for X-raywhere m=1, since 69% of the Gauss events are located in this region.4-The difference between the X-ray centroid and the noise centroidgives the X-ray peak (Slide 3).5-Fit the X-ray peaks vs pixel number with a linear function (Slide 4) toobtain the CTI.
0100200300400501
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3 | 438 | 1,413 | {"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-2015-35 | latest | en | 0.763831 |
https://blogs.sas.com/content/graphicallyspeaking/2021/03/15/sas-graphs-for-r-programmers-bar-charts/ | 1,701,616,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00753.warc.gz | 171,489,546 | 16,952 | This is another in my series of blog posts where I take a deep dive into converting customized R graphs into SAS graphs. Today we'll be working on bar charts ...
And to give you a hint about what data I'll be using this time, here's a picture from a SAS break room, that my buddy John took. Yes, we have free M&Ms in the break room - the rumor is true! But, they only restock them once a week, and if the co-workers on your floor like M&Ms, then they're gone by the end of the day (which is probably for the best - otherwise there would be a lot of overweight people at SAS, LOL).
Sometimes when you're teaching a graphing technique, it's best to use some simple (fun) data. Therefore, this time I've decided to use the frequency counts of the various colors in a packet of M&Ms. Here are the R and SAS bar charts of the data, followed by an explanation and comparison of the R and SAS code:
### My Approach
I will be showing the R code (in blue) first, and then the equivalent SAS code (in red) that I used to create both of the bar charts. Note that there are many different ways to accomplish the same things in both R and SAS - and the code I show here isn't the only (and probably not even the 'best') way to do things. If you know of a better/simpler way to code it, feel free to share your suggestion in the comments!
Also, I don't include every bit of code here in the blog post (in particular, things I've already covered in my previous posts). I include links to the full R and SAS programs at the bottom.
### The Data
Since this example uses a very small amount of pre-summarized data, I just include it in the code. Here's how I did it in R. Note that the data is in 'random' order (neither alphabetical, nor ascending/descending numeric order).
mnm_color count
Green 99
Red 86
Blue 102
Orange 73
Yellow 54
Brown 77
")
And here's how I did it in SAS:
data my_data;
length mnm_color \$10;
input mnm_color Count;
datalines;
Green 99
Red 86
Blue 102
Orange 73
Yellow 54
Brown 77
;
run;
### Sorted Bar Chart
Depending on the nature of the data, and the answers you want to get from the graph, you might want to order the bars alphabetically, in ascending/descending numeric order, or a special custom order. In this case, I'd like the bars in descending numeric order.
In R, I use ggplot and the geom_bar() function to draw the bar chart, and the reorder() function to sort bars on-the-fly based on the values in the count variable. The negative sign before count tells it to sort them in descending order.
my_plot <- ggplot(my_data, aes(x=reorder(mnm_color,-count),y=count,
fill=mnm_color,label=count,text=calculated_percent)) +
geom_bar(color="#777777",width=.7,stat="identity") +
In SAS, I use Proc Sort to sort the data by descending count, and then Proc SGplot's vbarparm to draw the bar chart (it draws the bars in data-order).
proc sort data=my_data out=my_data;
by descending count;
run;
proc sgplot data=my_data noborder noautolegend;
vbarparm category=mnm_color response=count / group=mnm_color
groupdisplay=cluster barwidth=0.80;
### Coloring the Bars
In some graphs, the colors don't really matter - as long as they look good together, and are easily distinguishable. But in this case, I want the bar colors to represent the M&M colors ... therefore I can't leave things to chance. In both R and SAS, I could hard-code a color list, and specify the colors in a specific order such that each bar will get the desired color ... but if I ever want to re-run this code with slightly different data, then that hard-coded color-mapping might not be in the necessary order with the new/different data. Therefore I want to specify the colors in such a way that the blue M&M count is always guaranteed to get the blue color, and so on.
In R, I set up a color palette called 'pal', and then told the scale_fill_manual() function to use that palette to color the bars.
pal <- c(
"Blue" = "#4cbbe6",
"Green" = "#74e059",
"Red" = "#d22515",
"Orange" = "#fbb635",
"Yellow" = "#f4f25f",
"Brown" = "#5d242a"
)
scale_fill_manual(values=pal,limits=names(pal)) +
In SAS, I create an attribute map dataset, and specify the fillcolor for each of the M&M text values. And then in Proc SGplot, I use the dattrmap option to point to that dataset. And then in the vbarparm, I tell it which attribute id to use (in this case there's only one id in the dataset, but you could have multiple ones to control various different aspects of the graph).
data myattrs;
length value linecolor markercolor \$100;
id="someid";
linecolor="gray99";
fillcolor="cx4cbbe6"; value="Blue"; output;
fillcolor="cx74e059"; value="Green"; output;
fillcolor="cxd22515"; value="Red"; output;
fillcolor="cxfbb635"; value="Orange"; output;
fillcolor="cxf4f25f"; value="Yellow"; output;
fillcolor="cx5d242a"; value="Brown"; output;
run;
proc sgplot data=my_data noborder noautolegend dattrmap=myattrs;
vbarparm category=mnm_color response=count / group=mnm_color attrid=someid
groupdisplay=cluster barwidth=0.80;
### Values on Bars
When there's only a small amount of data in a graph, I often like to show the data values (numbers) right there on the graph. That way the user doesn't have to work too hard, and visually guess/interpolate the values, based on the axes and gridlines. In this case, I want to show the frequency count inside the bar, and the percent (which I'll have to calculate) outside the bar. This will help the user more easily answer questions such as "What percent of the M&Ms were green?"
In R I use the mutate function() to calculate the percent values:
my_data <- my_data %>% mutate(calculated_percent = count/sum(count))
And then the following two lines add the text to the bars. Note that the geom_label (text labels inside the bar) allows me to specify a fill color behind the text, and an alpha transparency for that fill. The fill helps guarantee that the label will be easy to read, and the transparency helps it 'blend' in with the graph.
geom_label(size=3.2,vjust=1.0,fontface="bold",fill=alpha(c("white"),0.7)) +
geom_text(size=3.2,vjust=-.50,fontface="bold",aes(label=percent(calculated_percent,.1))) +
In SAS, I use Proc SQL to calculate the percent values, and then a data step to calculate a custom position value. I could have done both in the SQL, but I think the code is easier to follow this way.
proc sql noprint;
create table my_data as
select unique *, count/sum(count) format=percent7.1 as calculated_percent
from my_data;
quit; run;
data my_data; set my_data;
run;
I then use vbarparm's datalabel option to add the percent values above the bars, and the text command to add the count inside the bar at the 'adjusted_position' I calculated earlier.
vbarparm category=mnm_color response=count / group=mnm_color attrid=someid
datalabel=calculated_percent datalabelattrs=(size=11pt color=gray33 weight=bold)
groupdisplay=cluster barwidth=0.80;
strip position=bottom backfill fillattrs=(color=white transparency=.3)
textattrs=(size=11pt color=gray33 weight=bold);
### Final Clean Up
In most default graphs, I find the axes a bit too 'busy' and crowded. I like to customize my graphs to de-emphasize (or eliminate) the things that aren't important, and emphasize the things that are important.
In R's geom_bar() chart, there's a tick mark centered under each bar, and a vertical grid line going from that tick mark to the top of the graph. Why does a bar chart need that?!? Why is it the default? I get rid of the tick mark and grid line with the following commands.
theme(axis.ticks=element_blank()) +
theme(panel.grid.major.x=element_blank()) +
Let's also get rid of the default gray color behind the graph, and also eliminate the legend and the minor tick marks:
theme_bw() +
theme(legend.position="none") +
theme(panel.grid.minor=element_blank()) +
In a graph like this with a very short label on the Y axis, I like to position the label at the top of the axis, rather than in the side/margin area - this saves more space for the 'data' part of the graph. I don't think R lets me move the Y axis label into this exact position (I can move it to the top, but it's still in the left margin area). Therefore I get rid of the axis label altogether, and use a left-justified 'subtitle' to fake a Y axis label in the desired position:
labs(x=NULL,y=NULL) +
labs(subtitle="Count") +
theme(plot.subtitle=element_text(color="#555555",face="plain",hjust=-.06,size=11,margin=margin(0,0,12,0))) +
In SAS, I get rid of the tick marks with the xaxis noticks options:
xaxis display=(nolabel noticks);
I use the noautolegend option to get rid of the legend:
proc sgplot data=my_data noborder noautolegend dattrmap=myattrs;
And the SAS yaxis has a simple built-in labelpos=top to get the legend label in the exact position I like:
yaxis display=(noticks noline) labelpos=top grid gridattrs=(color=graydd);
### Candy Break!
If you made it through all that, I think you deserve to treat yourself to a handful of M&Ms! 🙂
### My Code
Here is a link to my complete R program that produced the R bar chart.
Here is a link to my complete SAS program that produced the SAS bar chart.
If you have any comments, suggestions, corrections, or observations - I'd be happy to hear them in the comments section!
Share | 2,325 | 9,255 | {"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-2023-50 | longest | en | 0.931771 |
https://www.teachoo.com/13616/402/Ex-1.1--4-iii/category/Ex-1.1/ | 1,722,995,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00598.warc.gz | 806,628,993 | 20,865 | Ex 1.1
Chapter 1 Class 9 Number Systems
Serial order wise
### Transcript
Ex 1.1, 4 State whether the following statements are true or false. Give reasons for your answers. (iii) Every rational number is a whole number. Fractions like 3/5 , 1/2 etc. are not whole number but are rational numbers Hence false | 82 | 309 | {"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-2024-33 | latest | en | 0.811401 |
http://er.yuvayana.org/regular-expression-in-theory-of-computation-solved-examples-part-2/ | 1,524,350,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945459.17/warc/CC-MAIN-20180421223015-20180422003015-00572.warc.gz | 102,726,766 | 22,326 | This is 2nd Part of Regular expression in theory of computation solved examples. You can also read Regular expression in theory of computation solved examples Part – 1.
## 1. Construct the regular expression for all strings in which all runs of a’s has lengths that are multiple of three, over input alphabets ∑ = {a, b, c}.
Solution:
We have the input alphabets ∑ = {a, b, c}.
Here, the resultant regular expression will denote the set of all string in which all runs of a’s has length that are multiple of three, i.e., the length of a’s in the string varies as 3 × 0 = 0, 3 × 1 = 3, 3 × 2 = 6, 3 × 3 = 9, 3 × 4 = 12,…. and soon.
To fix this problem, we first write the regular expression which represents the set of all strings over the given ∑. It is-
(a + b + c)*
But, the given problem requires the length of ‘a’ which is multiple of three.
Thus, the regular expression which fulfill this requirement of the given problem can be written as-
(aaa, + b + c)*
Hello I am Er Parag Verma. I am tech blogger, Professor and Entrepreneur. I am on the mission to change the pattern of learning to make it easy, valuable and advance. | 304 | 1,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-17 | longest | en | 0.927666 |
https://codegolf.stackexchange.com/questions/144146/frequency-distribution-of-multiple-dice-rolls/144147 | 1,580,119,492,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00238.warc.gz | 375,598,416 | 47,966 | # Frequency Distribution of Multiple Dice Rolls
Given two positive integers a and b, output the frequency distribution of rolling a b-sided die a times and summing the results.
A frequency distribution lists the frequency of each possible sum if each possible sequence of dice rolls occurs once. Thus, the frequencies are integers whose sum equals b**a.
## Rules
• The frequencies must be listed in increasing order of the sum to which the frequency corresponds.
• Labeling the frequencies with the corresponding sums is allowed, but not required (since the sums can be inferred from the required order).
• You do not have to handle inputs where the output exceeds the representable range of integers for your language.
• Leading or trailing zeroes are not permitted. Only positive frequencies should appear in the output.
## Test Cases
Format: a b: output
1 6: [1, 1, 1, 1, 1, 1]
2 6: [1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1]
3 6: [1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1]
5 2: [1, 5, 10, 10, 5, 1]
6 4: [1, 6, 21, 56, 120, 216, 336, 456, 546, 580, 546, 456, 336, 216, 120, 56, 21, 6, 1]
10 10: [1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92368, 167860, 293380, 495220, 810040, 1287484, 1992925, 3010150, 4443725, 6420700, 9091270, 12628000, 17223250, 23084500, 30427375, 39466306, 50402935, 63412580, 78629320, 96130540, 115921972, 137924380, 161963065, 187761310, 214938745, 243015388, 271421810, 299515480, 326602870, 351966340, 374894389, 394713550, 410820025, 422709100, 430000450, 432457640, 430000450, 422709100, 410820025, 394713550, 374894389, 351966340, 326602870, 299515480, 271421810, 243015388, 214938745, 187761310, 161963065, 137924380, 115921972, 96130540, 78629320, 63412580, 50402935, 39466306, 30427375, 23084500, 17223250, 12628000, 9091270, 6420700, 4443725, 3010150, 1992925, 1287484, 810040, 495220, 293380, 167860, 92368, 48620, 24310, 11440, 5005, 2002, 715, 220, 55, 10, 1]
5 50: [1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410, 135751, 148995, 163185, 178365, 194580, 211876, 230300, 249900, 270725, 292825, 316246, 341030, 367215, 394835, 423920, 454496, 486585, 520205, 555370, 592090, 630371, 670215, 711620, 754580, 799085, 845121, 892670, 941710, 992215, 1044155, 1097496, 1152200, 1208225, 1265525, 1324050, 1383746, 1444555, 1506415, 1569260, 1633020, 1697621, 1762985, 1829030, 1895670, 1962815, 2030371, 2098240, 2166320, 2234505, 2302685, 2370746, 2438570, 2506035, 2573015, 2639380, 2704996, 2769725, 2833425, 2895950, 2957150, 3016881, 3075005, 3131390, 3185910, 3238445, 3288881, 3337110, 3383030, 3426545, 3467565, 3506006, 3541790, 3574845, 3605105, 3632510, 3657006, 3678545, 3697085, 3712590, 3725030, 3734381, 3740625, 3743750, 3743750, 3740625, 3734381, 3725030, 3712590, 3697085, 3678545, 3657006, 3632510, 3605105, 3574845, 3541790, 3506006, 3467565, 3426545, 3383030, 3337110, 3288881, 3238445, 3185910, 3131390, 3075005, 3016881, 2957150, 2895950, 2833425, 2769725, 2704996, 2639380, 2573015, 2506035, 2438570, 2370746, 2302685, 2234505, 2166320, 2098240, 2030371, 1962815, 1895670, 1829030, 1762985, 1697621, 1633020, 1569260, 1506415, 1444555, 1383746, 1324050, 1265525, 1208225, 1152200, 1097496, 1044155, 992215, 941710, 892670, 845121, 799085, 754580, 711620, 670215, 630371, 592090, 555370, 520205, 486585, 454496, 423920, 394835, 367215, 341030, 316246, 292825, 270725, 249900, 230300, 211876, 194580, 178365, 163185, 148995, 135751, 123410, 111930, 101270, 91390, 82251, 73815, 66045, 58905, 52360, 46376, 40920, 35960, 31465, 27405, 23751, 20475, 17550, 14950, 12650, 10626, 8855, 7315, 5985, 4845, 3876, 3060, 2380, 1820, 1365, 1001, 715, 495, 330, 210, 126, 70, 35, 15, 5, 1]
• Can we assume that b is at least 2? (Or if not, what should the frequency list for sums of a 1-sided die look like?) – Misha Lavrov Oct 1 '17 at 3:32
• May we have leading or trailing zeroes? – xnor Oct 1 '17 at 4:56
# Octave, 38 bytes
@(a,b)round(ifft(fft((a:a*b<a+b)).^a))
Try it online!
### Explanation
Adding independent random variables corresponds to convolving their probability mass functions (PMF), or multiplying their characteristic functions (CF). Thus the CF of the sum of a independent, identically distributed variables is given by that of a single variable raised to the power of a.
The CF is essentially the Fourier transform of the PMF, and can thus be computed via a FFT. The PMF of a single b-sided die is uniform on 1, 2, ..., b. However, two modifications are required:
• 1 is used instead of the actual probability values (1/b). This way the result will be de-normalized and will contain integers as required.
• Padding with zeros is needed so that the FFT output has the appropriate size (a*b-a+1) and the implicit periodic behaviour assumed by the FFT doesn't affect the results.
Once the characteristic function of the sum has been obtained, an inverse FFT is used to compute the final result, and rounding is applied to correct for floating-point inaccuracies.
### Example
Consider inputs a=2, b=6. The code a:a*b<a+b builds a vector with b=6 ones, zero-padded to size a*b-a+1:
[1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0]
Then fft(...) gives
[36, -11.8-3.48i, 0.228+0.147i, -0.949-1.09i, 0.147+0.321i, -0.083-0.577i, -0.083+0.577i, 0.147-0.321i, -0.949+1.09i, 0.228-0.147i, -11.8+3.48i]
One can almost recognize the sinc function here (Fourier transform of a rectangular pulse).
(...).^a raises each entry to a and then ifft(...) takes the inverse FFT, which gives
[1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1]
Although the results in this case are exactly integers, in general there may be relative errors of the order of 1e-16, which is why round(...) is needed.
• I really Impressed! – rahnema1 Oct 2 '17 at 18:08
• @rahnema1 Signal processing for the win! – Luis Mendo Oct 2 '17 at 19:26
# Mathematica, 29 bytes
Tally[Tr/@Range@#2~Tuples~#]&
Just generates all possible dice rolls, takes their totals, then counts. Each frequency comes labeled with its value.
# Mathematica, 38 bytes
CoefficientList[((x^#2-1)/(x-1))^#,x]&
Expands (1+x+x^2+...+x^(a-1))^b and takes the coefficients of x. Since 1+x+x^2+...+x^(a-1) is the generating function for a single die roll and products correspond to convolutions - adding values of dice - the result gives the frequency distribution.
Thanks to Lynn for the Cartesian product trick. -11 bytes thanks to many Haskell tricks from Funky Computer Man, -2 bytes from naming, -2 bytes thanks to Laikoni. Golfing suggestions are welcome! Try it online!
import Data.List
g x=[1..x]
a!b=map length$group$sort$map sum$mapM g$b<$g a
Ungolfed
import Data.List
rangeX x = [1..x]
-- sums of all the rolls of b a-sided dice
diceRolls a b = [sum y | y <- mapM rangeX $fmap (const b) [1..a]] -- our dice distribution distrib a b = [length x | x <- group(sort(diceRolls a b))] # Pyth - 10 bytes Just takes all possible dice combinations by taking the cartesian product of [1, b], a times, summing, and getting the length of each sum group. lM.gksM^SE # 05AB1E, 8 bytes LIãO{γ€g Try it online! ## How? LIãO{γ€g - Full program. L - Range [1 ... input #1] I - Input #2. ã - Cartesian Power. O - Map with sum. { - Sort. γ - Group consecutive equal elements. €g - Get the length of each # R, 58 bytes function(a,b)table(rowSums(expand.grid(rep(list(1:b),a)))) Try it online! # R, 52 bytes function(a,b)Re(fft(fft(a:(a*b)<a+b)^a,T)/(a*b-a+1)) Try it online! A port of @Luis Mendo's Octave solution, fft(z, inverse=T) unfortunately returns the unnormalized inverse FFT, so we have to divide by the length, and it returns a complex vector, so we take only the real part. • well played - payback for yesterday's cmdscale I figure :-) – flodel Oct 3 '17 at 0:20 • @flodel hah! I'm actually going to award you a bounty for that one :) – Giuseppe Oct 3 '17 at 12:55 • You were not joking! So generous of you! I enjoy seeing (and learning from) your answers, I will pay it back quickly! – flodel Oct 5 '17 at 0:27 ## SageMath, 40 bytes lambda a,b:reduce(convolution,[[1]*b]*a) Try it online convolution computes the discrete convolution of two lists. reduce does what it says on the tin. [1]*b is a list of b 1s, the frequency distribution of 1db. [[1]*b]*a makes a nested list of a copies of b 1s. # Python 2 + NumPy, 56 bytes lambda a,b:reduce(numpy.convolve,[[1]*b]*a) import numpy Try it online! I've included this solution with the above one, since they're essentially equivalent. Note that this function returns a NumPy array and not a Python list, so the output looks a bit different if you print it. numpy.ones((a,b)) is the "correct" way to make an array for use with NumPy, and thus it could be used in place of [[1]*b]*a, but it's unfortunately longer. # Jelly, 5 bytes ṗS€ĠẈ Try it online! Note that this takes the arguments in reverse order. ## How? ṗS€ĠL€ - Full program (dyadic) | Example: 6, 2 ṗ - Cartesian Power (with implicit range) | [[1, 1], [1, 2], ... , [6, 6]] S€ - Sum each | [2, 3, 4, ... , 12] Ġ - Group indices by values | [[1], [2, 7], [3, 8, 13], ... , [36]] L€ - Length of each group | [1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1] Alternative solutions: ṗZSĠL€ ṗZSµLƙ ṗS€µLƙ # Python 2, 102 91 bytes lambda b,a:map(map(sum,product(*[range(a)]*b)).count,range(b*~-a+1)) from itertools import* Try it online! # Haskell, 61 bytes g x=[1..x] a#b=[sum[1|m<-mapM g$b<$g a,sum m==n]|n<-[a..a*b]] Try it online! Use as a#b. Partly based on Sherlock9's Haskell answer. # MATL, 9 bytes :Z^!Xs8#u Same approach as Maltysen's Pyth answer. Inputs are in reverse order. Try it online! # Pari/GP, 28 bytes a->b->Vec(((x^b-1)/(x-1))^a) Try it online! • As far as I can tell, this is the shortest solution that definitely doesn't run out of memory on any of the provided test cases. – Misha Lavrov Oct 1 '17 at 16:09 # Perl 5, 53 bytes $g=join',',1..<>;map$r[eval]++,glob"+{$g}"x<>;say"@r"
Try it online!
Input format:
b
a
## JavaScript (ES6), 94 bytes
f=(n,m,a=[1],b=[])=>n?[...Array(m)].map((_,i)=>a.map((e,j)=>b[j+=i]=(b[j]|0)+e))&&f(n-1,m,b):a
<div oninput=o.textContent=f(+n.value,+m.value).join\n><input id=n type=number min=0 value=0><input id=m type=number min=1 value=1><pre id=o>1
Limited by 32-bit integer overflow, but floats could be used instead at a cost of 1 byte.
• Umm... this only takes one input – Herman L Oct 1 '17 at 9:06
• @HermanLauenstein Sorry, I somehow completely overlooked that part of the question... will fix shortly. – Neil Oct 1 '17 at 9:18
# J, 25 24 21 20 bytes
3 :'#/.~,+//y$i.{:y' Try it online! Initially I incremented the [0..n-1] list to get [1..n] but apparently it’s not necessary. • Nice answer. Here's a tacit version for same number of bytes: #/.~@,@(+///)@$i.@{:. Seems like there should be a way to shave it down a bit more making the verb dyadic, but I wasn't able to do it. – Jonah Oct 1 '17 at 16:35
• @Jonah you have an extra / in +// – FrownyFrog Oct 1 '17 at 16:53
• Actually, you're right. It just happens to work both ways. I guess that solution saves a byte then :) – Jonah Oct 1 '17 at 17:14
# Javascript (ES6), 89 bytes
b=>g=a=>a?(l=[..."0".repeat(b-1),...g(a-1)]).map((_,i)=>eval(l.slice(i,i+b).join+)):[1]
Takes input in currying syntax in reverse order f(b)(a)
f=b=>g=a=>a>0?(l=[..."0".repeat(b-1),...g(a-1)]).map((_,i)=>eval(l.slice(i,i+b).join+)):[1]
r=_=>{o.innerText=f(+inb.value)(+ina.value)}
<input id=ina type=number min=0 onchange="r()" value=0>
<input id=inb type=number min=1 onchange="r()" value=1>
<pre id=o></pre>
# Actually, 13 12 bytes
-1 byte thanks to Mr. Xcoder. Try it online!
R∙♂Σ;╗╔⌠╜c⌡M
Ungolfed
Implicit input: b, a
R∙ ath Cartesian power of [1..b]
♂Σ Get all the sums of the rolls, call them dice_rolls
;╗ Duplicate dice_rolls and save to register 0
╔ Push uniquify(dice_rolls)
⌠ ⌡M Map over uniquify(dice_rolls), call the variable i
╜ Push dice_rolls from register 0
c dice_rolls.count(i)
Implict return
• You do not need the @, do you? – Mr. Xcoder Oct 1 '17 at 13:12
• As a side note, I found an interesting alternative: R∙♂Σ╗╜╔⌠╜c⌡M – Mr. Xcoder Oct 1 '17 at 16:54
# AWK, 191 bytes
Outputs frequencies as a vertical column.
func p(z){for(m in z)S[z[m]]++
for(i=$1;i<=$1*$2;i++)print S[i]}func t(a,b,z,s){if(a){if(R++)for(n in z)for(i=0;i++<b;)s[n,i]=z[n]+i else for(i=0;i++<b;)s[i]=i t(--a,b,s)}else p(z)}{t($1,$2)} Try it online! Adding 6 more bytes allows for multiple sets of inputs. func p(z,S){for(m in z)S[z[m]]++ for(i=$1;i<=$1*$2;i++)print S[i]}func t(a,b,z,s){if(a){if(R++)for(n in z)for(i=0;i++<b;)s[n,i]=z[n]+i
else for(i=0;i++<b;)s[i]=i
t(--a,b,s)}else p(z)}{R=0;t($1,$2)}
Try it online!
## Clojure, 86 bytes
#(sort-by key(frequencies(reduce(fn[r i](for[y(range %2)x r](+ x y 1)))[0](range %))))
An example:
(def f #(...))
(f 5 4)
([5 1] [6 5] [7 15] [8 35] [9 65] [10 101] [11 135] [12 155] [13 155] [14 135] [15 101] [16 65] [17 35] [18 15] [19 5] [20 1])
# C (gcc), 142 bytes
i,j,k;int*f(a,b){int*r=malloc(sizeof(int)*(1+a*~-b));r[0]=1;for(i=1;i<=a;i++)for(j=i*~-b;j>=0;j--)for(k=1;k<b&k<=j;k++)r[j]+=r[j-k];return r;}
Try it online!
• sizeof(int)? Really? – orlp Oct 1 '17 at 20:04
• @orlp environment-dependent, you know – Leaky Nun Oct 1 '17 at 20:11
• It's allowed for a C program to assume a particular architecture. As long as it works on at least one machine. Furthermore, 8 would work on any architecture, overallocating a bit but that's ok. – orlp Oct 1 '17 at 22:44
• r[0]=1;for(i=1;i<=a;i++)for(j=i*~-b; -> for(i=r[0]=1;i<=a;)for(j=i++*~-b; for -2 bytes. – Kevin Cruijssen Oct 2 '17 at 13:49
# Julia, 43 bytes
f(n,d)=reduce(conv,repmat([ones(Int,d)],n))
Try it online! | 5,296 | 13,844 | {"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} | 2.828125 | 3 | CC-MAIN-2020-05 | latest | en | 0.612681 |
https://www.pgrmseducation.com/ncert-solutions-for-class-6-maths-chapter-8-exercise-8-1/ | 1,660,548,468,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572161.46/warc/CC-MAIN-20220815054743-20220815084743-00613.warc.gz | 793,554,757 | 38,442 | # NCERT Solutions For Class 6 Maths Chapter 8 Exercise 8.1
Ncert Solutions For Class 6 Maths Chapter 8 Decimals Exercise 8.1 pdf:-This Chapter is Very Interesting And Easy. Students get Enjoyed while Solving The Questions of this Exercise. This is the first Exercise of This Chapter Decimals. This Chapter is also important in terms of basics clarity because the concept ofĀ Decimals is Further use in your Higher classes. If you Clear Decimals Concept
in 6th class Then you get very helpful in your next Higher Standard.
Exercise 8.1Ā Class 6 maths NCERT solutions Chapter 8 Decimals pdf download:-
Ā
### Ncert Solution for Class 6 Maths Chapter 8 Decimals Exercise 8.1 Textbook Solutions:-
CLASS-6 Ex. 8.1 Decimals In Image Format
Important TopicsĀ Which will Cover in Chapter 8 Decimals Exercise 8.1 Are:-
• Place Value Table
• Fractions Into Decimals
• Decimals Into Fractions
• Showing Decimals on Number Line
• Word Problems
• etc.
### Tips Which Are Helpful In This Exercise 8.1:-
• Give Your 100% Concentrations while Solving The Questions Of This Exercise Because small 1 mistake can Wrong Your Whole Question. And Waste your precious time to solve it again.
• While Solving Decimals Questions:- In Addition, Subtraction, Multiplication, Solve the Question With Decimals And Then Put the Decimals In in Answer which Will Increase your speed and Accuracy of Solving Problems. and Also Save Your Time and Solve More Questions
• Carefully Put Decimals On Number line and Check Again After Putting. Which Will Decrease the Chance Of Mistakes And Help to Achieve good Marks in Exam.
Ā
Introduction:–
Savita and Shama were going to market to buy some stationary items.
Savita said, āI have 5 rupees and 75 paiseā. Shama said, āI have 7 rupees
and 50 paiseā.
They knew how to write rupees and paise using decimals.
So Savita said, I have Rs 5.75 and Shama said,
āI have Rs 7.50ā.
Have they written correctly?
We know that the dot represents a decimal point.
with decimals.
Tenths
Ravi and Raju measured the lengths of their pencils. Raviās pencil was
7 cm 5mm long and Rajuās pencil was 8 cm 3 mm long. Can you express
these lengths in centimetre using decimals?
We know that 10 mm = 1 cm
Ā
Let us recall what we have learnt earlier.
If we show units by blocks then one unit is
one block, two units are two blocks and so
on. One block divided into 10 equal parts
means each part is
1
10 (one-tenth) of a unit, 2 parts show 2 tenths and 5
parts show 5 tenths and so on. A combination of 2 blocks and 3 parts
(tenths) will be recorded as : | 690 | 2,561 | {"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-2022-33 | longest | en | 0.848811 |
http://slideplayer.com/slide/5265623/ | 1,596,964,260,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738523.63/warc/CC-MAIN-20200809073133-20200809103133-00269.warc.gz | 96,466,811 | 23,908 | # Take out physics supplies. Label a clean sheet “Projectiles” Make a line down the center to form 2 columns. On the right, list what you know about projectiles.
## Presentation on theme: "Take out physics supplies. Label a clean sheet “Projectiles” Make a line down the center to form 2 columns. On the right, list what you know about projectiles."— Presentation transcript:
Take out physics supplies. Label a clean sheet “Projectiles” Make a line down the center to form 2 columns. On the right, list what you know about projectiles. On the left, list what you think you know, or your associations with projectiles.
Projectile Demo: 1. Sketch the path. 2. Does this example describe linear 1 dimensional motion? Explain. 3. Is there any acceleration? Explain. 4. What force(s) are acting on the object? 5. Where during its flight is the velocity the slowest? 6. Where during its flight is the velocity the fastest?
Projectiles objects move vertically & horizontally, 2 dimensional, at same time. The only force acting on them is gravity! They accelerate in the vertical direction. They have constant forward velocity in the (horizontal direction).
For projectiles launched horizontally – (no initial vertical velocity) the projectile moves forward & down at the same time. The forward & vertical v are separate – they occur independently. The Resultant v is the vector sum of the perpendicular X(horz) & Y(vector) vector components.
Horizontal Launch Demo How does the launch velocity affect the time in the air?
Once launched it’s X (forward or horizontal) v stays constant. What about the Y (vertical) component? The Y v is accelerated motion with a = g. The Y v increases as the object falls. What about as it rises?
Demo Horz launch vs drop. Strobe position vs. Time 6 seconds. https://www.youtube.com/watch?v=z24_ihi kEqQ&list=PL3B2111CE2F7C797Bhttps://www.youtube.com/watch?v=z24_ihi kEqQ&list=PL3B2111CE2F7C797B
Cannonballs
Shoot with no gravity, object moves at const. v forever.
Trajectory with gravity. Object drops as it falls.
Can solve 2 motion equations to find resultant distance, velocity.
Resultant trajectory (path) = parabola.
Launch w/ no vertical v i V horiz stays const. V vert starts at zero & increases.
Drop from plane no v iy v horiz same as plane. Object stays below plane as it falls.
Running off a cliff
A cannon fires a cannonball parallel to the ground. What factors control trip time for a horizontal launch? Demo Height Only PHET Launch http://phet.colorado.edu/sims/proj ectile-motion/projectile- motion_en.html
For a horizontal launch the total trip time is controlled by the launch height !
What variable is always the same for horizontal and vertical projectile motion?
The only variable that is always the same in the X and Y direction is the trip time!
Projectile – object moving in 2 dimensions with constant horizontal velocity and accelerating vertically with gravity. Trajectory – the path of the projectile. Parabola – the shape of the trajectory. Range – the horizontal distance traveled. Vocabulary
Horizontal Launch Problems Horizontal Launch v is constant use v = d/t. Accl g in vertical. Use an acceleration equation.
1. A cannonball is shot horizontally from a 125 m cliff. Its horizontal velocity is 75 m/s. List the known & unknown horizontal & vertical variables. How long was the cannonball in the air? What was the range of the cannonball?
2. An airplane traveling parallel to the ground at 100 m/s drops a package from 3000-m. Calculate the time to hit the ground. How far in front of the target must the package be dropped? t = 25 s. d = 2500 m
3. A unicorn runs off a high diving board with a horizontal velocity of 2.8 m/s and lands in the water 7.3 meters in front of the board. How high is the platform? 33 m
Hwk MC packet and pg 102 Text
Do Now: For a projectile launched at a constant speed, what variable(s) determine the total trip time? For a projectile launched at a constant speed, what variable(s) determine how far forward (range) the projectile will travel? height v x, trip time
Projectiles Launched at Angles Resultant Velocity
The initial velocity is forward & up.
The velocity has a horizontal & vertical component at every point - including the initial launch velocity. If we know v x & v y, at a point, we can calculate the total resultant v, from Pythagorean v 2 = v x 2 + v y 2.
Resolution of Resultant to Components Any 2-d vectors (velocity, force, etc) can be described as the sum of perpendicular vectors v 2 = v x 2 + v y 2. Often, instead of adding vector components to give resultant, we take resultant & resolve it (break it ) into perpendicular components.
If plane takes off at an angle it has a forward and upward velocity component like a projectile. It is useful to break or “resolve” the velocity vector into horizontal & vertical components. Make a rough sketch of its velocity vector relative to the ground.
Vector a can be broken down, or resolved into 2 perpendicular components: a y & a x.
Ex 1: How fast must a car be moving to stay beneath a plane taking off at 105 km/h at 25 o to the ground? Need horizontal component of plane v. v x = v cos . v x = 105 km/h cos 25 = 95 km/h
b. What is the vertical plane velocity from the previous example? v y = v sin km/h sin 25 o 44 km/h
Ex 2: The landing speed of the space shuttle is 99.7 m/s. If the shuttle is landing at an angle of 15 o to the horizontal: a. How fast is it descending? b. What is its horizontal velocity? a. 25.8 m/s b. 96.3 m/s
Hwk pg Text pg 94 #2-7. Finish Horz Projectile Prac Sheet.
Need Rifle & Ammunition Bookbags between you & neighbors.
What assumptions can you make??
Velocity Vectors V 2 = v x 2 + v y 2 v iy ± 0. v x = Vcos . v iy = V sin .
For projectiles launched at angles, the v iy IS NOT zero. The total resultant v is a combination of v x & v y. Find v tot from V 2 = v x 2 + v y 2. Given v tot we resolve the initial total v into X and Y components. v x = v cos v iy = v sin .
Other Assumptions needed. v x = constant velocity v = d/t. Vertical (Y) motion is accelerated a = g. v y top = 0. This can be v f or v i. t up = t down. v iy = - v fy. Same altitude. v y up = v y down. Same altitude.
Ex: A cannonball is launched at a 30 o angle with a vertical velocity of 20 m/s and a horizontal velocity of 80 m/s. Make a rough vector sketch of the cannonball. Calculate its initial resultant velocity. Calculate the time to reach the high point in its trajectory. If it lands at the same height as it was launched, calculate its horizontal range.
1. A cannonball is launched at a 55 o angle to the horizontal at a velocity of 150 m/s. It lands at the same height from which it was fired. Make a sketch of path of the cannonball and on your sketch make vector arrows to show the horizontal and vertical velocity, and acceleration at 3 different points. Find the initial horizontal & vertical velocity. Find the maximum height. Find the total time of the trip. How far horizontally did it travel? (range).
v iy = 150 m/s (sin 55) = 123 m/s v x = 150 m/s (cos 55) = 86 m/s Max height v f 2 = v i 2 + 2ad 771 m= h. Tot trip time t = v/a (123m/s) - (- 123 m/s) /-9.81 m/s 2. 25 s. 2156 m.
Finish Projectiles the Sequel Film Mechanical Universe “Falling Bodies” Hewitt Angle Projectile Launch http://www.youtube.com/watch?v=H-Y4PcV_mto
Hanging Monkey Problem
Hwk Rev Book pg 37 “Try It” Prb AND Text pg 114 #26, 28, and pg 117 #58a.
Download ppt "Take out physics supplies. Label a clean sheet “Projectiles” Make a line down the center to form 2 columns. On the right, list what you know about projectiles."
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https://blogs.mathworks.com/cleve/2016/08/08/bug-report-revives-interest-in-svd-option-of-eigshow/?s_tid=blogs_rc_1 | 1,680,366,036,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950110.72/warc/CC-MAIN-20230401160259-20230401190259-00118.warc.gz | 177,910,890 | 31,835 | # Bug Report Revives Interest in SVD Option of “Eigshow”
A few days ago we received email from Mike Hennessey, a mechanical engineering professor at the University of St. Thomas in St. Paul, Minnesota. He has been reading my book “Numerical Computing with MATLAB” very carefully. Chapter 7 is about “Eigenvalues and Singular Values” and section 10.3 is about one of my all-time favorite MATLAB demos, eigshow. Mike discovered an error in my description of the svd option of eigshow that has gone unnoticed in the over ten years that the book has been available from both the MathWorks web site and SIAM.
### Contents
#### Eigshow
The program eigshow has been in the MATLAB demos directory for many years. I wrote a three-part series of posts about eigshow in this blog three years ago, but I’m happy to write another post now.
The svd option of eigshow invites you to use your mouse to move the green vector x and make A*x perpendicular to A*y. The animated gif above simulates that motion. The following figure is the desired final result.
As you move x, the vector y follows along, staying perpendicular to x. The two trace out the green unit circle. The default matrix A is shown in the title. (If you run your own eigshow you can change the matrix by editing the text box in the title.)
A = [1 3; 4 2]/4
A =
0.2500 0.7500
1.0000 0.5000
The blue vectors Ax and Ay are the images of x and y under the multiplicative mapping induced by A. As you move x and y around the unit circle Ax and Ay sweep out the blue ellipse. And when you stop with Ax perpendicular to Ay, they turn out to be the major and minor axes of the ellipse.
#### NCM
Numerical Computing with MATLAB, which is known to its friends as simply NCM, was published over 10 years ago, in 2004. An electronic edition is available from MathWorks and a print edition from SIAM. In chapter 7, the singular value decomposition, the SVD, of a real matrix, $A$, is defined as the product
$$A = U \Sigma V^T$$
In the simplest case where we assume $A$ is square, $U$ and $V$ are orthogonal and $\Sigma$ is diagonal.
This is illustrated by figure 10.3 in NCM, which is the same as the final figure here. It shows the action with the svd option on the default matrix. The explanation is given by the first sentence on page 7.
The vectors x and y are the columns of U in the SVD, the vectors
Ax and Ay are multiples of the columns of V, and the lengths of
the axes are the singular values.
Sounds OK, right? This is the sentence that Mike Hennessey questioned. Now that I call your attention to it, you should be able to spot the error right away. I’ll tell you what it is at the end of this post.
#### help eigshow
Here is the relevant paragraph from the help entry for eigshow . It does not get into the same trouble as NCM.
In svd mode, the mouse moves two perpendicular unit vectors, x and y.
The resulting A*x and A*y are plotted. When A*x is perpendicular to
A*y, then x and y are right singular vectors, A*x and A*y are
multiples of left singular vectors, and the lengths of A*x and A*y
are the corresponding singular values.
#### SVD
Here’s how I often think about the SVD. Take the definition,
$$A = U \Sigma V^T$$
Multiply both sides on the right by $V$.
$$A V = U \Sigma$$
The diagonal matrix $\Sigma$ is on the right so that the singular values can multiply the columns of $U$. When we write this out column by column, we have
$$A v_j = \sigma_j u_j, \ \ j = 1,\ …, n$$
A little bit more manipulation leads to
$$A^T u_j = \sigma_j v_j, \ \ j = 1,\ …, n$$
#### Example
Let’s see how this works out with the default 2-by-2 matrix.
A
A =
0.2500 0.7500
1.0000 0.5000
Compute the SVD.
[U,S,V] = svd(A)
U =
-0.5257 -0.8507
-0.8507 0.5257
S =
1.2792 0
0 0.4886
V =
-0.7678 0.6407
-0.6407 -0.7678
The singular values are on the diagonal of S.
sigma = diag(S)
sigma =
1.2792
0.4886
These are the lengths of the two blue vectors.
#### Dominant singular vector
It turns out that the components of $v_1$, the right singular vector corresponding to the largest singular value of an (irreducible) matrix with nonnegative elements all have the same sign. They could be all positive or all negative — take your pick. MATLAB happens to always pick negative. So, both components of V(:,1) are negative. But we stopped eigshow with a positive x. So we flip the sign of V(:,1).
x = -V(:,1)
y = V(:,2)
x =
0.7678
0.6407
y =
0.6407
-0.7678
These two look like the two green vectors in the final figure. And these look like the two blue vectors.
Ax = A*x
Ay = A*y
Ax =
0.6725
1.0881
Ay =
-0.4156
0.2569
Now check that the SVD relationship works.
AV = A*V
USigma = U*S
AV =
-0.6725 -0.4156
-1.0881 0.2569
USigma =
-0.6725 -0.4156
-1.0881 0.2569
#### The Correction
What was the error in NCM that Mike Hennessey reported and that prompted me to write this post? The matrices U and V should be switched. The sentence should read
The vectors x and y are the columns of V in the SVD, the vectors
Ax and Ay are multiples of the columns of U, and the lengths of
the axes are the singular values.
Every time I work with the SVD I have to think carefully about U and V. Which is which? Maybe writing this post will help.
#### Reference
Cleve Moler, Numerical Computering With MATLAB, 2004.
MathWorks, <https://www.mathworks.com/moler/chapters.html>,
Published with MATLAB® R2016a
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https://www.jiskha.com/display.cgi?id=1346392363 | 1,503,436,126,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112682.87/warc/CC-MAIN-20170822201124-20170822221124-00655.warc.gz | 927,181,122 | 4,303 | Math
posted by .
Draw a diagram to solve this problem: Ajax is 8 km due west of Oshawa. Uxbridge is 6 km NW of Oshawa. How far is it from Ajax to Uxbridge? Explain whether you have enough information to solve this problem?
• Math -
Draw the -8km vector from the origin on the neg. x-axis.
Draw the 6km vector from the origin 45o
west of north(135o).
Du = 6km @ 135o.
Du = 6*cos135 + i6*sin135
Du = -4.24 + i4.24 = Dist. from Oshawa
to Uxbridge.
Da = -8 km=Dist. from Oshawa to Ajax.
D = Du - Da
D = (-4.24+i4.24) - (-8)
D = 3.76 + i4.24
D^2 = (3.76)^2 + (4.24)^2 = 32.12
D = 5.67 km.
• Math -
post it.
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### Explore BrainMass
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Phil and Susan are married, filing a joint return. The couple has two dependent children. Susan has wages of \$34,000 in 2008. Phil does not work due to a disability, but he is a buyer and seller of stocks on the Internet. He generally buys and holds for long-term gain, but occasionally gets in and out of a stock quickly. The couple's 2008 stock transactions are detailed below. In addition, they have \$, 2300 of qualifying dividends.
Item Date Acquired Date Sold Cost Sales Price
Black stock 11/10/07 3/12/08 \$2K \$5K
Blue stock 12/13/06 5/23/08 \$36K \$32K
Puce stock 12/14/03 7/14/08 \$13K \$14,5K
Ecru stock 6/29/07 5/18/08 \$26K \$27K
Red stock 5/15/07 10/18/08 \$67K \$67,800
Gray stock 4/23/06 10/18/08 \$89K \$88,200
What is Phil and Susan's AGI?
#### Solution Preview
First, let us compute their net short- or long-term capital gain or loss:
Item Short-term Long-term
Black stock (\$3,000)
Blue stock ...
#### Solution Summary
Phil and Susan are married, filing a joint return. The couple has two dependent children. Susan has wages of \$34,000 in 2008. Phil does not work due to a disability, but he is a buyer and seller of stocks on the Internet. He generally buys and holds for long-term gain, but occasionally gets in and out of a stock quickly. The couple's 2008 stock transactions are detailed below. In addition, they have \$, 2300 of qualifying dividends.
Item Date Acquired Date Sold Cost Sales Price
Black stock 11/10/07 3/12/08 \$2K \$5K
Blue stock 12/13/06 5/23/08 \$36K \$32K
Puce stock 12/14/03 7/14/08 \$13K \$14,5K
Ecru stock 6/29/07 5/18/08 \$26K \$27K
Red stock 5/15/07 10/18/08 \$67K \$67,800
Gray stock 4/23/06 10/18/08 \$89K \$88,200
What is Phil and Susan's AGI?
\$2.19 | 582 | 1,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-34 | latest | en | 0.891572 |
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### Fair Games and 50% Win Chances
I’ll take it as an assumption in the rest of this article that a fair game is one where each player has a 50% chance of winning. We also sometimes call such a situation a “good match” or say that the game will be good if we believe such a state of affairs prevails. We also tend to view negatively the opposite condition, wherein one player has a huge advantage over the other and hence where we expect the probability of that player losing is very low (implying the probability of the other player losing is very high).
These considerations aren’t limited to two player competitive games. If we are playing a single player, digital or otherwise interactive game, we call that game “fair” when we have about a 50% chance of winning. We would call a game where our chance of winning is ~1% unfair or badly designed, and where our chance of winning is ~99% boring or badly designed.
At first glance this seems to imply a contradictory attitude, one illustrated by recalling that we also call a coin flip “fair” if there is a 50% chance of the coin landing on either face. If the purpose of a game is to determine which player is the better player, how can it be that we seem to also want the outcome of the game to be as random as possible (such that for good matches, each player has a 50% chance of winning). It would appear that good games have random outcomes and that seems to contradict their apparent purpose in measuring how well a player plays.
(NB. The account is a little harder to render in the case of single player interactive systems. However, it seems paradoxical that a player would engage with a system with the intent of winning when the outcome could equivalently be determined by the toss of a coin).
### Resolution
I don’t think this is a genuine paradox, of course: when we say a game is fair, what we are saying is that the outcome isn’t random, but that it depends, sensitively, on which player makes the better sequence of moves in response to the other player. Why sensitively? Well, when two players are closely matched the the outcome of the game, if the win probability for either player is 50%, should depend very sensitively on how well each player actually plays. In particular, close matches come down to one or two critical mistakes or strokes of brilliance to tip the scales in one direction.
(This is particularly true because of another property of games (approximate reversibility) which I believe games must also have, but which I don’t discuss here.)
So it isn’t really surprising that we can resolve this merely apparent contradiction about games. But the resolution points us towards another important argument:
Because the outcome of a good game should depend sensitively on the moves of the player, the randomness present in a good game should be minimal or not present at all. Why? Because if the outcome of a game depends sensitively on the moves the player makes, then it also must necessarily depend sensitively on random influences on the game state. Why? For outcomes to depend sensitively on a move implies that each move a player makes is carefully tuned for the game state, which they have correctly appreciated in order to make the right move. But if the game state changes randomly, then a good move might be turned into a bad move by a random change in the game state.
(It is possible to imagine random changes to the game state which don’t change the quality of moves. But if this is the case, then these changes to the game state are _extraneous_ to playing the game and may as well be removed).
### Conclusion
To restate the argument:
1. we believe games should be fair, which is to say that a given player should have a 50% change of winning
2. this is because we want games to be sensitive tests of the quality of play of the given player, where the outcome depends sensitively on moves. We don’t want the game itself to be actually be random in the sense that the outcome is extraneous to the game itself.
3. Random elements (which are necessarily extraneous to the game in their origin) reduce the sensitivity of the win condition on the specific moves made by a player
4. Hence, good games should have minimal random elements.
This argument puts game designers in a difficult position. For designers of multiplayer games, they must make sure that the game’s rules don’t advantage particular players or add the appropriate handicap if they do. This turns out to be difficult. In Chess, for instance, white has a slight win chance, although the precise probability is unknown. Typically, for a new game without a long history of play, it will be very hard to determine whether such a bias exists and what size it might be.
With the rise of computers and single player strategy games a different set of design concerns manifests. The temptation in single player game design is to use random elements to provide variety for a gameplay system which may not have the strategic depth furnished by the presence of a second rational player. It is hard to imagine a deterministic single player game with the same initial conditions each play that can stand up to repeated play.
I think the way forward here is to randomize the initial conditions of any such game subject to the constraint that a given initial condition preserves the win 50% rate (perhaps based on artificial intelligence play or some other way of characterizing win chance) and then to make play from that point forward completely deterministic.
# The Ethics of Game Design
In the next week or so, I’ll be on the Dinofarm Games Community Podcast talking about the ethics of game design. My baby is just one week old, though! So I might not have been as coherent there as I wanted to be. As such, I thought I’d collect a few notes here while they were still in my head.
As a preamble: there are lots of ethical implications of games that I don’t discuss here. Particularly social ones: since games often depict social and cultural situations (like novels, plays or television shows) similar ethical concerns operate for games as for those artifacts. Here I’m specifically interested in those special ethical questions associated with games as interactive systems.
The question I’m interested in is: “What are the ethical obligations of a game designer, particularly to the player?” In a way, this is an old question in a new disguise, recognizable as such since the answer tends to dichotomize in a familiar way: is the game designer supposed to give the player what they want or is she supposed to give the player that which is good for them?
Let’s eliminate some low hanging fruit: if we design a game which is addictive, in the literal sense, I think most people will agree that we’ve committed an ethical lapse. There are a few folks out there with unusual or extreme moral views who would argue that even a game with bona fide addictive qualities isn’t morally problematic, but to them I simply say we’re operating with a different set of assumptions. However, the following analysis should hopefully illuminate exactly why we consider addictive games problematic as well as outline a few other areas where games ethical impact is important.
I think the most obvious place to start with this kind of analysis is to ask whether games are leisure activity, recreation or whether they provide a practical value. By leisure activity I mean any activity which we perform purely for pleasure, by recreation, I mean an activity that is performed without an immediate practical goal but which somehow improves or restores our capacity to act on practical goals, and by practical value, I mean something which immediately provides for a concrete requirement of living.
Its a little unclear where games fall into this rubric. It is easiest to imagine that games are purely leisure activities. This fits the blurb provided by the wikipedia article and also dovetails, broadly, with my understanding of games in public rhetoric. Categorizing games as purely leisure activities seems to justify a non-philosophical attitude about them: what is the point of worrying about the implications of that which is, at a fundamental level, merely a toy¹?
Point number one is that even toys, which have no practical purpose but to provide fun, are subject to some broad ethical constraints. It isn’t implausible to imagine that we could implant an electrode directly into a person’s brain such that the application of a small current to that electrode would produce, without any intervening activity, the sensation of fun. We could then give the person a button connected to that electrode and allow them to push it. This is technically an interactive system, perhaps even a highly degenerate game. It is certainly providing the player with the experience of fun, directly. However, its likely that a person so equipped would forego important practical tasks in favor of directly stimulating the experience of fun. If we gradually add elements between button presses and the reward or between the electrodes and the reward circuitry, we can gradually transform this game into any interactive system we could imagine. Clearly, at some point, the game might lose its property that it overwhelms the player’s desire to perform practical tasks. That line is the line between ethical and non-ethical game design.
In other words, game designers subscribing to the leisure theory of games are still obligated, perhaps counter-intuitively, to make their games sufficiently unfun that they don’t interfere with the player’s practical goals.
We have two interpretations of game value: the recreational and the practical interpretations.
Of these, the idea of the game as recreation may be closest to what is often discussed on the Dinofarm Discord channel. Its also frequently the narrative used to justify non-practical games. You’ve likely heard or even used the argument that digital games can improve hand-eye coordination or problem solving skills. This interpretation rests on their existing an operational analogy between the skills required to play a game and those required to perform practical tasks. There is a lot of literature on whether such a link exists and what form or forms it takes.
If no such link exists we can rubbish this entire interpretation of games, so its more interesting to imagine the opposite (as it least seems to sometimes be the case). When a link exists the value proposition for a game is: this game provides, as a side effect of play, a practical benefit. Why the phrase “as a side effect of play?” Because, if the purpose of the game is to provide the practical benefit, then we must always compare our game against some practical activity which might provide more of that same benefit than an equivalent effort directed towards non-game activity.
To choose a particularly morally dubious example, we might find that playing Doom improves firing range scores for soldiers. But shouldn’t we compare that to time spent simply practicing on the firing range? Without some further argumentative viscera, this line of thinking seems to lead directly to the conclusion that if games are recreation, we might always or nearly always find some non-game activity which provides a better “bang” for our buck.
Elaborating on this line of argument reveals what the shape of the missing viscera might be. Why is it plausible that we could find some non-game activity that works as well or better than any given game at meeting a practical end? Because games must devote some of their time and structure to fun and, as such, seem to be less dense in their ability to meet a concrete practical goal. In Doom, for instance, there are a variety of mechanics in the game which make it an exciting experience which don’t have anything to do with the target fixation behavior we are using to justify our game.
But we can make an argument of the following form: a purely practical activity which results the improvement of a skill requires an amount of effort. That effort might be eased by sweetening the activity with some fun elements, converting it to a game, allowing less effort for a similar gain of skill.
On this interpretation the ethical obligation of the game designer is to ensure that whatever skill they purport to hone with their game is developed for less effort than the direct approach. If they fail to meet this criteria, then they fail to provide the justification for their game.
The final interpretation we need to consider is that games themselves provide a direct, practical, benefit. I think this is a degenerate version of the above interpretation. It turns out to be difficult to find examples of this kind of game, but they do exist. Consider Fold.it, a game where player activity helps resolve otherwise computationally expensive protein folding calculations.
In this kind of game the developer has a few ethical obligations. The first is to make sure that the fun the game provides is sufficient compensation for the work the player has done or to otherwise make sure the player’s play is given with informed consent. For instance, if we design a game that gives player’s fun to solve traveling salespeople problems which, for some reason, we are given a cash reward for solving, a good argument can be made that, unless the game is exceptionally fun, we’re exploiting our player base. If the game were really so fun as to justify playing on its own terms, why wouldn’t we simply be playing it ourselves?
Game designers of this sort also need to make sure that there isn’t a more efficient means to the practical end. Since the whole purpose of the game is to reach a particular end, if we discover a more efficient way to get there, the game is no longer useful.
I think there is probably much more to say on this subject but I had a baby a week ago and three hours of sleep last night, so I think I will float this out there in hopes of spurring some discussion.
#### The Dinofarm Community Interpretation
At the end of the podcast we decided on a very specific definition of games (from an ethical standpoint). We (myself and users Hopenager and Redless) decided games could be described as a kind of leisure whose purpose is to produce the feeling of pleasure associated with learning. Since this is a leisure interpretation, we aren’t concerned directly with practical value, which I think is square with the way we typically think of games. However, as a leisure interpretation we need a theory of how games operate in the context of the player’s larger goals.
Let’s sketch one. What circumstances transpire in a person’s life where they have the desire for the pleasure associated with learning but are unable to pursue that desire in productive terms? One possibility is fatigue: after working on productive activities, a person might have an excess of interest in the experience of learning but a deficit of energy to pursue those productive activities. In that situation, a game can satisfy the specific desire with a lower investment of energy (which could mean here literal energy or just lower stress levels – games, since they aren’t practical, are typically less stressful than similar real world situations).
Once the game is completed, the desire ought to be satisfied but not stimulated, allowing the player to rest and then pursue practical goals again.
Again, there are probably other possible ways of situation ethical games in this interpretation, but I think this is a compelling one: games should satisfy, but not stimulate, the desire to learn, and only in those situations where that desire might not be more productively used, as is in the case of mental exhaustion or the need to avoid stress.
Games shouldn’t have a “loop” which intends to capture the player’s attention permanently. Indeed, I think ethical games should be designed to give up the attention of the player fairly easily, so they don’t distract from practical goals.
And them’s my thoughts on the ethics of game design.
¹: Note that there is a loose correspondence between our rubric and The Forms. Toys, roughly, seem to be objects of leisure, puzzles and contests are arguably recreation, and games are, potentially, at least, objects of real practical value. Maybe this is the interpretation of games is the one underlying “gamification” enthusiasts.
# Goals, Anti-Goals and Multi-player Games
In this article I will try to address Keith Burgun‘s assertion that games should have a single goal and his analysis of certain kinds of goals as trivial or pathological. I will try to demonstrate that multi-player games either reduce to single player games or necessitate multiple goals, some of which are necessarily the sorts of goals which Burgun dismisses as trivial. I’ll try to make the case that such goals are useful ideas for game designers as well as being necessary components of non-trivial multi-player games.
(Note: I find Keith Burgun’s game design work very useful. If you are interested in game design and have the money, I suggest subscribing to his Patreon.)
# Notes on Burgun’s Analytical Frame
## The Forms
Keith Burgun is a game design philosopher focused on strategy games, which he calls simply games. He divides the world of interactive systems into four useful forms:
1. toys – an interactive system without goals. Discovery is the primary value of toys.
2. puzzle – bare interactive system plus a goal. Solving is the primary value of the puzzle.
3. contests – a toy plus a goal all meant to measure performance.
4. games – a toy, plus a goal, plus obfuscation of game state. The primary value is in synthesizing decision making heuristics to account for the obfuscation of the game state.
A good, brief, video introduction to the forms is available here. Burgun believes a good way to construct a game is to identify a core mechanism, which is a combination of a core action, a core purpose, and a goal. The action and purpose point together towards the goal. The goal, in turn, gives meaning to the actions the player can take and the states of the interactive system.
## On Goals
More should be said on goals, which appear in many of the above definitions. Burgun has a podcast which serves as a good long form explication of many of his ideas. There is an entire episode on goals here. The discussion of goals begins around the fifteen minute mark.
Here Burgun provides a related definition of games: contests of decision making. Goals are prominent in this discussion: the goal gives meaning to actions in the game state.
Burgun raises a critique of games which feature notions of second place. He groups such goals into a category of non-binary goals and gives us an example to clarify the discussion: goals of the form “get the highest score.”
His analysis of the poorness of this goal is that it seems to imply a few strange things:
1. The player always gets the highest score they are capable of because the universe is deterministic.
2. These goals imply that the game becomes vague after the previous high score is beaten, since the goal is met and yet the game continues.
The first applies to any interactive system at all, so isn’t a very powerful argument, as I understand it. Take a game with the rules of Tetris except that the board is initialized with a set of blocks already on the board. The player receives a deterministic sequence of blocks and must clear the already present blocks, at which point the game ends. This goal is not of the form “find the highest score” or “survive the longest” but the game’s outcome is already determined by the state of the universe at the beginning of the game. From this analysis we can conclude that if (1) constitutes a downside to the construction of a goal, it doesn’t apply uniquely to “high score” style goals.
(2) is more subtle. While it is true that in the form suggested, these rules leave the player without guidelines after the goal is met, I believe that in many cases a simple rephrasing of the goal in question resolves this problem. Take the goal:
`G`: Given the rules of Tetris, play for the highest score.
Since Tetris rewards you for clearing more lines at once and since Tetris ends when a block becomes fixed to the board but touches the top of the screen, we can rephrase this goal as:
`G'`: Do not let the blocks reach the top of the screen.
This goal is augmented by secondary goals which enhance play: certain ways of moving away from the negative goal `G'` are more rewarding than others. Call this secondary goal `g`: clear lines in the largest groups possible. Call `G'` and goals like it “anti-goals.”
This terminology implies the definition.
If a goal is a particular game state towards which the player tries to move, an anti-goal is a particular state which the player is trying to avoid. Usually anti-goals are of the form “Do not allow X to occur” Where X is related to a (potentially open ended) goal.
Goals of the “high score” or “survive” variety are (or may be) anti-goals in disguise. Rephrased properly, they can be conceived of in anti-goal language. Of course there are good anti-goals and bad ones, just as there are good goals and bad goals. However, I would argue that the same criteria applies to both types of goals: a good (anti) goal is just one which gives meaning to the actions a person is presented with over an interactive system.
# Multi-Player Games and Anti-Goals
I believe anti-goals can be useful game design, even in the single player case. In another essay I may try to make the argument that anti-goals must be augmented with mechanics which tend to move the player towards the anti-goal against which players must do all the sorts of complex decision making which produces value for players.
However, there is a more direct way of demonstrating that anti-goals are unavoidable aspects of games, at least when games are multi-player. This argument also demonstrates that games with multiple goals are in a sense inevitable, at least in the case of multi-player games. First let me describe what I conceive of as a multi-player game.
`multi-player game`: A game where players interact via an interactive system in order to reach a goal which can only be attained by a single player.
The critical distinction I want to make is that a multi-player game is not just two or more people engaged in separate contests of decision making. If there are not actions mediating the interaction of players via the game state then what is really going on is many players are playing many distinct games. A true multi-player game must allow players to interact (via actions).
In a multi-player game, players are working towards a win state we can call `G`. However, in the context of the mechanics which allow interaction they are also playing against a (set of) anti-goals `{A}`, one for each player besides themselves. These goals are of the form “Prevent player X from reaching goal `G`“. Hence, anti-goals are critical ingredients to successful multi-player game design and are therefore useful ideas for game designers. Therefore, for a game to really be multi-player then there must be actions associated with each anti-goal `{A}`.
An argument we might make at this point is that if players are playing for `{A}` and not explicitly for `G` then our game is not well designed (for instance, it isn’t elegant or minimal). But I believe any multi-player game where a player can pursue `G` and not concern herself with `{A}`, even in the presence of game actions which allow interaction, is a set of single player games in disguise. If we follow our urge to make `G` the true goal for all players at the expense of `{A}` then we may as well remove the actions which intermediate between players and then we may as well be designing a single player game whose goal is `G`.
So, if we admit that multi-player games are worth designing, then we also admit that at least a family of anti-goals are worth considering. Note that we must explicitly design the actions which allow the pursuit of `{A}` in order to design the game. Ideally these will be related and work in accord with the actions which facilitate `G` but they cannot be identical to those mechanics without our game collapsing to the single player case. We must consider `{A}` actions as a separate (though ideally related) design space.
# Summary
I’ve tried to demonstrate that in multi-player games especially, anti-goals, which are goals of the for “Avoid some game state”, are necessary, distinct goal forms worth considering by game designers. The argument depends on demonstrating that a multi-player game must contain such anti-goals or collapse to a single player game played by multiple people but otherwise disconnected.
In a broader context, the idea here is to get a foot in the door for anti-goals as rules which can still do the work of a goal, which is to give meaning to choices and actions in an interactive system. An open question is whether such anti-goals are useful for single player games, whether they are useful but only in conjunction with game-terminating goals, or whether, though useful, we can always find a related normal goal which is superior from a design point of view. Hopefully, this essay provides a good jumping off point for those discussions.
# On Inform 7, Natural Language Programming and the Principle of Least Surprise
I’ve been pecking away at Inform 7 lately on account of its recently acquired Gnome front end. For those not in the know, Inform (and Inform 7) is a text adventure authoring language. I’ve always been interested in game programming but never had the time (or more likely the persistence of mind) to develop one of any sophistication myself. Usually in these cases one lowers the bar, and as far as interactive media goes, you can’t get much lower, complexity wise, than text adventures.
Writing a game in Inform amounts to describing the world and it’s rules in terms of a programming language provided by Inform. The system then collects the rules and descriptions and creates a game out of them. Time was, programming in Inform used to look like:
``````Constant Story "Hello World";
Include "Parser";
Include "VerbLib";
[ Initialise;
location = Living_Room;
"Hello World"; ];
Object Kitchen "Kitchen";
Object Front_Door "Front Door";
Object Living_Room "Living Room"
with
description "A comfortably furnished living room.",
n_to Kitchen,
s_to Front_Door,
has light;
``````
Which is recognizably a programming language, if a bit strange and domain specific. These days, writing Inform looks like this: (from my little project):
``````"Frustrate" by "Vincent Toups"
Ticks is a number which varies.
Ticks is zero.
When play begins:
Now ticks is 1.
The Observation Room is a room. "The observation room cold and
surreal. Stars dot the floor underneath thick, leaded glass, cutting
across it with a barely perceptible tilt. This room seems to have been
adapted for storage, and is filled with all sorts of sub-stellar
detritus, sharp in the chill and out of place against the slowly
rotating sky. Even in the cold, the place smells of dust, old wood
finish, and mildew. [If ticks is less than two] As the sky cuts its
way across the milky way, the whole room seems to tilt. You feel
dizzy.[else if ticks is less than four]The plane of the galaxy is
sinking out of range and the portal is filling with the void of
space. It feels like drowning.[else if ticks is greater than 7]The
galactic plane is filling the floor with a powdering of
stars.[else]The observation floor looks out across the void of space.
You avert your eyes from the floor.[end if]"
Every turn: Now ticks is ticks plus one.
Every turn: if ticks is 10:
decrease ticks by 10.
``````
As you can see, the new Inform adopts a “natural language” approach to programming. As the Inform 7 website puts it
[The] Source language [is] modelled closely on a subset of English, and usually readable as such.
Also reproduced in the Inform 7 manual is the following quote from luminary Donald Knuth:
Programming is best regarded as the process of creating works of literature, which are meant to be read… so we ought to address them to people, not to machines. (Donald Knuth, “Literate Programming”, 1981)
Which better than anything else illustrates the desired goal of the new system: Humans are not machines! Machines should accommodate our modes of expression rather than forcing us to accommodate theirs! If it wasn’t for the unnaturalness of programming languages, the logic goes, many more people would program. The creation of interactive fiction means to be inclusive, so why not teach the machine to understand natural language?
This is a laudable goal. I really think the future is going to have a lot more programmers in it, and a primary task of language architects is to design programming languages which “regular” people find intuitive and useful. For successes in that arena see Python, or Smalltalk or even Basic. Perhaps these languages are not the pinnacle of intuitive programming environments but whatever that ultimate language is, I doubt seriously it will look much like Inform 7.
This is unfortunate, because reading Inform 7 is very pleasant, and the language is even charming from time to time. Unfortunately, it’s very difficult to program in1, and I say that as something of a programming language aficionado. It’s true that creating the basic skeleton of a text adventure is very easy, but even slightly non-trivial extensions to the language are difficult to intuitively get right. For instance, the game I am working on takes place on a gigantic, hollowed out natural satellite, spinning to provide artificial gravity. The game begins in a sort of observation bubble, where the floor is transparent and the stars are visible outside. Sometimes this observation window should be pointing into the plane of the Milky Way, but other times it should be pointing towards the void of space because the station’s axis of rotation is parallel to the plane of the galaxy. The description of the room should reflect these different possibilities.
Inform 7 operates on a turn based basis, so it seems like it should be simple enough to create this sort of time dependent behavior by keeping track of time but it was frustrating to figure out how to “tell” the Inform compiler what I wanted.
First I tried joint conditionals:
`````` When the player is in the Observation Room and
the turn is even, say: "The stars fill the floor."
``````
But this resulted in an error message. Maybe the system doesn’t know about “evenness” so I tried:
`````` When the player is in the Observation Room and
the turn is greater than 3, say "The stars fill the floor."
``````
(Figuring I could add more complex logic later).
Eventually I figured out the right syntax, which involved creating a variable and having a rule set its value each turn and a separate rule reset the value with the periodicity of the rotation of the ship, but the process was very frustrating. In Python the whole game might look, with the proper abstractions, like:
``````
while not game.over():
game.describe_location(player.position);
if (player.position == 'The Observation Room' and
game.turn() % 10):
print "The stars fill the floor."
``````
Which is not perhaps as “englishy” as the final working Inform code (posted near the beginning of this article) but is much more concise and obvious.
But that isn’t the reason the Python version is less frustrating to write. The reason is the Principle of Least Surprise, which states, roughly, that once you know the system, the least surprising way of doing things will work. The problem with Inform 7 is that “the system” appears to the observer to be “written English (perhaps more carefully constructed that usual)”. This produces in the coder a whole slew of assumptions about what sorts of statements will do what kind of things and as a consequence, you try a lot of things which, according to your mental model, inexplicably don’t work.
It took me an hour to figure out how to make what amounts to a special kind of clock and I had the benefit of knowing that underneath all that “natural English” was a (more or less) regular old (prolog flavored) programming environment. I can’t imagine the frustration a non-programmer would feel when they first decided to do something not directly supported or explained in the standard library or documentation.
That isn’t the only problem, either. Natural english is a domain specific language for communicating between intelligent things. It assumes that the recepient of the stream of tokens can easily resolve ambiguities, invert accidental negatives (pay attention, people do this all the time in speech) and tell the difference between important information and information it’s acceptable to leave ambiguous. Not only are computers presently incapable of this level of deduction/induction, but generally speaking we don’t want that behavior anyway: we are programming to get a computer to perform a very narrowly defined set of behaviors. The implication that Inform 7 will “understand you” in this context is doubly frustrating. And you don’t want it to “understand,” you want it to do exactly.
A lot of this could be ameliorated by a good piece of reference documentation, spelling out in exact detail the programmatic environment’s behavior. Unfortunately, the bundled documentation is a big tutorial which does a poor job of delineated between constructs in the language and elements of it. It all seems somewhat magical in the tutorial, in other words, and the intrepid reader, wishing to generalize on the rules of the system, is often confounded.
Nevertheless, I will probably keep using it. The environment is clean and pleasant, and the language, when you begin to feel out the classical language under the hood, is ok. And you can’t beat the built in features for text based games. I doubt that Inform 7, though, will seriously take off. Too many undeliverable promises.
1 This may make it the only “Read Only” programming language I can think of.
# Quick Thoughts about Interactive Fiction
I’ve recently started a podcast called Text Adventure Purgatory wherein myself and several friends play and talk about Text Adventures/Interactive Fiction. Doing so has crystalized, in my mind, a few thoughts have been in mere fluid suspension in the back of my head about games and fun in general.
“A Theory of Fun for Game Design,” by Raph Koster asserts the following basic premise: fun is learning. This predicts that if a game offers to you a system which you can learn, then you will have fun playing it up until you have exhausted either the system or your capacity to continue learning about it. It’s silly to suggest that this theory covers everything that is fun or everything we might want to assert is a game (this kind of idealism is counterproductive in any context, if you ask me), but it is, I would argue, a useful one.
What is learning, anyway? I think neuroscience and contemporary machine learning techniques (which are inspired by and inspire neuroscience) can provide us with a useful model of the process: learning is an optimization problem which attempts to map inputs onto “desired” outputs or outcomes. Eg: the pixels (and their history) on a screen are mapped by our brains into a series of button presses which result in Mario reaching the end of the screen, where he touches the flag pole. Better than just describing the process, we now have a reasonable idea of how it happens too, and how to imitate the process in software.
There are lots of techniques for the latter, but they basically boil down to optimizing an objective function (the mapping from input to output) by exploring the input space, finding, and following trends in the output space. That is, start with a naive model, take some characteristic input data, apply the model to it, measure the outcome, make small changes to the model to improve the outcome (lots of strategies for this step), repeat until the model behaves well enough for your purposes. In the brain this happens by adjusting synaptic weights (and other physiological properties) of the neurons in question. In computerized learning systems this occurs by modifying the numerical parameters of the model.
Now we are ready for the point of this reflection: text adventures and interactive fiction provide too sparse a set of inputs and outputs to meaningfully train a system for playing. They (generally, I’m sure exceptions exist or attempt to exist) don’t provide a rich enough state space for learning, and hence they aren’t fun in the way that “A Theory of Fun” proposes we interpret that word.
What do I mean by “too sparse?” I mean, for one thing, that for any state in the game I can specify some non-perverse measure of similarity and value for that measure which has the following property: there will be no neighboring states included within that boundary. This is in contrast to games which involve simulated motion in space, which is, for the purposes of our discussion, continuous (that computers actually only simulate discrete spaces is not really material to the discussion: they are discrete spaces of sufficient granularity that our brains perceive them to be continuous).
For instance, there is a state in Deadline, the infocom game we played for several episodes in TAP, wherein the player character has discovered several pieces of broken china in the rose garden near the balcony of the library in which a murder has taken place. We arrive at this state only and exactly when a particular sequence of events (amounting, in isolation, to a few turns in the right order) has occured. There is nothing to refine about the process of reaching this state: either you perform the sequence of actions that produce this outcome or you do not so perform them.
A bit of reflection reveals how much in contrast this is with more typical videogames: in Super Mario Brothers, for instance, there are effectively an infinite number of ways to touch (to specify a single instance) the final flagpole in each level. As we vary the exact moment we press the jump botton, where we jump from, how long we hold it, how long we have run before, we refine the final state of interest and can find a solution which maximizes our height on the pole. There is a continuum of input states and output states (and a clear way of measuring our success) which allows those learning circuits (to use a drastically oversimplifying colloquialism) to grab onto something.
When playing a text adventure, in contrast, we essentially have nothing to do but explore, often by brute force, the state space the game gives us branch by branch until we find the final state. This is not usually fun, and using the context clues embedded in the text rarely helps: they can be either obtuse, in which case we are in the first strategy, or obvious, in which case there isn’t much to do but follow their instructions and traverse the graph. This problem is exacerbated by the fact that text adventures present themselves to us as text, creating the illusion of a rich, detailed world where, computationally, the exact opposite is true: everything reduces to a set of nodes connected by edges. Labeling more than one of those edges as “ending” the game helps a little: we can repeat the experience and land at different ending nodes by virtue of knowledge obtained on previous playthroughs, but we are still jumping for discrete state to discrete state, connected by discrete edges of low cardinality.
This isn’t a dig at interactive fiction: it is a way of explaining why it doesn’t “play like” other kinds of videogames, despite sharing a medium (computers). Novels, for instance, are even more restricted than interactive fiction: they proceed only and exactly in one way and come to life only and exactly as we read them.
Maybe these reflections tell us what we already know: that interactive fiction is more literature than game and that we should look elsewhere than traditional videogame experiences for an interpretive strategy which will allow us to discuss interactive fiction meaningfully. | 8,219 | 40,266 | {"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-2018-17 | latest | en | 0.977293 |
https://web2.0calc.com/questions/geometry_80548 | 1,709,052,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.79/warc/CC-MAIN-20240227153053-20240227183053-00597.warc.gz | 611,721,724 | 5,555 | +0
# geometry
0
187
1
When a circle has its radius increased by 1 cm, its area increases by 43 cm^2. To the nearest integer, how many centimeters was the radius of the circle before it was increased?
Jul 28, 2022
#1
+126978
+1
Let r be the original radius
We have that
New area - Old area = 43 cm^2
pi (r + 1)^2 - pi r^2 = 43
pi ( r^2 + 2r + 1) - pi r^2 = 43
r^2 + 2r + 1 - r^2 = 43 / pi
2r + 1 = 43/pi
2r = 43/ pi - 1
r = (43/ pi - 1) / 2 ≈ 6 cm
Jul 28, 2022 | 228 | 505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-10 | latest | en | 0.762374 |
https://ask.csdn.net/math | 1,532,036,442,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591296.46/warc/CC-MAIN-20180719203515-20180719223515-00169.warc.gz | 598,703,321 | 8,949 | math 55个问题
2018.04.22 19:52来自 weixin_42042460 悬赏 40C
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JavaScript canvas Math 报错 帮忙看看 感谢!
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KFC -Z+W
Problem Description Welcome to KFC! We love KFC, but we hate the looooooooooong queue. Z is a programmer; h...
0 | 505 | 1,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-30 | longest | en | 0.671092 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-7-section-7-1-integration-by-parts-7-1-exercises-page-476/27 | 1,590,994,009,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00441.warc.gz | 740,017,491 | 12,663 | ## Calculus: Early Transcendentals 8th Edition
Published by Cengage Learning
# Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 476: 27
#### Answer
$\frac{-\ln 5 -1}{5} +1$
#### Work Step by Step
First, solve for the indefinite integral: let $u=\ln R$ and $dv=R^{-2} \ dR$. Then $du= \frac{1}{R}$ and $v=-\frac {1}{R}$. Apply integration by parts. $$\int \frac{\ln R}{R^2}= \frac{-\ln R}{R}- \int \frac{-1}{R^2} \ dR$$ $$=\frac{-\ln R}{R} - \frac{1}{R} = \frac{- \ln R -1}{R}$$ Evaluate at $R=5,1$ $$\frac{- \ln R -1}{R} \Bigg]^5_1= \frac{-\ln 5 -1}{5} +1$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 273 | 745 | {"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} | 4.1875 | 4 | CC-MAIN-2020-24 | latest | en | 0.615211 |
https://www.physicsforums.com/threads/finding-the-amplitude.281628/ | 1,508,254,707,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822116.0/warc/CC-MAIN-20171017144041-20171017164041-00735.warc.gz | 965,470,989 | 15,258 | # Finding the amplitude
1. Dec 26, 2008
### Niles
1. The problem statement, all variables and given/known data
Hi all.
I have the following harmonic function:
$$V(t)=A\cos(\omega t)\exp(-Ct),$$
where C is a constant, and A is the amplitude. I need to find the time t, where the amplitude is A/2. This gives me:
$$V(t)=A\cos(\omega t)\exp(-Ct) = \frac{A}{2},$$
but how do I solve this equation?
Sincerely,
Niles.
2. Dec 26, 2008
### HallsofIvy
Staff Emeritus
Well, the obvious first step is to cancel the "A"s: $cos(\omega t)e^{-Ct}= 1/2$. Next, I think I would write the cosine in exponential form: $cos(\omega t)= (e^{it}+ e^{-it})/2$ so $cos(\omega t)e^{-Ct}= (e^{(-C+ i\omega)t}+ e^{(C-i\omega)t})2= 1/2$
3. Dec 27, 2008
### Niles
Ahh, great.
If I was given a function on the form:
$$V(t)=(A\cos(\omega t)+B\sin(\omega t)\exp(-Ct),$$
then writing the sines and cosines as exponentials would be the way to go too. But am I even correct to say that the time t when the amplitude of the oscillation of V(t) is half of the original amplitude is when V(t) = A/2, where A is the amplitude?
Last edited: Dec 27, 2008
4. Dec 27, 2008
### HallsofIvy
Staff Emeritus
Yes, you said "find the time find the time t, where the amplitude is A/2". If the initial amplitude is A, then half of it is A/2.
5. Dec 27, 2008
### TaiwanCountry
Need numerical solve. | 459 | 1,368 | {"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} | 3.90625 | 4 | CC-MAIN-2017-43 | longest | en | 0.827298 |
http://mathandmultimedia.com/2012/09/22/number-word-problems-3/ | 1,721,491,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00404.warc.gz | 19,675,918 | 13,990 | # Math Word Problems: Solving Number Problems Part 3
This is the third part of the Solving Number Problems Series. The first part can be read here and the second part can be read here. In this post I will continue worked examples using problems which are slightly more complicated than the problems in the previous two parts. Without further ado, lets start with the seventh problem in the series.
PROBLEM 7
Twice a number added to $18$ is $5$ times that number. What is the number?
Solution
In the two previous posts, we have learned that if $n$ is a number, then twice that number is $2n$. So, twice a number added to $5$ is represented by $2n + 5$. Now, that number, the $2n + 5$ is five times that number of $5n$. So, we can now set up the equation
$2n + 18 = 5n$.
If we solve for $n$, we have $n = 6$
Twice $6$ added to $18$ is $30$. Five times $6$ is $30$. So, we are correct.
PROBLEM 8
Four times the sum of a number and $3$ is $96$. What is the number?
Solution
This one is quite tricky. Some interpret this as $4n + 3 = 96$. But that is not correct. If you read it carefully, the equation $4n + 3 = 96$ is “four times a number added to three.” However, what we want is four times the sum of a number and 3. So, if we let $n$ be the number, then the sum of a number and 3 is $n + 3$. Four times the sum of a number and 3 is there $4(n+3)$. Now, we set up the equation
$4(n + 3) = 96$.
Solving, we have
$4(n + 3) = 4n + 12 = 96$.
This gives us $4n = 84$ and $n = 21$.
Four times the sum of $21$ and $3$ is four times $24$ which is equal to $96$. So, we are correct.
PROBLEM 9
The sum of three numbers is $105$. The first number is $3$ less than the second number. The third number is four times the second number.
Solution
As you can observe, the second number has no description, so we let it be $n$. The first number is less than the second number by $3$, so the first number is $n - 3$. The third number is four times the second number or $4n$.
Their sum $(n - 3) + n + 4n$ is $105$, so
$(n-3) + n + 4n = 105$.
Simplifying, we have
$6n - 3= 105$
$6n = 108$
So $n = 18$ and the first number is $15$. The third number is $72$.
Checking the Ansewer
Left as an excersise.
***
You have probably noticed that in solving word problems, it is important to accurately convert phrases into algebraic expressions/equations and vice versa. The word “is” for instance is the same as “equal.” Of course, these keywords are help us in understanding the problem, but we should remember to understand the problem as a whole. In the fourth part (last part) of this series, we will discuss more complicated problems. | 773 | 2,640 | {"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": 45, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-30 | latest | en | 0.936968 |
https://www.coursehero.com/file/68836584/Lesson-4-Lab-Planetary-Orbit-Simulator-Worksheet1pdf/ | 1,638,768,643,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00019.warc.gz | 775,065,245 | 51,549 | # Lesson 4 Lab - Planetary Orbit Simulator Worksheet(1).pdf -...
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Name: NAAP Planetary Orbit Simulator 1/8 Lesson 4 Lab - Planetary Orbit Simulator Background Material Answer the following questions after reviewing the “Kepler's Laws and Planetary Motion” and “Newton and Planetary Motion” background pages. (1 point) Draw a line connecting each law on the left with a description of it on the right. (1 point) When written as P 2 = a 3 Kepler's 3rd Law (with P in years and a in AU) is applicable to a) any object orbiting our sun. b) any object orbiting any star. c) any object orbiting any other object. (1 point) The ellipse to the right has an eccentricity of about a) 0.25 b) 0.5 c) 0.75 d) 0.9 (1 point) For a planet in an elliptical orbit to “sweep out equal areas in equal amounts of time” it must a) move slowest when near the sun. b) move fastest when near the sun. c) move at the same speed at all times. Kepler’s 1 st Law Kepler’ s 2 nd Law Kepler’ s 3 rd Law Newton ’s 1 st Law planets orbit the sun in elliptical paths planets with large orbits take a long time to complete an orbit planets move faster when close to the sun only a force acting on an object can change its motion Michael Cruz
NAAP Planetary Orbit Simulator 2/8 d) have a perfectly circular orbit. (1 point) If a planet is twice as far from the sun at aphelion than at perihelion, then the strength of the gravitational force at aphelion will be ____________ as it is at perihelion. a) four times as much b) twice as much c) the same d) one half as much e) one quarter as much Kepler’s 1 st Law If you have not already done so, launch the NAAP Planetary Orbit Simulator . Open the Kepler’s 1 st Law tab if it is not already (it’s open by default). Enable all 5 check boxes. The white dot is the “simulated planet”. One can click on it and drag it around. Change the size of the orbit with the semimajor axis slider. Note how the background grid indicates change in scale while the displayed orbit size remains the same. Change the eccentricity and note how it affects the shape of the orbit. Be aware that the ranges of several parameters are limited by practical issues that occur when creating a simulator rather than any true physical limitations. We have limited the semi-major axis to 50 AU since that covers most of the objects in which we are interested in our solar system and have limited eccentricity to 0.7 since the ellipses would be hard to fit on the screen for larger values. Note that the semi-major axis is aligned | 637 | 2,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-49 | latest | en | 0.862959 |
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• 286 centimeters is 2.86 meters.
• tk10npubl tk10ncanl
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The Colour Code for Resistors (and Capacitors)
Resistance is measured in Ohms (Ω) 1k means 1000Ω and 1M means 1000000Ω The number of Ohms resistance is (usually) indicated by a series of coloured bands on the resistor. Most resistors have four coloured bands as shown above. To find the resistance, start with the coloured band which is nearest to the end of the resistor. The first two bands tell us the first two digits of the number of Ohms. The third band tells us the power of 10 by which the first two digits are multiplied. So, in the example illustrated above, the resistance is: 2 (red) 7 (violet) multiplied by 104 (yellow). That is, 27×104 or 270000 or 270k. For very low resistances, there are two other colours for the third band: silver for -1 and gold for -2. So, for example, if we replaced the yellow band with silver, we would have 27×10-1 or 2.7Ω The last colour (again, silver or gold) indicates the tolerance or precision of the manufacturing process. Silver means ±10% and gold ±5% Some capacitors also use this system in which case the colour code usually gives the capacitance in pico-Farads (1pF = 10-12F) and the last band indicates the voltage rating of the capacitor. More commonly, the capacitance is written on the component, though often still in a slightly coded way. For example, a capacitor of 4700pF could be labelled, 4700pF (no surprise there) or just 4700 or 4.7n (for nano, 10-9) or 4n7.
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# Although the lesser cornstalk borer is widely distributed,
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Although the lesser cornstalk borer is widely distributed, [#permalink]
### Show Tags
07 Jun 2004, 13:52
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Question Stats:
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Although the lesser cornstalk borer is widely distributed, control of them is necessary only in the South.
(A) the lesser cornstalk borer is widely distributed, control of them is
(B) widely distributed, measures to control the lesser cornstalk borer are
(C) widely distributed, lesser cornstalk borer control is
(D) the lesser cornstalk borer is widely distributed, measures to control it are
(E) it is widely distributed, control of the lesser cornstalk borer is
[Reveal] Spoiler: OA
If you have any questions
New!
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### Show Tags
07 Jun 2004, 14:10
singh_satya wrote:
Although the lesser cornstalk borer is widely distributed, control of them is necessary only in the South.
(A) the lesser cornstalk borer is widely distributed, control of them is
(B) widely distributed, measures to control the lesser cornstalk borer are
(C) widely distributed, lesser cornstalk borer control is
(D) the lesser cornstalk borer is widely distributed, measures to control it are
(E) it is widely distributed, control of the lesser cornstalk borer is
Thanks
Satya
Between A and D ...
I will go with D
A. Incorrect: "them" is not correct
B,C,E misplaced modifier
D. it correctly refers to "the lesser cornstalk borer "
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07 Jun 2004, 14:21
Agree with Bono on this one.
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### Show Tags
08 Jun 2004, 14:33
lesser cornstalk borer -> amount -> collective singular noun
it refers to above antecedent.
Between (A) and (D) (D) wins because of subject-verb agreement
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10 Jun 2004, 21:16
srijay007 wrote:
Between A and D ...
I will go with D
B,C,E misplaced modifier
if B has misplaced modifier, then D too entails the same error. isn't it?
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### Show Tags
10 Jun 2004, 22:21
D.
A is a probable but control of "them" does not agree with the singular. (as explained by srijay)
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11 Jun 2004, 06:14
(D) it is, (A) has a subject-verb agreement error.
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### Show Tags
05 Nov 2004, 19:39
Although the lesser cornstalk borer is widely distributed, control of them is necessary only in the South.
(A) the lesser cornstalk borer is widely distributed, control of them is
(B) widely distributed, measures to control the lesser cornstalk borer are
(C) widely distributed, lesser cornstalk borer control is
(D) the lesser cornstalk borer is widely distributed, measures to control it are
(E) it is widely distributed, control of the lesser cornstalk borer is
Pls explain your choices and POE.
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05 Nov 2004, 20:26
1
This post was
BOOKMARKED
Picked D
(A) the lesser cornstalk borer is widely distributed, control of them is
them refers to the less cornstalk borer and should be singular
(B) widely distributed, measures to control the lesser cornstalk borer are
widely distrbuted should modify the lesser cornstalk borer not measures,
(C) widely distributed, lesser cornstalk borer control is
incorrectly implies that it's the lesser cornstalk borer control that is only needed in the South
(D) the lesser cornstalk borer is widely distributed, measures to control it are
makes the most sense.
(E) it is widely distributed, control of the lesser cornstalk borer is
again, widely distributed should modify the lesser cornstalk borer
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### Show Tags
06 Nov 2004, 22:22
D
A - Pronoun problem - "them"
B,C - Change the meaning
E - "it" in the begining is awkward
_________________
Thanks !
Aspire
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10 Nov 2004, 05:28
Agree with D.
B is wrong because 'Although widely distributed,...' after the comma you are forced to ask what is widely distributed? Surely not 'measures' it is the 'lesser cornstalk borer'. soi just after the comma the 'lesser cornstalk borer' would have been appropriate. This is the modifier problem that nzgmat has outlined.
However, Look at C - it addresses the modifier issue. But...introduces a new error - which is it means something else altogether. Rightly pointed out by nzgmat.
E is wrong. What is 'it' here - 'lesser cornstalk borer' or 'control'
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### Show Tags
05 Dec 2004, 16:06
Although the lesser cornstalk borer is widely distributed, control of them is necessary only in the South.
(A) the lesser cornstalk borer is widely distributed, control of them is
(B) widely distributed, measures to control the lesser cornstalk borer are
(C) widely distributed, lesser cornstalk borer control is
(D) the lesser cornstalk borer is widely distributed, measures to control it are
(E) it is widely distributed, control of the lesser cornstalk borer is
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05 Dec 2004, 17:19
Go with B, it's the only one that makes any sense to me.
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06 Dec 2004, 08:10
Will go with D as well.
B & E have misplaced modifier problem.
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01 Jun 2005, 09:24
98. Although the lesser cornstalk borer is widely distributed, control of them is necessary only in the South.
(A) the lesser cornstalk borer is widely distributed, control of them is
(B) widely distributed, measures to control the lesser cornstalk borer are
(C) widely distributed, lesser cornstalk borer control is
(D) the lesser cornstalk borer is widely distributed, measures to control it are
(E) it is widely distributed, control of the lesser cornstalk borer is
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01 Jun 2005, 11:58
gmat2me2 wrote:
B looks OK
I vote for D.
B - sounds as if measures are widely distributed.
HMTG.
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01 Jun 2005, 14:41
Looks D to me too.
A use of them is wrong
B seems to indicate the measure are widely distributed
C same as B, misplaced modifier
D joins DC with IC
E is also weird, where it seems to refer to the controls.
So my pick is D.
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01 Jun 2005, 20:17
I prefer D too. we need a clause after although not a pharse.
Although the lesser cornstalk borer is widely distributed, measures to control it are necessary only in the South.
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04 Sep 2005, 13:48
Fuqua wrote:
98. Although the lesser cornstalk borer is widely distributed, control of them is necessary only in the South.
(A) the lesser cornstalk borer is widely distributed, control of them is
(B) widely distributed, measures to control the lesser cornstalk borer are
(C) widely distributed, lesser cornstalk borer control is
(D) the lesser cornstalk borer is widely distributed, measures to control it are
(E) it is widely distributed, control of the lesser cornstalk borer is
Why is E wrong? It has the DC, IC structure. Doesnt D change the meaning of the sentence? How are we to know that "measures" are needed [as opposed to just "control"]? The "It" in AC E isnt ambiguous because it can only refer to lesser cornstalk borer.
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04 Sep 2005, 17:32
D on this one..
first of we need to make sure what is the subject here...its broder cornstalk...D put it in place properly...
04 Sep 2005, 17:32
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5 Although the lesser cornstalk borer is widely distributed, 11 22 Apr 2012, 03:07
17 Although the lesser cornstalk borer is widely distributed, 12 22 Apr 2012, 03:04
distribution 2 07 Feb 2010, 09:27
1 Although the lesser cornstalk borer is widely distributed, 6 18 Apr 2008, 08:53
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# How can I schedule a rule to run every 2 weeks on a particular day?
Nick Menere Community Leader Oct 15, 2017
When scheduling a rule, I can set an interval every 14 days but I can't set the day I want it to execute.
If I use a Cron expression to schedule, you can't specify every 2 weeks (every fortnight).
Is there a way to run the rule every 2 weeks on a Monday at 9am?
#### 1 accepted
Nick Menere Community Leader Oct 15, 2017
In order to do this, you will need to use Cron to run the rule every Monday at 9am and then use a condition to check that it is an even (or odd) week of the year.
You condition (the first thing after the trigger) will look like:
What this is doing is:
• Rendering the current week of the year
• Finding the remainder when you divide it by 2
• If this is 0, it must be an even week, otherwise it is an odd week.
The formulae is:
`{{#=}}{{#now}}w{{/}} % 2{{/}}`
@Nick Menere it didn't work to me, however this comparison works fine:
It checks if the week number is even.
`{{#=}}{{now.weekOfYear}}/2 - ROUND({{now.weekOfYear}}/2,0){{/}}`
Hi It doesn't work for me. Can ypu help me
@Diego Quesada the simple way to validate math expressions is to print them in a comment. For this I recommend to create a rule for tests (that you execute manually or something like that) and add an action to add comment to the issue.
Put the expression above in the comment and run the rule and see the results.
The week number today is even (week #32), so the result should be zero.
If you need more help please share printscreen.
Will be like this?
yes, now run the rule manually in an issue and see the results. | 539 | 2,084 | {"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-2020-50 | latest | en | 0.892391 |
https://www.physicsforums.com/threads/how-does-the-vector-laplacian-come-about.930068/ | 1,521,770,259,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00794.warc.gz | 857,798,232 | 20,013 | # I How does the Vector Laplacian come about?
1. Oct 29, 2017
### yosimba2000
So the Laplacian of a scalar is divergence of the gradient of a scalar field, and it comes out to the double derivative of the field in X, Y, and Z.
My book says the Laplacian of a vector field is the double derivative of the X component of the field with respect to X, the double derivative of the Y component with respect to Y, and same with Z.
I'm not sure how this comes about from the definition of the Scalar Laplacian.
2. Oct 29, 2017
### NFuller
The most general definition of the Laplacian is
$$\nabla^{2}\mathbf{T}=\nabla\cdot\nabla\mathbf{T}$$
where $\mathbf{T}$ is some tensor, which could be a scalar, vector, matrix, etc. The gradient of $\mathbf{T}$ in Cartesian coordinates is
$$\nabla\mathbf{T}=\left[\nabla T_{x}\;\;\nabla T_{y}\;\;\nabla T_{z}\right]$$
where
$$\nabla T_{x}=\begin{bmatrix}\frac{\partial}{\partial x}T_{x} \\ \frac{\partial}{\partial y}T_{x} \\ \frac{\partial}{\partial z}T_{x} \end{bmatrix}$$
and so on for $T_{y}$ and $T_{z}$. The dot product $\nabla\cdot\nabla\mathbf{T}$ is just
$$\nabla\cdot\nabla\mathbf{T}=\left[\nabla\cdot\nabla T_{x}\;\;\nabla\cdot\nabla T_{y}\;\;\nabla\cdot\nabla T_{z}\right]$$
Thus if $\mathbf{T}$ is a scalar then the Laplacian returns a scalar, if $\mathbf{T}$ is a vector then the Laplacian returns a vector, and so on.
3. Oct 30, 2017
### yosimba2000
How would you do ∇T if T is a vector? I thought gradient only acts on a scalar field.
Just examining derivatives wrt X and components in X-hat direction to make it simpler.
So gradient of scalar takes the derivative of field with respect to X, then tacks on the X-hat direction.
So say vector T is only made of Tx in X direction.
How would you do the gradient of Tx in X-hat direction? It seems like you would do derivative of Tx in the X-hat direction wrt to X, then tack on another X-hat direction, but that doesn't make any sense.
4. Oct 30, 2017
### Metmann
I think you should view it within the calculus of differential forms. The Laplacian for any form (and hence also for vector fields, which are in one-to-one correspondence to 1-forms, and for functions, which are $0$-forms) is per definition given by
$$\Delta = d \circ \delta + \delta \circ d \, : \Omega^p(M) \rightarrow \Omega^p(M)$$, where $\delta$ is the codifferential given by $\delta = (-1)^a* d *$ where I don't know the exact expression for $a$ at the moment, but it depends on the dimension of the underlying space and the signature of the metric. $*$ is the hodge dual, which maps $p$-forms to $(n-p)$-forms, where $n$ is the dimension of the underlying space.
Applied to a function, $\Delta$ gives you the usual scalar Laplacian (if $M=\mathbb{R}^n$), applied to one-forms and identifying one-forms and vectors gives you the vector-Laplacian (also if $M = \mathbb{R}^n$).
PS: This is in fact just the same NFuller wrote, but in a different representation, where you do not have to apply some nabla to a tensor. The advantage of the form-approach is, that it is valid for arbitrary (regular) coordinate systems and also for curved spacetimes.
Last edited by a moderator: Oct 30, 2017
5. Oct 30, 2017
### vanhees71
The vector Laplacian has to be handled with some care. In Cartesian components you simply have
$$\Delta A^j=\partial_k \partial^k A^j=\delta^{ik} \partial_i \partial_k A^j.$$
It's not as easy in general coordinates (even not in general orthogonal curvilinear coordinates). To define it unambiguously in a covariant way we note that in Cartesian coordinates
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A},$$
and thus it's safe to define generally
$$\Delta \vec{A}=\vec{\nabla} (\vec{\nabla} \cdot \vec{A}) - \vec{\nabla} \times (\vec{\nabla} \times \vec{A}),$$
where the right-hand side has a well-defined covariant meaning.
6. Oct 30, 2017
### yosimba2000
Sorry Metmann, that is way beyond me.
7. Oct 30, 2017
### NFuller
Let's assume $\mathbf{T}$ has the form $\mathbf{T}=T_{x}\hat{x}+T_{y}\hat{y}+T_{z}\hat{z}$. The gradient is
$$\nabla\mathbf{T}=\begin{bmatrix}\frac{\partial}{\partial x}T_{x} & \frac{\partial}{\partial x}T_{y} & \frac{\partial}{\partial x}T_{z} \\ \frac{\partial}{\partial y}T_{x} & \frac{\partial}{\partial y}T_{y} & \frac{\partial}{\partial y}T_{z} \\ \frac{\partial}{\partial z}T_{x} & \frac{\partial}{\partial z}T_{y} & \frac{\partial}{\partial z}T_{z} \end{bmatrix}$$
So just as the gradient of a scalar produces a vector, the gradient of a vector produces a matrix. Then the Laplacian is
$$\nabla\cdot\nabla\mathbf{T}=\begin{bmatrix}\frac{\partial^{2}}{\partial x^{2}}T_{x} + \frac{\partial^{2}}{\partial y^{2}}T_{x} + \frac{\partial^{2}}{\partial z^{2}}T_{x} \\ \frac{\partial^{2}}{\partial x^{2}}T_{y} + \frac{\partial^{2}}{\partial y^{2}}T_{y} + \frac{\partial^{2}}{\partial z^{2}}T_{y} \\ \frac{\partial^{2}}{\partial x^{2}}T_{z} + \frac{\partial^{2}}{\partial y^{2}}T_{z} + \frac{\partial^{2}}{\partial z^{2}}T_{z} \end{bmatrix}$$
Which is just a vector, as expected. As stated by Metmann, this is only valid in Cartesian coordinates and there are more general ways to describe this. However, this is probably the simplest to understand.
8. Oct 31, 2017
### yosimba2000
Thanks. How do we know this, though? Is it just definition?
9. Oct 31, 2017
### NFuller
That's just how it's defined. Although this is technically an abuse of notation, I like to think of the gradient as a vector
$$\nabla=\begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}$$
such that the gradient of a vector is the outer product
$$\nabla(\mathbf{T}^{T})=\begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}\begin{bmatrix}T_{x} & T_{y} & T_{z}\end{bmatrix}=\begin{bmatrix}\frac{\partial}{\partial x}T_{x} & \frac{\partial}{\partial x}T_{y} & \frac{\partial}{\partial x}T_{z} \\ \frac{\partial}{\partial y}T_{x} & \frac{\partial}{\partial y}T_{y} & \frac{\partial}{\partial y}T_{z} \\ \frac{\partial}{\partial z}T_{x} & \frac{\partial}{\partial z}T_{y} & \frac{\partial}{\partial z}T_{z} \end{bmatrix}$$
10. Oct 31, 2017
### yosimba2000
I see, so the definition of the Laplacian of a Scalar is different than the definition of the Laplacian of a Vector, but they use the same notation?
11. Oct 31, 2017
### NFuller
The definition also works with a scalar variable such as $\mathbf{T}=\phi$
$$\nabla\phi=\begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}\begin{bmatrix}\phi\end{bmatrix}=\begin{bmatrix}\frac{\partial\phi}{\partial x} \\ \frac{\partial\phi}{\partial y} \\ \frac{\partial\phi}{\partial z} \end{bmatrix}$$
This produces a vector where the components are just the spatial derivatives of the scalar function $\phi$, just as one would expect.
12. Oct 31, 2017
### Metmann
The definition is in fact the same, the notation of usual vector analysis is just not good enough to display this fact ;)
I admit that my post might have been a bit over top, but consider the following (all valid only for $\mathbb{R}^n$): You may know the differential of a function: $df = \sum_i \frac{\partial f}{\partial x^i}dx^i$. This is nothing but the gradient on scalar fields, when identifiying $a_i dx^i \leftrightarrow (a^0,\ldots,a^n)^T$ with $a_i = a^i$.
Now, if you have a vectorfield $(X^0,\ldots,X^n)$ and identify it with $X= \sum_i X_i dx^i$ with $X^i=X_i$, than $dX = \sum_{i,j} \frac{\partial X^i}{\partial x^j}dx^j\wedge dx^i$, which is nothing but a generalization of the differential on functions. The "inverse" of $d$ is $\delta$ given for vectors as $\delta (\sum_i X_i dx^i) = \sum_i \frac{\partial X_i}{\partial x^i}$, which is nothing but the divergence of $\vec{X}$. Then $\Delta f = (d + \delta)^2f =\delta(df) + d(\delta f) = \sum_i \frac{\partial^2 f}{\partial (x^i)^2}$, where $\delta f = 0$. There is also a formula for the codifferential for objects of the type $a_{ij}dx^j \wedge dx^i$ (derived from the general definition given in my previous post). Then you can define the Laplacian for vectors exactly the same way as for functions, giving you $\Delta X = \sum_{ij} \frac{\partial^2 X_i}{\partial (x^j)^2}dx^i$ which is nothing but the Vector-Laplacian, i.e. scalar Laplacian in each component.
In fact you get: $d(\delta(X)) = \nabla(\nabla \cdot \vec{X}) = grad(div(\vec{x}))$ and $\delta (d(X)) = -\nabla \times (\nabla \times \vec{X}) =- rot(rot(\vec{X}))$, which is what vanhees71 gave.
The same way (with the same definition) you get a Laplacian for all tensors.
Last edited by a moderator: Oct 31, 2017
13. Nov 1, 2017
### vanhees71
I always prefer the index calculus (aka Ricci calculus), from which it becomes clear how the various components "naturally" transform under rotations (for 3D "classical" vector analysis in Euclidean space). The gradient is a co-vector, i.e., it has lower indices $\nabla_j=\partial_j$ wrt. to a Cartesian (co-)basis. Only because in Euclidean space wrt. a orthonormal (Cartesian) basis the metric reads $g_{ij}=\delta_{ij}$ the gradient looks like vector components (and this holds true of course only wrt. Cartesian bases). | 2,970 | 9,285 | {"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} | 3.78125 | 4 | CC-MAIN-2018-13 | longest | en | 0.825794 |
https://rdrr.io/cran/DstarM/man/estQdf.html | 1,713,104,407,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00101.warc.gz | 459,442,810 | 7,536 | # estQdf: Estimate quantiles of distribution In DstarM: Analyze Two Choice Reaction Time Data with the D*M Method
## Description
Estimate quantiles of distribution
## Usage
`1` ```estQdf(p, x, cdf) ```
## Arguments
`p` A vector of probabilities. `x` The x-axis values corresponding to the cumulative distribution function. `cdf` A cumulative distributions function, i.e. output of `estCdf`.
## Details
Quantiles are obtained in the following manner. For p = 0 and p = 1, the minimum and maximum of x is used. For other probabilities the quantiles are obtained via `q[i] = uniroot(x, cdf - p[i])\$root`. Y values are interpolated via `approxfun`.
## Value
Quantiles of cumulative distribution function(s). If the input was a matrix of cumulative distributions functions, a matrix of quantiles is returned.
## Examples
```1 2 3 4 5 6 7 8``` ```x = seq(-9, 9, .1) # x-grid d = dnorm(x) # density functions p = seq(0, 1, .2) # probabilities of interest cEst = estCdf(d) # estimate cumulative distribution functions qEst = estQdf(p = p, x = x, cdf = cEst) # estimate quantiles plot(x, cEst, bty = 'n', las = 1, type = 'l', ylab = 'Probability') # plot cdf abline(h = p, v = qEst, col = 1:6, lty = 2) # add lines for p and for obtained quantiles points(x = qEst, y = p, pch = 18, col = 1:6, cex = 1.75) # add points for intersections ```
DstarM documentation built on Aug. 29, 2020, 1:06 a.m. | 424 | 1,402 | {"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-2024-18 | latest | en | 0.709008 |
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A128889 a(n) = (2^(n^2) - 1)/(2^n - 1). 4
1, 5, 73, 4369, 1082401, 1090785345, 4432676798593, 72340172838076673, 4731607904558235517441, 1239150146850664126585242625, 1298708349570020393652962442872833, 5445847423328601499764522166702896582657, 91355004067076339167413824240109498970069278721 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS a(n) is prime for n in A156585. Conjecture: gpf(a(n)) = gpf(Phi(n,2^n)), where Phi(n,2^n) = A070526(n). - Thomas Ordowski, Feb 16 2014 The conjecture fails at n = 26, where 3340762283952395329506327023033 > 215656329382891550920192462661. Next counterexample for n = 30, but no odd counterexamples found so far. - Charles R Greathouse IV, Feb 17 2014 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..50 FORMULA a(n) = Sum_{k=1..n} 2^((n-k)*n). - Enrique Pérez Herrero, Feb 23 2009 MAPLE a:=n->(2^(n^2)-1)/(2^n-1): seq(a(n), n=1..13); MATHEMATICA f[n_] := (2^(n^2) - 1)/(2^n - 1); Array[f, 12] F[n_] := Plus @@ Table[2^((n - i)*n), {i, 1, n}] (* Enrique Pérez Herrero, Feb 23 2009 *) Table[(2^(n^2) - 1)/(2^n - 1), {n, 1, 20}] (* Vincenzo Librandi, Feb 18 2014 *) PROG (PARI) a(n)=(2^n^2-1)/(2^n-1) \\ Charles R Greathouse IV, Feb 17 2014 CROSSREFS Cf. A051156, A119408, A156585. Sequence in context: A334282 A317341 A012640 * A131958 A051156 A092826 Adjacent sequences: A128886 A128887 A128888 * A128890 A128891 A128892 KEYWORD nonn AUTHOR Leroy Quet, Apr 19 2007 EXTENSIONS More terms from Robert G. Wilson v and Emeric Deutsch, Apr 22 2007 STATUS approved
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Last modified December 1 14:10 EST 2021. Contains 349430 sequences. (Running on oeis4.) | 818 | 2,333 | {"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-2021-49 | latest | en | 0.607165 |
http://jondotcomdotorg.net/category/uncategorized/page/311/ | 1,531,910,661,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590127.2/warc/CC-MAIN-20180718095959-20180718115959-00067.warc.gz | 198,800,484 | 24,849 | ## 99 Clojure Problems (46, 49)
I had some difficulty with the “Logic and Codes” section, since neither Prolog or Scala are really all that lisp like.
Here’s my answer for 46. Many thanks to Brian Carper for responding to my StackOverflow question. It really helped me to reformulate this question with more of a clojure style than a Scala or Prolog style.
```; P46 (**) Truth tables for logical expressions.
(comment "Define predicates and/2, or/2, nand/2, nor/2, xor/2, impl/2 and equ/2 (for logical equivalence) which succeed or fail according to the result of their respective operations; e.g. and(A,B) will succeed, if and only if both A and B succeed. Note that A and B can be Prolog goals (not only the constants true and fail).")
(comment "A logical expression in two variables can then be written in prefix notation, as in the following example: and(or(A,B),nand(A,B)).")
(defn not_ [a] (if a false true))
(defn and_ [a b] (if a (if b true false) false))
(defn or_ [a b] (if a true (if b true false)))
(defn nand_ [a b] (not_ (and_ a b)))
(defn nor_ [a b] (not_ (or_ a b)))
(defn xor_ [a b] (or_ (and_ a (not_ b)) (and_ (not_ a) b)))
(defn impl_ [a b] (or_ (not_ a) (and_ a b)))
(defn equ_ [a b] (not_ (xor_ a b)))
(comment "Now, write a predicate table/3 which prints the truth table of a given logical expression in two variables.")
(defn table [f]
(doseq [a '(true false)
b '(true false)]
(println a "\t" b "\t" (f a b))
))```
I skipped number 47 (turning the functions into operators) because clojure doesn’t really deal with infix notation at all. It seems really foreign to me to try to write (a and b). I can see doing this in ruby
`a.and(b)`
or f#
`a |> and b`
but not clojure.
I skipped number 48, because I couldn’t figure out how to count the parameter “arity” of the function being passed in. Even with the number of parameters being passed in, I wasn’t able to figure out how to pass a list as the parameters to a function. (Basically converting (func ‘(a b c)) to (func a b c) with a simple bit of code.
Here’s my answer to number 49, which was mercifully easy after messing with 48 for a little while.
```; P49 (**) Gray code.
(comment "An n-bit Gray code is a sequence of n-bit strings constructed according to certain rules. For example,")
(comment "n = 1: C(1) = ['0','1']. ")
(comment "n = 2: C(2) = ['00','01','11','10'].")
(comment "n = 3: C(3) = ['000','001','011','010','110','111','101','100'].")
(defn gray [n]
(loop [nleft n
combos (list (list ))]
(if (zero? nleft)
(map #(apply str %) combos)
(recur (dec nleft) (concat (map #(conj %1 1) combos) (map #(conj %1 0) combos)) ))))
```
I did not do result caching. I couldn’t figure out where to start with it, and this code is so devilishly simple that I can’t figure out where I would store anything for future use.
## Follow These Instructions Exactly as they Are Written
My wife found this on ONTD, and I think I tracked it down to Creepy Pasta or a Tumblr.
Somewhere in West Philadelphia, you will find an old basketball court with a single ball lying in the middle. Pick it up and start shooting hoops. After a while, a small group of hooligans will approach you and challenge you to a fight, which you must accept.
After the fight, you must go home and relay the events to your mother. She will then inform you that you have an aunt and uncle living in one of the districts of Los Angeles, and out of fear, she will send you to live there for an indefinite period of time.
With your bags packed, go to the street corner, and whistle for a cab. The cab that will pull up will bear the word FRESH on the license plate, and upon closer inspection, novelty fuzzy dice will hang in the mirror. Although you will suddenly realize that cabs like these are extremely hard to find, do not bear any thought to it. At this point you MUST point out in front of the car and say ‘Yo homes to Bel Air’. You will stop in front of a mansion, and it will be sometime between 7 and 8 o’clock, even though it will feel like you’ve been traveling mere seconds. Get your luggage out and say ‘Yo homes, smell ya later!’, but do NOT turn back to face the cabby. Walk up to the door, look over your shoulder once, and then knock on the door three times.
If you follow these instructions, your life will get flip-turned upside-down.
Spooky stuff, eh?
Like most of my internet social applications, I expect that this tumblr account may eventually lose my interest.
Like most of my internet social applications, I expect that this tumblr account may eventually lose my interest. | 1,210 | 4,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-30 | longest | en | 0.826176 |
www.churchsalary.com | 1,718,574,384,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00190.warc.gz | 641,414,498 | 16,107 | Jump directly to the Content
Church Growth Series
## Don’t Expect Exponential: Churches and Salaries Grow Logarithmically
Church and salary growth follow counterintuitive patterns.
Image: Image: Illustration by Vasil Nazar and Aaron Hill / Source Images: Boris Zhitkov / Getty
The pitch is usually simple: “If everyone invites three people, our church will grow exponentially.” Some churches adopt a more realistic “each one, bring one” model. Regardless of the exact words, churches usually set growth goals based on the assumption Jesus taught a “multiplication model,” which always yields exponential growth.
While short-term exponential growth may be possible, ChurchSalary’s research indicates that sustainable, real-world growth almost always follows a logarithmic (or logistic) curve as new members become harder to attract and assimilate as church organizations become increasingly complex. This logarithmic trend impacts not just churches but salary growth for pastors as well.
Even the best methods at bypassing this trend—church planting and/or multisite churches—will inevitably encounter constraints.
#### Two trends
If you analyze the relationship of church size to average senior pastor salary as well as to the frequency of congregations, two logarithmic (or power) growth curves emerge.
What is happening here? And what can these curves tell us about growth and staffing in the church?
#### Three types of growth
There are three basic types of growth.Like every good sermon, we are condensing information here. Technically, there are a host of different mathematical formulas that predict or explain growth. We will discuss some of these other types of growth below.
Exponential growth starts slow, then rapidly increases.
During the pandemic, scientists explained that the Omicron variant had a terrifying growth rate or “R naught” of seven — i.e., every sick person will infect, on average, seven other people. Despite these predictions, the infection rate of each variant eventually slowed down. Why?
Linear growth remains constant over time.
Linear growth is constant; it neither accelerates nor decelerates. In the real world, linear trends over a long period are hard to find. Even seemingly simple relationships such as church budget to attendance don’t follow a linear pattern.
Logarithmic growth starts fast, then slows down.
Logarithmic growth is the inverse of exponential. One way to conceive of a logarithm is that the amount of time it takes to achieve the same result grows larger. For example, the average salary of a senior pastor doubles as the church grows from 30 to 300 people, but in order for their salary to double a second time the church must grow from 300 to over 3,000 people.
While logarithmic growth may not seem very intuitive, we encounter it every day.
#### We experience the world logarithmically
Image: Wikipedia
Everybody notices 10 visitors when your attendance is only 10 people, but can you tell the difference between 120 and 110 people without counting?
Logarithms best explain how we perceive the world through our senses. For example, the Weber-Fechner law explains that the just-noticeable difference between two stimuli—extra weight, louder noise, brighter light—follows a logarithmic scale. This is one of the reasons why we use log scales to measure a host of natural phenomena: sound intensity (decibels), star brightness (lumens), earthquakes (Richter scale), wind intensity (Beaufort scale), mineral hardness (Mohs scale), and acidity (pH scale).
In fact, the Weber-Fechner law explains how we perceive attendance growth in the church, especially in the past. Everyone remembers fondly that year when the church grew from 100 to 200 people, but nobody notices if the congregation grows from 2,000 to 2,100 this year. Why? Even though both growth spurt added 100 souls, the first one feels 10 times more significant.
This law of logarithmic perception even explains how time appears to pass by faster and faster as we age. As Maximilian Kiener explains in this amazing visualization:
“When you are 4 weeks old, a week is a quarter of your life. By the end of your first year, a week is only a fiftieth of your life. By the time you turn 50, a whole year will be a fiftieth of your life.”
#### Logistic growth mixes exponential and logarithmic
In truth, size and salaries in the church probably follow what’s called a “logistic growth model”—instead of a logarithmic or power model.
Logistic growth has nothing to do with logistics; instead, it is a combination of exponential and logarithmic change. It is marked by an initial period of slow change followed by rapid exponential expansion that transitions into slower logarithmic change. The resulting graph creates an S-shape. Depending on where you zoom into a logistic graph, if ignore the surrounding data, you can draw incorrect conclusions about your current situation.
Logistic growth models (or LGMs) are used to explain growth in a wide variety of fields: artificial intelligence, economics, tumor growth, crop yields, population growth, and more. In fact, a logistic model best explains the transmission and fatality of COVID-19 due to certain constraints.
##### Constraints
God designed constraints and limitations into the world: space is limited and food only grows so fast. Were it not for these constraints, Earth would long ago have become a soup of viruses and bacteria. As a result, every group of living things faces constraints in the form of limited resources, space, and time.
In the church, these constraints appear in the form of limited seats, staff, stuff (money and resources), and time. You need empty seats, a larger staff, more volunteers, and bigger budget (as well as more time) to both fuel and sustain growth.
Unfortunately, each constraint is usually coupled with other constraints. This creates a feedback loop or cycle that slows down growth in even the most dynamic churches.
• You need empty seats to grow.
• If you add new members, you need more staff to care for them.
• To hire more staff, you need more money.
• But budget growth usually lags behind numerical growth (tithing is taught or caught).
• To create more empty seats, you often need a bigger facility, which costs more money.
This feedback loop of constraints repeats for churches, ad infinitum. It can even create an expansion and contraction cycle at churches. The only “easy” way to temporarily bypass these constraints is by starting a new congregation (church planting) or a new campus (multisite churches). But even these approaches simply delay the inevitable because all churches face two additional obstacles: complexity and competition.
##### Complexity
As a church grows, its staffing, budget, and structures become exponentially more complex.
Every small group pastor learns in seminary that relationships increase exponentially as group size increases Technically, this is a power function, but you get the point.—X = [N * (N-1) /2]. A small group of 10 people creates 45 potential relationships—enough for a part-time Sunday School or small group leader to manage. However, a congregation of 100 people creates 4,950 potential heartaches, conflicts, and celebrations that only a full-time pastor and team of volunteers can hope to tackle. By the time a church reaches 1,000 people the number of potential relationships explodes to an eye-popping 499,500!
This formula applies to staffing as well. For example, even though you might assume managing 21 employees versus 3 is 7 times more complex, according to this formula it may actually be 70 times worse.
Complexity is also the reason why senior pastor salaries follow a logarithmic (or power) trend. As the church grows, every additional dollar must be divided into smaller and smaller slices to pay for more staff and more programs or events. This means that at certain inflection points or key complexity hurdles, senior pastors may need to sacrifice immediate salary growth for long-term church growth through staff and ministry expansion.
##### Competition
The final constraint that all churches face is competition for congregants. Pastors have a term for this—“sheep stealing.” Even though very few pastors set out to steal sheep, the reality is that finding, converting, and discipling lost people is exponentially harder than assimilating already-discipled believers.
Competition among churches in a community is a function of supply and demand. How many churches (supply) and how many churchgoers are present in the community (demand)? Churches situated in counties or cities experiencing population growth will find it easier to grow as a steady stream of “ready-made” churchgoers move into town. However, exponential growth rates simply are not possible in communities experiencing negative or stagnant population growth.
#### Closing thoughts
The same pattern of logistic growth likely occurs every time the Gospel is introduced into a new country or people group: slow initial growth that eventually explodes into rapid exponential expansion that transitions into slower logarithmic growth. The pace of growth that a church can expect will depend on where they fall in this lifecycle.
Church growth campaigns are great. Asking for a raise because you generated growth at your church is not a bad thing. Jesus is clear, “the worker deserves his/her wage.” But every church leader must bear in mind that the constraints of resources, staffing, complexity, and competition are conspiring to slow down both the growth of both your church and your salary.
Church planting (and/or multisite churches) can be fruitful and lead to rapid organizational growth because they are able to bypass some of these constraints. However, no matter how well designed they are, these new strategies cannot bypass growth constraints forever.
Future planning at your church should acknowledge and anticipate the logarithmic or logistic growth rate changes that pop up everywhere in ChurchSalary’s data—from a decrease in per person giving to key staffing hurdles.
This content is designed to provide accurate and authoritative information in regard to the subject matter covered. It is published with the understanding that the publisher is not engaged in rendering legal, accounting, or other professional service. If legal advice or other expert assistance is required, the services of a competent professional person should be sought. "From a Declaration of Principles jointly adopted by a Committee of the American Bar Association and a Committee of Publishers and Associations."
Due to the nature of the U.S. legal system, laws and regulations constantly change. The editors encourage readers to carefully search the site for all content related to the topic of interest and consult qualified local counsel to verify the status of specific statutes, laws, regulations, and precedential court holdings.
ChurchSalary is made possible through funding from the Lilly Endowment Inc. As part of Lilly's "National Initiative to Address Economic Challenges Facing Pastoral Leaders," ChurchSalary—and our parent, Church Law & Tax—is committed to helping church leaders and pastors develop an atmosphere of healthy financial stewardship, especially in the area of church staff compensation. | 2,229 | 11,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-26 | longest | en | 0.941422 |
https://number.academy/1667064 | 1,721,612,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00388.warc.gz | 367,521,917 | 11,507 | # Number 1667064 facts
The even number 1,667,064 is spelled 🔊, and written in words: one million, six hundred and sixty-seven thousand and sixty-four, approximately 1.7 million. The ordinal number 1667064th is said 🔊 and written as: one million, six hundred and sixty-seven thousand and sixty-fourth. The meaning of the number 1667064 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1667064. What is 1667064 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,667,064 in other units
The decimal (Arabic) number 1667064 converted to a Roman number is (M)(D)(C)(L)(X)(V)MMLXIV. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
1667064 seconds equals to 2 weeks, 5 days, 7 hours, 4 minutes, 24 seconds
1667064 minutes equals to 3 years, 5 months, 1 week, 2 days, 16 hours, 24 minutes
### Codes and images of the number 1667064
Number 1667064 morse code: .---- -.... -.... --... ----- -.... ....-
Sign language for number 1667064:
Number 1667064 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 ...
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#### Is Prime?
The number 1667064 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 1667064 are 2 * 2 * 2 * 3 * 7 * 9923
The factors of 1667064 are
1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168, 9923, 19846, 29769, 39692, 1667064. show more factors ...
Total factors 32.
Sum of factors 4763520 (3096456).
#### Powers
The second power of 16670642 is 2.779.102.380.096.
The third power of 16670643 is 4.632.941.530.172.358.656.
#### Roots
The square root √1667064 is 1291,148326.
The cube root of 31667064 is 118,572531.
#### Logarithms
The natural logarithm of No. ln 1667064 = loge 1667064 = 14,326575.
The logarithm to base 10 of No. log10 1667064 = 6,221952.
The Napierian logarithm of No. log1/e 1667064 = -14,326575.
### Trigonometric functions
The cosine of 1667064 is -0,988699.
The sine of 1667064 is 0,149912.
The tangent of 1667064 is -0,151625.
## Number 1667064 in Computer Science
Code typeCode value
1667064 Number of bytes1.6MB
Unix timeUnix time 1667064 is equal to Tuesday Jan. 20, 1970, 7:04:24 a.m. GMT
IPv4, IPv6Number 1667064 internet address in dotted format v4 0.25.111.248, v6 ::19:6ff8
1667064 Decimal = 110010110111111111000 Binary
1667064 Decimal = 10010200210010 Ternary
1667064 Decimal = 6267770 Octal
1667064 Decimal = 196FF8 Hexadecimal (0x196ff8 hex)
1667064 BASE64MTY2NzA2NA==
1667064 SHA384d8f93c5ae7d06a51432b7599a013831283f4d41c87cd23a3d6ef005d89895bb8c2b2f98c1076e4397c63daee387ab994
More SHA codes related to the number 1667064 ...
If you know something interesting about the 1667064 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1667064
### Character frequency in the number 1667064
Character (importance) frequency for numerology.
Character: Frequency: 1 1 6 3 7 1 0 1 4 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 1667064, the numbers 1+6+6+7+0+6+4 = 3+0 = 3 are added and the meaning of the number 3 is sought.
## № 1,667,064 in other languages
How to say or write the number one million, six hundred and sixty-seven thousand and sixty-four in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.667.064) un millón seiscientos sesenta y siete mil sesenta y cuatro German: 🔊 (Nummer 1.667.064) eine Million sechshundertsiebenundsechzigtausendvierundsechzig French: 🔊 (nombre 1 667 064) un million six cent soixante-sept mille soixante-quatre Portuguese: 🔊 (número 1 667 064) um milhão, seiscentos e sessenta e sete mil e sessenta e quatro Hindi: 🔊 (संख्या 1 667 064) सोलह लाख, सरसठ हज़ार, चौंसठ Chinese: 🔊 (数 1 667 064) 一百六十六万七千零六十四 Arabian: 🔊 (عدد 1,667,064) مليون و ستمائة و سبعة و ستون ألفاً و أربعة و ستون Czech: 🔊 (číslo 1 667 064) milion šestset šedesát sedm tisíc šedesát čtyři Korean: 🔊 (번호 1,667,064) 백육십육만 칠천육십사 Danish: 🔊 (nummer 1 667 064) en millioner sekshundrede og syvogtredstusindfireogtreds Hebrew: (מספר 1,667,064) מיליון שש מאות שישים ושבעה אלף שישים וארבע Dutch: 🔊 (nummer 1 667 064) een miljoen zeshonderdzevenenzestigduizendvierenzestig Japanese: 🔊 (数 1,667,064) 百六十六万七千六十四 Indonesian: 🔊 (jumlah 1.667.064) satu juta enam ratus enam puluh tujuh ribu enam puluh empat Italian: 🔊 (numero 1 667 064) un milione e seicentosessantasettemilasessantaquattro Norwegian: 🔊 (nummer 1 667 064) en million seks hundre og sekstisyv tusen og sekstifire Polish: 🔊 (liczba 1 667 064) milion sześćset sześćdziesiąt siedem tysięcy sześćdziesiąt cztery Russian: 🔊 (номер 1 667 064) один миллион шестьсот шестьдесят семь тысяч шестьдесят четыре Turkish: 🔊 (numara 1,667,064) birmilyonaltıyüzaltmışyedibinaltmışdört Thai: 🔊 (จำนวน 1 667 064) หนึ่งล้านหกแสนหกหมื่นเจ็ดพันหกสิบสี่ Ukrainian: 🔊 (номер 1 667 064) один мільйон шістсот шістдесят сім тисяч шістдесят чотири Vietnamese: 🔊 (con số 1.667.064) một triệu sáu trăm sáu mươi bảy nghìn lẻ sáu mươi bốn Other languages ...
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## Comment
If you know something interesting about the number 1667064 or any other natural number (positive integer), please write to us here or on Facebook.
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,013 | 5,890 | {"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} | 3.109375 | 3 | CC-MAIN-2024-30 | latest | en | 0.805268 |
https://answers.yahoo.com/question/index?qid=20070625170518AABwBgt | 1,603,524,868,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882102.31/warc/CC-MAIN-20201024051926-20201024081926-00600.warc.gz | 202,997,114 | 32,196 | Anonymous
# What is the slope of the line that passes through the points (3, -2) and (-1,0)?
What is the slope of the line that passes through the points (3, -2) and (-1,0)?
negative one-half
one-half
2
-2
Relevance
• alpha
Lv 7
negative one-half
Slope is defined as rise over run. Rise being the vertical distance the line travels and run being the horizontal distance. When plotted on a graph and traveling left to right this line drops two points, so the rise is -2. The run is 4. In fractional terms the slope is -2/4, or -1/2. -1 devided by 2 gives you the answer of -0.5, or negative one-half.
• b
Lv 5
Negative 1/2. You always do y2-y1 / x2-x1. So, 0--2 is 2 (add). -1-3 is -4. Then rise over run (y over x). The slope is 2/-4. This is -1/2.
• Como
Lv 7
m = (0 + 2) (-1 - 3)
m = 2 / (- 4)
m = (-1 / 2) ie negative one half. | 283 | 845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-45 | latest | en | 0.880292 |
https://ned.ipac.caltech.edu/level5/Sept01/Sandage/Sand2_4.html | 1,660,067,327,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00059.warc.gz | 384,181,820 | 2,984 | 2.4. The Volume V(r) in Robertson-Walker Spaces
One supposes that the space that describes any real universe must be homogeneous and isotropic. Otherwise, the notion of extension as applied to material bodies would have a complicated meaning. By this is meant that any material body, transported to any region of the space, must be transformed into itself without tearing or buckling. Such spaces are congruent. They can be rotated into themselves by a coordinate transformation without shear. This is not true for nonhomogeneous, non-isotropic spaces.
Robertson (1929, 1935) and Walker (1936) verified that the most general expression for the geometrical interval dl2 between two points in a space of constant curvature with coordinates r,, , and r + dr, + d, and + d is
(7)
where k is the sign of the space curvature (+1 for k > 0, 0 for k = 0, -1 for k < 0). Various coordinate transformations give a variety of equivalent forms (e.g. McVittie 1956, 1965, Misner et al. 1973). Equation 7 is particularly convenient in deriving the various relations between the observable parameters and the geometry in the standard model.
The r, , numbers in Equation 7 are comoving coordinates. They are fixed (constant) for all time for a given galaxy. They are also dimensions. The factor R(t) is a scale factor (dimensions of length) that is a function of time in an expanding or contracting manifold. R(t) is independent of r, , and in a congruent space (one of constant curvature).
The volume enclosed within the space from the origin (r = 0, put at the observer) and the coordinate value r is
(8)
For k = 1 this integrates to
(9)
for k = - 1 to
(10)
and for k = 0 to
(11)
Note that r is not the interval distance from the origin to the point r, , . The manifold distance in the space described by Equation 7 is
(12)
without loss of generality by rotating the coordinate system into the = = 0 plane. Hence, the coordinate r in Equations 9-11 is given in terms of the ratio of the measured distance l to the scale factor R (i.e. l /R), just as in Equations 2 and 4. (There should be no confusion about the changed definitions of l and r between Equations 2, 4, and 8-12). Explicitly,
(13)
Substituting Equation 13 into Equations 9, 10, and 11, and expanding for clarity to appreciate the dependence on the curvature, gives
(14)
By analogy with Equations 2 and 4, kR-2 is called the curvature of the space. Note that if k = 0, Equation 14 (i.e. Equation 11 also) gives the Euclidean volume.
The series expansion in Equation 14 illustrates the principle of the galaxy count-volume test. (In practice, of course, the exact equations are used.) If the distances l to a sample of galaxies were known and if a complete count of galaxies within this distance could be obtained, the curvature kR-2 could be measured by the excess (or deficiency) of the counts from an l3 dependence. The distance l and hence the coordinate value of r (from Equation 13) are related to the redshift via the standard theory, to be developed in the next sections. | 741 | 3,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-33 | latest | en | 0.909579 |
https://robotics.stackexchange.com/questions/5111/redundant-arm-path-planning-and-trajectory-following/5119 | 1,624,556,935,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00552.warc.gz | 438,873,972 | 40,685 | # redundant arm path planning and trajectory following
I have a 7dof robotic arm and a set of end effector trajectories in cartesian space I need it to follow.
How do I deal with the redundancy in the arm when planning to follow these trajectories both with and without obstacle avoidance?
• How are you generating these trajectories? What is the rest of your (software) system like? – Ben Dec 7 '14 at 23:28
• I am removing parts 2 and 3 of this question, since the answers would be opinion-based (grounds for closing the question entirely). – Ian Dec 11 '14 at 19:57
Controlling the arm with the Jacobian along an end-effector trajectory is one way to do it. However, this will not give you obstacle avoidance. Although you could manually check for collisions at each time step, but i imagine this could get ugly. Additionally, this is a gradient technique, so you will be constrained to the arm configurations close to the current one. But if you do want to go this route, I find the notation in the book: Robotics: Modelling, Planning and Control by Bruno Siciliano, Lorenzo Sciavicco, Luigi Villani, Giuseppe Oriolo a little easier to understand / use:
$$\dot{q}=J^{\dagger}v_{e}+(I-J^{\dagger}J)\dot{q_o}$$ where: $$J^{\dagger} = J^T(JJ^T)^{-1}$$ is the right-pseudo-inverse of the Jacobian. (This is needed because you are under-constrained and have more joints than the 6 DoF space of end-effector motion). There are other ways to get the pseudoinverse, but this method works well for arm Jacobians. $J$ is obviously the Jacobian, which changes based on the arm configuration, so it is really $J(q)$, and must be recalculated when the arm moves. $v_e$ is the end-effector velocity vector. The first three elements are the Cartesian translation velocity in the global frame, and the last three are rotational velocities. $I$ is the identity matrix. $\dot{q}$ is the joint velocities, and $\dot{q_o}$ are joint velocities to move the arm in its null space (again, which you have since you are under-constrained). I have never found this very useful, so I usually zero it out which greatly simplifies the equation.
To use this formulation, you will need a linear algebra package, which there are many available, and you will also need to calculate the geometric Jacobian of your arm. This is a topic for another question.
NOTE: I typically only use this method for small motions or when i know the arm is in no danger of collision.
To expand on my other answer:
I think the more standard way to solve this problem is to use an arm planner. These planners will have some model of the environment which constitutes obstacles. You give the planner start and goal arm configurations (i.e. joint angles), and it figures out how to best move the arm between them while avoiding obstacles. Note that you will typically have start and goal end-effector poses. and because you have a redundant arm, there are many arm configurations that can satisfy them. So you must first do some IK and find the configuration that is best for you based on some heuristic.
Some notable arm planners / toolkits:
• OpenRave - Typically Linux based, but i believe can be cross complied.
• ROS MoveIt! - again, ROS is typically Linux based, but i think has some limited cross-compile support, but i am not sure about MoveIt.
• Matlab Robotics Toolkit - matlab based, so works on Windows, but is not C++.
• Orocos - Typically Linux based, but i believe can be cross complied. Not sure if it has arm planning actually.
• Trajopt - no Windows support
Underneath these most of these libraries is a planner such as RRT. If you want to do planning yourself, you should check out:
1. The planner will handle the arm's redundancy without issue. (besides increasing the search space)
2. There are many planner algorithms. Too many to list here. For example, this is the list of planners available in OMPL. But if you use an off the shelf planner, it will output a trajectory in joint space which is easy to follow with standard control techniques.
3. Libraries and some "do it yourself" code listed above
To solve the redundancy problem, you can use the Resolved Motion Rate technique. It consist of finding the articular position using the iterative equation $$\Delta \theta = J^\dagger T \Delta x + J^\dagger (I-T) h$$ With
• $J^\dagger$ the pseudoinverse of the jacobian matrix,
• $T$ the projector that defines the null space of the task to accomplish,
• $\Delta x$ the difference in cartesian position between $x(t_{i-1})$ and $F(\theta(t_i))$ ($F(.)$ being the forward kinematics),
• h being the opposite of the gradient of the cost function to minimize.
Typically, we have the cost function $z(\theta)$ and the inverse of the gradient $h = - \nabla z$.
You can find more information and examples on the article from Huo L. and Baron L.
Do you mean you have an end-effector trajectory? Because typical arm planners (such as the one in OpenRAVE) output joint space trajectories. So they tell you how to move each joint, even for 7 DoF arms. If so, you should clarify the question because I think when you say you have a trajectory for your robot arm, it is assumed it is a joint space trajectory.
• It is ok if joint space trajectories are the output, the input is end-effector trajectories. – Andrew Hundt Dec 8 '14 at 23:43 | 1,243 | 5,316 | {"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} | 3.5625 | 4 | CC-MAIN-2021-25 | latest | en | 0.923152 |
https://metanumbers.com/1528504000000 | 1,601,493,073,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127397.84/warc/CC-MAIN-20200930172714-20200930202714-00552.warc.gz | 481,503,229 | 8,233 | ## 1528504000000
1,528,504,000,000 (one trillion five hundred twenty-eight billion five hundred four million) is an even thirteen-digits composite number following 1528503999999 and preceding 1528504000001. In scientific notation, it is written as 1.528504 × 1012. The sum of its digits is 25. It has a total of 17 prime factors and 280 positive divisors. There are 575,385,600,000 positive integers (up to 1528504000000) that are relatively prime to 1528504000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 25
• Digital Root 7
## Name
Short name 1 trillion 528 billion 504 million one trillion five hundred twenty-eight billion five hundred four million
## Notation
Scientific notation 1.528504 × 1012 1.528504 × 1012
## Prime Factorization of 1528504000000
Prime Factorization 29 × 56 × 17 × 11239
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 17 Total number of prime factors rad(n) 1910630 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,528,504,000,000 is 29 × 56 × 17 × 11239. Since it has a total of 17 prime factors, 1,528,504,000,000 is a composite number.
## Divisors of 1528504000000
280 divisors
Even divisors 252 28 14 14
Total Divisors Sum of Divisors Aliquot Sum τ(n) 280 Total number of the positive divisors of n σ(n) 4.0424e+12 Sum of all the positive divisors of n s(n) 2.51389e+12 Sum of the proper positive divisors of n A(n) 1.44371e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23633e+06 Returns the nth root of the product of n divisors H(n) 105.873 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,528,504,000,000 can be divided by 280 positive divisors (out of which 252 are even, and 28 are odd). The sum of these divisors (counting 1,528,504,000,000) is 4,042,396,694,160, the average is 144,371,310,50.,571.
## Other Arithmetic Functions (n = 1528504000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 575385600000 Total number of positive integers not greater than n that are coprime to n λ(n) 4495200000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56555689164 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 575,385,600,000 positive integers (less than 1,528,504,000,000) that are coprime with 1,528,504,000,000. And there are approximately 56,555,689,164 prime numbers less than or equal to 1,528,504,000,000.
## Divisibility of 1528504000000
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 5 0 7
The number 1,528,504,000,000 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1528504000000)
Base System Value
2 Binary 10110001111100001111100000001111000000000
3 Ternary 12102010100001221100220021
4 Quaternary 112033201330001320000
5 Quinary 200020334011000000
6 Senary 3130103545223224
8 Octal 26174174017000
10 Decimal 1528504000000
12 Duodecimal 208298a91b14
16 Hexadecimal 163e1f01e00
20 Vigesimal 2je2ha0000
36 Base36 ji6ntekg
## Basic calculations (n = 1528504000000)
### Multiplication
n×i
n×2 3057008000000 4585512000000 6114016000000 7642520000000
### Division
ni
n⁄2 7.64252e+11 5.09501e+11 3.82126e+11 3.05701e+11
### Exponentiation
ni
n2 2336324478016000000000000 3571081309945368064000000000000000000 5458412066576734867296256000000000000000000000000 8343204677410805551601796481024000000000000000000000000000000
### Nth Root
i√n
2√n 1.23633e+06 11519.2 1111.9 273.435
## 1528504000000 as geometric shapes
### Circle
Radius = n
Diameter 3.05701e+12 9.60387e+12 7.33978e+24
### Sphere
Radius = n
Volume 1.49585e+37 2.93591e+25 9.60387e+12
### Square
Length = n
Perimeter 6.11402e+12 2.33632e+24 2.16163e+12
### Cube
Length = n
Surface area 1.40179e+25 3.57108e+36 2.64745e+12
### Equilateral Triangle
Length = n
Perimeter 4.58551e+12 1.01166e+24 1.32372e+12
### Triangular Pyramid
Length = n
Surface area 4.04663e+24 4.20856e+35 1.24802e+12
## Cryptographic Hash Functions
md5 c0a74cb64aafb186617f253f5348502d 373245c0c658fa5fbfd4062c2c97552d97589fae 079da404f6e5a75062a2aa6361872b937a555c974f2c6aab55fb40a2c7c40be4 b2c22383d47800fef865dad46291456f5cbc42f856762127c7fd74ba74d08df918b484464b971465d89dd480ebce51b9da8a621ca4a2bc372a4e2dad420a0efa 71fb29ed1d1021bf77d27e34c96af6d4316d021c | 1,745 | 4,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-40 | latest | en | 0.778159 |
https://webelements.com/selenium/atom_sizes.html | 1,722,932,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00489.warc.gz | 508,226,668 | 7,519 | โธโธ
• ๐ฌ๐ง Selenium
• ๐บ๐ฆ ะกะตะปะตะฝ
• ๐จ๐ณ ็ก
• ๐ณ๐ฑ Selenium
• ๐ซ๐ท Sélénium
• ๐ฉ๐ช Selen
• ๐ฎ๐ฑ ืกืื ืืื
• ๐ฎ๐น Selenio
• ๐ฏ๐ต ใปใฌใณ
• ๐ต๐น Selênio
• ๐ช๐ธ Selenio
• ๐ธ๐ช Selen
• ๐ท๐บ ะกะตะปะตะฝ
# Selenium - 34Se: radii of atoms and ions
One measure of size is the element-element distance within the element. It is not always easy to make sensible comparisons between the elements however as some bonds are quite short because of multiple bonding (for instance the O=O distance in O2 is short because of the the double bond connecting the two atoms. The bond length in SeSe is: 232.1pm.
There are several other ways ways to define radius for atoms and ions. Follow the appropriate hyperlinks for literature references and definitions of each type of radius. All values of radii are given in picometres (pm). Conversion factors are:
• 1 pm = 1 × 10‑12 metre (meter)
• 100 pm = 1 Ångstrom
• 1000 pm = 1 nanometre (nm, nanometer)
The size of neutral atoms depends upon the way in which the measurement is made and the environment. Follow the appropriate hyperlinks for definitions of each radius type. The term "atomic radius" is not particularly helpful although its use is widespread. The problem is its meaning, which is clearly very different in different sources and books. Two values are given here, one is based upon calculations and the other upon observation - follow the appropriate link for further details.
The following are calculated values of valence shell orbital radii, Rmax
Table: valence shell orbital radii for selenium.
s orbital 82.4 1.55697
p orbital 96.8 1.82990
d orbital 24.1 0.455678
f orbital - -
### References
The Rmax values for neutral gaseous element valence orbitals are abstracted from reference 1.
1. J.B. Mann, Atomic Structure Calculations II. Hartree-Fock wave functions and radial expectation values: hydrogen to lawrencium, LA-3691, Los Alamos Scientific Laboratory, USA, 1968.
This table gives some ionic radii. In this table, geometry refers to the arrangment of the ion's nearest neighbours. Size does depend upon geometry and environment. For electronic configurations, where it matters, the values given for octahedral species are low spin unless stated to be high spin. The terms low spin and high spin refer to the electronic configurations of particular geomtries of certain d-block metal ions. Further information is available in inorganic chemistry textbooks, usually at Level 1 or First Year University level. For definitions of ionic radius and further information, follow the hypertext link.
Se(VI)4-coordinate, tetrahedral42
Se(IV)6-coordinate, octahedral64
Se(VI)6-coordinate, octahedral56
This table shows Pauling radii for selenium
Se(I) 66
Se(VI)42
Se(-I)232
Se(-II)198 | 820 | 2,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.759483 |
https://id.scribd.com/document/373885098/Introduction-to-Physical-Chemistry-Lecture-3 | 1,568,953,503,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573827.2/warc/CC-MAIN-20190920030357-20190920052357-00206.warc.gz | 524,597,714 | 78,996 | Anda di halaman 1dari 5
# Introduction to Physical Chemistry – Lecture 3
## I. LECTURE OVERVIEW of constant density, we have P = ρgh, where ρ is the
density of the substance. Therefore,
In this lecture, we will discuss more general equations
of state than just the Ideal Gas Law. P
ρ= ⇒
gh
101, 325N/m2
II. A QUICK NOTE ON PARTIAL PRESSURES
ρmercury = =
9.8m/s2 × 0.76m
kgm s2
The concept of partial pressure is applied to mixtures 13, 604 2 = 13, 604kg/m3 =
s m4
of gases. Consider two gases, A and B, in some box, at
kg 1000g 1m3
a total pressure P . If there are nA moles of the first 13, 604 3 = 13.604g/cm3 (3)
gas, and nB moles of the second, then the total number m kg 106 cm3
of moles is n = nA + nB , and the mole fraction of A is
and so the density of mercury is 13.604 grams per
xA = nA /n, and the mole fraction of B is xB = nB /n.
milliliter.
We then define the partial pressure of A to be pA =
xA P , and the partial pressure of B to be pB = xB P .
More generally, if a gas is composed of a mixture of n1
moles of substance 1, n2 moles of substance 2, etc., then IV. EQUATIONS OF STATE AND REAL GASES
the partial pressure of substance i is given by pi = xi P ,
where xi = ni /(n1 + n2 + · · · + nN ) is the mole fraction A. What is an equation of state?
of substance i.
For an ideal gas, the partial pressure is a physical quan-
The Ideal Gas Law is what is known as an Equation
tity, in that it gives the pressure of that gas in the system.
of State. That is, it provides a relation between various
To see this, note from the ideal gas law that,
thermodynamic variables associated with the gas. In our
ni RT case, the Ideal Gas Law relates the pressure P , volume
pi = (1) V , temperature T , and number of moles n to one another
V
in a closed form expression.
while if n is the total number of moles in the system, For real gases, the ideal gas law will generally not hold.
then, This is due to the fact that molecules are not point-
particles, but in fact occupy space. Secondly, and more
nRT importantly, the particles constituting a gas will gener-
P = (2)
V ally interact with each other. The stronger these inter-
molecular interactions, the more the behavior of the gas
so that pi /P = ni /n = xi . will deviate from the ideal gas law.
For real gases, however, intermolecular interactions
means that the total pressure is not the sum of the par-
tial pressures. This is a complicated issue, though, and
we will deal with it a bit later. B. Compression factor
## One way to measure the deviation of a gas from ideal
III. TWO ADDITIONAL MEASURES OF gas behavior is via a compression factor, denoted Z. If
PRESSURE we let V̄ = V /n denote the specific volume of a gas (the
volume per mole of gas), then the ideal gas law reads,
We should introduce two measures of pressure that we
did not cover in the previous two lectures: The Torr and pV̄ = RT (4)
mmHg.
The Torr is defined as 1/760 atm. The mmHg is the If we define,
amount of pressure exerted by a column of mercury 1 mm
high. It turns out that 1 Torr is almost exactly equal to pV̄
1 mmHg. Z= (5)
RT
As an exercise, we can compute the density of mercury
based on the information given. We know that, to a good then Z = 1 for an ideal gas. Therefore, deviations from
approximation, a column of mercury 760 mm high exerts Z may be used to measure deviation from ideal gas be-
a pressure of 1 atm = 101, 325 Pa at its base. For a fluid havior.
2
## C. Virial equations of state To account for intermolecular interactions, we note
that the pressure acting on the walls of the container
Now, at a given temperature, and for a given number of depends on both the frequency of collisions of gas parti-
moles, we expect that the pressure uniquely specifies the cles with the walls, and on the strength of the individual
specific volume (as the pressure increases, the density in- collisions. Each effect is roughly proportional to the mo-
creases and hence the specific volume decreases). There- lar concentration n/V , so the overall pressure is reduced
fore, V̄ = V̄ (p, T ), so that Z = pV̄ (p, T )/(RT ) = Z(p, T ) by an amount proportional to the square of the molar
(that is, at a given temperature T , specifying p uniquely concentration. The final equation of state is then,
determines Z). nRT n RT a
We can now expand Z in a Taylor series in p, to obtain, P = − a( )2 = − 2 (11)
V − nb V V̄ − b V̄
Z(p, T ) = 1 + B1 (T )p + B2 (T )p2 + . . . (6) The van der Waals equation of state may be re-arranged
where the first term is 1 because as p → 0, we expect to into the following form:
recover ideal gas behavior, so that Z → 1. n2
But this gives, (P + a )(V − nb) = nRT (12)
V2
pV̄ = RT (1 + B1 (T )p + B2 (T )p2 + . . . ) (7) 2
The quantity a Vn 2 is known as the internal pressure of
which is known as a virial equation of state. The terms the gas.
B1 , B2 , etc. are known as virial coefficients of the gas. Note that when the temperature is high, the term
An alternative form of the virial expansion is to con- −a/V̄ 2 in the van der Waals equation may be neglected.
sider pressure as a function of specific volume. Instead Furthermore, when the concentration of gas molecules is
of writing an expression p = p(V̄ , T ), however, we write low, so that V̄ is large, then the molar volume b can be
p = p(1/V̄ , T ). The reason for this is that we want the neglected, giving the ideal gas equation of state. There-
first term in our virial expansion to be 1. As 1/V̄ → 0, we fore, the van der Waals equation reduces to the ideal gas
have that V̄ → ∞, which means that the specific volume equation of state under the appropriate conditions.
is becoming infinitely large. Physically, this corresponds We will come back to this equation a little bit later.
to a highly dilute gas with large intermolecular distances.
Such a gas is expected to behave ideally.
Therefore, following the same reasoning as before, we E. Critical constants
have Z = Z(1/V̄ , T ), giving,
When a gas is cooled at constant pressure, its density
P V̄ = RT (1 + C1 (T )/V̄ + C2 (T )/V̄ 2 + . . . ) (8) increases, and its specific volume decreases. The average
energy of the gas particles also decreases, so that the
gas particles move more slowly. Eventually, the specific
D. Van der waals equation of state
volume of the gas reaches a point where the distances
between the gas particles are sufficiently small, and the
One of the earliest attempts to modify the ideal gas energy of the gas particles is sufficiently low, that the gas
equation of state to account for the properties of real particles become bound in multimolecular associations
gases was done by van der Waals. Van der Waals sought with one another. At this point the gas is no longer a
to incorporate two effects into an equation of state: (1) gas, but rather has either condensed into a liquid or solid.
The fact that gas molecules are not point particles, but Such a transformation is known as a phase transition.
take up space themselves. (2) The additional fact that At the temperature where condensation occurs, say
gas molecules generally exert weak attractive forces on into a liquid, the gas and liquid phases are in equilibrium.
one another at long distances. Due to intermolecular interactions (collisions, long-range
If we start with the ideal gas equation of state in the forces), some of the molecules in the liquid state will in-
following form, variably enter the gas phase. Thus, the liquid has a cer-
nRT tain vapor pressure at the given temperature, which is
P = (9) simply the pressure of the gas with which the liquid is in
V
thermodynamic equilibrium.
then the van der Waals equation may be derived as fol- The boiling point of a liquid at a given external pressure
lows: We let b denote the molar volume of the gas, that is simply the temperature at which the vapor pressure of
is, the total amount of space that a mole of gas particles the liquid is equal to the external pressure. The external
occupy themselves. Then the volume that the gas actu- gas pressure is no longer sufficiently strong to keep the
ally has available is not V , rather it is V − nb. So the molecules in the liquid state, and the molecules simply
first modification to the ideal gas law is to replace V in push outward against the external pressure source and
the previous equation with V − nb. This gives, enter the gas phase.
nRT RT Let us now try a different approach to achieving gas-
P = = (10) liquid transitions. Instead of cooling a gas at constant
V − nb V̄ − b
3
## pressure, let us compress it at constant temperature.
Eventually, the pressure will equal the vapor pressure
of the gas at the given temperature, and the gas will
liquefy. Basically, the gas molecules become compressed
sufficiently close together that the intermolecular attrac-
tions become sufficiently strong to produce a phase tran-
sition into a bound, liquid state.
Let us consider this liquification process as a function
of temperature. For every temperature T , there is a crit-
ical pressure Pvl (T ) at which the vapor to liquid transi-
tion takes place. At this boundary, the liquid and gas
states are characterized by different densities, or equiv-
alently, specific volumes. The gas has a specific volume
V̄g = V̄g (T, Pvl (T )), while the liquid has a specific volume
V̄l = V̄l (T, Pvl (T )). Clearly, V̄l < V̄g .
Now, as we increase the temperature, the gas parti- FIG. 1: Illustration of the van der Waals isotherms.
cles will have more energy, hence a greater pressure will
be required to force them together into the liquid state.
This means that Pvl is an increasing function of T . Fur- der Waals equation in the form,
thermore, we expect that, since the gas particles have
more energy with increasing temperature, they will on RT a
P = − (13)
average have to be forced closer together to induce a gas V̄ − b V̄ 2
to liquid transition. Therefore, we expect V̄g (T, Pvl (T ))
to decrease with increasing temperature. we can plot P as a function of V̄ for various temperatures
We also expect V̄l (T, Pvl (T )) to increase with increas- (see Figure 1). At high temperatures, the term −a/V̄ 2 is
ing temperature, since increasing pressure should have negligible, and we essentially have the ideal gas law, mod-
only a weak effect on liquid density, but increasing tem- ified slightly by the presence of the molar volume term b.
perature increases the average thermal motions of the At high V̄ , however, this term may also be neglected.
liquid molecules, and hence leads to a reduction in den- At lower temperatures, the term −a/V̄ 2 is no longer
sity. negligible. Because this function increases from −∞ at
V̄ = 0 to 0 as V̄ → ∞, and because RT /(V̄ −b) decreases
Eventually, we will reach a temperature, denoted Tc ,
from ∞ at V̄ = b to 0 as V̄ → ∞, then at lower temper-
at which V̄g (Tc , Pvl (Tc )) = V̄l (Tc , Pvl (Tc )). At this tem-
atures P first decreases to a local minimum as a function
perature, the densities of the liquid and gas phases are
of V̄ , increases to a local maximum, and then decreases
identical. Therefore, compressing the gas at this temper-
ature leads to no discernible phase transition. At temper-
atures T > Tc , the gas is so hot that to liquefy it would If we let V̄min denote where the locally minimum pres-
require compressing the gas to an extent that would make sure, Pmin is attained, and V̄max denote where the locally
the “gaseous” state more dense than the “liquid” state. maximum pressure, Pmax is attained, then the region be-
Clearly, this makes no sense, so that at temperatures tween V̄min and V̄max is physically unrealizable. This is
greater than Tc , there is no distinction between gas and because it is a region where the specific volume and pres-
liquid, and no gas-liquid phase transition is observed. A sure are both increasing, while physically we must have
gas in this temperature regime is known as supercritical. that the specific volume decreases as pressure increases.
The region between Pmin and Pmax , with P > 0, is in-
Given Tc , we can define Pc = Pvl (Tc ), and V̄c =
teresting because there are three distinct specific volumes
V̄g (Tc , Pvl (Tc )) = V̄l (Tc , Pvl (Tc )).
V̄ giving rise to the given pressure P . One of these, as
We should point out that gases such as oxygen, ni- we have seen, lies in the physically unrealizable region.
trogen, and hydrogen are supercritical at room temper- However, for a P > 0, Pmin , P < Pmax , there exists a
ature. That is why these gases cannot be liquefied by V̄1 < V̄min for which P = P (V̄1 ), where P (V̄1 ) is evalu-
compressing them at room temperature. They must first ated using the van der Waals equation. This is because
be cooled, then compressed. P (V̄ ) is decreasing from ∞ to Pmin as V̄ increases from
b to V̄min . Also, there exists a V̄2 > V̄max for which
P = P (V̄2 ). This is because P (V̄ ) is decreasing from
Pmax to 0 as V̄ increases from V̄max to ∞.
F. Critical constants in the van der Waals equation
The two physically realizable solutions V̄1 < V̄2 in the
region known as the van der Waals loops, corresponds
The van der Waals equation is interesting because it to a vapor-liquid equilibrium. We see then how, at suf-
predicts vapor-liquid equilibria, in contrast to the ideal ficiently low temperatures, the van der Waals equation
gas law. We illustrate what we mean: Writing the van gives rise to phase transitions.
4
We know, however, that at sufficiently high tempera- variables. In other words, two gases at the same reduced
tures, the van der Waals equation behaves similarly to temperature and volume will have the same reduced pres-
the ideal gas equation, for which no phase transitions are sure.
possible. Intuitively, the motivation for this hypothesis is that
The transition from the low temperature behavior of two gases with the same set of reduced variables have
the van der Waals equation to the high temperature be- an equal level of deviation from ideal gas behavior. The
havior must therefore occur by the steady disappearance idea is that the values of the critical constants set up
of the van der Waals loops as the temperature increases. natural “length” scales characterizing the behavior of the
This means that V̄min and V̄max should converge as T gas. Rescaling the parameters associated with the gas in
approaches Tc . Finally, at Tc , V̄min = V̄max , so that this terms of these natural length scales gives a set of dimen-
point is neither a local minimum or maximum, but rather sionless parameters that effectively hides these “length”
a point where both dP/dV̄ and d2 P/dV̄ 2 vanish. scales. Therefore, it is not unreasonable to assume that
Differentiating the van der Waals equation of state real gases may exhibit universal behavior when expressed
twice, we obtain, in terms of the reduced variables.
In reality, the law of corresponding states only works
dP RT 2a for gases composed of spherical, non-polar molecules.
=− + 3
dV̄ (V̄ − b)2 V̄ For non-spherical or polar molecules, the principle fails,
d2 P 2RT 6a sometimes badly.
2
= 3
− 4 (14)
dV̄ (V̄ − b) V̄
Setting both derivatives equal to 0 and solving, we ob- V. AN ALTERNATE DERIVATION OF THE
tain, V̄ = 3b, T = 8a/(27bR), and P = a/(27b2 ), and VAN DER WAALS EQUATION
so,
We conclude this lecture by providing a plausible, al-
V̄c = 3b ternative derivation to the van der Waals equation of
8a state. The idea is as follows: Because of intermolecular
Tc =
27bR interactions, the individual gas particles, denoted A, may
a form transient pairwise associations with one another, so
Pc = (15)
27b2 that the gas may be regarded as a chemically reacting
Of course, different equations of state will yield differ- system with the pair of chemical reactions,
ent values for the critical constants. A + A → A2
In concluding this subsection, we should point out that
A2 → A + A (16)
the van der Waals equation gives a value of 3/8 = 0.375
for the critical compression factor, defined by Zc = where the first reaction has a second-order rate constant
pc Vc /(RTc ). This value turns out to be somewhat larger of kf , and the second reaction has a first-order rate con-
than what is measured for a variety of real gases. Never- stant of kr . We then have,
theless, it turns out that Zc is fairly constant at around
0.3 for many gases. This certainly suggests that the van d[A]
= −kf [A]2 + 2kr [A2 ]
der Waals equation captures key physical properties as- dt
sociated with real gases. d[A2 ] 1
= kf [A]2 − kr [A2 ] (17)
dt 2
G. The law of corresponding states At steady-state, we have kf [A]2 = 2kr [A2 ], so that
2[A2 ]/[A]2 = K ≡ kf /kr .
The law of corresponding states is a misnomer, in that Now, the total concentration of gas particles A is given
it is not a derivable law, but rather an educated guess, by [A] + 2[A2 ]. By conservation of mass, this quantity is
backed by experiment, that approximates the behavior a constant, given by [A]0 . We therefore have that,
of real gases. [A]0 − [A]
The idea is as follows: Different gases will generally =K (18)
[A]2
have different values of Tc , Pc , and V̄c . This follows from
the different characteristic particle sizes and strength of Now, define x = [A]0 − [A], so that [A] = [A]0 − x. Then,
the interparticle interactions. However, for a given gas, x is the solution to the quadratic,
we can define reduced variables as follows: At a temper-
ature T , define Tr = T /Tc . The reduced variables Pr and 0 = x2 − (2[A]0 + 1/K)x + [A]20 (19)
V̄r are defined similarly.
The law of corresponding states then argues that there so that,
is a universal equation of state for real gases when the 1 p
equation of state is expressed in terms of the reduced x = [A]0 + (1 − 1 + 4K[A]0 ) (20)
2K
5
where the alternative solution is not realistic since it gives Therefore, intermolecular interactions have reduced
x > [A]0 . the effective molar concentration [A]0 = n/V of gas par-
Now, we are ultimately considering a gas with weak ticles by an amount (K/2)(n/V )2 .
concentration of particle-pairs A2 should be small, which
means that K is small. This allows us to Taylor-expand
Applying the ideal gas law using this reduced molar
the square root term out to second-order in K, giving,
concentration, we obtain,
p
1 + 4K[A]0 ≈ 1 + 2K[A]0 − 2K 2 [A]20 (21)
so that,
x ≈ K[A]20 (22)
## The total concentration of particles is then, nRT RT K n 2
P = − ( ) (24)
V 2 V
1
[A] + [A2 ] = [A]0 − ([A]0 − [A]) + 2[A2 ]
2
1
= [A]0 − ([A]0 − [A]) + ([A]0 − [A])
2
1
= [A]0 − ([A]0 − [A])
2
K 2 The introduction of the factor b has been discussed pre-
= [A]0 − [A]0 (23) viously. Clearly, in this case, we have a = RT K/2.
2 | 5,070 | 18,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-39 | latest | en | 0.895219 |
http://upscfever.com/upsc-fever/en/data/en-exercises-11.html | 1,553,402,673,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203326.34/warc/CC-MAIN-20190324043400-20190324065400-00331.warc.gz | 202,517,241 | 9,806 | • Purpose of Statistics Package Exercises : The Probability & Statistics course focuses on the processes you use to convert data into useful information. This involves
1. Collecting data,
2. Summarizing data, and
3. Interpreting data.
• In addition to being able to apply these processes, you can learn how to use statistical software packages to help manage, summarize, and interpret data. The statistics package exercises included throughout the course provide you the opportunity to explore a dataset and answer questions based on the output using R, Statcrunch, TI Calculator, Minitab, or Excel. In each exercise, you can choose to view instructions for completing the activity in R, Statcrunch, TI Calculator, Minitab, or Excel, depending on which statistics package you choose to use.
• The statistics package exercises are an extension of activities already embedded in the course and require you to use a statistics package to generate output and answer a different set of questions.
1. To download R, a free software environment for statistical computing and graphics, go to: https://www.r-project.org/ This link opens in a new tab and follow the instructions provided.
• Using R
1. Throughout the statistics package exercises, you will be given commands to execute in R. You can use the following steps to avoid having to type all of these commands in by hand:
2. Highlight the command with your mouse.
3. On the browser menu, click "Edit," then "Copy."
4. Click on the R command window, then at the top of the R window, click "Edit," then "Paste."
5. You may have to press to execute the command.
• R Version
1. The R instructions are current through version 3.2.5 released on April 14, 2016. Instructions in these statistics package exercises may not work with newer releases of R.
2. For help with installing R for MAC OS X or Windows click here
• The purpose of this activity is to give you guided practice in checking whether the conditions that allow us to use the two-sample t-test are met.
• Background : A researcher wanted to study whether or not men and women differ in the amount of time they watch TV during a week. In each of the following cases, you'll have to decide whether we can use the two-sample t-test to test this claim or not.
• 1. A random sample of 400 adults was chosen (191 women and 209 men). At the end of the week, each of the 400 subjects reported the total amount of time (in minutes) that he or she watched TV during that week.
• R Instructions
1. If you feel that you need to look at the two samples using histograms, you can open R with the data set preloaded by right-clicking here and choosing "Save Target As" to download the file to your computer. Then find the downloaded file and double-click it to open it in R.
2. The data have been loaded into the data frame tv2 . The two variables in the data frame are time.men and time.women .
3. Create two histograms to view the men's and women's data by modifying the following commands to add appropriate labels/titles.
1. hist(tv2$time.men) hist(tv2$time.women)
1. Explanation :
(i) Since the 400 subjects were chosen at random, we can assume that the two samples are independent. (ii) Since the sample sizes (191 and 209) are large, we can proceed with the two-sample t-test regardless of whether the populations are normal or not (and, thus, there is no need to look at the data using a histogram). In conclusion, we can reliably use the two-sample t-test in this case.
• A random sample of 50 married couples was chosen, which was split into a sample of 50 men and a sample of 50 women. At the end of the week, each of the 100 subjects reported the total amount of time (in minutes) that he or she watched TV during that week.
• R Instructions
1. If you feel that you need to look at the two samples using histograms, you can open R with the data set preloaded by right-clicking here and choosing "Save Target As" to download the file to your computer. Then find the downloaded file and double-click it to open it in R.
2. The data have been loaded into the data frame tv4 . The two variables in the data frame are time.men and time.women .
3. Create two histograms to view the men's and women's data by modifying the following commands to add appropriate labels/titles.
1. hist(tv4$time.men) hist(tv4$time.women)
1. Explanation :
(i) This is a case where the two samples are not independent. Since each subject in one sample is linked (by marriage) to a subject in the other sample, these samples are dependent. The two-sample t-test is therefore not appropriate in this case.
• The purpose of this activity is to give you guided practice in carrying out the two-sample t-test, and to show you how to use software to aid in the process.
• Background A study was conducted at a large state university in order to compare the sleeping habits of undergraduate students to those of graduate students. Random samples of 75 undergraduate students and 50 graduate students were chosen and each of the subjects was asked to report the number of hours he or she sleeps in a typical day. The thought was that since undergraduate students are generally younger and party more during their years in school, they sleep less, on average, than graduate students. Do the data support this hypothesis? The following figure summarizes the problem:
• Note that we defined:
1. μ1 the mean number of hours undergraduate students sleep in a typical day
2. μ2 the mean number of hours graduate students sleep in a typical day
• Comment: Before we move on to carry out the test, it is important to realize that in the two-sample problem, the data can be provided in three possible ways:
• (i) Sample data in one column, and another column that indicates which sample the observation belongs to. Recall that this is the way the data were given in our leading example (looks vs. personality score and gender):
• Note that essentially, one column contains the explanatory variable, and one contains the response.
• (ii) Sample data in different columns data from each of the two samples appear in a column dedicated to that category. As you'll see, this is the way the data are provided in this example:
• (iii) Summarized data we are not given the actual data, but just the data summaries: sample sizes, sample means and sample standard deviations of both samples. Recall that in our second example, the data were given in this format.
• R Instructions
1. To carry out the test, open R with the data set preloaded by right-clicking here and choosing "Save Target As" to download the file to your computer. Then find the downloaded file and double-click it to open it in R.
2. The data have been loaded into the data frame sleep . The two variables in the data frame are undergraduate and graduate .
3. To carry out the t-test, enter the command:
1. t.test(sleep$undergraduate,sleep$graduate,alternative = "less")
4. Note: Using R, when we used t.test() for a one-sample t-test in a previous activity, we specified a one-sample data set, a hypothetical mean, and an alternative hypothesis.
5. To perform a two-sample t-test, we use the same command, t.test() , but specify two sample data sets and an alternative hypothesis.
6. If the data set were structured so the sample data is in one column (called sleep ) and another column that indicates which sample the observation belongs to (called student.type ), then the command would be
1. t.test(sleep~student.type, alternative="less")
1. Explanation :
The test statistic is t = -1.2304 and the p-value is 0.1106
1. Explanation :
The p-value is not small (in particular, it is larger than 0.05), indicating that it is still reasonably likely (probability 0.111) to get data like those observed, or even more extreme data, under the null hypothesis (i.e., assuming that undergraduate and graduate students have the same mean sleeping hours). Therefore, the data do not provide evidence to reject Ho, and we cannot conclude that undergraduate students sleep less, on average, than graduate students.
• The purpose of this activity is to give you guided practice in carrying out the paired t-test and to teach you how to obtain the paired t-test output using statistical software. Here is some background for the historically important data that we are going to work with in this activity.
• Background: Gosset's Seed Plot Data
• William S. Gosset was employed by the Guinness brewing company of Dublin. Sample sizes available for experimentation in brewing were necessarily small, and new techniques for handling the resulting data were needed. Gosset consulted Karl Pearson (1857-1936) of University College in London, who told him that the current state of knowledge was unsatisfactory. Gosset undertook a course of study under Pearson and the outcome of his study was perhaps the most famous paper in statistical literature, "The Probable Error of a Mean" (1908), which introduced the t distribution.
• Since Gosset was contractually bound by Guinness, he published under a pseudonym, "Student," hence the t distribution is often referred to as Student's t distribution.
• As an example to illustrate his analysis, Gosset reported in his paper on the results of seeding 11 different plots of land with two different types of seed: regular and kiln-dried. There is reason to believe that drying seeds before planting will increase plant yield. Since different plots of soil may be naturally more fertile, this confounding variable was eliminated by using the matched pairs design and planting both types of seed in all 11 plots.
• The resulting data (corn yield in pounds per acre) are as follows:
• We are going to use these data to test the hypothesis that kiln-dried seed yields more corn than regular seed. Here is a figure that summarizes this problem:
• Because of the nature of the experimental design (matched pairs), we are testing the difference in yield.
• Note that the differences were calculated: regular − kiln-dried.
• R Instructions
2. The data have been loaded into the data frame seed . Enter the command
1. seed
3. to see the data. The variables in the data frame are regular.seed and kiln.dried.seed .
4. To carry out the paired t-test, use the following command:
1. t.test(seed$regular.seed, seed$kiln.dried.seed, alternative="less", paired=TRUE)
5. The mean of the differences is provided in the output in addition to the other pertinent information. Notice that the order of the variables indicates the order of the difference calculation (position 1 - position 2).
1. Explanation :
The test statistic is -1.69 and the p-value is .061, indicating that there is a 6.1% chance of obtaining data like those observed (or even more extremely in favor of the alternative hypothesis) had there really been no difference between regular and kiln-dried seeds (as the null hypothesis claims). Even though the p-value is quite small, it is not small enough if we use a significance level (cut-off probability) of .05. This means that even though the data show some evidence against the null hypothesis, it isn't quite strong enough to reject it. We therefore conclude that the data do not provide enough evidence that kiln-dried seeds yield more corn than regular seeds. Comment: While it is true that at the .05 significance level, our p-value is not small enough to reject Ho, it is "almost small enough." In other words, this is sort of a "borderline case" where personal interpretation and/or judgment is in order. You can stick to the .05 cut-off as we did above in our conclusion, but you might decide that .061 is small enough for you, and that the evidence that the data provide is strong enough for you to believe that indeed kiln-dried seeds yield more corn. This is the beauty of statistics ... there is no "black or white," and there is a lot of room for personal interpretation
• The purpose of this activity is to give you guided practice in carrying out the ANOVA F-test and to teach you how to obtain the ANOVA F-test's output using statistical software.
• Background: Critical Flicker Frequency (CFF), and Eye Color There is various flickering light in our environment; for instance, light from computer screens and fluorescent bulbs. If the frequency of the flicker is below a certain threshold, the flicker can be detected by the eye. Different people have slightly different flicker "threshold" frequencies (known as the critical flicker frequency, or CFF). Knowing the critical threshold frequency below which flicker is detected can be important for product manufacturing as well as tests for ocular disease. Do people with different eye color have different threshold flicker sensitivity? A 1973 study This link opens in a new tab ("The Effect of Iris Color on Critical Flicker Frequency," Journal of General Psychology [1973], 91–95) obtained the following data from a random sample of 19 subjects.
• Do these data suggest that people with different eye color have different threshold sensitivity to flickering light? In other words, do the data suggest that threshold sensitivity to flickering light is related to eye color?
• Comment: We recommend that before starting, you create for yourself a figure that summarizes this problem, similar to the figures that we presented for the examples that we used in this part.
• R Instructions
2. The data have been loaded into the data frame flicker . Enter the command
1. flicker
3. to see the data. The two variables in the data frame are color and cff .
4. Now use R to create side-by-side boxplots of CFF by eye color, and supplement them with the descriptive statistics of CFF by eye color. Use the output to check whether the conditions that allow us to safely use the ANOVA F-test are met.
5. To do this in R, enter the commands:
1. boxplot(flicker$cff~flicker$color,xlab="Eye Color",ylab="CFF")
tapply(flicker$cff, flicker$color, mean)
tapply(flicker$cff, flicker$color, sd)
1. Explanation :
Let's check the conditions: (i) We are told that the sample was chosen at random, so the three eye-color samples are independent. (ii) The sample sizes are quite low, but the boxplots do not display any extreme violation of the normality assumption in the form of extreme skewness or outliers. (iii) We can assume that the equal population standard deviation condition is met, since the rule of thumb is satisfied (1.843 / 1.365 is less than 2) In summary, we can safely proceed with the ANOVA F-test.
• R Instructions
1. For the next question, we need to carry out the ANOVA F-test using R. To do this, we use the aov() command. Similar to the lm() command, the aov() command produces more output than we need, so we will save the output to a variable name and then use other commands to extract the information of interest. We choose here a generic name, cff.aov , but any name would work within the code as long as it is used throughout.
1. cff.aov=aov(cff~color,flicker)
2. Now we can extract the ANOVA table from cff.aov using either summary() or anova() , which both return the same result.
1. summary(cff.aov)
anova(cff.aov)
3. Note: For more advanced analysis of assumptions for ANOVA, we can use functions such as
1. plot(cff.aov,1)
4. for residual plots,
1. plot(cff.aov,2)
5. for normal QQ plots, and
1. residuals(cff.aov)
6. to extract the residuals.
1. Explanation :
The test statistic F is 4.8 (which is quite large), and the p-value is .023, indicating that it is unlikely (probability of .023) to get data like those observed assuming that CFF is not related to eye color (as the null hypothesis claims). Since the p-value is small (in particular, smaller than .05), we have enough evidence in the data to reject Ho and conclude that the mean CFFs in the three eye-color populations are not all the same. In other words, we can conclude that CFF is related to eye color. | 3,518 | 15,888 | {"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} | 3.078125 | 3 | CC-MAIN-2019-13 | latest | en | 0.895829 |
https://kerodon.net/tag/02HJ | 1,660,033,352,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00061.warc.gz | 342,370,094 | 4,206 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Corollary 4.6.6.12. Let $X$ be a Kan complex and let $x \in X$ be a vertex. The following conditions are equivalent:
$(1)$
The vertex $x$ is initial when viewed as an object of the $\infty$-category $X$.
$(2)$
The vertex $x$ is final when viewed as an object of the $\infty$-category $X$.
$(3)$
The Kan complex $X$ is contractible.
In particular, these conditions are independent of the choice of vertex $x \in X$.
Proof. If the Kan complex $X$ is contractible, then the projection map $X_{x/} \rightarrow X$ is a trivial Kan fibration (Corollary 4.3.7.19), so the object $x \in X$ is initial by virtue of Proposition 4.6.6.11. Conversely, if the projection map $X_{x/} \rightarrow X$ is a trivial Kan fibration, then it is a homotopy equivalence (Proposition 3.1.6.10). Since the Kan complex $X_{x/}$ is contractible (Corollary 4.3.7.14), it follows that $X$ is contractible. This proves the equivalence of $(1)$ and $(3)$; the equivalence of $(2)$ and $(3)$ follows by a similar argument. $\square$ | 338 | 1,084 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-33 | latest | en | 0.656982 |
http://support.sas.com/documentation/cdl/en/grstatproc/65235/HTML/default/n1qa8zljn0alajn1bc41ve27q6rv.htm | 1,718,544,575,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00014.warc.gz | 31,508,276 | 5,795 | # Example 2: Defining Dynamic Variables
Features: DYNAMIC Statement Sample library member: SGREND1
This example uses dynamic variables to set values within the StatGraph template. By using dynamic variables to set the variable names, variable labels, and other parameters, the StatGraph template can be used with different data sets.
The first PROC SGRENDER statement generates a graph for the SASHELP.HEART data set.
The second PROC SGRENDER statement generates multiple graph for the CARS data set by using BY grouping.
## Program Template and Heart Data
```proc template;
define statgraph distribution;
dynamic VAR VARLABEL TITLE NORMAL _BYLINE_;
begingraph;
entrytitle TITLE;
entrytitle _BYLINE_;
layout lattice / columns=1 rows=2 rowgutter=2px
rowweights=(.9 .1) columndatarange=union;
columnaxes;
columnaxis / label=VARLABEL;
endcolumnaxes;
layout overlay / yaxisopts=(offsetmin=.035);
layout gridded / columns=2 border=true autoalign=(topleft topright);
entry halign=left "Nobs";
entry halign=left eval(strip(put(n(VAR),8.)));
entry halign=left "Mean";
entry halign=left eval(strip(put(mean(VAR),8.2)));
entry halign=left "StdDev";
entry halign=left eval(strip(put(stddev(VAR),8.3)));
endlayout;
histogram VAR / scale=percent;
if (exists(NORMAL))
densityplot VAR / normal( );
endif;
fringeplot VAR / datatransparency=.7;
endlayout;
boxplot y=VAR / orient=horizontal;
endlayout;
endgraph;
end;
run;```
```proc sgrender data=sashelp.heart template=distribution;
dynamic var="cholesterol" varlabel="Cholesterol (LDL)" normal="yes"
title="Framingham Heart Study";
run;
title;```
## Program Description
Create the Statgraph template.
```proc template;
define statgraph distribution;
dynamic VAR VARLABEL TITLE NORMAL _BYLINE_;
begingraph;
entrytitle TITLE;
entrytitle _BYLINE_;
layout lattice / columns=1 rows=2 rowgutter=2px
rowweights=(.9 .1) columndatarange=union;
columnaxes;
columnaxis / label=VARLABEL;
endcolumnaxes;
layout overlay / yaxisopts=(offsetmin=.035);
layout gridded / columns=2 border=true autoalign=(topleft topright);
entry halign=left "Nobs";
entry halign=left eval(strip(put(n(VAR),8.)));
entry halign=left "Mean";
entry halign=left eval(strip(put(mean(VAR),8.2)));
entry halign=left "StdDev";
entry halign=left eval(strip(put(stddev(VAR),8.3)));
endlayout;
histogram VAR / scale=percent;
if (exists(NORMAL))
densityplot VAR / normal( );
endif;
fringeplot VAR / datatransparency=.7;
endlayout;
boxplot y=VAR / orient=horizontal;
endlayout;
endgraph;
end;
run;```
Generate the first graphics output from the template using the SASHELP.HEART data set. The DYNAMIC statement defines dynamic variables in the template.
```proc sgrender data=sashelp.heart template=distribution;
dynamic var="cholesterol" varlabel="Cholesterol (LDL)" normal="yes"
title="Framingham Heart Study";
run;
title;```
## Program for Grouped Cars Data
```proc sort data=sashelp.cars out=cars;
by origin;
run;```
```proc sgrender data=cars template=distribution;
by origin;
dynamic var="weight" varlabel="Weight in LBS"
title="Distribution of Vehicle Weight";
run;
title;```
## Program Description
Sort the SASHELP.CARS data set. The data set must be sorted by the same variable that the following PROC SGRENDER block uses in its BY statement.
```proc sort data=sashelp.cars out=cars;
by origin;
run;```
Generate the second graphics output from the template using the WORK.CARS data set. The BY statement generates multiple graphs for each value of the BY variable. The DYNAMIC statement defines dynamic variables in the template.
```proc sgrender data=cars template=distribution;
by origin;
dynamic var="weight" varlabel="Weight in LBS"
title="Distribution of Vehicle Weight";
run;
title;``` | 956 | 3,697 | {"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-2024-26 | latest | en | 0.338941 |
https://www.stata.com/features/overview/power-analysis-for-linear-regression-models/ | 1,726,878,017,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00026.warc.gz | 900,504,034 | 11,514 | Home / Products / Features / Power analysis for linear regression
<- See Stata's other features
Highlights
• Three methods for linear regression (LR)
• Slope test in a simple LR
• R2 test in a multiple LR
• Partial-correlation test in a multiple LR
• Multiple values of parameters
• Automatic and customizable tables
• Automatic and customizable graphs
Stata's power command performs power and sample-size analysis (PSS). Its features include PSS for linear regression.
As with all other power methods, the methods allow you to specify multiple values of parameters and to automatically produce tabular and graphical results.
Linear regression
Stata's power command provides three PSS methods for linear regression.
power oneslope performs PSS for a slope test in a simple linear regression. It computes one of the sample size, power, or target slope given the other two and other study parameters. See [PSS-2] power oneslope.
power rsquared performs PSS for an R2 test in a multiple linear regression. An R2 test is an F test for the coefficient of determination (R2). The test can be used to test the significance of all the coefficients, or it can be used to test a subset of them. In both cases, power rsquared computes one of the sample size, power, or target R2 given the other two and other study parameters. See [PSS-2] power rsquared.
power pcorr performs PSS for a partial-correlation test in a multiple linear regression. A partial-correlation test is an F test of the squared partial multiple correlation coefficient. The command computes one of the sample size, power, or target squared partial-correlation coefficient given the other two and other study parameters. See [PSS-2] power pcorr.
Here, we demonstrate PSS for an R2 test of a subset of coefficients in a multiple linear regression.
Consider a test of the significance of two covariates in a multiple linear regression adjusting for three other covariates. We will call the two covariates the tested covariates and the three others control covariates. The reduced model with the control covariates has an R2 of 0.1, and the full model with all five covariates has an R2 of 0.2. We want to compute the required sample size for the two-sided R2 test to achieve 80% power with a 5% significance level—power rsquared defaults.
. power rsquared 0.1 0.2, ntested(2) ncontrol(3)
Performing iteration ...
Estimated sample size for multiple linear regression
F test for R2 testing subset of coefficients
Ho: R2_F = R2_R versus Ha: R2_F != R2_R
Study parameters:
alpha = 0.0500
power = 0.8000
delta = 0.1250
R2_R = 0.1000
R2_F = 0.2000
R2_diff = 0.1000
ncontrol = 3
ntested = 2
Estimated sample size:
N = 81
We need 81 observations.
Suppose that we want to investigate the impact of the effect size on the required sample size. We plot below the sample-size curve as a function of the R2 values of the full model.
. power rsquared 0.1 (0.2(0.1)0.5), ntested(2) ncontrol(3) graph
As the R2 of the full model increases, the required sample size decreases. When the R2 is closer to 0.2, the curve is steeper.
Tell me more
Learn more about Stata's power, precision, and sample size features.
Read more about PSS for linear regression: | 813 | 3,277 | {"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 | 3 | CC-MAIN-2024-38 | latest | en | 0.783833 |
http://www.mypearsonstore.com/bookstore/introduction-to-digital-signal-processing-coursesmart-0131394096 | 1,441,425,915,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645378542.93/warc/CC-MAIN-20150827031618-00033-ip-10-171-96-226.ec2.internal.warc.gz | 598,476,961 | 8,406 | # Introduction to Digital Signal Processing, CourseSmart eTextbook
Published Date: Sep 5, 2012
More Product Info
## Description
Designed for the undergraduate discrete-time signal processing course
Discrete-Time Signal Processing covers the information that the undergraduate electrical computing and engineering student needs to know about DSP. Core material, with necessary theory and applications, is presented in Chapters 1-7. Four unique chapters that focus on advanced applications follow the core material. MATLAB® is heavily emphasized throughout the book. Most applications have an accompanying lab or sequence of homework problems that have a lab component.
1.1 What is a digital filter?
The analog circuit analysis
A digital filter replacement
1.2 Overview of Analysis and Design
The Analysis Process
The Design Process
CHAPTER 2 Discrete-Time Signals
2.0 Introduction
2.1 Discrete-Time Signals and Systems
Unit Impulse and Unit Step Functions
Related operations
2.2 Transformations of Discrete-Time Signals
Time Transformations
Amplitude Transformations
2.3 Characteristics of Discrete-Time Signals
Even and Odd Signals
Signals Periodic in n
Signals Periodic in &
2.4 Common Discrete-Time Signals
2.5 Discrete-Time Systems
2.6 Convolution for Discrete-Time Systems
Impulse representation of discrete-time signals
Convolution
Properties of convolution
Power gain
Chapter Summary
CHAPTER 3 Frequency Domain Concepts
3.0 Introduction
3.1 Orthogonal Functions and Fourier Series
The Exponential Fourier Series
Discrete Fourier Series
3.2 The Fourier Transform
Definition of the Fourier Transform
Properties of the Fourier Transform
Fourier Transforms of Periodic Functions
3.3 The Discrete-Time Fourier Transform
The Discrete-Time Fourier Transform (DTFT)
Properties of the Discrete-Time Fourier Transform
Discrete-Time Fourier Transforms of Periodic Sequences
3.4 Discrete Fourier Transform
Shorthand Notation for the DFT
Frequency resolution of the DFT
3.5 Fast Fourier Transform
Decomposition-in-Time Fast Fourier Transform Algorithm
Applications of the Discrete / Fast Fourier Transform
Calculation of Fourier Transforms
Convolution Calculations with the DFT/FFT
Linear Convolution with the DFT
Computational Efficiency
3.6 The Laplace Transform
Properties of the Laplace transform
Transfer functions
Frequency response of continuous-time LTI systems
3.7 The z-Transform
Definitions of z-Transforms
z-Transforms
Regions of Convergence
Inverse z-Transforms
z-Transform Properties
LTI System Applications
Transfer Functions
Causality
Stability
Invertibility
Discrete-Time Fourier Transform–z-transform Relationship
Frequency Response Calculation
Chapter Summary
Chapter 4 Sampling and Reconstruction
4.1 Sampling Continuous-Time Signals
Impulse Sampling
Shannon’s sampling theorem
Practical sampling
4.2 Anti-aliasing Filters
Low pass analog Butterworth filters
A low pass Butterworth analog filter has a transfer function given by Switched-capacitor filters
Oversampling
4.3 The Sampling Process
Errors in the sampling process
4.4 Analog to Digital Conversion
Conversion techniques
Successive Approximation Converter
Flash Converter
Sigma-Delta Conversion
Error in A/D conversion process
Dither
4.5 Digital to Analog Conversion
D/A conversion techniques
4.6 Anti-Imaging Filters
Chapter 5 FIR Filter Design and Analysis
5.1 Filter Specifications
5.2 Fundamentals of FIR Filter Design
Linear phase and FIR filters
Conditions for linear phase in FIR filters
Restrictions Imposed by Symmetry
Window Functions and FIR Filters
High pass, band pass, and band stop filters
Kaiser Window
Dolph-Chebyshev window
5.4 Frequency Sampling FIR filters
5.5 The Parks-McClellan Design Technique for FIR filters
5.6 Minimum Phase FIR filters
5.7 Applications
Moving Average FIR Filter
Comb Filters
Differentiators
Hilbert Transformers
5.8 Summary of FIR Characteristics
Chapter 6 Analysis and Design of IIR Filters
6.1 Fundamental IIR design Using the Bilinear Transform
Example 6.1
6.2 Stability of IIR Filters
6.3 Frequency transformations
6.4 Classic IIR filters
The Butterworth Filter
Chebyshev Filters
Inverse Chebyshev filter
Elliptic Filters
Summary of Classic IIR Filters
Invariant Impulse Response
6.5 Poles and Zeros in the z-Plane for IIR Filters
Summary of pole and zero locations for IIR filters
6.6 Direct Design of IIR Filters
Design by pole/zero placement
Design of resonators and notch filters of second order
Numerical Direct Design — Pade method
Numerical Direct Design — Prony's method
Numerical Direct Design — Yule-Walker method
6.7 Applications of IIR Filters
All Pass Filters
IIR Moving Average Filters
IIR Comb Filters
Inverse Filters
Chapter Summary
Chapter 7 Sample Rate Conversion
7.1 Integer Decimation
Frequency spectrum of the down sampled signal
7.2 Integer Interpolation
7.3 Conversion by a Rational Factor
7.4 FIR Implementation
Decimation filters
Interpolation filters
7.5 Narrow Band Filters
7.6 Conversion by an Arbitrary Factor
Hold interpolation
Linear Interpolation
7.7 Bandpass Sampling
7.8 Oversampling in Audio Applications
Chapter Summary
Chapter 8 Realization and Implementation of Digital Filters
8.1 Implementation Issues
8.2 Number Representation
Two’s Complement
Sign/Magnitude
Floating point representation
8.3 Realization Structures
FIR Structures
IIR Structures
State Space Representation
8.4 Coefficient Quantization Error
8.5 Output Error due to Input Quantization
8.6 Product Quantization
8.7 Quantization and Dithering
8.8 Overflow and Scaling
8.9 Limit Cycles
8.10 DSP on Microcontrollers
Microcontroller Characteristics for DSP
Implementation in C
FIR Implementation in C
IIR Implementation in C
Speed optimization
Chapter 9 Digital Audio Signals
9.1 The Nature of Audio Signals
9.2 Audio File Coding
Pulse Code Modulation
Differential Pulse Code Modulation
9.3 Audio File Formats
Lossless file format examples
Lossless compressed format examples
Lossy compressed format examples
9.4 Audio Effects
Oscillators and signal generation
Delay
Flanging
Chorus
Tremolo and Vibrato
Reverberation
The Doppler Effect
Equalizers
Chapter Summary
Chapter 10 Introduction to Two-Dimensional Digital Signal Processing
10.1 Representation of Two-Dimensional Signals
Properties of Two-Dimensional Difference Equations
10.2 Two-Dimensional Transforms
The Z-Transform in Two Dimensions
The two-dimensional Discrete Fourier Transform
Properties of the 2D DFT
The Two Dimensional DFT and Convolution
The Two-Dimensional DFT and Optics
The Discrete Cosine Transform in Two Dimensions
10.3 Two-Dimensional FIR Filters
Window method
Frequency Sampling in Two-Dimensions
Transform methods
Applying FIR Filters to Images
Chapter Summary
Chapter 11 Introduction to Wavelets
11.1 Overview
11.2 The Short Term Fourier Transform
11.3 Wavelets and the Continuous Wavelet Transform
The HAAR Wavelet
The Daubechies Wavelet
Other Wavelet Families
11.4 Interpretation of the Wavelet Transform Data
11.5 The Undecimated Discrete Wavelet Transform
11.6 The Discrete Wavelet Transform
Chapter Summary
APPENDIX A Analog Filter Design
A.1 Analog Butterworth Filters
A.2 Analog Chebyschev Filters
A.4 Analog Elliptic Filters
A.5 Summary of analog filter characteristics
APPENDIX B Bibliography
APPENDIX C Background Mathematics
C.1. Summation Formulas for Geometric Series
C.2. Euler’s Relation
C.3. Inverse Bilateral Z-Transforms by Partial Fraction Expansion
C.4. Matrix Algebra
C.5 State Variable Equations
APPENDIX D MATLAB® User Functions and Commands
D.1. MATLAB User Functions
D.2. MATLAB Commands
## Purchase Info ?
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Introduction to Digital Signal Processing, CourseSmart eTextbook
Format: Safari Book
\$89.99 | ISBN-13: 978-0-13-139409-4 | 1,856 | 8,194 | {"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} | 3.078125 | 3 | CC-MAIN-2015-35 | latest | en | 0.75579 |
http://www.topperlearning.com/forums/home-work-help-19/what-does-it-mean-for-a-circle-or-a-triangle-to-be-circumscr-mathematics-circles-46849/reply | 1,498,676,779,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323730.30/warc/CC-MAIN-20170628185804-20170628205804-00560.warc.gz | 679,266,676 | 40,233 | Question
Wed March 09, 2011 By: Srijan Sood
# What does it mean for a circle or a triangle to be circumscribed , circumscribing , inscribed , inscribing etc.
Wed March 09, 2011
Dear Student,
In geometry, the circumscribed circle or circumcircle of a polygon is a circle which passes through all the vertices of the polygon. The center of this circle is called the circumcenter.
In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is called the triangle's incenter.
Hope this clears the concept.
Regards
Team Topperlearning
Related Questions
Sat March 25, 2017
# iN THE FIG, THE COMMON TANGENT, AB AND CD TO TWO CIRCLES WITH CENRES O, AND O' INTERSECT AT E. PROVE THAT THE POINTS O, E, O' ARE COLLINEAR
Sat March 25, 2017 | 229 | 856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-26 | latest | en | 0.877737 |
https://www.basic-mathematics.com/printable-mean-median-mode-range-worksheets.html | 1,713,839,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00787.warc.gz | 577,285,109 | 3,477 | Mean worksheet
Name _____________________ Date:_____________________
Find the mean of each data set.
1. 5, 8, 4, 2 2. 7, 2, 3
Mean: ________ Mean: ________
3. 1, 9, 3, 10, 4, 9 4. 90, 10, 200
Mean: ________ Mean: ________
5. Find the mean of the following data set and then explain a quick way to find the answer.
9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
Mean: _________
Explanation:
6. Find two numbers whose mean is 15
First number: ________ Second number: _________
7. A man lost 3, 1, and 5 pounds. Find the number of pounds the man lost on average.
8. Find x if the mean of 6, 4, and x is 7.
x: _________
Median worksheet
Name _____________________ Date:_____________________
Find the median of each data set.
1. 6, 1, 7 2. 9, 8, 6
Median: ________ Median: ________
3. 5, 10, 2, 0, 8 4. 5, 7, 11, 3, 1, 3
Median: ________ Median: ________
5. If the number between 8 and 9 is the median, what is that number?
Median: ________
6. 4, 6, 8, 2 7. 5, 7, 11, 3, 1, 3
Median: ________ Median: ________
8. In a small company, the annual salaries of the employees are 51200, 50000, 53000, 48500, 60000, 51000, 52000, 58000, 49000, 48100, and 61000. What is the median annual salary?
Median: _________
Mode worksheet
Name _____________________ Date:_____________________
Find the mode of each data set.
1. 8, 5, 6, 4, 2, 8
Mode: ___________
2. 10, 12, 8, 14, 1, 12, 15, 3, 6
Mode: ___________
3. Orange, lime, apple, orange, pear, kiwi
Mode: ___________
4. 13, 2, 6, 8, 4, 25, 2, 10, 3, 11, 25
Mode: ___________
5. Pencil, iPhone, book, pen, book, eraser, ruler, iPhone
Mode: ____________
Range worksheet
Name _____________________ Date:_____________________
Find the range of each data set.
1. 7, 1, 4, 8, 4, 9
Range: ___________
2. 10, 2, 3, 6, 12, 7, 4
Range: ___________
3. 12, 15, 9, 7, 6, 16, 17, 11
Range: ___________
4. 50, 40, 35, 60, 35, 45, 70, 45
Range: ___________
5. 150, 120, 100, 130, 200, 175, 160, 140
Range: ___________
Mean median mode range worksheet
Name _____________________ Date:_____________________
1. 6, 6, 3 2. 1, 3
Mean: ________ Mean: ________
3. 5, 6, 3, 6, 4, 0 4. 12, 12, 12, 12
Mean: ________ Mean: ________
Median: ________ Median: ________
Mode: ________ Mode: ________
5. In a small company, the annual salaries of the employees are 51200, 50000, 53000, 48500, 60000, 51000, 52000, 58000, 49000, 48100, and 61000. What is the range?
Range: __________
6. Create a list with 4 different numbers that has a mean of 3.
______ _______ ________ _________
7. What is the median and range for the list you created in question 6.?
Median: ____________
Range: ____________
8. 75, 70, 74, 80, 60, 70, 81, 70, 85, 74, 75, 68, 74, 65, 90
Mode: __________________ | 1,191 | 3,178 | {"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-2024-18 | latest | en | 0.683746 |
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2-Guest
## How to adjust decimal places in a parameter
The 0.015000 radius comes from a parameter. How do I adjust this so that it has fewer decimal places. I can't type anything else in the &MIN_RADIUS box (such as [.3]), and if I enter text manually it doesn't recognize that I'm trying to pull a parameter.
1 ACCEPTED SOLUTION
Accepted Solutions
12-Amethyst
(To:ANDERSON)
Yes, the problem is that your parameter is of type "Real Number", and the sketch text tool works better when displaying parameters of type "String".
And "converting" real numbers into strings in Creo is possible but difficult because PTC never provided the "RTOS" (Real to String) function. They do give you the "ITOS" (integer to string) function, which can be manipulated to give you what you the result you want but you will have to write relations to do it. Maybe you can find them on this new forum by searching for "RTOS".
This is one that works well for me:
```RN = FLOOR((MIN_RADIUS+(5/10^(DEC_PLACES+1))),DEC_PLACES)
IF FLOOR(RN) == 0
ELSE
ENDIF
```
Note:
MIN_RADIUS = the parameter that is of type REAL NUMBER
MIN_RAD_STRING = parameter of type STRING that is the rounded number
DEC_PLACES = parameter of type INTEGER which value specifies the # of decimal places in the produced MIN_RAD_STRING
6 REPLIES 6
12-Amethyst
(To:ANDERSON)
Yes, the problem is that your parameter is of type "Real Number", and the sketch text tool works better when displaying parameters of type "String".
And "converting" real numbers into strings in Creo is possible but difficult because PTC never provided the "RTOS" (Real to String) function. They do give you the "ITOS" (integer to string) function, which can be manipulated to give you what you the result you want but you will have to write relations to do it. Maybe you can find them on this new forum by searching for "RTOS".
This is one that works well for me:
```RN = FLOOR((MIN_RADIUS+(5/10^(DEC_PLACES+1))),DEC_PLACES)
IF FLOOR(RN) == 0
ELSE
ENDIF
```
Note:
MIN_RADIUS = the parameter that is of type REAL NUMBER
MIN_RAD_STRING = parameter of type STRING that is the rounded number
DEC_PLACES = parameter of type INTEGER which value specifies the # of decimal places in the produced MIN_RAD_STRING
5-Regular Member
(To:psobejko)
Hi psobejko,
So i found that a simplified version of your method still works:
`EXTRACT(ITOS((RN-FLOOR(RN)+1)*10^(DEC_PLACES)),2,DEC_PLACES)`
you could just use
`ITOS((RN-FLOOR(RN))*10^(DEC_PLACES))`
the syntax would than be:
```RN = FLOOR((MIN_RADIUS+(5/10^(DEC_PLACES+1))),DEC_PLACES)
IF FLOOR(RN) == 0
ELSE
ENDIF```
The EXTRACT function just extracted the values behind the comma (AKA decimals) as a string, which is what ITOS ((RN-FLOOR(RN))*10^(DEC_PLACES)) will do also.
For EXAMPLE:
RN=2,594956
DEC_PLACES=2
RN-FLOOR(RN) equals a real number with only decimals: 0,594956
Multiplying that number by 10^(DEC_PLACES) will give you a real number with x=(DEC_PLACES) decimals "promoted" to whole numbers: 59,4956
Using ITOS will extract only the WHOLE numbers, which is what we want:
ITOS(RN-FLOOR(RN)*10^(DEC_PLACES))=59
21-Topaz I
(To:ANDERSON)
I've used relations to do this also, but sometimes the number I want is a negative number, making the floor function provide unexpected results ( floor ( -5.1 ) = -6, etc). The code I need to use is more complicated to handle this possibility. Also, I've noticed that ITOS rounds the value it is using, for example ITOS ( 12.6 ) = "13", so I don't need to "adjust" the integer I'm converting to get a rounded value.
--- Variables ---
realnum : the real number that will be represented by the string
stringer : the resultant string
numplaces : the number of decimal places in the final result
--- Relations Code ---
if realnum < 0
stringer = "-" + ITOS ( floor ( ABS ( realnum ) ) )
else
stringer = ITOS ( floor ( realnum ) )
endif
stringer = stringer + "."
stringer = stringer + EXTRACT ( ITOS ( 10^numplaces * ( 1 + ABS ( realnum ) - floor ( ABS ( realnum ) ) ) ), 2, numplaces )
Kind of fun if you're doing this for one number, but this stuff gets tedious if you have a lot of numbers you want to use in notes and such.
12-Amethyst
(To:KenFarley)
I want to do the same... except the parameter is in the weld symbol. 45.000° What I want is 45° Ill mess around w/ your relations techniqe. I like it BTW but those darn SW users are guna laugh.
Bart Brejcha
Design-engine.com
6-Contributor
(To:ANDERSON)
Annotate>Format>Decimal Places enter number of decimal places and select the parameter. You might have to click 2 or 3 times to select the parameter.
As displayed in drawing after changing decimal places:
4-Participant
(To:Pieface)
I noticed that there is a quick way to change/add decimal places in a real number parameter by using text "&PARAMETER[.3]" in the drawing. The number in brackets is the number of decimal places you get (in this case, 3). If you need only 1 decimal place, for instance, just replace the .3 in the brackets with .1. | 1,391 | 5,207 | {"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} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.879885 |
https://mathspace.co/textbooks/syllabuses/Syllabus-453/topics/Topic-8419/subtopics/Subtopic-111619/?activeTab=interactive | 1,642,687,870,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00596.warc.gz | 448,027,580 | 53,770 | ## Interactive practice questions
Solve for the two possible values of $x$x:
$10x^2+40x=0$10x2+40x=0
Write all solutions in fraction form, on the same line separated by commas.
Easy
Approx a minute
Solve for $x$x:
$11x^2=7x$11x2=7x
Solve for $x$x:
$81x^2-64=0$81x264=0
Solve the equation $4x^2+8x-32=0$4x2+8x32=0. | 132 | 322 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-05 | longest | en | 0.676748 |
https://web2.0calc.com/questions/squares_11 | 1,659,896,761,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00738.warc.gz | 560,113,624 | 5,524 | +0
# Squares
0
30
1
In the figure below, ABDC, EFHG, and ASHY are all squares; and AB = EF = 1 and AY = 6. What is the area of quadrilateral DYES?
Jun 28, 2022
#1
+2222
+1
The area of the total region is 6 x 6 = 36.
The area of the 2 squares is 1 + 1 = 2.
The area of the triangle outside of DYES but inside the big square is 2.5 ((5 x 1) /2)
There are 4 of these, so the total area of the triangles is 10.
Can you tell me the area now?
Jun 28, 2022 | 169 | 461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-33 | longest | en | 0.929977 |
https://www.physicsforums.com/threads/determining-number-of-bits-in-mar.669968/ | 1,532,343,286,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596204.93/warc/CC-MAIN-20180723090751-20180723110751-00189.warc.gz | 957,799,213 | 12,841 | # Homework Help: Determining Number of Bits in MAR
Tags:
1. Feb 6, 2013
### SpiffWilkie
I'm trying to determine how many bits are required in the memory buffer register and in the memory address register given certain memory systems.
For example, given the below system, how many bits are needed in MAR, and MBR if the memory is word addressable and how many bits if the memory is byte-addressable.
64K x 32
From what I understand the bits in the MBR are equal to the number of bits in the memory unit, or 32 for this example. I'm having a hard time grasping the MAR concept, though.
If I'm using word address memory, would it just be 16 bits (26 x 210 = 64K)? I guess, I'm asking if the unit size is irrelevant in that case.
If the memory is byte-addressable, would it be 18 bits? 216 x 22 (since each unit is 4 bytes?)
Thanks for any insight.
2. Feb 7, 2013
### Staff: Mentor
All of your numbers look correct to me.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 258 | 1,010 | {"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-2018-30 | latest | en | 0.907027 |
https://www.studymode.com/essays/Thermal-Physics-Ib-Dp-Lab-1318583.html | 1,652,841,061,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00381.warc.gz | 1,210,843,586 | 12,732 | # Thermal Physics Ib Dp Lab
Powerful Essays
DP Physics IA Thermal physics: Specific Heat Capacity of Metals Introduction:
In this experiment we are going to measure the specific heat capacity of a unknown metal. To measure the specific heat capacity we will heat up the metal to certain temperature and release the metal in beaker filled with water. By knowing the mass and temperature of water filled in beaker, we will be able to calculate the specific capacity of unknown metal by change in temperature of beaker willed with water.
Hypothesis:
I guess that in this experiment we will get specific heat capacity little bit lower that the actual specific heat capacity of the metal. This is because there will be a lot of energy loss during the experiment. For example there will be energy loss between taking the metal out of heater and placing it inside the water, and energy transfer to beaker while the temperature is rising.
Theory:
Variables:
Independent variable: specific heat capacity of unknown metal
Dependent variable: temperature rise of water
Controlled variables: law of physics
Method:
1, place the unknown metal into the beaker of water.
2, heat the metal and beaker of water with Bunsen burner. (do not let the metal touch the glass)
3, take out the metal when it is heated enough.
4, place it in another beaker filled with water which was kept at constant temperature.
5, measure temperature of water filled in beaker after few minutes
6, repeat 1~5 4 times
DATA TABLE 1: Collected data needed for calculation | Trial 1 | Trial 2 | Trial 3 | Trial 4 | Temp. of the water in the calorimeter (℃)±0.5 | 24.0 | 24.0 | 24.0 | 26.0 | Mass of the water in the calorimeter(g) ±0.05 | 92.41 | 91.40 | 92.26 | 92.62 | Mass of the metal (g) ±0.05 | 99.85 | 100.06 | 99.85 | 100.06 | Initial temp. of the heated metal in beaker (℃) ±0.5 | 97.0 | 97.8 | 95.8 | 96.0 | Final temp. of the metal and water(℃) ±0.5 | 32.0 | 26.0 | 29.0 | 28.0 | Change in temp. of the water
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Satisfactory Essays | 1,882 | 7,393 | {"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} | 3.46875 | 3 | CC-MAIN-2022-21 | latest | en | 0.87036 |
https://gilkalai.wordpress.com/category/combinatorics/page/4/ | 1,427,580,759,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131297831.4/warc/CC-MAIN-20150323172137-00013-ip-10-168-14-71.ec2.internal.warc.gz | 948,460,557 | 33,803 | # Erdős’ Birthday
Paul Erdős was born on March 26, 1913 2013 a hundred years ago. This picture (from Ehud Friedgut’s homepage) was taken in September ’96 in a Chinese restaurant in Warsaw, a few days before Paul Erdős passed away. The other diners are Svante Janson, Tomasz Łuczack and Ehud Friedgut. Erdős’ influence is felt everywhere in combinatorics, mathematics as a whole, and this blog as well. (A few more links: my most decorated MO answer is about Erdős, a recent heated discussion on the “two cultures in mathematics,” a new post on Erdős discrepancy problem on GLL, and, most important, a link to Erdős centennial conference, in Budapest July 1-5, 2013. Join the celebration!)
Do not hesitate to contribute a comment!
# F ≤ 4E
## 1. E ≤ 3V
Let G be a simple planar graph with V vertices and E edges. It follows from Euler’s theorem that
## E ≤ 3V
In fact, we have (when V is at least 3,) that E 3V – 6.
To see this, denote by F the number of regions or faces determined by G (in other words, the number of connected components in the complement of the embedded graph). Euler’s theorem asserts that
E – V + F = 2
## V – E + F = 2
and now note that every face must have at least three edges and every edge is contained in two faces and therefore $2E \ge 3F$, so 6=3V – 3E + 3F ≤ 3V – 3E +2E.
## 2. F ≤ 4E
Now let K be a two-dimensional simplicial complex and suppose that K can be embedded in $R^4$. Denote by E the number of edges of K and by F the number of 2-faces of K.
Here is a really great conjecture:
Conjecture:
## F ≤ 4E
A weaker version which is also widely open and very interesting is:
For some absolute constant C,
## F ≤ C E
Remarks: The conjecture extends to higher dimensions. If K is an r-dimensional simplicial complex that can be embedded into $R^{2r}$ then the conjecture is that
$f_r(K) \le C_rf_{r-1}(K),$
Where $C_r$ is a constant depending on r. Here $f_i(K)$ is the number of i-dimensional faces of K. A stronger statement is that $C_r= r+2$. The conjecture also extends to polyhedral complexes and more general form of complexes. In the conjecture ‘embed’ refers to a topological embedding.
# Lionel Pournin found a combinatorial proof for Sleator-Tarjan-Thurston diameter result
I just saw in Claire Mathieu’s blog “A CS professor blog” that a simple proof of the Sleator-Tarjan-Thurston’s diameter result for the graph of the associahedron was found by Lionel Pournin! Here are slides of his lecture “The diameters of associahedra” and link to the paper with the same title “The diameters of associahedra.” The original proof was based on hyperbolic volume computations and was quite difficult. (Here is an earlier post on the associahedron and an earlier mention of a connection found by Dehornoy with the Thompson group.)
# Happy Birthday Ron Aharoni!
Ron Aharoni, one of Israel’s and the world’s leading combinatorialists celebrated his birthday last month. This is a wonderful opportunity to tell you about a few of the things that Ron did mainly around matching theory.
## Menger’s theorem for infinite graphs
### Hall’s marriage theorem
Hall marriage theorem (Philip Hall, 1935) gives a necessary and sufficient condition for a perfect matching in bipartite graphs. Suppose that you have a set A of n men and a set B of n women and a list of pairs of men and women that know each other. A perfect matching is a bijection from A to B which matches every man to a woman he knows.
Hall’s marriage theorem asserts that a necessary and sufficient condition for a perfect matching is that every set S of men knows together at least |S| women.
This is an extremely important theorem and the starting point for a wonderful matching theory. It is a primary example of combinatorial duality. Other theorems of this kind are Menger’s theorem on connectivity in graphs, Dilworth’s theorem (1950) on covering posets with chains, the max-flow min-cut theorem (1956), and quite a few more.
### Menger’s theorem
Menger Theorem (Karl Menger, 1927). Let G be a finite graph and let x and y be two distinct vertices. Then the minimum number of edges whose removal disconnects x and is equal to the maximum number of pairwise edge-disjoint paths from x to y.
### Infinite Menger
Ron Aharoni and Eli Berger proved the following theorem (here is a link to the arxived version):
Aharoni and Berger Theorem (2005): Given two sets of vertices, A and B, in a (possibly infinite) digraph, there exists a family P of disjoint A to B paths, and a separating set consisting of the choice of precisely one vertex from each path in P.
# The Quantum Debate is Over! (and other Updates)
### Quid est noster computationis mundus?
Nine months after is started, (much longer than expected,) and after eight posts on GLL, (much more than planned,) and almost a thousand comments of overall good quality, from quite a few participants, my scientific debate with Aram Harrow regarding quantum fault tolerance is essentially over. Some good food for thought, I hope. As always, there is more to be said on the matter itself, on the debate, and on other”meta” matters, but it is also useful to put it now in the background for a while, to continue to think about quantum fault tolerance in the slow pace and solitude, as I am used to, and also to move on in other fronts, which were perhaps neglected a little.
Here are the links to the eight posts: My initial post “Perpetual Motion of The 21st Century?” was followed by three posts by Aram. The first “Flying Machines of the 21st Century?” the second “Nature does not conspire” and the third “The Quantum super-PAC.” We had then two additional posts “Quantum refutations and reproofs” and “Can you hear the shape of a quantum computer?.” Finally we had two conclusion posts: “Quantum repetition” and “Quantum supremacy or quantum control?
### EDP 22-27
We had six new posts on the Erdos Discrepancy Problem over Gowers’s blog (Here is the link to the last one EDP27). Tim contributed a large number of comments and it was interesting to follow his line of thought. Other participants also contributed a few comments. One nice surprise for me was that the behavior of the hereditary discrepancy for homogeneous arithmetic progression in {1,2,…,n} was found by Alexander Nikolov and Kunal Talwar. See this post From discrepancy to privacy and back and the paper. Noga Alon and I showed that it is ${\tilde{\Omega}(\sqrt{\log n})}$ and at most ${n^{O(\frac{1}{\log\log n})}}$, and to my surprise Alexander and Kunal showed that the upper bound is the correct behavior. The argument relies on connection with differential privacy.
### Möbius randomness and computation
After the $AC^0$-prime number theorem was proved by Ben Green, and the Mobius randomness of all Walsh functions and monotone Boolean function was proved by Jean Bourgain, (See this MO question for details) the next logical step are low degree polynomials over Z/2Z . (The Walsh functions are degree 1 polynomials.) The simplest case offered to me by Bourgain is the case of the Rudin-Shapiro sequence. (But for an ACC(2) result via Razborov-Smolensky theorem we will need to go all the way to polynomial of degree polylog.) I asked it over MathOverflaw. After three months of no activity I offered a bounty of 300 of my own MO-reputations. Subsequently, Terry Tao and Ben Green discussed some avenues and eventually Tao solved the problem (and earned the 300 reputation points). Here is a very recent post on Möbius randomness on Terry Tao’s blog.
### Influences on large sets
In my post Nati’s Influence I mentioned two old conjectures (Conjecture 1 dues to Benny Chor and Conjecture 2) about influence of large sets on Boolean functions. During Jeff Kahn’s visit to Israel we managed to disprove them both. The disproof is inspired by an old construction of Ajtai and Linial.
### Tel Aviv, New Haven, Jerusalem
Last year we lived for a year in Tel Aviv which was a wonderful experience: especially walking on the beach every day and feeling the different atmosphere of the city. It is so different from my Jerusalem and still the people speak fluent Hebrew. I am now in New Haven. It is getting cold and the fall colors are getting beautiful. And it also feels at home after all these years. And next week I return to my difficult and beautiful (and beloved) Jerusalem.
# Looking Again at Erdős’ Discrepancy Problem
Over Gowers’s blog Tim and I will make an attempt to revisit polymath5. Last Autumn I prepared three posts on the problems and we decided to launch them now. The first post is here. Here is a related MathOverflow question. Discrepancy theory is a wonderful theory and while I am not an expert we had several posts about it here. (This post on Beck-Fiala and related matters; and this news item on Beck’s 3-permutation conjecture.) I am aware of at least one important recent development in the theory that I did not report. My posts go around the problem and do not attack it directly but I hope people will have a chance to think about the problem again and perhaps also about polymathing again. Meanwhile, polymath7 (over the polymath blog) is in a short recess but I hope a new research thread will start soon.
# Tokyo, Kyoto, and Nagoya
Near Nagoya: Firework festival; Kyoto: with Gunter Ziegler; with Takayuki Hibi, Hibi, Marge Bayer, Curtis Green and Richard Stanly; Tokyo: Peter Frankl; crowded crossing. Added later: Mazi and I at the same restaurant taken by Stanley.
I just returned from a trip to Japan to the FPSAC 2012 at Nagoya and a workshop on convex polytopes in Kyoto. As in my first visit to Osaka in 1999 I found Japan stunning, and this time I was able to share the experience with my wife.
Kyoto: The week before FPSAC there was a workshop at RIMS devoted to convex polytopes. Some highlights: A classical result by Venkov and McMullen characterizes polytopes whose translates tile the Euclidean d-space. Sinai Robins talked about his work with Nick Gravin and Dima Shiryaev about k-tilings (every point is covered k times). Not a full characterization yet but already some impressive results. Gunter Ziegler talked about his work with Karim Adiprasito disproving an old conjecture by Shephard which asserts that there are only finitely many projectively unique d-polytopes for every dimension d. This is false above dimension 81. Eran Nevo talked about his solution with Satoshi Murai of the generalized lower bound conjecture and some subsequent works on triangulated manifolds. There were several talks relating Ehrhard polynomials of polytopes with enumerative combinatorics, and several talks on algebraic geometry, toric varieties, Fano varieties, and mirror symmetry. Here are the slides of my lecture entitled open problems on convex polytopes I’d love to see solved (but some further explanations for some parts are needed). I hope to return to some of the topics of the workshop in a later post.
Nagoya: FPSAC (Formal power series and algebraic combinatorics) is a central annual combinatorial meeting with special emphasis on enumerative and algebraic combinatorics. This year it was the 24th such event although I could have swear that I took part in an FPSAC in Montreal already in 1985 but apparently this was a conference with a similar flavor and different name. Much is going on in enumerative and algebraic combinatorics. Cluster algebras, miraculously discovered a decade ago by Fomin and Zelevinsky are going strong in combinatorics as well as in other areas. Alternating sign matrices continue to offer amazing problems and answers. There were quite a few lectures on representation theory, symmetric functions, random matrices, and also on relations of enumerative combinatorics with hyperplane arrangements with polytopes and with physics. There were lectures involving massive computations and new computerized method and packages for symbolic computations were described. Here are the slides of my lecture entitled Discrete isoperimetry: problems, results, applications and methods. The program page now contains slides for most lectures.
### What are alternating sign matrices?
I suppose that you all know what a permutation matrix is (an n by n matrix with 0,1 entries and one non zero entry in each row and each column) and alternating sign matrices are amazing generalization of permutation matrices discovered by William Mills, David Robbins, and Howard Rumsey. (See also this article by Bressoud and Propp) They are integral n by n matrices with 0,1 and -1 entries. In every row and every columns the non zero aliments (and there must be at least one such element) should alternate between 1 and -1 and start and end with a ‘1’. Alternating sign matrices are in one to one correspondence with monotone triangle. Thos are triangles of integers starting with a row: 1,2,…,n. With the following rules: (a) all rows are strictly increasing; (b) every element in row i is weakly between the two elements above it.
The amazing thing is that the number of alternating sign matrices of order n is precisely
## 1!4!7!…(3n-2)!/n!(n+1)!…(2n-1)!
This was a conjecture by Mills, Robbins and Rumsey and it took over a decade until it was proved by Doron Zeilberger. Later Greg Kuperberg found a short proof and by now several proofs are known. If you have some information or ideas on alternating sign matrices that you would like to share or some questions about them, or about anything else, please contribute a comment.
# Celebrations in Sweden and Norway
### Celebrations for Endre, Jean and Terry
Anders Bjorner presents the 2012 Crafoord Prize in Mathematics
I am in Sweden for two weeks to work with colleagues and to take part in two celebrations. Jean Bourgain and Terence Tao are the 2012 laureates of the Crafoord Prize in mathematics which was awarded last Tuesday at Lund. Along with them the 2012 Crafoord Prize in Astronomy was awarded to Reinhand Genzel and Andrea Ghez. I took part in the symposium entitled “From chaos to harmony” to celebrate the event.
Next Friday the Swedish Royal academy will celebrate with a mini-symposium in honor of the 2012 Abel prize winner Endre Szeméredi. (Here are the slides of my future talk looking at and around the Szeméredi-Trotter theorem. Please alert me of mistakes if you see them.) The Abel prize symposium and ceremony in Oslo are Tuesday (Today! see the picture above) and Wednesday of this week.
Congratulations again to Jean, Terry and Endre for richly deserved awards.
### Crafoord days at Lund
Owing to the passing of Count Carl Johan Bernadotte af Wisborg, H.M. King Carl XVI Gustaf was unable to attend the Crafoord Days 2012. The prizes were presented by Margareta Nilsson, daughter of the Donors, Holge and Anna-Greta Crafoord. Ms Nilsson’s kind hospitality, deep devotion to science, culture and other noble social causes, and moving childhood memories shared at the dinner, have led Reinhard Genzel in his moving speech on behalf of the winners in Astronomy to refer to Margareta Nilsson by the words: “You were our King these past two days!”.
The one day symposium itself was very interesting, and so were the four prize lectures on Tuesday morning. In a few days The videos of the two days’ lectures will be are posted here. Here are the slides of my talk on analysis of Boolean functions, featuring, among other things a far-reaching conjectural extension of a recent theorem by Hamed Hatami.
### Gothenburg and Stockholm
From Lund I continued to a short visit of Gothenburg hosted by Jeff Steif with whom I share much interest in noise sensitivity and many other things. I then continued to Stockholm where I visit Anders Björner who is a long-time collaborator and friend since the mid eighties. For me this is perhaps the twelfth visit to Stockholm and it is always great to be here.
We will celebrate on this blog these exciting events with a rerun of the classic, much-acclaimed piece by Christine Björner on the Golden room and the golden mountain.
Speakers at Crafoord symposium, (from right to left) Carlos Kenig, Ben Green, Jean Bourgain, Terry Tao, me and Michael Christ. Copyright: Crafoord foundation.
Update(Oct 2014): Here is a picture of me and Jean at IHES 1988
# Galvin’s Proof of Dinitz’s Conjecture
## Dinitz’ conjecture
The following theorem was conjectured by Jeff Dinitz in 1979 and proved by Fred Galvin in 1994:
Theorem: Consider an n by n square table such that in each cell (i,j) you have a set $A_{i,j}$ with n or more elements. Then it is possible to choose elements $a_{ij}$ from $A_{ij}$ such that the chosen elements in every row and in every colummn are distinct.
Special case: if all $A_{ij}$ are the same set, say {1,2,…,n} then this is possible. Such a choice is called a Latin square for example $a_{ij}=i+j({\rm mod} n)$.
1 2 3 4
2 3 4 1
3 4 1 2
4 1 2 3
Here are a few remarks before we go ahead with Galvin’s proof:
1) The Theorem is about graph-coloring from lists of colors, or briefly list-coloring. Given a graph G with n vertices and a list of colors $A_i$ for vertex $i$ we ask for a proper coloring of the graph such that vertex i is colored by an element of $A_i$. (A proper coloring is a coloring such that two djacent vertices have different colors.)
The list chromatic number of a graph G is the smallest number k such that G can be colored from any lists of colors, where every $A_i$ has k elements.
2) We consider the graph G whose vertices are the pairs (i,j) and two vertices are adgacent if one of their coordinates agree. This graph is the line graph of the complete bipartite graph $K_{n,n}$. The line graph L(H) of a graph H is the graph whose vertices correspond to the edges of H and two vertices are adjacent if the corresponding edges share a vertex.
3) Vizing proved that the chromatic number of the line graph L(H) is at most one more than the maximum degree of H. He conjectured that this is also true for list coloring. A bolder version asserts that for every line-graph G the list chromatic number equals the chromatic number. Dinits’s conjecture is the special case (of the bolder version) where G is the line graph of the complete bipartite graph $K_{n,n}$. The proof demonstrates Vizing’s conjecture for the line graph of every bipartite graph.
4) Shortly before Galvin, Jeannette Janssen gave an amazing proof for a slightly weaker statement where in each square you allow n+1 colors. (She proved the Dinitz conjecture for rectangular arrays.) Janssen applied a theorem of Alon and Tarsi, who used the “polynomial method” to obtain a powerful necessary condition for list coloring.
5) Here are now (or when I will be able to find them later) links to Galvin’s paper, to Janssen’s paper, to Alon-Tarsi’s paper, to expositions by Barry Cipra and by Doron Zeilberger and to a short self-contained version of the proof by Tomaž Slivnik.
## Galvin’s proof
Galvin’s ingenious proof of Dinitz’ the conjecture combines two known theorems which are quite easy themselves.
For a directed graph G an induced subgraph is a subgraph spanned by a subset of the set of vertices of G. A kernel in a directed graph is an independent and absorbant set of vertices. I.e., it is a set of vertices W such that there is no edge between any two vertices in W, and from every vertex not in W there is an edge directed from it to a vertex in W.
Lemma 1 (Bondy, Boppana, Siegal): (mentioned e.g. in the paper by Alon and Tarsi)
# Fractional Sylvester-Gallai
Avi Wigderson was in town and gave a beautiful talk about an extension of Sylvester-Gallai theorem. Here is a link to the paper: Rank bounds for design matrices with applications to combinatorial geometry and locally correctable codes by Boaz Barak, Zeev Dvir, Avi Wigderson, and Amir Yehudayoff.
### Sylvester-Gallai
The Sylvester-Gallai Theorem: Let X be a finite set of n points in an eulidean space such that for every two distinct points $x,y \in X$ the line through $x$ and $y$ contains a third point $z \in X$. Then all points in $X$ are contained in a line.
I heard about this result when I took Benjy Weiss’s mathematics course for high-school students in 1970/1. a The Sylvester-Gallai theorem was the last question marked with (*) in the first week’s homework. In one of the next meetings Benjy listened carefully to our ideas on how to prove it and then explained to us why our attempts of proving it are doomed to fail: What we tried to do only relied on the very basic incidence axioms of Euclidean geometry but the Sylvester-Gallai theorem does not hold for finite projective planes. (Sylvester conjectured the result in 1893. The first proof was given by Mechior in 1940 and Gallai proved it in 1945.)
### My MO question
Befor describing the new results let me mention my third ever MathOverflow question that was about potential extensions of the G-S theorem. The question was roughly this:
Suppose that V is an r dimensional variety embedded into n space so that if the intersection of every j-dimensional subspace with V is full dimensional then this intersection is “complicated”. Then $n$ cannot be too large.
I will not reproduce the full question here but only a memorable remark made by Greg Kuperberg:
If you claimed that Gil is short for Gilvester (which is a real first name although rare), then you could say that any of your results is the “Gilvester Kalai theorem”. – Greg Kuperberg Nov 24 2009 at 5:13
### The result by Barak, Dvir, Wigderson and Yehudayoff
Theorem: Let X be a finite set of n points in an Euclidean space such that for every point $x \in X$ the number of $y, y\in X,y \ne x$ such that the line through $x$ and $y$ contains another point of $X$ is at least $\delta (n-1)$. Then
$\dim (Aff(X))\le 13/\delta^2$
### Some remarks:
1) The proof: The first ingredient of the proof is a translation of the theorem into a question about ranks of matrices with a certain combinatorial structure. The next thing is to observe is that when the non zero entries of the matrix are 1’s the claim is simple. The second surprising ingredient of the proof is to use scaling in order to “tame” the entries of the matrix.
2) The context – locally correctable codes: A $q$-query locally correctable $(q,\delta)$-code over a field $F$ is a subspace of $F^n$ so that, given any element $\tilde y$ that disagrees with some $y \in C$ in at most $\delta n$ positions and an index $i$, $1 \le i \le n$ we can recover $y_i$ with probability 3/4 by reading at most $q$ coordinates of $\tilde y$. The theorem stated above imply that, for two queries, over the real numbers (and also over the complex numbers), such codes do not exist when $n$ is large. Another context where the result is of interest is the hot area of sum product theorems and related questions in the geometry of incidences.
3) Some open problems: Is there a more detailed structure theorem for configurations of points satisfying the condition of the theorem? Can the result be improved to $\dim (Aff(X))=O(1/\delta )$? Can a similar result be proved on locally correctable codes with more than two queries? This also translates into an interesting Sylvester-Gallai type question but it will require, Avi said, new ideas. | 5,786 | 23,212 | {"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": 46, "codecogs_latex": 0, "wp_latex": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2015-14 | longest | en | 0.904465 |
http://www.askphysics.com/tag/solar-system/ | 1,603,990,078,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904834.82/warc/CC-MAIN-20201029154446-20201029184446-00089.warc.gz | 120,643,654 | 24,156 | Home » Posts tagged 'Solar System'
# Tag Archives: Solar System
## Why moon does not fall to Earth?
The moon does not fall to earth because of its tangential velocity. The following video explains it well.
## Is the gravitational force between moon & earth are equal ?
Is the gravitational force between moon & earth are equal ?
If the question is- “Is the force exerted by earth on moon is equal to the force exerted by moon on earth?” then the answer is “Yes
I hope that more questions will follow.
• Comparison of charge and mass
## Why does the sun appear to rise and set though it is actually not moving?
Why does the sun appear to rise and set though it is actually not moving?
First of all I would like to remind you that it is incorrect to say that “sun is not moving”. Sun is also moving. But the question here is, “Sun appears to rise and set not because of its motion”.
Sunset and sunrise are caused by the rotation of earth. When earth rotates, the position of sun appears to change and it seems as if the sun is revolving around the earth; just like we feel that the ground is revolving around us when we are looking out from a rotating merry-go-round.
## Parallax Method
How to use it to calculate the distance between two astronomical objects?
Parallax is the change in the apparent position of an object due to change iv view point.
The parallax method is a way of measuring distances of far-away objects.
## Parallax Method to determine the distance of stars
WHAT IS THE WAY TO MEASURE THE DISTANCE OF DISTANT OBJECTS LIKE STARS WITH THE HELP OF PARALLAX METHOD
Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines.
Astronomers use the principle of parallax to measure distances to celestial objects including to the Moon, the Sun, and to stars beyond the Solar System.
See the animation below to understand the parallax method for determining the stellar distances
http://spiff.rit.edu/classes/phys240/lectures/parallax/parallax.html
is it the same method that we use in trigonometry ?
Parallax is a different concept, though we are using the principle of trigonometry to solve problems based on it.
### Hits so far @ AskPhysics
• 2,240,930 hits | 514 | 2,340 | {"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} | 3.375 | 3 | CC-MAIN-2020-45 | longest | en | 0.930775 |
https://isabelle.in.tum.de/repos/testboard/rev/8893e0ed263a | 1,638,893,422,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00449.warc.gz | 381,823,870 | 8,267 | author paulson Thu, 03 Jun 2021 10:47:20 +0100 changeset 74052 8893e0ed263a parent 74049 26c0ccf17f31 child 74053 56f31baaa837
src/HOL/Analysis/Convex_Euclidean_Space.thy file | annotate | diff | comparison | revisions src/HOL/Analysis/Derivative.thy file | annotate | diff | comparison | revisions src/HOL/Analysis/Linear_Algebra.thy file | annotate | diff | comparison | revisions src/HOL/Analysis/Path_Connected.thy file | annotate | diff | comparison | revisions src/HOL/Complex_Analysis/Complex_Singularities.thy file | annotate | diff | comparison | revisions src/HOL/Complex_Analysis/Contour_Integration.thy file | annotate | diff | comparison | revisions src/HOL/Deriv.thy file | annotate | diff | comparison | revisions src/HOL/Limits.thy file | annotate | diff | comparison | revisions
```--- a/src/HOL/Analysis/Convex_Euclidean_Space.thy Mon May 31 20:27:45 2021 +0000
+++ b/src/HOL/Analysis/Convex_Euclidean_Space.thy Thu Jun 03 10:47:20 2021 +0100
@@ -1658,6 +1658,31 @@
unfolding islimpt_def by blast
qed
+lemma islimpt_Ioc [simp]:
+ fixes a :: real
+ assumes "a<b"
+ shows "x islimpt {a<..b} \<longleftrightarrow> x \<in> {a..b}" (is "?lhs = ?rhs")
+proof
+ show "?lhs \<Longrightarrow> ?rhs"
+ by (metis assms closed_atLeastAtMost closed_limpt closure_greaterThanAtMost closure_subset islimpt_subset)
+next
+ assume ?rhs
+ then have "x \<in> closure {a<..<b}"
+ using assms closure_greaterThanLessThan by blast
+ then show ?lhs
+ by (metis (no_types) Diff_empty Diff_insert0 interior_lessThanAtMost interior_limit_point interior_subset islimpt_in_closure islimpt_subset)
+qed
+
+lemma islimpt_Ico [simp]:
+ fixes a :: real
+ assumes "a<b" shows "x islimpt {a..<b} \<longleftrightarrow> x \<in> {a..b}"
+ by (metis assms closure_atLeastLessThan closure_greaterThanAtMost islimpt_Ioc limpt_of_closure)
+
+lemma islimpt_Icc [simp]:
+ fixes a :: real
+ assumes "a<b" shows "x islimpt {a..b} \<longleftrightarrow> x \<in> {a..b}"
+ by (metis assms closure_atLeastLessThan islimpt_Ico limpt_of_closure)
+
lemma connected_imp_perfect_aff_dim:
"\<lbrakk>connected S; aff_dim S \<noteq> 0; a \<in> S\<rbrakk> \<Longrightarrow> a islimpt S"
using aff_dim_sing connected_imp_perfect by blast```
```--- a/src/HOL/Analysis/Derivative.thy Mon May 31 20:27:45 2021 +0000
+++ b/src/HOL/Analysis/Derivative.thy Thu Jun 03 10:47:20 2021 +0100
@@ -3281,6 +3281,10 @@
unfolding C1_differentiable_on_eq
by (rule continuous_intros | simp add: continuous_at_imp_continuous_on differentiable_imp_continuous_within)+
+lemma C1_differentiable_on_of_real [derivative_intros]: "of_real C1_differentiable_on S"
+ unfolding C1_differentiable_on_def
+ by (smt (verit, del_insts) DERIV_ident UNIV_I continuous_on_const has_vector_derivative_of_real has_vector_derivative_transform)
+
definition\<^marker>\<open>tag important\<close> piecewise_C1_differentiable_on
(infixr "piecewise'_C1'_differentiable'_on" 50)
@@ -3298,7 +3302,7 @@
C1_differentiable_on_def differentiable_def has_vector_derivative_def
intro: has_derivative_at_withinI)
-lemma piecewise_C1_differentiable_compose:
+lemma piecewise_C1_differentiable_compose [derivative_intros]:
assumes fg: "f piecewise_C1_differentiable_on S" "g piecewise_C1_differentiable_on (f ` S)" and fin: "\<And>x. finite (S \<inter> f-`{x})"
shows "(g \<circ> f) piecewise_C1_differentiable_on S"
proof -
@@ -3334,7 +3338,7 @@
unfolding C1_differentiable_on_eq continuous_on_eq_continuous_within
using differentiable_at_withinI differentiable_imp_continuous_within by blast
-lemma C1_differentiable_on_empty [iff]: "f C1_differentiable_on {}"
+lemma C1_differentiable_on_empty [iff,derivative_intros]: "f C1_differentiable_on {}"
unfolding C1_differentiable_on_def
by auto
@@ -3356,7 +3360,7 @@
done
qed
-lemma piecewise_C1_differentiable_cases:
+lemma piecewise_C1_differentiable_cases [derivative_intros]:
fixes c::real
assumes "f piecewise_C1_differentiable_on {a..c}"
"g piecewise_C1_differentiable_on {c..b}"
@@ -3444,12 +3448,21 @@
qed
-lemma piecewise_C1_differentiable_neg:
+lemma piecewise_C1_differentiable_const [derivative_intros]:
+ "(\<lambda>x. c) piecewise_C1_differentiable_on S"
+
+lemma piecewise_C1_differentiable_scaleR [derivative_intros]:
+ "\<lbrakk>f piecewise_C1_differentiable_on S\<rbrakk>
+ \<Longrightarrow> (\<lambda>x. c *\<^sub>R f x) piecewise_C1_differentiable_on S"
+ by (force simp add: piecewise_C1_differentiable_on_def continuous_on_scaleR)
+
+lemma piecewise_C1_differentiable_neg [derivative_intros]:
"f piecewise_C1_differentiable_on S \<Longrightarrow> (\<lambda>x. -(f x)) piecewise_C1_differentiable_on S"
unfolding piecewise_C1_differentiable_on_def
by (auto intro!: continuous_on_minus C1_differentiable_on_minus)
assumes "f piecewise_C1_differentiable_on i"
"g piecewise_C1_differentiable_on i"
shows "(\<lambda>x. f x + g x) piecewise_C1_differentiable_on i"
@@ -3466,10 +3479,26 @@
qed
-lemma piecewise_C1_differentiable_diff:
+lemma piecewise_C1_differentiable_diff [derivative_intros]:
"\<lbrakk>f piecewise_C1_differentiable_on S; g piecewise_C1_differentiable_on S\<rbrakk>
\<Longrightarrow> (\<lambda>x. f x - g x) piecewise_C1_differentiable_on S"
+lemma piecewise_C1_differentiable_cmult_right [derivative_intros]:
+ fixes c::complex
+ shows "f piecewise_C1_differentiable_on S
+ \<Longrightarrow> (\<lambda>x. f x * c) piecewise_C1_differentiable_on S"
+ by (force simp: piecewise_C1_differentiable_on_def continuous_on_mult_right)
+
+lemma piecewise_C1_differentiable_cmult_left [derivative_intros]:
+ fixes c::complex
+ shows "f piecewise_C1_differentiable_on S
+ \<Longrightarrow> (\<lambda>x. c * f x) piecewise_C1_differentiable_on S"
+ using piecewise_C1_differentiable_cmult_right [of f S c] by (simp add: mult.commute)
+
+lemma piecewise_C1_differentiable_on_of_real [derivative_intros]:
+ "of_real piecewise_C1_differentiable_on S"
+ by (simp add: C1_differentiable_imp_piecewise C1_differentiable_on_of_real)
+
end```
```--- a/src/HOL/Analysis/Linear_Algebra.thy Mon May 31 20:27:45 2021 +0000
+++ b/src/HOL/Analysis/Linear_Algebra.thy Thu Jun 03 10:47:20 2021 +0100
@@ -688,6 +688,9 @@
shows "linear f \<Longrightarrow> f differentiable net"
by (metis linear_imp_has_derivative differentiable_def)
+lemma of_real_differentiable [simp,derivative_intros]: "of_real differentiable F"
+ by (simp add: bounded_linear_imp_differentiable bounded_linear_of_real)
+
subsection\<^marker>\<open>tag unimportant\<close> \<open>We continue\<close>
@@ -1056,7 +1059,7 @@
lemma norm_le_infnorm:
fixes x :: "'a::euclidean_space"
shows "norm x \<le> sqrt DIM('a) * infnorm x"
- unfolding norm_eq_sqrt_inner id_def
+ unfolding norm_eq_sqrt_inner id_def
proof (rule real_le_lsqrt[OF inner_ge_zero])
show "sqrt DIM('a) * infnorm x \<ge> 0"
@@ -1085,7 +1088,7 @@
(is "?lhs \<longleftrightarrow> ?rhs")
proof (cases "x=0")
case True
- then show ?thesis
+ then show ?thesis
by auto
next
case False
@@ -1095,9 +1098,9 @@
norm x * (norm y * (y \<bullet> x) - norm x * norm y * norm y) = 0)"
using False unfolding inner_simps
by (auto simp add: power2_norm_eq_inner[symmetric] power2_eq_square inner_commute field_simps)
- also have "\<dots> \<longleftrightarrow> (2 * norm x * norm y * (norm x * norm y - x \<bullet> y) = 0)"
+ also have "\<dots> \<longleftrightarrow> (2 * norm x * norm y * (norm x * norm y - x \<bullet> y) = 0)"
using False by (simp add: field_simps inner_commute)
- also have "\<dots> \<longleftrightarrow> ?lhs"
+ also have "\<dots> \<longleftrightarrow> ?lhs"
using False by auto
finally show ?thesis by metis
qed
@@ -1125,7 +1128,7 @@
shows "norm (x + y) = norm x + norm y \<longleftrightarrow> norm x *\<^sub>R y = norm y *\<^sub>R x"
proof (cases "x = 0 \<or> y = 0")
case True
- then show ?thesis
+ then show ?thesis
by force
next
case False
@@ -1206,7 +1209,7 @@
by (auto simp: insert_commute)
next
case False
- show ?thesis
+ show ?thesis
proof
assume h: "?lhs"
then obtain u where u: "\<forall> x\<in> {0,x,y}. \<forall>y\<in> {0,x,y}. \<exists>c. x - y = c *\<^sub>R u"
@@ -1250,7 +1253,7 @@
proof
assume "\<bar>x \<bullet> y\<bar> = norm x * norm y"
then show "collinear {0, x, y}"
- unfolding norm_cauchy_schwarz_abs_eq collinear_lemma
+ unfolding norm_cauchy_schwarz_abs_eq collinear_lemma
by (meson eq_vector_fraction_iff nnz)
next
assume "collinear {0, x, y}"```
```--- a/src/HOL/Analysis/Path_Connected.thy Mon May 31 20:27:45 2021 +0000
+++ b/src/HOL/Analysis/Path_Connected.thy Thu Jun 03 10:47:20 2021 +0100
@@ -207,6 +207,9 @@
unfolding pathstart_def reversepath_def pathfinish_def
by auto
+lemma reversepath_o: "reversepath g = g \<circ> (-)1"
+ by (auto simp: reversepath_def)
+
lemma pathstart_join[simp]: "pathstart (g1 +++ g2) = pathstart g1"
unfolding pathstart_def joinpaths_def pathfinish_def
by auto```
```--- a/src/HOL/Complex_Analysis/Complex_Singularities.thy Mon May 31 20:27:45 2021 +0000
+++ b/src/HOL/Complex_Analysis/Complex_Singularities.thy Thu Jun 03 10:47:20 2021 +0100
@@ -18,6 +18,11 @@
shows "is_pole g b"
using is_pole_cong assms by auto
+lemma is_pole_shift_iff:
+ fixes f :: "('a::real_normed_vector \<Rightarrow> 'b::real_normed_vector)"
+ shows "is_pole (f \<circ> (+) d) a = is_pole f (a + d)"
+ by (metis add_diff_cancel_right' filterlim_shift_iff is_pole_def)
+
lemma is_pole_tendsto:
fixes f::"('a::topological_space \<Rightarrow> 'b::real_normed_div_algebra)"
shows "is_pole f x \<Longrightarrow> ((inverse o f) \<longlongrightarrow> 0) (at x)"```
```--- a/src/HOL/Complex_Analysis/Contour_Integration.thy Mon May 31 20:27:45 2021 +0000
+++ b/src/HOL/Complex_Analysis/Contour_Integration.thy Thu Jun 03 10:47:20 2021 +0100
@@ -68,8 +68,8 @@
text\<open>Show that we can forget about the localized derivative.\<close>
lemma has_integral_localized_vector_derivative:
- "((\<lambda>x. f (g x) * vector_derivative g (at x within {a..b})) has_integral i) {a..b} \<longleftrightarrow>
- ((\<lambda>x. f (g x) * vector_derivative g (at x)) has_integral i) {a..b}"
+ "((\<lambda>x. f (g x) * vector_derivative p (at x within {a..b})) has_integral i) {a..b} \<longleftrightarrow>
+ ((\<lambda>x. f (g x) * vector_derivative p (at x)) has_integral i) {a..b}"
proof -
have *: "{a..b} - {a,b} = interior {a..b}"
@@ -78,8 +78,8 @@
qed
lemma integrable_on_localized_vector_derivative:
- "(\<lambda>x. f (g x) * vector_derivative g (at x within {a..b})) integrable_on {a..b} \<longleftrightarrow>
- (\<lambda>x. f (g x) * vector_derivative g (at x)) integrable_on {a..b}"
+ "(\<lambda>x. f (g x) * vector_derivative p (at x within {a..b})) integrable_on {a..b} \<longleftrightarrow>
+ (\<lambda>x. f (g x) * vector_derivative p (at x)) integrable_on {a..b}"
lemma has_contour_integral:```
```--- a/src/HOL/Deriv.thy Mon May 31 20:27:45 2021 +0000
+++ b/src/HOL/Deriv.thy Thu Jun 03 10:47:20 2021 +0100
@@ -708,6 +708,23 @@
(\<lambda>x. f x * g x) differentiable (at x within s)"
unfolding differentiable_def by (blast intro: has_derivative_mult)
+lemma differentiable_cmult_left_iff [simp]:
+ fixes c::"'a::real_normed_field"
+ shows "(\<lambda>t. c * q t) differentiable at t \<longleftrightarrow> c = 0 \<or> (\<lambda>t. q t) differentiable at t" (is "?lhs = ?rhs")
+proof
+ assume L: ?lhs
+ {assume "c \<noteq> 0"
+ then have "q differentiable at t"
+ using differentiable_mult [OF differentiable_const L, of concl: "1/c"] by auto
+ } then show ?rhs
+ by auto
+qed auto
+
+lemma differentiable_cmult_right_iff [simp]:
+ fixes c::"'a::real_normed_field"
+ shows "(\<lambda>t. q t * c) differentiable at t \<longleftrightarrow> c = 0 \<or> (\<lambda>t. q t) differentiable at t" (is "?lhs = ?rhs")
+ by (simp add: mult.commute flip: differentiable_cmult_left_iff)
+
lemma differentiable_inverse [simp, derivative_intros]:
fixes f :: "'a::real_normed_vector \<Rightarrow> 'b::real_normed_field"
shows "f differentiable (at x within s) \<Longrightarrow> f x \<noteq> 0 \<Longrightarrow>```
```--- a/src/HOL/Limits.thy Mon May 31 20:27:45 2021 +0000
+++ b/src/HOL/Limits.thy Thu Jun 03 10:47:20 2021 +0100
@@ -11,6 +11,20 @@
imports Real_Vector_Spaces
begin
+text \<open>Lemmas related to shifting/scaling\<close>
+ fixes a::"'a::group_add" shows "range ((+) a) = UNIV"
+
+lemma range_diff [simp]:
+ fixes a::"'a::group_add" shows "range ((-) a) = UNIV"
+
+lemma range_mult [simp]:
+ fixes a::"real" shows "range ((*) a) = (if a=0 then {0} else UNIV)"
+ by (simp add: surj_def) (meson dvdE dvd_field_iff)
+
+
subsection \<open>Filter going to infinity norm\<close>
definition at_infinity :: "'a::real_normed_vector filter"
@@ -1461,6 +1475,28 @@
for a d :: "real"
by (simp add: filter_eq_iff eventually_filtermap eventually_at_filter filtermap_nhds_shift[symmetric])
+lemma filterlim_shift:
+ fixes d :: "'a::real_normed_vector"
+ assumes "filterlim f F (at a)"
+ shows "filterlim (f \<circ> (+) d) F (at (a - d))"
+ unfolding filterlim_iff
+proof (intro strip)
+ fix P
+ assume "eventually P F"
+ then have "\<forall>\<^sub>F x in filtermap (\<lambda>y. y - d) (at a). P (f (d + x))"
+ using assms by (force simp add: filterlim_iff eventually_filtermap)
+ then show "(\<forall>\<^sub>F x in at (a - d). P ((f \<circ> (+) d) x))"
+ by (force simp add: filtermap_at_shift)
+qed
+
+lemma filterlim_shift_iff:
+ fixes d :: "'a::real_normed_vector"
+ shows "filterlim (f \<circ> (+) d) F (at (a - d)) = filterlim f F (at a)" (is "?lhs = ?rhs")
+proof
+ assume L: ?lhs show ?rhs
+ using filterlim_shift [OF L, of "-d"] by (simp add: filterlim_iff)
+qed (metis filterlim_shift)
+
lemma at_right_to_0: "at_right a = filtermap (\<lambda>x. x + a) (at_right 0)"
for a :: real
using filtermap_at_right_shift[of "-a" 0] by simp``` | 4,356 | 13,726 | {"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-2021-49 | latest | en | 0.655759 |
https://the-algorithms.com/algorithm/sudoku | 1,669,820,890,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00819.warc.gz | 608,644,848 | 27,432 | #### Sudoku
R
G
a
T
J
R
and 5 more contributors
```"""
Given a partially filled 9×9 2D array, the objective is to fill a 9×9
square grid with digits numbered 1 to 9, so that every row, column, and
and each of the nine 3×3 sub-grids contains all of the digits.
This can be solved using Backtracking and is similar to n-queens.
We check to see if a cell is safe or not and recursively call the
function on the next column to see if it returns True. if yes, we
have solved the puzzle. else, we backtrack and place another number
in that cell and repeat this process.
"""
from __future__ import annotations
Matrix = list[list[int]]
# assigning initial values to the grid
initial_grid: Matrix = [
[3, 0, 6, 5, 0, 8, 4, 0, 0],
[5, 2, 0, 0, 0, 0, 0, 0, 0],
[0, 8, 7, 0, 0, 0, 0, 3, 1],
[0, 0, 3, 0, 1, 0, 0, 8, 0],
[9, 0, 0, 8, 6, 3, 0, 0, 5],
[0, 5, 0, 0, 9, 0, 6, 0, 0],
[1, 3, 0, 0, 0, 0, 2, 5, 0],
[0, 0, 0, 0, 0, 0, 0, 7, 4],
[0, 0, 5, 2, 0, 6, 3, 0, 0],
]
# a grid with no solution
no_solution: Matrix = [
[5, 0, 6, 5, 0, 8, 4, 0, 3],
[5, 2, 0, 0, 0, 0, 0, 0, 2],
[1, 8, 7, 0, 0, 0, 0, 3, 1],
[0, 0, 3, 0, 1, 0, 0, 8, 0],
[9, 0, 0, 8, 6, 3, 0, 0, 5],
[0, 5, 0, 0, 9, 0, 6, 0, 0],
[1, 3, 0, 0, 0, 0, 2, 5, 0],
[0, 0, 0, 0, 0, 0, 0, 7, 4],
[0, 0, 5, 2, 0, 6, 3, 0, 0],
]
def is_safe(grid: Matrix, row: int, column: int, n: int) -> bool:
"""
This function checks the grid to see if each row,
column, and the 3x3 subgrids contain the digit 'n'.
It returns False if it is not 'safe' (a duplicate digit
is found) else returns True if it is 'safe'
"""
for i in range(9):
if grid[row][i] == n or grid[i][column] == n:
return False
for i in range(3):
for j in range(3):
if grid[(row - row % 3) + i][(column - column % 3) + j] == n:
return False
return True
def find_empty_location(grid: Matrix) -> tuple[int, int] | None:
"""
This function finds an empty location so that we can assign a number
for that particular row and column.
"""
for i in range(9):
for j in range(9):
if grid[i][j] == 0:
return i, j
return None
def sudoku(grid: Matrix) -> Matrix | None:
"""
Takes a partially filled-in grid and attempts to assign values to
all unassigned locations in such a way to meet the requirements
for Sudoku solution (non-duplication across rows, columns, and boxes)
>>> sudoku(initial_grid) # doctest: +NORMALIZE_WHITESPACE
[[3, 1, 6, 5, 7, 8, 4, 9, 2],
[5, 2, 9, 1, 3, 4, 7, 6, 8],
[4, 8, 7, 6, 2, 9, 5, 3, 1],
[2, 6, 3, 4, 1, 5, 9, 8, 7],
[9, 7, 4, 8, 6, 3, 1, 2, 5],
[8, 5, 1, 7, 9, 2, 6, 4, 3],
[1, 3, 8, 9, 4, 7, 2, 5, 6],
[6, 9, 2, 3, 5, 1, 8, 7, 4],
[7, 4, 5, 2, 8, 6, 3, 1, 9]]
>>> sudoku(no_solution) is None
True
"""
if location := find_empty_location(grid):
row, column = location
else:
# If the location is ``None``, then the grid is solved.
return grid
for digit in range(1, 10):
if is_safe(grid, row, column, digit):
grid[row][column] = digit
if sudoku(grid) is not None:
return grid
grid[row][column] = 0
return None
def print_solution(grid: Matrix) -> None:
"""
A function to print the solution in the form
of a 9x9 grid
"""
for row in grid:
for cell in row:
print(cell, end=" ")
print()
if __name__ == "__main__":
# make a copy of grid so that you can compare with the unmodified grid
for example_grid in (initial_grid, no_solution):
print("\nExample grid:\n" + "=" * 20)
print_solution(example_grid)
print("\nExample grid solution:")
solution = sudoku(example_grid)
if solution is not None:
print_solution(solution)
else:
print("Cannot find a solution.")
``` | 1,457 | 3,482 | {"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-2022-49 | latest | en | 0.672874 |
https://www.slideserve.com/graceland/cfd-techniques-quiz-powerpoint-ppt-presentation | 1,575,746,340,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540501887.27/warc/CC-MAIN-20191207183439-20191207211439-00424.warc.gz | 872,572,050 | 13,489 | CFD Techniques Quiz
1 / 30
# CFD Techniques Quiz - PowerPoint PPT Presentation
## CFD Techniques Quiz
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. CFD Techniques Quiz
2. What is the weight of an adult hippo skin? • 0.5 tons • 1 ton • 2 tons 10
3. What type of equation is this if the flow is supersonic? • elliptic • parabolic • hyperbolic 10
4. What is X in the definition below?The X is the difference between the PDE and the finite difference representation • Rounding error • Goodness of fit • Truncation error • Convergence measure • Marching error 10
5. Consider the finite difference approximation to the first derivative below. Which statement is false? • If the grid spacing is halved the error goes down by 1/4 • This is a central difference approximation • This form might be useful close to a wall • This form might be useful near the edge of the simulation 10
6. What is X in the description below?A X is one for which errors from any source (round-off, truncation, mistakes) are not permitted to grow in the sequence of numerical procedures as the calculation proceeds from one marching step to the next. • consistent numerical scheme • stable numerical scheme • convergent numerical scheme • second-order numerical scheme 10
7. What is X in the description below?A solution to a finite difference representation is said to be X if it approaches the true solution to the PDE having the same initial and boundary conditions as the mesh is refined. • consistent • stable • convergent • Ill-posed 10
8. What is X in the description below?By X, we mean those errors that arise as a result of the rounding to a finite number of digits in the arithmetic operations. • truncation • round-off errors • systematic errors • happy accidents 10
9. What is X in the definition below?A finite difference representation is said to be X if we can show that the difference between the PDE and its finite difference representation vanishes as the mesh is refined. • consistent • convergent • stable • first order • most excellent 10
10. Which statement is false? • Parabolic equations have degenerate characteristics • The initial value problem is ill-posed for elliptic problems • Discontinuities can appear in elliptic problems • The unsteady heat conduction equation is an example of a parabolic problem 10
11. What is X in the definition below? X error is the error in the solution to the PDE subject to the given initial values and boundary conditions caused by replacing the continuous problem by a discrete one and is defined as the difference between the exact solution of the PDE (round-off free) and the exact solution to the finite difference equations (round-off free). • Discretisation • Truncation • Round-off • Integration 10
12. Which statement is false? • The Courant number has no units • If you double the sound speed, the Courant number doubles • If you halve the space grid the Courant number doubles • If you halve the time grid, the Courant number doubles 10
13. Which term would NOT be required in a von Neumann stability analysis of the equation below? Hint: 1. 2. 3. 4. 5. 10
14. The amplitude in a von Neumann stability test is found to have the following modulus. What can we say about the finite difference representation? • It’s unconditionally stable • It’s conditionally stable • It’s conditionally unstable • It’s unconditionally unstable 10
15. For what value of B in the equation below does the equation become parabolic? • 2 • 4 • 8 • 16 • 32 10
16. What is the equilibrium value of A in the square grid used to solve Laplace’s equation in 2-d? • 2.3 • 2.4 • 2.5 • 2.6 • 2.7 10
17. What is the value of A in the grid below? • 1.5 • 0.5 • 1 • 0 • -1 10
18. What is the value of B in the grid below? • 1.5 • 0.5 • 1 • 0 • -1 10
19. What type are the boundary conditions in the previous question? • Dirichlet • Neumann • Mixed 10
20. Which of these is the correct expression for the iteration at point 3 in an explicit (Jacobi) method? • 1. • 2. • 3. • 4. 10
21. Which of these is the correct expression for the iteration at point 2 in an implicit (Gauss-Seidel) method? • 1. • 2. • 3. • 4. 10
22. Which of these is a characteristic of the (wave) equation below? 1. 2. 3. 4. 5. 10
23. A solution of this equation is Which way is the wave travelling and what is its speed? • Left to right (toward positive x) speed 2 • Right to left (toward negative x) speed 2 • Left to right (toward positive x) speed 0.5 • Right to left (toward positive x) speed 0.5 10
24. Which of these is not a solution of the equation? 1. 2. 3. 4. 5. 10
25. Which of the following statements is false? • Dispersion is caused by odd derivative terms in a PDE truncation • Diffusion is caused by odd derivative terms in a PDE truncation 10
26. A CFD simulation of a shock develops ‘waviness’ as pictured below? This is evidence for what in the solution? • Dispersion • Diffusion 10
27. Which of these is NOT an implicit scheme? 1. 2. 3. 4. 10
28. Which of these is NOT a point at the entry to the duct (x=0) joined to the point P by 45 degree characteristics (i.e. lines of slope -1 and +1) ? • (0,0.6) • (0,-0.6) • (0,0.4) • (0,-0.4) 10 | 1,416 | 5,271 | {"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} | 3.421875 | 3 | CC-MAIN-2019-51 | latest | en | 0.851046 |
http://nrich.maths.org/public/leg.php?code=-290&cl=3&cldcmpid=5335 | 1,480,839,353,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541220.56/warc/CC-MAIN-20161202170901-00468-ip-10-31-129-80.ec2.internal.warc.gz | 216,913,087 | 7,031 | # Search by Topic
#### Resources tagged with History of mathematics similar to How Long Is the Cantor Set?:
Filter by: Content type:
Stage:
Challenge level:
### There are 25 results
Broad Topics > History and Philosophy of Mathematics > History of mathematics
### Negative Numbers
##### Stage: 3
A brief history of negative numbers throughout the ages
### Ishango Bone
##### Stage: 2, 3, 4 and 5 Short Challenge Level:
Can you decode the mysterious markings on this ancient bone tool?
### The Development of Algebra - 2
##### Stage: 3, 4 and 5
This is the second article in a two part series on the history of Algebra from about 2000 BCE to about 1000 CE.
### The History of Negative Numbers
##### Stage: 3, 4 and 5
This article -useful for teachers and learners - gives a short account of the history of negative numbers.
### History of Trigonometry - Part 2
##### Stage: 3, 4 and 5
The second of three articles on the History of Trigonometry.
### The History of Trigonometry- Part 1
##### Stage: 3, 4 and 5
The first of three articles on the History of Trigonometry. This takes us from the Egyptians to early work on trigonometry in China.
### Babylon Numbers
##### Stage: 3, 4 and 5 Challenge Level:
Can you make a hypothesis to explain these ancient numbers?
### Circles, Circles Everywhere
##### Stage: 2 and 3
This article for pupils gives some examples of how circles have featured in people's lives for centuries.
### Keeping it Safe and Quiet
##### Stage: 2, 3, 4 and 5
Simon Singh describes PKC, its origins, and why the science of code making and breaking is such a secret occupation.
### Proof: A Brief Historical Survey
##### Stage: 4 and 5
If you think that mathematical proof is really clearcut and universal then you should read this article.
### The Secret World of Codes and Code Breaking
##### Stage: 2, 3 and 4
When you think of spies and secret agents, you probably wouldn’t think of mathematics. Some of the most famous code breakers in history have been mathematicians.
### History of Trigonometry - Part 3
##### Stage: 3, 4 and 5
The third of three articles on the History of Trigonometry.
### The Development of Algebra - 1
##### Stage: 3, 4 and 5
This is the first of a two part series of articles on the history of Algebra from about 2000 BCE to about 1000 CE.
### Galley Division
##### Stage: 4 Challenge Level:
Can you explain how Galley Division works?
### Ancient Astronomical Terms
##### Stage: 3, 4 and 5
Some explanations of basic terms and some phenomena discovered by ancient astronomers
### From A Random World to a Rational Universe
##### Stage: 2, 3 and 4
In the time before the mathematical idea of randomness was discovered, people thought that everything that happened was part of the will of supernatural beings. So have things changed?
### Magic Squares II
##### Stage: 4 and 5
An article which gives an account of some properties of magic squares.
### The Four Colour Theorem
##### Stage: 3 and 4
The Four Colour Conjecture was first stated just over 150 years ago, and finally proved conclusively in 1976. It is an outstanding example of how old ideas can be combined with new discoveries. prove. . . .
### Maths in the Victorian Classroom
##### Stage: 2 and 3
What was it like to learn maths at school in the Victorian period? We visited the British Schools Museum in Hitchin to find out.
### A Brief History of Time Measurement
##### Stage: 2, 3, 4 and 5
Noticing the regular movement of the Sun and the stars has led to a desire to measure time. This article for teachers and learners looks at the history of man's need to measure things.
### History of Morse
##### Stage: 2, 3, 4 and 5
This short article gives an outline of the origins of Morse code and its inventor and how the frequency of letters is reflected in the code they were given.
### Shaping the Universe I - Planet Earth
##### Stage: 3 and 4
This article explores ths history of theories about the shape of our planet. It is the first in a series of articles looking at the significance of geometric shapes in the history of astronomy.
### Shaping the Universe II - the Solar System
##### Stage: 3 and 4
The second in a series of articles on visualising and modelling shapes in the history of astronomy. | 1,023 | 4,290 | {"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} | 3.265625 | 3 | CC-MAIN-2016-50 | longest | en | 0.864317 |
http://math.stackexchange.com/questions/367551/solving-linear-matrix-equation-problem | 1,469,772,440,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.19/warc/CC-MAIN-20160723071029-00146-ip-10-185-27-174.ec2.internal.warc.gz | 166,758,397 | 17,080 | # solving linear matrix equation problem
Given two matrices $X$ and $Y$, with $Y$ invertible. Suppose that
$$X=YZY^{-1}.$$
so
$$Z=Y^{-1}XY.$$
In what order should I do the corresponding matrix multiplications to compute $Z$ ?
Thanks.
-
You seem to already have solved for $Z$ by multiplying both sides by $Y^{-1}$ from the left and by $Y$ from the right. – Mårten W Apr 20 '13 at 18:49
Matrix multiplication is associative, that is given matrices $A,B$ and $C$,
$$A(BC)=(AB)C.$$ | 144 | 486 | {"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 | 3 | CC-MAIN-2016-30 | latest | en | 0.92027 |
http://oeis.org/A099305 | 1,481,046,819,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541950.13/warc/CC-MAIN-20161202170901-00257-ip-10-31-129-80.ec2.internal.warc.gz | 207,896,874 | 4,189 | This site is supported by donations to The OEIS Foundation.
Annual Appeal: Please make a donation (tax deductible in USA) to keep the OEIS running. Over 5000 articles have referenced us, often saying "we discovered this result with the help of the OEIS".
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A099305 Number of solutions of the equation (n+k)' = n' + k', with 1 <= k <= 2n, where n' denotes the arithmetic derivative of n. 4
1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 2, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 2, 2, 1, 3, 1, 3, 1, 2, 3, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 1, 2, 1, 3, 2, 3, 2, 2, 2, 2, 3, 3, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 2, 2, 1, 3, 3, 3, 1, 2, 2, 3, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Observe that when n and c*n have the same parity, a(c*n) >= a(n) for all integers c. For even n, there are always at least two solutions, k=n/2 and k=2n. For odd n, k=2n is always a solution. a(A258138(n)) = n and a(m) != n for m < A258138(n). - Reinhard Zumkeller, May 21 2015 REFERENCES See A003415 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 MATHEMATICA dn[0]=0; dn[1]=0; dn[n_]:=Module[{f=Transpose[FactorInteger[n]]}, If[PrimeQ[n], 1, Plus@@(n*f[[2]]/f[[1]])]]; Table[lst={}; k=0; While[k<2n, k++; While[k<=2n && dn[n]+dn[k] != dn[n+k], k++ ]; If[dn[n]+dn[k]==dn[n+k], AppendTo[lst, k]]]; Length[lst], {n, 100}] PROG (Haskell) a099305 n = a099305_list !! (n-1) a099305_list = f 1 \$ h 1 empty where f x ad = y : f (x + 1) (h (3 * x + 1) ad) where y = length [() | k <- [1 .. 2 * x], let x' = ad ! x, ad ! (x + k) == x' + ad ! k] h z = insert z (a003415 z) . insert (z+1) (a003415 (z+1)) . insert (z+2) (a003415 (z+2)) -- Reinhard Zumkeller, May 21 2015 CROSSREFS Cf. A003415 (arithmetic derivative of n), A099304 (least k > 0 such that (n+k)' = n' + k'). Cf. A258138. Sequence in context: A244145 A086435 A266226 * A033109 A235644 A175096 Adjacent sequences: A099302 A099303 A099304 * A099306 A099307 A099308 KEYWORD nonn AUTHOR T. D. Noe, Oct 12 2004 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc. | 1,041 | 2,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2016-50 | longest | en | 0.576106 |
https://www.turito.com/ask-a-doubt/Mathematics-choose-the-correct-multiplication-sentence-this-array-show-4-9-36-4-10-4-10-40-4-9-qc14dff | 1,713,359,569,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817153.39/warc/CC-MAIN-20240417110701-20240417140701-00097.warc.gz | 924,652,587 | 41,378 | Mathematics
Easy
Question
# Choose the correct multiplication sentence this array show.
Hint:
## The correct answer is: 4 × 9 = 36
### In the question there is an array using which we have to find the multiplicative expression of that array.The number of rows = 4The number of columns = 9The multiplicative expression= 4 × 9 = 36.So, the multiplicative expression of the given array will be 4 × 9 = 36.Therefore, the correct option is d, i.e., 4 × 9 = 36.
For simple multiplication we have to remember the multiplication table of one-digit numbers and for two-digit number we can find it using the multiplication table of one-digit numbers. Here, we have to find the product of one digit number. | 170 | 701 | {"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} | 4.03125 | 4 | CC-MAIN-2024-18 | latest | en | 0.804196 |
https://docsbay.net/description-of-activity-steps-with-transitions-for-teacher | 1,638,295,630,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359065.88/warc/CC-MAIN-20211130171559-20211130201559-00248.warc.gz | 292,921,659 | 6,109 | Description of Activity (Steps with Transitions) for Teacher
Description of activity (steps with transitions) for Teacher.
Set up:
1. Review the attached practice questions (entitled RelayEquationsEnvelope). Whether you choose to use these questions, or select others to meet the needs of your students, make enough copies of the questionsso that you have one set for every four students. (It is helpful to have the sets printed and in envelopes, organized for student teams. In the heat of the competition you want to make sure you have/keep everything in order.)
2. Before students arrive, arrange the classroom desks/chairs in groups to be suitable for a relay game. Student will be passing a paper between four group members – they could be inseated in rows and pass the papers back or in a circle and pass the paper clockwise, according to your preference.
3. Students will be competing against each other, so label/identify teams in a way that will be easily distinguishable throughout the game. Depending on the behavioral and academic needs or your students, you may choose to group heterogeneously (with mixed ability levels) or allow students to sit according to their preference.
4. Display the relay roles and instructions on the board, or individually at student seats. They are as follows:
Student #1/Molar Mass Mogul: At the word “go”, Student #1 from each group will quickly walk to the front of the room to obtain their sample problem. In a blue pencil, this student writes his/her name and computes the molar mass of each substance. (Students could be instructed to make a menu - see resource 29179 for more information. The menu is seen in the sample provided.)Once this task is completed, Student #1 will pass the paper and problem to Student #2. Student #1 must then remain in their seats until it is their turn again.
Student #2/ Dimensional Analysis Demon: Using a green pencil,this student writes her/name and sets up the dimensional analysis process (per your expectations of how to properly set up a stoichiometry problem) to convert necessary items in the problem (not solve). They may also correct work written by the previous student. Once satisfied with their work, student #2 will pass the problem to student #3.
Student #3:/Calculator King/Queen: With a red pencil, this student writes his/her name and does the math work to solve the dimensional analysis problem set up by Student #2. They may also correct work written by the previous students. Once satisfied with their work, student #3 will pass the problem to student #4.
Student #4/“Units Guru”: Using an orange pencil, this student writes his/her name and crosses out all cancelling units within the dimensional analysis and checks the entire series of work for proper units as well as satisfying significant digits. They may also correct work written by the previous students. Once satisfied with their work, student #4 will take their group’s answer to the instructor to be assessed.
1. If the answer is incorrect, student #4 may take the problem and their work to their team to discuss possible problems and corrections to be made. Once the team has come to a conclusion and a new answer, student #4 may resubmit the question again.
1. If the answer is correct, student #4 will be given another question to be completed.
Once the student #4 receives the new problem for their team, he/she becomes the new student #1. Every student in that group then assumes the role that had come behind them. (Here is where you could have the colored pencils rotate instead of the students physically move.) Note: The students are responsible to perform transitions between each problem. Students who do not transition are disqualified or otherwise penalized.
The group that completes the most problems correctly in the given class time (or quickest to complete all problems provided) will be deemed the winner. Provide acknowledgement for the winning team as fits for your class.
Other rules to consider and post:
Can students talk to each other? (If so, to whom and how loudly?) What should they do if their colored pencil breaks? What does the winner get?
During Class
1. As students enter the class, ask them to sit in seats quietly and read the instructions.Once class has begun, read through the instructions with the class.
2.Project a sample problem on the board – one is provided, but you may prefer one of your own. It is very helpful for students to see you work through the sample, one color at a time, so they can see the progression as the problem is completed by the team. (The final product is very colorful, but the colors build as the problem sheet goes through the roles in the team.)As you’re working through the problem, be very clear about what role you’re showing and the color you’re using. You may also want to mention that the sample problem is simpler than the problems they will be working on during the game.
3.Give students anothersimple sample problem and allow them to practice the game’s procedures. Address any misconceptions regarding rules at this time.Clarity of the rules is important because students will want to win. (Again mention how it is an “easier” problem just meant to practice the parameters of the relay.)
4.When you are satisfied that students understand your expectations for the relay game, play until a preset time or until all questions have been answered correctly.
After Class
Assess students’ level of understanding using the questions that they have submitted to you. You can easily see the work of each student, because they have written their name using the color of their work. | 1,144 | 5,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-49 | latest | en | 0.948171 |
http://gmatclub.com/forum/ds-gmatprep-united-nations-31282.html?kudos=1 | 1,484,951,291,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280888.62/warc/CC-MAIN-20170116095120-00366-ip-10-171-10-70.ec2.internal.warc.gz | 118,305,670 | 44,257 | DS: GMATPREP United Nations : DS Archive
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# DS: GMATPREP United Nations
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29 Jun 2006, 10:29
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I cant figure out how to solve this pb, especially in 1,5mn...
In a certain year the UN's total expenditures were 1.6 billion. Of this amount, 67,8 % was paid by the 6 highest-contributing countries, and the balance was paid by the remaining 153 countries. Was country X among the 6 highest-contributing countries?
1. 56% of the total expenditures was paid by the 4 highest contributing countries, each of which paid more than country X.
2. Country X paid 4.8 % of the total expenditures
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Re: DS: GMATPREP United Nations [#permalink]
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29 Jun 2006, 10:44
karlfurt wrote:
I cant figure out how to solve this pb, especially in 1,5mn...
In a certain year the UN's total expenditures were 1.6 billion. Of this amount, 67,8 % was paid by the 6 highest-contributing countries, and the balance was paid by the remaining 153 countries. Was country X among the 6 highest-contributing countries?
1. 56% of the total expenditures was paid by the 4 highest contributing countries, each of which paid more than country X.
2. Country X paid 4.8 % of the total expenditures
appears (E) at first shot.
I. clearly insufficient. Doesnt tell us anything abt X. X can be the 5th or 6th one, or can be among the remaining 153
II. insufficient again. doesnt tell us whether X can be in top 6 or not. Imagine, each country in top 6 contributing 67.8/6, which is > 4.8 or imagine one country contributing 60% and then X can be among top 6. Not sufficient.
Combining doesnt make it any clear.
We have 11.8% to be divided among 2 countries. Its possible that 2 countries contribute 11.8/2 leaving X in thelot of 153, or one country contributing 7% and remaining by X (making it into top 6)
(E) is hte best choice.
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29 Jun 2006, 11:47
I am also getting E.
Lets say expenditure is 100
Six countries paf 67.8
St1:56 is paid by top 4 countries. Nothing can be said about X, it may be number 5,6,7 etc.... or last : INSUFF
St2: 4.8 is paid by X. X could be anywhere (But not #1). : INSUFF
Combined: X could be anywhere starting from 5 to last....
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30 Jun 2006, 01:09
E
1) 56% is contributed by top4 countries. We donot know anything about country X contribution hence Not suff
2)Country X 's contribution is 4 %. This is Not suff.
Both
Remaining two countries contribute 13odd percent. X has a contribuion of 4%. It can be one of the member and it may not be. hence not suff
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30 Jun 2006, 05:31
OA E. Thanks for your nice explanations.
30 Jun 2006, 05:31
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# DS: GMATPREP United Nations
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,261 | 4,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-04 | latest | en | 0.914006 |
http://www.lmfdb.org/NumberField/8.0.12895281.1 | 1,571,514,206,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697760.44/warc/CC-MAIN-20191019191828-20191019215328-00491.warc.gz | 290,828,179 | 7,690 | # Properties
Label 8.0.12895281.1 Degree $8$ Signature $[0, 4]$ Discriminant $3^{6}\cdot 7^{2}\cdot 19^{2}$ Root discriminant $7.74$ Ramified primes $3, 7, 19$ Class number $1$ Class group Trivial Galois Group $C_2^2 \wr C_2$ (as 8T18)
# Related objects
Show commands for: Magma / SageMath / Pari/GP
magma: R<x> := PolynomialRing(Rationals()); K<a> := NumberField(R![1, 3, 5, 0, -3, -6, 2, 0, 1]);
sage: x = polygen(QQ); K.<a> = NumberField(x^8 + 2*x^6 - 6*x^5 - 3*x^4 + 5*x^2 + 3*x + 1)
gp: K = bnfinit(x^8 + 2*x^6 - 6*x^5 - 3*x^4 + 5*x^2 + 3*x + 1, 1)
## Normalizeddefining polynomial
$$x^{8}$$ $$\mathstrut +\mathstrut 2 x^{6}$$ $$\mathstrut -\mathstrut 6 x^{5}$$ $$\mathstrut -\mathstrut 3 x^{4}$$ $$\mathstrut +\mathstrut 5 x^{2}$$ $$\mathstrut +\mathstrut 3 x$$ $$\mathstrut +\mathstrut 1$$
magma: DefiningPolynomial(K);
sage: K.defining_polynomial()
gp: K.pol
## Invariants
Degree: $8$ magma: Degree(K); sage: K.degree() gp: poldegree(K.pol) Signature: $[0, 4]$ magma: Signature(K); sage: K.signature() gp: K.sign Discriminant: $$12895281=3^{6}\cdot 7^{2}\cdot 19^{2}$$ magma: Discriminant(K); sage: K.disc() gp: K.disc Root discriminant: $7.74$ magma: Abs(Discriminant(K))^(1/Degree(K)); sage: (K.disc().abs())^(1./K.degree()) gp: abs(K.disc)^(1/poldegree(K.pol)) Ramified primes: $3, 7, 19$ magma: PrimeDivisors(Discriminant(K)); sage: K.disc().support() gp: factor(abs(K.disc))[,1]~ This field is not Galois over $\Q$. This is not a CM field.
## Integral basis (with respect to field generator $$a$$)
$1$, $a$, $a^{2}$, $a^{3}$, $a^{4}$, $a^{5}$, $\frac{1}{3} a^{6} + \frac{1}{3} a^{5} - \frac{1}{3} a^{4} + \frac{1}{3} a^{3} - \frac{1}{3} a^{2} + \frac{1}{3} a + \frac{1}{3}$, $\frac{1}{3} a^{7} + \frac{1}{3} a^{5} - \frac{1}{3} a^{4} + \frac{1}{3} a^{3} - \frac{1}{3} a^{2} - \frac{1}{3}$
magma: IntegralBasis(K);
sage: K.integral_basis()
gp: K.zk
## Class group and class number
Trivial group, which has order $1$
magma: ClassGroup(K);
sage: K.class_group().invariants()
gp: K.clgp
## Unit group
magma: UK, f := UnitGroup(K);
sage: UK = K.unit_group()
Rank: $3$ magma: UnitRank(K); sage: UK.rank() gp: K.fu Torsion generator: $$-\frac{4}{3} a^{7} + \frac{1}{3} a^{6} - 3 a^{5} + 8 a^{4} + a^{3} - a^{2} - \frac{11}{3} a - \frac{1}{3}$$ (order $6$) magma: K!f(TU.1) where TU,f is TorsionUnitGroup(K); sage: UK.torsion_generator() gp: K.tu[2] Fundamental units: $$\frac{1}{3} a^{6} - \frac{2}{3} a^{5} + \frac{2}{3} a^{4} - \frac{11}{3} a^{3} + \frac{8}{3} a^{2} + \frac{4}{3} a + \frac{4}{3}$$, $$\frac{1}{3} a^{7} - \frac{4}{3} a^{6} + a^{5} - 5 a^{4} + 7 a^{3} + a^{2} + \frac{2}{3} a - \frac{8}{3}$$, $$\frac{1}{3} a^{7} - a^{6} + \frac{4}{3} a^{5} - \frac{13}{3} a^{4} + \frac{19}{3} a^{3} - \frac{4}{3} a^{2} + a - \frac{10}{3}$$ magma: [K!f(g): g in Generators(UK)]; sage: UK.fundamental_units() gp: K.fu Regulator: $$6.06288932638$$ magma: Regulator(K); sage: K.regulator() gp: K.reg
## Galois group
$C_2^2\wr C_2$ (as 8T18):
magma: GaloisGroup(K);
sage: K.galois_group(type='pari')
gp: polgalois(K.pol)
A solvable group of order 32 The 14 conjugacy class representatives for $C_2^2 \wr C_2$ Character table for $C_2^2 \wr C_2$
## Intermediate fields
Fields in the database are given up to isomorphism. Isomorphic intermediate fields are shown with their multiplicities.
## Sibling fields
Galois closure: data not computed Degree 8 siblings: data not computed Degree 16 siblings: data not computed
## Frobenius cycle types
$p$ Cycle type 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 ${\href{/LocalNumberField/2.4.0.1}{4} }^{2}$ R ${\href{/LocalNumberField/5.4.0.1}{4} }^{2}$ R ${\href{/LocalNumberField/11.4.0.1}{4} }^{2}$ ${\href{/LocalNumberField/13.2.0.1}{2} }^{4}$ ${\href{/LocalNumberField/17.4.0.1}{4} }^{2}$ R ${\href{/LocalNumberField/23.4.0.1}{4} }^{2}$ ${\href{/LocalNumberField/29.4.0.1}{4} }^{2}$ ${\href{/LocalNumberField/31.2.0.1}{2} }^{4}$ ${\href{/LocalNumberField/37.2.0.1}{2} }^{4}$ ${\href{/LocalNumberField/41.2.0.1}{2} }^{4}$ ${\href{/LocalNumberField/43.2.0.1}{2} }^{4}$ ${\href{/LocalNumberField/47.4.0.1}{4} }^{2}$ ${\href{/LocalNumberField/53.4.0.1}{4} }^{2}$ ${\href{/LocalNumberField/59.2.0.1}{2} }^{4}$
In the table, R denotes a ramified prime. Cycle lengths which are repeated in a cycle type are indicated by exponents.
magma: p := 7; // to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
magma: idealfactors := Factorization(p*Integers(K)); // get the data
magma: [<primefactor[2], Valuation(Norm(primefactor[1]), p)> : primefactor in idealfactors];
sage: p = 7; # to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
sage: [(e, pr.norm().valuation(p)) for pr,e in K.factor(p)]
gp: p = 7; \\ to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
gp: idealfactors = idealprimedec(K, p); \\ get the data
gp: vector(length(idealfactors), j, [idealfactors[j][3], idealfactors[j][4]])
## Local algebras for ramified primes
$p$LabelPolynomial $e$ $f$ $c$ Galois group Slope content
$3$3.4.3.1$x^{4} + 3$$4$$1$$3$$D_{4}$$[\ ]_{4}^{2} 3.4.3.1x^{4} + 3$$4$$1$$3$$D_{4}$$[\ ]_{4}^{2}$
$7$7.2.0.1$x^{2} - x + 3$$1$$2$$0$$C_2$$[\ ]^{2} 7.2.0.1x^{2} - x + 3$$1$$2$$0$$C_2$$[\ ]^{2}$
7.4.2.1$x^{4} + 35 x^{2} + 441$$2$$2$$2$$C_2^2$$[\ ]_{2}^{2} 1919.2.0.1x^{2} - x + 2$$1$$2$$0$$C_2$$[\ ]^{2}$
19.2.1.1$x^{2} - 19$$2$$1$$1$$C_2$$[\ ]_{2} 19.2.0.1x^{2} - x + 2$$1$$2$$0$$C_2$$[\ ]^{2}$
19.2.1.1$x^{2} - 19$$2$$1$$1$$C_2$$[\ ]_{2}$ | 2,412 | 5,503 | {"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": 2, "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} | 3.109375 | 3 | CC-MAIN-2019-43 | latest | en | 0.271138 |
https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH105/April_2013/Question_01_(b) | 1,726,253,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00812.warc.gz | 579,003,177 | 11,186 | # Science:Math Exam Resources/Courses/MATH105/April 2013/Question 01 (b)
MATH105 April 2013
Other MATH105 Exams
### Question 01 (b)
Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty.
Find an equation for the level curve of ${\displaystyle \displaystyle f(x,y)=3xy^{2}+2y-1}$ that goes through the point ${\displaystyle \displaystyle (1,-2)}$.
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result. | 392 | 1,665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 12, "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-2024-38 | latest | en | 0.956626 |
http://mathhelpforum.com/differential-geometry/229237-cohen-sutherland-line-clipping-problem-print.html | 1,529,812,633,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866191.78/warc/CC-MAIN-20180624024705-20180624044705-00115.warc.gz | 202,165,382 | 3,657 | # Cohen-Sutherland Line Clipping Problem
• May 31st 2014, 05:49 PM
lamentofking
Cohen-Sutherland Line Clipping Problem
Hello,
I'm trying to learn how to do the Cohen-Sutherland Line Clipping algorithm by hand. I have an example 2D space below:
Attachment 31053
Now if I understand correctly, the goal is to clip everything outside of the rectangle correct? The problem I'm having is the math approach. For example the segment AD. I know the code for A and D is 0000 and 1001 respectively. Now the logical and of the two endpoints needs to not be 0 to be rejected which means if the logical and is 1 then then segment is accepted? So the logical and of AD is 0 so trivial accept correct? I have a feeling it should be rejected because the point D is outside of the viewport (rectangle).
• Jun 1st 2014, 05:22 PM
johng
Re: Cohen-Sutherland Line Clipping Problem
Hi,
At the very least, you need psuedo code for the algorithm. Rather than provide such, I just excerpted a Java implementation from one of my graphics programs. I hope you can read Java or at least some "C like" language. There are some subtle points in the algorithm. For example, in computing a clipped point, a division by 0 is never performed. First is a diagram of a window and a line to be clipped. Following is the Java code.
http://i60.tinypic.com/mlm9w6.png
Code:
``` /* Following is an implementation of the Cohen Sutherland clip algorithm against a window in the "world" with lower left coordinates (wx0,wy0) and upper right corner (wx1,xy1). If the line is completely outside the window, just return; Otherwise, draw the clipped line. */ public void drawLine(Graphics2D g2,double x1,double y1,double x2,double y2) { int c1,c2,c; double x,y; // code for point P1=(x1,y1) c1=x1<wx0 ? 1 : x1>wx1 ? 2 : 0; if (y1<wy0) c1 |=4; else if (y1>wy1) c1 |=8; // code for P2=(x2,y2) c2=x2<wx0 ? 1 : x2>wx1 ? 2 : 0; if (y2<wy0) c2 |=4; else if (y2>wy1) c2 |=8; while ((c1|c2)!=0) { if ((c1&c2) != 0) return; // both points completely outside window c=(c1!=0) ? c1 : c2; if ((c&1)!=0 || (c&2)!=0) { // point to left or right of window x=((c&1)!=0) ? wx0 : wx1; y=y1+(y2-y1)/(x2-x1)*(x-x1); } else { // point above or below wintdow y=((c&4)!=0) ? wy0 : wy1; x=x1+(x2-x1)/(y2-y1)*(y-y1); } if (c==c1) { x1=x; y1=y; c1=(x1<wx0) ? 1 : (x1>wx1) ? 2 : 0; if (y1<wy0) c1 |=4; else if (y1>wy1) c1 |=8; } else { x2=x; y2=y; c2=(x2<wx0) ? 1 : (x2>wx1) ? 2 : 0; if (y2<wy0) c2 |=4; else if (y2>wy1) c2 |=8; } } // if we get to here, a line needs to be drawn // finally draw the line in Java: code omitted }``` | 922 | 2,795 | {"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} | 3.359375 | 3 | CC-MAIN-2018-26 | latest | en | 0.830244 |
https://biology.stackexchange.com/questions/14848/gene-distance-and-recombination-frequency-mechanism?noredirect=1 | 1,718,382,815,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861567.95/warc/CC-MAIN-20240614141929-20240614171929-00637.warc.gz | 115,406,658 | 37,579 | # Gene distance and Recombination Frequency: Mechanism
Why is the recombination frequency higher if the genes are farther apart?
• Alan, thanks for your excellent analogy. We're talking about the probability of a random event happening at some place on the string.To me the key in the analogy is the probability of this random event happening. Whether that's "cutting", or "tied-up", or even "melted together", the random event, in my mind, is affected by how many marks there are on the string, and how far apart they are relative to each other. This probability calculation is the recombination frequency. Commented Oct 27, 2017 at 3:18
The genetic phenomenon referred to as recombination reflects the process of crossing over which occurs during meiosis. Crossing over creates an exchange of genetic information between homologous chromosomes. To a first approximation cossing over events take place at random positions along the aligned chromosomes. Consequently the further two loci are apart, the more likely that there will be a crossing over event between them. Thus the recombination frequency can be used to measure the distance between two genetic loci (or genes).
Supplement added in response to comment from OP
An analogy
Imagine a piece of string. It has 6 marks on it (1-6 on the diagram) which divide it into 7 intervals (A-G). We are going to cut the string at one of the marks. We throw a die to decide which position to cut.
What is the probability that our cut will separate A and G? It's 100% since all single cuts will do this.
What is the probability that our cut will separate A and D? It's 50% since single cuts at 1,2 or 3 will do this.
What is the probability that our cut will separate A and B? It's 16.7% (1/6) since only a cut at position 1 will do this and the probability of throwing a 1 on our die is 1/6.
Think of the intervals (A-G) as genes and the cut sites (1-6) as possible crossing over events. If two homologous chromosomes are going to undergo a single crossover somewhere at random, then the closer together are the genes (intervals), the less likely is it that the crossover (cut) will take place between them. This assumes of course that the site of crossing over is chosen at random.
In the model presented here, if the order of the letters was scrambled and didn't know what it was but we were told the frequencies of occurrence of the separation of the intervals with random cutting we could deduce the order of the letters.
• To me the statement "Consequently the further two loci are apart, the more likely that there will be a crossing over event between them" is still not obvious. It is the core of my question. Please elaborate further. Commented Feb 5, 2014 at 16:06
• Thank you very much for the effort. However, I am not yet clear, sorry for this. Crossing-over involves genes physically being 'tied up' temporarily. I fail to see how is cutting analogous to it? Why should chances of such 'tying up' increased if the genes are farther apart? Commented Feb 6, 2014 at 1:25
• I don't know what you mean by being 'tied up'. Crossing over involves a physical interaction between two aligned chromosomes at a randomly chosen point. The further that two genetic loci are apart then the more likely it is that the randomly chosen point of crossing over will lie between those loci. I really can't think of any other way to explain this, so I hope someone else comes along and gives you the explanation you are looking for. Commented Feb 6, 2014 at 10:36
• @DurgaDatta, i absolutely got it why you're not understanding. You simply don't understand how the crossing over happens on a visual level. youtube.com/watch?v=jd3H0pgVCr8. Minute 1:06 . If you pay atention to this frame, you'll see that, had the cut been a little over the loci on the top, the meiosis would yield the "same" old gametes - at least in terms of these loci. But, since the cut was between the two loci, we're having recombined gametes, like Ab and aB. Thats why one would say that if two loci are very further apart, crossing over between them is very likely. Commented Aug 4, 2021 at 16:48 | 968 | 4,124 | {"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} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.948407 |
http://www.cfd-online.com/Forums/fluent/133325-boundary-condition-high-pressure-burner-print.html | 1,444,323,463,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737896527.58/warc/CC-MAIN-20151001221816-00172-ip-10-137-6-227.ec2.internal.warc.gz | 465,792,780 | 3,594 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- FLUENT (http://www.cfd-online.com/Forums/fluent/)
- - Boundary Condition of a High Pressure Burner (http://www.cfd-online.com/Forums/fluent/133325-boundary-condition-high-pressure-burner.html)
WJXu April 13, 2014 15:31
Boundary Condition of a High Pressure Burner
Hi, all. I have a question about the computation of reactive flow of a small burner, which is used in the lab. The burner has two inlets, one for air and the other for fuel--CH4. The pressure in most part of the burner is 5atm. And the outlet is a small converging nozzle which will expand the gas to sonic state.
1. I have some confusion about boundary conditions to apply. I have already tried multiple combinations.
a). PressureInet and PressureOutlet, this could not reach the mass flow rate measured in the experiment.
b). MassFlowInlet and PressureOutlet, this could not reach the desired pressure
c). VelocityInlet and PressureOutlet, this would be the same as 2.
Any one have similar experience?
2. Because of converging nozzle, the flow is transferring form incompressible flow to compressible flow. Therefore I think I need to model the gas as ideal gas, which allows density variation. But this setting would give out the warning: Boundary mach number exceeds maximum limit on pressure outlet==0.98. However, when I set the gas to be incompressible ideal gas (maybe reasonable), there's no such warning. Could anybody know how to get rid of this??
WJXu April 13, 2014 17:17
WJXu April 14, 2014 03:22
Up, help? Any body?
WJXu April 14, 2014 14:13
plz..........................................
AbbasRahimi April 14, 2014 15:59
Quote:
Originally Posted by WJXu (Post 485862) Hi, all. I have a question about the computation of reactive flow of a small burner, which is used in the lab. The burner has two inlets, one for air and the other for fuel--CH4. The pressure in most part of the burner is 5atm. And the outlet is a small converging nozzle which will expand the gas to sonic state. 1. I have some confusion about boundary conditions to apply. I have already tried multiple combinations. a). PressureInet and PressureOutlet, this could not reach the mass flow rate measured in the experiment. b). MassFlowInlet and PressureOutlet, this could not reach the desired pressure c). VelocityInlet and PressureOutlet, this would be the same as 2. Any one have similar experience? 2. Because of converging nozzle, the flow is transferring form incompressible flow to compressible flow. Therefore I think I need to model the gas as ideal gas, which allows density variation. But this setting would give out the warning: Boundary mach number exceeds maximum limit on pressure outlet==0.98. However, when I set the gas to be incompressible ideal gas (maybe reasonable), there's no such warning. Could anybody know how to get rid of this?? I am finishing my thesis, and it's kind of urgent. Any advice and help is highly appreciated. Thank you and please help..:mad::mad::mad:
Try this: Set the inlet BCs to mass flow rate and set the pressure outlet to zero and also tick the target mass flow rate for outlet boundary. Although this way may over-specify the problem but smt it helps.
WJXu April 14, 2014 21:08
Quote:
Originally Posted by AbbasRahimi (Post 486095) Try this: Set the inlet BCs to mass flow rate and set the pressure outlet to zero and also tick the target mass flow rate for outlet boundary. Although this way may over-specify the problem but smt it helps.
Thanks a lot, I will try this
shk12345 April 21, 2014 10:53
Quote:
Originally Posted by AbbasRahimi (Post 486095) Try this: Set the inlet BCs to mass flow rate and set the pressure outlet to zero and also tick the target mass flow rate for outlet boundary. Although this way may over-specify the problem but smt it helps.
The solution provided by Abbas is quite good and that should work out.
You may also try with mass flow inlet, pressure outlet without target mass flow inlet using ideal gas law.
This may provide some problem with convergence.
Let me know if you require any other help
macfly April 21, 2014 15:09
Hi WJXu,
Sorry in advance, I'm not bringing any solution to your problem. But I would like to know: what combustion model do you use?
I have to model furnace high pressure natural gas burners that work at velocities 300-500 m/s. I'm not actually modeling the burners, my model starts at the burner inlet where I impose a mass-flow-inlet in order to obtain the desired velocity. I'm not really caring about matching the pressure at the burner inlet, should I? The mass flow of the burner inlet is negligeable compared to the air mass flow in the furnace.
WJXu April 21, 2014 19:25
Quote:
Originally Posted by macfly (Post 487376) Hi WJXu, Sorry in advance, I'm not bringing any solution to your problem. But I would like to know: what combustion model do you use? I have to model furnace high pressure natural gas burners that work at velocities 300-500 m/s. I'm not actually modeling the burners, my model starts at the burner inlet where I impose a mass-flow-inlet in order to obtain the desired velocity. I'm not really caring about matching the pressure at the burner inlet, should I? The mass flow of the burner inlet is negligeable compared to the air mass flow in the furnace.
Hi?
I am comparing Eddy dissipation model and Finite rate/Eddy dissipation model both, which show similar results with some discrepancy about the location of highest temperature region. I am simulating CH4/Air combustion.
I don't know the model and I don't know what you are simulating. In my case, I believe the pressure is important, because this pressure is what measured in the experiment. And in my case, there is pressure variation.
Actually I think both of the two, pressure and mass flow rates are important. The mass flow rate determines the equivalence ratio which apparent affects the reactions. The pressure, especially in pressurized burner/furnace should also play a role in the reaction. Confusing...:confused:
Now what I did is use pressure inlet/pressure outlet boundary. The mass flow rate deviate a little from measurements, but still acceptable.
WJXu April 21, 2014 19:28
Quote:
Originally Posted by AbbasRahimi (Post 486095) Try this: Set the inlet BCs to mass flow rate and set the pressure outlet to zero and also tick the target mass flow rate for outlet boundary. Although this way may over-specify the problem but smt it helps.
Thanks for you advice. The mass flow rate and pressure outlet boundary cannot maintain the pressure inside. This may be because of the pressure drop at the burner exit--a converging nozzle.
All times are GMT -4. The time now is 12:57. | 1,605 | 6,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2015-40 | longest | en | 0.919185 |
https://brainmass.com/chemistry/physical-chemistry/calculating-volume-pressure-density-131744 | 1,620,500,714,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00512.warc.gz | 167,852,164 | 75,096 | Explore BrainMass
# Calculating Volume, Pressure & Density
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
1. A sample of CO2 gas at 22°C and 7.71 atm has a volume of 6.0 L. If the pressure decreases to 5.42 atm, what is the volume? (Round your answer to one decimal place)
2. 120 mL of ethyl alcohol, C2H5OH, is poured into a soft drink bottle at room temperature. The ethyl alcohol weighs 32.4 g and the bottle is heated to 100°C. What is the pressure after heating? (Round your answer to the nearest whole number)
3. Calculate the density, in g/L, of SO2 at 75°C and 1.33 bar. (Round your answer to two decimal places)
https://brainmass.com/chemistry/physical-chemistry/calculating-volume-pressure-density-131744
#### Solution Summary
The solution provides applications of ideal gas law to find the volume, pressure, and density (see attachment).
\$2.49 | 258 | 1,013 | {"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-2021-21 | latest | en | 0.898136 |
http://www.jimloy.com/puzz/3triangl.htm | 1,369,483,898,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705948348/warc/CC-MAIN-20130516120548-00087-ip-10-60-113-184.ec2.internal.warc.gz | 539,834,445 | 2,618 | My Three Triangle Puzzle
What do these three triangles have in common, besides a side of seven? You might want to think about it before you go on to the next paragraph.
Comments: Unless you recognize the second and third triangles, the above question seems a little difficult to answer. The areas are all different, and the perimeters are all different. Most of the angles are different. Could the upper angle in all three triangles be the same (60 degrees)? It looks like that might be the answer, but how do we prove that? The Pythagorean Theorem comes to mind (let's ignore the law of cosines for now). How could we possibly apply the Pythagorean theorem?
If those angles are 60 degrees, then making a copy of the middle triangle, next to the rightmost triangle, would produce an equilateral triangle. See the drawing on the right. It looks like a triangle. Can we prove that the right side is a straight line? This is the same puzzle as the above, but from a different perspective. Does that help us any?
Solution: I guess the easiest way of solving our problem is to look at the equilateral triangle on the left, and show that the two appropriate triangles are congruent with those in the previous diagram. By the Pythagorean Theorem, the altitude (h) of the equilateral triangle is 4sqr(3) (4 times the square root of 3). A further application of the Pythagorean Theorem shows that the hypotenuse (x) of the skinny right triangle (with sides h, 1, and x) is indeed 7. And so the two appropriate triangles are congruent in the two diagrams (by SSS (see Congruence Of Triangles, Part I)). The appropriate supplementary angles are also congruent. And so the rightmost line in the upper right diagram is indeed a straight line. And incidentally, the angles that we thought might be 60 degrees are indeed 60 degrees, which answers our original question.
Above, I mentioned the law of cosines, which is that in any triangle ABC, c^2=a^2+b^2-2ab(cos(C)) (with c^2 meaning c squared). The cosine of 60 degrees is 1/2. We could have applied this formula to the original triangles, and shown that the cosine of the upper angles was 1/2 in all three triangles. That would have been too easy.
One further diagram, I've redrawn all three triangles on the same base. And we see that all five vertices are on a circle. This is the result of one of Euclid's theorems: An angle inscribed in a circle is half the central angle of the inscribed arc. So all inscribed angles with the same arc are congruent angles.
On the left, we see some more triangles that have integer-length sides. The leftmost angle is 60 degrees (there is an equilateral triangle in the diagram).
Let's call the following Jim's Theorem (sort of like the Pythagorean Theorem, but Jim is 30 degrees short of a right angle): In a triangle in which C=60 degrees, c^2=a^2+b^2-ab. The proof follows directly from the law of cosines. Then we have pairs of Jim triples (see Pythagorean Triples), ignoring those that are multiples of smaller Jim triples:
a,b,c (C=60 degrees) a,b,c (C=60 degrees) 1,1,1 3,8,7 5,8,7 7,15,13 8,15,13 5,21,19 16,21,19 11,35,31 24,35,31 7,40,37 33,40,37 13,48,43 35,48,43 16,55,49 39,55,49 9,65,61 56,65,61 32,77,67 45,77,67 17,80,73 63,80,73 40,91,79 51,91,79 11,96,91 85,96,91 19,99,91 80,99,91
Each pair fits together to form an equilateral triangle. | 896 | 3,344 | {"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-2013-20 | latest | en | 0.93708 |
https://www.dataunitconverter.com/terabyte-per-day-to-tebibit-per-second | 1,716,628,941,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00613.warc.gz | 621,115,155 | 17,419 | # TB/Day to Tibps → CONVERT Terabytes per Day to Tebibits per Second
expand_more
info 1 TB/Day is equal to 0.0000842124723863822442513925057870370364 Tibps
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
## Terabytes per Day (TB/Day) Versus Tebibits per Second (Tibps) - Comparison
Terabytes per Day and Tebibits per Second are units of digital information used to measure storage capacity and data transfer rate.
Terabytes per Day is a "decimal" unit where as Tebibits per Second is a "binary" unit. One Terabyte is equal to 1000^4 bytes. One Tebibit is equal to 1024^4 bits. There are 0.137438953472 Terabyte in one Tebibit. Find more details on below table.
Terabytes per Day (TB/Day) Tebibits per Second (Tibps)
Terabytes per Day (TB/Day) is a unit of measurement for data transfer bandwidth. It measures the number of Terabytes that can be transferred in one Day. Tebibits per Second (Tibps) is a unit of measurement for data transfer bandwidth. It measures the number of Tebibits that can be transferred in one Second.
## Terabytes per Day (TB/Day) to Tebibits per Second (Tibps) Conversion - Formula & Steps
The TB/Day to Tibps Calculator Tool provides a convenient solution for effortlessly converting data rates from Terabytes per Day (TB/Day) to Tebibits per Second (Tibps). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Terabyte) and target (Tebibit) data units.
Source Data Unit Target Data Unit
Equal to 1000^4 bytes
(Decimal Unit)
Equal to 1024^4 bits
(Binary Unit)
The conversion from Data per Day to Second can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Terabytes per Day (TB/Day) to Tebibits per Second (Tibps) can be expressed as follows:
diamond CONVERSION FORMULA Tibps = TB/Day x (8x10004) ÷ 10244 / ( 60 x 60 x 24 )
Now, let's apply the aforementioned formula and explore the manual conversion process from Terabytes per Day (TB/Day) to Tebibits per Second (Tibps). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Tebibits per Second = Terabytes per Day x (8x10004) ÷ 10244 / ( 60 x 60 x 24 )
STEP 1
Tebibits per Second = Terabytes per Day x (8x1000x1000x1000x1000) ÷ (1024x1024x1024x1024) / ( 60 x 60 x 24 )
STEP 2
Tebibits per Second = Terabytes per Day x 8000000000000 ÷ 1099511627776 / ( 60 x 60 x 24 )
STEP 3
Tebibits per Second = Terabytes per Day x 7.2759576141834259033203125 / ( 60 x 60 x 24 )
STEP 4
Tebibits per Second = Terabytes per Day x 7.2759576141834259033203125 / 86400
STEP 5
Tebibits per Second = Terabytes per Day x 0.0000842124723863822442513925057870370364
Example : By applying the previously mentioned formula and steps, the conversion from 1 Terabytes per Day (TB/Day) to Tebibits per Second (Tibps) can be processed as outlined below.
1. = 1 x (8x10004) ÷ 10244 / ( 60 x 60 x 24 )
2. = 1 x (8x1000x1000x1000x1000) ÷ (1024x1024x1024x1024) / ( 60 x 60 x 24 )
3. = 1 x 8000000000000 ÷ 1099511627776 / ( 60 x 60 x 24 )
4. = 1 x 7.2759576141834259033203125 / ( 60 x 60 x 24 )
5. = 1 x 7.2759576141834259033203125 / 86400
6. = 1 x 0.0000842124723863822442513925057870370364
7. = 0.0000842124723863822442513925057870370364
8. i.e. 1 TB/Day is equal to 0.0000842124723863822442513925057870370364 Tibps.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Terabytes per Day to Tebibits per Second using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Terabyte ?
A Terabyte (TB) is a decimal unit of digital information that is equal to 1,000,000,000,000 bytes (or 8,000,000,000,000 bits) and commonly used to measure the storage capacity of computer hard drives, flash drives, and other digital storage devices. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of Tebibyte (TiB) is used instead.
arrow_downward
#### What is Tebibit ?
A Tebibit (Tib or Tibit) is a binary unit of digital information that is equal to 1,099,511,627,776 bits and is defined by the International Electro technical Commission(IEC). The prefix 'tebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'terabit' (Tb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Terabytes per Day (TB/Day) to Tebibits per Second (Tibps)
Apply the formula as shown below to convert from 1 Terabytes per Day (TB/Day) to Tebibits per Second (Tibps).
A B C
1 Terabytes per Day (TB/Day) Tebibits per Second (Tibps)
2 1 =A2 * 7.2759576141834259033203125 / ( 60 * 60 * 24 )
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Terabytes per Day (TB/Day) to Tebibits per Second (Tibps) Conversion
You can use below code to convert any value in Terabytes per Day (TB/Day) to Terabytes per Day (TB/Day) in Python.
terabytesperDay = int(input("Enter Terabytes per Day: "))
tebibitsperSecond = terabytesperDay * (8*1000*1000*1000*1000) / (1024*1024*1024*1024) / ( 60 * 60 * 24 )
print("{} Terabytes per Day = {} Tebibits per Second".format(terabytesperDay,tebibitsperSecond))
The first line of code will prompt the user to enter the Terabytes per Day (TB/Day) as an input. The value of Tebibits per Second (Tibps) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Terabytes(TB) are there in a Tebibit(Tibit)?expand_more
There are 0.137438953472 Terabytes in a Tebibit.
#### What is the formula to convert Tebibit(Tibit) to Terabyte(TB)?expand_more
Use the formula TB = Tibit x 10244 / (8x10004) to convert Tebibit to Terabyte.
#### How many Tebibits(Tibit) are there in a Terabyte(TB)?expand_more
There are 7.2759576141834259033203125 Tebibits in a Terabyte.
#### What is the formula to convert Terabyte(TB) to Tebibit(Tibit)?expand_more
Use the formula Tibit = TB x (8x10004) / 10244 to convert Terabyte to Tebibit.
#### Which is bigger, Terabyte(TB) or Tebibit(Tibit)?expand_more
Terabyte is bigger than Tebibit. One Terabyte contains 7.2759576141834259033203125 Tebibits.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 2,012 | 6,654 | {"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} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.789123 |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_4 | 1,701,378,464,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00756.warc.gz | 138,291,976 | 11,569 | # 1986 AIME Problems/Problem 4
## Problem
Determine $3x_4+2x_5$ if $x_1$, $x_2$, $x_3$, $x_4$, and $x_5$ satisfy the system of equations below.
$2x_1+x_2+x_3+x_4+x_5=6$
$x_1+2x_2+x_3+x_4+x_5=12$
$x_1+x_2+2x_3+x_4+x_5=24$
$x_1+x_2+x_3+2x_4+x_5=48$
$x_1+x_2+x_3+x_4+2x_5=96$
## Solution
Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$. Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$, so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$.
## Solution 2
Subtracting the first equation from every one of the other equations yields \begin{align*} x_2-x_1&=6\\ x_3-x_1&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*} Thus \begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ 6x_1+156&=6\\ x_1&=-25 \end{align*} Using the previous equations, $$3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}$$
~ Nafer | 488 | 991 | {"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": 19, "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} | 4.8125 | 5 | CC-MAIN-2023-50 | latest | en | 0.339815 |
https://en.academic.ru/dic.nsf/enwiki/736081 | 1,600,773,724,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00749.warc.gz | 375,397,622 | 14,877 | # Four-square cipher
Four-square cipher
The Four-square cipher is a manual symmetric encryption technique. It was invented by famous French cryptographer Felix Delastelle.
The technique encrypts pairs of letters ("digraphs"), and thus falls into a category of ciphers known as polygraphic substitution ciphers. This adds significant strength to the encryption when compared with monographic substitution ciphers which operate on single characters. The use of digraphs makes the four-square technique less susceptible to frequency analysis attacks, as the analysis must be done on 676 possible digraphs rather than just 26 for monographic substitution. The frequency analysis of digraphs is possible, but considerably more difficult - and it generally requires a much larger ciphertext in order to be useful.
Using four-square
The four-square cipher uses four 5 by 5 matrices arranged in a square. Each of the 5 by 5 matrices contains the letters of the alphabet (usually omitting "Q" or putting both "I" and "J" in the same location to reduce the alphabet to fit). In general, the upper-left and lower-right matrices are the "plaintext squares" and each contain a standard alphabet. The upper-right and lower-left squares are the "ciphertext squares" and contain a mixed alphabetic sequence.
To generate the ciphertext squares, one would first fill in the spaces in the matrix with the letters of a keyword or phrase (dropping any duplicate letters), then fill the remaining spaces with the rest of the letters of the alphabet in order (again omitting "Q" to reduce the alphabet to fit). The key can be written in the top rows of the table, from left to right, or in some other pattern, such as a spiral beginning in the upper-left-hand corner and ending in the center. The keyword together with the conventions for filling in the 5 by 5 table constitute the cipher key. The four-square algorithm allows for two separate keys, one for each of the two ciphertext matrices.
As an example, here are the four-square matrices for the keywords "example" and "keyword." The plaintext matrices are in lowercase and the ciphertext matrices are in caps to make this example visually more simple:
a b c d e E X A M P f g h i j L B C D F k l m n o G H I J K p r s t u N O R S T v w x y z U V W Y Z K E Y W O a b c d e R D A B C f g h i j F G H I J k l m n o L M N P S p r s t u T U V X Z v w x y z
Algorithm
To encrypt a message, one would follow these steps:
* Split the payload message into digraphs. ("HELLO WORLD" becomes "HE LL OW OR LD")
* Find the first letter in the digraph in the upper-left plaintext matrix. a b c d e E X A M P f g "h" i j L B C D F k l m n o G H I J K p r s t u N O R S T v w x y z U V W Y Z K E Y W O a b c d e R D A B C f g h i j F G H I J k l m n o L M N P S p r s t u T U V X Z v w x y z
* Find the second letter in the digraph in the lower-right plaintext matrix. a b c d e E X A M P f g "h" i j L B C D F k l m n o G H I J K p r s t u N O R S T v w x y z U V W Y Z K E Y W O a b c d "e" R D A B C f g h i j F G H I J k l m n o L M N P S p r s t u T U V X Z v w x y z
* The first letter of the encrypted digraph is in the same row as the first plaintext letter and the same column as the second plaintext letter. It is therefore in the upper-right ciphertext matrix. a b c d e E X A M P f g h i j L B C D "F" k l m n o G H I J K p r s t u N O R S T v w x y z U V W Y Z K E Y W O a b c d e R D A B C f g h i j F G H I J k l m n o L M N P S p r s t u T U V X Z v w x y z
* The second letter of the encrypted digraph is in the same row as the second plaintext letter and the same column as the first plaintext letter. It is therefore in the lower-left ciphertext matrix. a b c d e E X A M P f g h i j L B C D F k l m n o G H I J K p r s t u N O R S T v w x y z U V W Y Z K E "Y" W O a b c d e R D A B C f g h i j F G H I J k l m n o L M N P S p r s t u T U V X Z v w x y z
Using the four-square example given above, we can encrypt the following plaintext:
Plaintext: he lp me ob iw an ke no bi Ciphertext: FY GM KY HO BX MF KK KI MD
Here is the four-square written out again but blanking all of the values that aren't used for encrypting the first digraph "he" into "FY"
- - - - - - - - - - - - h - - - - - - F - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Y - - - - - - e - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
As can be seen clearly, the method of encryption simply involves finding the other two corners of a rectangle defined by the two letters in the plaintext digraph. The encrypted digraph is simply the letters at the other two corners, with the upper-right letter coming first.
Decryption works the same way, but in reverse. The ciphertext digraph is split with the first character going into the upper-right matrix and the second character going into the lower-left matrix. The other corners of the rectangle are then located. These represent the plaintext digraph with the upper-left matrix component coming first.
Four-square cryptanalysis
Like most pre-modern era ciphers, the four-square cipher can be easily cracked if there is enough text. Obtaining the key is relatively straightforward if both plaintext and ciphertext are known. When only the ciphertext is known, brute force cryptanalysis of the cipher involves searching through the key space for matches between the frequency of occurrence of digrams (pairs of letters) and the known frequency of occurrence of digrams in the assumed language of the original message.
Cryptanalysis of four-square generally involves pattern matching on repeated monographs. This is only the case when the two plaintext matrices are known. A four-square encipherment usually uses standard alphabets in these matrices but it is not a requirement. If this is the case, then certain words will always produce single-letter ciphertext repeats. For instance, the word MI LI TA RY will always produce the same ciphertext letter in the first and third positions regardless of the keywords used. Patterns like these can be cataloged and matched against single-letter repeats in the ciphertext. Candidate plaintext can then be inserted in an attempt to uncover the ciphertext matrices.
Unlike the Playfair cipher, a four-square cipher will not show reversed ciphertext digraphs for reversed plaintext digraphs (e.g. the digraphs AB BA would encrypt to some pattern XY YX in Playfair, but not in four-square). This, of course, is only true if the two keywords are different. Another difference between four-square and Playfair which makes four-square a stronger encryption is the fact that double letter digraphs will occur in four-square ciphertext.
By all measures, four-square is a stronger system for encrypting information than Playfair. However, it is more cumbersome because of its use of two keys and preparing the encryption/decryption sheet can be time consuming. Given that the increase in encryption strength afforded by four-square over Playfair is marginal and that both schemes are easily defeated if sufficient ciphertext is available, Playfair has become much more common.
A good tutorial on reconstructing the key for a four-square cipher can be found in chapter 7, "Solution to Polygraphic Substitution Systems," of [http://www.umich.edu/~umich/fm-34-40-2/ Field Manual 34-40-2] , produced by the United States Army.
* Topics in cryptography
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Four square (disambiguation) — Four square can mean: *Four square, a ball game for all ages *Four square cipher in classical cryptography *Four Square (company) is a division of Mars, Incorporated *Four Square supermarkets in New Zealand *Four Square Writing Method, an… … Wikipedia
• Two-square cipher — The Two square cipher is a manual symmetric encryption technique. It was developed to ease the cumbersome nature of the large encryption/decryption matrix used in the four square cipher while still being slightly stronger than the Playfair cipher … Wikipedia
• Square — may mean:Mathematics*Square (algebra), to multiply a number or other quantity by itself **Perfect square **Square matrix **Square number **Square root*Square (geometry), a polygon with four equal sides and angles **Unit square*Square wave, a… … Wikipedia
• Trifid cipher — In classical cryptography, the trifid cipher is a cipher invented around 1901 by Felix Delastelle, which extends the concept of the bifid cipher to a third dimension, allowing each symbol to be fractionated into 3 elements instead of two. That is … Wikipedia
• Bifid cipher — In classical cryptography, the bifid cipher is a cipher which combines the Polybius square with transposition, and uses fractionation to achieve diffusion. It was invented around 1901 by Felix Delastelle. Operation First, a mixed alphabet… … Wikipedia
• Playfair cipher — The Playfair cipher or Playfair square is a manual symmetric encryption technique and was the first literal digraph substitution cipher. The scheme was invented in 1854 by Charles Wheatstone, but bears the name of Lord Playfair who promoted the… … Wikipedia
• Classical cipher — A cipher is a means of concealing a message, where letters of the message are substituted or transposed for other letters, letter pairs, and sometimes for many letters. In cryptography, a classical cipher is a type of cipher that was used… … Wikipedia
• Substitution cipher — In cryptography, a substitution cipher is a method of encryption by which units of plaintext are replaced with ciphertext according to a regular system; the units may be single letters (the most common), pairs of letters, triplets of letters,… … Wikipedia
• Caesar cipher — The action of a Caesar cipher is to replace each plaintext letter with one fixed number of places down the alphabet. This example is with a shift of three, so that a B in the p … Wikipedia
• Book cipher — A book cipher is a cipher in which the key is some aspect of a book or other piece of text; books being common and widely available in modern times, users of book ciphers take the position that the details of the key is sufficiently well hidden… … Wikipedia | 2,461 | 10,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-40 | latest | en | 0.897212 |
https://www.physicsforums.com/threads/ac-current-source.818121/ | 1,527,012,075,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864837.40/warc/CC-MAIN-20180522170703-20180522190703-00154.warc.gz | 828,493,448 | 17,722 | # AC Current source
Tags:
1. Jun 8, 2015
### SanDiegan
Hi everybody,
As part of my research at UCSD, we are trying to measure Hall resistance of some materials. And to do that we need to send an AC current. So I wondered how we could easily build an ac current source, probably based on op-amp. I know there is the Howland circuit (current pump), but I am not sure it can be used to get an AC current as well.
The specifications of the current we need are the following: constant ~µA RMS, frequency range ~1kHz, sine wave. The impedance of the material we are testing is about ~mΩ.
Yours
Last edited: Jun 8, 2015
2. Jun 8, 2015
### meBigGuy
Do you have a sine wave source from a signal generator. If so, a bipolar current source like the howland is the correct answer.
You can cheat, in that a 1V sine wave source in series with a 1 megohm resistor driving a milli-ohm load is very nearly a 1uA constant current source to 1 part in 10^9. (but maybe you need better than that?)
3. Jun 8, 2015
### SanDiegan
We can have a sine wave source, but I cannot find any information anywhere clearly stating the Howland circuit can be used to generate an AC current.
And actually, we would like to avoid using a resistor, but yes that would have been an easy solution!
4. Jun 8, 2015
### meBigGuy
The howland circuit will generate a current that follows the reference voltage. In order to generate AC without coupling capacitors you need to have a bipolar power supply for the opamp (like +5 and -5 volts). Since the opamp is supplying/sinking the current, it can go positive and negative.
You probably want to use a high quality cmos opamp to avoid bias currents.
I'm surprised that I can't find a google image of the simplest possibility (if your load need not be grounded).
Opamp drives the load in series with a sense resistor to ground. Sense resistor/load node feeds back to - input of the opamp. Reference voltage is applied to + input.
So the reference voltage appears across the sense resistor, causing a reference current through the load.
Again, this requires a + and - supply referenced to ground.
Sometimes op-amps have crossover distortion when they drive high resistances. You can use a resistor from the output of the opamp to the negative supply to "bias" the output into class A operation.
5. Jun 9, 2015
### SanDiegan
Okay now I understand how it could generate an AC current. Does the attached picture correspond to your circuit / can it be used in my case?
And last question how do you choose the power supply for the op-amp and the resistors? I know how to choose R3,R4,R5 (cf. the circuit) since they set the current amplitude, but what about the others?
Thank you!
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7.7 KB
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140
6. Jun 9, 2015
### tech99
The resistance under test is only milliohms, so it is very easy to obtain constant current by just placing a resistor in series with either a signal generator or a simple oscillator (such as a Wien Bridge type). For instance, a resistor of 100k would enable a 1 volt SG to pass 10uA through your device under test, and would swamp out the milliohm fluctuations of the device itself. I suggest also placing a capacitor in series to eliminate the chance of a DC component.
7. Jun 9, 2015
### meBigGuy
The circuit you drew cannot generate an AC signal in the load. The transistor will only source current to the load. It cannot sink current (draw current out of ground through the load). Take out the transistor and connect the opamp directly to the R4 R5 junction. If you need the load grounded, just follow the howland design notes. http://www.ti.com/lit/an/snoa474a/snoa474a.pdf. Otherwise try the simple circuit I described.
8. Jun 10, 2015
### Jeff Rosenbury
Naively I would go with tech99's solution as simple and effective.
However, I don't know what Hall resistance is? I understand the Hall effect. Are you trying to measure the resistance of the material side to side with a working Hall effect sensor?
If tech99's solution is too simple for your needs, you might consider a current mirror as a current source. These can give multiple, identical (with hand matched betas) currents. I could see where multiple, identical current sources might be useful with Hall measurements. You could for example insert a current in a working circuit and extract the same current from the other side of the sensor. (You may need to build some sort of photo transistor current mirror if different voltages are a problem.)
9. Jun 10, 2015
### SanDiegan
Hi All,
Thank you for all your contribution. I am going to try few circuits based on op-amp, and also with only resistors in series first before trying anything more complicated. We want an AC current just to be able to modulate it and filter the signal with the lockIn in order to cancel the noise.
@JeffRosenBury The Hall resistance is simply related to the Hall voltage you can get according to the current you send. Actually to analyse and compare the materials, the Hall resistivity is more relevant (just have to take into account the dimensions of the sample).
10. Jun 10, 2015
### meBigGuy
Like the one in post #2?
11. Jun 11, 2015
### Jeff Rosenbury
Perhaps.
Sorry, I read your first sentence and wandered off to look up the Howland. I should have kept reading. | 1,261 | 5,293 | {"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-2018-22 | latest | en | 0.918308 |
https://arxiv.org/abs/1401.1526 | 1,566,329,875,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315558.25/warc/CC-MAIN-20190820180442-20190820202442-00416.warc.gz | 368,499,608 | 6,730 | cs.CR
# Title:Additional Constructions to Solve the Generalized Russian Cards Problem using Combinatorial Designs
Abstract: In the generalized Russian cards problem, we have a card deck $X$ of $n$ cards and three participants, Alice, Bob, and Cathy, dealt $a$, $b$, and $c$ cards, respectively. Once the cards are dealt, Alice and Bob wish to privately communicate their hands to each other via public announcements, without the advantage of a shared secret or public key infrastructure. Cathy should remain ignorant of all but her own cards after Alice and Bob have made their announcements. Notions for Cathy's ignorance in the literature range from Cathy not learning the fate of any individual card with certainty (weak $1$-security) to not gaining any probabilistic advantage in guessing the fate of some set of $\delta$ cards (perfect $\delta$-security). As we demonstrate, the generalized Russian cards problem has close ties to the field of combinatorial designs, on which we rely heavily, particularly for perfect security notions. Our main result establishes an equivalence between perfectly $\delta$-secure strategies and $(c+\delta)$-designs on $n$ points with block size $a$, when announcements are chosen uniformly at random from the set of possible announcements. We also provide construction methods and example solutions, including a construction that yields perfect $1$-security against Cathy when $c=2$. We leverage a known combinatorial design to construct a strategy with $a=8$, $b=13$, and $c=3$ that is perfectly $2$-secure. Finally, we consider a variant of the problem that yields solutions that are easy to construct and optimal with respect to both the number of announcements and level of security achieved. Moreover, this is the first method obtaining weak $\delta$-security that allows Alice to hold an arbitrary number of cards and Cathy to hold a set of $c = \lfloor \frac{a-\delta}{2} \rfloor$ cards. Alternatively, the construction yields solutions for arbitrary $\delta$, $c$ and any $a \geq \delta + 2c$.
Subjects: Cryptography and Security (cs.CR); Combinatorics (math.CO) Cite as: arXiv:1401.1526 [cs.CR] (or arXiv:1401.1526v2 [cs.CR] for this version)
## Submission history
From: Colleen Swanson [view email]
[v1] Tue, 7 Jan 2014 22:18:05 UTC (31 KB)
[v2] Tue, 5 Aug 2014 17:41:40 UTC (27 KB) | 557 | 2,336 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-35 | latest | en | 0.921587 |
http://mathhelpforum.com/geometry/122522-areas-sector.html | 1,527,423,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868248.78/warc/CC-MAIN-20180527111631-20180527131631-00427.warc.gz | 186,205,052 | 9,573 | # Thread: Areas of a sector
1. ## Areas of a sector
Please note if I am incorrect
Question: There is a sector of a circle with centrepoint O. The radius of this circle is 23cm. The angle of the sector is 165 degrees. Work out the area of the sector correct to two significant figures.
I did $\displaystyle (\frac{\pi *23^2}{180})*165$
From which I got (thinking about the significant figs) 1500.
2. Hello, Mukilab!
Your answer is twice the correct answer.
Area of a sector: .$\displaystyle A \:=\:{\color{red}\tfrac{1}{2}}r^2\theta$
3. theta? What does that mean? I don't understand it outside of trigonometry.
Also what do you mean by twice the correct answer?
4. theta is the angle in radians | 193 | 705 | {"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} | 3.390625 | 3 | CC-MAIN-2018-22 | latest | en | 0.888326 |
https://lykelozabed.allianceimmobilier39.com/enter-your-math-problem-algebra-36109vp.html | 1,628,113,726,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155188.79/warc/CC-MAIN-20210804205700-20210804235700-00341.warc.gz | 365,774,866 | 4,607 | # Enter your math problem algebra
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This overhelmed him and he felt ashamed because he didn't "get it" right away like he always had. The following words in a sentence indicate that Adding is taking place.
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Oh, one more thing. But then we ended up with information on the three girls rows down on the first matrix. 1. Enter the problem either using the symbols or by starting with an example. 2.
You can make sure you entered your problem correctly by clicking the Show button next to Math allianceimmobilier39.com will show your problem in the format you’re used to seeing. Photomath is the #1 app for math learning; it can read and solve problems ranging from arithmetic to calculus instantly by using the camera on your mobile device.
With Photomath, learn how to approach math problems through animated steps and detailed instructions or check your homework for any printed or handwritten problem. After spending countless hours trying to understand my homework night after night, I found Algebrator. Most other programs just give you the answer, which did not help me when it come to test time, Algebrator helped me through each problem step by step.
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For questionssolve the problem and enter your answer in the grid on the answer sheet. College Algebra Math problem. Be sure to reduce your first term. Part 2 or 2: (6 pts) Use that model to determine the times at which the ball is 36 feet in the air. | 924 | 4,575 | {"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} | 3.03125 | 3 | CC-MAIN-2021-31 | longest | en | 0.945222 |
http://www.math.uni.edu/~campbell/mdm/cake.html | 1,506,436,068,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696182.97/warc/CC-MAIN-20170926141625-20170926161625-00614.warc.gz | 496,693,772 | 4,433 | # Two person fair division
## You cut, I choose
The standard method for dividing a piece of cake: you cut I choose, is fair in the sense that it is proportionate (i.e., each person gets at least 1/2 of the cake by their own evaluation) and it is envy free (i.e., neither person would prefer the other person's portion). Yet it is not fair, I would much prefer to have you cut, which gives me the choice, rather than cut the cake and give you the choice. The reason is that cake is not homogeneous, your and my relative preferences for cake, frosting, and aesthetic appearance may be different. The cutter divides the cake into two portions which he perceives as equal, but the chooser may not perceive them as equal. Hence the cutter gets half the cake (by his perception), but the chooser may get more than half the cake (by his perception). For example, if a cake were half chocolate and half white, and the cutter was indifferent to chocolate versus white cake, the cutter might cut the cake down the middle producing one chocolate and one white piece; in that case, if the chooser preferred chocolate cake, he would be much happier with the chocolate piece than with the white piece.
Conversely, if the preferences of the chooser (or both parties) are known, the advantage is shifted to the cutter. For example, if the chooser prefers frosting, while the cutter likes both frosting and cake equally well, the cutter will put enough excess frosting in the smaller portion to entice the chooser to take it. Hence the chooser will get just over half the value of the cake (by the chooser's values) (the cutter must make sure the chooser chooses the proper portion), while the cutter will get substantially over half the value of the cake (by the cutter's values). Using the example of the cake which is half chocolate and half white, if the cutter is indifferent between chocolate and white cake but the chooser is allergic to chocolate (and the cutter knows that fact); the cutter could cut the cake into 1/3 white (i.e., 2/3 of the white half) versus 1/6 white with 1/2 chocolate. The chosser would choose the 1/3 white, and the cutter would end up with 2/3 of the original cake.
## Quantified preferences and trading to pareto optimality
Assume you like chocolate cake twice as much as white (i.e., you have equal preference for one square of chocolate or two squares of white). Assume also that your sibling likes white cake thrice as much as chocolate (i.e., your sibling has equal preference for one square of white or three squares of chocolate). The Pareto optimal divisions are as described above (either you have no white or your sibling has no chocolate). But what apportionment would ensue from trading after your mother's "fair" division?
Since you like chocolate equal to twice as much white cake, you would be willing to give up your 1/4 white cake for one-eighth chocolate cake, resulting in your sibling having all the white cake (one-half cake) and one-eighth chocolate cake. But your sibling would be willing to give up chocolate cake for one-third as much white cake, so if you offered one-twelfth white cake (1/3 of the 1/4 your mother gave you), you could get all your sibling's chocolate cake (one-quarter cake) with the ultimate result you would have all the chocolate cake (one-half cake) plus 1/6 white cake, while your sibling had only 1/3 white cake. Hence even with an initial division, there will in general not be a unique Pareto optimal division which ensues from it. How much you can get will depend on whether you know your siblings preferences, and how good you are at bargaining. Any division between the two extremes calculated is possible.
Exercise: How would Jack Sprat divide a roast if he did not know his wife's preferences? How would he divide the roast if he knew his wife's preferences?
If you knew the chooser wanted exactly equal amounts of cake and frosting, how would you divide a cake?
If you preferred chocolate cake thrice as much as white, and your sibling preferred white cake quatrice as much as chocolate, what are the Pareto optimal divisions of a cake which is 1/2 chocolate and 1/2 white? Which of those partitions could ensue from an initial division giving each of you 1/4 chocolate and 1/4 white cake?
## The case of discrete objects
If the matter to be divided is discrete rather than a continuum, the chooser has the obvious advantage. For example, if there are 5 marbles to be divided, the fairest division is 2 and 3, and the chooser gets three. If the objects are not the same, the advantage may not be so clear. For example, if there is a red shirt, a green shirt, red pants, and green pants, the "cutter" decides according to his pleasure whether the wardrobes will be mixed or matched; if the "cutter" decides on mixed wardrobes, the chooser canot get green pants with a green shirt.
A more common implementation is alternating choices, in which case the first choice is a definite advantage (unless compensated by the next two choices going to the second individual, or some other caompensatory mechanism).
## If opponent's preferences known
If you know your opponents preferences, you may have a significant advantage under alternate choices. Assume a family estate consists of A tailcoat (T), a wedding gown (W), a pair of cufflinks ((C), a necklace (N), a samovar (S), and a mirror (M).Annabelle has the first choice, and the preference order WNSMTC, Alfred has the second choice and the preference TCSMNW. Aternating choices should provide Annabelle with WNS and Alfred with TCM. But if Alfred knows Annabelle does not want cufflinks, he can put them at the bottom of his preference and end up with TSC while Annabelle gets WNM. (Of course if Annabelle knew Alfred's preferences, she could put the wedding gown at the bottom of her list ... .)
1) I read my web page on two person fair division (cake cutting) (above) and thought it could be improved. Hence I have drafted the following (below) which I hope clarifies some of it.
Cake cutting:
The key to many of the problems entailing cutting a cake with two flavors (or dividing a pizza which is half mushroom and half sausage) is to evaluate the cake in terms of a single flavor (like converting Swiss Francs and Euros to U. S. Dollars). [You could have a similar problem dividing up nickels and dimes where your younger brother prefers nickels because they are larger, but you prefer dimes because they are worth more.]
For example, if you like chocolate cake three times as much as vanilla, then you are indifferent between one square of chocolate or three squares of vanilla. If you have three squares of chocolate and three squares of vanilla, to you it would be the same as three squares of chocolate and one square of chocolate (i.e., four squares of chocolate) or nine squares of vanilla and three squares of vanilla (i.e., twelve squares of vanilla). If you had 1/4 chocolate cake and 1/4 vanilla cake, it would be the same to you as having 1/4 chocolate cake and (1/3)(1/4)=1/12 chocolate cake (i.e., 1/4 + 1/12 = 1/3 chocolate cake) or 3/4 vanilla cake and 1/4 vanilla cake (i.e., 1 vanilla cake).
Assuming the 3:1 preference ratio above, this lets you determine whether you prefer 1/4 chocolate and 1/4 vanilla to 1/6 chocolate and 2/5 vanilla. Using chocolate to evaluate the portions, 1/4 chocolate and 1/4 vanilla is equivalent to 1/3 chocolate as calculated above. 1/6 chocolate and 2/5 vanilla is equivalent to 1/6 chocolate and (1/3)(2/5) chocolate equals 1/6 + 2/15 = 9/30 = 3/10 chocolate. Since .3 < .33 (= 1/3), you prefer 1/4 chocolate and 1/4 vanilla to 1/6 chocolate and 2/5 vanilla.
If you were offered 1/6 chocolate and 3/5 vanilla, since 1/6 + (1/3)(3/5) = 11/30 =.37 > .33, you would prefer 1/6 chocolate and 3/5 vanilla to 1/4 chocolate and 1/4 vanilla.
This also lets you evaluate the most you (or your opponent) could get by trading. If you start with 1/4 chocolate and 1/4 vanilla, you will not accept anything you value at less than 1/3 chocolate (assuming the 3:1 conversion ratio above). Therefore your opponent (who prefers vanilla) could never get more than the entire cake less 1/3 chocolate (i.e., 1/2 chocolate and 1/2 vanilla less 1/3 chocolate equals 1/6 chocolate and 1/2 vanilla) | 1,957 | 8,270 | {"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-2017-39 | latest | en | 0.972067 |
https://www.physicsforums.com/threads/first-and-second-derivative-assistance-please.468949/ | 1,620,868,637,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00552.warc.gz | 986,310,731 | 14,893 | # First and second derivative assistance please?
## Homework Statement
r = 2/(2 - cos ($$\pi$$*t))
N/A
## The Attempt at a Solution
Hello everyone, first I would just like to say (which is obvious since i'm asking :tongue: ) That it's been a long long time since i've had to do derivatives, hence my total cluelessness with this. I've had to integrate frequently in my engineering courses, but this is a problem that arose in my dynamics class...and embarrassingly enough, i'm not even sure where to begin.
Could someone help me out? I'm going to be unable to proceed with my problem (it involves radius of curvatures) until I get the first and second derivatives of this bad boy. Thanks in advance!
Last edited:
berkeman
Mentor
## Homework Statement
r = 2/(2 - cos ($$\pi$$*t))
N/A
## The Attempt at a Solution
Hello everyone, first I would just like to say (which is obvious since i'm asking :tongue: ) That it's been a long long time since i've had to do derivatives, hence my total cluelessness with this. I've had to integrate frequently in my engineering courses, but this is a problem that arose in my dynamics class...and embarrassingly enough, i'm not even sure where to begin.
Could someone help me out? I'm going to be unable to proceed with my problem (it involves radius of curvatures) until I get the first and second derivatives of this bad boy. Thanks in advance! | 334 | 1,393 | {"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} | 2.546875 | 3 | CC-MAIN-2021-21 | longest | en | 0.970502 |
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• Recently I found a site that contains the equations to create the graphs for the electron orbitals (s, p, d, f...) but when I use Graphing Calculator to try to
Message 1 of 2 , Jan 18, 2010
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Recently I found a site that contains the equations to create the graphs for the electron orbitals (s, p, d, f...) but when I use Graphing Calculator to try to recreate the orbital graphs shown, I don't get the same results. The site is...
http://www.uky.edu/~holler/html/orbitals_2.html
The graphs are shown by chosing the orbital in question from the left column, the equations are shown by clicking on the equations reference on top.
I am making an assumption, which may be the problem... the graph would be of the form r = f(theta, phi), so I pick the phi from column 1, and theta from column 2. If I only use column 1 (the phi column) then I get shapes that look like the 0 value orbital for each state (I take the absolute value to get the other half for some of them). But I can't get the theta sections to work, they give me very wierd surfaces that don't look at all like an orbital.
Okay, what am I doing wrong here?
• I m going to take the easy way out here because I ve got about 49 things that need doing before classes begin. Have a gander at the Mathematical Physics
Message 1 of 2 , Jan 18, 2010
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I'm going to take the easy way out here because I've got about 49 things that need doing before classes begin. Have a gander at the "Mathematical Physics" section of
<http://web.lemoyne.edu/~craigda/Physics/Sims/Files/filelibrary.html>.
I've posted a gc file that plots spherical harmonics (which is what those images of orbitals really are.)
For a REALLY cool application check out
<http://daugerresearch.com/orbitals/index.shtml>
David
On Jan 18, 2010, at 11:21 AM, gjmcclure wrote:
> Recently I found a site that contains the equations to create the graphs for the electron orbitals (s, p, d, f...) but when I use Graphing Calculator to try to recreate the orbital graphs shown, I don't get the same results. The site is...
>
> http://www.uky.edu/~holler/html/orbitals_2.html
>
> The graphs are shown by chosing the orbital in question from the left column, the equations are shown by clicking on the equations reference on top.
>
> I am making an assumption, which may be the problem... the graph would be of the form r = f(theta, phi), so I pick the phi from column 1, and theta from column 2. If I only use column 1 (the phi column) then I get shapes that look like the 0 value orbital for each state (I take the absolute value to get the other half for some of them). But I can't get the theta sections to work, they give me very wierd surfaces that don't look at all like an orbital.
>
> Okay, what am I doing wrong here?
>
>
David Craig
<http://web.lemoyne.edu/~craigda/>
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Solver programs for the Satisfiability Modulo Theories (SMT). The SMT problem is a decision problem for logical formulas with combinations of background theories expressed in classical first-order logic with equality.
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# Design Irrigation System II
## Presentation on theme: "Design Irrigation System II"— Presentation transcript:
Design Irrigation System II
Asher Azenkot
Local Head Losses The local head loss due to a local disturbance in water flow is proportional to the head velocity. K - Coefficient
Hydraulic Valve Local Head Loss
Filter Local Head Loss
Metzer drip line Water flow velocity reduced gradually along the lateral pipe
Local head loss in “head connector”
m m3/h
Local head loss Flow rate Local head loss m
Example: A 12" valve (K = 2.5) is installed in 1,250 meters long pipe (12” and C = 130). What is the total head loss due to the valve and the pipe when the water flow rate is = 100, and 400 m3/h. The pipe cross section area is:
Continue:
Continue: If an 8" valve is replaced the 12", what will be the new total head loss?
Lateral Pipes A lateral pipe is characterized by a continuous decline in water discharge along the pipe. The flow rate starts at Qu (m3/h) at the upstream end and ends up with a q1 (m3/h) downstream. (Lateral pipe is abide by: 1. A same size of pipe, 2. even distance between outlets, 3. a same outlet (sprinkler or emitter) flow rate. The calculation of the head loss is done in two steps: The head loss is calculated by assuming the pipe is plain The outcome is multiplied by the coefficient F Qu = n*q 3q 2q q D Sl 95 m
Coefficient F 1. F1 to be used when the distance from the lateral inlet to the first outlet is Sl meters. 2. F2 to be used when the first outlet is near the lateral inlet. 3. F3 to be used when the distance from the lateral inlet to the first outlet is Sl/2 meters.
Characteristics of a Lateral pipe
The sprinkler pressure along the lateral pipe decline faster along the first 40% of the length than afterwards (figure 2). The sprinkler flow rate along the lateral pipe declines faster along the first 40% of the length (figure 1). The location of the sprinkler (or emitter) with the average pressure and flow rate is 40% away from the lateral’s inlet. Three quarter of the lateral head loss takes place along the first two fifth sections (40%).
Fig. 1: Flow rate reduction in a plain pipe and in a lateral with sprinklers.
Fig. 2: Head loss and percent of head loss
Head loss Calculation Along lateral
Select a suitable sprinkler or emitter with a required Hs, qs and sl from a catalogue (figure 3). The number of sprinklers (n) along the lateral is determined by (L/sl). The discharge rate at the lateral inlet is determined by (Qu = n x qs ). The lateral diameter (D) should comply with maximum head loss of 20%. The head loss along a lateral (Qu, q, D and L) is computed by: Assuming the lateral pipe is plain and. The outcome is multiplied by F factor.
Fig. 3 - Naan 233
Head loss in drip lateral pipe
A modified Hazen-Williams head loss equation: HL = head loss along a lateral drip line L = lateral length (m) D = internal diameter (m) N = number of emitters q = average emitter flow rate (m3/h) C = Hazen-Williams coefficient ( for polyethylene pipe with ID < 16 mm) F = 0.37 for more than 20 emitters
Hydro P.C. & Hydro P.C.N.D - 1.2* L/H MAXIMUM RECOMMENDED DRIPLINE LENGTH (m) PIPE DIAMETER -16/13.8 (OD/ID)
Number Of Mamkad spinklers
Mamkad mini-sprinkler nozzles
Example: A flat field, 360 x 360 m, is irrigated with a hand moved aluminum lateral pipe (C = 140). The water source to the lateral pipe is from a sub-main, which crosses the center of the field. The selected sprinklers are Naan 233/92 with a nozzle size of 4.5 mm, pressure of 25 m (hs) and flow rate (qs) of 1.44 m3/hr. The space between the sprinklers is 12 meters apart, and the location of the first sprinkler is 6 meters away from lateral inlet. The riser height is 0.8 m and diameter of 3/4".
Answer: lateral 360 m Submain
The number of sprinklers on the lateral is The length of the lateral (l) is l = (14 sprinkler x 12 m apart) + 6 m = 174 meters
Continue The inlet flow rate of the lateral is
Qu = 15 (sprinklers) x 1.44 m3/h = 21.6 m3/h The maximum allowed head loss (20%) throughout the field is For a plain 2" aluminum pipe - the hydraulic gradient out of Hazen Williams is: J = ‰
Continue: The head loss in a 2" (plain) aluminum pipe is as follows:
The F factor for 15 sprinklers is F15 = 0.363 For a 3" aluminum pipe - the hydraulic gradient out of a table or ruler is: J = 26.2‰
Continue: The head loss in a 3" (plain) aluminum pipe is as follows:
The F factor for 15 sprinklers is F15 = 0.363 The difference = 3.34 meters head loss which will be used as the head loss for the sub- main pipe.
A Lateral Inlet Pressure
3q 2q D Sl Qu = n*q
A Lateral Inlet Pressure
The pressure head at the lateral inlet (hu) is determined by: hu - lateral inlet pressure head hs - pressure head of selected sprinkler hf - head loss along lateral riser – the length (height) of the riser - local head loss (incurred between laleral pipe and sprinkler)
Local head loss Flow rate Local head loss m
Example: Following the previous example, what is the inlet pressure?
hf = 1.66 meters riser height = 0.8 meters hs = 25 meters
Inlet pressure in case of a Lateral pipe Laid out on a Slop
The inlet pressure of a lateral pipe which laid out along a slope is as follows: hu the lateral inlet pressure hs pressure head of selected sprinkler hf head loss along lateral riser - riser height - adjustment for an upward slope - adjustment for an downward slope
Example: Following the previous example, but this time with: a. 2% downward slope, or b. 2% upward slope. The difference elevation between the two ends is as follows: a. 2% downward slope
Cont. The pressure by the last sprinkler is as follows:
The head loss between lateral inlet and last sprinkler is:
Cont. 25.3m 27.12m 360 m Δhf=3.2m 20%=5m P=30.3m P=28.5m Sub-main 28.5 – 25.3 m = 3.2 m is taken place along the sub-main pipe. Therefore, the pressure at the head of the field is 28.5 m.
Continue: b. 2% upward slope
The total head loss throughout the lateral pipe is: 5.14 meters are just the permitted 20% head loss. Therefore, nothing is left for the sub-main. In this case, pressure regulators should be installed in every lateral inlets or selecting a wider pipe.
Maximum Permitted Head loss
Distribution of water and pressure
“The 20% rule” In order to maintain up to 10% difference in flow rate between sprinklers or emitter within a sub-plot, then the pressure difference inside the plot should be less than 20%. or Q - flow rate C – coefficient, which depends on a nozzle type A - cross section area of a nozzle H - pressure head X - exponent which depends on the flow pattern.
Pressure Vs. Flow Pressure m Liter/hr
O-tif Flow rate Vs. Pressure
Brown 2lph Black 4lph Green 8lph Purple 16lph
3/8" Rondo Nozzles - Flow rate Vs. Pressure
Supertif - flow rate Vs. Pressure
Ram & Agriplas Flow rate Vs Pressure
Flow rate Vs Pressure
Temperature Vs CV (tapes drip-line)
Example: What is the expected difference discharge between the two ends of a lateral sprinkler? When the hydraulic gradient along a lateral pipe is 20%. The flow rate of a sprinkler is as follows:
Continue: Two identical sprinklers have a same coefficient:
The difference in flow rate between the two ends is 10% (within 20% rule), once the exponent is 0.5.
TUFFTIF Dripper Flow Rate Table
Q=LPH H=m. %DH - MAXIMUM ALLOWABLE HEAD LOSS DIFFERENCE. ALONG A LATERAL (AS PERCENT OF THE WORKING PRESSURE. TO KEEP FLOW RATE DIFFERENCEC ≤ 10% OF WORKING FLOW RATE.
Tufftif performance chart
Example for a micro sprinklers
A polyethylene lateral pipe, grade 4, has 10 micro-sprinklers at 10 meters apart, while the first is only one half way. The flow rate of the selected sprinkler is qs = 120 l/h at hs = 20 meters. The riser’s height is 0.15 meter (can be ignored). What is the required pipe for the lateral pipe? qs=120l/h hs=20m 0.15 m Q=1.2m3/h D = ? mm 10 m 95 m
Continue: n = 10 micro-sprinklers
length (L) = (9 sprinkler x 10 m) + 5 m = 95 meters F10 = 0.384
Continue: The maximum allowable in the field is as follows:
For a 20 mm polyethylene pipe grade 4 (ID mm), the hydraulic gradient found out of a slide ruler or monograph Q = 1.2 m3/h is J = 18.5%. 6.74 meters is exceeding the allowable 4 meters (20%)
Continue: The hydraulic gradient for a 25 mm P.E. pipe (ID 21.2 mm) and Q = 1.2 m3/h is J = 5.8%. The head loss of 2.1 meters is less than the allowable 4 m (20%). The maximum allowed head loss along the manifold is 4 m m = 1.9 meters.
Continue: The required pressure by the lateral inlet pipe is as follows:
Design an Irrigation System
Option 1 - The rule of 20% is applied to all the outlets (either sprinklers or drips) on the same subplot. Any excess pressure over 20% between the subplots is controlled by flow pressure regulators. P=< 20%
Lay out of drip line without pressure regulator
Drip line inlet without pressure regulator
Design an Irrigation System
Option 2 - The rule of 20% is applied to a single lateral, and pressure regulators control the pressure difference between the laterals. P =< 20%
Inlet with pressure regulator
Design an Irrigation System
Option 3 - The difference pressure along a lateral pipe exceeds the 20% head loss by any desired amount, and the excess pressure should be reduced by pressure or flow regulators in each emitters or sprinklers. P > 20%
Example: Ten micro-sprinklers are installed along a plastic lateral pipe (grade 4) at 10 m (32.8 ft) apart (the first sprinkler is 5 meters). The flow rate of the selected sprinkler is qs = 120 l/h (0.5 GPM), at a pressure of hs = 20 meters. The riser height is m (which can be ignored). What is the appropriate lateral pipe diameter, if the field is designed and abided by options 1, 2 and 3? n = 10, L = 95 m F10 =0.384
Continue: Option 1: For 20 mm - The hydraulic gradient For a 20 mm P.E. pipe and Q = 1.2 m3/h is J = 18.5%. 6.72 meters exceed the allowable 4 meters (20%)
Continue: For 25 mm - The hydraulic gradient For a 25 mm P.E. pipe and Q = 1.2 m3/h is J = 5.8%. The head loss difference = 1.89 m, which is available for the manifold head loss. The inlet lateral pressure is:
Continue: Option 2: If the allowable pressure variation along the lateral pipe is 4 meters, then 25 mm P.E. pipe is too much and 20 mm pipe too small. Therefore, a combination of the two pipes can be used. The design procedure for the combined lateral pipe is: Try first D = 25 mm along 35 meters (n = 4) and D = 20 mm along 60 meters (n = 6)
Continue: Compute the head loss for a pipe D = 25, L = 95 m, n = 10 and Q = 1.2 m3/h (from previous calculation which it was found 2.11 m) Compute the head loss for D = 25, L = 60 m, n = 6 and F6 = 0.458
Continue: From a table or a slide ruler - the hydraulic gradient for D = 25 mm and Q = 0.72 is J = 2.4%. The head loss for D = 25 mm and 35 meter long with four sprinklers is 2.11 m = 1.47 meter L = 35 m h = h95 – h60 L = 60 m h = 0.64 L = 95 m h = 2.11 mm
Continue: Compute the head loss in D = 20 mm, L = 60 meters, n = 6 and F6 = and From tables or a slide ruler the hydraulic gradient for D = 20 mm and Q = 0.72 is J = 7.6%.
Continue: The total head loss along the combined 25 and 20 mm lateral is as follows: Since 3.5 m is too less than 4.0 meters. Therefore, it is possible to try a shorter 25 mm pipe with a length of 25 m and n = 3 and a longer 20 mm diameter pipe along 70 m and n = 7. The previous procedure should be repeated. The new head loss is 4.5 meters, which exceeds the limit of 4 meters - (20% rule).
Continue: The lateral inlet pressure requirement is as follows:
Continue: Option 3: The lateral pipe is design either with flow or pressure regulators in every - sprinkler. The laterals diameter can be reduced to 20 mm or even further to 16 mm. In case of 20 mm diameter pipe, the head loss is meters (see Option 1). Therefore, the pressure requirement at the last lateral inlet is: The lateral inlet pressure - the entire head loss is added to the required sprinkler pressure.
Example: A manifold was installed along the center of a rectangular field (100 x 100 m). The lateral pipes were hooked up to the two sides of the manifold pipe. The difference in elevation between the center and the end of the field is 2 meters (either positive or negative). Each lateral pipe has eight l/hr micro-sprinklers at 6 meters apart and the pressure (hs) is 25 meters. What is the required diameter of the lateral pipess, if the system is designed and abides by option 1?
Answer: The lateral's head loss along the two sides of the manifold should be close enough (in away that the total head loss due to the difference in elevation and friction on both sides of the manifold should be almost the same). 6m 100 m 4% slope
Continue: The maximum head loss between the sprinklers throughout the field is: For a 20 mm (ID = 16.6 mm) lateral pipe on the two sides: n = 8 F8 = 0.394 L = (7 sprinklers x 6 m) + 3 m = 45 m
Continue: The hydraulic gradient for Q = 0.96 m3/hr and D = 20 mm (ID = 16.6 mm) is J = % The inlet pressure on the downward slope lateral is:
Continue: The pressure by the last sprinkler is
The pressure difference between the two ends is
Continue: The inlet pressure by the lateral upward is:
The pressure by the last sprinkler is The head loss along the upward lateral is 27.65 m m = 4.25 m, which is less than 5 m - 20% rule
Continue: The pressure requirement for the upward laterals inlet is m and for downward laterals inlet is only 25.6 m. The head loss along the upward laterals is 4.25 m, almost all the permitted 20% (5 m). Therefore either: the upward lateral will be increased to 25 mm or more, the manifold can be reallocated to a higher position. (or pressure regulators should be installed by the lateral inlets,)
Continue: When 20 mm lateral pipes are in used, the values of hu for both sides of the manifold vary by = 2.05 m . To avoid this difference (hu) in the pressure, the upward 20 mm laterals can be replaced by 25 mm. The inlet pressure (hu) for 25 mm is 26.5 m. Therefore, the the difference inlet pressure for both sides of the manifold will be less, only = 0.8 m. Less expensive alternative is by reallocating the manifold away from the center of the field to a higher point. That way, 6 sprinklers will be on the upward side and 10 sprinklers on the downward laterals.
Continue: Downward laterals:
D = 20 mm n = 10 L = (9 sprinkler x 6 m) + 3 m = 57 m Q = 1.2 m3/hr F10 = J = 18.5%
Continue: The difference elevation is as follows:
The pressure at the lateral inlet is as follows:
Continue: The pressure head by the last lateral sprinkler is
The head loss along the downward lateral is
Continue: Upward laterals:
n = 6 F6 = L = (5 sprinklers x 6) + 3 = 33 m Q = 0.72 m3/hr D=20 mm (ID=16.6mm) J = 7.6% The elevation difference is for a slope of 4% is:
Continue: The pressure head at the lateral inlet is:
The pressure head at the last lateral sprinkler is
Continue: The head loss along the upward lateral is
The values of hu for both sides are 27.1 m and m which is practically the same. The maximum head loss is 2.3 m, so 2.7 meters are available as a head loss for the manifold.
Design of a manifold pipe
The manifold is a pipe with multiple outlets with the same space between the outlets, therefore the manifold is designed the same way as a lateral.
Example: A fruit tree plot (96 x 96 m) is designed for irrigation with a solid set system. A manifold is laid throughout the center of the field. The whole plot is irrigated simultaneously. The flow rate of the selected micro-sprinkler is qs = 0.11 m3/hr at a pressure (hs) of 2.0 atmosphere. The space between the micro-sprinklers along the lateral is 8 meters (26.24 ft.) and between the laterals is 6 meters (19.68 ft.). What is the required diameter of the pipes? (The local head loss is 10% of the total head loss and is taken in account).
Answer: qs=.11m3/h Ps=20m lateral 6 m 8m 96 m manifold
Continue: The maximum allowable pressure head variation is
The number of micro-sprinkler on every lateral is
Continue: F6= 0.405 L = (5 sprinklers x 8 m) + 4 m = 44 m
Q = 0.11 m3/hr x 6 sprinklers = 0.66 m3/hr For 16 mm P.E. lateral pipe (ID = 12.8 mm) The hydraulic gradient for 16 mm P.E. pipe (ID=12.8mm) and Q = 0.66 m3/hr is J = 22.3%
Continue: The head loss along 16 mm lateral pipe (including 10% local head loss) is as follows: 4.36 m head loss exceeds the allowable 4 meter (20%). So we have to try the head loss for 20 mm P.E. lateral pipe.
Continue: The hydraulic gradient for 20 mm P.E. pipe (ID = 16mm) and Q = 0.66 m3/hr is J = 6.5% The head loss in 20-mm lateral pipe (including 10 local head loss) is
Continue: 1.3 m head loss is less than 4 m (20%) and can be selected as a lateral. The lateral inlet pressure is: The water pressure at the last micro-sprinkler on the lateral pipe is, as follows:
Continue: Manifold Design: The number of laterals is
F32 = L = (31 laterals x 6 m) + 3 = 93 m Q = 32 x 0.66 = 21.1 m3/hr
Continue: The hydraulic gradient for 63 mm P.E. pipe (ID = mm) and Q = 21.1 m3/hr is J = 7.2% The head loss in 63 mm P.E. pipe (including 10% local head loss) is as follows:
Continue: The pressure by the manifold inlet is as follows:
The pressure by the last lateral inlet is as follows: The maximum pressure in the entire plot is at the first lateral inlet meters (2.3 atmosphere)
Continue: The minimum pressure throughout the system is at the last sprinkler on the last lateral, which is as follows: = m The pressure difference between the first and last sprinkler is as follows: = 4.06 m (i.e. just above 4 meters (20%))
Distribution of water and pressure
Distribution of water and pressure
manifold p max. p min <20% haverage q max qaverage q min <10% main pipe lateral
Designing of Irrigation System
Considerations: soil, topography, water supply and quality, kind of crops, climate. Soil – infiltration rate, field capacity, (the lighter the soil is - a higher advantage to drip system). המטרה Topography – the steeper the terrain - a higher advantage to drip system. Water supply – availability (time), pressure and quantity Water quality – salinity (chlorine, SAR, B, heavy metal or any other toxic), hardness, Fe, Mn, total suspended material and type. Crop – as the root system shallower a higher advantage to micro irrigation system (closing spacing). Price – as the expected income is relative higher - a better water distribution system is an advantage. Crop – Layout of the crop and type. Climate – evaporation, wind pattern, crop protection (high or low temperature)
Continue: Farm schedule. Water application: Working time.
Crop related activity – such as chemical application, harvesting, weeds control and so on. Water application: Estimate water application depth at each irrigation cycle. Determine the peak period of daily water consumption. Determine the frequency of water supply.
Continue: Irrigation system:
Consider several alternative types of irrigation systems. Determine the sprinklers or emitters spacing, discharge, nozzle sizes, water pressure. המטרה טפטוף Determine the minimum number of sprinklers or emitters (or a size of subplot) which must be operated simultaneously.
Continue: Irrigation layout:
Divide the field into sub-plots according to the crops, availability of water and number of shifts (in one complete irrigation cycle). Determine the best layout of main and laterals. Determine the required lateral size. Determine the size of a main pipe. Select a pump.
Continue Prepare plans, schedules, and instructions for a proper layout and operation. Prepare a schematic diagram for each set of sub-mains or manifolds which can operate simultaneously. Prepare a diagram to show the discharge, pressure requirement, elevation and pipe length. Select appropriate pipes, starting at the downstream end and ending up by the water source.
Combination of pipes The total head loss along 300 meters PVC (grade 6) pipe is 15 m, with a flow rate of 180 m3/h. Which size of pipes are required? Q=180 m3/h 300 m h=15m
Answer The hydraulic gradient for 160 mm (ID 150.2 mm) is: 3.4%
Therefore, the head loss for 300 m long pipe is: 10.2 m (too big pipe) The hydraulic gradient for 140 mm (ID mm) is: 6.4% Therefore, the head loss for 300 m long pipe is: 19.2 m (too small pipe). Therefore, a combination of the two can make it.
Cont. Q=150 m3/h 300 m L 300-L 140 m 160 mm PVC pipe m 140 mm PVC pipe
Example: A flat field with two plots, each plot is divided into six subplots. The selected system for this field is drip irrigation. The flow rate in each subplot is 21 m3/hr and the pressure requirement to the sub- main inlets is 25 meters. The interval of water supply is every three days and only one shift a day. Therefore, two subplots in each plot must be irrigated simultaneously. The main pipes are made of PVC (C = 150) and are buried 0.6 meters deep. The local head losses is up to 10% of the longitudinal head losses. The pump pressure is 50 meters and with a flow rate of 84m3/h (local head loss due primary filter and others pump attached accessories is 10m).
Continue: 1 3 5 1’ 3’ 5’ 50m 96m 96m D C B A 2 4 6 250m E C’ B’ A’
50m 96m 96m D C B A 250m 1’ ’ 5’ E C’ B’ A’ 120m 2’ ’ 50m 96m 96m Q=84m3/h F Pump
Pressure requirement Qu=21m3 Hu=25 m Qu=84m3 Hu=40m
Continue: The sequence of water application is as follows:
First day - 1, 5, 1' and 5' plots Second day - 2, 6, 2' and 6' plots Third day - 3, 4, 3' and 4' plots
Continue: The diagram for the first and second day of water supply is: D Q=21m3/h hu=25.6m 2Q=42m3/h C B A L=50m L=192m 250m 1’ ’ 5’ Q=21m3/h E 2Q=42m3/h C’ B’ A’ 120m 2’ ’ L=50m L=192 F Pump (Q=84m3/H= 50m)
Continue: The diagram for the third day of water supply is: hu=25.6m D Q=0m3/h 2Q=42m3/h C B A L=146m 250m 1’ ’ 5’ Q=0m3/h E 2Q=42m3/h C’ B’ A’ 2’ ’ 120m L=146m F Pump (Q=84m3/h, 40m)
Main pipes’ diagram for first and second day
84 m3/h 300 m 42 m3/h 192m 21m3/h
Main pipes’ diagram for third day (Case 2)
B F 120 m 84 m3/h 396 m 42 m3/h
Continue: The following table presents the the head loss (including 10%) for local head loss for selected pipe:
Continue: Case 1: Design system for the first and second day.
hu m (including the depth of the main pipe 0.6 m) pump pressure - 40 m Total head loss = = 14.4 m Head loss for the selected pipe: A-C 192m Q = 21 m3/hr 3" pipe = 4.5 m C-D 50m Q = 42 m3/hr 4" pipe = 1m D-E 250m Q = 42 m3/hr 4" pipe = 5.2 m The pressure by E is : = 36.3 m. The head loss available for E-F = 40 – 36.3 = 3.7 m
Pressure diagram for case 1
Continue: For E - F (pump) section:
4" pipe is too small (8.8 m head loss, 6.7%), on the other hand 5" pipe is too much (3 m head loss, 2.3%). Therefore, a combination of the two is selected for E-F section. L (4") = 13.7 m L (5“) = 120 – 13.7 = m
Case 2 Hu-25.6 B D 146m 42 m3 250m 42 m3 E F Hu-40m
Continue Case 2: Design system for the third day:
hu – 25.6 m (including the depth of a main pipe 0.6 m) After pump pressure - 40 m Maximum head loss = 40 – 25.6 = 14.4 m Selected pipes – B-D Q = 42 m3/hr 4" pipe = 3 m D-E Q = 42 m3/hr 4" pipe = 5.2 m The pressure head by E tee is =33.8 m. The head loss available for E-F = 40 – 33.8 = 6.2 m
Case 2 42 m3 H-28.6m Hu-25.6 B D 4” 146m 4” 250m H-33.8m E F Hu-40m
Continue B D E A combination of pipes 3” and 4” for DB section with
The pressure in case 2 at point E is 33.8 m which is lower than in case 2 which was 36.3 m (pressure different of 2.5 m). . Therefore, the pipe for DB should be reduced, in a way that the pressure at a point D should be a same as case 1 which is m. 42 m3 H-31.1m Hu-25.6 B D 146m 250m E A combination of pipes 3” and 4” for DB section with 5.6 m head loss should be selected.
Continue Section DB: the hydraulic gradient for 3” pipe with 42 m3 is 7.5% and for 4” is 1.9% 3” pipe is = m and for 4” is m (The actual length should take in account the commercial pipe length)
Case 2 H-31.2m Hu-25.6 4” 3” B D 106.25m 39.75 m 42 m3 42 m3 4” 250m E In this situation the size of pipes DE and EF are a same as in case 1. In case 2, the required pipes are smaller than case 1 (DB). (The selected pipe in case 1 are too small to maintain the pressure requirement in case 2)
Pipe diagram for case 1 & 2 D C B A E F Hu-25.6m 4” 106.75m 3” 135.25m
Head loss in case 1 For section A – C (192m 21m3/h):
For section C – D (42m3/h) 4” 50 m m For section D – E (42 m3/h): 4” 250 m m For section E – F (120m 84m3/h): 4” 13.7 m m 5: m m The total head loss from A to F is m
Pressure diagram for case 1
H-30.1 H-29.1m 25.6m D 4” 104.6m 3” 137.4m C B A 4”250m E H-35.3m H-39m F
Head loss in case 2 For section D – B (1146m 42m3/h):
For section D – E (42 m3/h): 4” 250 m m For section E – F (120m 84m3/h): 4” 13.7 m m 5”: m m The total head loss from A to F is m
Pressure diagram for case 2
H-31.1 H-28.9m 25.6m 4” m D 3” 39.75m C B 4”250m E H-36.3m H-40m צריך להוסיף מפה דומה ל- Case 2 F
Diagram for E – A’ case 2 D C B A E B’ F H-31.1 H-28.9m 25.6m
The pipe for E – A’ for case 1
42m3/h C’ A’ Q – 21m3/h H-35.3m H-25.6m 50 m 192 m The head loss E – A’: 35.3 – 25.6 = 9.7 m The length of E – C’ is 50 m 42 m3/h for 3” pipe the head loss is 4.1 m The length of C’ A’ is 192 m 21m3/h the required head loss for this section is: 9.7 – 4.1 = 5.6m The hydraulic gradient for 2” is 15% and for 3” is 2.1% and combination of the two will make it.
Continue The hydraulic gradient for 2” is 15% and for 3” is 2.1% and combination of the two will make it. 8.2 m 2” pipe, and m 3” pipe. The head loss for E-C’ 3” pipe (42 m3/h is 4.1 m. The total head loss for E-A’ is 9.7 m. E C’ A’ 42m3/h Q – 21m3/h 2” 8.2m 3” 233.8 m
The pipe for E – B’ for case 2
42m3/h B’ H-36.3m H-25.6m 146 m The head loss E – B’: 36.3 – 25.6 = 10.7 m The length of E – B’ is 146 m 42 m3/h The head loss for 3” is too much 12.1 m and for 4” is too small 3 m A combination of the two will make it, the hydraulic gradient for 3” is 7.5% and for 4” is 1.9%.
The combination pipes for E – B’
3” pipe m 4” pipe m
Diagram for E – A’ for case 2
H-30.1 H-29.1m 25.6m 4” 104.6m D 3” 137.4m C B A 4”250m 25.6m B’ E H-35.3m A’ 3” 211.9m 4” 21.9m 2” 8.2m H-39m F Since the size of the pipe E-A’ increased, therefore we have to reconsider the pipe size for EA’ in case 1.
Section E-A’ case 1 E A’ The head loss E – B’: 35.3 – 25.6 = 9.7 m
3” 211.9m H-35.3m 4” 21.9m 2” 8.2m The head loss E – B’: 35.3 – 25.6 = 9.7 m The length of E – C’ is 50 m 42 m3/h The head loss for 4” 21.9 m is 0.5 m The head loss for 3” m is 2.3 m The length of C’ – A’ is 192 m 21 m3/h The head loss for 3” m is 4.3 m The head loss for 2” 8.2 m is 1.4 m The total head loss is = 8.5 m Therefore, the length of 2” pipe can be extended a little bit by (to increase the head loss): 9.7 – 8.5 = 1.2 m:
Continue The length of B – A is 96 m with 21 m3/h
The current head loss is: for 3” m is m for 2” 8.2 m is m The total head loss should be increased by 1.2 m = 6.9 m The hydraulic gradient for 2” and 21m3/h is 15% and for 3” pipe is 2.1% The length of 2” pipe is 17.4 m and 3” pipe is m
Final Diagram for E – A’ D C B A C’ E A’ F
H-30.1 H-29.1m 25.6m D 4” 104.6m 3” 137.4m C B A 4”250m C’ 25.6m B’ E H-35.3m A’ 3” 202.7m 4” 21.9m 2” 17.4m 4” 5” H-39m F To overcome the pressure differences, pressure regulator should be installed by every risers.
To overcome the pressure differences, pressure regulator should be installed by every risers.
צריך להוסיף את בחינה של גודל צינורות E-A’ עבור case2
אשר נדרש לקטרים אחרים
Case 2 H-31.2m Hu-25.6 4” 3” B D 104.6m 41.4 m 42 m3 250m E
Continue The pipe selection for F - A is as follows: A-D 3" 242 meters
D-E 4" 250 meters E-F 6" 120 meters
Continue The pipe selection for the section E-A' for case 1 is as follows: Hu=38m C’ B’ A’ E Hu=25.6 50 m 42m3/h 192m 21m3/h
Continue The head loss is 38.0 - 25.6 = 12.4 m Selection of pipes -
A'-C' Q = 21 m3/hr 2" pipe = 7.7 m The head loss available for E-C' = 4.7 m
Continue 2" pipe is too small (6.5 m), on the other hand 3" pipe is too much (2.4 m). Therefore, a combination of the two pipes is required for E- C' section. The two pipes which required for E- C section are: L = 33 m L(3") = 33 meters and 2" pipe 27 meters
Continue The design for the section E-A' for case 2 is as follows: A’
H=38m A’ B’ C’ E 146m 42m3/h QEB' = 42 m3/hr 2" pipe = 7.7 m The hydraulic gradient for E-B’ section is = 12.4 m
Continue 2" pipe is too small (18.9 m), on the other hand 3" pipe is too much (6.9m) Therefore, a mix of the two is required for F-B', section. L = 95 L (3") = 95 m and 2" pipe 51 m
Continue: The selected pipe for E-B' section will be as in case 2, 95 meters 3" pipe and 51 meters 2" pipe. Pressure regulators on some of the risers should be considered to compensate the total head loss over 20% (including the head loss inside each plot).
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• Sep 13th 2012, 01:14 PM
onaslowtrain
light intensity equation
I have an equation I need help solving that deals with light intensity.
Ez = E0 * e(-k*z) is the formula with all the variables.
The equation I need to solve is Ez = 25,000 lux * e (.12 m^-1 * 5m) where (e) is exponential function.
Thanks to all.
• Sep 13th 2012, 01:41 PM
skeeter
Re: light intensity equation
note that ...
$0.12m^{-1} \cdot 5m = 0.6$
• Sep 13th 2012, 02:17 PM
onaslowtrain
Re: light intensity equation
given that light intensity declines exponentially with depth, how does that work with the above equation. do i take e^.6 m and subtract that from 25,000 lux?
• Sep 13th 2012, 04:52 PM
Wilmer
Re: light intensity equation
Quote:
Originally Posted by onaslowtrain
given that light intensity declines exponentially with depth, how does that work with the above equation. do i take e^.6 m and subtract that from 25,000 lux?
Huh? 25000 * e^.6 = ~45553 | 290 | 963 | {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-34 | longest | en | 0.901028 |
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// zNumbers | 742 | 2,637 | {"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} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.917466 |
https://mathematica.stackexchange.com/questions/80951/how-to-properly-use-conditionals-inside-ndsolve | 1,627,610,597,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153899.14/warc/CC-MAIN-20210729234313-20210730024313-00424.warc.gz | 410,112,079 | 38,625 | # How to properly use conditionals inside NDSolve
Currently I am working with Mathematica to learn the program, and I'm confused what exactly is wrong with this code.
s = NDSolve[{if[t < 1, Derivative[1][Ca][T] == -10*Ca[T] + 2 T,
Derivative[1][Ca] == -10*Ca[T]],
Derivative[1][Cb][T] == (10*Ca[T]) - 0.192, Ca[0] == 0,
Cb[0] == 0}, {Ca[T], Cb[T]}, {T, 0, 10}]
What I'm trying to do is set a conditional for the Ca derivative so that if T < 1 then it goes in the first derivative, but if not, it'll go in the second derivative.
Can anyone help me with this? Thanks.
• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – Michael E2 Apr 26 '15 at 23:52
• You can format inline code and code blocks by selecting it and clicking the {} button above the edit window. The edit window help button ? is also useful for learning how to format your questions and answers. – Michael E2 Apr 26 '15 at 23:53
• Please note that precise and accurate syntax is essential for any programming language. In Mathematica, built-in symbols and functions start with capital letters, such as If. It is better to avoid starting your own variable and function names with a capital (i.e. avoid T, but especially built-in functions like D and N). – Michael E2 Apr 27 '15 at 0:03
Use Piecewise for discontinuous right-hand sides and coefficients. If, with a capital I, more a programming construct than an algebraic/functional one. NDSolve does discontinuity processing, which improves error estimation and step size when done accurately; using Piecewise helps with that.
s = NDSolve[{ | 512 | 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": 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} | 3.15625 | 3 | CC-MAIN-2021-31 | latest | en | 0.916177 |
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# PHYSICS C: MECHANICS
SECTION I
Time45 minutes
35 Questions
Directions: Each of the questions or incomplete statements below is followed by five suggested answers or
completions. Select the one that is best in each case and place the letter of your choice in the corresponding box on
## Note: To simplify calculations, you may use g = 10 m/ s2 in all problems.
1. Which of the following graphs of position d 2. A ball is thrown straight up from a point 2 m
versus time t corresponds to motion of an object above the ground. The ball reaches a maximum
in a straight line with positive acceleration? height of 3 m above its starting point and then
falls 5 m to the ground. When the ball strikes the
(A) ground, what is its displacement from its starting
point?
(A) Zero
(B) 8 m below
(C) 5 m below
(D) 2 m below
(E) 3 m above
(B)
3. What do acceleration and velocity have in
common?
(A) Both are scalars.
(B) Both are vectors.
(C) Both are measured in units of distance
divided by time.
(D) Both are measured in units of distance
(C) divided by time squared.
(E) They are different names for the same
quantity.
## 4. Two projectiles are launched with the same initial
speed from the same location, one at a 30 angle
and the other at a 60 angle with the horizontal.
(D) They land at the same height at which they were
launched. If air resistance is negligible, how do
the projectiles respective maximum heights, H30
and H60 , and times in the air, T30 and T60 ,
compare with each other?
Maximum Height Time in Air
## (E) (A) H30 > H60 T30 > T60
(B) H30 > H60 T30 < T60
(C) H30 = H60 T30 = T60
(D) H30 < H60 T30 > T60
(E) H30 < H60 T30 < T60
## GO ON TO THE NEXT PAGE.
-3-
5. An object of mass 100 kg is initially at rest on a Questions 7-8
horizontal frictionless surface. At time t = 0, a
horizontal force of 10 N is applied to the object
for 1 s and then removed. Which of the following
is true of the object at time t = 2 s if it is still on
the surface?
(A) It is at the same position it had at t = 0, since
a force of 10 N is not large enough to move
such a massive object.
(B) It is moving with constant nonzero
acceleration.
(C) It is moving with decreasing acceleration. A rock is thrown from the edge of a cliff with an
(D) It is moving at a constant speed. initial velocity u0 at an angle q with the horizontal
(E) It has come to rest some distance away from as shown above. Point P is the highest point in the
the position it had at t = 0. rocks trajectory and point Q is level with the starting
point. Assume air resistance is negligible.
6. Several forces act on an object, but the object is in
equilibrium. Which of the following statements 7. Which of the following correctly describes the
about the object must be true? horizontal and vertical speeds and the acceleration
I. It has zero acceleration. of the rock at point P ?
II. The net force acting on it is zero. Horizontal Vertical
III. It is at rest. Speed Speed Acceleration
IV. It is moving with constant velocity.
(A) u0 cos q 0 g
(A) I and II
(B) I and III (B) 0 0 g
(C) I and IV (C) u0 cos q 0 0
(D) II and III (D) u0 cos q u0 sin q g
(E) II and IV
(E) 0 u0 cos q 0
## 8. Which of the following correctly describes the
horizontal and vertical speeds and the acceleration
of the rock at point Q ?
Horizontal Vertical
Speed Speed Acceleration
(A) u0 cos q 0 g
(B) 0 0 g
(C) u0 cos q 0 0
(D) u0 cos q u0 sin q g
(E) 0 u0 cos q 0
-4- | 973 | 3,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-10 | latest | en | 0.912486 |
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# MGMAT SC vs Powerscore Critical Reasoning Bible
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MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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23 Apr 2012, 15:19
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hi All,
ive currently reached the CR section of the MGMAT strategy guides and have a query. i also have the Powerscore Critical Reasoning bible (heard very good reviews).
I was wondering if i should be choosing one of them to start preparing. Im currently revising the basics and my CR is at best 60%-70% on most of my tests.
which one would you recommend? Should i use both one after another or just pick one of them?
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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23 Apr 2012, 17:43
We all know that powerscore CR is the best bet, i would recommend you to go for it.
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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25 Apr 2012, 23:23
Powerscore CR is good but it is very time consuming . If you don't have time just go through the MGMAT CR book and the OG questions
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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27 Apr 2012, 08:33
If you have sometime, try PowerScore. It definitely helped me. My accuracy was ridiculously low on CRs. I feel drastic improvement in the way I approach CR questions after only 1 read of the CR Bible. (You may find it a bit dry in the beginning, but dont give up) Best of Luck.
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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27 Apr 2012, 08:35
See this one too: critical-reasoning-guides-manhattan-vs-powerscore-67389.html?fl=similar
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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27 Apr 2012, 13:00
^^ Thanks for the link BB
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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28 Apr 2012, 02:47
Power Score bible has worked miracles for my hit rate
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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28 Apr 2012, 13:11
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Although I'm a tad late for this discussion and my opinion mostly repeats what others have stated, the Powerscore CR Bible is fantastic. In fact, I would say it's my favorite GMAT guidebook.
While some may note that it's a large book, that's because the book is so comprehensive. If you are running low on time, then my suggestion is to somehow make time. It's worth it!
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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28 Apr 2012, 15:12
Hey!! Thanks a lot everyone!
I think I will start with the powerscore CR then. BTW, I've heard that there are a lot of changes to the MGMAT CR book in the 5th version as compared to 4th ver.
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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03 Jul 2012, 22:47
I think powerscore CR and MGMAT CR are almost close (I have both of them), but powerscore CR holds edge.
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink]
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03 Jul 2012, 23:02
I went through the 5th edition MGMAT CR. I don't see many changes. They have done away with those T-diagrams but the focus is still a lot on diagramming
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Re: MGMAT SC vs Powerscore Critical Reasoning Bible [#permalink] 03 Jul 2012, 23:02
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# MGMAT SC vs Powerscore Critical Reasoning Bible
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File Name: parallel circuits problems and answers .zip
Size: 16919Kb
Published: 10.05.2021
In this circuit, three resistors receive the same amount of voltage 24 volts from a single source. The answers to this question may seem paradoxical to students: the lowest value of resistor dissipates the greatest power. Math does not lie, though.
Solved examples with detailed answer description, explanation are given and it would be easy to understand. Here you can find objective type Electronics Series-Parallel Circuits questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided.
## Series And Parallel Circuits Worksheet Answer Key Pdf
Refer to Figure 5 A. The terms series circuit and parallel circuit are sometimes used, but only the simplest of circuits are entirely one type or the other. About the worksheets This booklet contains the worksheets that you will be using in the discussion section of your course. Select a circuit. For this sentence the form of the words is present tense verb. Key Stage 3 worksheet on series and parallel circuits.
Total power in a parallel circuit is the sum of the power consumed on the individual branches. Chapter 07 Series-Parallel Circuits Source And now c we are left with R in parallel with R 3. What would happen if they were wired in series? Some circuits include both series and parallel parts. Identify a series and parallel circuit, state the rules for series circuits, apply the rules to a circuit and calculate resistance, explain why and apply to more complex circuits.
## Parallel DC Circuits Practice Worksheet With Answers
The total. Read Free Series And Parallel Circuits Problems Answers solving series parallel circuits solving series parallel circuits by Ron Call 7 years ago 8 minutes, 3 seconds , views solving , series parallel , combination , circuits , for electronics, to find resistances, voltage drops, and currents. The most important thing to keep in mind in such calculations is that resistors in series carry exactly the same current and that resistors in parallel have exactly the same voltage across them. A simple circuit is one in which there is a single voltage source and a single resistance. Representing most real world circuits, these circuits are connected in series as well as in parallel.
Determine the total voltage (electric potential) for each of the following circuits below. 13V In a parallel circuit, there is more than one loop or Questions 6 and 7 refer to the following: The diagram to the right represents an electric circuit.
## Electronics - Series-Parallel Circuits
The meter movement itself can handle up to 1 mA. The currents into a junction flow along two paths. One current is 4 A and the other is 3 A.
Two elements are in parallel if they are connected between the same pair of notes. If each element is in parallel with every other element, it is called a parallel circuit. Example 2: For the following circuit, find the total resister value.
### Simple Series And Parallel Circuits Problems And Answers Pdf
Solved examples with detailed answer description, explanation are given and it would be easy to understand. Here you can find objective type Electronics Parallel Circuits questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided. You can easily solve all kind of Electronics questions based on Parallel Circuits by practicing the objective type exercises given below, also get shortcut methods to solve Electronics Parallel Circuits problems. What is the total power loss if 2 k and 1 k parallel-connected resistors have an I T of 3 mA?
The rate of increase in reliability with each additional component decreases as the number of components increases. You will discuss with your lab partner ways of connecting a circuit to measure the voltage of the battery and light the light bulbs by connecting different types of circuits. Find the total resistance! This circuit is neither simple series nor simple parallel. The circuit elements cannot be disconnected from each other, and other circuit components can only be connected at points A and B. This physics video tutorial explains how to solve series and parallel circuits. Take the real part of the solution as your answer at the end.
Answer: D. Justification: When resistors are in series, we simply add up the resistances to get the total resistance. When resistors are in parallel however, we add.
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In this circuit, three resistors receive the same amount of voltage 24 volts from a single source. The answers to this question may seem paradoxical to students: the lowest value of resistor dissipates the greatest power. Math does not lie, though. Challenge your students to recognize any mathematical patterns in the respective currents and power dissipations. What will happen to the brightness of the light bulb if the switch in this circuit is suddenly closed? This question illustrates a disparity between the ideal conditions generally assumed for theoretical calculations, and those conditions encountered in real life.
Differentiate between creepers and climbers.
In this circuit, three resistors receive the same amount of voltage 24 volts from a single source. The answers to this question may seem paradoxical to students: the lowest value of resistor dissipates the greatest power. Math does not lie, though. Challenge your students to recognize any mathematical patterns in the respective currents and power dissipations. What will happen to the brightness of the light bulb if the switch in this circuit is suddenly closed?
Some of the worksheets for this concept are Kindle file format series and parallel, Series and parallel circuits, Series and parallel circuits work, Series parallel circuits, 9 14 work, Series and parallel circuits, Electricity unit, Electrical circuits 11 14 student revision notes. Vout is unconnected. Now, you can use your VIRP table to analyze the circuit.
На лице привратника появилась обиженная гримаса, словно Беккер чем-то его оскорбил. - Рог aqui, senor. - Он проводил Беккера в фойе, показал, где находится консьерж, и поспешил исчезнуть.
### Related Posts
3 Response
1. Agnelo P.
Related Problem Ifa resistor is connected from the bottom end ofR3 to the top end of R5 in Figure Solution Note that R2and R3 are in parallel in this circuit.
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Programming language concepts paradigms and models pdf madina book 1 pdf english
3. Rochelle M.
Parallel DC Circuits Practice Worksheet With Answers. Basic Electricity. PDF Version. Question 1. In this circuit, three resistors. | 1,368 | 6,740 | {"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} | 3.828125 | 4 | CC-MAIN-2022-21 | latest | en | 0.893574 |
https://joningram.org/questions/Trigonometry/989990 | 1,709,532,395,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00692.warc.gz | 325,716,302 | 4,344 | # Convert to a Simplified Fraction square root of 64
Convert to a Simplified Fraction square root of 64
Rewrite as .
Pull terms out from under the radical, assuming positive real numbers.
Do you know how to Convert to a Simplified Fraction square root of 64? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.
### Name
Name one billion four hundred eleven million one hundred ninety-five thousand two hundred six
### Interesting facts
• 1411195206 has 16 divisors, whose sum is 2913435648
• The reverse of 1411195206 is 6025911141
• Previous prime number is 31
### Basic properties
• Is Prime? no
• Number parity even
• Number length 10
• Sum of Digits 30
• Digital Root 3
### Name
Name five hundred eighty-three million five hundred thousand three hundred thirty-four
### Interesting facts
• 583500334 has 4 divisors, whose sum is 875250504
• The reverse of 583500334 is 433005385
• Previous prime number is 2
### Basic properties
• Is Prime? no
• Number parity even
• Number length 9
• Sum of Digits 31
• Digital Root 4
### Name
Name one billion three hundred ninety-four million five hundred fifty-three thousand eight hundred seventy-one
### Interesting facts
• 1394553871 has 8 divisors, whose sum is 1457748864
• The reverse of 1394553871 is 1783554931
• Previous prime number is 571
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 10
• Sum of Digits 46
• Digital Root 1 | 392 | 1,495 | {"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-2024-10 | latest | en | 0.763342 |
https://paperzz.com/doc/9407927/area-circumference-circles-blank | 1,660,176,581,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571222.74/warc/CC-MAIN-20220810222056-20220811012056-00118.warc.gz | 413,453,113 | 5,693 | ### Area Circumference Circles blank
```Name: ________________
Date: _________
Aim: How do we find the area & circumference of a circle?
Do Now: Define the following and label on the circle.
Diameter___________________________________
Circumference_______________________________
1. The diameter of a circle is 8 inches.
2. The radius of a circle is 15 cm.
Find the diameter.
Geo
Complete the following chart using the information from the class circleactivity:
Diameter Circumference
Diameter _______ = Circumference
50
40
30
20
10
Circumference formulas :
π = _________
Practice:
Find the circumference of a circle when :
1. the radius is 4, in terms of π .
2. the diameter is 14,
to the nearest tenth.
3. Find the radius of a circle, to the
nearest tenth, when the
circumference is 40 .
4. Find the diameter of a circle when
the circumference is 20 π .
Area of a Circle
Area
Area of
Rectangle
Substitute
r and C
Plug in
Circumference
Formula
Solve
Area of a Circle Formula
Practice:
Find the area of a circle when...
4. the radius is 4, to the nearest tenth.
6. the diameter is 12, to the
nearest hundredth.
8. If the circumference is 12π,
5. the diameter is 14, in terms of π.
7. the radius is 12, in terms of π.
9. If the area is 64π, find the diameter.
11. If the circumference is 10π, find the area, in terms of π.
12. If the area is 9π, find the circumference, in terms of π. 13. The side of a square is 20 cm long. What is the circumference of the circle when …
a. the circle is inscribed within the square?
b. the square is inscribed within the circle?
14. A designer needs to create prefectly circular necklaces. The necklaces each need
to have a radius of 8 inches. What is the largest number of necklaces that can be
made from 2000 inches of wire?
15. The stained glass window below is a semi circle.
To the nearest tenth, what is the distance around the window?
16. The cost of painting the circular traffic sign shown below is \$2.75 per
square foot. How much, to the nearest cent, will it cost to paint the sign if its
diameter is 30 inches?
Sum it Up
• The area of a circle is πr .
• The circumference of a circle is πd or 2πr.
2
``` | 580 | 2,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-33 | latest | en | 0.815897 |
https://cybergeeksquad.co/2021/10/longest-and-subarray-codechef-oct.html | 1,702,121,479,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100909.82/warc/CC-MAIN-20231209103523-20231209133523-00628.warc.gz | 219,520,977 | 51,068 | # October Long Challenge Longest AND Subarray ANDSUBAR
## Longest AND Subarray Solution
You are given an integer N. Consider the sequence containing the integers 1,2,…,N in increasing order (each exactly once). Find the length of the longest subarray in this sequence such that the bitwise AND of all elements in the subarray is positive.
Input Format
• The first line contains T denoting the number of test cases. Then the test cases follow.
• Each test case contains a single integer N on a single line.
Output Format
For each test case, output on a single line the length of the longest subarray that satisfy the given property.
Constraints
• 1≤T≤105
• 1≤N≤109
• Subtask 1 (100 points): Original constraints
Sample Input 1
``````5
1
2
3
4
7``````
Sample Output 1
``````1
1
2
2
4``````
Explanation
Test case 1: The only possible subarray we can choose is [1].
Test case 2: We can’t take the entire sequence [1,2] as a subarray because the bitwise AND of 1 and 2 is zero. We can take either [1] or [2] as a subarray.
Test case 3: It is optimal to take the subarray [2,3] and the bitwise AND of 2 and 3 is 2.
Test case 4: It is optimal to take the subarray [4,5,6,7] and the bitwise AND of all integers in this subarray is 4.
### SOLUTION
Program Python: Longest AND Subarray Solution in Python
``````for _ in range(int(input())):
n=int(input())
if n==1:
print(1)
continue
tmp=1
while(tmp*2<=n):
tmp*=2
st=n-tmp+1
if n==tmp:
print(tmp//2)
else:
print(max(st,tmp//2))`````` | 458 | 1,493 | {"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-2023-50 | latest | en | 0.488491 |
http://mathequalslove.blogspot.ca/2013/06/literal-equations-scavenger-hunt.html | 1,500,850,097,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424623.68/warc/CC-MAIN-20170723222438-20170724002438-00294.warc.gz | 209,505,932 | 27,618 | Math = Love: Literal Equations Scavenger Hunt
## Wednesday, June 5, 2013
### Literal Equations Scavenger Hunt
I was browsing through all of the pictures I took over my first year of teaching, and I realized that I had never posted about the scavenger hunt I did with my Algebra 1 students. During our unit on linear equation, my Algebra 1 students really struggled with converting equations from standard form to slope-intercept form. So, I decided to save our unit on systems of equations until later in the school year. After covering exponent rules, polynomials, absolute value, and factoring quadratics, it was finally time to jump into systems. I had also skipped the literal equations section from the first chapter of our textbook.
We ended up spending half a week on literal equations. I emphasized both solving famous formulas for a specific variable and converting an equation in standard form to slope-intercept form. Instead of giving a test over this mini-unit, I gave a short, two-question quiz over rearranging formulas for a specific variable, and we did a scavenger hunt to demonstrate mastery over converting equations between various forms.
To create the scavenger hunt, I folded eight sheets of colored paper in half. On the front of each half-sheet, I wrote an equation in standard form. Inside each half-sheet, I wrote the equivalent slope-intercept form of a *different* equation. I also wrote a letter of the alphabet next to each answer. Then, I taped these papers in random places around the classroom. I made sure to write down the order of the answers on a scrap of paper so I could quickly spot-check their answers.
Front of Scavenger Hunt Card
Inside of Scavenger Hunt Card
For this activity, I let students work in pairs, but I required each student to show their own work on a sheet I created to help them organize their work. Next to each answer and completed work, students copied the corresponding letter of the alphabet. Though each pair started with a different letter, I could still check their answers. (The fact that I could do this did blow a few of my students' minds.)
My students really enjoyed this activity. It got them out of their seats and moving around. I don't think I do that enough in my classroom. I really liked hearing the conversations that started when students couldn't find a corresponding answer card after rearranging an equation. Often, they would have the right answer, but it was written in a different way from the answer card. The fact that x/4 and 1/4 x mean the same thing really confused some of my students. Or, if students had a very similar answer to a card found around the room, they would often work together to rework their problem. Usually, they had just made a mistake with their signs or addition/subtraction.
Once students find the answer card to their previous question, they solve the new question. The process continues until the students return to their starting place.
Because my students couldn't move on with the activity until they found each answer around the room, they were much more careful and focused. Their questions were more specific, and almost every student completed the assigned activity during the 50-minute class period. I think I only had 3 students who chose to not participate and work as hard as they should. | 706 | 3,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-30 | latest | en | 0.954073 |
https://www.jiskha.com/display.cgi?id=1328996804 | 1,500,850,010,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424623.68/warc/CC-MAIN-20170723222438-20170724002438-00325.warc.gz | 806,765,761 | 3,496 | # math
posted by .
A telephone company's goal is to have no more than 4 monthly line failures on any 100 miles of line. The company currently experiences an average of 5 monthly line failures per 50 miles of line. Let x denote the number of monthly line failures per 100 miles of line. Assuming x has a Poisson distribution: | 74 | 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} | 2.59375 | 3 | CC-MAIN-2017-30 | longest | en | 0.921568 |
http://ssinglekh.sbh-online.xyz/how-many-pieces-full-sheet-cake.html | 1,576,180,643,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540545146.75/warc/CC-MAIN-20191212181310-20191212205310-00002.warc.gz | 135,403,534 | 3,734 | # How many pieces full sheet cake
Many full
## How many pieces full sheet cake
How many pieces full sheet cake. But how many pieces of cake can you get from a half or full sheet pan? how Continue Reading. Sheet cakes sometimes are the finale of how long food loaded meals and guests are too full to attempt a decent piece. About CakeCentral.
Full Sheet Cake servesIf you’ re planning to bake a sheet cake at home, 18″ x 24″ the most common size cake pan pieces is 9″ x pieces 13″ ( slightly larger than the one- quarter sheet cake offered by bakeries) Other popular sheet cake sizes are half- and quarter- sheet cakes. The typical question when you finally get things sorted out and have decided on a simple buffet of stress free easy to organize party food choices usually concerns the celebration cake. How many pieces full sheet cake. What Are Half Sheet Cake Sizes? In my area the many measurements are; I found this on Yahoo - " A full sheet cake is 18" x 24" will serve 70 to 80 people. A quarter sheet is 9" x13" a half sheet is 13" x18" a full sheet is 18" x26". Sheet cakes are usually 2 how inches in height. Sheet cakes typically come in quarter half full sheets.
If the sheet cake has three layers, then there are between 115 to 158 pieces of cake. 1 cm), 2 many in pieces × 2 in ( 5. Because of the difference in available pan sizes, half sheet cakes come in a variety of sizes. The number of servings each cake has is dependent on the serving size. In turn if this cake is the many only food for a casual farewell party many at work people will often take bigger portions. I normally just ask people how many servings they need. We have a simple equation to help you calculate the many perfect amount of cake for your guests.
have three standard sizes of sheet cakes available. Most bakeries in the U. This produces 96 slices from a full- sized sheet cake , 48 from a half- sized cake 24 from a quarter- pieces sheet cake. A two- layer full sheet cake can serve between 80 to 111 full people. com how is how the world' s largest cake community for cake how decorating professionals and enthusiasts. how if your making a cake out of many cake box, how many boxes do you need to make full size sheet? Most pieces bakeries suggest many a range such as " 90 to 100, " to allow for second portions erratic cutting. Wedding- sized slices many can be as small as 1- by 2- pieces inch, yielding up to 192 portions. That means a 1/ 2 sheet should be 12" x 18" and serve 35 to full 40 people I think.
inches of Cake Apiece ( 2 how pieces x 2). A Full Sheet Cake Feeds 96: 4 Sq. 1/ pieces 4 Sheet Cake 18″ x 24″, 9″ x 12″, serves pieces 18- how 40 Full Sheet Cake, serves/ 2 Sheet Cake, 11″ x 15″ serves 48- 108. Consider the number of guests you want to serve with your cake cut it accordingly. Sheet cakes easily slice into a few different sizes. A 1/ 4 sheet cake is 9" x 12". Feb 02 every one , · oversized 1/ 2 sheet: 12x18 undersized full sheet: 15xx15' s together) full sheet: 16xx16' s many together) Of course, every bakery is different but the dimensions add how up perfectly many with these sizes. 1 cm) 2 pieces in full × 3 in ( 5.
A half sheet cake , is a large rectangular- shaped cake that is normally served at a party , 1/ 2 pieces sheet cake any time a crowd is expected. A full sheet cake measures 16 inches by 24 inches. How pieces can the answer be improved? Sheet cakes are often used to feed crowds at weddings and other large events. Just how many people does a sheet cake feed. Typically, a sheet cake slice is either 1 in × 2 in ( 2.
## Full sheet
How Big Is a Quarter Sheet Cake? | Dimensions Info. A full sheet cake is 18 inches by 24 inches and serves approximately 80 people. A full sheet cake can be cut into 64 2- inch- by- 3- inch slices or 96 2- inch- by- 2- inch pieces. Sheet Cake Cutting And Servings Guide Per several requests to post this here, here it is.
``how many pieces full sheet cake``
This is the guide that I created to determine how many servings per each sheet cake and a diagram of how I would recommend it being cut. sheet; You May Also Be Interested In. | 1,039 | 4,121 | {"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} | 3.015625 | 3 | CC-MAIN-2019-51 | latest | en | 0.960846 |
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