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http://spmath81609.blogspot.com/2010/02/nheas-ratio-scribepost-2210.html | 1,531,772,145,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589455.35/warc/CC-MAIN-20180716193516-20180716213516-00606.warc.gz | 340,112,635 | 20,320 | ## Tuesday, February 2, 2010
### Nhea's Ratio Scribepost, 2/2/10
7. A herd of 100 caribou was moved to a new location. The population increased by 10% the first year and then increased by 20% the second year.
a) Find the population after the second year.
First year - 100 / 10 = 10 = 10%
-100 + 10 = 110
Second year - 110 / 10 = 11 x 2 = 22 = 20%
-110 + 122 = 132
b) Explain why there was not a 30% increase in population over the two years
The increase was not by 30% because the 20% was worked in after the 10 percent was already placed, making it 132.
11. A student is awarded a \$1000 scholars
hip and places it in an account that pays 3% simple interest per year.
a) What is the total value of the scholarship
amount at the end of the second year?
First Year - 1000 / 10 = 100 / 10 = 10 x 3 =30 = 3 %
-1000 + 30 = \$1030
Second year -
1030 / 10 = 103 / 10 = 10.30 x 3 =30.90 = 3 %
- 1030 + 30.90 = \$1060.90
b) What is the single percent increase in value of the scholarship after two years?
(3 % = 30) + (3 % = 30.90) = 60.90 / 10 = 6.09% after two years | 364 | 1,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-30 | latest | en | 0.953027 |
https://number.academy/1000066 | 1,702,195,981,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00769.warc.gz | 469,061,729 | 11,847 | Number 1000066 facts
The even number 1,000,066 is spelled 🔊, and written in words: one million and sixty-six, approximately 1.0 million. The ordinal number 1000066th is said 🔊 and written as: one million and sixty-sixth. The meaning of the number 1000066 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1000066. What is 1000066 in computer science, numerology, codes and images, writing and naming in other languages
What is 1,000,066 in other units
The decimal (Arabic) number 1000066 converted to a Roman number is (M)LXVI. Roman and decimal number conversions.
Weight conversion
1000066 kilograms (kg) = 2204745.5 pounds (lbs)
1000066 pounds (lbs) = 453627.0 kilograms (kg)
Length conversion
1000066 kilometers (km) equals to 621413 miles (mi).
1000066 miles (mi) equals to 1609451 kilometers (km).
1000066 meters (m) equals to 3281017 feet (ft).
1000066 feet (ft) equals 304824 meters (m).
1000066 centimeters (cm) equals to 393726.8 inches (in).
1000066 inches (in) equals to 2540167.6 centimeters (cm).
Temperature conversion
1000066° Fahrenheit (°F) equals to 555574.4° Celsius (°C)
1000066° Celsius (°C) equals to 1800150.8° Fahrenheit (°F)
Time conversion
(hours, minutes, seconds, days, weeks)
1000066 seconds equals to 1 week, 4 days, 13 hours, 47 minutes, 46 seconds
1000066 minutes equals to 2 years, 3 weeks, 1 day, 11 hours, 46 minutes
Codes and images of the number 1000066
Number 1000066 morse code: .---- ----- ----- ----- ----- -.... -....
Sign language for number 1000066:
Number 1000066 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
Mathematics of no. 1000066
Multiplications
Multiplication table of 1000066
1000066 multiplied by two equals 2000132 (1000066 x 2 = 2000132).
1000066 multiplied by three equals 3000198 (1000066 x 3 = 3000198).
1000066 multiplied by four equals 4000264 (1000066 x 4 = 4000264).
1000066 multiplied by five equals 5000330 (1000066 x 5 = 5000330).
1000066 multiplied by six equals 6000396 (1000066 x 6 = 6000396).
1000066 multiplied by seven equals 7000462 (1000066 x 7 = 7000462).
1000066 multiplied by eight equals 8000528 (1000066 x 8 = 8000528).
1000066 multiplied by nine equals 9000594 (1000066 x 9 = 9000594).
show multiplications by 6, 7, 8, 9 ...
Fractions: decimal fraction and common fraction
Fraction table of 1000066
Half of 1000066 is 500033 (1000066 / 2 = 500033).
One third of 1000066 is 333355,3333 (1000066 / 3 = 333355,3333 = 333355 1/3).
One quarter of 1000066 is 250016,5 (1000066 / 4 = 250016,5 = 250016 1/2).
One fifth of 1000066 is 200013,2 (1000066 / 5 = 200013,2 = 200013 1/5).
One sixth of 1000066 is 166677,6667 (1000066 / 6 = 166677,6667 = 166677 2/3).
One seventh of 1000066 is 142866,5714 (1000066 / 7 = 142866,5714 = 142866 4/7).
One eighth of 1000066 is 125008,25 (1000066 / 8 = 125008,25 = 125008 1/4).
One ninth of 1000066 is 111118,4444 (1000066 / 9 = 111118,4444 = 111118 4/9).
show fractions by 6, 7, 8, 9 ...
Calculator
1000066
Is Prime?
The number 1000066 is not a prime number. The closest prime numbers are 1000039, 1000081.
Factorization and factors (dividers)
The prime factors of 1000066 are 2 * 47 * 10639
The factors of 1000066 are 1, 2, 47, 94, 10639, 21278, 500033, 1000066.
Total factors 8.
Sum of factors 1532160 (532094).
Powers
The second power of 10000662 is 1.000.132.004.356.
The third power of 10000663 is 1.000.198.013.068.287.488.
Roots
The square root √1000066 is 1000,032999.
The cube root of 31000066 is 100,0022.
Logarithms
The natural logarithm of No. ln 1000066 = loge 1000066 = 13,815577.
The logarithm to base 10 of No. log10 1000066 = 6,000029.
The Napierian logarithm of No. log1/e 1000066 = -13,815577.
Trigonometric functions
The cosine of 1000066 is -0,945715.
The sine of 1000066 is 0,324998.
The tangent of 1000066 is -0,343654.
Properties of the number 1000066
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
Number 1000066 in Computer Science
Code typeCode value
1000066 Number of bytes976.6KB
Unix timeUnix time 1000066 is equal to Monday Jan. 12, 1970, 1:47:46 p.m. GMT
IPv4, IPv6Number 1000066 internet address in dotted format v4 0.15.66.130, v6 ::f:4282
1000066 Decimal = 11110100001010000010 Binary
1000066 Decimal = 1212210211111 Ternary
1000066 Decimal = 3641202 Octal
1000066 Decimal = F4282 Hexadecimal (0xf4282 hex)
1000066 BASE64MTAwMDA2Ng==
1000066 MD57b38d321f82e6bf1b807a35c21637422
1000066 SHA1b34efab33ba028b59d32e2697bc7bee1de13f840
1000066 SHA224f419bb0ae6c47235006d97d26aa548d49eeeb7df08260f31b432bc48
1000066 SHA25670c25db59688ca7c0b3d7880e48735ce7ab0740484e4f2f418cd05bb5c1a9405
More SHA codes related to the number 1000066 ...
If you know something interesting about the 1000066 number that you did not find on this page, do not hesitate to write us here.
Numerology 1000066
Character frequency in the number 1000066
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 4 6 2
Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1000066, the numbers 1+0+0+0+0+6+6 = 1+3 = 4 are added and the meaning of the number 4 is sought.
№ 1,000,066 in other languages
How to say or write the number one million and sixty-six in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.000.066) un millón sesenta y seis German: 🔊 (Nummer 1.000.066) eine Million sechsundsechzig French: 🔊 (nombre 1 000 066) un million soixante-six Portuguese: 🔊 (número 1 000 066) um milhão e sessenta e seis Hindi: 🔊 (संख्या 1 000 066) दस लाख, छयासठ Chinese: 🔊 (数 1 000 066) 一百万零六十六 Arabian: 🔊 (عدد 1,000,066) واحد مليون و ستة و ستون Czech: 🔊 (číslo 1 000 066) milion šedesát šest Korean: 🔊 (번호 1,000,066) 백만 육십육 Dutch: 🔊 (nummer 1 000 066) een miljoen zesenzestig Japanese: 🔊 (数 1,000,066) 百万六十六 Indonesian: 🔊 (jumlah 1.000.066) satu juta enam puluh enam Italian: 🔊 (numero 1 000 066) un milione e sessantasei Norwegian: 🔊 (nummer 1 000 066) en million og seksti-seks Polish: 🔊 (liczba 1 000 066) milion sześćdziesiąt sześć Russian: 🔊 (номер 1 000 066) один миллион шестьдесят шесть Turkish: 🔊 (numara 1,000,066) birmilyonaltmışaltı Thai: 🔊 (จำนวน 1 000 066) หนึ่งล้านหกสิบหก Ukrainian: 🔊 (номер 1 000 066) один мільйон шістдесят шість Vietnamese: 🔊 (con số 1.000.066) một triệu lẻ sáu mươi sáu Other languages ...
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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. | 2,493 | 7,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.693726 |
https://www.dataunitconverter.com/mebibit-per-hour-to-kibibyte-per-hour/8 | 1,708,699,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00464.warc.gz | 746,493,076 | 16,131 | Mibit/Hr to KiB/Hr - 8 Mibit/Hr to KiB/Hr Conversion
expand_more
Input Mebibits per Hour (Mibit/Hr) - and press Enter.
Mibit/Hr
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
label_important RESULT sentiment_satisfied_alt
8 Mibit/Hr =1,024 KiB/Hr
( Equal to 1.024E+3 KiB/Hr )
content_copy
Calculated as → 8 x 1024 ÷ 8 smart_display Show Stepsexpand_more
Below chart table shows the amount of data that can be transferred at a constant speed of 8 Mibit/Hr in various time frames.
Transfer RateAmount of Data can be transferred
@ 8 Mibit/Hrin 1 Second0.2844444444444444444444444444444444443648 Kibibytes
in 1 Minute17.0666666666666666666666666666666666665984 Kibibytes
in 1 Hour1,024 Kibibytes
in 1 Day24,576 Kibibytes
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toc Table of Contents
Mebibits per Hour (Mibit/Hr) to Kibibytes per Hour (KiB/Hr) Conversion - Formula & Steps
The Mibit/Hr to KiB/Hr Calculator Tool provides a convenient solution for effortlessly converting data rates from Mebibits per Hour (Mibit/Hr) to Kibibytes per Hour (KiB/Hr). 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 (Mebibit) and target (Kibibyte) data units.
Source Data Unit Target Data Unit
Equal to 1024^2 bits
(Binary Unit)
Equal to 1024 bytes
(Binary Unit)
The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Mebibit to Kibibyte in a simplified manner.
÷ 1024
÷ 1024
x 1024
x 1024
Based on the provided diagram and steps outlined earlier, the formula for converting the Mebibits per Hour (Mibit/Hr) to Kibibytes per Hour (KiB/Hr) can be expressed as follows:
diamond CONVERSION FORMULA KiB/Hr = Mibit/Hr x 1024 ÷ 8
Now, let's apply the aforementioned formula and explore the manual conversion process from Mebibits per Hour (Mibit/Hr) to Kibibytes per Hour (KiB/Hr). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Kibibytes per Hour = Mebibits per Hour x 1024 ÷ 8
STEP 1
Kibibytes per Hour = Mebibits per Hour x 128
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By applying the previously mentioned formula and steps, the conversion from 8 Mebibits per Hour (Mibit/Hr) to Kibibytes per Hour (KiB/Hr) can be processed as outlined below.
1. = 8 x 1024 ÷ 8
2. = 8 x 128
3. = 1,024
4. i.e. 8 Mibit/Hr is equal to 1,024 KiB/Hr.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Mebibits per Hour to Kibibytes per Hour using any of the programming language such as Java, Python, or Powershell.
Unit Definitions
What is Mebibit ?
A Mebibit (Mib or Mibit) is a binary unit of digital information that is equal to 1,048,576 bits and is defined by the International Electro technical Commission(IEC). The prefix 'mebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'megabit' (Mb). 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.
- Learn more..
arrow_downward
What is Kibibyte ?
A Kibibyte (KiB) is a binary unit of digital information that is equal to 1024 bytes (or 8,192 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'kibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'kilobyte' (KB). 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.
- Learn more..
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Excel Formula to convert from Mebibits per Hour (Mibit/Hr) to Kibibytes per Hour (KiB/Hr)
Apply the formula as shown below to convert from 8 Mebibits per Hour (Mibit/Hr) to Kibibytes per Hour (KiB/Hr).
A B C
1 Mebibits per Hour (Mibit/Hr) Kibibytes per Hour (KiB/Hr)
2 8 =A2 * 128
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
Python Code for Mebibits per Hour (Mibit/Hr) to Kibibytes per Hour (KiB/Hr) Conversion
You can use below code to convert any value in Mebibits per Hour (Mibit/Hr) to Mebibits per Hour (Mibit/Hr) in Python.
mebibitsperHour = int(input("Enter Mebibits per Hour: "))
kibibytesperHour = mebibitsperHour * 128
print("{} Mebibits per Hour = {} Kibibytes per Hour".format(mebibitsperHour,kibibytesperHour))
The first line of code will prompt the user to enter the Mebibits per Hour (Mibit/Hr) as an input. The value of Kibibytes per Hour (KiB/Hr) is calculated on the next line, and the code in third line will display the result.
Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,398 | 4,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.702785 |
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This web site contains a set of curriculum materials for middle school and high school on the topic of magnetism. Topics include magnet basics, electromagnets, magnetic fields, superconductors, magnetic levitation, and applications of magnetic properties. It is organized sequentially, with a concept-building approach. Each unit is supported with videos and hands-on experiments, as well as links to background information.
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Faraday's Law, Lenz's Law, bar magnets, dipole, field lines, maglev, magnetic field lines, magnetic poles, magnetism experiments, magnetism labs, magnetism unit, magnetosphere, magnets
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### AAAS Benchmark Alignments (2008 Version)
#### 4. The Physical Setting
4D. The Structure of Matter
• 6-8: 4D/M9. Materials vary in how they respond to electric currents, magnetic forces, and visible light or other electromagnetic waves.
4G. Forces of Nature
• 3-5: 4G/E2. Without touching them, a magnet pulls on all things made of iron and either pushes or pulls on other magnets.
• 6-8: 4G/M3. Electric currents and magnets can exert a force on each other.
• 9-12: 4G/H5ab. Magnetic forces are very closely related to electric forces and are thought of as different aspects of a single electromagnetic force. Moving electrically charged objects produces magnetic forces and moving magnets produces electric forces.
• 9-12: 4G/H5c. The interplay of electric and magnetic forces is the basis for many modern technologies, including electric motors, generators, and devices that produce or receive electromagnetic waves.
• 9-12: 4G/H7. Electric currents in the earth's interior give the earth an extensive magnetic field, which we detect from the orientation of compass needles.
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R. Hoadley, (2005), WWW Document, (http://www.coolmagnetman.com/magindex.htm).
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### Featured By
Physics Front
Jul 10 - Sep 30, 2022 | 1,161 | 4,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.864924 |
https://www.perlmonks.org/index.pl/?node_id=785293 | 1,590,374,028,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387155.10/warc/CC-MAIN-20200525001747-20200525031747-00328.warc.gz | 861,406,419 | 6,704 | ### Re: accurately rounding numbers for percentages
by Trimbach (Curate)
on Aug 02, 2009 at 22:24 UTC ( #785293=note: print w/replies, xml ) Need Help??
in reply to accurately rounding numbers for percentages
What you're asking is not possible. Anytime you round a number you're going to introduce error, how much error will depend on how much you're rounding. Add enough errors together and your total will always be off from the "expected" total (in this case 100%).
The only way around this is to go ahead and round the individual entries to whole numbers for display, but when calculating the total don't add the rounded entries, add the unrounded entries, and then round the result for display, if you want.
Gary Blackburn
Trained Killer
• Comment on Re: accurately rounding numbers for percentages
Replies are listed 'Best First'.
Re^2: accurately rounding numbers for percentages
by derekn (Initiate) on Aug 02, 2009 at 22:53 UTC
So i'm gonna have to live with "37%, 23%, 9%, 16%, 16%" (rounded values) equalling 101, even though it SHOULD equal 100%?
That's not what Trimbach said, by a long shot.
If you add the UNrounded numbers percentages, they should total 100% (except for the fact that you'll sometimes run into value/count pairs that are rounded at the end of whatever length decimal value you use: 100/6, for example).
But, for cases such as I infer yours is, a quite standard and commonly accepted practice is to include the disclaimer "Totals may not equal 100% because of rounding."
Update: For clarity (in light of OP's next reply), s/numbers/percentages/ at strikeout above.
Hmm, I don't know that I made myself clear, so I will go into more detail to make sure we are on the same page. The users vote for which search engine they used to find site. (Google, Yahoo, Bing) The votes are stored in database as total votes for EACH search engine (Google=30,Yahoo=23,Bing=13) Then, I am figuring out the percentage of each search engine used by dividing by total number of votes (\$searchEngine/\$totalVotes*100) which of course does not always yield nice results. This number (\$google,\$yahoo,\$bing) is then rounded. The problem of course is that sometimes the 3 different percentages don't add up to 100%. Derek
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Notices? | 656 | 2,663 | {"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-2020-24 | longest | en | 0.919235 |
http://www.kineticmediaworks.com/rirxfgz/fluid-mechanics-pressure-measurement-ppt-c648a4 | 1,627,853,022,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154219.62/warc/CC-MAIN-20210801190212-20210801220212-00267.warc.gz | 61,784,099 | 14,491 | fluid mechanics pressure measurement ppt
See Fig.7.73. Academic year. Results indicate that the cavity pressure drop prior to surface closure is an order of magnitude greater than previously assumed. Suppose the pipe contains water, and mercury is used as the measuring liquid. . ii. . General Energy Equation 8. Figure shows one of the many types of mechanical pressure gauges in use today. Pressure measurement operations will be conducted by using U-tube and inclined tube manometers. ¦ Fi 0 Viscous/Inviscid Steady/Unsteady Compressible/ Incompressible ¦ Fi! Flow measurement B. Fluid properties C. Fluid statics D. Energy, impulse, and momentum equations E. Pipe and other internal flow 7% of FE Morning Session Up to 15% of FE Afternoon Session Afternoon (Depends on Discipline) A. Bernoulli equation and mechanical energy balance B. Hydrostatic pressure INTI International University. Water, Oils, Alcohols, etc. In the English system of units, pressure is expressed as "psi" or lbf/in 2. Fluid mechanics is a discipline within the broad field of applied mechanics concerned with the behaviour of liquids and gases at rest or in motion. Fluid mechanics itself is also divided into several categories. . The local atmospheric pressure is 15 psia. Fluid Mechanics 9-10a Fluid Measurements Pitot Tube – measures flow velocity • The static pressure of the fluid at the depth of the pitot tube (p0) must be known. Share. Pressure distribution for a fluid at rest A force balance in the z direction gives: For an infinitesimal element (Dz 0) Incompressible fluid Liquids are incompressible i.e. Zain Chughtai. LAB REPORT EXPERIMENT # 1 PRESSURE MEASUREMENT PNGE 211: AN INTRODUCTION TO FLUID MECHANICS Made By: Nabeel Ahmed Khan Submitted To: Doruk Alp Date of Submission: 15/04/16 OBJECTIVE The purpose of this experiment is to demonstrate different pressure measurement methods. . Remote sensing is often more convenient than putting a measuring device into a system, such as a person’s artery. Fluid Mechanics Overview Gas Liquids Statics Dynamics Air, He, Ar, N 2, etc. … pressure of a fluid stays constant, the gauge pressure of the same fluid will vary as atmospheric pressure changes. The undiminished transmission of pressure through a fluid allows precise remote sensing of pressures. Pressure always acts inward normal to any surface (even imaginary surfaces as in a control volume). FIGURE 3–2 Two basic pressure gages. Selection of pressure gauges depends upon the location and type of fluid flow. PRESSURE MEASUREMENT Pressure is an important variable in fluid mechanics and many instruments have been devised for its measurement. Fluid Mechanics Introduction Video Lecture From Properties of Fluid Chapter of Fluid Mechanics Subject For All Students. For example, when a car drives up a mountain (atmospheric air pressure decreases), the (gauge) tire pressure goes up. . . Fluids are a collection of randomly arranged molecules held together by by weak cohesive forces. . . If the pitot tube is to be used to measure the velocity of a liquid in a pipe, then we must adopt some method to know the static pressure head H. For instance, we may use a pitot tube and a vertical piezometer tube and measure the difference of the liquid levels in the two tubes. . . FERC FERC. . ASSIGNME NT/ TEST/PROJ ECT Assignment I Test I. COMPLETI ON DATE* … Thought Questions
Why is the electricity produced at the bottom of dams?
When you catch a deep-sea fish, why does its eyes pop-out?
Why do your ears pop on an airplane or up in the mountains?
Chapter 23 | 785 | 3,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-31 | latest | en | 0.873934 |
https://www.crazy-numbers.com/en/46659 | 1,713,619,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00243.warc.gz | 624,319,771 | 3,429 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 46659: mathematical and symbolic properties | Crazy Numbers
Discover a lot of information on the number 46659: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 46659
Is 46659 a prime number? No
Is 46659 a perfect number? No
Number of divisors 8
List of dividers
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1, 3, 103, 151, 309, 453, 15553, 46659
Sum of divisors 63232
Prime factorization 3 x 103 x 151
Prime factors
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3, 103, 151
## How to write / spell 46659 in letters?
In letters, the number 46659 is written as: Forty-six thousand six hundred and fifty-nine. And in other languages? how does it spell?
46659 in other languages
Write 46659 in english Forty-six thousand six hundred and fifty-nine
Write 46659 in french Quarante-six mille six cent cinquante-neuf
Write 46659 in spanish Cuarenta y seis mil seiscientos cincuenta y nueve
Write 46659 in portuguese Quarenta e seis mil seiscentos cinqüenta e nove
## Decomposition of the number 46659
The number 46659 is composed of:
1 iteration of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4
2 iterations of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9 | 597 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.744914 |
http://clarvis.co.uk/Design_and_Technology/Exercise_4_Series_Parallel.html | 1,611,342,927,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531335.42/warc/CC-MAIN-20210122175527-20210122205527-00370.warc.gz | 19,891,936 | 2,730 | Ohm’s Law
Teaching and Learning Resources
Exercise 4: A Series / Parallel Circuit
1. 1.Build the circuit as shown. The resistors should be of these values: R1 = 22 Ohms, R2 = 10 Ohms and R3 = 15 Ohms. The supply voltage is 6V.
1. 2.Using Ohm’s Law, calculate what you would expect the current flowing through each resistor to be and the total current.
1. 3.Calculate what voltage should be dropped across each resistor.
1. 4.Measure all of the voltages and currents and check that they match what you would expect.
5. Challenge: Replace all of the resistors with three “unknown” values. Using your wit an guile, determine what values the resistors must be. | 172 | 663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-04 | latest | en | 0.90018 |
https://en.m.wikiversity.org/wiki/PlanetPhysics/Categorical_Algebra | 1,606,704,127,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00011.warc.gz | 277,178,109 | 10,702 | # PlanetPhysics/Categorical Algebra
\newcommand{\sqdiagram}[9]{$\displaystyle \diagram #1 \rto^{#2} \dto_{#4}& \eqno{\mbox{#9}}$ }
### An Outline of Categorical Algebra
This topic entry provides an outline of an important mathematical field called categorical algebra ; although specific definitions are in use for various applications of categorical algebra to specific algebraic structures, they do not cover the entire field. In the most general sense, categorical algebras -- as introduced by Mac Lane in 1965 -- can be described as the study of representations of algebraic structures, either concrete or abstract, in terms of categories, functors and natural transformations.
In a narrow sense, a categorical algebra is an associative algebra, defined for any locally finite category and a commutative ring with unity. This notion may be considered as a generalization of both the concept of group algebra and that of an incidence algebra, much as the concept of category generalizes the notions of group and partially ordered set.
### Extensions of Categorical Algebra
• Thus, ultimately, since categories are interpretations of the axiomatic \htmladdnormallink{theories of abstract category {http://planetphysics.us/encyclopedia/Formula.html} (ETAC)}, so are categorical algebras. The general definition of representation introduced above can be still further extended by introducing supercategorical algebras as interpretations of ETAS, as explained next.
• Mac Lane (1976) wrote in his Bull. AMS review cited here as a verbatim quotation: \emph{On some occasions I have been tempted to try to define what algebra is, can, or should be - most recently in concluding a survey [72] on Recent advances in algebra. But no such formal definitions hold valid for long, since algebra and its various subfields steadily change under the influence of ideas and problems coming not just from logic and geometry, but from analysis, other parts of mathematics, and extra mathematical sources. The progress of mathematics does indeed depend on many interlocking, unexpected and multiform developments.}
### Basic Definitions
An algebraic representation is generally defined as a \htmladdnormallink{morphism {http://planetphysics.us/encyclopedia/TrivialGroupoid.html} ${\displaystyle \rho }$ from an abstract algebraic structure ${\displaystyle {\mathcal {A}}_{S}}$ to a concrete algebraic structure ${\displaystyle A_{c}}$}, a Hilbert space ${\displaystyle {\mathcal {H}}}$, or a family of linear operator spaces.
The key notion of representable functor was first reported by Alexander Grothendieck in 1960.
Thus, when the latter concept is extended to categorical algebra, one has a representable functor ${\displaystyle S}$ from an arbitrary category ${\displaystyle {\mathcal {C}}}$ to the category of sets ${\displaystyle Set}$ if ${\displaystyle S}$ admits a functor representation defined as follows. A functor representation of ${\displaystyle S}$ is defined as a pair, ${\displaystyle ({R},\phi )}$, which consists of an object ${\displaystyle R}$ of ${\displaystyle {\mathcal {C}}}$ and a family ${\displaystyle \phi }$ of equivalences $\displaystyle \phi (C): \Hom_{\mathcal{C}}(R,C) \cong S(C)$ , which is natural in C, with C being any object in ${\displaystyle {\mathcal {C}}}$. When the functor ${\displaystyle S}$ has such a representation, it is also said to be represented by the object ${\displaystyle R}$ of ${\displaystyle {\mathcal {C}}}$. For each object ${\displaystyle R}$ of ${\displaystyle \mathbf {C} }$ one writes $\displaystyle h_{R}: \mathcal{C} \lra Set$ for the covariant $\displaystyle \Hom$ --functor $\displaystyle h_{R}(C)\cong \Hom_{\mathcal{C}}(R,C)$ . A representation ${\displaystyle (R,\phi )}$ of ${\displaystyle {S}}$ is therefore a \htmladdnormallink{natural equivalence {http://planetphysics.us/encyclopedia/IsomorphismClass.html} of functors}:
${\displaystyle \phi :h_{R}\cong {S}~.}$
The equivalence classes of such functor representations (defined as natural equivalences) directly determine an algebraic (groupoid) structure.
## References
1. Saunders Mac Lane: Categorical algebra., Bull. AMS , 71 (1965), 40-106., Zbl 0161.01601, MR 0171826,
2. Saunders Mac Lane: Topology and Logic as a Source of Algebras., Bull. AMS , 82 , Number 1, 1-36, January 1, 1976. | 1,054 | 4,319 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "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": 5} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.894238 |
https://ozzmaker.com/success-with-a-balancing-robot-using-a-raspberry-pi/?replytocom=382 | 1,582,797,477,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146681.47/warc/CC-MAIN-20200227094720-20200227124720-00470.warc.gz | 495,485,606 | 27,165 | # Success with a Balancing Robot using a Raspberry Pi
I have had success with getting PiBBOT to balance.
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When I saw my first two wheel balancing robot I was very fascinated. And after receiving my Raspberry Pi, I decided to build one myself.
PiBBOT (Pi Balancing roBOT) is my first successful balancing robot. And it has room for improvement and extra functionality.
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When building PiBBOT, I had a few roadblocks I needed to overcome;
1. I originally had the Anker as the power source for both Pi and the motors, however the amperage was too low.
2. Original H-Bridge not powerful enough for my motors
3. Gyro calculation off by 15 degrees
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The TFT displays the angles from the accelerometer, gyro, complementary filter and power to the motors.
The buttons are to turn the motors on and off and to reset the gyro.
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What next?
• Incorporate the wheel encoders for better balance
• Build a sturdier frame
• Implement direction control
• The able to self right itself after falling over
• Room mapping
• Retrieve a can of beer from the fridge
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PiBBOT consists of these components;
TFT; 2.2″ 18-bit color TFT LCD display
RF Receiver :RF M4 Receiver – 315MHz
Pololu;
Motors; 9.7:1 Metal Gearmotor 25Dx48L mm with 48 CPR Encoder
Wheels; Pololu Wheel 90x10mm
IMU; MinIMU-9 v2 Gyro, Accelerometer, and Compass (L3GD20 and LSM303DLHC).
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Anker;
Battery; Anker® Astro3 10000mAh 5V / 9V / 12V 2A Dual USB Output External Battery
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Tenergy;
Battery; 7.2V Tenergy 3800mAh Flat NiMH High Power
## Theory
The basics behind a balancing robot is based on the Inverted Pendulum concept. The goal is to have a control algorithm called Proportional Integral Derivative (PID) to keep the robot balanced by trying to keep the wheels under the center of gravity. Eg. If the robot leans forwards, the wheels spin forward trying correct the lean.
One axis of an accelerometer and one axis from a gyroscope are used to measure the current angle and the rate of rotation. A well timed looped is needed to keep track of everything.
Calculations are then done to provide power via PWM to the motors and in the right direction to keep the robot upright.
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## Angle Measurement
To measure the angles, I use the MinIMU-9 v2 Gyro, Accelerometer, and Compass inertial measurement unit(IMU) from Pololu.
This is a digital IMU and can be read using I2C. It is also powered off the Raspberry Pi 3v pin.
Here you can see the 3 individual I2C addresses;
pi@raspberrypi ~ \$ sudo /usr/sbin/i2cdetect -y 1
0 1 2 3 4 5 6 7 8 9 a b c d e f
00: — — — — — — — — — — — — —
10: — — — — — — — — — 19 — — — — 1e —
20: — — — — — — — — — — — — — — — —
30: — — — — — — — — — — — — — — — —
40: — — — — — — — — — — — — — — — —
50: — — — — — — — — — — — — — — — —
60: — — — — — — — — — — — 6b — — — —
70: — — — — — — — —
pi@raspberrypi ~ \$
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When using the IMU to calculate the angle, readings from both the gyro and accelerometer are needed, which are then combined. This is because using either on their own will result in inaccurate readings.
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Here is why;
Gyro – A gyro measures the rate of rotation, which has to be tracked over time to calculate the current angle. This tracking <and noise> causes the gyro to drift. However, gyros are good at measuring quick sharp movements.
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Accelerometer – Accelerometers are used to sense both static (e.g. gravity) and dynamic (e.g. sudden starts/stops) acceleration. They don’t need to be tracked like a gyro and can measure the current angle at any given time. Accelerometers however are very noisy and are only useful for tracking angles over a long period of time.
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The gyro and accelerometer have a number of sensitivity levels. For the gyro I am using a measurement range of 250dps, which has a sensitivity of is 0.00875 dps/LSB. And the accelerometer I am using 8 g full scale, 4 mg/LSB and FS = 10.
Later I will discuss how to combine the angles from both sensors.
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## Converting the RAW gyro and accelerometer values to something useful.
The gyro I use is a L3GD20
For the gyro, we need to work out how fast it is rotating in degrees per second(dps). We then need to track the angle deviation from point zero. Point zero is when the PiBBOT is upright and balanced. We then need to track this over time.
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Gyro Rotation Rate;
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rate_gyr_x = (float) gyr_x_raw * G_GAIN
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gyr_x_raw = raw data acquired from the gyro X axis.
G_GAIN = 0.00875, which is based off the sensitivity level used by the gyro.
rate_gyr_x = the rate of rotation per second.
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Gyro Angle Tracking
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gyroXangle+=rate_gyr_x * DT
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DT = loop period.
gyroXangle = is the current X angle calculated from the gyro X data.
I have DT set top 0.02, which is 20ms. This is how long it takes to complete one cycle of the main loop. This loop period has to be constant and accurate, otherwise your gyro will drift.
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Accelerometer Angle
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The accelerometer I use is LSM303DLHC
The X angle can be calculated by using trigonometry and the raw values from the accelerometer Y and Z axis. This is done using the Atan2 function to return the principal value of the tangent of Y and Z, expressed in radians.
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We add π to the radians so that we get a result between 0 and 2. We then convert the radians to degrees by multiplying the radians by 57.29578 (180/π).
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AccXangle = (float) (atan2(acc_y_raw,acc_z_raw)+M_PI)*RAD_TO_DEG
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M_PI = 3.14159265358979323846
RAD_TO_DEG = 57.29578 , 1 radian = 57.29578 degrees
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## Combining the angles
Once you have the both X angles, you will have to combined them to overcome the gyro drift and the accelerometer noise.
We can do this by using a filter, which will trust the gyro for short periods of time and the accelerometer for longer periods of time.
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There are two filers you could use, the Kalman Filter or the Complementary Filter. I used the Complementary filter as it is simple to understand and less CPU intensive. The Kalman filter is way to far complex and would be very processor intensive.
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Complementary filter;
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Current angle = 98% x (current angle + gyro rotation rate) + (2% * Accelerometer angle)
or
CFangleX=AA*(CFangleX+rate_gyr_x*DT) +(1 – AA) * AccXangle;
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AA = 0.98 Complementary filter constant
CFangleX is our final angle which will be used to balance PiBBOT.
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In summary, our main piece of code for angles is;
```rate_gyr_x = (float) gyr_x_raw * G_GAIN;
AccXangle = (float) (atan2(acc_y_raw,acc_z_raw)+M_PI)*RAD_TO_DEG;
CFangleX=AA*(CFangleX+rate_gyr_x*DT) +(1 - AA) * AccXangle;```
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## Proportional Integral Derivative (PID)
A PID control algorithm is used to balance the robot by driving the motors based on the tilt.
The proportional(P) term of the PID is based on the current angle difference or error from Point Zero (where robot balances) multiplied by the P gain. The P gain is a number we need to tweak to get the right amount of proportional control.
The proportional control will try and correct the balance based on the current error from point zero.
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The portion of code for the P term is;
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Pterm = KP * CFangleX;
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The integral (I) term of the PID is based on the current angle difference or error from Point Zero multiplied by the I gain, which is that accumulated over time. The integral control assist in balancing the robot if it is moving.
The portion of code for the I term is;
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iTerm += KI * CFangleX;
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The derivative (D) term of the PID is based on the current rate of rotation. It is used to dampen the response as the robot reaches Zero Point. Without the derivative control, the robot will overshoot and than start to oscillate.
The portion of code for the D term is;
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dTerm = KD * (CFangleX – lastAngle);
lastAngle = CFangleX;
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The addition of all the PID values will be used to drive the motors in the right direction and with the right power;
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output = Pterm + iTerm + dTerm;
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Getting the PID values correct(KP,KI,KD) is the hardest part of building PiBBOT.
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The PID values need to be manually tuned;
1. Set KI and KD to zero.
2. Set KP high enougth so that the motors drive the wheels under the robot in the direction it is falling. The robot should overshoot Point Zero a little bit and then start to oscillate.
3. Reduce KP by about 10% to so you get just below the oscillation.
4. Increase KI. This will help the robot reach Point Zero faster and will also start to oscillate. Try and get a value so that the robot just oscillates
5. Increase KD to dampen the oscillation and until the robot balances.
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It is best to change these values while trying to balance to see the results in real time, a potentiometer could be used for this. I have used the RF module to increase or decrease the values.
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## 36 thoughts on “Success with a Balancing Robot using a Raspberry Pi”
1. Grant Johnston says:
This is great, its a really good use of a very small Pi. The BBC have an article about a Segway style platform with an Ipad on the top that you can remotely control. This could be a DIY version.
Are you going to package up the code….please.
1. cheers.
Regarding the code, I will post it.. maybe under another entry on how to read the values from the IMU.
I still need to clean it up a bit.
2. Rob says:
You have followed a very similar path to my own project earwig robot . I agree the Kalman is overkill and the complementary does just as good a job in less time . My mix was about 98/2 also.
So encoders next. …you will have a lot of backlash in the gearbox so the encoder can show the velocity spike . you could filter in SW but I prefer to use belt drive with maxon coreless DC motors . They are the DB’s, world class stall torque which is off the blocks acceleration, and if you can gain >1g acceleration you can lift to the verticle. if your wheels grip, carpet will hold over 1 coifficient of friction so >1g forward accel can be achived, iso standards all =1. ,. I think mapping is the great horizon, the raspberry pi chip i like because its got loads of memory ontop of the CPU, you can develope on it . the error can never end on angle it should have velocity=0 as a goal.
1. thanks for the comments.
I have seen earwig, and it balances very well.. and very precise.
I think a may start to look at new motors. What model number are your Maxon motors?
3. Jay K says:
Hello, this is a great project.
Luckily I am also working on interfacing the Raspberry Pi with the same gyro that you have used.I am also using an accelermeter from Sparkfun(https://www.sparkfun.com/products/10955). To combine the accelerometer and the gyroscope values, I am using the same Complementary filter. The problem where I am stuck is that time DT is coming out to be quite high. So all my calculations are generating undesired results. This is my code (https://github.com/smithakamat/Hipi_Accelerometer)….It would be great if you could share your code, so I could know where am I going wrong. I have to submit this project in school in a few days, and I am stuck in the gyro angle part.
4. Jayz says:
Hi this is great explanation. Can you post your code about angle calculation from IMU.
Thanks
5. Christian Sievers says:
Hallo, i see a problem in your calculations for the closed loop. There is no sample time in your equations. For example it should be iTerm += KI * CFangleX * sampletime. But to be hornest, I don’t know how to realize a constant sampling on the pi. May be there is the possibility of using realtime linux.
1. The sample time is not really needed in the PID equation.
Sample time is only needed to get the gyro rate of rotation, you can see this in;
gyroXangle+=rate_gyr_x * DT.
I somewhat disagree with your comment regarding sample time. A real-time Unix OS would differently be better; however I am able to get my code to run at a constant 20ms loop. It actually runs at 6-7ms, but I add a delay to get to 20ms. The gyro drift is only about 2 degrees every 15 secs, this is very, very, very good from a \$35 computer. And the complementary filter cancels this drift anyway.
If the gyro was drifting 2 degrees every 3 seconds, then I would look for another platform.
In the video, there is some instability, however this is from motor backlash.
6. Larry Cuba says:
I don’t see a mention of the structural system. What’s that ErectorSet-like system that you’re using?
1. Larry Cuba says:
terrific. thanks.
7. ARM says:
hi,
how to use that code for arduino users?.
thank you
8. Johnny says:
check this out http://www.engineers-excel.com/Apps/PID_Simulator/Equations.htm
(You can see engineers actually use kI= 1/I rather than kI=I it might help your tuning that way round, easier 10 or 100 rather than 0.1 or 0.001)
that ref suggests integral term something like [excuse my syntax]
Iterm=Ki*(Iold + CfAngleX*tstep)
Iold=Iold + CfAngleX*tstep
or I guess equivalently
Iold=Iold+CfAngleX*tstep
Iterm=Ki*Iold
…where tstep is the time step.
When tstep is constant, like you had, it can be ignored and then effectively becomes absorbed into KI
…and I think is a little different from your formula
iTerm += KI * CFangleX;
My version only applies KI to the ‘new’ integral, yours saves it as part of the old integral.
I’m not quite sure whether that will make much difference in reality, but it will affect the numerical value of KI you end up with.
9. What is the chassis made of? I mean those plates with many holes for screws, and all these things. Couldn’t find them in shops.
10. And do the holes and spacings between them fit all the things you connected with screws? I mean the pololu motor brackets and screw holes on the PCBs.
1. I had no issues with the motor brackets, but I had to improvise when connecting the PCB
11. HoaNT says:
Great work!
I dont know if PID is right if we put some object to one side of the balancing bot. Do we need to tune PID values again?
Thank you so much.
12. Can you explain why you took he full 2000 dps scale on the gyro ? I assume 1000 dps would be more than enough but would come with the advantage of a
a higher resolution.
Nice one. May I get the circuit diagram for all connections? I’m still new in using Raspberry Pi
14. zack says:
the output of the PID will attach to the speed of the motor right? do you get the output value in range of 0-255?
1. the output I use wasd -128 to +128. Anything in the negative was reverse.
15. trungvn says:
hi, can you give me program PID for self robot?? Thank you
!
16. nicola says:
Hi , good work , but……
your floor is very helpfull …… if yuo try on table the results will not the same…..
Nicola.
17. foo says:
Awesome!
Thanks for explanation and especially the code!
18. Lenny says:
What motor driver are you using and how do you control your motor? SoftPWM, HardwarePWM or with I2C (saw your github code). Thanks!
19. Harry says:
Hello Guys,
I have decided to build this robot to learn electronics/programming.
got a couple of questions. Is this possible to make this robot using the sense hat?
What did you use to build the frame of the robot?
Thanks heaps
20. Hi!
This post has been really helpful! Very well explained.
Im building from scratch a raspberry pi based quadcopter and now I want to code my PID controller.
My doubt is about how to control the loop times, frequency,…
How did you do in your project?
Thanks!!
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https://www.traditionaloven.com/tutorials/distance/convert-decimeter-dm-to-light-hours-lh.html | 1,596,921,209,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738351.71/warc/CC-MAIN-20200808194923-20200808224923-00297.warc.gz | 870,760,793 | 17,444 | Convert dm to lh - lhr | decimeter to light-hours
# length units conversion
## Amount: 1 decimeter (dm) of length Equals: 0.000000000000093 light-hours (lh - lhr) in distance
Converting decimeter to light-hours value in the length units scale.
TOGGLE : from light-hours into decimeters in the other way around.
## length from decimeter to light-hour conversion results
### Enter a new decimeter number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
How many light-hours are in 1 decimeter? The answer is: 1 dm equals 0.000000000000093 lh - lhr
## 0.000000000000093 lh - lhr is converted to 1 of what?
The light-hours unit number 0.000000000000093 lh - lhr converts to 1 dm, one decimeter. It is the EQUAL length value of 1 decimeter but in the light-hours distance unit alternative.
dm/lh - lhr length conversion result From Symbol Equals Result Symbol 1 dm = 0.000000000000093 lh - lhr
## Conversion chart - decimeters to light-hours
1 decimeter to light-hours = 0.000000000000093 lh - lhr
2 decimeters to light-hours = 0.00000000000019 lh - lhr
3 decimeters to light-hours = 0.00000000000028 lh - lhr
4 decimeters to light-hours = 0.00000000000037 lh - lhr
5 decimeters to light-hours = 0.00000000000046 lh - lhr
6 decimeters to light-hours = 0.00000000000056 lh - lhr
7 decimeters to light-hours = 0.00000000000065 lh - lhr
8 decimeters to light-hours = 0.00000000000074 lh - lhr
9 decimeters to light-hours = 0.00000000000083 lh - lhr
10 decimeters to light-hours = 0.00000000000093 lh - lhr
11 decimeters to light-hours = 0.0000000000010 lh - lhr
12 decimeters to light-hours = 0.0000000000011 lh - lhr
13 decimeters to light-hours = 0.0000000000012 lh - lhr
14 decimeters to light-hours = 0.0000000000013 lh - lhr
15 decimeters to light-hours = 0.0000000000014 lh - lhr
Convert length of decimeter (dm) and light-hours (lh - lhr) units in reverse from light-hours into decimeters.
## Length, Distance, Height & Depth units
Distance in the metric sense is a measure between any two A to Z points. Applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
# Converter type: length units
First unit: decimeter (dm) is used for measuring length.
Second: light-hour (lh - lhr) is unit of distance.
QUESTION:
15 dm = ? lh - lhr
15 dm = 0.0000000000014 lh - lhr
Abbreviation, or prefix, for decimeter is:
dm
Abbreviation for light-hour is:
lh - lhr
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing decimeters and light-hours ( dm vs. lh - lhr ) measures exchange.
2. for conversion factors between unit pairs.
3. work with length's values and properties.
To link to this length decimeter to light-hours online converter simply cut and paste the following.
The link to this tool will appear as: length from decimeter (dm) to light-hours (lh - lhr) conversion.
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https://learn.careers360.com/ncert/question-take-a-thick-white-sheet-fold-the-paper-once-draw-two-line-segments-of-different-lengths-as-shown-in-fig-312-cut-along-the-line-segments-and-open-up-you-have-the-shape-of-a-kite-fig-313-has-the-kite-any-line-symmetry-fold-both-the-diago/ | 1,638,406,381,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00569.warc.gz | 435,102,178 | 115,107 | # Take a thick white sheet. Fold the paper once. Draw two line segments of different lengths as shown in Fig 3.12. Cut along the line segments and open up. You have the shape of a kite (Fig 3.13). Has the kite any line symmetry? Show that and are congruent. What do we infer from this? Fold both the diagonals of the kite. Use the set-square to check if they cut at right angles. Are the diagonals equal in length? Verify (by paper-folding or measurement) if the diagonals bisect each other. By folding an angle of the kite on its opposite, check for angles of equal measure. Observe the diagonal folds; do they indicate any diagonal being an angle bisector? Share your findings with others and list them. A summary of these results are given elsewhere in the chapter for your reference.
Kite has symmetry along AC diagonal. and are congruent and equal triangles.
Diagonals AC and BD are of different lengths.
Diagonals bisect each other.
The two diagonals AC and BD bisect .
## Related Chapters
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https://softkeys.uk/blogs/blog/how-to-use-absolute-reference-in-excel | 1,718,959,310,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00614.warc.gz | 461,669,496 | 59,971 | Blog
# How to Use Absolute Reference in Excel?
Excel is a powerful tool used by many companies to assist with data analysis, tracking and forecasting. One of the core features of Excel is the ability to use absolute references when creating formulas and functions. In this article, we’ll look at how to use absolute references in Excel, as well as some common scenarios where they can be used. With this knowledge, you can take your Excel skills to the next level and get more value out of your data.
## What is an Absolute Reference in Excel?
An absolute reference in Excel is a cell reference that remains the same throughout a formula, even when the formula is moved or copied to a different cell. It is indicated by inserting a dollar sign (\$) in front of the row and column references of the cell. This ensures that the cell reference remains the same, regardless of where the formula is located.
Absolute references are very useful when you want to refer to the same cell or range of cells in multiple formulas. For example, if you want to use a certain cell’s value in several formulas, you can use an absolute reference to refer to that cell in each formula.
## How to Create an Absolute Reference in Excel
Creating an absolute reference in Excel is very easy. All you need to do is insert a dollar sign (\$) in front of the row and column references for the cell you want to reference. For example, if you wanted to reference cell B4, you would enter the following: \$B\$4.
You can also use absolute references for ranges of cells. For example, to reference a range of cells from B4 to B6, you would enter the following: \$B\$4:\$B\$6.
### Using Absolute References in Formulas
Once you have created an absolute reference, you can use it in a formula just like any other cell reference. For example, if you had a formula that added the values in B4 and B5, you could use an absolute reference to refer to those cells in the formula. The formula would look like this: =\$B\$4+\$B\$5.
Using absolute references in formulas is especially useful when you need to refer to the same cell or range of cells in multiple formulas. This way, if you move or copy the formulas, the reference will remain the same.
### Using Absolute References in Functions
You can also use absolute references in functions such as SUM and AVERAGE. For example, if you wanted to calculate the average of the values in B4 to B6, you could use the following formula: =AVERAGE(\$B\$4:\$B\$6).
Using absolute references in functions is very useful when you want to calculate the same range of cells in multiple formulas. This way, if you move or copy the formulas, the reference will remain the same.
## Using Absolute References in Charts
Absolute references can also be used when creating charts in Excel. For example, if you wanted to create a chart that displays the values in B4 to B6, you could use the following formula: =\$B\$4:\$B\$6.
Using absolute references in charts is especially useful when you need to refer to the same range of cells in multiple charts. This way, if you move or copy the charts, the reference will remain the same.
## Using Absolute References in Conditional Formatting
Absolute references can also be used in conditional formatting rules. For example, if you wanted to apply a rule to the values in B4 to B6, you could use the following formula: =\$B\$4:\$B\$6.
Using absolute references in conditional formatting rules is especially useful when you want to apply the same rule to multiple cells or ranges of cells. This way, if you move or copy the rules, the reference will remain the same.
## Using Absolute References in Macros
Absolute references can also be used in macros. For example, if you wanted to refer to the values in B4 to B6 in a macro, you could use the following formula: =\$B\$4:\$B\$6.
Using absolute references in macros is especially useful when you need to refer to the same range of cells in multiple macros. This way, if you move or copy the macros, the reference will remain the same.
## Top 6 Frequently Asked Questions
### What is an Absolute Reference in Excel?
An absolute reference in Excel is a cell address that doesn’t change when a formula is copied or moved to another cell. When a cell reference is made absolute, the cell reference is preceded with a dollar sign (\$) which locks the reference to a specific cell. This is useful when referring to cells in a formula, so that the cell reference remains the same even when the formula is moved around.
### What are the Benefits of using an Absolute Reference?
Using an absolute reference in Excel has many benefits. It allows formulas to be copied and moved without having to update the cell references, which can save time and effort. It also ensures that the formula is always referencing the same cells, which can help to prevent errors. Additionally, absolute references can make formulas easier to read and understand, as they clearly indicate which cells are being referenced.
### How do You Create an Absolute Reference?
To create an absolute reference in Excel, you need to place a dollar sign (\$ ) in front of the column and row of the cell you are referencing. For example, if you want to reference cell A1, you would use the absolute reference \$A\$1. You can also make only one part of the reference absolute by using only one dollar sign. For example, if you want to reference the column A but allow the row to change, you would use the absolute reference \$A1.
### What are Relative and Mixed References?
Relative references in Excel are the opposite of absolute references. They do not contain any dollar signs and the cell reference will change when the formula is copied or moved. A mixed reference contains a combination of both absolute and relative references. For example, if you want to reference the row of a cell but allow the column to change, you would use the mixed reference A\$1.
### When Should You Use an Absolute Reference?
Absolute references should be used when you want to ensure that the formula always references the same cell, even when it is copied or moved. This is especially useful when you are referencing cells in a formula that are located outside of the current range of cells.
### What are some Examples of Absolute References?
Some examples of absolute references include \$A\$1, \$B2, and \$C\$3. In each of these examples, the column and row references are locked, meaning that they will not change when the formula is copied or moved.
### How To Use Relative & Absolute Cell References In Excel
In conclusion, we can see that using absolute references in Excel is a great way to make sure that your formulas are accurate and up to date. Absolute references enable you to create formulas that remain constant, no matter how much you move the data around. Knowing how to use them is essential for any Excel user, from beginner to advanced. With a few simple steps, you can quickly and easily incorporate absolute references into your work.
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# What are the first five common multiples of four?
Wiki User
2016-11-29 15:36:18
The first five multiples of 4 are 4, 8, 12, 16 and 20. For them to be common, they need to be compared to another set of multiples.
Wiki User
2016-11-29 15:36:18
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http://swagatam.expertscolumn.com/article/how-balanced-microphone-preamplifier-circuits-work | 1,508,345,206,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823016.53/warc/CC-MAIN-20171018161655-20171018181655-00858.warc.gz | 344,316,518 | 13,708 | The article discusses how a balanced microphone preamplifier circuit functions and also provides a simple circuit which may be easily built at home and used for personal entertainment in conjunction with your existing power amplifier unit.
A balanced microphone preamplifier system may be defined as an amplifier circuit or a device consisting of two discrete inputs where only the difference in the applied signals is amplified. Therefore this amplifier is also termed as differential amplifier.
Now for understanding the working of such differential amplifiers, let’s consider the diagram provided below.
The diagram shows an Opamp configuration, consisting of two resistors R1 (@ inverting
i/p) and R4 (@non-inv i/p) connected to its two inputs, one resistor R2 (9 times R1 and R4) is connected across the non-inverting input and the output of the opamp, and another resistor R3(same as R2) connected across the inverting input and ground.
For keeping things simpler, let’s assume the gain of the opamp to be 9, such that R1=R4= 1 and R2=R3=9. The precise units are irrelevant here as ae are only concerned with the ratio values.
Let’s also assume the end of
R1 to be at 0 potential and that of R4 to be at 100mV If we consider the opamp to be an ideal one, we may expect two things from the set configuration.
No current will be taken inside the inputs and the inputs will adjust such that the output shows a zero potential. The above terms means that there must be a 100mV intact across R4 and subsequently a potential of 900mV across the feedback resistor R2.
This will lead to an output of -900mV. Now if we reverse the input voltage conditions such that end of R1 is held at +100mV and the end of R4 at 0, then the voltage at the inverting input can be calculated as:
V at end of R1 * R3/R1+R4 = 90mV
READ MORE at AVENUE SOUND BLOGS
### Article Written By Swagatam
Swagatam is a blogger at Expertscolumn.com
Last updated on 26-07-2016 109 | 468 | 1,960 | {"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-2017-43 | longest | en | 0.909326 |
https://www.physicsforums.com/threads/two-signs-for-rate-of-change-of-angle-in-polar-coordinates.794071/ | 1,618,753,793,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00437.warc.gz | 1,050,865,637 | 20,210 | # Two Signs for Rate of Change of Angle in Polar Coordinates
## Homework Statement
I didn't know if this was considered "advanced" physics, but it's an intermediate classical mechanics course so I'll just post my question here. Basically, if you have a cardioid ##r(\theta)=k(1+\cos(\theta))##, you can show that the ##\dot{\theta}=\frac{v}{\sqrt{2kr}}##. That means for a given ##r## with ##v## constant, the rate of change of the angle is both positive and negative. But what does this actually mean?
## Homework Equations
Description of (2D) motion in Polar Coordinates
## The Attempt at a Solution
I was thinking it could possibly have something to do with the cosine function being even (i.e. ##\cos(-\theta) = \cos(\theta)## but I don't understand the implications.
mfb
Mentor
What is v? The velocity along your curve? A non-zero v does not allow r to be constant then.
How is the change in angle "both positive and negative"? The sign of the change in angle depends on the direction of v, of course.
What is v? The velocity along your curve? A non-zero v does not allow r to be constant then.
How is the change in angle "both positive and negative"? The sign of the change in angle depends on the direction of v, of course.
It's a constant speed ##v## along the curve. ##r## isn't constant. The graph of ##r## is a cardioid.
Chestermiller
Mentor
It's a constant speed ##v## along the curve. ##r## isn't constant. The graph of ##r## is a cardioid.
If ##\vec{r}## is the position vector drawn from the origin to the moving particle traveling at constant velocity v along the curve, then we can write:
$$\vec{r}(θ)=r(θ)\vec{i}_r(θ)$$
where ##\vec{i}_r(θ)## is the unit vector in the radial direction at angular position θ. The time derivative of this position vector is the the velocity vector of the particle along the curve:
$$\vec{v}=\frac{d\vec{r}(θ)}{dt}=\frac{d[r(θ)\vec{i}_r(θ)]}{dt}$$
Do you know how to take the time derivative of the right hand side of this equation to determine the velocity vector?
Chet
If ##\vec{r}## is the position vector drawn from the origin to the moving particle traveling at constant velocity v along the curve, then we can write:
$$\vec{r}(θ)=r(θ)\vec{i}_r(θ)$$
where ##\vec{i}_r(θ)## is the unit vector in the radial direction at angular position θ. The time derivative of this position vector is the the velocity vector of the particle along the curve:
$$\vec{v}=\frac{d\vec{r}(θ)}{dt}=\frac{d[r(θ)\vec{i}_r(θ)]}{dt}$$
Do you know how to take the time derivative of the right hand side of this equation to determine the velocity vector?
Chet
$$\frac{d[r(θ)\vec{i}_r(θ)]}{dt} = \frac{dr(\theta)}{d\theta}\frac{d\theta}{dt}\vec{i}_r(θ) + r(\theta)\frac{d\vec{i}_r(θ)}{d\theta}\frac{d\theta}{dt}$$
Chestermiller
Mentor
$$\frac{d[r(θ)\vec{i}_r(θ)]}{dt} = \frac{dr(\theta)}{d\theta}\frac{d\theta}{dt}\vec{i}_r(θ) + r(\theta)\frac{d\vec{i}_r(θ)}{d\theta}\frac{d\theta}{dt}$$
Right. Are you also aware that
$$\frac{d\vec{i}_r(θ)}{dθ}=-\vec{i}_θ$$
So, substitute this and your equation for r(θ) into the equation for velocity, and also factor the dθ/dt. What do you get for the velocity vector?
Chet
Right. Are you also aware that
$$\frac{d\vec{i}_r(θ)}{dθ}=-\vec{i}_θ$$
So, substitute this and your equation for r(θ) into the equation for velocity, and also factor the dθ/dt. What do you get for the velocity vector?
Chet
Shouldn't that be positive? $$\frac{d\hat{r}}{dθ}=\hat{\theta}$$
Chestermiller
Mentor
Shouldn't that be positive? $$\frac{d\hat{r}}{dθ}=\hat{\theta}$$
Oooops. You're right. Senior Moment. OK, Please continue.
Chet
Then, $$\vec{v}(\theta)=\dot{\theta}[-k\sin(\theta)\hat{r}+k(1+\cos(\theta))\hat{\theta}]$$
Chestermiller
Mentor
Then, $$\vec{v}(\theta)=\dot{\theta}[-k\sin(\theta)\hat{r}+k(1+\cos(\theta))\hat{\theta}]$$
OK. Nice. Now also factor out the k, and then take the dot product of ##\vec{v}## with itself to get the square of its magnitude v2. What do you get?
Chet
##v^2=k^2\dot{\theta^2}[\sin^2(\theta)+1+2cos(\theta)+\cos^2(\theta)]=k^2\dot{\theta^2}[2+2\cos(\theta)]=2k\dot{\theta^2}r##. I did all of this before we did the problem (this is how I got the expression for ##\dot{\theta}## in the original post).
Chestermiller
Mentor
##v^2=k^2\dot{\theta^2}[\sin^2(\theta)+1+2cos(\theta)+\cos^2(\theta)]=k^2\dot{\theta^2}[2+2\cos(\theta)]=2k\dot{\theta^2}r##. I did all of this before we did the problem (this is how I got the expression for ##\dot{\theta}## in the original post).
Oh. I was confused. I thought that is what you were trying to show.
Now I see that you were saying that dθ/dt has to be both positive and negative, and you were wondering how that can be. It's not positive and negative at the same time, right. If v is pointing clockwise, then it is one sign, and if v is pointing counter clockwise, then it's the other sign. Is that what the issue was? I guess mfb already said that in post #2.
Chet
But ##\dot{\theta}## is a magnitude. How can this magnitude be negative?
Chestermiller
Mentor
Suppose you associate the minus sign with the unit vector in the θ direction, rather than with the magnitude ##\dot{θ}##. So you have the vector ##\dot{θ}(-\vec{i}_θ)##. Does that work for you?
EDIT: What I really meant to say is that, in the case where the object is moving in the negative θ direction, we express the θ component of velocity as ##\vert r\dot{θ}\vert (-\vec{i}_θ)##.
Chet
Last edited: | 1,689 | 5,412 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-17 | longest | en | 0.900707 |
https://community.oracle.com/thread/2563411 | 1,511,090,510,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805541.30/warc/CC-MAIN-20171119095916-20171119115916-00280.warc.gz | 596,331,845 | 22,608 | 4 Replies Latest reply on Jul 25, 2013 12:37 AM by Frank Kulash
# ROUND and TRUNC function
Guys,
About the query below, how is the output and how SQL treats that?
SELECT TRUNC(ROUND(156,00,-1),-1)
FROM DUAL
Thank you!
• ###### 1. Re: ROUND and TRUNC function
Don't be afraid of breaking Oracle by actually trying things.
How is the output when you execute it?
• ###### 2. Re: ROUND and TRUNC function
Hi rp0428,
Let me correct it. I know the output is 160, but wants to understand how it's done. I read the oracle documentation and understand how ROUND and TRUNC works, but I can't understand the result.
Thank you!
• ###### 3. Re: ROUND and TRUNC function
Don't be afraid of breaking Oracle by actually trying things.
Break it into pieces and look at each piece one at a time.
Remove the TRUNC operation and see what you get. Do you understand why you get that result? If not, post it and we can help explain it.
But you need to make the effort yourself.
See the SQL Language doc. It has an example just like yours.
http://docs.oracle.com/cd/B28359_01/server.111/b28286/functions142.htm
>
`ROUND` returns `n` rounded to `integer` places to the right of the decimal point. If you omit `integer`, then `n` is rounded to zero places. If `integer` is negative, then `n` is rounded off to the left of the decimal point.
The following example rounds a number one digit to the left of the decimal point:
`SELECT ROUND(15.193,-1) "Round" FROM DUAL; Round ---------- 20 `
>
Which means your number, 156,00, will get rounded one digit to the left of the decimal point. So 156 gets rounded to 160.00 since the '6' is equal to or larger than '5' it gets rounded up.
That leaves a result of 160.00. The TRUNC operation does something similar excepts instead of rounding one place left it truncates. But since one place left already has a '0' there is nothing for TRUNC to do so you still have 160.00
• ###### 4. Re: ROUND and TRUNC function
Hi,
eb816488-cca6-45a3-9b74-db582530b8a7 wrote:
Hi rp0428,
Let me correct it. I know the output is 160, but wants to understand how it's done. I read the oracle documentation and understand how ROUND and TRUNC works, but I can't understand the result.
Thank you!
If you've read the manual, then you can ask a question like "Why does this query return 160? I expected it to return ... because, in the section on ROUND, the manual says ... so I expect ROUND to return ... but when I run ROUND by itself, I get ...". That way, you can get an answer that really focuses on what you need to know. | 672 | 2,562 | {"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-2017-47 | latest | en | 0.869965 |
https://blogs.msmvps.com/richardsiddaway/2012/05/28/some-thoughts-on-hash-tables/ | 1,643,142,745,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00209.warc.gz | 205,783,048 | 10,374 | ## Some thoughts on hash tables
Posted by: | May 28, 2012 | 2 Comments |
I need to test the existence of an element in a hash table.
First create a hash table
\$myht = DATA {
ConvertFrom-StringData -StringData @’
1 = aaaaaaa
2 = bbbbbbb
3 = ccccccc
4 = ddddddd
5 = eeeeeee
‘@
}
I tend to create them using ConvertFrom-StringData as it simplifies typing, I have a template for this and its easy to see what entries I’m making
lets see what happens
"1 = \$(\$myht[‘1’])"
"6 = \$(\$myht[‘6’])"
1 = aaaaaaa
6 =
The first test works as we would expect but interestingly we don’t get an error message when we try to access an element that doesn’t exist
So we can build on that
"1","6" | foreach {
if (\$(\$myht[\$_])){
"\$_ = \$(\$myht[\$_])"
}
else {
}
}
a simple logic test shows if the element exists or not and we can branch from there
Life gets a bit more interesting when we are dealing with boolean and NULL values. In this case I built hash table with three values as shown, true, false and null
\$myht2 = @{"1"=\$true; "2"=\$false; "3"=\$null}
if we run the same if statement
"1","2", "3", "4" | foreach {
if (\$myht2[\$_]){
"\$_ = \$(\$myht2[\$_])"
}
else {
}
}
we get this
1 = True
which isn’t quite true
So lets try a switch statement
"1","2", "3", "4" | foreach {
\$x = \$_
switch (\$myht2[\$_]){
\$true {"\$x = \$(\$myht2[\$x])"}
\$false {"\$x is false"}
\$null {"\$x is null"}
}
}
1 = True
2 is false
3 is null
4 is null
Answers for 1,2 & 3 are OK but we can’t differentiate between an element that is null or a non-existent element.
The answer is that we need to think carefully about the data in our hash tables and how we test for existance
under: PowerShell Basics
1. By: Stephen Mills on May 28, 2012 at 8:12 pm
I always use the ContainsKey method. It avoids all the problems you mentioned. If it is there, it returns true, otherwise false.
1..6 | % { “\$_ Exists \$(\$MyHT.ContainsKey(“\$_”))” }
1 Exists True
2 Exists True
3 Exists True
4 Exists True
5 Exists True
6 Exists False
1..4 | % { “\$_ Exists \$(\$MyHT2.ContainsKey(“\$_”))” }
1 Exists True
2 Exists True
3 Exists True
4 Exists False
2. By: Chris Warwick on May 29, 2012 at 9:03 am
Hi Richard,
Try the .ContainsKey() method:
PS> \$ht=@{“1″=\$true;”2″=\$false;”3″=\$null}
PS> ‘1’,’2′,’3′,’4’|%{\$myht2.containskey(\$_)}
True
True
True
False
PS>
hth!
Cheers,
Chris | 771 | 2,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-05 | latest | en | 0.751769 |
https://nl.mathworks.com/matlabcentral/fileexchange/29851-runge-kutta-4th-order-ode | 1,590,694,210,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399830.24/warc/CC-MAIN-20200528170840-20200528200840-00587.warc.gz | 466,316,472 | 20,929 | File Exchange
## Runge Kutta 4th order ode
version 1.4.0.0 (1.32 KB) by Judah S
### Judah S (view profile)
solves ode using 4th order Runge Kutta method
401 Downloads
Updated 16 Jan 2013
View License
This code defines an existing function and step size which you can change as per requirement.
P.S: This code has no new feature compared to existing codes available online. Intention behind posting this very simple code is to help students understand the concept and solve assignments.
### Cite As
Judah S (2020). Runge Kutta 4th order ode (https://www.mathworks.com/matlabcentral/fileexchange/29851-runge-kutta-4th-order-ode), MATLAB Central File Exchange. Retrieved .
Soumyadeep
Yaman Yucel
### Yaman Yucel (view profile)
It does not work for y(0) = 0 initial condition
Nidhi Menon
Abraham
### Abraham (view profile)
Great work! What about a code for Runge Kutta method for second order ODE. Something of this nature:
d^2y/dx^2 + 0.6*dy/dx 0.8y = 0
Thank you
math16
Sumith YD
Dogba Djaze
masoud rahmani
Tareque Hossain
### Tareque Hossain (view profile)
how can i solve SIR model using RK4 method in matlab? can you write the code please
faiz islam
### faiz islam (view profile)
sir can you assist me ,that how we can apply 4th order Runge kutta method for 4 coupled equation?
dx/dt=−ax − eω + yz
dy/dt= by + xz
dz/dt= cz + fω − xy
dω/dt = dω – gz
a = 50, b =−16, c = 10, d = 0.2, e = 10, f = 16, g = 0.5
Step size 0.001 .
regards
faiz
Ali Abbas
### Ali Abbas (view profile)
@Shahzaib Asif Very helpful program.JazakAllah
Stephanie Valerio
### Stephanie Valerio (view profile)
How do I run/call to this code?
Hamza Fokraoui
### Hamza Fokraoui (view profile)
for this function : f'''' - f*f''' + 4*g = 0
where i need to insert it in this code?
thank you
Shahzaib Asif
### Shahzaib Asif (view profile)
function RK4(f,a,x0,y0,h)
% Runge Kutta Method 4th Order
% function @(x,y) e.g. f=@(x,y)(x+y);
% a = the point up to which you obtain the results
% x0 = initial condition of x
% y0 = initial condition of y
% step size
x = x0:h:a;
y(1) = y0;
for i=1:(length(x)-1)
k1 = f(x(i),y(i));
k2 = f(x(i)+0.5*h,y(i)+0.5*h*k1);
k3 = f((x(i)+0.5*h),(y(i)+0.5*h*k2));
k4 = f((x(i)+h),(y(i)+k3*h));
y(i+1) = y(i) + (1/6)*(k1+2*k2+2*k3+k4)*h;
end
y(:)
%Shahzaib Asif (zaibi7402)
%shahzaib.7402@gmail.com
Chris FUNG
clear coding
Christoph
Ying
Ying
### Ying (view profile)
Very good to learn. Thanks.
Arun
Pi Ting
excellent work
Ido
### Ido (view profile)
Excellent program,
very helpful.
### Updates
14 Jan 2013 1.4.0.0 Just an update 25 Jan 2012 1.1.0.0 Nothing much
##### MATLAB Release Compatibility
Created with R2010a
Compatible with any release
##### Platform Compatibility
Windows macOS Linux | 938 | 2,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-24 | longest | en | 0.757697 |
https://linguistics.stackexchange.com/questions/25757/what-statistical-tests-can-be-used-to-check-for-lexicalisation-effects-in-a-judg/25797 | 1,670,099,354,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00849.warc.gz | 403,951,221 | 37,308 | # What statistical tests can be used to check for lexicalisation effects in a judgement task?
In a run-of-the-mill judgement rating task where participants have to rate sentences on a Likert scale (e.g. 1 to 6) and that is constructed using a Latin square design, what statistical tests can be used to check for lexicalisation effects and which is the most commonly used?
Update:
By lexicalisation, I mean checking the different realisations of each condition. So if there's a variable 'transitivity' giving two conditions, i.e. transitive and intransitive, I want to test whether any of the choices of transitive verbs behave in a different way to the other choices of verbs in that condition. Can it be shown just using t-tests?
• Mann-Whitney U test, Chi square. This is because of the data produced by Likert judgements. How you test for lexicalisation effects depends entirely on your hypotheses: what does lexicalisation mean in your particular case, and what effect would you expect it to exert on the judgements? Without working out the hypotheses first you can't "test" anything. Aug 23, 2017 at 10:13
• @FlorianBreit Thanks. By lexicalisation, I mean checking the different realisations of each condition. So if there's a variable 'transitivity' giving two conditions, i.e. transitive and intransitive, I want to test whether any of the choices of transitive verbs behave in a different way to the other choices of verbs in that condition. Can it be shown just using t-tests? Is an items analysis needed? Aug 23, 2017 at 11:26
• In principle I think yes, and I'd go with the MWU here because it's difficult to establish that the conditions for the t-test are met (but if you can, go ahead and use a regular t-test). The test can show that it is likely that the two groups (transitive and intransitive in this case) behave differently and exclude to a degree that this difference is due to chance. No test can show that they are in fact different though, and more importantly, that the underlying cognitive division is the driver of the difference, those things are up to your arguments for the experimental choices. Aug 23, 2017 at 12:27
• Thanks. I've updated the question to include the example. If you put your comments as an answer, I'll mark it as the accepted answer. Aug 23, 2017 at 13:53 | 530 | 2,309 | {"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} | 2.875 | 3 | CC-MAIN-2022-49 | latest | en | 0.928892 |
https://skepticsplay.blogspot.com/2011/05/one-universes-favorite-digit.html?showComment=1304460393889 | 1,721,102,967,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00318.warc.gz | 465,977,441 | 18,401 | ## Monday, May 2, 2011
### One: the universe's favorite digit
Out of all the digits, from zero to nine, one is the most common. This has to do with the log scale.
The log scale captures an important fact that is true of many quantities in life. Take money for instance. If you have one dollar, then earning another dollar is great because you've doubled your money! If you have a million dollars, earning another dollar does not make much of a difference. Small changes matter less the more you already have.
This is true on a log scale too. On a log scale, 1 is the same distance from 2 as 100 is from 200. The higher you go up, the more the numbers all get smooshed together. What does that mean for the digits from zero to nine?
In the above picture, I show a log scale. And on that scale, I highlighted in blue all the regions where 1 is the first digit of the number. You should see that the blue regions cover more than one tenth of the log scale. In fact, they cover about 30%. And so, if we pick numbers randomly on the log scale, about 30% of those numbers will have 1 as their first digit.
Just for fun, let's apply this concept on the fundamental constants of nature. I will compare two hypotheses:
1. The fundamental constants of nature are distributed on a log scale. About 30% of the constants will have 1 as their first digit, 18% will have 2 as their first digit, and so on.
2. Each digit from 1 to 9 is equally likely to be the first digit of the fundamental constants of nature.
Of course, it could be that both hypotheses are wrong. It's difficult to say that these constants are really random. To talk about randomness, we'd need a whole collection of universes with different sets of fundamental constants so we can do some statistical analysis. But clearly we have only one universe. Luckily this one universe has 26 fundamental constants (as far as we know), which may be a big enough collection of numbers to do some statistical analysis.
The caveat is that there is more than one way to choose our 26 fundamental constants! For instance, the mass of the u quark in planck units could be considered a fundamental constant. But instead of this constant, we could use a different constant: the ratio of the mass of the u quark to the mass of the electron. It's possible I could bias my results by "choosing" the fundamental constants that confirm my hypothesis.
Therefore, I will not choose which constants to use. I will simply use the list compiled by John Baez and David Black.* Note that 6 of the constants are unknown, so we'll have to make do with the remaining 20.
*Baez offers two sets of equivalent constants for the CKM matrix. I just used the first set.
Results
The frequency of first digits more resembles hypothesis 1 than hypothesis 2. I also tried doing some Bayesian analysis. If our prior belief is that the likelihood of each hypothesis is 1:1, then this evidence increases the ratio to 15.8:1. In other words, if we previously thought they were equally likely, then after seeing this evidence, hypothesis 1 is almost 16 times as likely. As far as evidence goes, this is fairly weak evidence, but you're not going to get much better with only 20 fundamental constants.
Does this really mean fundamental constants are randomly distributed on a log scale? Gee, I don't know. Probably not really. What's your interpretation?
Eduard said...
What about mathematical constants like Pi, e, Gamma, plastic number, golden section, ... volume to surface ratios of polyhedra, ratios of of cutting length CL to side length in minimal CL problems?
miller said...
Oh Eduard, great question! When I have the chance, I'll try to apply the analysis to Wikipedia's list of mathematical constants.
miller said...
Okay, here are the results:
[Digit]/[frequency]/[expected frequency]
1/16/12.6
2/9/7.4
3/4/5.2
4/1/4.1
5/3/3.3
6/4/2.8
7/2/2.4
8/2/2.1
9/1/1.9
Bayesian analysis shows that this is 840,000 times more likely than the hypothesis that all first digits are equally likely.
Rain said...
I agree that 1 is the universe's favourite digit, but it hardly has anything to do with the log scale, in my opinion (although 0 is, i believe, the universe's other favourite digit, and log_b(1)=0 for every complex number b... which is rather cool). In fact, there is nothing special about the particular log operator you're using (log_10); we (humans) only use it because having ten fingers and ten toes makes counting in a decimal system easier for us than in any other system (e.g. hexadecimal, tertiary, binary...); if you plotted log_n with n any integer other than 10, the percentage of numbers beginning with 1 would not be 30% (it would still be greater than the percentage of numbers beginning with any other digit, though).
I believe 1 and 0 are the universe's favourite digits because of their beautiful mathematical properties, not because of the values of what we call the fundamental constants in a particularly comfortable (for us) measuring system (mks). In fact, i would argue that the so-called "natural system of units", where speed and not length is a fundamental quantity and c (the speed of light), h or hbar (Planck's constant), epsilon0 (vacuum permittivity), mu0 (vacuum permeability) and G (Newton's constant) are all valued 1, is the universe's preferred measuring system; it makes much more sense to call the speed of light 1 and calculate all other speeds from there than to call Earth's circumference 40,000 and calculate all other lengths from there. Same goes for Planck's constant and for every other "fundamental constant" there is (and earlier, when i said that only we call them such, i was referring to the fact that other intelligent species in the universe probably have a different set of "fundamental constants"; it all depends on which fundamental quantities (e.g. length, time, speed, area, power, pressure, force, illuminance, electric charge, etc) we are comfortable with using (mks has, among others, length, time and mass, while the "natural system" has, among others, speed, time and mass, for instance).
But on to my point. Leaving 0 aside for another discussion, 1 has many great properties. No matter which measuring system or fundamental quantities you're using, 1 is always the unit, the base number from which all others are constructed, the first integer, the first intuitive or "counting" number. We define everything in terms of 1; we define 2 as 1+1, 3 as 2+1, and so on; we define the multiplicative inverse of a number A as that number whose product with A equals 1; we are able to tell which of two quantities is larger by dividing them and comparing the result with 1 (or subtracting them and comparing with 0, but i said i'd leave 0 aside for now); and so on.
Although someone might argue that that's just the way languages evolved on this planet, it seems natural to make a distinction between 1 and all other numbers: 1 is singular, while all integers from 2 onwards are plural (in some (or all?) slavic languages, there is also a distinction between the numbers 2-5 and the numbers 6-9, but 1 is still special on its own).
(continued on next comment)
Rain said...
This comment has been removed by the author.
Rain said...
(continued from previous comment)
On top of that, there are many mathematical properties involving the numbers 0, 1, pi, e and phi (or multiples of them). Off the top of my head, (da/db)(db/dc)(dc/da)=-1 for any functions a,b,c (da/db indicates the partial derivative of a with respect to b while c remains constant) and e^(i phi)=-1.
I'm sure i could think of some other reasons why 1 is special, but i'm terribly busy this week and i only stopped to read today's article.
In short, i'd say the reasons 1 is one of the universe's favourite digits are more mathematical and conceptual than physical. And, so my opinion doesn't seem biased, i'm a physics student, not a maths one. ;)
One last note: I would agree with Eduard in that 1 isn't the universe's only favourite number (there's pi, e and phi (the golden ratio) as well), but remember that we are talking about digits, not numbers.
miller said...
Alex,
You should check out John Baez's discussion of what constitutes a fundamental constant of nature. I actually didn't use any constants like c, h-bar, or G. I only used unitless constants.
Rain said...
Ah! I see i've made a bit of a fool of myself.
Even though i now see what you mean (and those constants, if i'm reading the table correctly, don't depend on measuring system because they're ratios), i maintain that the fact that 30% of them begin with 1 is only because we use a decimal counting system.
I'll illustrate this with the number pi. In a decimal system, pi equals roughly 3.14159265, so the first digit of pi is 3. However, in a binary counting system pi would be a little over 11 and in a tertiary system pi would be a little over 10; in both of these cases the first digit of pi is 1. This is probably not the best example, since writing nonintegers in nondecimal systems is complicated (at least i don't know how to do it), but i hope you get my point.
Still, it is interesting to note that, at least in a decimal system, 1 takes up just about as much space as you predicted with your log_10 theory.
(Have you noticed that most important numbers in physics (other than 0 and 1) are irrational? I doubt this has anything to do with the cardinality of irrationals as opposed to that of rationals, so it's an interesting phenomenon. It's unrelated to this discussion on first digits, but i noticed it when looking at the fundamental constants of the standard model and thinking about the numbers Eduard mentioned.)
miller said...
Alex,
I agree that the 30% figure has to do with the base 10 system we're using. Under any other base, the frequencies would be different. And there's nothing fundamental about base 10.
I think if a physical constant appeared to be rational, we wouldn't bother giving it a name, because it already has one! Or maybe it's because there are just so many more irrational numbers than rational numbers. Or maybe they really are rational, and we just haven't determined them to sufficient precision.
Rain said...
On second thought, Miller, you're right; it is because there are more irrationals than rationals. The values of (non-unitless) physical constants depend on our definitions of the base units (the metre, the second, etc), as i myself mentioned earlier, so actual physical constants are never multiples of our units; add to that the fact that pi and e, which appear everywhere, are irrationals, and irrational constants are bound to appear much more often than rational ones. I should have seen this the moment i thought about how many irrational constants there are.
There's also the first thing you said; 3 appears all over the place, but we don't call it the "number-of-quarks-per-hadron constant" because it's already called "3".
What do you think?
miller said...
I was trying to think of a constant earlier that was just a rational number, but couldn't think of one. "Number of quarks per hadron"--I like that! | 2,611 | 11,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.917637 |
https://unicyclist.com/t/unicycle-push-along/125283 | 1,709,173,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474775.80/warc/CC-MAIN-20240229003536-20240229033536-00627.warc.gz | 599,106,894 | 10,272 | Unicycle push-along?
I did ask this before, but embedded in my other thread and no-one picked it up. But it’s a serious question. Since my kids learned to ride bikes there as been something of a revolution in little-kid learning with the introduction of the pedal less crank less push-along bikes, propelled by feet on the ground. The plus is that kids get to feel the balance without having to coordinate it with the balance-disturbing activity of pedalling. So when they get a pedal bike, they only have to learn to pedal. So my question is, would this work with a unicycle, or does the physics of one wheel, with its lack of forward backward stability mean that you couldn’t scoot it with your feet? If I could get hold of an old kids unicycle I would try it, but I don’t fancy trying to take the cranks off mine!
Would that work ?
Edit: Oh no I just realised. You need to practice taking your cranks off for when you get more unicycles.
I thought of trying it with just the pedals off, but I can’t get the seat post low enough unless I cut it down. Though, I did think that the shorter post from my 24" might do it if I swapped it over. I have no idea if they are the same diameter, though.
I just about know how to take pedals off, having put them on in the first place, but no idea at all about cranks!
No way as a beginner. In order to control the forward/backward balance on a uni you need to have the possibility of accelerating/braking the wheel. This is normaly done by pedaling. Once a pedal less uni would be slightly in front (or back) of your center of gravity, it will be accelerated by you body weight and just fly away
Basically it is possible as coasting shows, but it is much harder to control than normal riding. If you can do this trick you do not need to modify your uni, just put your feet on the crown when having enough speed.
You would have feet on the ground, though? But I accept it may not be feasible, which is why I thought I would ask rather than try it<G>.
With your feet on the ground, you can not bring the uni under your center of gravity as you have no control over it.
Pedalling is integral to unicycling so I don’t think a trainer uni with no cranks would help.
However I have seen little bikes with a long handle out the back, clamped on under the seat. The adult trainer holds the handle to stabilise the bike while the kid learns.
This could work with a small unicycle. Clearly there is a limit to the size of the child that the adult could support so it would only work with really small children.
A set of handlebars on the end of the training handle would help.
BTW I have read posts about very small children (about three years old) learning to ride a uni. One had not even learnt to ride a bike until later.
So it would be like a wheelbarrow
I personally don’t think there is a shortcut to learn unicycling, but this guy in one of my favourite videos is beeing very creative:
For a child, I don’t think there would be any point in trying to stabilise the unicycle itself with handles etc - better to just hold the child’s hand. I was just trying to work out whether there was any way to split the balance from the pedalling for a child or adult learner.
As has already been said - fundamentally the answer is no, because on a unicycle (unlike a bicycle) the pedalling is integral to the balancing.
To look at the idea of the balance bike from a more fundamental perspective, what it is doing is breaking down the components of riding a bike so that the child only has to learn one at a time. Whilst you can’t quite do the same thing with a uni, you can (and most people do) learn by breaking it down a bit. For example one of the standard things is to ride along a railing, which not only removes the need to balance side to side, it also helps a lot with the fore aft balance and allows you to learn the pedalling and how to balance using that.
Unimyra, that thing is insane! What does it even do? I can’t work it out!
A sort of cantilever principle, I suspect one gets a decent hold on the handle, and when pressure is applied either forward or backward, the seatpost/frame corrects itself to suit the rider balance. I certainly would not entertain one, especially as a learner. It could well be a case of UPD impalement, plus the weight is ridiculous. But, it may work, who knows… ??
If you need training wheels to get the feel for uni, removing the pedals is not going to help.
You can try the ski poles, but not are not great in the long run. Others have experimented with supermarket trolleys and got something out of it. As long as you let it go before you can only do it with the trolley…
Pedalling uphill helped me. It seemed to slow everything down a bit. Just a slight uphill though. I had a rail on my left for that. I could slowly use less and less rail
Lower tire pressures helped at first as well. It made the unicycle less twitchy at first. I then slowly raised the pressures and moved to a flatter surface.
Bottom line is you just have to keep falling off until it clicks - and it does. Then you just fall off less and less!!
My son learned on rough grass and found a flat surface hard to ride on at first.
What a clever idea; a stillstand cheater device! Or it might more usefully be a stillstand training aid.
When you move the lever, it shifts your weight to the front and rear. Never tried anything like it, so it’s hard to tell if it would make for a useful training aid. I get the impression that the guy in the video has just completed it, so is showing what the mechanism does, but isnt’ (yet?) riding the unicycle. I’d like to try one.
If you are going to teach children, don’t let them hold you. Hold their wrist instead, so that they can not drag you down. The last thing you want is a 40lb flailing child trying to hold themselves upright with your finger, either you end up falling on them, or bad things happen to your finger.
Also, that still stand trainer looks very interesting, though very time consuming to make.
Some insurance company actuaries studied old people and broken hips. The only correlation they were able to make was: using a cane increased the likelihood of someone breaking their hip.
A cane is supposed to help, and some people rely on it, but the moment it stops working, bad things can happen. I spent a fair amount of time holding onto a wall or fence as a beginner. In retrospect, I think that increased my chances of falling into the fence or the wall. Once I improved and stopped using the wall, returning to the wall made me feel like I’d forgotten how to ride. The wall was psyching me out.
I ride alongside a fence practicing wheel walking (beginner stages). There are some nice side-to-side hip stretches that can be done while holding onto the wall. For beginners learning how to sit on the seat and turn the pedals, the wall is a good thing. But for learning to actually ride, holding onto something means that both arms can no longer flail madly.
Whenever I read that someone ‘didn’t learn until they ditched the fence’ I am reminded of people who say ‘Why is it that my keys are always in the last place that I look’…
Which is not to say it is wrong, just that I think that at a certain point you are ready to ditch the fence, but that doesn’t mean that ditching it earlier would necessarily have been better.
Didn’t know about the cane making things worse, but it makes sense. I’m an advocate of not using poles/canes as it doesn’t take much to become good at riding with them. And the second (less than a second actually) you throw them to ride “bare”, you fall. I did that when I first learnt, 20 odd years ago, never did any good.
The launching in space is the thing. Scary, but probably the only way. But only once you have a good feeling of the front/back balance (with a fence or a wall). When I learnt again, a couple of years ago, I had a high railing around the building where I was staying. It took me a while to let go of the railing, and I was always trying to stay within grabbing distance. I could have stayed there forever, thank God it had a portion - about 15 to 20ft - without the railing. It became the launch point where I could let go from A to B for a few revolutions. Always good to have a grabbing point you can aim at, like in that great learning video from my favorite scot uni riders:
I made a lot of progress once I voluntarily decided to veer frankly off the railing when I would let go of it. But it took a lot more unplanned dismounts too! | 1,927 | 8,523 | {"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-2024-10 | latest | en | 0.97143 |
https://kidsworksheetfun.com/practice-molarity-and-molality-worksheet-with-answers/ | 1,695,370,181,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00082.warc.gz | 388,977,101 | 25,489 | # Practice Molarity And Molality Worksheet With Answers
Work in groups on these problems. What is the molarity of the solution.
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Molarity Chemistry Tutorial Jessica Fail Really Need To Keep This Up In My Mind Over The Summer Chemistry Lessons Chemistry Classroom Chemistry | 990 | 4,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.823712 |
https://www.jiskha.com/questions/433559/a-70kg-man-stands-on-a-bathroom-scale-in-an-elevator-as-the-elevator-starts-moving-the | 1,610,720,718,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00059.warc.gz | 947,724,766 | 4,715 | # physics
A 70kg man stands on a bathroom scale in an elevator. As the elevator starts moving, the scale reads 90kg
Is the elevator acceleration up or down, is the elevator moving up or down?
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1. i think you mean it's weight (units in N), because mass is constant.
anyway, since the weight increased, the elevator accelerates and moves up. :)
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2. 👎
2. Force up on man - weight of man = m a up
F scale - m g = m a
Fscale = 90*9.8
90*9.8 - 70*9.8 = 70 a
a = 20*9.8 /70
the acceleration is up
since we started from rest, we will be moving up but have not gained any speed yet
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Apparent Weight. A physics student weighing 600 N stands on a bathroom scale in an elevator. As the elevator starts moving, the scale reads 420 N . A )Find the magnitude of the acceleration of the elevator. B ) Find the direction | 858 | 3,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-04 | latest | en | 0.88061 |
https://api-project-1022638073839.appspot.com/questions/how-do-you-differentiate-v-r-4-3pir-3 | 1,723,698,898,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00355.warc.gz | 76,975,121 | 5,738 | How do you differentiate V(r)=4/3pir^3?
$V ' \left(r\right) = 4 \pi {r}^{2}$
differentiate using the $\textcolor{b l u e}{\text{power rule}}$
$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(a {x}^{n}\right) = n a {x}^{n - 1}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\Rightarrow V ' \left(r\right) = 3 \times \frac{4}{3} \pi {r}^{2} = 4 \pi {r}^{2}$ | 208 | 479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-33 | latest | en | 0.454124 |
https://chemistry.stackexchange.com/questions/31075/how-to-calculate-molecular-dipole-moment-from-a-known-wavefunction | 1,718,356,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00528.warc.gz | 159,003,976 | 45,232 | # How to calculate molecular dipole moment from a known wavefunction?
Say I have a molecular wavefunction as a set of molecular orbitals and want to calculate the molecule's dipole moment, but don't know how! I searched a lot but couldn't find any practical example.
$$\psi _i=\sum ^N_{i=1}C_i\mathrm e^{-\alpha _ir^2}$$
• You could use the variational theorem to determine the molecular orbital coefficients of the molecular orbitals (one electron wavefunctions). If you let you molecular wavefunctions be a linear superposition of basis atomic wavefunctions $\Psi =\sum c_i\psi_i$, with the orbital coefficients you can understand key properties of your molecule. You can rationalise trends in bond polarity too, that can't be explained with other theories! Commented May 9, 2015 at 1:19
The necessary formal derivation has already been nicely done by AngusTheMan. I'll start from the last equation:
$$\langle \mu_{z} \rangle = \langle \Psi | \hat{\mu}_{z} | \Psi \rangle$$
where $\Psi$ is the variational wavefunction; it can be any molecular state. It's important that it's variational, otherwise the expectation value approach is not exact. So, this works for SCF, CI, and MCSCF wavefunctions, but extra derivatives need to be taken for Moller-Plesset and coupled cluster wavefunctions. More work needs to be done for multideterminental wavefunctions like CI and MCSCF, but the complexity is no different for a single state in each wavefunction. There may be some MO space partitioning I'm neglecting that's required for MCSCF, so I'll restrict my work to a single-determinental wavefunction.
Expand the wavefunction as a linear combination of molecular orbitals (MOs)
$$\Psi = \sum_{i} \psi_{i},$$
where each molecular orbital is a linear combination of atomic orbitals (AOs)
$$\psi_{i} = \sum_{\mu} C_{\mu i} \phi_{\mu},$$
where $C_{\mu i}$ is the MO coefficient matrix, so our expectation value now looks like this:
$$\langle \mu_{z} \rangle = \sum_{i}^{\textrm{occ MOs}} \sum_{\mu\nu}^{\textrm{AOs}} C_{\mu i} C_{\nu i} \langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle.$$
The indices $\mu,\nu$ run over all AOs, and the index $i$ runs over the occupied MOs. There's only one index because this is a one-electron operator. I'm also neglecting any complex values here, since we almost always work with real-valued AOs and MO coefficients.
We do one last rearrangement. Replace the MO coefficients with the density matrix
$$P_{\mu\nu} = \sum_{i}^{\textrm{occ MOs}} C_{\mu i} C_{\nu i}$$
to give the first explicit "working equation":
$$\langle \mu_{z} \rangle = \sum_{\mu\nu}^{\textrm{AOs}} P_{\mu\nu} \langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle$$
I say first for two reasons: we usually try and avoid explicit loops like this, and the expression can be broken down further, depending on what molecular properties are of interest; I'll be more clear about this later. Inside the sum there are two terms:
• The density matrix $P_{\mu\nu}$, which comes from a converged SCF calculation.
• The integral of the dipole operator over two basis functions, $\langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle$. Atomic orbitals are represented as atom-centered basis functions. These can be calculated once and at any time, since the quantities here don't change over the course of a calculation.
Each of these terms is represented as a matrix. Since the index $\mu$ runs along the rows and $\nu$ runs along the columns for each matrix, "contraction" involves either a matrix product followed by the trace, or an elementwise product followed by an accumulation sum over all matrix elements. There are other details one needs to be careful about, such as what units the result should be in, which changes the prefactor (programs work internally in atomic units), and what the origin for the dipole operator is, but that's really it.
Well, sort of. I'm actually treating some of the program internals as a black box. If you're familiar with Hartree-Fock, it should be clear where $P_{\mu\nu}$ comes from, but what about the integral? For a general expectation value $\langle A \rangle$ with its corresponding operator, where does $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ come from? If it's already available in the code, then you call a wrapper function that then calls the integral engine to do all the nasty work, and you get back a tidy matrix without having to worry about the details. If $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ isn't present, depending on the complexity of $\hat{A}$, there can be a non-trivial amount of derivation required for the working integral equation, followed by the implementation.
Ignoring any possible contraction of primitive basis functions, expand $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ using the definition of $\phi$ in Cartesian coordinates:
$$\phi(\mathbf{r}; \mathbf{A}, \mathbf{a}, \zeta) = (x-A_x)^{a_x} (y-A_y)^{a_y} (z-A_z)^{a_z} e^{-\zeta |\mathbf{r} - \mathbf{A}|^2}$$
where $\mathbf{r} = (x, y, z)$ is the electron position, $\mathbf{A} = (A_x, A_y, A_z)$ is the position of the basis function (almost always atom-centered), and $\mathbf{a} = (a_x, a_y, a_z)$ are the angular momenta for each coordinate, with $l_{\textrm{max}} = a_x + a_y + a_z$ total angular momentum of the basis function. $(0,0,0)$ is an s-function, $(1,1,0)$ and $(0,0,2)$ are d-functions, and so on.
Forming the integral more explicitly gives
$$\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle = \int\int d\mathbf{r}_1 d\mathbf{r}_2 \left[ (x_1-A_x)^{a_x} (y_1-A_y)^{a_y} (z_1-A_z)^{a_z} e^{-\zeta_a |\mathbf{r}_1 - \mathbf{A}|^2} \right] \\ \times\left[ \hat{A} \right]\left[ (x_2-B_x)^{b_x} (y_2-B_y)^{b_y} (z_2-B_z)^{b_z} e^{-\zeta_b |\mathbf{r}_2 - \mathbf{B}|^2} \right]$$
Before going any further, $\hat{A}$ must be defined. If $\hat{A} = 1$, this becomes an overlap integral. The dipole operator in the z-direction is given by $\hat{A} = \hat{\mu}_{z} = -ez = -e(z_3 - C_z)$, where $z_3$ is the integration coordinate and $C_z$ is the origin of the dipole in the z-direction, usually taken to be zero. Everything is kept in atomic units until after the integral/density contraction, so drop the prefactor $-e$. We can now generalize this to an arbitrary Cartesian multipole moment operator,
$$\hat{A} = \mathfrak{M}(\mathbf{r}_3) = (x_3 - C_x)^{c_x} (y_3 - C_y)^{c_y} (z_3 - C_z)^{c_z}$$
where $(c_x, c_y, c_z)$ determine the coordinate of each multipole, and their sum is the total multipole order; for example, $(1,0,0), (0,1,0), (0,0,1)$ are the x-, y-, and z-directions of the dipole operator. The Cartesian moment operator looks just like a Gaussian basis function where $\zeta = 0$.
Once the form of an operator has been derived, it needs to be implemented as part of an integral package, each of which implement one or more algorithms for computing integrals. Each algorithm is named after the authors of the paper in which they were introduced, and are usually abbreviated. For example, the first one I know of is the Taketa, Huzinaga, O-Ohata paper (THO, DOI: 10.1143/JPSJ.21.2313), where explicit working equations are given for 2-center overlap, 2-center kinetic energy, 2-center electron-nuclear attraction, and 4-center electron repulsion integrals. A working implementation can be found in the PyQuante package. I made an IPython notebook translation of the code snippets on the front page of the official documentation.
Other, more complex algorithms are from the Pople-Hehre (PH), McMurchie-Davidson (MD, DOI: 10.1016/0021-9991(78)90092-X), Obara-Saika (OS, DOI: 10.1063/1.450106), Dupuis-Rys-King (DRK), and Head-Gordon-Pople (HGP) papers. I'm sure I'm neglecting some, including the seminal paper by Boys which introduced the use of Gaussian functions as a substitute for Slater-type functions in basis sets. A good review of these algorithms is found in a Peter Gill paper (DOI: 10.1016/S0065-3276(08)60019-2); he is the original author of the integral code in both Gaussian and Q-Chem.
To bring these things full circle, I wrote some code a few months ago to calculate the dipole moment using pyquante2 and a wrapper that calls an implementation of the Obara-Saika recursive integral algorithm. You can find it here with some comparisons to "industrial" quantum programs.
• Thanks all. I really recommend reading the Obara-Saika paper if you're at all interested in integral evaluation. The DALTON package (not sure what the primary integral algorithm is) has an impressive list of one-electron integrals that can be calculated, many of which correspond directly (expectation value times some constants) to molecular properties. Commented Sep 20, 2015 at 1:29
The dipole moment $\mu$ of a molecule is a measure of charge distribution in the molecule and the polarity formed by the nuclei and electron cloud.
We can perturb our system with an external electric field $\vec E$ and gauge the response of the electron cloud and nuclei by the polarisability, i.e how much the dipole moment changes. In practice the nuclei might be so heavy that their motion is not perturbed, while electrons being light are very mobile. If we imagine that the external field is caused by some other species, and that it itself is not changing, we can call this external constant electric field $\vec E$ at least over the volume of the molecule we are considering. Imagine for arbitrary book keeping that we point it down the $z$ axis. We could also investigate how the dipole moment changes with bond vibrations to discuss IR spectroscopy or if the polarisability changes during a vibration to give Raman spectroscopy.
We can use perturbation theory to expand the wavefunction and the molecular energy in terms of small perturbations of the field. We start by Taylor expanding the energy and molecular wavefunction in terms of the electric field which acts as the perturbation parameter. $$E(\vec E)=E^0+\bigg(\frac{\partial E}{\partial \vec E}\bigg)_0\vec E+\bigg(\frac{\partial ^2E}{\partial \vec E^2}\bigg)_0\frac{\vec E^2}{2!}+\bigg(\frac{\partial ^3E}{\partial \vec E^3}\bigg)_0\frac{\vec E^3}{3!}+\dots$$ $$\psi(\vec E)=\psi^0+\bigg(\frac{\partial \psi}{\partial \vec E}\bigg)_0\vec E+\bigg(\frac{\partial ^2\psi}{\partial \vec E^2}\bigg)_0\frac{\vec E^2}{2!}+\bigg(\frac{\partial ^3\psi}{\partial \vec E^3}\bigg)_0\frac{\vec E^3}{3!}+\dots$$ If we use the notation that the wavefunction derivatives are given by; $$\bigg(\frac {1}{i!}\bigg)\frac{\partial ^i\psi}{\partial \vec E^i}=\psi ^{(i)}$$ The Hamiltonian for such a system under the influence of an electric field in the $z$ direction is $\hat H(\vec E)$. $$\hat H(\vec E)=\hat H^0-\vec E\hat \mu _z$$ Where $\hat \mu _z$ is the dipole moment operator that is a summation of the charges of the nuclei and electrons in the molecule. This is the result of Hellmann-Feynman theorem of the energy and the electric field, e.g. $\hat H(\vec E)=\hat H^0 +\hat H^1(\vec E)$, $$\frac{dE}{d\vec E}=\bigg\langle \frac{d\hat H}{d\vec E}\bigg\rangle=\bigg\langle \frac{d(-\mu _z\vec E)}{d\vec E}\bigg\rangle$$
The time-independent Schrödinger equation is now, $$\hat H(\vec E)\psi(\vec E)=E(\vec E)\psi (\vec E)$$ With an energy; $$E(\vec E)=\big \langle \psi (\vec E)\big|\hat H(\vec E)\big|\psi (\vec E)\big \rangle=\big\langle \psi ^{(0)}+\psi ^{(1)}\vec E+\psi ^2 \vec E^2+ \dots \big|\hat H-\vec E\hat \mu _z \big|\psi ^{(0)}+\psi ^{(1)}\vec E+\psi^2\vec E^2+\dots \big \rangle$$
With a little algebra and use of $E^{(0)}=\langle \psi ^{(0)}|\hat H^0 |\psi ^{(0)}\rangle$, as well as the Hermitian properties of the Hamiltonian. $$E(\vec E)=E^{(0)}+2\vec E\big \langle \psi ^{(1)}\big|\hat H^{(0)}\big|\psi ^{(0)}\rangle -\vec E\big\langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big \rangle +\mathcal O(\vec E^2)$$ Using the Schrödinger equation $H\psi =E\psi$ and pulling the scalar energy out of the integral; $$\big\langle \psi ^{(1)}\big|\hat H^0\big|\psi ^{(0)}\rangle =E^{(0)}\big\langle \psi ^{(1)}\big|\psi ^{(0)}\big \rangle$$ Since $\langle \phi |\psi \rangle=0$, $$E(\vec E)=E^{(0)}-\vec E\big \langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big\rangle$$ Therefore the expectation value of the dipole moment along the $z$ axis for a molecular state $\psi ^{(0)}$ is $\langle \mu _z \rangle$; $$\langle \mu _z\rangle =\big\langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big\rangle$$ To understand the strength of the interaction that causes the transition between the states $\psi ^{(0)}$ and $\psi ^{(1)}$ we use the transition dipole moment which is basically exactly the same but there is a wavefunction from both of the states involved, initial and final!
If you were to repeat this process but retain higher orders (Messy!) you would get (2nd order) the polarisability of the molecule which in essence is the susceptibility of the electron cloud to change with respect to an external electric field (so how the dipole moment changes. Third order would give the hyperpolarisability etc., etc...
As I said, you could also approach this from a really different angle by interpreting the molecular orbital diagram and using computational chemistry (so variational principle etc) to find the molecular orbital coefficients! That would give you a good idea of what is going on!
• Hrm. This is a quite thorough formal elaboration of the theory involved, but it doesn't actually describe how one would implement practically the dipole moment calculation. (At least, I am no closer to understanding how I would code a dipole moment calculation given a set of MO coefficients and basis functions.) Commented Sep 3, 2015 at 11:57
• This whole derivation is unnecessary. Per definition the dipole moment is the expectation value of the dipole moment operator with the given wave-function.
– Greg
Commented Sep 3, 2015 at 14:38
• Thank you both for your comments, I will take this into account and update my answer shortly. I agree with @Brian, this answer does not give a non-specialist sufficient information on how to perform calculations to obtain values for molecules etc. However, I do not feel it is obvious that the dipole moment is the expectation value to someone who is new to this material and not as knowledgable as others. It is my experience that undergrads can struggle in this area if they can not see where something comes from. That's why I always like to give people a little background material. :) Commented Sep 3, 2015 at 15:39
• @Greg For those like myself (background in chemical engineering, and not applied mathematics / quantum physical chemistry; but with an interest in a general understanding of the inner workings of quantum computation) it is perhaps not quite as sad...? I very much appreciate the efforts of both AngusTheMan and pentavalentcarbon to lay out the details. Also, frankly: isn't this sort of exposition the entire purpose of StackExchange? Commented Sep 22, 2015 at 13:56
• @Greg Practical quantum computation was the topic of the original question -- so, no, it's not at all off-topic. Also, if you're looking for mathematically rigorous developments from axiom to theorem or whatever, you're on the wrong SE site. There are Math.SE and MathOverflow for that. Commented Sep 22, 2015 at 14:11
An often used approach, especially for semi-empirical methods, is to use a set of atom-centered charges (most often the Mulliken charges) to calculate the dipole moment. In that case, the molecular dipole moment is given by:
$\vec{\mu} = \sum_a \vec{r} \times q_a$
Note that if the system has a finite charge, this equation has a dependence on the position of the molecular system relative to the origin of the coordinate space. Most program use the center of mass or center of nuclear charge as origin in such cases. | 4,390 | 15,756 | {"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": 11, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-26 | latest | en | 0.869501 |
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https://nbviewer.ipython.org/github/diktat/prml/blob/master/4.1.1-discriminant-functions-two-classes.ipynb | 1,660,855,041,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00685.warc.gz | 360,594,814 | 32,338 | In [1]:
%pylab inline
def myshow(mytitle=None):
ax = axes()
ax.spines['left'].set_position('zero');
ax.spines['right'].set_color('none')
ax.spines['bottom'].set_position('zero');
ax.spines['top'].set_color('none')
ax.spines['left'].set_smart_bounds(True);
ax.spines['bottom'].set_smart_bounds(True)
ax.set_aspect('equal');
if mytitle is not None:
title(mytitle)
show()
Populating the interactive namespace from numpy and matplotlib
In [2]:
# we see how the vector w define the orientation of the decision surface
# and how w_0 define the location of the decision surface
w_0 = -1.
w_length = .8
print("angle of the vector w: %.1f degrees"%degrees(w_angle))
print("normal distance from the origin to the deciison surface: %.1f"%(-w_0/w_length))
w = array([cos(w_angle), sin(w_angle)]) * w_length
print("w_1: %f w_2: %f"%(w[0],w[1]))
x = linspace(-5,5,100)
# since w^Tx + w_0 = x1*w1 + x2*w2 + w_0 = 0 we can rearrange terms to get
# x2 = -(x1*w1 + w_0) / w2
x = column_stack((x, -(x*w[0] + w_0)/w[1]))
plot([0,w[0]],[0,w[1]], linewidth=2, color='g') # w vector
plot(x[:,0],x[:,1],'r') # decision surface
myshow("w is the green vector, the decision surface is the red line")
angle of the vector w: 60.0 degrees
normal distance from the origin to the deciison surface: 1.2
w_1: 0.400000 w_2: 0.692820
In [3]:
# we see how y(pointx) gives a signed measure of the perpendicular
# distance distancex of the point pointx from the decision surface
pointx = array([2.5,2.5])
distancex = (dot(w,pointx) + w_0) / w_length
print("perpendicular distance r of the point x from the decision surface: %f"%distancex)
# x_perp is found by rearraging terms in (4.6) on page 182
x_perp = pointx - distancex*(w/w_length)
plot([x_perp[0],pointx[0]],[x_perp[1],pointx[1]],'g', ls='dashed')
plot([pointx[0]],[pointx[1]],'bo') # arbitrary point x
plot(x[:,0],x[:,1],'r') # decision surface
myshow('x is the blue point, the decision surface is the red line, the perpendicular distance is the dashed green line')
perpendicular distance r of the point x from the decision surface: 2.165064 | 674 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-33 | latest | en | 0.640149 |
https://python.astrotech.io/numpy/about.html | 1,606,892,309,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00394.warc.gz | 445,788,242 | 7,583 | # 1.1. Numpy¶
NumPy is a library for the Python programming language, adding support for large, multi-dimensional arrays and matrices, along with a large collection of high-level mathematical functions to operate on these arrays.
## 1.1.1. Rationale¶
NumPy is the fundamental package for scientific computing with Python. It contains among other things:
• a powerful N-dimensional array object
• tools for integrating C/C++ and Fortran code
• useful linear algebra, Fourier transform, and random number capabilities
Besides its obvious scientific uses, NumPy can also be used as an efficient multi-dimensional container of generic data. Arbitrary data-types can be defined. This allows NumPy to seamlessly and speedily integrate with a wide variety of databases.
## 1.1.2. Installation¶
Code 1.25. Installation
$pip install numpy Code 1.26. Upgrade $ pip install --upgrade numpy
Code 1.27. Check version
import numpy as np
np.__version__
# '1.19.1'
## 1.1.3. Performance¶
Code 1.28. Results with Jupyter and %%timeit -n 1_000_000 -r 10
import numpy as np
np.arange(0, 100, step=2, dtype=float)
# 756 ns ± 10.3 ns per loop (mean ± std. dev. of 10 runs, 1000000 loops each)
np.array(range(0, 100, 2), dtype=float)
# 8.28 µs ± 364 ns per loop (mean ± std. dev. of 10 runs, 1000000 loops each)
np.array([x for x in range(0, 100) if x % 2 == 0], dtype=float)
# 9.76 µs ± 324 ns per loop (mean ± std. dev. of 10 runs, 1000000 loops each)
np.array([float(x) for x in range(0, 100) if x % 2 == 0])
# 12.7 µs ± 195 ns per loop (mean ± std. dev. of 10 runs, 1000000 loops each)
np.array([float(x) for x in range(0, 100, 2)])
# 8.35 µs ± 196 ns per loop (mean ± std. dev. of 10 runs, 1000000 loops each)
np.array([x for x in range(0, 100, 2)], dtype=float)
# 5.89 µs ± 77 ns per loop (mean ± std. dev. of 10 runs, 1000000 loops each)
## 1.1.4. References¶
NP1
Alex Chabot-Leclerc. Introduction to numerical computing with numpy: broadcasting rules. 2019. URL: https://youtu.be/ZB7BZMhfPgk?t=5142.
NP2
Alex Chabot-Leclerc. Introduction to numerical computing with numpy: visualizing multi-dimensional arrays. 2019. URL: https://youtu.be/ZB7BZMhfPgk?t=5142.
NP3
Wikipedia. Normal distribution. 2019. URL: https://en.wikipedia.org/wiki/Normal_distribution.
NP4
Wikipedia. Pearson correlation coefficient. 2019. URL: https://en.wikipedia.org/wiki/Pearson_correlation_coefficient.
NP5
Wikipedia. Poisson distribution. 2019. URL: https://en.wikipedia.org/wiki/Poisson_distribution.
NP6
Wikipedia. Polynomial. 2019. URL: https://en.wikipedia.org/wiki/Polynomial.
NP7
Wikipedia. Continuous uniform distribution. 2020. URL: https://en.wikipedia.org/wiki/Uniform_distribution_(continuous). | 814 | 2,705 | {"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": 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} | 2.796875 | 3 | CC-MAIN-2020-50 | longest | en | 0.554311 |
http://www.reddit.com/r/anime/comments/1py7ce/a_wild_kyubey_appears/ | 1,406,374,467,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997901076.42/warc/CC-MAIN-20140722025821-00181-ip-10-33-131-23.ec2.internal.warc.gz | 1,054,978,942 | 43,748 | [–]http://myanimelist.net/animelist/Aruseus493 47 points48 points
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This might be fun.
• TWGOK - Katsuragi Keima (yes he is a guy) wishes to live in the realm of games. -> Manga to Visual Novel Adaption with same story. Meta huh? :-P
Okay, now let's get to a serious example.
Infinite Stratos
• Charl wishes that Ichika isn't so dense.
• The very fabric of the universe unfolds and re-aligns with the appropriate reasoning.
• Ichika starts dating a harem of girls (excluding Houki because she is /u/EcchiMasterV2's waifu)
• However, Ichika decides that he isn't in love with the girls and becomes a hardcore sis-con abandoning his harem.
• Charl becomes depressed and the school falls apart as Ichika was a shining ray of hope for all the girls.
[–]http://myanimelist.net/animelist/PumpkynPye 70 points71 points
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Charl wishes that Ichika isn't so dense.
This is a catastrophic wish in a harem series.
Pye's Inverse Square Law of Harem Stability (academic paper forthcoming) states that the denser the MC in a harem anime, the more stable his harem. Should the MC ever drift above below a certain critical density (referred to as the School Days Density, SDD), the harem collapses. The formula for computing the SDD (and thus for computing a given Harem's Ease of Maintenance, EM) follows:
``````SDD = Sum[(Y/T) * (S + L)]
Where:
Y = Yandere factor
S = Sisicon factor
L = Lolicon factor
T = Tsundere factor
``````
In the case of IS, the total summed Yandere factor is moderately high, the Siscon factor is significant, the Lolicon factor not quite so significant, and the Tsundere factor is through the roof. Given the number of girls in the harem contributing to the SDD, I expect it to at least be above average.
(EDIT: I screwed up my formula and corrected it)
[–]http://myanimelist.net/animelist/Aruseus493 4 points5 points
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Where does Seitokai no Ichizon fit into this?
[–]http://myanimelist.net/animelist/PumpkynPye 24 points25 points
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I haven't watched Seitokai, but from what I'm reading, it's a parody of the usual harem trope. The correct approach is to apply an additional factor P, which ranges from 0 to 1, making the equation the following:
``````SDD = P(Sum[((Y/T) * (S + L))])
Y = Yandere factor
T = Tsundere factor
S = Sisicon factor
L = Lolicon factor
P = Parodidity of the series, the ratio of seriousness to parodyness of the series
``````
(In my haste to get to class on time, I fucked up the formula in my first post, so I corrected it here).
Other series requiring unique factors include Monogatari, which while already noteworthy for its alarmingly high Lolicon and Sisicon factors, has been shown to need an additional empirical factor Tooth, tentatively coined the Toothbrush factor, in order to realistically measure the harem's stability. When combined with the series' very significant Yandere factor and distinct lack of Tsundere elements, the ending of Monogatari Second Season Very minor implicational spoiler should have come as no surprise to anybody.
[–] 15 points16 points
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Tagged as Harem Scientist
I question how much your basic formula makes sense, but I'll be damned if your terms and written explanation doesn't make actual sense.
[–]http://myanimelist.net/animelist/PumpkynPye 8 points9 points
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The formula makes sense as long as all four main terms (excluding P) range between 1 and infinity: The presence of yandere characters is destabilizing, while the presence of tsundere characters is stabilizing, hence (Y/T). The presence of "risque" elements like lolis or siblings in the harem is likely to upset the other girls and is thus also destabilizing, hence *(S + L).
I worked with models like this in ecology for a few years while at university :P I'll be damned if any mathematical model I write isn't at least sensible on the surface.
I am also honored by your tag.
[–] 5 points6 points
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Wait lol ya I get it now. That's hilariously logical
Shouldn't MC denseness be a variable?
[–]http://myanimelist.net/animelist/PumpkynPye 6 points7 points
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No, the purpose of the equation is to compute the SDD, a density threshold that the MC can't drift above lest...well...see School Days. This means the larger the result of that formula, the more inherently unstable the harem, and thus the more dense the MC has to be to maintain it. This threshold is independent of the MC's actual denseness, and is a property of the harem.
[–] 6 points7 points
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I see I see. Thank you for enlightening me professor, I look forward to buying your book.
[–]http://myanimelist.net/animelist/PumpkynPye 11 points12 points
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When Maximum Dumb will Save your Cerebrum: The Harem Theorem and Other Sums, coming to stores near you.
Yeah I'm never posting here again, don't worry.
[–]http://anilist.co/user/2276/Arbalor 5 points6 points
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So how do we find the magic spot where the MC picks a girl and it doesn't go school days
[–]http://myanimelist.net/animelist/PumpkynPye 4 points5 points
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I'm not sure there necessarily exists such a point for all harems. For instance, a harem consisting of 5 dangerously violent yandere is obviously a game the MC can't win if he ever tries to make a choice. At best, he can maybe survive the harem by being so massively dense that he not only doesn't notice anyone's romantic interest, but doesn't notice their murderous impulses either (as predicted by my theory).
On the other hand, suppose a theoretical harem consisting of only one dangerously violent yandere and 3-5 shy tsundere. If the MC chooses one of the tsundere, it's obviously game over, the yandere is going to fucking kill him. But if he chooses the yandere...that might work, right?
This looks like it would require a different (and substantially more complex) model. I'll make that my next paper.
[–] 1 point2 points
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Check out Shuffle! If you want something like that, it may be somewhat boring but at least the mc isn't a thick headed idiot who doesn't know the difference between sadness and a sandwich.
[–] 1 point2 points
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Sorry if I am miss reading your directions, does the number value of Mc density proportional or inversely proportional to their density.
[–]http://myanimelist.net/animelist/PumpkynPye 1 point2 points
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The answer is "proportional", but you need to read the following line from my first post:
Should the MC ever drift above a certain critical density (referred to as the School Days Density, SDD), the harem collapses
Should the MC ever drift below a certain critical density (referred to as the School Days Density, SDD), the harem collapses
Now I think everything works. Unless I made another dumb mistake somewhere. Which, honestly, seems likely at this point.
[–]http://myanimelist.net/animelist/Aruseus493 2 points3 points
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I really would recommend Seitokai no Ichizon because it might end up outside your formula. ;-) Gotta say that I love all the thought you put into this. I've been using it on just a bunch of series for the fun of it. :-P
[–]http://myanimelist.net/animelist/PumpkynPye 2 points3 points
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Ha! I'm glad someone found it entertaining :P
You may need clear-cut definitions for those factors to really use it, but I'm sure approximating works fine.
I'll be sure to check out Seitokai, since you recommend it. Despite my...uh..."expertise"...in the field, I haven't ever been big on the harem genre, but a series that falls outside my theory? This I must see!
EDIT: Just watched the first episode...that was so cute!! I think the size of their eyes is...uh...worrisome, but this is such a sweet show! If it continues like this, I can ignore the rather ridiculous eye size choice, it's really enjoyable.
[–]http://myanimelist.net/animelist/ctom42 1 point2 points
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What does the C stand for?
[–]http://myanimelist.net/animelist/PumpkynPye 4 points5 points
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Yeah sorry, I screwed up my formula the first time because I was short on time. I corrected it about a minute ago.
"C" was an outdated factor called the "caring factor", designed to measure the willingness of the harem to harm the MC, either physically or emotionally, and thus their willingness to dissolve the harem. It has since been incorporated into the calculation of the Tsundere factor and is no longer necessary
[–]http://angelicshaft.myanimelist.net 1 point2 points
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So where would a harem like OreShura land? Sure it pokes fun at the genre, but it also does it's own things that result in a lopsided harem.
[–]http://myanimelist.net/animelist/PumpkynPye 2 points3 points
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See here for my caveat on harem parody series.
I can't believe I actually just referred someone to further "research" that already answered their question. Maybe I should actually develop this theory.
[–]http://angelicshaft.myanimelist.net 1 point2 points
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Oh I saw that, but it's more along the lines that the MC becomes less dense, but in the end just sucks at words.. So harem stays together, but he's aware of all.
[–]http://myanimelist.net/animelist/PumpkynPye 1 point2 points
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Oh, I haven't actually watched that series. Ironically enough, I'm not actually into the harem genre all that much.
I guess that acting dense has the same effect as actually being dense...but how stable to you really think that is long-term? My theory predicts a harem collapse, but it doesn't necessarily have to happen right away :P
[–]http://myanimelist.net/animelist/postblitz[S] 2 points3 points
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sick buuuuuuuuurn on all the IS girls! that would make for a killer twist on the series, probably more so than.. that popular siscon series i don't want to spoil.
your first wish is half-way adapted in a brief ED.. but you probably knew that already. i wonder what real consequences would Keima suffer from in there to mangle his dream.. probably /u/unknownaus 's version of things but in virtual reality.
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This scenario would enable me to swoop in and comfort my waifu during her time of sorrow.
[–]http://myanimelist.net/animelist/Aruseus493 2 points3 points
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If you're referring to Charl, feel free. I have dibs on Lin. :-P
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I most certainly am. You must be one of the elusive Lin fans.
[–]http://myanimelist.net/animelist/Aruseus493 1 point2 points
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Just don't tell anyone. :-P Also, never speak that you saw me here. No one must ever know of our existence. ;-)
[–]http://myanimelist.net/animelist/postblitz[S] 1 point2 points
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[–]http://myanimelist.net/animelist/Aruseus493 1 point2 points
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I choose Nichijou. Yuuko will probably wish for something dumb from Kyubey, like never having to do homework ever.
She constantly brags about never having to do homework in front of her friends, until it's revealed that Mai made the same wish to Kyubey. Mio and Nano track down Kyubey to get wishes. Nano is conflicted, and doesn't know whether Hakase will be mad or not after she becomes a girl. Unfortunately Kyubey tells her that robots don't have souls, and that he can't make a contract with her. She cries. Mio wishes for Sasahara to be her boyfriend.
Now that the three main characters have made wishes, they backfire. Yuuko fails every test she takes because she doesn't study or do any schoolwork at all. Her Mom beats her for this. Mai still studies and does her homework anyway in order to shock Yuuko. Mio has nothing in common with Sasahara after getting to know him, and she realizes their relationship is due to her one sided attraction to him.
[–]http://myanimelist.net/animelist/postblitz[S] 8 points9 points
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i like this. well thought out in inner-series, especially Nano's nature. cheers!
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Thanks. You sure like gifs, don't you?
[–]http://myanimelist.net/animelist/postblitz[S] 4 points5 points
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gifs make it easier for the reader to understand my state of mind when writing things. i have a tendency to write 'aggressive' text apparently.. even though i'm usually zen.
the only time i'm pissed off for real is when i curse a lot.. a LOT!
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Fascinating
I may have to try this style of posting. I do have quite a few unused gifs in my dropbox after all.
Although, there is the possibility that I'll be looking to use my gif collection at any place at any time, regardless of appropriateness or relevance
[–]http://myanimelist.net/animelist/postblitz[S] 5 points6 points
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you're doing good so far, just remember to post in /r/anime with anime only gifs!
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[–]http://myanimelist.net/animelist/postblitz[S] 9 points10 points
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I humbly accept
Yeah, I had only three saved anime gifs left, and this one fit the most.
[–]http://myanimelist.net/profile/ChazzU 18 points19 points
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Basically, this post will spoil Shinsekai Yori so be careful.
Shinsekai Yori
1. Saki meets Kyuubey right after she escaped from Shun's world and left him to die there.
2. Saki wishes Shun back to life
3. Shun comes back to life
4. However, Shun is still a Karma Demon as Saki never asked for that to be changed. She now has to witness how Shun not only pollutes the water in and around the holy barrier, but also kills off nearly all animals and is the source of multiple diseases swiftly spreading across the village and its people.
After seeing the person she loves but had to leave to die also seeing seen resurrected, she also has to witness how he once again gets Tainted Cats sent after him because he is a liability to the village, and the world in general. Overwhelmed by the sheer number of Tainted Cats but also realising he can't do anything good without doing more harm instead, he lets the Tainted Cats ambush him. Thus, Saki losing Shun once again.
5. Saki now turns into a Fiend, filled with grief and guilt but also mourning over Shuns second death, and turns onto the village in a fit of rage. After brutally murdering everyone who couldn't escape their houses and flee to the Tempel of Purity she recovers her senses and commits suicide because she does not believe she can atone for her sins in any other way.
Because I didn't actually want to see something happy happen anyway.
[–]http://myanimelist.net/animelist/postblitz[S] 1 point2 points
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well.. that was... surprisingly satisfying. when exactly does die? he was after all..
[–]http://myanimelist.net/profile/ChazzU 2 points3 points
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[–]http://myanimelist.net/animelist/loldamar 17 points18 points
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Ok I'll do a couple:
Steins;Gate
I doubt Okabe would ever trust a weird fluffy bunny thing for any "help", he's way too beyond paranoid. "The organization is trying to attack my mind with the words of cute furry creatures, they shan't succeed."
I could maybe see kurisu being tricked into it, if Kyubey plays his cards right concerning her issues with her dad.
Mayuri is pretty much PMMM tier retarded so I could easily see her being tricked into it. Though the phone wave would be able to undo it all, as long as her making the contract isn't a fixed point on the attractor field, if so its an entirely new can of worms shit that Okabe has to deal with. Things would play out relatively the same.
Death Note II: The return of Kyubey
L and Light seem like the kind of people that'd be highly suspicious of a strange creature promising magical powers. Fairly certain L at the very least would see right through it. Though "Magical Girl L" is certainly a tantalizing thought to consider. Misa would probably most prone to getting tricked. Her wish? "I wish Light would luv me 5ver". I wonder what effects the Death Note would have on witches/Kyubey. I also wonder if Ryuk and Kyubey would recognize each other. Funnily enough, I doubt Kyubey would change the plot much.
Attack on Kyubey
Humanity is pretty fucked as is, I really don't think its necessary to add Kyubey to the mix. Wishes are fairly obvious: "I wish all Titans were dead", "I wish Eren would love me forever", "I wish I had more good food to eat", not really sure what Armin would wish for. This story would end very badly, titans on the outside AoT spoilers with witches on the inside?
ggwp
Cowboy Kyubey
uh....does not compute
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Cowboy Kyubey
Kyubey appears to Ed. She wishes for a sandwich. She never falls to despair because her wish was a delicious sandwich. Now she's an expert hacker slash magical girl.
[–] 4 points5 points
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This post, I like it.
I like the thoughts on death note. Just focusing on L and Light, there's no way they would let them screwed over without a fight. "Want a wish?" would be met by both with extreme skepticism and overwhelming caution.
An interaction with Okabe and a little furry wish granter is a funny thought, and ya Mayuri would take that bait in a heartbeat haha.
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. I wonder what effects the Death Note would have on witches/Kyubey.
They would die? but Kyubey would reappear
[–]http://myanimelist.net/animelist/loldamar 2 points3 points
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IIRC the death note can only be used on "humans". So that would certainly confuse things.
[–] 8 points9 points
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Fullmetal Alchemist: Brotherhood
Already a Faustian story, to an extent, so I'm not going to do this one.
Shakugan no Shana
Kyubey would probably appear to Yuji during his angsty period in SnS Second, when he felt powerless and useless, and make a contract with him so he could help Shana. Yuji, now a pseudo-flame-haze would battle alongside Shana and Margery for a while, but he would begin to see the fight as a losing battle Shakugan no Shana Second/Final Spoilers From then on, the plot would be like the last season, but with Shakugan no Shana Final
Kyubey would appear to a lovelorn Taiga and get her to make a contract. She would ask for Kitamura to fall in love with her, which would make her very happy for a while, until she would inevitable begin to feel the love is not real, and fall into despair. Magical girl Taiga would be funny to see, at least.
Tengen Toppa Gurren Lagann
Hmm. Kyubey would probably appear to Simon after , and offer to . Of course, Kamina would hate that, and Simon would regret it immediately, sinking into despair. Then, Kamina would then punch him in the face, , and Simon would Giga Drill Break Kyubey with Sorairo Days playing in the background.
[–]http://myanimelist.net/animelist/loldamar 4 points5 points
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Fullmetal Alchemist: Kyubeyhood
FTFY
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Monogatari Series, specifically Nekomonogatari: Kuro and Kabukimonogatari Spoilers
[–]http://myanimelist.net/animelist/postblitz[S] 6 points7 points
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Indeed. I hope you don't mind me "stealing" your SnK/VVV/Madoka gifs in exchange for some Touma/Raildex gifs?
[–]http://myanimelist.net/animelist/postblitz[S] 1 point2 points
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usually the gifs aren't mine anyway.. but the VVV ones i've made myself, especially the damn captions and "liberal scene edits" which took forever to make. it's so easy to just make any gif from any scene if you don't alter them in any way.
not like i'd care anyway, in the end it's all sunrise's shit. it'd be funny to see one gif i know i made god knows where..
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Oh wow, went back to episode 16 thread to find em > Akira Best Girl
Very nice gifs, especially the upvote/downvote ones!
I was talking about the VVV ones here which are amazing showcases of Valvrave animation.
Edit: Yep, I added a bunch of the Gundam shows to PTW, planning to watch that one first
[–]http://myanimelist.net/animelist/postblitz[S] 1 point2 points
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if you want some SUPERB Sunrise quality animation: Mobile Suit Gundam : 08th MS Team is my absolute favourite so far. i have this preconceived notion that pre-CG stuff are superior
gonna wait for the weekend to pick up on some gifmaking, i have some killer ideas on captions and using scenes.. just lack the time since i'm very busy @ work.
[–]http://myanimelist.net/animelist/awpaca 4 points5 points
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I'm drooling. This was such a pleasure to ponder on and super relevant as well. Not to mention was still an entire possibility at one point even without the QB wish.
You have won all of my tears and love now.
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[–]http://myanimelist.net/animelist/awpaca 2 points3 points
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the Love was reluctantly accepted
Awh, I should tag you as Tsundere-chan!
[–]http://myanimelist.net/animelist/postblitz[S] 21 points22 points
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i guess gifs alone won't cut it so to encourage posting i'll provide an example:
• Chihaya from Chihayafuru, runs into Kyubey and asks to become better at Karuta then starts winning and overpowering people left and right.
• her newfound skill easily causes her to win and their teamgames would lose more often as she won't need her fighting spirit, thus it keeps the team from being motivated
• Taichi would get discouraged from playing karuta cause he'd never be able to defeat her and begrudgingly he'd go back to playing the girls and school.
• Arata would keep playing karuta then retire because the title of master and queen would both belong to Chihaya and he'd end up with Shinobu as #2 underdogs.
• Chihaya would end up lonely at the top and eventually quit karuta because she can't play seriously with anyone anymore. the queen retiring as such would cause a rapid decline in the sport. Chihaya will likely end up at a newsstand, selling papers for a living, dejected of her success burying her beloved sport.
[–]http://myanimelist.net/animelist/Icekracker 6 points7 points
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You're a monster.
...well done.
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Jesus Christ.
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I'm not really creative enough to contribute myself but I hope this thread gets popular, your concept is nice.
[–]http://myanimelist.net/animelist/postblitz[S] 8 points9 points
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thanks for the support.
you could contribute yourself if you really wanted. it's actually really simple since it doesn't have to be a serious wish or something in-line with the story
..like idaknow.. you've seen steins;gate, how about Haru wishing Kurisu had ample breasts or FMA's Edward wishing he were taller! ..and then he gets hit by a horse or something and Kurisu gets overly groped by Mayushii so she decides to quit science and become a model, thus Haru losing both girls from the team..
just make sure after the characters exalt in their magnificence, you break them down with something equal but negative.
yin and yang can be fun, not just gruesome.
[–] 6 points7 points
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K-ON: Azusa wishes that the band would practice together more instead of drinking tea and eating cake all the time.
This gives the others an insane drive to debut at Budokan. They start practicing super hard - like Haruki from White Album 2 level hard, except forever until they get to Budokan. The band members become insanely good with the combination of latent talent and hardcore practice, but Azusa cannot keep up with the schedule. She starts to fall behind and feels even worse when Yui passes her in skill. She ends up leaving the group because she feels like she's holding them back - they're so focused on Budokan that they hardly notice her leave. A few months later she sees a flier to see Ho-kago Tea Time at Budokan. The series ends with her thinking, "That could have been me... :'("
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[–]http://myanimelist.net/animelist/postblitz[S] 0 points1 point
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never seen K-On but i can imagine what all that meant. was beautiful in its composed sadness
[–]http://myanimelist.net/animelist/Redkrimson 12 points13 points
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you can't choose Puella Magi Madoka Magica
Okay... fine.
Well lets see, #2 is... Black Lagoon. Shit, that's pretty Urobutchery already. There's no challenge in that.
Okay, #3 is Code Geass. Uhhh... well...
Number 4, Fate/Zero... ...fuck.
You know what, I don't think I can play this game. All my favorite anime are already cynical tragedies.
[–]http://myanimelist.net/animelist/postblitz[S] 15 points16 points
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you could very well ignore my suggestion and turn the protagonists into magical girls or something.. i'm sure it would twist the cynical tragedy into an unusual one
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I like what you are doing 'round these parts. I'll be posting here later today when I think of something good.
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One Piece
Luffy punches Kyubey in the face and forces him to become his nakama since the easy route is not the fun one.
But seriously
Sora no Otoshimono
Tomo gets his wish to become the ultimate playboy (or Sora would stop karate chopping him) and he gets all the girls.
However with this Tomo becomes extremely depressed because there is no more excitement, the threat of punishment if caught... no more adrenaline in seeking girls anymore.
Tomo becomes extremely depressed and withdraws from society, closeting himself in his room reminiscing about days when he used to get punished for his crimes against women.
The angeloids unable to take him out of his depression also become depressed and lethargic, not doing anything and staying at home. Same goes for Sora.
In the end the Angeloids in ever escalating attempts to Tomo cheer up will accidentally destroy the world.
The End.
[–]http://myanimelist.net/animelist/postblitz[S] 0 points1 point
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nakama, angeloids.. i need to contact the mothership for this one
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Good one
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I haven't gotten around to seeing Madoka yet (I know, I know) but I think I get the idea. Sounds interesting.
I'm going to pick Rider from Fate/Zero.
--Minor spoilers from here on out--
Essentially we can simply imagine that Rider wins the grail and gets his wish, the result should be the same either way. In any case Rider's wish is to be made corporeal once more and immortal so he can ooze greatness forever. Rider doesn't mention immortality specifically in the anime I don't believe, but it's clear his ideals encourage such a wish.
I like this now because I gotta ask... would this backfire on him? Could it? Obviously we can imagine bad things happen. Over the years as he conquers and exhibits greatness to endlessly inspire his followers he would lose many, but he knows this and in fact already knows what it's like.
The thing about Rider is that he is seemingly immune to being dragged down by harsh realities. His nature of conquest can not be stamped down, not even by infinity!
TL;DR- Rider makes worrisome wish for immortality but it doesn't backfire, because Rider can't be beaten down by mere infinity.
Rider's kinda like Kamina I think, but instead of ignoring limitations he accepts them and achieves greatness anyways. Because he's mother*cking Alexander the Great and he rides a lightning chariot.
[–]http://myanimelist.net/animelist/postblitz[S] 4 points5 points
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much as you may try, the faustian deal's yang cannot be overcome.
if Rider were to choose to become living, he'd quickly see his dreams shattered as nobody in the modern world cares for his violent behavior outside of politics. assuming he'd even get there and not be dragged into an insane asylum or abused by the judicial system for his numerous petty crimes.
so he's in office and he quickly finds out that the barbarous ways he used to achieve popularity and support don't work anymore. bogged down with paperwork he'll soon subject many to unfair work ethics to do the work for him as he plots to provoke nations into a war. the jewish lobby in washington decides he's a threat and conspires his removal as a ruthless dictator. months later he's found in some hole, remembering the good old days when slaughtering opposing armies was all down to the men in the battlefield instead of tech and scheming.
there are no lightning chariots as mortals in the modern world, unless he plays video games.. in that universe he's a NEET.
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noooo
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Trigun!
Upon witnessing another horrific event resulting in the deaths of countless innocents, Vash finally relents and wishes that no one dies, ever. Magically all those people didnt die! Happy day!
Vash makes a cool exit, only to be followed by the incessantly irritating Bernardelli Insurance girls. Legato shortly makes his appearance and captures the girls. He makes impossible demands of Vash , but Vash won't do it. It goes against every moral hes ever had! So Legato gets to work torturing the girls. It's gruesome, Meryl is a pretzel, Milly is showing her ass and her breasts at the same time. Vash is weeping, and after many hours, Meryl wimpers out "Please Vash... kill us...". After seeing her grotesque state, he is overcome with pity, and readies his gun. Legato is surprised, he had finally broken Vash the Stampede, and tightens his grip on the girls.
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...
:(
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Fuck that got dark.
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Ouran High School Host Club
Haruhi wishes she could repay her debt without being in the troublesome host club (a la, part time job or whatever). Kyubey instead makes her family RIDICULOUSLY wealthy.
So wealthy, that in her school of grandeur and affluence, she becomes massively popular and high-school-famous. She decides that the host club is to blame for all of this, and uses her wealth and newfound power to rail against them, eventually tearing the club apart. Tamaki spends the rest of his life crying in a corner. The twins begin having extreme separation anxiety that interferes with any normal life they could have had. Honey never grows up, preferring to spend his time drinking tea with stuffed rabbits, Mad Hatter style. Mori becomes a fighting champion and very aggressive, though still quiet. She hates every minute of the vapid attention and wasted money and becomes a very cynical hermit.
I have to go now but I was thinking Kyouya gets Baker-acted by his families police force by his brother for being crazy obsessed with taking over the business. No thoughts for Renge. Anyone?
[–] 8 points9 points
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Seems intersting . Lets give this a try and see how it goes
Keima from The World God Only Knows makes a wish to become better at capturing girls. This is so its easier for him to capture the girls that have the loose souls so he has time for this games.
However what he doesn't realise is this skill applies to all girls which include the girls in his game. He then starts to finish off games as quickly as possible which at first seems fun but then realises that its allbecome too easy and pointless
Now since his reason for living (i.e playing games) has been taking away from him he falls into a deep depression and then falls into a coma after an OD on anti depressants
(Yes i know the ending is a bit out there but lets be honest Urobuchi has done a lot worse)
[–]http://myanimelist.net/animelist/postblitz[S] 1 point2 points
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hah, i chuckled.. i can certainly imagine Keima doing something of the sort but your tale leaves out one thing: Elsie! she'd probably intervene on his depression - but maybe Keima's skill inadvertedly activates and seduces her without wanting to thus breaking her genuine support.
he could very well wish for his bracelet to come off.. and then he'd remain trapped in the game world forever!
[–] 1 point2 points
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I did think about Elsie but i reckon she would intervene by convincing him to go to a psychologist to which he get prescribed anti depressants to which he soon becomes addicted.
Otherwise as you said he could just end up seducing her.
[–]http://myanimelist.net/animelist/ctom42 1 point2 points
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So I thought I would do this for Umineko no Naku Koro Ni, but anyone who has read the VN knows that huge VN spoilers
So my next thought was Steins;Gate, but that seems too easy.
So I guess I am going to go with Monster. But instead of the protagonist I'm going to have Kyubey show up for Nina/Anna. She wishes to be able to kill her brother. She is then able to do so, but in the process he corrupts her. monster spoilers Inspector Lunge realizes Johan is real when his body is found, thus clearing Tenma of his crimes, however Lunge now thinks that Tenma was the one to kill Johan. Meanwhile Anna grows into an even bigger monster than Johan ever was. The reason being that more spoilers
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bleach. after the hollow that killed ichigos mother escapes ichigo wishes for more power. the kyubey gives it to him...by making his hollow take complete control over him. ichigos hollow ends up killing everyone he cares about.
[–] 1 point2 points
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Nerima Magi Brothers
• Hideki would wish for the concert dome. Though on the surface, his dream is attained, fame drives him and his bandmates apart. Mako, her material wants now satisfied forever, ignores him and embarks on several relationships with fellow celebrities. Their relationship deteriorates, and their infighting prompts Ichigo to leave the band and start up his own act with the pandas. Unable to replace him, the Nerima Daikon Brothers split up. PMMM spoilers ensue.
• Mako would wish for Dom Perignon, and she gets Dom Perignon. Then the next time she hankers for something, PMMM spoilers ensue, and she goes Charlotte on the whole cast.
• Ichigo doesn't have any want or problem warranting a wish until late in the series, when he would wish for Mako and Hideki to reconcile. They do, probably in some intensely OOC, squicky way, and in a way that alienates both of them from Ichigo. Without a full band, Mako and Hideki are unable to defeat Oizumi, they and Nerima get "developed," and, well, you get the idea.
• Widgett would wish to be able to capture the Nerima Daikon Brothers. She succeeds (alienating Pandaikon in the process) and is rapidly promoted up in the police, where she uncovers the extent of its corruption and its yakuza ties. Disillusionment, and PMMM spoilers, ensue.
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Hi OP can you list the name of all the manga you used as GIF please ?
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Most of them are from Yuru Yuri. And it's an anime.
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I can't extrapolate things like this very well so I leave it up to you guys to see what happens when Sorata from Sakurasou wishes to become a better game designer.
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Gatchaman Crowds.
Since Hajime is more a force of nature than a main character (or a Mary Sue if you didn't like her), the only thing I know about her wish is that she'd figure out how to stop it from backfiring, and probably manage to make the world a better place in the process. | 9,554 | 39,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2014-23 | latest | en | 0.926448 |
https://mathwithbaddrawings.com/2021/05/24/when-a-trillion-dice-are-no-better-than-a-dozen/?replytocom=226875 | 1,669,563,407,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00055.warc.gz | 430,492,298 | 28,219 | # When a trillion dice are no better than a dozen.
For my forthcoming book Math Games with Bad Drawings, I considered including a classic dice game called “Drop Dead.” It narrowly missed the cut. Strike one: the lousy name. Strike two: it’s a game of pure chance, with no room for decision-making. And strike three: well, that name again.
Still, I want to share the game here, because it teaches a useful lesson about the mathematics of risk.
On your turn, you roll five dice, and score their sum, with one big exception: 2’s and 5’s are fatal. They immediately drop dead and are removed from play. Not only that, but whenever any 2’s or 5’s appear, the other dice are worthless; you score no points for the roll. Then, whether you scored points or not, you roll all remaining dice again, and continue repeating this process until all five dice have dropped dead. Play for a set number of turns per player (say four), after which the highest total score wins.
Here’s a sample turn that lasted for six rolls, scoring a total of 15 points.
Notice that on my first roll, I didn’t score any points. Die #4 (by coming up with a two) negated the collective efforts of Die #1, Die #2, Die #3, and Die #5.
That kind of failure is common. You’ll score on your opening roll just 13% of the time. A single 2 or 5 suffices to spoil the party, and with five potential party spoilers, few parties remain unspoiled. You thus wind up scoring most of your points with just one or two dice remaining, because smaller “parties” are more likely to go off successfully.
This leads to our larger theme, and the lesson that interests me: Starting with extra dice barely helps. In fact, there’s little benefit past your eighth die, and almost none past your twelfth.
It seems weird. An extra die can’t hurt, can it? Best-case scenario, it adds to your score, and worst-case scenario, it comes up 2 or 5, at which point you discard it, and wind up right back where you started.
Well, sure, it can’t hurt. But past a certain point, it doesn’t much help, either. Each die has a 1-in-3 risk of dropping dead. Compounded many times, that becomes a virtual guarantee: somebody is going to spoil the party. With just twenty dice, the probability of entirely avoiding 2’s and 5’s is just 0.03%, roughly your lifetime chance of being struck by lightning.
Let’s say you begin with 5 quadrillion dice, enough to blanket the state of West Virginia. Seems like you should score tons of points, right? Nope. Roll after roll, about 1/3 of your dice will spoil the party. This will repeat a hundred times in succession, your score stuck on zero, until finally, with just a few dice remaining, you begin to score points. (About 17 points, on average.)
5,000,000,000,000,000 dice. 17 points.
The moral: Don’t design systems where everything needs to go right. If your machine is doomed by one broken part; if your party is spoiled by one late guest; if your game plan crumbles when one player strays out of position; then you’ve got yourself a problem. A lot of problems, actually: one per component. Crowds are good for some things, but achieving unanimity is not one of them.
By the way, if you want to turn this into an actual game, Joe Kisenwether has a good idea: You may start with as many dice as you want, but your turn ends immediately after your 5th roll. Thus, you want to pick enough dice that you don’t run out (1 or 2 is probably too few) but not so many that you waste early rolls on scoring zero (so 20 is too many).
Puzzle: in this version, what’s the optimal number of dice?
## 18 thoughts on “When a trillion dice are no better than a dozen.”
1. Linus says:
What’s your expected score if you start with infinity dice? (i.e. the limit of your expected score as the number of dice goes to infinity.)
Is it true that for all n, starting with n+1 dice is at least as good than starting with n?
1. If you *truly* have infinite dice, then I believe the expected value is impossible to compute, since the game never ends. (Each roll, infinite dice drop dead, but infinite other dice still remain; so you never score, but you also never have to stop playing.)
But asymptotically, as the number of dice goes to infinity, your score seems to level off around 17.3. I don’t have a proof but I’m pretty confident (since as n grows, the chance of scoring decays exponentially, while the potential score grows linearly, and the number of extra chances to score grows even slower than that).
I’m also almost certain that for all n, the expected score with n+1 dice is at least as high as the expected score with n dice, but I haven’t proved it to my satisfaction.
1. Erin says:
Actually, if you had infinite dice, an infinite amount of possibilities will occur, which means you will score an infinite amount of points in an infinite amount of rounds.
So the expected score would be infinity.
2. Steven Stowers says:
How did you determine the “average score per round” numbers?
1. As I recall, I did something like this:
Define f(n) as the average score with n dice
Calculate f(1) directly from the fact that f(1) = (2/3)*[3.5 + f(1)]
Calculate f(2) similarly by writing f(2) in terms of f(1) and f(2)
Continue iterating
1. Chrissy says:
This is why probability and I are always at odds. I spent a good amount of time today playing with this using CoCalc for calculations, thinking about maybe using Markov chains, creating a probability matrix, looking at its powers, then kind of starting over finding the expected value of 1 die via a series of nested geometric series (and got 7), then going up to 2 dice and finding its expected value through another series of nested geometric series (and got 11.2).
Now I see this comment, try it out this way, and sure enough in 3 minutes total, got f(1)=7 and f(2)=11.2.
I admit, some of my earlier explorations were necessary for understanding how to write the iterated formula, what probabilities go with what average values, etc.
But this is as it ever was with me and probability. There’s usually a better way of looking at the problem, but it’s not the way I begin to approach it.
1. Chrissy says:
I think I’ve got my little average value program running correctly. But just to check, the average value for 5 dice is actually 16.06, not 16.6, right? Everything else checks out, including average value for 12 on up being 17.24 or smaller, so I think I’m doing it correctly.
It’s a neat little game/question!
3. Quadari says:
I believe the answer is six dice.
The probability that you actually get to score in any round is (2/3)^N where N is the number of die you have in that round. If you score, each alive die gives you an expected score of 3.5. (.25x(1+3+4+6)).
So, for example, if you start round 1 with 6 die, then there is an 8.78% chance that you actually score points in the first round. If you score points, your expected score will be 6×3.5=21. Therefore, your unconditional expected score in the first round is .0878×21 = 1.844.
Now, let’s think about round 2. You only actually make it to round two if at least one die in round one is not a 2 or a 5. Said another way, you don’t make it to round 2 if all of the die are a 2 or 5. I.e., you don’t make it to round 2 with probability (1/3)^N (where N is the number of dice). Therefore you do make it to round two with probability 1-(1/3)^N. If you do make it to round two, the expected number of dice that you will have is 2/3rds of the number you had in round one. (Since, on average, 1/3rd of them die.)
E.g., if you started round 1 with six dice, then you’d expect two of them to die, leaving you with four expected dice in round 2. Now just repeat the calculations described above:
There’s a 99.86% chance that any dice survive to round 2. If you get to round 2 you expect to have 4 dice. You expect to be able to actually score points in round two (2/3)^4 = 19.75% of the time. If you do score, each of the 4 dice is expected to give you 3.5 points. So your overall expected score in round 2 is 0.9986×0.1975x4x3.5=2.76.
Rinse and repeat!
You can then model this out for five rounds and add the expected score in each round. I believe that you’ll find the maximum expected score occurs when you start with six dice.
1. Ed says:
I gave up with the Markhov chains and just ran a simulation. I found that 5 dice had the highest expected value, and on average had higher winning percentages against both 4 and 6 dice, even though that percentage is less than 50% in both cases, because of ties.
4. I want to preface this comment by clarifying that I am a “language person”, not a mathematician. I love reading your blog though, because arithmetic is fun, and mathematics fascinating. Your writing is clear and typically you find some enlightening topics and have great teaching stories. I even bought your first book in hardcover. .
That said, this game looks neat. If two-thirds of reason that you are not including it, is really because you don’t like the name, know that you can change the name. If you are publishing a book about math games, it is your prerogative to give this a name that you think is better. Since it will then be published with this name, that becomes the name, and it may carry into common usage.
I propose: “Eighty-six 2s and 5s”.
Definition of the slang term 86 follows, from Merriam Webster:
“Eighty-six is slang meaning ‘to throw out,’ ‘to get rid of,’ or ‘to refuse service to.’ It comes from 1930s soda-counter slang meaning that an item was sold out. There is varying anecdotal evidence about why the term eighty-six was used, but the most common theory is that it is rhyming slang for nix.”
By the way, you have an uncharacteristic language error in your post. You say, “Here’s a sample turns that lasted for six rolls”. “A sample” is singular, but you pluralized “turn”. I can imagine how it occurred, since dice and rolls are both plural, but it threw me for a moment. ☺️
Keep up the great work in both mathematics teaching and blog writing!
1. Hi Millie, thanks for the thoughtful comment!
Good call on the language error, though you’re far too generous to call it “uncharacteristic”! I’m constantly fiddling with sentences, so most likely I pluralized (or singular-ized) one part of the sentence and neglected to follow through.
I like your new name for Drop Dead! I did rename a few games for the book (e.g., “Taxman” became “Tax Collector”) although in general I tried to remain loyal to existing names (e.g., I don’t like the name “Amazons,” but the game is well-known under that title, so I stuck with it). Anyway, the real issue was that every other game in the book gives players meaningful decisions, so this purely random game wouldn’t quite have fit.
Anyway, thanks again for your kind reading!
5. eromer42 says:
An extension of Joe Kisenwether’s what-is-the-optimal-number-of dice question: does the answer change if instead of playing “Eighty-six 2s and 5s” you play “Eighty-six 1s and 4s” (or any of the other possible versions of the game (there are 15 possible versions in all) )?
1. Ooh, interesting! Need to think through how changing the expected value per successful roll would affect this analysis…
6. Noun the Adjective says:
I wrote a MATLAB program to simulate the game at the end of the post, with only rolling 5 rounds. I *think* it’s working right. With large numbers of trials, 5 seems to win slightly over 4 and 6 (100,000 trials). It was very interesting to write – my first time actually using MATLAB to solve something.
Here’s the program (I don’t know how to upload files):
% This script’s purpose is to simulate the math game from the end of
% mathwithbaddrawings.com/2021/05/24/when-a-trillion-dice-are-no-better-than-a-dozen/
%
% The game is: roll n dice. score their sum. but if any comes up 2 or 5,
% remove that die, and the score for the round doesn’t count. roll the
% remaining dice. continue until no dice are left. for this version, Ben
% Orlin calculated that after even n = 12 dice, any more dice barely
% matter, with the average score pretty much 17.2. this is because with
% lots of dice, you’re almost definitely going to get at least one 2 or 5
% each round until you’re down to a few dice.
%
% At the end of the post was this variation: pick how many dice you start
% with, but you only get five roll-rounds.
%
% The purpose of this script is to simulate playing this starting with m to
% n dice, with x trials.
%
% In the future, I might extend this program so you can change which rolls
% lead to ditching the round (e.g., instead of 2 and 5, 1 and 4, or just 3)
% based on a commenter’s suggestion (eromer42).
m = input(‘Min dice to start with? (Put 1 probably)’);
n = input(‘Max dice to start with?’);
x = input(‘Number of trials for each number of dice?’);
averages = zeros(1,n-m+1); % This will store the average score for each number of dice
for i = m:n % this does the calculating, calling the below funtions
averages((i-m)+1) = averageScore(i,x);
end
[bestScore,bestDice] = maxk(averages,1);
bestDice = bestDice + m-1;
averages
fprintf(‘For numbers of dice from %.0f to ‘,m);
fprintf(‘%.0f with ‘,n);
fprintf(‘%.0f trials per number of dice,\n’,x);
fprintf(‘the highest average score was with %.0f dice,’,bestDice);
fprintf(‘with an average score of %.2f\n’,bestScore);
function score = rollDice(p) % This function tests a number of dice, p, once
a = p; % a is a placeholder for how many dice are left
score = 0; % this will store the running score
for j = 1:5
v = zeros(1,a); % this will store the rolls in each round of rolling
for i = 1:a % rolls each die once; stores in vector v
v(i) = randi(6);
end
k = find(v==2|v==5);
v(k) = [];
a = a-length(k);
if isempty(k) == true
score = score + sum(v);
end
end
end
function avg = averageScore(p,x) % p is number of dice, x is number of trials
% this function conducts the specified number of trials for the current
% number of dice and calculates the average
scores = zeros(1,x); % this will store each trial score for the current number of dice
for i = 1:x
scores(i) = rollDice(p);
end
avg = mean(scores);
end
7. Wes says:
I concur with Chrissy that the expected value for 5 dice is (approximately) 16.06, not 16.6. Precisely, it is 837242/52117. | 3,632 | 14,234 | {"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-2022-49 | latest | en | 0.934765 |
https://convertoctopus.com/551-days-to-hours | 1,659,897,194,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00279.warc.gz | 201,766,603 | 8,148 | ## Conversion formula
The conversion factor from days to hours is 24, which means that 1 day is equal to 24 hours:
1 d = 24 hr
To convert 551 days into hours we have to multiply 551 by the conversion factor in order to get the time amount from days to hours. We can also form a simple proportion to calculate the result:
1 d → 24 hr
551 d → T(hr)
Solve the above proportion to obtain the time T in hours:
T(hr) = 551 d × 24 hr
T(hr) = 13224 hr
The final result is:
551 d → 13224 hr
We conclude that 551 days is equivalent to 13224 hours:
551 days = 13224 hours
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 hour is equal to 7.5620084694495E-5 × 551 days.
Another way is saying that 551 days is equal to 1 ÷ 7.5620084694495E-5 hours.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that five hundred fifty-one days is approximately thirteen thousand two hundred twenty-four hours:
551 d ≅ 13224 hr
An alternative is also that one hour is approximately zero times five hundred fifty-one days.
## Conversion table
### days to hours chart
For quick reference purposes, below is the conversion table you can use to convert from days to hours
days (d) hours (hr)
552 days 13248 hours
553 days 13272 hours
554 days 13296 hours
555 days 13320 hours
556 days 13344 hours
557 days 13368 hours
558 days 13392 hours
559 days 13416 hours
560 days 13440 hours
561 days 13464 hours | 402 | 1,534 | {"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 | latest | en | 0.83142 |
https://programsandcourses.anu.edu.au/course/STAT6046 | 1,713,532,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00369.warc.gz | 435,387,307 | 12,248 | • Offered by Rsch Sch of Finance, Actuarial Studies & App Stats
• ANU College ANU College of Business and Economics
• Classification Transitional
• Course subject Statistics
• Areas of interest Actuarial Studies, Finance, Statistics
• Course convener
• Jacie Liu
• Mode of delivery In Person
• Co-taught Course
• Offered in First Semester 2024
Second Semester 2024
Financial Mathematics (STAT6046)
This course provides an introduction to the valuation of cash flows. Topics include: compound interest functions; valuation of annuities certain; loans repayable by instalments; comparison of value and yield of cash flow transactions; valuation of fixed interest securities, with and without tax on interest and capital gains; duration and volatility of securities; introduction to concept of immunisation and matching; consumer credit contracts.
## Learning Outcomes
Upon successful completion, students will have the knowledge and skills to:
1. Define and describe in detail the use of cash flow models, simple and compound rates of interest and discount as well as compare and distinguish between nominal and effective rates of interest and discount;
2. Describe in detail the various types of annuities and perpetuities and use them to solve financial transaction problems;
3. Derive equations of value and various tools like linear interpolation & annuity tables;
4. Communicate the difference in capital budgeting decision tools like Net Present Values, Internal Rates of Return and Discounted Payback Periods;
5. Analyse in depth basic fixed interest financial transactions like Loan Valuation, Fixed Interest securities (eg. Bonds) and employ the skills developed in this course to evaluate such transactions. Incorporate the effects of taxation on such financial transactions;
6. Demonstrate advanced knowledge of the term structure of interest rates and its applications in forward and spot rates; and
7. Define in detail the interest rate risk in terms of duration and convexity of fixed interest products, using this to define immunisation and assess its use in mitigating interest rate risk.
## Indicative Assessment
1. Typical assessment may include, but is not restricted to: exams, assignments, quizzes, presentations and other assessment as appropriate (100) [LO 1,2,3,4,5,6,7]
The ANU uses Turnitin to enhance student citation and referencing techniques, and to assess assignment submissions as a component of the University's approach to managing Academic Integrity. While the use of Turnitin is not mandatory, the ANU highly recommends Turnitin is used by both teaching staff and students. For additional information regarding Turnitin please visit the ANU Online website.
Students are expected to commit 130 hours of work in completing this course. This includes time spent in scheduled classes and self-directed study time.
Not applicable
## Requisite and Incompatibility
To enrol in this course you must have completed STAT7055 or be enrolled in the Master of Statistics, Master of Actuarial Studies or Master of Actuarial Practice. Incompatible with STAT2032.
## Prescribed Texts
Information about the prescribed textbook will be available via the Class Summary.
## Fees
Tuition fees are for the academic year indicated at the top of the page.
Commonwealth Support (CSP) Students
If you have been offered a Commonwealth supported place, your fees are set by the Australian Government for each course. At ANU 1 EFTSL is 48 units (normally 8 x 6-unit courses). More information about your student contribution amount for each course at Fees
Student Contribution Band:
1
Unit value:
6 units
If you are a domestic graduate coursework student with a Domestic Tuition Fee (DTF) place or international student you will be required to pay course tuition fees (see below). Course tuition fees are indexed annually. Further information for domestic and international students about tuition and other fees can be found at Fees.
Where there is a unit range displayed for this course, not all unit options below may be available.
Units EFTSL
6.00 0.12500
## Course fees
Domestic fee paying students
Year Fee
2024 \$4440
International fee paying students
Year Fee
2024 \$6360
Note: Please note that fee information is for current year only.
## Offerings, Dates and Class Summary Links
ANU utilises MyTimetable to enable students to view the timetable for their enrolled courses, browse, then self-allocate to small teaching activities / tutorials so they can better plan their time. Find out more on the Timetable webpage.
The list of offerings for future years is indicative only.
Class summaries, if available, can be accessed by clicking on the View link for the relevant class number.
### First Semester
Class number Class start date Last day to enrol Census date Class end date Mode Of Delivery Class Summary
4041 19 Feb 2024 26 Feb 2024 05 Apr 2024 24 May 2024 In Person View
### Second Semester
Class number Class start date Last day to enrol Census date Class end date Mode Of Delivery Class Summary
9114 22 Jul 2024 29 Jul 2024 31 Aug 2024 25 Oct 2024 In Person N/A | 1,081 | 5,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.891807 |
http://www.conservapedia.com/index.php?title=Correlation&diff=603066&oldid=600201 | 1,500,691,970,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423842.79/warc/CC-MAIN-20170722022441-20170722042441-00305.warc.gz | 387,559,543 | 10,115 | # Difference between revisions of "Correlation"
Correlation describes the relationship of two factors to one another (see cause and effect).. In common usage, it denotes an association of one variable with another in quite general terms; for example, one might say, "success is correlated with hard work". In mathematics, however, and in science and engineering, which make use of mathematical concepts, correlation is a technical term with a precise definition.
Correlation must be distinguished from causation (see article on correlation is not causation). When one factor changes and another factor changes with it, there is usually a direct relationship between the two factors, observed as a correlation. Alternatively, both changed could be the result of changes in a third factor. For example, the prices of two unrelated goods might increase during a period of inflation; the two price rises are correlated with each other but neither has caused the other.
Once a correlation is established, scientists may conduct research to determine causation. Are respiration deaths causing air pollution, or is it the other way around? It is easy to determine that sickness among the elderly does not cause air pollution. Rather, it is chemicals like sulfur dioxide (typically from coal burning power plants) which are the culprits. Cities and states measure the amount of pollutants in the air and epidemiologists can use these data, comparing them to the number of people who develop respiratory diseases.
Regulations which restrict air pollution are made on the basis of these correlations, and on the cause and effect relationships which the correlations help scientists to discover. However, activists have sometimes created false correlations by selective use of data. [1]
## Formal definition
This section is at the level of advanced high school maths (e.g. A-level or Baccalaureat) and can be skipped by general readers.
The correlation coefficient, also known as Pearson's r, is a statistical measure of association between two continuous variables. It is defined as:
r= ΣZxZy/n
Where: Zx= the Z-score of the independent variable X, Zy= the Z-score of the dependent variable Y, and n= the number of observations of variables X and Y.
Thus, Pearson's r is the arithmetic mean of the products of the variable Z-scores. The Z-scores used in the correlation coefficient must be calculated using the population formula for the standard deviation and thus:
Zx= (X-Mx)/SDx
Where: Mx= the arithmetic mean of the variable x and SDx= the standard deviation of the variable x.
SDx= Σ(X-Mx)2/n
Where: n= the number of observations of variables x and y
Zy= (Y-My)/SDy
Where: My= the arithmetic mean of the variable Y and SDy= the standard deviation of the variable Y.
SDy= Σ(Y-My)2/n
Where: n= the number of observations of variables X and Y
Pearson's r varies between -1 and +1. A value of zero indicates that no association is present between the variables, a value of +1 indicates that the strongest possible positive association exists and a value of -1 indicates that there is the strongest possible negative association. In a positive relationship, as variable X increases in value, variable Y will also increase in value (e.g. as number of hours worked (X) increases, weekly pay (Y) also increases). In contrast, in a negative relationship, as variable X increases in value, variable Y will decrease in value (e.g. as number of alcoholic beverages consumed (X) increases, score on a test of hand-eye coordination (Y) decreases).
It is important to note that while a correlation coefficient may be calculated for any set of numbers there are no guarantees that this coefficient is statistically significant. That is to say, without statistical significance we cannot be certain that the computed correlation is an accurate reflection of reality. Significance may be assessed using a variant on the standard Student's t-test defined as:
t= (r)√(n-2)/√(1-r2)
The results of this test should be evaluated against the standard t-distribution with degrees of freedom equal to n-2. As usual in significance tests, a result described as significant (i.e. a low value of P in the t test) means that the chance of getting a correlation coefficient at least as large (whichever the direction) as that observed, if there is in fact no correlation, is small. A statistically significant result can therefore arise if the actual correlation is strong, even if the dataset is small, or if the actual correlation is weak but many data are observed.
It is important to note that the correlation coefficient should not be calculated when either of the variables is not continuous. That is, when they do not vary continuously and have a meaningful zero. As such, correlating a dichotomous variable (e.g. sex) with a ratio variable (e.g. IQ) is inappropriate and will return uninterpretable results.
Additionally, correlation estimates a linear relationship between X and Y. Thus, an increase in variable X is assumed to exert the same influence on Y across all values of X and Y.
## Correlation and Causation
Main article: Correlation is not causation
Correlation is a linear statistic and makes no mathematical distinction between independent and dependent variables. As a consequence, it is impossible to assert causation based on a correlation even if that correlation is statistically significant. Therefore, when a significant correlation has been identified it is possible that X -> Y (i.e. X causes Y), Y -> X (i.e. Y causes X), or that X <- Z -> Y (i.e. Both X and Y are caused by an additional variable, Z). Caution must be exercised in asserting causation using correlation and claims to this effect must be viewed with considerable skepticism.
## References
Aron, Arthur, Elaine N. Aron and Elliot J. Coups. 2008. Statistics for the Behavioral and Social Sciences: A Brief Course. 4/e. Upper Saddle River, New Jersey: Prentice Hall. | 1,252 | 5,953 | {"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-2017-30 | longest | en | 0.957704 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-3-section-3-1-systems-of-linear-equations-in-two-variables-exercise-set-page-190/33 | 1,585,642,266,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500331.13/warc/CC-MAIN-20200331053639-20200331083639-00081.warc.gz | 874,807,095 | 13,280 | ## Intermediate Algebra for College Students (7th Edition)
To solve the given problem we will have to plug the simplest equation into the other equation and then solve for the other variable. As we are given that $x-2y=4$ or, $x=2y+4$ Now, $2(2y+4)-4y=5$ or, $4y+8-4y=5$ or, $8 \neq 5$ Hence, No solution. | 103 | 306 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-16 | latest | en | 0.819621 |
http://www.jiskha.com/display.cgi?id=1339123373 | 1,495,571,443,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607702.58/warc/CC-MAIN-20170523202350-20170523222350-00524.warc.gz | 539,474,034 | 4,031 | # Physics Equations of Motion
posted by on .
Along a straight road through town, there are three speed-limit signs. They occur in the following order: 66 , 31 , and 19 mi/h, with the 31 -mi/h sign located midway between the other two. Obeying these speed limits, the smallest possible time tA that a driver can spend on this part of the road is to travel between the first and second signs at 66 mi/h and between the second and third signs at 31 mi/h. More realistically, a driver could slow down from 66 to 31 mi/h with a constant deceleration and then do the same thing from 31 to 19 mi/h. This alternative requires a time tB. Find the ratio tB/tA
• Physics Equations of Motion - ,
90.5x10^3 :)
• Physics Equations of Motion - ,
90.5x10^3 doesn't sound reasonable since it's a ratio but i went ahead and tried it and it didn't work i believe you have to combine the equations of motion i'm just not sure how to do it! | 236 | 925 | {"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-2017-22 | latest | en | 0.960976 |
http://support.hfm.io/1.8/api/semigroupoids-5.3.2/Data-Semigroup-Traversable.html | 1,611,228,868,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00292.warc.gz | 94,452,147 | 4,111 | semigroupoids-5.3.2: Semigroupoids: Category sans id
Data.Semigroup.Traversable
Description
# Documentation
class (Foldable1 t, Traversable t) => Traversable1 t where #
Minimal complete definition
Methods
traverse1 :: Apply f => (a -> f b) -> t a -> f (t b) #
sequence1 :: Apply f => t (f b) -> f (t b) #
Instances
# Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Par1 a -> f (Par1 b) #sequence1 :: Apply f => Par1 (f b) -> f (Par1 b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Complex a -> f (Complex b) #sequence1 :: Apply f => Complex (f b) -> f (Complex b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Min a -> f (Min b) #sequence1 :: Apply f => Min (f b) -> f (Min b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Max a -> f (Max b) #sequence1 :: Apply f => Max (f b) -> f (Max b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> First a -> f (First b) #sequence1 :: Apply f => First (f b) -> f (First b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Last a -> f (Last b) #sequence1 :: Apply f => Last (f b) -> f (Last b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Identity a -> f (Identity b) #sequence1 :: Apply f => Identity (f b) -> f (Identity b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Dual a -> f (Dual b) #sequence1 :: Apply f => Dual (f b) -> f (Dual b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Sum a -> f (Sum b) #sequence1 :: Apply f => Sum (f b) -> f (Sum b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Product a -> f (Product b) #sequence1 :: Apply f => Product (f b) -> f (Product b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> NonEmpty a -> f (NonEmpty b) #sequence1 :: Apply f => NonEmpty (f b) -> f (NonEmpty b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Tree a -> f (Tree b) #sequence1 :: Apply f => Tree (f b) -> f (Tree b) # Traversable1 (V1 :: Type -> Type) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> V1 a -> f (V1 b) #sequence1 :: Apply f => V1 (f b) -> f (V1 b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a0 -> f b) -> (a, a0) -> f (a, b) #sequence1 :: Apply f => (a, f b) -> f (a, b) # Traversable1 f => Traversable1 (Lift f) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> Lift f a -> f0 (Lift f b) #sequence1 :: Apply f0 => Lift f (f0 b) -> f0 (Lift f b) # Traversable1 f => Traversable1 (Rec1 f) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> Rec1 f a -> f0 (Rec1 f b) #sequence1 :: Apply f0 => Rec1 f (f0 b) -> f0 (Rec1 f b) # Traversable1 f => Traversable1 (Alt f) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> Alt f a -> f0 (Alt f b) #sequence1 :: Apply f0 => Alt f (f0 b) -> f0 (Alt f b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a -> f b) -> Join p a -> f (Join p b) #sequence1 :: Apply f => Join p (f b) -> f (Join p b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> IdentityT f a -> f0 (IdentityT f b) #sequence1 :: Apply f0 => IdentityT f (f0 b) -> f0 (IdentityT f b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a0 -> f b) -> Tagged a a0 -> f (Tagged a b) #sequence1 :: Apply f => Tagged a (f b) -> f (Tagged a b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> Reverse f a -> f0 (Reverse f b) #sequence1 :: Apply f0 => Reverse f (f0 b) -> f0 (Reverse f b) # # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> Backwards f a -> f0 (Backwards f b) #sequence1 :: Apply f0 => Backwards f (f0 b) -> f0 (Backwards f b) # (Traversable1 f, Traversable1 g) => Traversable1 (f :+: g) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> (f :+: g) a -> f0 ((f :+: g) b) #sequence1 :: Apply f0 => (f :+: g) (f0 b) -> f0 ((f :+: g) b) # (Traversable1 f, Traversable1 g) => Traversable1 (f :*: g) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> (f :*: g) a -> f0 ((f :*: g) b) #sequence1 :: Apply f0 => (f :*: g) (f0 b) -> f0 ((f :*: g) b) # (Traversable1 f, Traversable1 g) => Traversable1 (Product f g) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> Product f g a -> f0 (Product f g b) #sequence1 :: Apply f0 => Product f g (f0 b) -> f0 (Product f g b) # (Traversable1 f, Traversable1 g) => Traversable1 (Sum f g) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> Sum f g a -> f0 (Sum f g b) #sequence1 :: Apply f0 => Sum f g (f0 b) -> f0 (Sum f g b) # Traversable1 f => Traversable1 (M1 i c f) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> M1 i c f a -> f0 (M1 i c f b) #sequence1 :: Apply f0 => M1 i c f (f0 b) -> f0 (M1 i c f b) # (Traversable1 f, Traversable1 g) => Traversable1 (f :.: g) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> (f :.: g) a -> f0 ((f :.: g) b) #sequence1 :: Apply f0 => (f :.: g) (f0 b) -> f0 ((f :.: g) b) # (Traversable1 f, Traversable1 g) => Traversable1 (Compose f g) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f0 => (a -> f0 b) -> Compose f g a -> f0 (Compose f g b) #sequence1 :: Apply f0 => Compose f g (f0 b) -> f0 (Compose f g b) # Traversable1 g => Traversable1 (Joker g a) # Instance detailsDefined in Data.Semigroup.Traversable.Class Methodstraverse1 :: Apply f => (a0 -> f b) -> Joker g a a0 -> f (Joker g a b) #sequence1 :: Apply f => Joker g a (f b) -> f (Joker g a b) #
foldMap1Default :: (Traversable1 f, Semigroup m) => (a -> m) -> f a -> m # | 2,194 | 6,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.50017 |
http://us.metamath.org/mpeuni/blssps.html | 1,657,167,435,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00216.warc.gz | 56,796,642 | 10,384 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > blssps Structured version Visualization version GIF version
Theorem blssps 22450
Description: Any point 𝑃 in a ball 𝐵 can be centered in another ball that is a subset of 𝐵. (Contributed by NM, 31-Aug-2006.) (Revised by Mario Carneiro, 24-Aug-2015.) (Revised by Thierry Arnoux, 11-Mar-2018.)
Assertion
Ref Expression
blssps ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝐵 ∈ ran (ball‘𝐷) ∧ 𝑃𝐵) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ 𝐵)
Distinct variable groups: 𝑥,𝐵 𝑥,𝐷 𝑥,𝑃 𝑥,𝑋
Proof of Theorem blssps
Dummy variables 𝑟 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 blrnps 22434 . . 3 (𝐷 ∈ (PsMet‘𝑋) → (𝐵 ∈ ran (ball‘𝐷) ↔ ∃𝑦𝑋𝑟 ∈ ℝ* 𝐵 = (𝑦(ball‘𝐷)𝑟)))
2 elblps 22413 . . . . . . 7 ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) → (𝑃 ∈ (𝑦(ball‘𝐷)𝑟) ↔ (𝑃𝑋 ∧ (𝑦𝐷𝑃) < 𝑟)))
3 simpl1 1228 . . . . . . . . . . 11 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) → 𝐷 ∈ (PsMet‘𝑋))
4 simpl2 1230 . . . . . . . . . . 11 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) → 𝑦𝑋)
5 simpr 479 . . . . . . . . . . 11 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) → 𝑃𝑋)
6 psmetcl 22333 . . . . . . . . . . 11 ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑃𝑋) → (𝑦𝐷𝑃) ∈ ℝ*)
73, 4, 5, 6syl3anc 1477 . . . . . . . . . 10 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) → (𝑦𝐷𝑃) ∈ ℝ*)
8 simpl3 1232 . . . . . . . . . 10 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) → 𝑟 ∈ ℝ*)
9 qbtwnxr 12244 . . . . . . . . . . 11 (((𝑦𝐷𝑃) ∈ ℝ*𝑟 ∈ ℝ* ∧ (𝑦𝐷𝑃) < 𝑟) → ∃𝑧 ∈ ℚ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))
1093expia 1115 . . . . . . . . . 10 (((𝑦𝐷𝑃) ∈ ℝ*𝑟 ∈ ℝ*) → ((𝑦𝐷𝑃) < 𝑟 → ∃𝑧 ∈ ℚ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟)))
117, 8, 10syl2anc 696 . . . . . . . . 9 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) → ((𝑦𝐷𝑃) < 𝑟 → ∃𝑧 ∈ ℚ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟)))
12 qre 12006 . . . . . . . . . . 11 (𝑧 ∈ ℚ → 𝑧 ∈ ℝ)
13 simpll1 1255 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝐷 ∈ (PsMet‘𝑋))
14 simplr 809 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝑃𝑋)
15 simpll2 1257 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝑦𝑋)
16 psmetsym 22336 . . . . . . . . . . . . . . . 16 ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑃𝑋𝑦𝑋) → (𝑃𝐷𝑦) = (𝑦𝐷𝑃))
1713, 14, 15, 16syl3anc 1477 . . . . . . . . . . . . . . 15 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃𝐷𝑦) = (𝑦𝐷𝑃))
18 simprrl 823 . . . . . . . . . . . . . . 15 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑦𝐷𝑃) < 𝑧)
1917, 18eqbrtrd 4826 . . . . . . . . . . . . . 14 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃𝐷𝑦) < 𝑧)
20 simprl 811 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝑧 ∈ ℝ)
21 psmetcl 22333 . . . . . . . . . . . . . . . . . . 19 ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑃𝑋𝑦𝑋) → (𝑃𝐷𝑦) ∈ ℝ*)
2213, 14, 15, 21syl3anc 1477 . . . . . . . . . . . . . . . . . 18 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃𝐷𝑦) ∈ ℝ*)
23 rexr 10297 . . . . . . . . . . . . . . . . . . 19 (𝑧 ∈ ℝ → 𝑧 ∈ ℝ*)
2423ad2antrl 766 . . . . . . . . . . . . . . . . . 18 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝑧 ∈ ℝ*)
25 xrltle 12195 . . . . . . . . . . . . . . . . . 18 (((𝑃𝐷𝑦) ∈ ℝ*𝑧 ∈ ℝ*) → ((𝑃𝐷𝑦) < 𝑧 → (𝑃𝐷𝑦) ≤ 𝑧))
2622, 24, 25syl2anc 696 . . . . . . . . . . . . . . . . 17 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → ((𝑃𝐷𝑦) < 𝑧 → (𝑃𝐷𝑦) ≤ 𝑧))
2719, 26mpd 15 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃𝐷𝑦) ≤ 𝑧)
28 psmetlecl 22341 . . . . . . . . . . . . . . . 16 ((𝐷 ∈ (PsMet‘𝑋) ∧ (𝑃𝑋𝑦𝑋) ∧ (𝑧 ∈ ℝ ∧ (𝑃𝐷𝑦) ≤ 𝑧)) → (𝑃𝐷𝑦) ∈ ℝ)
2913, 14, 15, 20, 27, 28syl122anc 1486 . . . . . . . . . . . . . . 15 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃𝐷𝑦) ∈ ℝ)
30 difrp 12081 . . . . . . . . . . . . . . 15 (((𝑃𝐷𝑦) ∈ ℝ ∧ 𝑧 ∈ ℝ) → ((𝑃𝐷𝑦) < 𝑧 ↔ (𝑧 − (𝑃𝐷𝑦)) ∈ ℝ+))
3129, 20, 30syl2anc 696 . . . . . . . . . . . . . 14 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → ((𝑃𝐷𝑦) < 𝑧 ↔ (𝑧 − (𝑃𝐷𝑦)) ∈ ℝ+))
3219, 31mpbid 222 . . . . . . . . . . . . 13 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑧 − (𝑃𝐷𝑦)) ∈ ℝ+)
3320, 29resubcld 10670 . . . . . . . . . . . . . . 15 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑧 − (𝑃𝐷𝑦)) ∈ ℝ)
34 xrleid 12196 . . . . . . . . . . . . . . . . 17 ((𝑃𝐷𝑦) ∈ ℝ* → (𝑃𝐷𝑦) ≤ (𝑃𝐷𝑦))
3522, 34syl 17 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃𝐷𝑦) ≤ (𝑃𝐷𝑦))
3620recnd 10280 . . . . . . . . . . . . . . . . 17 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝑧 ∈ ℂ)
3729recnd 10280 . . . . . . . . . . . . . . . . 17 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃𝐷𝑦) ∈ ℂ)
3836, 37nncand 10609 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑧 − (𝑧 − (𝑃𝐷𝑦))) = (𝑃𝐷𝑦))
3935, 38breqtrrd 4832 . . . . . . . . . . . . . . 15 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃𝐷𝑦) ≤ (𝑧 − (𝑧 − (𝑃𝐷𝑦))))
40 blss2ps 22429 . . . . . . . . . . . . . . 15 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑃𝑋𝑦𝑋) ∧ ((𝑧 − (𝑃𝐷𝑦)) ∈ ℝ ∧ 𝑧 ∈ ℝ ∧ (𝑃𝐷𝑦) ≤ (𝑧 − (𝑧 − (𝑃𝐷𝑦))))) → (𝑃(ball‘𝐷)(𝑧 − (𝑃𝐷𝑦))) ⊆ (𝑦(ball‘𝐷)𝑧))
4113, 14, 15, 33, 20, 39, 40syl33anc 1492 . . . . . . . . . . . . . 14 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃(ball‘𝐷)(𝑧 − (𝑃𝐷𝑦))) ⊆ (𝑦(ball‘𝐷)𝑧))
42 simpll3 1259 . . . . . . . . . . . . . . 15 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝑟 ∈ ℝ*)
43 simprrr 824 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝑧 < 𝑟)
44 xrltle 12195 . . . . . . . . . . . . . . . . 17 ((𝑧 ∈ ℝ*𝑟 ∈ ℝ*) → (𝑧 < 𝑟𝑧𝑟))
4524, 42, 44syl2anc 696 . . . . . . . . . . . . . . . 16 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑧 < 𝑟𝑧𝑟))
4643, 45mpd 15 . . . . . . . . . . . . . . 15 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → 𝑧𝑟)
47 ssblps 22448 . . . . . . . . . . . . . . 15 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋) ∧ (𝑧 ∈ ℝ*𝑟 ∈ ℝ*) ∧ 𝑧𝑟) → (𝑦(ball‘𝐷)𝑧) ⊆ (𝑦(ball‘𝐷)𝑟))
4813, 15, 24, 42, 46, 47syl221anc 1488 . . . . . . . . . . . . . 14 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑦(ball‘𝐷)𝑧) ⊆ (𝑦(ball‘𝐷)𝑟))
4941, 48sstrd 3754 . . . . . . . . . . . . 13 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → (𝑃(ball‘𝐷)(𝑧 − (𝑃𝐷𝑦))) ⊆ (𝑦(ball‘𝐷)𝑟))
50 oveq2 6822 . . . . . . . . . . . . . . 15 (𝑥 = (𝑧 − (𝑃𝐷𝑦)) → (𝑃(ball‘𝐷)𝑥) = (𝑃(ball‘𝐷)(𝑧 − (𝑃𝐷𝑦))))
5150sseq1d 3773 . . . . . . . . . . . . . 14 (𝑥 = (𝑧 − (𝑃𝐷𝑦)) → ((𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟) ↔ (𝑃(ball‘𝐷)(𝑧 − (𝑃𝐷𝑦))) ⊆ (𝑦(ball‘𝐷)𝑟)))
5251rspcev 3449 . . . . . . . . . . . . 13 (((𝑧 − (𝑃𝐷𝑦)) ∈ ℝ+ ∧ (𝑃(ball‘𝐷)(𝑧 − (𝑃𝐷𝑦))) ⊆ (𝑦(ball‘𝐷)𝑟)) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟))
5332, 49, 52syl2anc 696 . . . . . . . . . . . 12 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ (𝑧 ∈ ℝ ∧ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟))) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟))
5453expr 644 . . . . . . . . . . 11 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ 𝑧 ∈ ℝ) → (((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟)))
5512, 54sylan2 492 . . . . . . . . . 10 ((((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) ∧ 𝑧 ∈ ℚ) → (((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟)))
5655rexlimdva 3169 . . . . . . . . 9 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) → (∃𝑧 ∈ ℚ ((𝑦𝐷𝑃) < 𝑧𝑧 < 𝑟) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟)))
5711, 56syld 47 . . . . . . . 8 (((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) ∧ 𝑃𝑋) → ((𝑦𝐷𝑃) < 𝑟 → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟)))
5857expimpd 630 . . . . . . 7 ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) → ((𝑃𝑋 ∧ (𝑦𝐷𝑃) < 𝑟) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟)))
592, 58sylbid 230 . . . . . 6 ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) → (𝑃 ∈ (𝑦(ball‘𝐷)𝑟) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟)))
60 eleq2 2828 . . . . . . 7 (𝐵 = (𝑦(ball‘𝐷)𝑟) → (𝑃𝐵𝑃 ∈ (𝑦(ball‘𝐷)𝑟)))
61 sseq2 3768 . . . . . . . 8 (𝐵 = (𝑦(ball‘𝐷)𝑟) → ((𝑃(ball‘𝐷)𝑥) ⊆ 𝐵 ↔ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟)))
6261rexbidv 3190 . . . . . . 7 (𝐵 = (𝑦(ball‘𝐷)𝑟) → (∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ 𝐵 ↔ ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟)))
6360, 62imbi12d 333 . . . . . 6 (𝐵 = (𝑦(ball‘𝐷)𝑟) → ((𝑃𝐵 → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ 𝐵) ↔ (𝑃 ∈ (𝑦(ball‘𝐷)𝑟) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ (𝑦(ball‘𝐷)𝑟))))
6459, 63syl5ibrcom 237 . . . . 5 ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝑦𝑋𝑟 ∈ ℝ*) → (𝐵 = (𝑦(ball‘𝐷)𝑟) → (𝑃𝐵 → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ 𝐵)))
65643expib 1117 . . . 4 (𝐷 ∈ (PsMet‘𝑋) → ((𝑦𝑋𝑟 ∈ ℝ*) → (𝐵 = (𝑦(ball‘𝐷)𝑟) → (𝑃𝐵 → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ 𝐵))))
6665rexlimdvv 3175 . . 3 (𝐷 ∈ (PsMet‘𝑋) → (∃𝑦𝑋𝑟 ∈ ℝ* 𝐵 = (𝑦(ball‘𝐷)𝑟) → (𝑃𝐵 → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ 𝐵)))
671, 66sylbid 230 . 2 (𝐷 ∈ (PsMet‘𝑋) → (𝐵 ∈ ran (ball‘𝐷) → (𝑃𝐵 → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ 𝐵)))
68673imp 1102 1 ((𝐷 ∈ (PsMet‘𝑋) ∧ 𝐵 ∈ ran (ball‘𝐷) ∧ 𝑃𝐵) → ∃𝑥 ∈ ℝ+ (𝑃(ball‘𝐷)𝑥) ⊆ 𝐵)
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 383 ∧ w3a 1072 = wceq 1632 ∈ wcel 2139 ∃wrex 3051 ⊆ wss 3715 class class class wbr 4804 ran crn 5267 ‘cfv 6049 (class class class)co 6814 ℝcr 10147 ℝ*cxr 10285 < clt 10286 ≤ cle 10287 − cmin 10478 ℚcq 12001 ℝ+crp 12045 PsMetcpsmet 19952 ballcbl 19955 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1988 ax-6 2054 ax-7 2090 ax-8 2141 ax-9 2148 ax-10 2168 ax-11 2183 ax-12 2196 ax-13 2391 ax-ext 2740 ax-sep 4933 ax-nul 4941 ax-pow 4992 ax-pr 5055 ax-un 7115 ax-cnex 10204 ax-resscn 10205 ax-1cn 10206 ax-icn 10207 ax-addcl 10208 ax-addrcl 10209 ax-mulcl 10210 ax-mulrcl 10211 ax-mulcom 10212 ax-addass 10213 ax-mulass 10214 ax-distr 10215 ax-i2m1 10216 ax-1ne0 10217 ax-1rid 10218 ax-rnegex 10219 ax-rrecex 10220 ax-cnre 10221 ax-pre-lttri 10222 ax-pre-lttrn 10223 ax-pre-ltadd 10224 ax-pre-mulgt0 10225 ax-pre-sup 10226 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1073 df-3an 1074 df-tru 1635 df-ex 1854 df-nf 1859 df-sb 2047 df-eu 2611 df-mo 2612 df-clab 2747 df-cleq 2753 df-clel 2756 df-nfc 2891 df-ne 2933 df-nel 3036 df-ral 3055 df-rex 3056 df-reu 3057 df-rmo 3058 df-rab 3059 df-v 3342 df-sbc 3577 df-csb 3675 df-dif 3718 df-un 3720 df-in 3722 df-ss 3729 df-pss 3731 df-nul 4059 df-if 4231 df-pw 4304 df-sn 4322 df-pr 4324 df-tp 4326 df-op 4328 df-uni 4589 df-iun 4674 df-br 4805 df-opab 4865 df-mpt 4882 df-tr 4905 df-id 5174 df-eprel 5179 df-po 5187 df-so 5188 df-fr 5225 df-we 5227 df-xp 5272 df-rel 5273 df-cnv 5274 df-co 5275 df-dm 5276 df-rn 5277 df-res 5278 df-ima 5279 df-pred 5841 df-ord 5887 df-on 5888 df-lim 5889 df-suc 5890 df-iota 6012 df-fun 6051 df-fn 6052 df-f 6053 df-f1 6054 df-fo 6055 df-f1o 6056 df-fv 6057 df-riota 6775 df-ov 6817 df-oprab 6818 df-mpt2 6819 df-om 7232 df-1st 7334 df-2nd 7335 df-wrecs 7577 df-recs 7638 df-rdg 7676 df-er 7913 df-map 8027 df-en 8124 df-dom 8125 df-sdom 8126 df-sup 8515 df-inf 8516 df-pnf 10288 df-mnf 10289 df-xr 10290 df-ltxr 10291 df-le 10292 df-sub 10480 df-neg 10481 df-div 10897 df-nn 11233 df-2 11291 df-n0 11505 df-z 11590 df-uz 11900 df-q 12002 df-rp 12046 df-xneg 12159 df-xadd 12160 df-xmul 12161 df-psmet 19960 df-bl 19963 This theorem is referenced by: blssexps 22452
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The Human Connectome Project aims to provide an unparalleled compilation of neural data, an interface to graphically navigate this data and the opportunity to achieve never before realized conclusions about the living human brain. | 735 | 3,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-24 | latest | en | 0.694573 |
https://www.coursehero.com/tutors-problems/Statistics-and-Probability/26903941-What-is-the-income-distribution-of-super-shoppers-A-supermarket-super/ | 1,656,614,499,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00042.warc.gz | 788,982,389 | 36,795 | Question
# What is the income distribution of super shoppers? A supermarket super shopperx
is defined as a shopper for whom at least 70% of the items purchased were on sale or purchased with a coupon. In the following table, income units are in thousands of dollars, and each interval goes up to but does not include the given high value. The midpoints are given to the nearest thousand dollars.Income range5-1515-2525-3535-4545-5555 or moreMidpoint 102030405060Percent of super shoppers20%14%21%17%18%10%
(d) Compute the standard deviation σ for the income of super shoppers (in thousands of dollars). (Enter a number. Round your answer to two decimal places.)
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onec aliqlamet, consectetlFusce dui lectlipiscing elit.lipiscing elit.le vel laoreet ala. Fusce dui lectlpulv | 369 | 1,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-27 | latest | en | 0.668206 |
https://computergraphics.stackexchange.com/questions/10791/pbr-understanding-the-right-part-of-split-sum-integration-of-specular-ibl | 1,713,913,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00588.warc.gz | 156,498,055 | 43,113 | # PBR : Understanding the right part of Split sum integration of specular IBL
This is the website I am learning from.
In the specular part of IBL this is the split sum approximation
Now the left part of the equation I have asked in This post and I have understood everything now this question is about the right part.
On simplifying the right part of the equation with the procedure shown in the website we get this
And on translating this to code we get this
1)What is the V vector? How did they compute that
2)Why have they assumed N=(0,0,1)?
3)why is NDotL=max(L.z,0)? How did they simplifying a dot product to an min/max Operation same question I have for NDotH
4)G_Vis=GVDoth/(NDotHNDotL) why the division by (NDOTH*NDOTL)? that was nowhere in the equation.
5)what is vector L? it looks like they have reflected vector V about H to get this but why?
These are just some of the questions I have come up with so far but if there are any other important parts I should not know about A general explanation on how they derived this code would be appreciated
1)What is the V vector? How did they compute that
It's the vector toward the camera (view vector), i.e. direction of the reflected ray. The lookup table they're building is parameterized in terms of NdotV, so they are working backward from NdotV to obtain a corresponding V vector. For purposes of doing this integration it doesn't matter what N and V are exactly, just that they have the correct NdotV. So, just to make things concrete, they assume that $$N = (0,0,1)$$ and V is aligned with the positive x-axis (i.e. in the positive xz half-plane). Then you have $$V = (\sin θ, 0, \cos θ)$$ where θ is the angle from N to V (this is just the equation for a circular arc from z down to x). Now $$N \cdot V = \cos θ$$, so by a standard trig identity $$\sin θ = \sqrt{1 - \cos^2 θ} = \sqrt{1 - (N \cdot V)^2}$$.
2)Why have they assumed N=(0,0,1)?
This whole thing is a pre-integration of the BRDF alone, it doesn't depend on any scene or lighting environment. So it doesn't matter where the BRDF is or how it is oriented in space, only the relative orientation between the N, L, and V vectors is important. So just for the sake of convenience they have chosen to fix N = (0, 0, 1), let V vary only in the xz half-plane as mentioned above, and let L vary over the whole hemisphere. To put it another way, they have chosen to set the coordinate system to be aligned with N and V.
3)why is NDotL=max(L.z,0)? How did they simplifying a dot product to an min/max Operation same question I have for NDotH
Since N = (0, 0, 1), dotting N with any vector just takes the z component of that vector. $$N \cdot L = 0L_x + 0L_y + 1L_z = L_z$$. (This is why it's convenient to assume a coordinate system in which N = (0, 0, 1).) Taking the max with 0 clamps it to the upper hemisphere, otherwise the dot product would go negative if the vector happens to be pointing below the hemisphere. This clamp is a standard thing to do when dealing with opaque BRDFs to avoid accidentally getting unwanted negative values in your calculations— although it is not really necessary here since they check if (NdotL > 0.0) immediately afterward.
4)G_Vis=GVDoth/(NDotHNDotL) why the division by (NDOTH*NDOTL)? that was nowhere in the equation.
It is a standard part of the microfacet BRDF formulation: $$f_r(p, w_i, w_o) = \frac{DFG}{4(\omega_o \cdot n)(\omega_i \cdot n)}$$ Note that $$\omega_o = V$$, and $$\omega_i = L$$, these are just different notations/names for the incoming and outgoing light vectors. However I think there is a typo in the code, the denominator should contain NdotV rather than NdotH. I am also not sure why there is a multiplication by VdotH in the numerator, or where the factor of 4 went, but perhaps they have accounted for it elsewhere. I would probably double-check the BRDF equations here against some other sources as I don't think it's entirely correct.
5)what is vector L? it looks like they have reflected vector V about H to get this but why?
As noted, L is the vector toward the incoming light. H is the "half-angle" vector exactly halfway between L and V, so reflecting V around H will give you L, or vice versa. The significance of H is that in microfacet BRDFs, it is the normal vector of those microfacets that are "active" for a given L and V. So, to importance-sample a BRDF, we sample a random normal vector from the NDF, and use that as H. Since H and V are specified, we then calculate L from those.
I recommend to ignore my post (however I consider papers that I've sent valueable) and read post of Nathan Reed, which is much better in explaining this problem!
I'm still a junior when it comes to computer graphics so you may want to take my opinion with a grain of salt:
To start with - I used to learn/implement stuff using website that you've sent (OpenGL) and I don't recommend that. After few months of learning I knew barely anything. Then I start reading original papers and implementing them on my own to better understand PBR, what N * G * F means in Cook-Torrance shading model and I got much better understanding of basics. I was recommend by many graphics programmers to avoid tutorial pages and straight read and implement papers. In case if you don't understand something in the paper you're reading, you found paper on subject that you don't understand and read about it.
1). As I said - avoid this site or use only to get interesting papers - here are original notes from [Karis2013] UE4 presentation - https://blog.selfshadow.com/publications/s2013-shading-course/karis/s2013_pbs_epic_notes_v2.pdf ; Please notice comment "/sin" "/cos" in listing at page 7 and usage of Pythagorean trigonometric identity. Maybe it's position on the hemishpere, I don't really want to dig in, I'm sorry. But I would recommend checking Epic Games repository on Github if you want to see how they're using it.
2.) In the same paper [Karis2013] there is listing at the page 4. and function ImportanceSampleGGX. It uses N to determine orientation of the world and returns either float3(0,0,1) or float3(1,0,0) based on N.z value.
3). $$N = (0, 0, 1)$$ $$L = (x, y, z)$$ $$dot(N, L) = (0 * x, 0 * y, 1 * z) = (0, 0, z) = L.z$$
http://graphicrants.blogspot.com/2013/08/specular-brdf-reference.html | 1,627 | 6,298 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-18 | latest | en | 0.953775 |
https://us.metamath.org/mpeuni/dirkertrigeqlem2.html | 1,720,991,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514638.53/warc/CC-MAIN-20240714185510-20240714215510-00759.warc.gz | 540,643,923 | 19,621 | Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > dirkertrigeqlem2 Structured version Visualization version GIF version
Theorem dirkertrigeqlem2 42728
Description: Trigonomic equality lemma for the Dirichlet Kernel trigonomic equality. (Contributed by Glauco Siliprandi, 11-Dec-2019.)
Hypotheses
Ref Expression
dirkertrigeqlem2.a (𝜑𝐴 ∈ ℝ)
dirkertrigeqlem2.sinne0 (𝜑 → (sin‘𝐴) ≠ 0)
dirkertrigeqlem2.n (𝜑𝑁 ∈ ℕ)
Assertion
Ref Expression
dirkertrigeqlem2 (𝜑 → (((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) / π) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / ((2 · π) · (sin‘(𝐴 / 2)))))
Distinct variable groups: 𝐴,𝑛 𝑛,𝑁 𝜑,𝑛
Proof of Theorem dirkertrigeqlem2
Dummy variable 𝑗 is distinct from all other variables.
StepHypRef Expression
1 1cnd 10629 . . . . . . . . 9 (𝜑 → 1 ∈ ℂ)
21halfcld 11874 . . . . . . . 8 (𝜑 → (1 / 2) ∈ ℂ)
3 fzfid 13340 . . . . . . . . 9 (𝜑 → (1...𝑁) ∈ Fin)
4 elfzelz 12906 . . . . . . . . . . . . 13 (𝑛 ∈ (1...𝑁) → 𝑛 ∈ ℤ)
54zcnd 12080 . . . . . . . . . . . 12 (𝑛 ∈ (1...𝑁) → 𝑛 ∈ ℂ)
65adantl 485 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → 𝑛 ∈ ℂ)
7 dirkertrigeqlem2.a . . . . . . . . . . . . 13 (𝜑𝐴 ∈ ℝ)
87recnd 10662 . . . . . . . . . . . 12 (𝜑𝐴 ∈ ℂ)
98adantr 484 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → 𝐴 ∈ ℂ)
106, 9mulcld 10654 . . . . . . . . . 10 ((𝜑𝑛 ∈ (1...𝑁)) → (𝑛 · 𝐴) ∈ ℂ)
1110coscld 15479 . . . . . . . . 9 ((𝜑𝑛 ∈ (1...𝑁)) → (cos‘(𝑛 · 𝐴)) ∈ ℂ)
123, 11fsumcl 15085 . . . . . . . 8 (𝜑 → Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴)) ∈ ℂ)
132, 12addcld 10653 . . . . . . 7 (𝜑 → ((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) ∈ ℂ)
148sincld 15478 . . . . . . 7 (𝜑 → (sin‘𝐴) ∈ ℂ)
15 dirkertrigeqlem2.sinne0 . . . . . . 7 (𝜑 → (sin‘𝐴) ≠ 0)
1613, 14, 15divcan4d 11415 . . . . . 6 (𝜑 → ((((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) · (sin‘𝐴)) / (sin‘𝐴)) = ((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))))
1716eqcomd 2807 . . . . 5 (𝜑 → ((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) = ((((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) · (sin‘𝐴)) / (sin‘𝐴)))
183, 14, 11fsummulc1 15135 . . . . . . . . 9 (𝜑 → (Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴)) · (sin‘𝐴)) = Σ𝑛 ∈ (1...𝑁)((cos‘(𝑛 · 𝐴)) · (sin‘𝐴)))
1914adantr 484 . . . . . . . . . . . 12 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘𝐴) ∈ ℂ)
2011, 19mulcomd 10655 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → ((cos‘(𝑛 · 𝐴)) · (sin‘𝐴)) = ((sin‘𝐴) · (cos‘(𝑛 · 𝐴))))
21 sinmulcos 42494 . . . . . . . . . . . 12 ((𝐴 ∈ ℂ ∧ (𝑛 · 𝐴) ∈ ℂ) → ((sin‘𝐴) · (cos‘(𝑛 · 𝐴))) = (((sin‘(𝐴 + (𝑛 · 𝐴))) + (sin‘(𝐴 − (𝑛 · 𝐴)))) / 2))
229, 10, 21syl2anc 587 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘𝐴) · (cos‘(𝑛 · 𝐴))) = (((sin‘(𝐴 + (𝑛 · 𝐴))) + (sin‘(𝐴 − (𝑛 · 𝐴)))) / 2))
23 1cnd 10629 . . . . . . . . . . . . . . . . 17 ((𝜑𝑛 ∈ (1...𝑁)) → 1 ∈ ℂ)
246, 23, 9adddird 10659 . . . . . . . . . . . . . . . 16 ((𝜑𝑛 ∈ (1...𝑁)) → ((𝑛 + 1) · 𝐴) = ((𝑛 · 𝐴) + (1 · 𝐴)))
2523, 9mulcld 10654 . . . . . . . . . . . . . . . . 17 ((𝜑𝑛 ∈ (1...𝑁)) → (1 · 𝐴) ∈ ℂ)
2610, 25addcomd 10835 . . . . . . . . . . . . . . . 16 ((𝜑𝑛 ∈ (1...𝑁)) → ((𝑛 · 𝐴) + (1 · 𝐴)) = ((1 · 𝐴) + (𝑛 · 𝐴)))
278mulid2d 10652 . . . . . . . . . . . . . . . . . 18 (𝜑 → (1 · 𝐴) = 𝐴)
2827oveq1d 7154 . . . . . . . . . . . . . . . . 17 (𝜑 → ((1 · 𝐴) + (𝑛 · 𝐴)) = (𝐴 + (𝑛 · 𝐴)))
2928adantr 484 . . . . . . . . . . . . . . . 16 ((𝜑𝑛 ∈ (1...𝑁)) → ((1 · 𝐴) + (𝑛 · 𝐴)) = (𝐴 + (𝑛 · 𝐴)))
3024, 26, 293eqtrrd 2841 . . . . . . . . . . . . . . 15 ((𝜑𝑛 ∈ (1...𝑁)) → (𝐴 + (𝑛 · 𝐴)) = ((𝑛 + 1) · 𝐴))
3130fveq2d 6653 . . . . . . . . . . . . . 14 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘(𝐴 + (𝑛 · 𝐴))) = (sin‘((𝑛 + 1) · 𝐴)))
3210, 9negsubdi2d 11006 . . . . . . . . . . . . . . . . 17 ((𝜑𝑛 ∈ (1...𝑁)) → -((𝑛 · 𝐴) − 𝐴) = (𝐴 − (𝑛 · 𝐴)))
3332eqcomd 2807 . . . . . . . . . . . . . . . 16 ((𝜑𝑛 ∈ (1...𝑁)) → (𝐴 − (𝑛 · 𝐴)) = -((𝑛 · 𝐴) − 𝐴))
3433fveq2d 6653 . . . . . . . . . . . . . . 15 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘(𝐴 − (𝑛 · 𝐴))) = (sin‘-((𝑛 · 𝐴) − 𝐴)))
3510, 9subcld 10990 . . . . . . . . . . . . . . . 16 ((𝜑𝑛 ∈ (1...𝑁)) → ((𝑛 · 𝐴) − 𝐴) ∈ ℂ)
36 sinneg 15494 . . . . . . . . . . . . . . . 16 (((𝑛 · 𝐴) − 𝐴) ∈ ℂ → (sin‘-((𝑛 · 𝐴) − 𝐴)) = -(sin‘((𝑛 · 𝐴) − 𝐴)))
3735, 36syl 17 . . . . . . . . . . . . . . 15 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘-((𝑛 · 𝐴) − 𝐴)) = -(sin‘((𝑛 · 𝐴) − 𝐴)))
3834, 37eqtrd 2836 . . . . . . . . . . . . . 14 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘(𝐴 − (𝑛 · 𝐴))) = -(sin‘((𝑛 · 𝐴) − 𝐴)))
3931, 38oveq12d 7157 . . . . . . . . . . . . 13 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘(𝐴 + (𝑛 · 𝐴))) + (sin‘(𝐴 − (𝑛 · 𝐴)))) = ((sin‘((𝑛 + 1) · 𝐴)) + -(sin‘((𝑛 · 𝐴) − 𝐴))))
409, 10addcld 10653 . . . . . . . . . . . . . . . 16 ((𝜑𝑛 ∈ (1...𝑁)) → (𝐴 + (𝑛 · 𝐴)) ∈ ℂ)
4140sincld 15478 . . . . . . . . . . . . . . 15 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘(𝐴 + (𝑛 · 𝐴))) ∈ ℂ)
4231, 41eqeltrrd 2894 . . . . . . . . . . . . . 14 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘((𝑛 + 1) · 𝐴)) ∈ ℂ)
4335sincld 15478 . . . . . . . . . . . . . 14 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘((𝑛 · 𝐴) − 𝐴)) ∈ ℂ)
4442, 43negsubd 10996 . . . . . . . . . . . . 13 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘((𝑛 + 1) · 𝐴)) + -(sin‘((𝑛 · 𝐴) − 𝐴))) = ((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 · 𝐴) − 𝐴))))
456, 9mulsubfacd 11094 . . . . . . . . . . . . . . 15 ((𝜑𝑛 ∈ (1...𝑁)) → ((𝑛 · 𝐴) − 𝐴) = ((𝑛 − 1) · 𝐴))
4645fveq2d 6653 . . . . . . . . . . . . . 14 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘((𝑛 · 𝐴) − 𝐴)) = (sin‘((𝑛 − 1) · 𝐴)))
4746oveq2d 7155 . . . . . . . . . . . . 13 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 · 𝐴) − 𝐴))) = ((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))
4839, 44, 473eqtrd 2840 . . . . . . . . . . . 12 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘(𝐴 + (𝑛 · 𝐴))) + (sin‘(𝐴 − (𝑛 · 𝐴)))) = ((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))
4948oveq1d 7154 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → (((sin‘(𝐴 + (𝑛 · 𝐴))) + (sin‘(𝐴 − (𝑛 · 𝐴)))) / 2) = (((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) / 2))
5020, 22, 493eqtrd 2840 . . . . . . . . . 10 ((𝜑𝑛 ∈ (1...𝑁)) → ((cos‘(𝑛 · 𝐴)) · (sin‘𝐴)) = (((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) / 2))
5150sumeq2dv 15055 . . . . . . . . 9 (𝜑 → Σ𝑛 ∈ (1...𝑁)((cos‘(𝑛 · 𝐴)) · (sin‘𝐴)) = Σ𝑛 ∈ (1...𝑁)(((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) / 2))
52 2cnd 11707 . . . . . . . . . . 11 (𝜑 → 2 ∈ ℂ)
53 peano2cnm 10945 . . . . . . . . . . . . . . 15 (𝑛 ∈ ℂ → (𝑛 − 1) ∈ ℂ)
546, 53syl 17 . . . . . . . . . . . . . 14 ((𝜑𝑛 ∈ (1...𝑁)) → (𝑛 − 1) ∈ ℂ)
5554, 9mulcld 10654 . . . . . . . . . . . . 13 ((𝜑𝑛 ∈ (1...𝑁)) → ((𝑛 − 1) · 𝐴) ∈ ℂ)
5655sincld 15478 . . . . . . . . . . . 12 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘((𝑛 − 1) · 𝐴)) ∈ ℂ)
5742, 56subcld 10990 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) ∈ ℂ)
58 2ne0 11733 . . . . . . . . . . . 12 2 ≠ 0
5958a1i 11 . . . . . . . . . . 11 (𝜑 → 2 ≠ 0)
603, 52, 57, 59fsumdivc 15136 . . . . . . . . . 10 (𝜑 → (Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) / 2) = Σ𝑛 ∈ (1...𝑁)(((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) / 2))
613, 57fsumcl 15085 . . . . . . . . . . 11 (𝜑 → Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) ∈ ℂ)
6261, 52, 59divrec2d 11413 . . . . . . . . . 10 (𝜑 → (Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) / 2) = ((1 / 2) · Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))))
6360, 62eqtr3d 2838 . . . . . . . . 9 (𝜑 → Σ𝑛 ∈ (1...𝑁)(((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) / 2) = ((1 / 2) · Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))))
6418, 51, 633eqtrd 2840 . . . . . . . 8 (𝜑 → (Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴)) · (sin‘𝐴)) = ((1 / 2) · Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))))
6564oveq2d 7155 . . . . . . 7 (𝜑 → (((1 / 2) · (sin‘𝐴)) + (Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴)) · (sin‘𝐴))) = (((1 / 2) · (sin‘𝐴)) + ((1 / 2) · Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))))
662, 12, 14adddird 10659 . . . . . . 7 (𝜑 → (((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) · (sin‘𝐴)) = (((1 / 2) · (sin‘𝐴)) + (Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴)) · (sin‘𝐴))))
672, 14, 61adddid 10658 . . . . . . 7 (𝜑 → ((1 / 2) · ((sin‘𝐴) + Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))) = (((1 / 2) · (sin‘𝐴)) + ((1 / 2) · Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))))
6865, 66, 673eqtr4d 2846 . . . . . 6 (𝜑 → (((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) · (sin‘𝐴)) = ((1 / 2) · ((sin‘𝐴) + Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))))
6968oveq1d 7154 . . . . 5 (𝜑 → ((((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) · (sin‘𝐴)) / (sin‘𝐴)) = (((1 / 2) · ((sin‘𝐴) + Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))) / (sin‘𝐴)))
7010sincld 15478 . . . . . . . . . . . . 13 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘(𝑛 · 𝐴)) ∈ ℂ)
7142, 70, 56npncand 11014 . . . . . . . . . . . 12 ((𝜑𝑛 ∈ (1...𝑁)) → (((sin‘((𝑛 + 1) · 𝐴)) − (sin‘(𝑛 · 𝐴))) + ((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))) = ((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))
7271eqcomd 2807 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) = (((sin‘((𝑛 + 1) · 𝐴)) − (sin‘(𝑛 · 𝐴))) + ((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))))
7372sumeq2dv 15055 . . . . . . . . . 10 (𝜑 → Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) = Σ𝑛 ∈ (1...𝑁)(((sin‘((𝑛 + 1) · 𝐴)) − (sin‘(𝑛 · 𝐴))) + ((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))))
7442, 70subcld 10990 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘((𝑛 + 1) · 𝐴)) − (sin‘(𝑛 · 𝐴))) ∈ ℂ)
7570, 56subcld 10990 . . . . . . . . . . 11 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) ∈ ℂ)
763, 74, 75fsumadd 15091 . . . . . . . . . 10 (𝜑 → Σ𝑛 ∈ (1...𝑁)(((sin‘((𝑛 + 1) · 𝐴)) − (sin‘(𝑛 · 𝐴))) + ((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))) = (Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘(𝑛 · 𝐴))) + Σ𝑛 ∈ (1...𝑁)((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))))
77 fvoveq1 7162 . . . . . . . . . . . 12 (𝑗 = 𝑛 → (sin‘(𝑗 · 𝐴)) = (sin‘(𝑛 · 𝐴)))
78 fvoveq1 7162 . . . . . . . . . . . 12 (𝑗 = (𝑛 + 1) → (sin‘(𝑗 · 𝐴)) = (sin‘((𝑛 + 1) · 𝐴)))
79 fvoveq1 7162 . . . . . . . . . . . 12 (𝑗 = 1 → (sin‘(𝑗 · 𝐴)) = (sin‘(1 · 𝐴)))
80 fvoveq1 7162 . . . . . . . . . . . 12 (𝑗 = (𝑁 + 1) → (sin‘(𝑗 · 𝐴)) = (sin‘((𝑁 + 1) · 𝐴)))
81 dirkertrigeqlem2.n . . . . . . . . . . . . 13 (𝜑𝑁 ∈ ℕ)
8281nnzd 12078 . . . . . . . . . . . 12 (𝜑𝑁 ∈ ℤ)
83 nnuz 12273 . . . . . . . . . . . . . 14 ℕ = (ℤ‘1)
8481, 83eleqtrdi 2903 . . . . . . . . . . . . 13 (𝜑𝑁 ∈ (ℤ‘1))
85 peano2uz 12293 . . . . . . . . . . . . 13 (𝑁 ∈ (ℤ‘1) → (𝑁 + 1) ∈ (ℤ‘1))
8684, 85syl 17 . . . . . . . . . . . 12 (𝜑 → (𝑁 + 1) ∈ (ℤ‘1))
87 elfzelz 12906 . . . . . . . . . . . . . . . 16 (𝑗 ∈ (1...(𝑁 + 1)) → 𝑗 ∈ ℤ)
8887zcnd 12080 . . . . . . . . . . . . . . 15 (𝑗 ∈ (1...(𝑁 + 1)) → 𝑗 ∈ ℂ)
8988adantl 485 . . . . . . . . . . . . . 14 ((𝜑𝑗 ∈ (1...(𝑁 + 1))) → 𝑗 ∈ ℂ)
908adantr 484 . . . . . . . . . . . . . 14 ((𝜑𝑗 ∈ (1...(𝑁 + 1))) → 𝐴 ∈ ℂ)
9189, 90mulcld 10654 . . . . . . . . . . . . 13 ((𝜑𝑗 ∈ (1...(𝑁 + 1))) → (𝑗 · 𝐴) ∈ ℂ)
9291sincld 15478 . . . . . . . . . . . 12 ((𝜑𝑗 ∈ (1...(𝑁 + 1))) → (sin‘(𝑗 · 𝐴)) ∈ ℂ)
9377, 78, 79, 80, 82, 86, 92telfsum2 15155 . . . . . . . . . . 11 (𝜑 → Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘(𝑛 · 𝐴))) = ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘(1 · 𝐴))))
94 1cnd 10629 . . . . . . . . . . . . . . . . . 18 (𝑛 ∈ (1...𝑁) → 1 ∈ ℂ)
955, 94pncand 10991 . . . . . . . . . . . . . . . . 17 (𝑛 ∈ (1...𝑁) → ((𝑛 + 1) − 1) = 𝑛)
9695eqcomd 2807 . . . . . . . . . . . . . . . 16 (𝑛 ∈ (1...𝑁) → 𝑛 = ((𝑛 + 1) − 1))
9796adantl 485 . . . . . . . . . . . . . . 15 ((𝜑𝑛 ∈ (1...𝑁)) → 𝑛 = ((𝑛 + 1) − 1))
9897fvoveq1d 7161 . . . . . . . . . . . . . 14 ((𝜑𝑛 ∈ (1...𝑁)) → (sin‘(𝑛 · 𝐴)) = (sin‘(((𝑛 + 1) − 1) · 𝐴)))
9998oveq1d 7154 . . . . . . . . . . . . 13 ((𝜑𝑛 ∈ (1...𝑁)) → ((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) = ((sin‘(((𝑛 + 1) − 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))
10099sumeq2dv 15055 . . . . . . . . . . . 12 (𝜑 → Σ𝑛 ∈ (1...𝑁)((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) = Σ𝑛 ∈ (1...𝑁)((sin‘(((𝑛 + 1) − 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))
101 oveq1 7146 . . . . . . . . . . . . . 14 (𝑗 = 𝑛 → (𝑗 − 1) = (𝑛 − 1))
102101fvoveq1d 7161 . . . . . . . . . . . . 13 (𝑗 = 𝑛 → (sin‘((𝑗 − 1) · 𝐴)) = (sin‘((𝑛 − 1) · 𝐴)))
103 oveq1 7146 . . . . . . . . . . . . . 14 (𝑗 = (𝑛 + 1) → (𝑗 − 1) = ((𝑛 + 1) − 1))
104103fvoveq1d 7161 . . . . . . . . . . . . 13 (𝑗 = (𝑛 + 1) → (sin‘((𝑗 − 1) · 𝐴)) = (sin‘(((𝑛 + 1) − 1) · 𝐴)))
105 oveq1 7146 . . . . . . . . . . . . . 14 (𝑗 = 1 → (𝑗 − 1) = (1 − 1))
106105fvoveq1d 7161 . . . . . . . . . . . . 13 (𝑗 = 1 → (sin‘((𝑗 − 1) · 𝐴)) = (sin‘((1 − 1) · 𝐴)))
107 oveq1 7146 . . . . . . . . . . . . . 14 (𝑗 = (𝑁 + 1) → (𝑗 − 1) = ((𝑁 + 1) − 1))
108107fvoveq1d 7161 . . . . . . . . . . . . 13 (𝑗 = (𝑁 + 1) → (sin‘((𝑗 − 1) · 𝐴)) = (sin‘(((𝑁 + 1) − 1) · 𝐴)))
109 1cnd 10629 . . . . . . . . . . . . . . . 16 ((𝜑𝑗 ∈ (1...(𝑁 + 1))) → 1 ∈ ℂ)
11089, 109subcld 10990 . . . . . . . . . . . . . . 15 ((𝜑𝑗 ∈ (1...(𝑁 + 1))) → (𝑗 − 1) ∈ ℂ)
111110, 90mulcld 10654 . . . . . . . . . . . . . 14 ((𝜑𝑗 ∈ (1...(𝑁 + 1))) → ((𝑗 − 1) · 𝐴) ∈ ℂ)
112111sincld 15478 . . . . . . . . . . . . 13 ((𝜑𝑗 ∈ (1...(𝑁 + 1))) → (sin‘((𝑗 − 1) · 𝐴)) ∈ ℂ)
113102, 104, 106, 108, 82, 86, 112telfsum2 15155 . . . . . . . . . . . 12 (𝜑 → Σ𝑛 ∈ (1...𝑁)((sin‘(((𝑛 + 1) − 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) = ((sin‘(((𝑁 + 1) − 1) · 𝐴)) − (sin‘((1 − 1) · 𝐴))))
11481nnred 11644 . . . . . . . . . . . . . . . 16 (𝜑𝑁 ∈ ℝ)
115114recnd 10662 . . . . . . . . . . . . . . 15 (𝜑𝑁 ∈ ℂ)
116115, 1pncand 10991 . . . . . . . . . . . . . 14 (𝜑 → ((𝑁 + 1) − 1) = 𝑁)
117116fvoveq1d 7161 . . . . . . . . . . . . 13 (𝜑 → (sin‘(((𝑁 + 1) − 1) · 𝐴)) = (sin‘(𝑁 · 𝐴)))
1181subidd 10978 . . . . . . . . . . . . . . . . 17 (𝜑 → (1 − 1) = 0)
119118oveq1d 7154 . . . . . . . . . . . . . . . 16 (𝜑 → ((1 − 1) · 𝐴) = (0 · 𝐴))
1208mul02d 10831 . . . . . . . . . . . . . . . 16 (𝜑 → (0 · 𝐴) = 0)
121119, 120eqtrd 2836 . . . . . . . . . . . . . . 15 (𝜑 → ((1 − 1) · 𝐴) = 0)
122121fveq2d 6653 . . . . . . . . . . . . . 14 (𝜑 → (sin‘((1 − 1) · 𝐴)) = (sin‘0))
123 sin0 15497 . . . . . . . . . . . . . . 15 (sin‘0) = 0
124123a1i 11 . . . . . . . . . . . . . 14 (𝜑 → (sin‘0) = 0)
125122, 124eqtrd 2836 . . . . . . . . . . . . 13 (𝜑 → (sin‘((1 − 1) · 𝐴)) = 0)
126117, 125oveq12d 7157 . . . . . . . . . . . 12 (𝜑 → ((sin‘(((𝑁 + 1) − 1) · 𝐴)) − (sin‘((1 − 1) · 𝐴))) = ((sin‘(𝑁 · 𝐴)) − 0))
127100, 113, 1263eqtrd 2840 . . . . . . . . . . 11 (𝜑 → Σ𝑛 ∈ (1...𝑁)((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) = ((sin‘(𝑁 · 𝐴)) − 0))
12893, 127oveq12d 7157 . . . . . . . . . 10 (𝜑 → (Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘(𝑛 · 𝐴))) + Σ𝑛 ∈ (1...𝑁)((sin‘(𝑛 · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))) = (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘(1 · 𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0)))
12973, 76, 1283eqtrd 2840 . . . . . . . . 9 (𝜑 → Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))) = (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘(1 · 𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0)))
130129oveq2d 7155 . . . . . . . 8 (𝜑 → ((sin‘𝐴) + Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))) = ((sin‘𝐴) + (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘(1 · 𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0))))
13127fveq2d 6653 . . . . . . . . . . . 12 (𝜑 → (sin‘(1 · 𝐴)) = (sin‘𝐴))
132131oveq2d 7155 . . . . . . . . . . 11 (𝜑 → ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘(1 · 𝐴))) = ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴)))
133132oveq1d 7154 . . . . . . . . . 10 (𝜑 → (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘(1 · 𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0)) = (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴)) + ((sin‘(𝑁 · 𝐴)) − 0)))
134133oveq2d 7155 . . . . . . . . 9 (𝜑 → ((sin‘𝐴) + (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘(1 · 𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0))) = ((sin‘𝐴) + (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴)) + ((sin‘(𝑁 · 𝐴)) − 0))))
135115, 1addcld 10653 . . . . . . . . . . . . . 14 (𝜑 → (𝑁 + 1) ∈ ℂ)
136135, 8mulcld 10654 . . . . . . . . . . . . 13 (𝜑 → ((𝑁 + 1) · 𝐴) ∈ ℂ)
137136sincld 15478 . . . . . . . . . . . 12 (𝜑 → (sin‘((𝑁 + 1) · 𝐴)) ∈ ℂ)
138137, 14subcld 10990 . . . . . . . . . . 11 (𝜑 → ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴)) ∈ ℂ)
139115, 8mulcld 10654 . . . . . . . . . . . . 13 (𝜑 → (𝑁 · 𝐴) ∈ ℂ)
140139sincld 15478 . . . . . . . . . . . 12 (𝜑 → (sin‘(𝑁 · 𝐴)) ∈ ℂ)
141 0cnd 10627 . . . . . . . . . . . 12 (𝜑 → 0 ∈ ℂ)
142140, 141subcld 10990 . . . . . . . . . . 11 (𝜑 → ((sin‘(𝑁 · 𝐴)) − 0) ∈ ℂ)
14314, 138, 142addassd 10656 . . . . . . . . . 10 (𝜑 → (((sin‘𝐴) + ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0)) = ((sin‘𝐴) + (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴)) + ((sin‘(𝑁 · 𝐴)) − 0))))
144143eqcomd 2807 . . . . . . . . 9 (𝜑 → ((sin‘𝐴) + (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴)) + ((sin‘(𝑁 · 𝐴)) − 0))) = (((sin‘𝐴) + ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0)))
14514, 137pncan3d 10993 . . . . . . . . . . 11 (𝜑 → ((sin‘𝐴) + ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴))) = (sin‘((𝑁 + 1) · 𝐴)))
146140subid1d 10979 . . . . . . . . . . 11 (𝜑 → ((sin‘(𝑁 · 𝐴)) − 0) = (sin‘(𝑁 · 𝐴)))
147145, 146oveq12d 7157 . . . . . . . . . 10 (𝜑 → (((sin‘𝐴) + ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0)) = ((sin‘((𝑁 + 1) · 𝐴)) + (sin‘(𝑁 · 𝐴))))
148137, 140addcomd 10835 . . . . . . . . . 10 (𝜑 → ((sin‘((𝑁 + 1) · 𝐴)) + (sin‘(𝑁 · 𝐴))) = ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴))))
149147, 148eqtrd 2836 . . . . . . . . 9 (𝜑 → (((sin‘𝐴) + ((sin‘((𝑁 + 1) · 𝐴)) − (sin‘𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0)) = ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴))))
150134, 144, 1493eqtrd 2840 . . . . . . . 8 (𝜑 → ((sin‘𝐴) + (((sin‘((𝑁 + 1) · 𝐴)) − (sin‘(1 · 𝐴))) + ((sin‘(𝑁 · 𝐴)) − 0))) = ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴))))
151130, 150eqtrd 2836 . . . . . . 7 (𝜑 → ((sin‘𝐴) + Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴)))) = ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴))))
152151oveq2d 7155 . . . . . 6 (𝜑 → ((1 / 2) · ((sin‘𝐴) + Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))) = ((1 / 2) · ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴)))))
153152oveq1d 7154 . . . . 5 (𝜑 → (((1 / 2) · ((sin‘𝐴) + Σ𝑛 ∈ (1...𝑁)((sin‘((𝑛 + 1) · 𝐴)) − (sin‘((𝑛 − 1) · 𝐴))))) / (sin‘𝐴)) = (((1 / 2) · ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴)))) / (sin‘𝐴)))
15417, 69, 1533eqtrd 2840 . . . 4 (𝜑 → ((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) = (((1 / 2) · ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴)))) / (sin‘𝐴)))
155 halfre 11843 . . . . . . . . . . . 12 (1 / 2) ∈ ℝ
156155a1i 11 . . . . . . . . . . 11 (𝜑 → (1 / 2) ∈ ℝ)
157114, 156readdcld 10663 . . . . . . . . . 10 (𝜑 → (𝑁 + (1 / 2)) ∈ ℝ)
158157, 7remulcld 10664 . . . . . . . . 9 (𝜑 → ((𝑁 + (1 / 2)) · 𝐴) ∈ ℝ)
159158recnd 10662 . . . . . . . 8 (𝜑 → ((𝑁 + (1 / 2)) · 𝐴) ∈ ℂ)
1602, 8mulcld 10654 . . . . . . . 8 (𝜑 → ((1 / 2) · 𝐴) ∈ ℂ)
161 sinmulcos 42494 . . . . . . . 8 ((((𝑁 + (1 / 2)) · 𝐴) ∈ ℂ ∧ ((1 / 2) · 𝐴) ∈ ℂ) → ((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) = (((sin‘(((𝑁 + (1 / 2)) · 𝐴) + ((1 / 2) · 𝐴))) + (sin‘(((𝑁 + (1 / 2)) · 𝐴) − ((1 / 2) · 𝐴)))) / 2))
162159, 160, 161syl2anc 587 . . . . . . 7 (𝜑 → ((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) = (((sin‘(((𝑁 + (1 / 2)) · 𝐴) + ((1 / 2) · 𝐴))) + (sin‘(((𝑁 + (1 / 2)) · 𝐴) − ((1 / 2) · 𝐴)))) / 2))
163115, 2, 8adddird 10659 . . . . . . . . . . . 12 (𝜑 → ((𝑁 + (1 / 2)) · 𝐴) = ((𝑁 · 𝐴) + ((1 / 2) · 𝐴)))
164163oveq1d 7154 . . . . . . . . . . 11 (𝜑 → (((𝑁 + (1 / 2)) · 𝐴) + ((1 / 2) · 𝐴)) = (((𝑁 · 𝐴) + ((1 / 2) · 𝐴)) + ((1 / 2) · 𝐴)))
165139, 160, 160addassd 10656 . . . . . . . . . . 11 (𝜑 → (((𝑁 · 𝐴) + ((1 / 2) · 𝐴)) + ((1 / 2) · 𝐴)) = ((𝑁 · 𝐴) + (((1 / 2) · 𝐴) + ((1 / 2) · 𝐴))))
1662, 2, 8adddird 10659 . . . . . . . . . . . . . 14 (𝜑 → (((1 / 2) + (1 / 2)) · 𝐴) = (((1 / 2) · 𝐴) + ((1 / 2) · 𝐴)))
16712halvesd 11875 . . . . . . . . . . . . . . 15 (𝜑 → ((1 / 2) + (1 / 2)) = 1)
168167oveq1d 7154 . . . . . . . . . . . . . 14 (𝜑 → (((1 / 2) + (1 / 2)) · 𝐴) = (1 · 𝐴))
169166, 168eqtr3d 2838 . . . . . . . . . . . . 13 (𝜑 → (((1 / 2) · 𝐴) + ((1 / 2) · 𝐴)) = (1 · 𝐴))
170169oveq2d 7155 . . . . . . . . . . . 12 (𝜑 → ((𝑁 · 𝐴) + (((1 / 2) · 𝐴) + ((1 / 2) · 𝐴))) = ((𝑁 · 𝐴) + (1 · 𝐴)))
171115, 1, 8adddird 10659 . . . . . . . . . . . 12 (𝜑 → ((𝑁 + 1) · 𝐴) = ((𝑁 · 𝐴) + (1 · 𝐴)))
172170, 171eqtr4d 2839 . . . . . . . . . . 11 (𝜑 → ((𝑁 · 𝐴) + (((1 / 2) · 𝐴) + ((1 / 2) · 𝐴))) = ((𝑁 + 1) · 𝐴))
173164, 165, 1723eqtrrd 2841 . . . . . . . . . 10 (𝜑 → ((𝑁 + 1) · 𝐴) = (((𝑁 + (1 / 2)) · 𝐴) + ((1 / 2) · 𝐴)))
174173fveq2d 6653 . . . . . . . . 9 (𝜑 → (sin‘((𝑁 + 1) · 𝐴)) = (sin‘(((𝑁 + (1 / 2)) · 𝐴) + ((1 / 2) · 𝐴))))
175163oveq1d 7154 . . . . . . . . . . 11 (𝜑 → (((𝑁 + (1 / 2)) · 𝐴) − ((1 / 2) · 𝐴)) = (((𝑁 · 𝐴) + ((1 / 2) · 𝐴)) − ((1 / 2) · 𝐴)))
176139, 160pncand 10991 . . . . . . . . . . 11 (𝜑 → (((𝑁 · 𝐴) + ((1 / 2) · 𝐴)) − ((1 / 2) · 𝐴)) = (𝑁 · 𝐴))
177175, 176eqtr2d 2837 . . . . . . . . . 10 (𝜑 → (𝑁 · 𝐴) = (((𝑁 + (1 / 2)) · 𝐴) − ((1 / 2) · 𝐴)))
178177fveq2d 6653 . . . . . . . . 9 (𝜑 → (sin‘(𝑁 · 𝐴)) = (sin‘(((𝑁 + (1 / 2)) · 𝐴) − ((1 / 2) · 𝐴))))
179174, 178oveq12d 7157 . . . . . . . 8 (𝜑 → ((sin‘((𝑁 + 1) · 𝐴)) + (sin‘(𝑁 · 𝐴))) = ((sin‘(((𝑁 + (1 / 2)) · 𝐴) + ((1 / 2) · 𝐴))) + (sin‘(((𝑁 + (1 / 2)) · 𝐴) − ((1 / 2) · 𝐴)))))
180179oveq1d 7154 . . . . . . 7 (𝜑 → (((sin‘((𝑁 + 1) · 𝐴)) + (sin‘(𝑁 · 𝐴))) / 2) = (((sin‘(((𝑁 + (1 / 2)) · 𝐴) + ((1 / 2) · 𝐴))) + (sin‘(((𝑁 + (1 / 2)) · 𝐴) − ((1 / 2) · 𝐴)))) / 2))
181162, 180eqtr4d 2839 . . . . . 6 (𝜑 → ((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) = (((sin‘((𝑁 + 1) · 𝐴)) + (sin‘(𝑁 · 𝐴))) / 2))
182148oveq1d 7154 . . . . . 6 (𝜑 → (((sin‘((𝑁 + 1) · 𝐴)) + (sin‘(𝑁 · 𝐴))) / 2) = (((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴))) / 2))
183140, 137addcld 10653 . . . . . . 7 (𝜑 → ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴))) ∈ ℂ)
184183, 52, 59divrec2d 11413 . . . . . 6 (𝜑 → (((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴))) / 2) = ((1 / 2) · ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴)))))
185181, 182, 1843eqtrrd 2841 . . . . 5 (𝜑 → ((1 / 2) · ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴)))) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))))
186185oveq1d 7154 . . . 4 (𝜑 → (((1 / 2) · ((sin‘(𝑁 · 𝐴)) + (sin‘((𝑁 + 1) · 𝐴)))) / (sin‘𝐴)) = (((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) / (sin‘𝐴)))
1878, 52, 59divcan2d 11411 . . . . . . . . 9 (𝜑 → (2 · (𝐴 / 2)) = 𝐴)
188187eqcomd 2807 . . . . . . . 8 (𝜑𝐴 = (2 · (𝐴 / 2)))
189188fveq2d 6653 . . . . . . 7 (𝜑 → (sin‘𝐴) = (sin‘(2 · (𝐴 / 2))))
1908halfcld 11874 . . . . . . . 8 (𝜑 → (𝐴 / 2) ∈ ℂ)
191 sin2t 15525 . . . . . . . 8 ((𝐴 / 2) ∈ ℂ → (sin‘(2 · (𝐴 / 2))) = (2 · ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2)))))
192190, 191syl 17 . . . . . . 7 (𝜑 → (sin‘(2 · (𝐴 / 2))) = (2 · ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2)))))
193189, 192eqtrd 2836 . . . . . 6 (𝜑 → (sin‘𝐴) = (2 · ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2)))))
194193oveq2d 7155 . . . . 5 (𝜑 → (((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) / (sin‘𝐴)) = (((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) / (2 · ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2))))))
195190sincld 15478 . . . . . . . 8 (𝜑 → (sin‘(𝐴 / 2)) ∈ ℂ)
196190coscld 15479 . . . . . . . 8 (𝜑 → (cos‘(𝐴 / 2)) ∈ ℂ)
19752, 195, 196mulassd 10657 . . . . . . 7 (𝜑 → ((2 · (sin‘(𝐴 / 2))) · (cos‘(𝐴 / 2))) = (2 · ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2)))))
1988, 52, 59divrec2d 11413 . . . . . . . . 9 (𝜑 → (𝐴 / 2) = ((1 / 2) · 𝐴))
199198fveq2d 6653 . . . . . . . 8 (𝜑 → (cos‘(𝐴 / 2)) = (cos‘((1 / 2) · 𝐴)))
200199oveq2d 7155 . . . . . . 7 (𝜑 → ((2 · (sin‘(𝐴 / 2))) · (cos‘(𝐴 / 2))) = ((2 · (sin‘(𝐴 / 2))) · (cos‘((1 / 2) · 𝐴))))
201197, 200eqtr3d 2838 . . . . . 6 (𝜑 → (2 · ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2)))) = ((2 · (sin‘(𝐴 / 2))) · (cos‘((1 / 2) · 𝐴))))
202201oveq2d 7155 . . . . 5 (𝜑 → (((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) / (2 · ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2))))) = (((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) / ((2 · (sin‘(𝐴 / 2))) · (cos‘((1 / 2) · 𝐴)))))
203159sincld 15478 . . . . . 6 (𝜑 → (sin‘((𝑁 + (1 / 2)) · 𝐴)) ∈ ℂ)
20452, 195mulcld 10654 . . . . . 6 (𝜑 → (2 · (sin‘(𝐴 / 2))) ∈ ℂ)
205160coscld 15479 . . . . . 6 (𝜑 → (cos‘((1 / 2) · 𝐴)) ∈ ℂ)
206195, 196mulcld 10654 . . . . . . . . 9 (𝜑 → ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2))) ∈ ℂ)
207193, 15eqnetrrd 3058 . . . . . . . . 9 (𝜑 → (2 · ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2)))) ≠ 0)
20852, 206, 207mulne0bbd 11289 . . . . . . . 8 (𝜑 → ((sin‘(𝐴 / 2)) · (cos‘(𝐴 / 2))) ≠ 0)
209195, 196, 208mulne0bad 11288 . . . . . . 7 (𝜑 → (sin‘(𝐴 / 2)) ≠ 0)
21052, 195, 59, 209mulne0d 11285 . . . . . 6 (𝜑 → (2 · (sin‘(𝐴 / 2))) ≠ 0)
211195, 196, 208mulne0bbd 11289 . . . . . . 7 (𝜑 → (cos‘(𝐴 / 2)) ≠ 0)
212199, 211eqnetrrd 3058 . . . . . 6 (𝜑 → (cos‘((1 / 2) · 𝐴)) ≠ 0)
213203, 204, 205, 210, 212divcan5rd 11436 . . . . 5 (𝜑 → (((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) / ((2 · (sin‘(𝐴 / 2))) · (cos‘((1 / 2) · 𝐴)))) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / (2 · (sin‘(𝐴 / 2)))))
214194, 202, 2133eqtrd 2840 . . . 4 (𝜑 → (((sin‘((𝑁 + (1 / 2)) · 𝐴)) · (cos‘((1 / 2) · 𝐴))) / (sin‘𝐴)) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / (2 · (sin‘(𝐴 / 2)))))
215154, 186, 2143eqtrd 2840 . . 3 (𝜑 → ((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / (2 · (sin‘(𝐴 / 2)))))
216215oveq1d 7154 . 2 (𝜑 → (((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) / π) = (((sin‘((𝑁 + (1 / 2)) · 𝐴)) / (2 · (sin‘(𝐴 / 2)))) / π))
217 picn 25055 . . . 4 π ∈ ℂ
218217a1i 11 . . 3 (𝜑 → π ∈ ℂ)
219 pire 25054 . . . . 5 π ∈ ℝ
220 pipos 25056 . . . . 5 0 < π
221219, 220gt0ne0ii 11169 . . . 4 π ≠ 0
222221a1i 11 . . 3 (𝜑 → π ≠ 0)
223203, 204, 218, 210, 222divdiv32d 11434 . 2 (𝜑 → (((sin‘((𝑁 + (1 / 2)) · 𝐴)) / (2 · (sin‘(𝐴 / 2)))) / π) = (((sin‘((𝑁 + (1 / 2)) · 𝐴)) / π) / (2 · (sin‘(𝐴 / 2)))))
224203, 218, 204, 222, 210divdiv1d 11440 . . 3 (𝜑 → (((sin‘((𝑁 + (1 / 2)) · 𝐴)) / π) / (2 · (sin‘(𝐴 / 2)))) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / (π · (2 · (sin‘(𝐴 / 2))))))
225218, 52, 195mulassd 10657 . . . . 5 (𝜑 → ((π · 2) · (sin‘(𝐴 / 2))) = (π · (2 · (sin‘(𝐴 / 2)))))
226218, 52mulcomd 10655 . . . . . 6 (𝜑 → (π · 2) = (2 · π))
227226oveq1d 7154 . . . . 5 (𝜑 → ((π · 2) · (sin‘(𝐴 / 2))) = ((2 · π) · (sin‘(𝐴 / 2))))
228225, 227eqtr3d 2838 . . . 4 (𝜑 → (π · (2 · (sin‘(𝐴 / 2)))) = ((2 · π) · (sin‘(𝐴 / 2))))
229228oveq2d 7155 . . 3 (𝜑 → ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / (π · (2 · (sin‘(𝐴 / 2))))) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / ((2 · π) · (sin‘(𝐴 / 2)))))
230224, 229eqtrd 2836 . 2 (𝜑 → (((sin‘((𝑁 + (1 / 2)) · 𝐴)) / π) / (2 · (sin‘(𝐴 / 2)))) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / ((2 · π) · (sin‘(𝐴 / 2)))))
231216, 223, 2303eqtrd 2840 1 (𝜑 → (((1 / 2) + Σ𝑛 ∈ (1...𝑁)(cos‘(𝑛 · 𝐴))) / π) = ((sin‘((𝑁 + (1 / 2)) · 𝐴)) / ((2 · π) · (sin‘(𝐴 / 2)))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 = wceq 1538 ∈ wcel 2112 ≠ wne 2990 ‘cfv 6328 (class class class)co 7139 ℂcc 10528 ℝcr 10529 0cc0 10530 1c1 10531 + caddc 10533 · cmul 10535 − cmin 10863 -cneg 10864 / cdiv 11290 ℕcn 11629 2c2 11684 ℤ≥cuz 12235 ...cfz 12889 Σcsu 15037 sincsin 15412 cosccos 15413 πcpi 15415 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2773 ax-rep 5157 ax-sep 5170 ax-nul 5177 ax-pow 5234 ax-pr 5298 ax-un 7445 ax-inf2 9092 ax-cnex 10586 ax-resscn 10587 ax-1cn 10588 ax-icn 10589 ax-addcl 10590 ax-addrcl 10591 ax-mulcl 10592 ax-mulrcl 10593 ax-mulcom 10594 ax-addass 10595 ax-mulass 10596 ax-distr 10597 ax-i2m1 10598 ax-1ne0 10599 ax-1rid 10600 ax-rnegex 10601 ax-rrecex 10602 ax-cnre 10603 ax-pre-lttri 10604 ax-pre-lttrn 10605 ax-pre-ltadd 10606 ax-pre-mulgt0 10607 ax-pre-sup 10608 ax-addf 10609 ax-mulf 10610 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2601 df-eu 2632 df-clab 2780 df-cleq 2794 df-clel 2873 df-nfc 2941 df-ne 2991 df-nel 3095 df-ral 3114 df-rex 3115 df-reu 3116 df-rmo 3117 df-rab 3118 df-v 3446 df-sbc 3724 df-csb 3832 df-dif 3887 df-un 3889 df-in 3891 df-ss 3901 df-pss 3903 df-nul 4247 df-if 4429 df-pw 4502 df-sn 4529 df-pr 4531 df-tp 4533 df-op 4535 df-uni 4804 df-int 4842 df-iun 4886 df-iin 4887 df-br 5034 df-opab 5096 df-mpt 5114 df-tr 5140 df-id 5428 df-eprel 5433 df-po 5442 df-so 5443 df-fr 5482 df-se 5483 df-we 5484 df-xp 5529 df-rel 5530 df-cnv 5531 df-co 5532 df-dm 5533 df-rn 5534 df-res 5535 df-ima 5536 df-pred 6120 df-ord 6166 df-on 6167 df-lim 6168 df-suc 6169 df-iota 6287 df-fun 6330 df-fn 6331 df-f 6332 df-f1 6333 df-fo 6334 df-f1o 6335 df-fv 6336 df-isom 6337 df-riota 7097 df-ov 7142 df-oprab 7143 df-mpo 7144 df-of 7393 df-om 7565 df-1st 7675 df-2nd 7676 df-supp 7818 df-wrecs 7934 df-recs 7995 df-rdg 8033 df-1o 8089 df-2o 8090 df-oadd 8093 df-er 8276 df-map 8395 df-pm 8396 df-ixp 8449 df-en 8497 df-dom 8498 df-sdom 8499 df-fin 8500 df-fsupp 8822 df-fi 8863 df-sup 8894 df-inf 8895 df-oi 8962 df-card 9356 df-pnf 10670 df-mnf 10671 df-xr 10672 df-ltxr 10673 df-le 10674 df-sub 10865 df-neg 10866 df-div 11291 df-nn 11630 df-2 11692 df-3 11693 df-4 11694 df-5 11695 df-6 11696 df-7 11697 df-8 11698 df-9 11699 df-n0 11890 df-z 11974 df-dec 12091 df-uz 12236 df-q 12341 df-rp 12382 df-xneg 12499 df-xadd 12500 df-xmul 12501 df-ioo 12734 df-ioc 12735 df-ico 12736 df-icc 12737 df-fz 12890 df-fzo 13033 df-fl 13161 df-seq 13369 df-exp 13430 df-fac 13634 df-bc 13663 df-hash 13691 df-shft 14421 df-cj 14453 df-re 14454 df-im 14455 df-sqrt 14589 df-abs 14590 df-limsup 14823 df-clim 14840 df-rlim 14841 df-sum 15038 df-ef 15416 df-sin 15418 df-cos 15419 df-pi 15421 df-struct 16480 df-ndx 16481 df-slot 16482 df-base 16484 df-sets 16485 df-ress 16486 df-plusg 16573 df-mulr 16574 df-starv 16575 df-sca 16576 df-vsca 16577 df-ip 16578 df-tset 16579 df-ple 16580 df-ds 16582 df-unif 16583 df-hom 16584 df-cco 16585 df-rest 16691 df-topn 16692 df-0g 16710 df-gsum 16711 df-topgen 16712 df-pt 16713 df-prds 16716 df-xrs 16770 df-qtop 16775 df-imas 16776 df-xps 16778 df-mre 16852 df-mrc 16853 df-acs 16855 df-mgm 17847 df-sgrp 17896 df-mnd 17907 df-submnd 17952 df-mulg 18220 df-cntz 18442 df-cmn 18903 df-psmet 20086 df-xmet 20087 df-met 20088 df-bl 20089 df-mopn 20090 df-fbas 20091 df-fg 20092 df-cnfld 20095 df-top 21502 df-topon 21519 df-topsp 21541 df-bases 21554 df-cld 21627 df-ntr 21628 df-cls 21629 df-nei 21706 df-lp 21744 df-perf 21745 df-cn 21835 df-cnp 21836 df-haus 21923 df-tx 22170 df-hmeo 22363 df-fil 22454 df-fm 22546 df-flim 22547 df-flf 22548 df-xms 22930 df-ms 22931 df-tms 22932 df-cncf 23486 df-limc 24472 df-dv 24473 This theorem is referenced by: dirkertrigeq 42730
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General Physics Discussion, Independent Research
Moderator: BenTheMan
### spinning vrs reciprocating mass...
The object is a sphere with a spindle through it's center in both cases.
----O----
The object is at rest and accelerates and then stops
and reverses direction.
Both oscillate. In one case back and forth and in the
second case around the spindle.
There is a small dot on the side of the sphere. A line drawn
through the dot is on a line that is perpendicular to the spindle. O.
<- ----O.---- > back and forth
@ ----O.---- @ rotate
In both cases the dot moves the same distance and during the same period of time.
Which requires more energy rotation or straight motion and is there a very simple equation that expresses this relationship?
(An equation wherein I could insert a value
representing the variables of mass and it would tell me how much force
[relative not absolute] is needed to transverse a standard distance such as 3 centimeters.)
Would this be a linear function? Twice the distance requires two times the energy?
Thanks.
JohnnyG
Posts: 1
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Question
# Calculate the work done during isothermal reversible expansion of one mole ideal gas from 10 atm to 1 atm at 300 K.
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Solution
## work done =−2.303nRTlog(P1P2)=−2.303×8.314×300log(110)=−2.303×8.314×300=−5.74kJ∴ work done will be 5.74 kJ.
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# Testing Method
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## Testing Method
(OP)
Hi All,
I'm working DOE project on leak test and tranfer the product to factory. We facing high leak failure due to design issue. Meanwhile waiting for new design change , the management want to increase the spec so that we have alot of passes. My question is as follows:-
Is there any statistical method that I can use instead of increase the spec.
I want to test the product using original spec vs increase spec. Is that method can tell us something?
Balan.G
### RE: Testing Method
Before condemning management for "changing the spec", do you understand the details of why the spec was set where it was? Obviously, changing the spec just to arbitrarily inflate the passing results will undermind the entire purpose of measuring the value. However, if the current spec limit has no emperical basis, then perhaps they are simply adjusting the limits to remove unnecassry part rejection (alpha rejection). Here is an example:
I am a pencil manufacturer, and I need to make pencils that are 8 inches in length. One of my final inpsection items is to measure the overall length of the product. However, I defined the length to be 8.000 +/- 0.001" . This implies that I need a piece of measurming equipment that has at least 0.0001" precision. That means I need to use a Coordinate Measurement Machine (CMM) to measure the LENGTH OF A PENCIL. Doesn't this seem like overkill? Perhaps, it makes more sense to define the length to be 8.0 +/- 0.1". This will allow me to use a digital caliper to measure the value. Furthermore, there is no real benefit to have a pencil that is guarenteed to be 8.000" when a pencil that is 8.1" will function just as well.
I don't know the specifics of your scenario, however, is it possible that they are simply redefining the length of the pencil to something more realistic? Do the current test limits guarentees that functionality of the end product is met?
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# Online Classes for GMAT
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30 Jan 2007, 14:35
Hi!
I just wanted to know if anyone has an idea of what online courses someone could take to get started with GMAT preparation. I am not opposed to spending some cash for this exercise. BTW i did take the GMAT last year and scored a 620 so there is a lot of inertia on my part to get going on the prep. I am hoping a course can help me get some focus
thanks
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30 Jan 2007, 14:53
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31 Jan 2007, 08:29
I wonder how it is rated against GMAXOnline?
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31 Jan 2007, 13:17
I have heard good things about Manhattan.
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31 Jan 2007, 19:26
I am doing Princeton Review's online course and would definitely recommend it over Kaplan. The online tools are great.
31 Jan 2007, 19:26
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# Online Classes for GMAT
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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®. | 771 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-22 | latest | en | 0.892359 |
https://math.stackexchange.com/questions/309954/integral-int-infty-infty-frac-coss-arctanax1x21a2x2s/1014962 | 1,575,666,855,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490972.13/warc/CC-MAIN-20191206200121-20191206224121-00395.warc.gz | 460,969,112 | 34,449 | # Integral $\int_{-\infty}^{\infty}\frac{\cos(s \arctan(ax))}{(1+x^2)(1+a^2x^2)^{s/2}}dx$
Prove that:
$$\int_{-\infty}^{\infty}{\cos\left(\vphantom{\Large A} s\ \arctan\left(\vphantom{\large A}ax\right)\right)\over \left(1 + x^{2}\right)\left(1 + a^{2}x^{2}\right)^{s/2}}\,{\rm d}x ={\pi \over \left(1 + a\right)^{s}}$$
where $a,s \in \mathbb{R}^{+}$.
This looks difficult. What would be a good start? Can we use the Residue Theorem?
• I would start with a substitution, $u = \arctan ax$ and then hit the resulting integral with residue calculus. – mrf Feb 21 '13 at 9:57
• @mrf: That's a good suggestion. I will try it. – Shobhit Bhatnagar Feb 21 '13 at 10:44
• First of all, I'd have a very serious and most probably unpleasant talk with the sadist that came up with this integral. Second, perhaps the identities $$\cos\arctan x=\frac{1}{\sqrt{1+x^2}}\,\,,\,\,\sin\arctan x=\frac{x}{\sqrt{1+x^2}}$$can help a little... – DonAntonio Feb 21 '13 at 13:53
• The result should be $\displaystyle{\large{\pi \over \left(\,1 + \color{#c00000}{\left\vert\,a\,\right\vert}\,\,\right)^{\,\,\,s}}}$. – Felix Marin Jun 28 '14 at 1:47
Assume that $a \ge 0$ and $s > -1$.
Using the principal branch of the logarithm, \begin{align} \text{Re} \ (1-iax)^{-s} &= \text{Re} \ (1+a^{2}x^{2})^{-s/2} \ e^{-is\arctan (-ax)} \\ &= \text{Re} \ (1+a^{2}x^{2})^{-s/2} \ e^{i s \arctan (ax) } \\ &= \frac{\cos (s \arctan ax)}{(1+a^{2}x^{2})^{s/2}} . \end{align}
Therefore,
$$\int_{-\infty}^{\infty} \frac{\cos (s \arctan ax)}{(1+x^{2})(1+a^{2}x^{2})^{s/2}} \ dx = \text{Re} \int_{-\infty}^{\infty} \frac{1}{(1+x^{2})(1-iax)^{s}} \ dx .$$
Now since $a \ge 0$, $\displaystyle f(z) = \frac{1}{(1+z^{2})(1-iaz)^{s}}$ is meromorphic in the upper half-plane.
And since $s>-1$, $\displaystyle \int f(z) \ dz$ will vanish along the upper half of the circle $|z|=R$ as $R \to \infty$.
So under these conditions,
\begin{align} \int_{-\infty}^{\infty} \frac{\cos (s \arctan ax)}{(1+x^{2})(1+a^{2}x^{2})^{s/2}} \ dx &= \text{Re} \ 2 \pi i \ \text{Res}[f(z),i] \\ &= \text{Re} \ 2 \pi i \lim_{z \to i} \frac{1}{(z+i)(1-iaz)^{s}} \\ &= \frac{\pi}{(1+a)^{s}} . \end{align}
I can't resist the temptation to answer this problem. The integral really makes me excited! ٩(˘◡˘)۶
Here is a non-contour/ residue approach. We have a classic result $$\int_{0}^{\infty} y^{s-1}\cos (a y)\,e^{-by}\,dy = \Gamma (s)\frac{\cos \left(\! s \arctan \left(\frac{a}{b}\right)\right)}{\left(a^2+b^2\right)^{s/2}},\quad\mbox{for}\quad \Re(s),\, a,\,b>0.\tag{1}$$ where it can easily be proved by using $$\Re\int_{0}^{\infty} y^{s-1}e^{-(b-ia)y}\,dyt = \Re\left[ \frac{\Gamma (s)}{(b-ia)^s}\right]\tag{2}$$ Putting $a=ax$ and $b=1$ to $(1)$, then dividing by $\dfrac{1}{1+x^2}$ and taking the integral both of sides, we have $$\int_{-\infty}^{\infty}\int_{0}^{\infty} \frac{y^{s-1}\cos (axy)\,e^{-y}}{1+x^2}\,dy\,dx = \Gamma (s)\int_{-\infty}^{\infty}\frac{\cos \left(\! s \arctan \left(ax\right)\right)}{(1+x^2)\left(1+a^2x^2\right)^{s/2}}\,dx\tag{3}$$ Changing the order of integration on the LHS which can be justified by using Fubini's theorem we have $$\int_{0}^{\infty}y^{s-1}\,e^{-y}\int_{-\infty}^{\infty} \frac{\cos (ay\,x)}{1+x^2}\,dx\,dy = \Gamma (s)\int_{-\infty}^{\infty}\frac{\cos \left(\! s \arctan \left(ax\right)\right)}{(1+x^2)\left(1+a^2x^2\right)^{s/2}}\,dx\tag{4}$$ Using another classic result $$\int_{-\infty}^{\infty} \frac{\cos nx}{1+x^2}\,dx=\pi e^{-n}\tag{5}$$ then we obtain \begin{align} \int_{0}^{\infty}y^{s-1}\,e^{-y}\int_{-\infty}^{\infty} \frac{\cos (ay\,x)}{1+x^2}\,dx\,dy &=\pi\int_{0}^{\infty}y^{s-1}\,e^{-(1+a)y}\,dy\\ &=\frac{\pi\,\Gamma (s)}{(1+a)^s}\tag{6} \end{align} Thus, by putting $(6)$ to $(4)$, we obtain $$\int_{-\infty}^{\infty}\frac{\cos \left(\! s \arctan \left(ax\right)\right)}{(1+x^2)\left(1+a^2x^2\right)^{s/2}}\,dx=\frac{\pi}{(1+a)^s}$$ which is the announced result.$\qquad\square$
This may be done with a contour integration. Consider the integral
$$\oint_C dz \frac{e^{i s \arctan{a z}}}{(1+z^2)(1+a^2 z^2)^{s/2}}$$
where $a>0$ and $C$ is a contour that is a semicircle in the upper half plane, except that it detours up just to the left of the imaginary axis to $z=i/a$, around that point, and the back down just to the right of the imaginary axis to the real axis, where it continues along the semicircle. This detour is needed to avoid the branch point at $z=i/a$.
In this way, note that there is a pole within $C$ at $z=i$ only when $a>1$. When $a<1$, then there is no pole within $C$ and the integral is zero. So consider the case when $a>1$. This contour integral is zero except along the real axis. To see this, note that along the outer semicircle of radius $R$, the integral is bounded by
$$\frac{R}{R^{2+s}} \left | \int_0^{\pi} d\theta e^{-s \mathrm{arctanh}(a R \sin{\theta})} \right | \sim \frac{1}{R^{1+s}}$$
which vanishes when $s>-1$. I will assume this for the result. The integrals along the vertical pieces cancel. The integral about the branch point $z=i/a$ may be parametrized as $z=i/a + t e^{i \phi}$, where $t \rightarrow 0$. Then we get a term for small $t$ as follows:
$$i (2 a)^{s/2} t^{1-(s/2)} \frac{e^{-s \mathrm{arctanh}(1)}}{1-1/a^2} \int_0^{2 \pi} d \phi e^{i (1-s/2) \phi} = 0$$
when $s \ne 2$. I will consider the case of $s=2$ later.
Then the integral along the real axis is equal $i 2 \pi$ times to the residue at the pole $z=i$. This is
$$i 2 \pi \frac{e^{-s \mathrm{arctanh}(a)}}{2 i (1-a^2)^{s/2}}$$
Use the fact that
$$\mathrm{arctanh}(a) = \frac{1}{2} \log{\left ( \frac{1+a}{1-a} \right )}$$
and we finally get
$$\int_{-\infty}^{\infty} dx \frac{e^{i s \arctan{a x}}}{(1+x^2)(1+a^2 x^2)^{s/2}} = \frac{\pi}{(1+a)^s}$$
Because the result of the complex conjugate requires using the contour in the lower half-plane, we get an identical result for this. We may then say that
$$\int_{-\infty}^{\infty} dx \frac{\cos{( s \arctan{a x})}}{(1+x^2)(1+a^2 x^2)^{s/2}} = \frac{\pi}{(1+a)^s}$$
when $s>-1$ and $a>1$.
When $s=2$, we may make a substitution of $u=\arctan{a x}$ and the integral becomes
$$2 a \int_0^{\pi/2} du \frac{\cos{2 u}}{a^2 + \tan^2{u}} = \frac{\pi}{(1+a)^2}$$
so the $s=2$ case is covered.
It does seem to me that $a<1$ should also be covered by this result, but I would need to rethink the contour. | 2,536 | 6,286 | {"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": 3, "equation": 6, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-51 | latest | en | 0.63607 |
https://www.scribd.com/document/176708931/EMC620S-Quiz2-2013-doc | 1,566,435,019,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316555.4/warc/CC-MAIN-20190822000659-20190822022659-00189.warc.gz | 967,849,762 | 58,528 | You are on page 1of 3
# POLYTECHNIC OF NAMIBIA
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## Department of Mechanical Engineering
Engineering Mechanics Rigid-body Dynamics (EMC620S)
Quiz #2 (implemented in Moodle) Total marks: 72
Issue date: 11-09-2013 @ 14:00 Due date: 13-09-2013 @ 17:00
1. This quiz tests your understanding of the concepts of mass moments of inertia. Enter your calculated values! All your units should be in Kg.m2 2. You must have AT LEAST 3 significant figures in your keyed-in entries. There's a 1% error margin on all answers to cater for round-off errors. 3. The quiz opens at 14:00 on Wednesday 11th September 2013 and closes on Friday 13th September 2013 at 17:00. 4. Let's crack it! :-)
Question 1
An aluminium machine element is shown below. It is required to calculate its dynamic properties based on the following: Part 1 is the rectangular parallelepiped formed by dimensions 2*{A} m x {B} m x {C} m; part 2 is a quarter-cylindrical piece of radius {A}m and height {C} m; part 3 is the semicircular hole with radius 0.5*{A} m and height{C} m. The origin of the xy-z coordinate is shown (density of aluminium is 2700 kg/m3).
Determine the mass moment of inertia of Part 1 with respect to the x-axis (Ixx) [4].
Question 2
Determine the mass moment of inertia of Part 1 with respect to the y-axis (Iyy) [4].
Question 3
Determine the mass moment of inertia of Part 1 with respect to the z-axis (Izz) [4].
Question 4
Determine the mass product of inertia of Part 1 with respect to the x-y plane (Ixy) [1].
Question 5
Determine the mass product of inertia of Part 1 with respect to the x-z plane (Ixz) [1].
Question 6
Determine the mass product of inertia of Part 1 with respect to the y-z plane (Iyz) [2].
Question 7
Determine the mass moment of inertia of Part 2 with respect to the x-axis (Ixx) [4].
Question 8
Determine the mass moment of inertia of Part 2 with respect to the y-axis (Iyy) [3].
Question 9
Determine the mass moment of inertia of Part 2 with respect to the z-axis (Izz) [4].
Question 10
Determine the mass product of inertia of Part 2 with respect to the x-y plane (Ixy) [2].
Question 11
Determine the mass product of inertia of Part 2 with respect to the x-z plane (Ixz) [2].
Question 12
Determine the mass product of inertia of Part 2 with respect to the y-z plane (Iyz) [2].
Question 13
Determine the mass moment of inertia of Part 3 with respect to the x-axis (Ixx) [4].
Question 14
Determine the mass moment of inertia of Part 3 with respect to the y-axis (Iyy) [4].
Question 15
Determine the mass moment of inertia of Part 3 with respect to the z-axis (Izz) [4].
Question 16
Determine the mass product of inertia of Part 3 with respect to the x-y plane (Ixy) [2].
Question 17
Determine the mass product of inertia of Part 3 with respect to the x-z plane (Ixz) [2].
Question 18
Determine the mass product of inertia of Part 3 with respect to the y-z plane (Iyz) [2].
Question 19
Determine the total mass moment of inertia of with respect to the x-axis (Ixx) [1].
Question 20
Determine the total mass moment of inertia of with respect to the y-axis (Iyy) [1]. -2-
Question 21
Determine the total mass moment of inertia of with respect to the z-axis (Izz) [1].
Question 22
Determine the total product of inertia of with respect to the x-y plane (Ixy) [1].
Question 23
Determine the total product of inertia of with respect to the x-z plane (Ixz) [1].
Question 24
Determine the total product of inertia of with respect to the y-z plane (Iyz) [1].
Question 25
Determine the moment of inertia of with respect to an arbitrary axis if the axis makes angles of 120, 60 and 45 with each of the x, y and z axes respectively (I) [3].
Question 26
Determine the first principal mass moment of inertia (I1) [4].
Question 27
Determine the second principal mass moment of inertia (I2) [4].
Question 28
Determine the third principal mass moment of inertia (I3) [4].
-3- | 1,198 | 4,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-35 | latest | en | 0.828126 |
http://www.math.ucsd.edu/~helton/MorasAlgorithm.html | 1,516,124,631,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886476.31/warc/CC-MAIN-20180116164812-20180116184812-00757.warc.gz | 516,040,184 | 2,322 | ## Mora's Algorithm
### A Noncommutative Gröbner Basis Package for Mathematica©
This is another package under development which makes use of the NCAlgebra package. What follows is a brief introduction to that package, but more information is available. The entire user manual is available in either dvi or postscript format. The entire source code for this package is also available via anonymous ftp.
This package joins with the package NCAlgebra.m to add powerful automatic methods for handling systems of equations in noncommuting variables. All commutative algebra packages contain commands based on Gröbner Bases and uses of them. For example in Mathematica the Solve command put systems of equations in a "cannonical" form which for simple systems readily yields a solution. Likewise the Mathematica Eliminate command tries to convert a system of equations in unknowns x(1), x(2), ..., x(n) to a "triangular" form in unknowns, that is, a new system of equations like
Here the polynomials {q(j) : j=1,..., k} generate the same ideal that the polynomials {p(j) : 1<= j <= k} do. Therefore, the set of solutions to the system the polynomial equations p(j) = 0 equals the set of solutions to the system of polynomial equations q(j) = 0. This canonical form greatly simplifies the task of solving systems of polynomial equations by allowing one to backsolve for x(j) in terms of {x(1),...,x(j-1)}.A brief mathematical discussion of Gröbner bases appears in the appendix in the NCGB document. The user who is not acquainted at all with Gröbner Basis should still be able to read and use much of the material which is contained within this document. In [FMora], c.f. [TMora], F. Mora described a version of the Gröbner basis algorithm which applied to noncommutative free algebras. We refer to this algorithm as Mora's algorithm. It too puts systems of equations into a "cannonical form" which we believe has considerable possibilities in the noncommutative case. This package implements the Mora algorithm and these applications as well. We think of this package as being useful for at least 4 things.
1. Simplifying complicated expressions
2. Making the process of discovering algebraic theorems and appealing formulas semi-automatic
3. Producing GB's in noncommuting situations
4. Finding small bases for ideals in a noncommuting algebra
We begin by giving simple examples which are sufficient to show how this package is used. They begin with applications and move to the GB algorithm itself. We then describe the Mathematica commands in subsequent chapters. We refer the user to [HW], [HWS1] and [HWS2] for which describe in detail our approach to simplification of noncommutative expressions. [HS] describes our approach to discovering theorems and formulas as does this document to some extent. Another document describes a version of Mora's algorithm which allows the generation of and the manipulation of infinite collections of polynomials (which are often unavoidable in the noncommutative setting). | 667 | 3,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-05 | latest | en | 0.914662 |
https://mirror.codeforces.com/blog/entry/78446 | 1,713,632,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00759.warc.gz | 355,426,443 | 20,146 | ### ABhinav2003's blog
By ABhinav2003, history, 4 years ago,
#include<bits/stdc++.h>
using namespace std;
int in=0 , ex;
int sub(int a[] , int len)
{
// if(len<1)
// return 0;
if(len <= 0 )
return 0;
ex = sub(a,len-1);
cout<<"EX "<<ex<<"\n";
if(a[len-1] >a[len-2])
{
int t = a[len-1]>a[len-2];cout<<"T "<<t<<"\n";
in = 1+sub(a,len-1);
}
cout<<"IN "<<in<<"\n";
int max1 = max(in,ex);
cout<<"MAX "<<max1<<"\n";
return max1;
}
int main()
{
int len;
cin>>len;
int a[len];
for(int i = 0;i<=len-1;i++)
{
cin>>a[i];
}
int m = sub(a , len);
cout<<m;
}
[**HERE IS THE PROBLEM**](https://mirror.codeforces.com/problemset/problem/702/A)
• 0
| Write comment?
» 4 years ago, # | +3 There are many flaws in this code. Base case is len <= 0 so, when len = 1, the index of a[len-2] will become negative. in is a global variable, you'll have to add in = 0 before the if statement. Even if you correct these flaws, the basic flaw is that this method will give you a subsequence instead of a subarray.For example, for the input 1 2 3 10 4 5 6, your output will be 6 (after correcting the mentioned 2 points).
• » » 4 years ago, # ^ | 0 PLZ CAN U CORRECT MY FLAW AND GIVE IT TO ME I CAN"T UNDERSTAND IT PROPERLY BRO PLZ HELP I AM struggling for about 2-3 days now and not able to do thisactually i have just started DP PLZ UNDERSTAND AND HELP ME JUST REWRITE THE FUNCTION WITH REMOVING ALL FLAWS
• » » » 4 years ago, # ^ | +12
• » » » 4 years ago, # ^ | +6 Recursive code which will give TLE#include #define M 100005 using namespace std; int sub(int a[], int len) { if(len <= 1) return 1; if(a[len-1] > a[len-2]) return sub(a, len-1) + 1; else return 1; } int main() { int len; cin>>len; int a[len], ans = 0; for(int i=0; i<=len-1; i++) { cin>>a[i]; } for(int i=0; i<=len-1; i++) { ans = max(ans, sub(a, i+1)); } cout< #define M 100005 using namespace std; int dp[M]; int sub(int a[], int len) { if(len <= 1) return 1; if(dp[len] != -1) return dp[len]; if(a[len-1] > a[len-2]) return dp[len] = sub(a, len-1) + 1; else return dp[len] = 1; } int main() { memset(dp, -1, sizeof dp); int len; cin>>len; int a[len], ans = 0; for(int i=0; i<=len-1; i++) { cin>>a[i]; } for(int i=0; i<=len-1; i++) { ans = max(ans, sub(a, i+1)); } cout< using namespace std; int main() { int n; cin>>n; int a[n], ans = 0, mx = 1; for(int i=0; i>a[i]; for(int i=1; i
• » » 2 months ago, # ^ | 0 I am having some problem with same question i have applied non dp version same as yours but i am still getting wrong ans in test 5 heres the code import java.util.Scanner;/** * A702 */ public class A702 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int a[] = new int[n]; int highest = 1; for (int i = 0; i < a.length; i++) { a[i] = sc.nextInt(); } int current = 1; for (int i = 1; i < a.length; i++) { if (a[i] > a[i - 1]) { current++; } else if ( current > highest) { highest = current; current = 1; } } if (current > highest) { highest = current; } System.out.println(highest); }}
• » » » 2 months ago, # ^ | ← Rev. 2 → 0 My first ever Java Submission on any website, for you. (Well actually 2nd, because 1st was compilation error.)Change: if (a[i] > a[i - 1]) { current++; } else { if ( current > highest) { highest = current; } current = 1; // This has to be done evertime you see non-increasing element. }
• » » » » 2 months ago, # ^ | 0 thanks man, yes i noticed that logical error after this post ,but thank you for your response i really appreciate it | 1,184 | 3,497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.418529 |
https://www.ginac.de/ginac.git/?p=ginac.git;a=commitdiff;h=e989719ca767691eb75b34785baaaed716ea2624;ds=sidebyside | 1,600,820,648,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00565.warc.gz | 876,302,256 | 8,209 | author Jens Vollinga Mon, 3 Nov 2008 14:50:31 +0000 (15:50 +0100) committer Jens Vollinga Mon, 3 Nov 2008 14:50:31 +0000 (15:50 +0100)
ginac/factor.cpp patch | blob | history
index f7dded5e1566e1bebe33470b6800c0e37583fc88..222e05fd55760cfd155c36ca0056060af5e1940e 100644 (file)
@@ -1,6 +1,6 @@
/** @file factor.cpp
*
/** @file factor.cpp
*
- * Polynomial factorization code (Implementation).
+ * Polynomial factorization code (implementation).
*
* Algorithms used can be found in
* [W1] An Improved Multivariate Polynomial Factoring Algorithm,
*
* Algorithms used can be found in
* [W1] An Improved Multivariate Polynomial Factoring Algorithm,
@@ -49,6 +49,7 @@ using namespace GiNaC;
#endif
#include <algorithm>
#endif
#include <algorithm>
+#include <cmath>
#include <list>
#include <vector>
using namespace std;
#include <list>
#include <vector>
using namespace std;
@@ -634,8 +635,206 @@ static void berlekamp(const umod& a, umodvec& upv)
}
}
}
}
+static umod rem_xq(int q, const umod& b)
+{
+ cl_univpoly_modint_ring UPR = b.ring();
+ cl_modint_ring R = UPR->basering();
+
+ int n = degree(b);
+ if ( n > q ) {
+ umod c = UPR->create(q);
+ c.set_coeff(q, R->one());
+ c.finalize();
+ return c;
+ }
+
+ mvec c(n+1, R->zero());
+ int k = q-n;
+ c[n] = R->one();
+ DCOUTVAR(k);
+
+ int ofs = 0;
+ do {
+ cl_MI qk = div(c[n-ofs], coeff(b, n));
+ if ( !zerop(qk) ) {
+ for ( int i=1; i<=n; ++i ) {
+ c[n-i+ofs] = c[n-i] - qk * coeff(b, n-i);
+ }
+ ofs = ofs ? 0 : 1;
+ DCOUTVAR(ofs);
+ DCOUTVAR(c);
+ }
+ } while ( k-- );
+
+ if ( ofs ) {
+ c.pop_back();
+ }
+ else {
+ c.erase(c.begin());
+ }
+ umod res = umod_from_modvec(c);
+ return res;
+}
+
+static void distinct_degree_factor(const umod& a_, umodvec& result)
+{
+ umod a = COPY(a, a_);
+
+ DCOUT(distinct_degree_factor);
+ DCOUTVAR(a);
+
+ cl_univpoly_modint_ring UPR = a.ring();
+ cl_modint_ring R = UPR->basering();
+ int q = cl_I_to_int(R->modulus);
+ int n = degree(a);
+ size_t nhalf = n/2;
+
+
+ size_t i = 1;
+ umod w = UPR->create(1);
+ w.set_coeff(1, R->one());
+ w.finalize();
+ umod x = COPY(x, w);
+
+ umodvec ai;
+
+ while ( i <= nhalf ) {
+ w = expt_pos(w, q);
+ w = rem(w, a);
+
+ ai.push_back(gcd(a, w-x));
+
+ if ( ai.back() != UPR->one() ) {
+ a = div(a, ai.back());
+ w = rem(w, a);
+ }
+
+ ++i;
+ }
+
+ result = ai;
+ DCOUTVAR(result);
+ DCOUT(END distinct_degree_factor);
+}
+
+static void same_degree_factor(const umod& a, umodvec& result)
+{
+ DCOUT(same_degree_factor);
+
+ cl_univpoly_modint_ring UPR = a.ring();
+ cl_modint_ring R = UPR->basering();
+ int deg = degree(a);
+
+ umodvec buf;
+ distinct_degree_factor(a, buf);
+ int degsum = 0;
+
+ for ( size_t i=0; i<buf.size(); ++i ) {
+ if ( buf[i] != UPR->one() ) {
+ degsum += degree(buf[i]);
+ umodvec upv;
+ berlekamp(buf[i], upv);
+ for ( size_t j=0; j<upv.size(); ++j ) {
+ result.push_back(upv[j]);
+ }
+ }
+ }
+
+ if ( degsum < deg ) {
+ result.push_back(a);
+ }
+
+ DCOUTVAR(result);
+ DCOUT(END same_degree_factor);
+}
+
+static void distinct_degree_factor_BSGS(const umod& a, umodvec& result)
+{
+ DCOUT(distinct_degree_factor_BSGS);
+ DCOUTVAR(a);
+
+ cl_univpoly_modint_ring UPR = a.ring();
+ cl_modint_ring R = UPR->basering();
+ int q = cl_I_to_int(R->modulus);
+ int n = degree(a);
+
+ cl_N pm = 0.3;
+ int l = cl_I_to_int(ceiling1(the<cl_F>(expt(n, pm))));
+ DCOUTVAR(l);
+ umodvec h(l+1, UPR->create(-1));
+ umod qk = UPR->create(1);
+ qk.set_coeff(1, R->one());
+ qk.finalize();
+ h[0] = qk;
+ DCOUTVAR(h[0]);
+ for ( int i=1; i<=l; ++i ) {
+ qk = expt_pos(h[i-1], q);
+ h[i] = rem(qk, a);
+ DCOUTVAR(i);
+ DCOUTVAR(h[i]);
+ }
+
+ int m = std::ceil(((double)n)/2/l);
+ DCOUTVAR(m);
+ umodvec H(m, UPR->create(-1));
+ int ql = std::pow(q, l);
+ H[0] = COPY(H[0], h[l]);
+ DCOUTVAR(H[0]);
+ for ( int i=1; i<m; ++i ) {
+ qk = expt_pos(H[i-1], ql);
+ H[i] = rem(qk, a);
+ DCOUTVAR(i);
+ DCOUTVAR(H[i]);
+ }
+
+ umodvec I(m, UPR->create(-1));
+ for ( int i=0; i<m; ++i ) {
+ I[i] = UPR->one();
+ for ( int j=0; j<l; ++j ) {
+ I[i] = I[i] * (H[i] - h[j]);
+ }
+ DCOUTVAR(i);
+ DCOUTVAR(I[i]);
+ I[i] = rem(I[i], a);
+ DCOUTVAR(I[i]);
+ }
+
+ umodvec F(m, UPR->one());
+ umod f = COPY(f, a);
+ for ( int i=0; i<m; ++i ) {
+ DCOUTVAR(i);
+ umod g = gcd(f, I[i]);
+ if ( g == UPR->one() ) continue;
+ F[i] = g;
+ f = div(f, g);
+ DCOUTVAR(F[i]);
+ }
+
+ result.resize(n, UPR->one());
+ if ( f != UPR->one() ) {
+ result[n] = f;
+ }
+ for ( int i=0; i<m; ++i ) {
+ DCOUTVAR(i);
+ umod f = COPY(f, F[i]);
+ for ( int j=l-1; j>=0; --j ) {
+ umod g = gcd(f, H[i]-h[j]);
+ result[l*(i+1)-j-1] = g;
+ f = div(f, g);
+ }
+ }
+
+ DCOUTVAR(result);
+ DCOUT(END distinct_degree_factor_BSGS);
+}
+
+static void cantor_zassenhaus(const umod& a, umodvec& result)
+{
+}
+
static void factor_modular(const umod& p, umodvec& upv)
{
static void factor_modular(const umod& p, umodvec& upv)
{
+ //same_degree_factor(p, upv);
berlekamp(p, upv);
return;
}
berlekamp(p, upv);
return;
}
@@ -736,8 +935,8 @@ static ex hensel_univar(const ex& a_, const ex& x, unsigned int p, const umod& u
umod sigmatilde = umod_from_ex(phi, x, R);
phi = expand(umod_to_ex(t, x) * c);
umod tautilde = umod_from_ex(phi, x, R);
umod sigmatilde = umod_from_ex(phi, x, R);
phi = expand(umod_to_ex(t, x) * c);
umod tautilde = umod_from_ex(phi, x, R);
- umod q = div(sigmatilde, w1);
- umod r = rem(sigmatilde, w1);
+ umod q = UPR->create(-1);
+ umod r = remdiv(sigmatilde, w1, q);
umod sigma = COPY(sigma, r);
phi = expand(umod_to_ex(tautilde, x) + umod_to_ex(q, x) * umod_to_ex(u1, x));
umod tau = umod_from_ex(phi, x, R);
umod sigma = COPY(sigma, r);
phi = expand(umod_to_ex(tautilde, x) + umod_to_ex(q, x) * umod_to_ex(u1, x));
umod tau = umod_from_ex(phi, x, R);
@@ -851,11 +1050,14 @@ struct ModFactors
static ex factor_univariate(const ex& poly, const ex& x)
{
static ex factor_univariate(const ex& poly, const ex& x)
{
+ DCOUT(factor_univariate);
+ DCOUTVAR(poly);
+
ex unit, cont, prim;
poly.unitcontprim(x, unit, cont, prim);
// determine proper prime and minimize number of modular factors
ex unit, cont, prim;
poly.unitcontprim(x, unit, cont, prim);
// determine proper prime and minimize number of modular factors
- unsigned int p = 3, lastp;
+ unsigned int p = 3, lastp = 3;
cl_modint_ring R;
unsigned int trials = 0;
unsigned int minfactors = 0;
cl_modint_ring R;
unsigned int trials = 0;
unsigned int minfactors = 0;
@@ -973,6 +1175,7 @@ static ex factor_univariate(const ex& poly, const ex& x)
}
}
}
}
+ DCOUT(END factor_univariate);
return unit * cont * result;
}
return unit * cont * result;
}
@@ -1043,59 +1246,37 @@ void change_modulus(umod& out, const umod& in)
void eea_lift(const umod& a, const umod& b, const ex& x, unsigned int p, unsigned int k, umod& s_, umod& t_)
{
DCOUT(eea_lift);
void eea_lift(const umod& a, const umod& b, const ex& x, unsigned int p, unsigned int k, umod& s_, umod& t_)
{
DCOUT(eea_lift);
- DCOUTVAR(a);
- DCOUTVAR(b);
- DCOUTVAR(x);
- DCOUTVAR(p);
- DCOUTVAR(k);
cl_modint_ring R = find_modint_ring(p);
cl_univpoly_modint_ring UPR = find_univpoly_ring(R);
umod amod = UPR->create(degree(a));
cl_modint_ring R = find_modint_ring(p);
cl_univpoly_modint_ring UPR = find_univpoly_ring(R);
umod amod = UPR->create(degree(a));
- DCOUTVAR(a);
change_modulus(amod, a);
umod bmod = UPR->create(degree(b));
change_modulus(bmod, b);
change_modulus(amod, a);
umod bmod = UPR->create(degree(b));
change_modulus(bmod, b);
- DCOUTVAR(amod);
- DCOUTVAR(bmod);
umod g = UPR->create(-1);
umod smod = UPR->create(-1);
umod tmod = UPR->create(-1);
exteuclid(amod, bmod, g, smod, tmod);
umod g = UPR->create(-1);
umod smod = UPR->create(-1);
umod tmod = UPR->create(-1);
exteuclid(amod, bmod, g, smod, tmod);
- DCOUTVAR(smod);
- DCOUTVAR(tmod);
- DCOUTVAR(g);
- DCOUTVAR(a);
-
cl_modint_ring Rpk = find_modint_ring(expt_pos(cl_I(p),k));
cl_univpoly_modint_ring UPRpk = find_univpoly_ring(Rpk);
umod s = UPRpk->create(degree(smod));
change_modulus(s, smod);
umod t = UPRpk->create(degree(tmod));
change_modulus(t, tmod);
cl_modint_ring Rpk = find_modint_ring(expt_pos(cl_I(p),k));
cl_univpoly_modint_ring UPRpk = find_univpoly_ring(Rpk);
umod s = UPRpk->create(degree(smod));
change_modulus(s, smod);
umod t = UPRpk->create(degree(tmod));
change_modulus(t, tmod);
- DCOUTVAR(s);
- DCOUTVAR(t);
cl_I modulus(p);
cl_I modulus(p);
- DCOUTVAR(a);
-
umod one = UPRpk->one();
for ( size_t j=1; j<k; ++j ) {
umod one = UPRpk->one();
for ( size_t j=1; j<k; ++j ) {
- DCOUTVAR(a);
umod e = one - a * s - b * t;
umod e = one - a * s - b * t;
- DCOUTVAR(one);
- DCOUTVAR(a*s);
- DCOUTVAR(b*t);
- DCOUTVAR(e);
e = divide(e, modulus);
umod c = UPR->create(degree(e));
change_modulus(c, e);
umod sigmabar = smod * c;
umod taubar = tmod * c;
e = divide(e, modulus);
umod c = UPR->create(degree(e));
change_modulus(c, e);
umod sigmabar = smod * c;
umod taubar = tmod * c;
- umod q = div(sigmabar, bmod);
- umod sigma = rem(sigmabar, bmod);
+ umod q = UPR->create(-1);
+ umod sigma = remdiv(sigmabar, bmod, q);
umod tau = taubar + q * amod;
umod tau = taubar + q * amod;
@@ -1109,8 +1290,6 @@ void eea_lift(const umod& a, const umod& b, const ex& x, unsigned int p, unsigne
s_ = s; t_ = t;
s_ = s; t_ = t;
- DCOUTVAR(s);
- DCOUTVAR(t);
DCOUT2(check, a*s + b*t);
DCOUT(END eea_lift);
}
DCOUT2(check, a*s + b*t);
DCOUT(END eea_lift);
}
@@ -1144,9 +1323,9 @@ umodvec univar_diophant(const umodvec& a, const ex& x, unsigned int m, unsigned
eea_lift(a[1], a[0], x, p, k, s, t);
ex phi = expand(pow(x,m) * umod_to_ex(s, x));
umod bmod = umod_from_ex(phi, x, R);
eea_lift(a[1], a[0], x, p, k, s, t);
ex phi = expand(pow(x,m) * umod_to_ex(s, x));
umod bmod = umod_from_ex(phi, x, R);
- umod buf = rem(bmod, a[0]);
+ umod q = UPR->create(-1);
+ umod buf = remdiv(bmod, a[0], q);
result.push_back(buf);
result.push_back(buf);
- umod q = div(bmod, a[0]);
phi = expand(pow(x,m) * umod_to_ex(t, x));
umod t1mod = umod_from_ex(phi, x, R);
umod buf2 = t1mod + q * a[1];
phi = expand(pow(x,m) * umod_to_ex(t, x));
umod t1mod = umod_from_ex(phi, x, R);
umod buf2 = t1mod + q * a[1];
@@ -1185,7 +1364,7 @@ struct make_modular_map : public map_function {
static ex make_modular(const ex& e, const cl_modint_ring& R)
{
make_modular_map map(R);
static ex make_modular(const ex& e, const cl_modint_ring& R)
{
make_modular_map map(R);
- return map(e);
+ return map(e.expand());
}
vector<ex> multivar_diophant(const vector<ex>& a_, const ex& x, const ex& c, const vector<EvalPoint>& I, unsigned int d, unsigned int p, unsigned int k)
}
vector<ex> multivar_diophant(const vector<ex>& a_, const ex& x, const ex& c, const vector<EvalPoint>& I, unsigned int d, unsigned int p, unsigned int k)
@@ -1247,10 +1426,8 @@ vector<ex> multivar_diophant(const vector<ex>& a_, const ex& x, const ex& c, con
for ( size_t i=0; i<r; ++i ) {
buf -= sigma[i] * b[i];
}
for ( size_t i=0; i<r; ++i ) {
buf -= sigma[i] * b[i];
}
- ex e = buf;
- e = make_modular(e, R);
+ ex e = make_modular(buf, R);
- e = e.expand();
DCOUTVAR(e);
DCOUTVAR(d);
ex monomial = 1;
DCOUTVAR(e);
DCOUTVAR(d);
ex monomial = 1;
@@ -1259,9 +1436,11 @@ vector<ex> multivar_diophant(const vector<ex>& a_, const ex& x, const ex& c, con
while ( !e.is_zero() && e.has(xnu) ) {
monomial *= (xnu - alphanu);
monomial = expand(monomial);
while ( !e.is_zero() && e.has(xnu) ) {
monomial *= (xnu - alphanu);
monomial = expand(monomial);
+ DCOUTVAR(monomial);
DCOUTVAR(xnu);
DCOUTVAR(alphanu);
ex cm = e.diff(ex_to<symbol>(xnu), m).subs(xnu==alphanu) / factorial(m);
DCOUTVAR(xnu);
DCOUTVAR(alphanu);
ex cm = e.diff(ex_to<symbol>(xnu), m).subs(xnu==alphanu) / factorial(m);
+ cm = make_modular(cm, R);
DCOUTVAR(cm);
if ( !cm.is_zero() ) {
vector<ex> delta_s = multivar_diophant(anew, x, cm, Inew, d, p, k);
DCOUTVAR(cm);
if ( !cm.is_zero() ) {
vector<ex> delta_s = multivar_diophant(anew, x, cm, Inew, d, p, k);
@@ -1272,8 +1451,8 @@ vector<ex> multivar_diophant(const vector<ex>& a_, const ex& x, const ex& c, con
sigma[j] += delta_s[j];
buf -= delta_s[j] * b[j];
}
sigma[j] += delta_s[j];
buf -= delta_s[j] * b[j];
}
- e = buf.expand();
- e = make_modular(e, R);
+ e = make_modular(buf, R);
+ DCOUTVAR(e);
}
}
}
}
}
}
@@ -1424,7 +1603,6 @@ ex hensel_multivar(const ex& a, const ex& x, const vector<EvalPoint>& I, unsigne
for ( size_t i=j-1; i<nu-1; ++i ) {
coef = coef.subs(I[i].x == I[i].evalpoint);
}
for ( size_t i=j-1; i<nu-1; ++i ) {
coef = coef.subs(I[i].x == I[i].evalpoint);
}
- coef = expand(coef);
coef = make_modular(coef, R);
int deg = U[m].degree(x);
U[m] = U[m] - U[m].lcoeff(x) * pow(x,deg) + coef * pow(x,deg);
coef = make_modular(coef, R);
int deg = U[m].degree(x);
U[m] = U[m] - U[m].lcoeff(x) * pow(x,deg) + coef * pow(x,deg);
@@ -1468,7 +1646,6 @@ ex hensel_multivar(const ex& a, const ex& x, const vector<EvalPoint>& I, unsigne
deltaU[i] *= monomial;
U[i] += deltaU[i];
U[i] = make_modular(U[i], R);
deltaU[i] *= monomial;
U[i] += deltaU[i];
U[i] = make_modular(U[i], R);
- U[i] = U[i].expand();
DCOUTVAR(U[i]);
}
ex Uprod = 1;
DCOUTVAR(U[i]);
}
ex Uprod = 1;
@@ -1477,13 +1654,10 @@ ex hensel_multivar(const ex& a, const ex& x, const vector<EvalPoint>& I, unsigne
}
DCOUTVAR(Uprod.expand());
DCOUTVAR(A[j-1]);
}
DCOUTVAR(Uprod.expand());
DCOUTVAR(A[j-1]);
- e = expand(A[j-1] - Uprod);
+ e = A[j-1] - Uprod;
e = make_modular(e, R);
DCOUTVAR(e);
}
e = make_modular(e, R);
DCOUTVAR(e);
}
- else {
- break;
- }
}
}
}
}
}
}
@@ -1740,8 +1914,8 @@ static ex factor_multivariate(const ex& poly, const exset& syms)
while ( true ) {
numeric trialcount = 0;
ex u, delta;
while ( true ) {
numeric trialcount = 0;
ex u, delta;
- unsigned int prime;
- size_t factor_count;
+ unsigned int prime = 3;
+ size_t factor_count = 0;
ex ufac;
ex ufaclst;
while ( trialcount < maxtrials ) {
ex ufac;
ex ufaclst;
while ( trialcount < maxtrials ) {
@@ -2005,10 +2179,13 @@ struct find_symbols_map : public map_function {
static ex factor_sqrfree(const ex& poly)
{
static ex factor_sqrfree(const ex& poly)
{
+ DCOUT(factor_sqrfree);
+
// determine all symbols in poly
find_symbols_map findsymbols;
findsymbols(poly);
if ( findsymbols.syms.size() == 0 ) {
// determine all symbols in poly
find_symbols_map findsymbols;
findsymbols(poly);
if ( findsymbols.syms.size() == 0 ) {
+ DCOUT(END factor_sqrfree);
return poly;
}
return poly;
}
@@ -2018,16 +2195,19 @@ static ex factor_sqrfree(const ex& poly)
if ( poly.ldegree(x) > 0 ) {
int ld = poly.ldegree(x);
ex res = factor_univariate(expand(poly/pow(x, ld)), x);
if ( poly.ldegree(x) > 0 ) {
int ld = poly.ldegree(x);
ex res = factor_univariate(expand(poly/pow(x, ld)), x);
+ DCOUT(END factor_sqrfree);
return res * pow(x,ld);
}
else {
ex res = factor_univariate(poly, x);
return res * pow(x,ld);
}
else {
ex res = factor_univariate(poly, x);
+ DCOUT(END factor_sqrfree);
return res;
}
}
// multivariate case
ex res = factor_multivariate(poly, findsymbols.syms);
return res;
}
}
// multivariate case
ex res = factor_multivariate(poly, findsymbols.syms);
+ DCOUT(END factor_sqrfree);
return res;
}
return res;
} | 7,310 | 17,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-40 | latest | en | 0.073681 |
https://s23.cs251.com/Recitations/Rec_08_Randomized_Algorithms/contents.html | 1,686,199,922,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654097.42/warc/CC-MAIN-20230608035801-20230608065801-00566.warc.gz | 531,064,851 | 10,521 | MODULE 9: Randomized Algorithms
Recitation 8
1
Review
• A randomized algorithm is an algorithm that has access to a random number generator. In this class we will allow randomized algorithms to call RandInt($$n$$) and Bernoulli($$p$$).
• Here are two interesting types of randomized algorithms:
• An algorithm $$A$$ is a $$T(n)$$-time Las Vegas algorithm if
• for every input $$x \in \Sigma^*$$, $$A(x)$$ returns the right answer with probability $$1$$, and
• for every input $$x \in \Sigma^*$$, $$\mathbb{E}[\text{number of steps A(x) takes}] \le T(|x|)$$.
• An algorithm $$A$$ is a $$T(n)$$-time Monte Carlo algorithm with error probability $$\varepsilon$$ if
• for every input $$x \in \Sigma^*$$, $$A(x)$$ gives the wrong answer with probability at most $$\varepsilon$$, and
• for every input $$x \in \Sigma^*$$, $$A(x)$$ has a worst-case running-time of at most $$T(|x|)$$.
• Note that above, we require correctness and running-time guarantees for every input $$x$$.
2
What happens in Las Vegas doesn’t stay in Monte Carlo
The expected number of comparisons that the Quicksort algorithm makes is at most $$2n \ln n$$ (which you can cite without proof — you might see a proof of this fact if you take 15-210). Describe how to convert this Las Vegas algorithm into a Monte Carlo algorithm with the worst-case number of comparisons being $$1000n \ln n$$. Give an upper bound on the error probability of the Monte Carlo algorithm.
Solution
First attempt: Run the Las Vegas algorithm, but if it performs $$1000 n\ln n$$ comparisons, we will stop the algorithm and declare failure. Let $$\boldsymbol{X}$$ be a random variable for the number of comparisons performed by a run of the Las Vegas algorithm. Our algorithm only fails if the Las Vegas algorithm tries to perform $$\ge 1000 n \ln n$$ comparisons. By Markov’s inequality, this occurs with probability $\Pr[\boldsymbol{X} \ge 1000 n \ln n] \le \frac{1}{500}.$ This is not really a great bound, so let’s try something else.
Second attempt: Now we will use boosting to achieve a much better bound for the error probability. Run the Las Vegas algorithm, but if it performs $$4n \ln n$$ comparisons, stop the algorithm and declare failure. Again, by Markov’s inequality, this occurs with probability at most $$1/2$$. We will do this independently $$250$$ times (so we have at $$1000n \ln n$$ comparisons in total). If any repetition succeeds, we can give its output and be correct. This way only fails if all $$250$$ repetitions fail. The events of failing in these repetitions are independent events. Therefore the overall error probability is $$(1/2)^{250} < 1/10^{75}$$.
3
Atlantic City to Monte Carlo
Suppose you are given a randomized algorithm that solves $$f : \Sigma^* \to \Sigma^*$$ in expected time $$T(n)$$ and with $$\varepsilon$$ probability of error (i.e., the algorithm gambles both with correctness and running time). Show that for any constant $$\varepsilon' > 0$$, there is a Monte Carlo algorithm computing $$f$$ with running time $$O(T(n))$$ and error probability $$\varepsilon + \varepsilon'$$.
Solution
Let $$A$$ be the algorithm given to us. We want to construct a Monte Carlo algorithm $$A'$$, which works as follows: run $$A$$ for $$T(n)/\varepsilon'$$ steps. If it halts, give its answer. If not, declare “failure”. We can fail in two ways: either the simulation of $$A$$ halts and gives the wrong answer, or the simulation of $$A$$ does not halt in $$T(n)/\varepsilon'$$ steps. The first happens with probability at most $$\varepsilon$$, and the second happens with probability at most $$\varepsilon'$$ (by Markov). Using the union bound, the error probability is at most $$\varepsilon + \varepsilon'$$.
Here is another way of thinking about the analysis. For any fixed input, the original algorithm induces a probability tree representing the computation. We know that at most $$\varepsilon'$$ fraction of the leaves are too deep (i.e., running time exceeds $$T(n)/\varepsilon'$$). And we know at most $$\varepsilon$$ fraction of the leaves give the wrong answer. By the union bound, at most $$\varepsilon+\varepsilon'$$ fraction of the leaves either give the wrong answer or are too deep. If the wrong leaves intersect with deep ones, that would be better, but in the worst case, wrong leaves and deep ones would be disjoint.
Technicality Alert: There is a small technical issue here. Algorithm $$A'$$ needs to be able to compute $$T(|x|)$$ from $$x$$ in $$O(T(|x|))$$ time. This is indeed the case for most $$T(\cdot)$$ that we care about.
4
Randomization Meets Approximation
Consider the MAX-3SAT problem where, given a CNF formula in which every clause has exactly 3 literals (with distinct variables), we want to find a truth assignment to the variables in the formula so that we maximize the number of clauses that evaluate to True.
Describe a polynomial-time randomized algorithm with the property that, given a 3CNF formula with $$m$$ clauses, it outputs a truth assignment to the variables such that the expected number of clauses that evaluate to True is $$\frac{7}{8}m$$.
Solution
The solution here is simply to pick the assignment to the variables randomly: For each variable, flip a coin. If it is heads, assign it the value True. If it is tails, assign it False.
Suppose we have $$m$$ clauses. Now let $$\boldsymbol{X}$$ be the number of clauses satisfied. We are interested in computing $$\mathbb{E}[\boldsymbol{X}]$$. Define the indicator random variable $$X_i$$ to be 1 if the $$i$$’th clause is satisfied, and 0 otherwise. Then $$\boldsymbol{X} = \sum_{i=1}^{m} \boldsymbol{X}_i$$ and so $$\mathbb{E}[\boldsymbol{X}] = \mathbb{E}[\sum_{i=1}^{m} \boldsymbol{X}_i] = \sum_{i=1}^{m} \mathbb{E}[\boldsymbol{X}_i]$$ where in the last equality, we used linearity of expectation. Note that for all $$i$$, $$\mathbb{E}[\boldsymbol{X}_i] = \Pr[\boldsymbol{X}_i = 1] = 7/8$$ (the only way a clause is not satisfied is when all the literals are False, which happens with probability $$(1/2)^3 = 1/8$$). So $$\mathbb{E}[\boldsymbol{X}] = \frac{7}{8}m$$.
Here it is important to note that the $$\boldsymbol{X}_i$$’s need not be independent! In fact, they can be perfectly correlated: consider a case where all the $$m$$ clauses are identical to each other. Then if you know that one clause is satisfied, you automatically know that all of the clauses must be satisfied. Yet, this does not change the expected number of satisfied clauses at all. And as you can see in the calculation above, linearity of expectation allows us to treat each clause independently, without worrying about potential dependencies among clauses. We know that $$\mathbb{E}[\boldsymbol{X}_i] = \Pr[\boldsymbol{X}_i = 1] = 7/8$$ no matter what is going on with other clauses.
5
(Extra) (Brain Teaser) Passive-Aggressive Passengers
Consider a plane with $$n$$ seats $$s_1, s_2, \ldots , s_n$$. There are $$n$$ passengers, $$p_1, p_2, \ldots, p_n$$ and they are randomly assigned unique seat numbers. The passengers enter the plane one by one in the order $$p_1, p_2,\ldots p_n$$. The first passenger $$p_1$$ does not look at their assigned seat and instead picks a uniformly random seat to sit in. All the other passengers, $$p_2, p_3, \ldots, p_n$$, use the following strategy. If the seat assigned to them is available, they sit in that seat. Otherwise they pick a seat uniformly at random among the available seats, and they sit there. What is the probability that the last passenger, $$p_n$$, will end up sitting in their assigned seat?
Solution
For ease of presentation, note that the random assignment of seats is a red herring — we can without loss of generality rearrange seat numbers so that $$p_i$$ is assigned to $$s_i$$.
We take a different perspective on the problem as follows: The passengers $$p_2, p_3$$, …, $$p_{n-1}$$ use the following strategy. Passenger $$p_i$$ goes to their assigned seat $$s_i$$. If it is occupied, they kick out the intruder (who will always be $$p_1$$). No matter what, $$p_i$$ sits in $$s_i$$. Then the intruder picks a uniformly random available seat to sit in. This is the same as the original problem, but the identities are changed around so that $$p_1$$ is the only passenger going around picking uniformly random seats.
This perspective makes the situation a little easier to think about. Now we want to find the probability that $$s_n$$ is unoccupied when $$p_n$$ comes to take a seat.
Notice that it only matters whether $$p_1$$ eventually sits in $$s_1$$ or in $$s_n$$. If $$p_1$$ sits elsewhere, they will be kicked out and have to repick their seat until eventually they choose $$s_1$$ or $$s_n$$. Then $$p_1$$ won’t move again until $$p_n$$ comes, at which point $$s_n$$ is unoccupied iff $$p_1$$ sits in $$s_1$$. Every time $$p_1$$ picks a seat, they have an equal chance of going to $$s_1$$ and of going to $$s_n$$, so the answer is $$1/2$$. | 2,356 | 8,888 | {"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": 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} | 3.703125 | 4 | CC-MAIN-2023-23 | latest | en | 0.784998 |
https://www.bestdaixie.com/Matlab%E4%BB%A3%E5%86%99+%7C++INFO20003++Query+Processing+and+Query+Optimisation | 1,718,207,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00442.warc.gz | 639,272,331 | 20,469 | # BEST代写-线上编程学术专家
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# Matlab代写 | INFO20003 Query Processing and Query Optimisation
### Matlab代写 | INFO20003 Query Processing and Query Optimisation
INFO20003 Semester 2, 2019
Assignment 3 – Query Processing and Query Optimisation
Due: 6:00pm Friday 18 October
Submission: Via LMS https://lms.unimelb.edu.au
out of 20 marks.
Question 1 (5 marks)
Consider two relations called Parts and Supply. Imagine that relation Parts has 60,000 tuples
and Supply has 150,000 tuples. Both relations store 50 tuples per page. Consider the following
SQL statement:
SELECT *
FROM Parts INNER JOIN Supply
ON Parts.PartID = Supply.PID;
We wish to evaluate an equijoin between Supply and Parts, with an equality condition
Parts.PartID = Supply.PID. There are 202 buffer pages available in memory for this operation.
Both relations are stored as (unsorted) heap files. Neither relation has any indexes built on it.
Consider the alternative join strategies described below and calculate the cost of each
alternative. Evaluate the algorithms using the number of disk I/O’s (i.e. pages) as the cost. For
each strategy, provide the formulae you use to calculate your cost estimates.
a) Page-oriented Nested Loops Join. Consider Parts as the outer relation. (1 mark)
b) Block-oriented Nested Loops Join. Consider Parts as the outer relation. (1 mark)
c) Sort-Merge Join. Assume that Sort-Merge Join can be done in 2 passes. (1 mark)
d) Hash Join. (1 mark)
e) What would be the lowest possible cost to perform this query, assuming that no
indexes are built on any of the two relations, and assuming that sufficient buffer space
is available? What would be the minimum buffer size required to achieve this cost?
Explain briefly. (1 mark)
2
Question 2 (5 marks)
Consider a relation with the following schema:
Employee (EmpID, firstname, lastname, department, salary)
The Employee relation has 1200 pages and each page stores 120 tuples. The department
attribute can take one of six values (“Marketing”, “Human Resource”, “Finance”, “Public
Relations”, “Sales and Distribution”, “Operation Management”) and salary can have values
between 100,000 and 500,000 ([100,000, 500,000]).
Suppose that the following SQL query is executed frequently using the given relation:
SELECT *
FROM Employee
WHERE salary > 300,000 AND department = ‘Marketing’;
Your job is to analyse the query plans and estimate the cost of the best plan utilizing the
information given about different indexes in each part.
a) Compute the estimated result size for the query, and the reduction factor of each filter.
(1 mark)
b) Compute the estimated cost of the best plan assuming that a clustered B+ tree index
on (department, salary) is the only index available. Suppose there are 300 index
pages. Discuss and calculate alternative plans. (1 mark)
c) Compute the estimated cost of the best plan assuming that an unclustered B+ tree
index on (salary) is the only index available. Suppose there are 200 index pages.
Discuss and calculate alternative plans. (1 mark)
d) Compute the estimated cost of the best plan assuming that an unclustered Hash index
on (department) is the only index available. Discuss and calculate alternative plans.
(1 mark)
e) Compute the estimated cost of the best plan assuming that an unclustered Hash index
on (salary) is the only index available. Discuss and calculate alternative plans.
(1 mark)
3
Question 3 (10 marks)
Consider the following relational schema and SQL query. The schema captures information
about employees, their departments and the projects they are involved in.
Employee (eid: integer, salary: integer, name: char(30))
Project (projid: integer, code: char(20), start: date, end: date, eid: integer)
Department (did: integer, projid: integer, budget: real, floor: integer)
Consider the following query:
SELECT e.name, d.projid
FROM Employee e, Project p, Department d
WHERE e.eid = p.eid AND p.projid = d.projid
AND e.salary < 300,000 AND p.code = ‘alpha 340’;
The system’s statistics indicate that there are 1000 different project code values, and salary
of the employees range from 100,000 to 500,000 ([100,000, 500,000]). There is a total of
60,000 projects, 5,000 employees and 20,000 departments in the database. Each relation fits
100 tuples in a page. Assume eid is a candidate key for Employee, projid is a candidate key
for Project, and did is a candidate key for the Department table. Suppose there exists a
clustered B+ tree index on (Project.projid) of size 200 pages and suppose there is a clustered
B+ tree index on (employee.salary) of size 10 pages.
a) Compute the estimated result size and the reduction factors (selectivity) of this query.
(2 marks)
b) Compute the cost of the plans shown below. Assume that sorting of any relation (if
required) can be done in 2 passes. NLJ is a Page-oriented Nested Loops Join.
Assume that 100 tuples of a resulting join between Employee and Project fit in a page.
Similarly, 100 tuples of a resulting join between Project and Department fit in a page.
If selection over filtering predicates is not marked in the plan, assume it will happen
on-the-fly after all joins are performed, as the last operation in the plan.
(8 marks, 2 marks per plan)
4
Formatting Requirements:
For each question, present an answer in the following format:
• Show the question number before each question. You do not need to include the text of the question
itself.
• Start Question 2 and Question 3 on a new page.
For each of the calculations, provide all the formulae you used to calculate your cost estimates, and
show your working, not only the result.
Employee
(Heap scan)
Project
(Heap scan)
Department
(Heap scan)
Project
(Heap scan)
Department
(Heap scan)
Employee
(Heap scan)
NLJ
NLJ
HJ
SMJ
[1] [2]
[3] [4]
Project
(Heap scan)
Employee
(Index scan (salary))
Department
(Heap scan)
NLJ
HJ
𝝈(salary <300,000)
Department
(Heap scan)
Project
(Index scan (projid))
Employee
(Heap scan)
SMJ
SMJ
5
Submission Process:
Submit a single PDF showing your answers to all questions to the Assessment page on LMS by 6pm on
the due date of Friday 18 October. Name your file ‘STUDENT_ID’.pdf, where STUDENT_ID corresponds
If you need an extension due to a valid (medical) reason, you will need to provide evidence to support
your request by 9pm, Thursday 17th of October. Medical certificates need to be at least two days in
length.
To request an extension:
1. Email Farah Khan (khanf1@unimelb.edu.au) from your university email address, supplying your
student ID, the extension request and supporting evidence. | 1,629 | 6,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.868826 |
https://www.mrexcel.com/board/threads/if-value-is-same-in-two-columns-then-display-a-third-value-from-another-column.1087008/ | 1,624,555,772,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00096.warc.gz | 794,393,143 | 19,686 | # IF value is same in two columns then display a third value from another column
#### Rick187
##### New Member
Hi,
Does anyone know which formula can achieve this:
If value in column A is the same as value in column B, then display the value from Column C
For example if Column A contains these values
C1 - AAA
C2 - BBB
Column B contains the same:
B1 - AAA
B2 - BBB
B3 - CCC
Column C contains
C1 - London
C2 - New York
Etc.
In Column D, I would like it to insert the value from Column C, so next to "AAA", I would like it to display "London"
Any help would be appreciated.
Thanks in advance.
### Excel Facts
How to calculate loan payments in Excel?
Use the PMT function: =PMT(5%/12,60,-25000) is for a \$25,000 loan, 5% annual interest, 60 month loan.
#### Tim_Excel_
##### Well-known Member
Code:
``=IF(A1=B1,C1,"")``
This is the simplest of formulas... Maybe consider googling the problem before opening a new thread?
#### Rick187
##### New Member
Unfortunately this doesn't seem to work when the columns are in different sheets.
I've tried this: =IF('Sheet1'!A:A=Sheet2!G:G,Sheet2!I:I,"")
#### Tim_Excel_
##### Well-known Member
Again, Google is your best friend... (although one could argue it isn't because of privacy reasons, but that's another topic)
Code:
``[COLOR=#333333] =IF([/COLOR]Sheet1![COLOR=#333333]A:A=Sheet2!G:G,Sheet2!I:I,"")[/COLOR]``
#### Rick187
##### New Member
ADVERTISEMENT
Again, Google is your best friend... (although one could argue it isn't because of privacy reasons, but that's another topic)
Code:
``[COLOR=#333333] =IF([/COLOR]Sheet1![COLOR=#333333]A:A=Sheet2!G:G,Sheet2!I:I,"")[/COLOR]``
This doesn't seem to work. Nothing displays when I use this formula.
#### Tim_Excel_
##### Well-known Member
It does work on my end when I use regular text and number values. I'm assuming there's some information missing here. What do these ranges contain?
#### Rick187
##### New Member
ADVERTISEMENT
It does work on my end when I use regular text and number values. I'm assuming there's some information missing here. What do these ranges contain?
They contain text.
In Sheet 1 If the value from A2 is contained in a range from another sheet (Sheet 2 Column G), then Display the corresponding Value from (Sheet 2 Column I)
So if I have the word "Cat" in the the first sheet in Cell A2, and in sheet 2 Column G I have 30 different words and one of them is "Cat", Next to the word "Cat" in "Column I" It says "London". I would like the word "London" to display in Sheet 1 cell B2.
seen post#7
Last edited:
#### Tim_Excel_
##### Well-known Member
That's a whole 'nother cookie
For this you will have to use the MATCH function. This will return the row number of the found value, or #N/A if the value was not found. You can then return the value of the cell next to it. Use the IFNA function to handle values that won't be found.
#### jtakw
##### Well-known Member
Hi,
Change/adjust cell references/range, add sheet name, as needed.
Change "No Match" to "" (Blank) or whatever you like, formula copied down:
<b></b><table cellpadding="2.5px" rules="all" style=";background-color: rgb(255,255,255);border: 1px solid;border-collapse: collapse; border-color: rgb(187,187,187)"><colgroup><col width="25px" style="background-color: rgb(218,231,245)" /><col /><col /><col /><col /><col /><col /><col /><col /><col /></colgroup><thead><tr style=" background-color: rgb(218,231,245);text-align: center;color: rgb(22,17,32)"><th></th><th>A</th><th>B</th><th>C</th><th>D</th><th>E</th><th>F</th><th>G</th><th>H</th><th>I</th></tr></thead><tbody><tr ><td style="color: rgb(22,17,32);text-align: center;">2</td><td style=";">Cat</td><td style=";">London</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style=";">Dog</td><td style="text-align: right;;"></td><td style=";">New York</td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">3</td><td style=";">Pig</td><td style=";">No Match</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style=";">Mouse</td><td style="text-align: right;;"></td><td style=";">Paris</td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">4</td><td style=";">Chicken</td><td style=";">No Match</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style=";">Cat</td><td style="text-align: right;;"></td><td style=";">London</td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">5</td><td style=";">Dog</td><td style=";">New York</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style=";">Horse</td><td style="text-align: right;;"></td><td style=";">California</td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">6</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style=";">Cow</td><td style="text-align: right;;"></td><td style=";">Washington</td></tr></tbody></table><p style="width:6.4em;font-weight:bold;margin:0;padding:0.2em 0.6em 0.2em 0.5em;border: 1px solid rgb(187,187,187);border-top:none;text-align: center;background-color: rgb(218,231,245);color: rgb(22,17,32)">Sheet558</p><br /><br /><table width="85%" cellpadding="2.5px" rules="all" style=";border: 2px solid black;border-collapse:collapse;padding: 0.4em;background-color: rgb(255,255,255)" ><tr><td style="padding:6px" ><b>Worksheet Formulas</b><table cellpadding="2.5px" width="100%" rules="all" style="border: 1px solid;text-align:center;background-color: rgb(255,255,255);border-collapse: collapse; border-color: rgb(187,187,187)"><thead><tr style=" background-color: rgb(218,231,245);color: rgb(22,17,32)"><th width="10px">Cell</th><th style="text-align:left;padding-left:5px;">Formula</th></tr></thead><tbody><tr><th width="10px" style=" background-color: rgb(218,231,245);color: rgb(22,17,32)">B2</th><td style="text-align:left">=IFERROR(<font color="Blue">LOOKUP(<font color="Red">2,1/SEARCH(<font color="Green">A2,G\$2:G\$6</font>),I\$2:I\$6</font>),"No Match"</font>)</td></tr></tbody></table></td></tr></table><br />
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Go back | 2,290 | 7,727 | {"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} | 3.21875 | 3 | CC-MAIN-2021-25 | latest | en | 0.881472 |
https://ask.cvxr.com/t/how-to-write-the-objective-function-in-matlab/11403 | 1,686,297,437,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655446.86/warc/CC-MAIN-20230609064417-20230609094417-00246.warc.gz | 128,308,647 | 4,841 | # How to write the objective function in Matlab?
First time use CVX.Find it hard to write the objective function in Matlab.Too many errors made me worried.Is there any way to write it with no errors?In the function, L is constant, you can set it as 30.While Xi is variable, so there are 30 variables.
For L = 2, that is neither convex not concave, even if the x_i's are restricted to be nonnegative.
The objective function can be written in “MATLAB” (for example, using YALMIP)., but can’t be written in CVX.
Yes, one of the constraints is Xi’s are all nonnegative.So this is a nonconvex problem?Can YALMIP solve this kind of problem?By the way, thank you for your reply and suggestion.
Is is neither convex nor concave, even if all the variables are nonnegative.
The objective function can be entered in CVX (and presuming the constraints also can be), either a local optimization solver, such as FMINCON can be used, or a global optimization solver, such as BARON or YALMIP’s own solver, BMIBNB can be used to attempt to solve to global optimality. I don’t know how difficult this problem will be to solve to global optimality to a small optimality gap tolerance using a global optimization solver when L = 30 (the constraints may also matter).
Got it, thanks a lot for your help. | 309 | 1,289 | {"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-2023-23 | latest | en | 0.918019 |
https://rememberingsomer.com/what-percent-of-200-is-25/ | 1,638,819,983,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00044.warc.gz | 553,875,926 | 5,132 | Detailed calculations below.
"Percent (%)" means "out of one hundred": 25% = 25 "out of one hundred", p% is check out p "percent", 25% = 25/100 = 25 ÷ 100
25% × 200 = 25/100 × 200 = (25 ÷ 100) × 200 = 25 × 200 ÷ 100 = 5,000 ÷ 100 = 50
100/100 = 100 ÷ 100 = 100% = 1 multiply a number by the portion 100/100, ... And also its value doesn"t change. N/100 = n%, any number.
Signs: % percent, ÷ divide, × multiply, = equal, / fraction bar, ≈ approximately equal; creating numbers: comma "," as thousands separator; suggest "." together a decimal mark;
% rememberingsomer.com online calculator: percent worth of numbers, amounts
Percent worth %... Of number: ... Amounts to what?
recent calculated number rememberingsomer.com
25% of 200 = 50 Sep 29 11:40 UTC (GMT) 21% the 1,133 = 237.93 Sep 29 11:40 UTC (GMT) 102% of 15.56 = 15.8712 Sep 29 11:40 UTC (GMT) 92.71% that 55,596 = 51,543.0516 Sep 29 11:40 UTC (GMT) 20% of 267.46 = 53.492 Sep 29 11:40 UTC (GMT) 995% the 500,000 = 4,975,000 Sep 29 11:40 UTC (GMT) 16.2% that 6.225 = 1.00845 Sep 29 11:40 UTC (GMT) 9.7% that 30,401 = 2,948.897 Sep 29 11:40 UTC (GMT) 0% of 131.87 = 0 Sep 29 11:40 UTC (GMT) 11% of 1,921 = 211.31 Sep 29 11:40 UTC (GMT) 32% the 4,072 = 1,303.04 Sep 29 11:40 UTC (GMT) 62% that 2.024 = 1.25488 Sep 29 11:40 UTC (GMT) 27,693% that 100 = 27,693 Sep 29 11:40 UTC (GMT) All individuals calculated numbers rememberingsomer.com
Percent
Percent is one hundredth that a provided amount, is an lot calculated with reference to a hundred, hundredth, 1 ÷ 100 = 1/100, ie. A portion with molecule equal to 1 and denominator come one hundred.The word comes from the Latin "per Centum", an interpretation "by the hundred". The latin word "Centum" way "100", for instance a century is 100 years. So, "percent" method "per 100".Percent way 1/100, 2 percent means 2/100, three percent means 3/100, and so on.Although a fraction, the percent wake up in writing without the denominator (100), yet only through the numerator, adhered to by the authorize %: 1 percent = 1%, 2 percent = 2%, 3 percent = 3%, and so on.
You are watching: What percent of 200 is 25
Percentage
Percentage is the "result of multiplying a amount by a specific percent".So 10 percent (10 ÷ 100 = 10/100 = 10%) out of 50 apples is 5 to apologize (10% that 50 = 10/100 × 50 = 500/100 = 5) - the 5 to apologize is the percentage.
When carry out we to speak percent and when percentage?
The indigenous percent (or the prize %) accompanies a specific number: roughly 60 percent (60%) of the people voted because that a change.The an ext general word portion is used without a number: the percentage of the human being that voted because that a readjust was approximately 60 percent (60%); Correct: 60 percent (60%) that the people; Incorrect: 60 portion of the people; Correct: a higher percentage the voters; Incorrect: a higher percent the voters; Note: In front of the word "percent" usage the number, don"t spell the out: "60 percent" is right, "sixty percent" is rather not. Always use figures for periods of civilization (He"s 43 years old), days (May 26), monetary quantities (\$80,000), rememberingsomer.com (60 percent) and ratios (2-to-3).
See more: How To Find Common Multiples Of 7 And 6, What Are The First 6 Multiples Of 7
rememberingsomer.com converted to decimal numbers:
1 percent, 1%, way 1 every 100, 1% = 1/100 = 0.01.50 percent, 50%, means 50 every 100, 50% = 50/100 = 1/2 (half) = 0.5.1 percent of 70 is: 1% of 70 = 1/100 × 70 = 7/10 = 0.750 percent that 70 is: 50% that 70 = 50/100 × 70 = 1/2 × 70 = 35. | 1,186 | 3,564 | {"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} | 3.734375 | 4 | CC-MAIN-2021-49 | latest | en | 0.842255 |
https://3enews.edublogs.org/2019/12/ | 1,716,914,235,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00714.warc.gz | 57,649,966 | 17,754 | # Our Week – December 20
3 E’s Inquiry Wonderland is Tuesday, January 14. Dress rehearsal from 2:15 – 2:45 and then in the evening from 5:30 to 6:15
Our week has been full of reading, writing, researching and math. Thank you for supporting our Secret Friend Project. The pieces of writing were magical and the gifts were truly thoughtful creations. It was so much fun to see the children, both give and receive their gifts.
Exploring Fractions
This month our calendar pattern is designed to help us discover more about fractions and equivalence. Through the pattern so far we have learned about halves, thirds, fourths, sixths and eighths. We’ve discovered how halves, fourths, and eights AND thirds and sixths help us understand equivalence. We’ve been reminded of what was learned last year, so we can build on that understanding.
The class has been learning a lot of about fractions in music with Mrs. Oliver too. They’ve been creating rhythms by combining whole, half, quarter and eighth notes. It’s not easy to do and takes a lot of concentration.
Global Geography
This week we’ve begun to learn about Earth’s diversity by exploring the continents. We’ve watched National Geographic’s Destination World videos of the continents and read a couple short books about each one. We’ve explored five of the continents so far. Part of this challenge has been listening and viewing carefully to remember the facts that are shared. Some of the children have developed interesting strategies for remembering. Others find this task to be a real challenge. It is a skill we are working to develop. Please encourage them if they need that boost.
Bits and Pieces –
We’ve completed most of the Humphrey book clubs. It fun to have those characters such a part of our class. Just before our Secret Friend exchange we read Humphrey’s gift giving tips from Winter According to Humphrey. He is indeed a wise hamster.
We’re about two thirds of the way through Mystery at Pine Lake. We are trying to figure out who put the board in the dam at Pine Lake. Cooper and Packrat have successfully created a nesting platform for the loons after the rising water flooded their fist nest. But will it be left alone? Who is responsible? We’ll find out after the break and begin a new round of mystery book clubs.
We’ve been exploring perimeters in math. It seems as though the children have a sound understanding of that concept. Soon we’ll be adding area.
We’ve also continued to work with rounding to the nearest 10 and nearest 100. It seems as though more of the children are clarifying their understanding. Sometimes rounding to the nearest 100 is still confusing.
# Our Week – December 13
It has been exciting to have a full week of school. It’s been a busy one. We chose a name and date for our classroom museum. We’re creating our museum displays. We’ve begun a study of global geography. We used this theme, combined with the “Choose Kind” initiative at NHS, to decorate our door – “3E is spreading kindness every step of the way.”
Our Inquiry Wonderland
This week we brainstormed a list of possible names for our museum. After a few days of voting for top fives, threes and then our final choice, we decided upon the name: Our Inquiry Wonderland. We also decided to add the by-line: Please, please come to our wonder-full museum to learn information you didn’t know before. It was an interesting process of compromise and negotiation.
We also chose the date and times for this event. There was no perfect time for everyone, but we hope that with advance notice, families might be able to rearrange scheduled events so that everyone is able to attend. The class decided they could all be ready to present the information on Tuesday, January 14. We’ll have our museum open that afternoon for a dress rehearsal and then again between 5:30 and 6:15. We hope all the children will be able to attend and that families will tour the Inquiry Wonderland, asking each of the children to share what they have learned. The children will share their research and talk about what inspired them to select their topic.
The goal of this project was to help children learn more about using a variety of information sources, to learn note-taking strategies and to be able to organize their information around subtopics to begin learning about paragraphing. In addition, they are learning to draft and revise their writing and to choose images that illustrate the main ideas.
When you come to the museum, you’ll see a wide variety of presentations. We’ve allowed and encouraged the children to work independently. We talked about final presentation and the need to attention to accuracy, details and conventions. Some children have been engaged throughout the whole process. They selected a topic they wanted to understand and learn more about. Others chose a topic they felt they already knew and did little new research. Some of the children have asked for teacher support for every step, while others have chosen to work with peer support. It is interesting to see how each child works through this learning process in his or her own unique way. There is a lot to attend to in a project like this and we will be able to see how they children learn from their choices when we move on to the next research project.
Global Geography and Mapping
This week we explored maps. We read about them. We learned that most maps have six basic elements: keys (or legends), a compass rose, a scale, a title, labels and symbols. We tried including all of the elements (excluding scale) to create maps of our yards. Next week we’ll be using our maps to lead us into some personal narratives telling about the places we play and the stories that grow from our games and imagination.
In addition to mapping we’ve been exploring the ways people divide up and label the places that we live. It is abstract and seems to feel random to the children. We live in neighborhoods, towns, counties, states, countries and continents. We are exploring continents, countries and climates. You may want to talk to your child about the poems and books we’ve been sharing. It is interesting to learn about our place in the world and it is interesting to understand how others live as well – the same, but different. Next month the children will be selecting a country to research and share with others.
Estimation and Rounding
We’ve been exploring the notion of rounding. We’ve been using a number line in the classroom to make sure we’re thinking about the even 100’s or 10’s our number is in between, before deciding to round down or up. Some children are confused about this, especially when rounding down. They are rounding to the 10 or 100 beyond. For example: 523 is being rounded down to 400, rather then 500. We’ll keep practicing. I’ve been asking the children to write the amounts their number is between before rounding. It takes time, and some of them are resistant, but it helps with understanding.
We’re using this skill with a Gram Challenge. In one of our math centers the children have been weighing common objects in the classroom. The children had an opportunity to weigh many different objects and to find combinations of objects that would weigh as close to 100 grams as possible. Now we have 7 days to find a combination of objects that weigh 1 kilogram. We began with a crayon. One crayon weighed 5 grams so Brady decided to place fifty crayons on the scale for an estimate of 250 grams. We’re excited to see what the next days bring.
Bits and Pieces –
• In Open Circle we’ve been discussing Positive Self Talk and how we can encourage ourselves to keep working through times of challenge and difficulty. We discussed the importance of staying positive and giving ourselves the opportunity to find a way through challenges.
• We’re thinking a lot about what is happening to the loons as we read The Mystery on Pine Lake. We’ve got a couple ideas about who is trying to get the loons to leave.
• Next week, we’ll complete all of the Humphrey book clubs. It has been a real commitment. The class deserves to feel proud of their work and their attention to details and summarizing. My guess is the Humphrey will stay a favorite character for quite some time.
• Next week is our Secret Friend Celebration. Please make sure the writing and artwork are wrapped according to the assignment directions and brought to school no later then Wednesday, December 18.
# Our Week – December 6
This will be a short note this week. As I write, we’ve only had two days in the classroom. We are continuing to learn about estimation, rounding. We’ve begun to learn about weighing with grams and fractions. We are continuing to research and are using our notes to create informational displays for a museum we hope to hold in early January. We are beginning to explore mapping and our world in a global geography unit.
Learning More About Place Value with Rounding
We’ve continued to explore of estimation. We hope to answer these questions: What is an estimate? Why do people estimate? When could we use estimation? Can we find examples where estimation is used outside of school?
As we make estimates, we will be learning more about place value and rounding up and down to the nearest ten and the nearest hundred. We’ve learned that to round you look at the digit to the right of the amount you are rounding and then use the rule – 5 and above round up, 4 and below round down. We’ll be developing this understanding over the next few weeks.
Exploring Fractions
This month our calendar pattern is designed to help us discover more about fractions and equivalence. Through the pattern so far we have learned about halves, thirds, and fourths. The class has been learning a lot of about fractions in music with Mrs. Oliver too. They’ve been creating rhythms by combining whole, half, quarter and eighth notes. It’s not easy to do and takes a lot of concentration.
SEL – Strengthening Relationships
In Open Circle we have been learning ways to strengthen relationships. We’ve been practicing giving and receiving compliments. We’ve been learning how to speak up for things we believe in or feel strongly about. We’ve been finding ways to share our opinions and to accept differences of opinion. Through role-playing activities we’ve learned the difference between advocating and telling – safety is a big part of our decision making process. We’ve practiced ways of being a supportive friend by encouraging and looking for the good in each person and each situation. We’ve spent time talking about what a cooperative classroom looks like, sounds like and feels like.
Bits and Pieces –
• We’ve learning about rubrics and we know what is expected for our inquiry project.
• We’ve finally discovered what the problem is in The Mystery of Pine Lake. We’ll be learning more about the elements of a mystery.
• We’re finishing our Humphrey book clubs. Three have finished and the last two will finish next week. Both the reading and the responsibility to a group have been challenging. The children deserve to feel proud of themselves for their dedication and effort.
• We were also able to finish a few blog posts this week. Thanks for checking them out and leaving comments. | 2,410 | 11,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-22 | latest | en | 0.93268 |
https://pandas.pydata.org/pandas-docs/version/0.23.4/computation.html | 1,721,224,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00852.warc.gz | 388,786,126 | 23,178 | Computational tools¶
Statistical Functions¶
Percent Change¶
Series, DataFrame, and Panel all have a method pct_change() to compute the percent change over a given number of periods (using fill_method to fill NA/null values before computing the percent change).
In [1]: ser = pd.Series(np.random.randn(8))
In [2]: ser.pct_change()
Out[2]:
0 NaN
1 -1.602976
2 4.334938
3 -0.247456
4 -2.067345
5 -1.142903
6 -1.688214
7 -9.759729
dtype: float64
In [3]: df = pd.DataFrame(np.random.randn(10, 4))
In [4]: df.pct_change(periods=3)
Out[4]:
0 1 2 3
0 NaN NaN NaN NaN
1 NaN NaN NaN NaN
2 NaN NaN NaN NaN
3 -0.218320 -1.054001 1.987147 -0.510183
4 -0.439121 -1.816454 0.649715 -4.822809
5 -0.127833 -3.042065 -5.866604 -1.776977
6 -2.596833 -1.959538 -2.111697 -3.798900
7 -0.117826 -2.169058 0.036094 -0.067696
8 2.492606 -1.357320 -1.205802 -1.558697
9 -1.012977 2.324558 -1.003744 -0.371806
Covariance¶
Series.cov() can be used to compute covariance between series (excluding missing values).
In [5]: s1 = pd.Series(np.random.randn(1000))
In [6]: s2 = pd.Series(np.random.randn(1000))
In [7]: s1.cov(s2)
Out[7]: 0.00068010881743108204
Analogously, DataFrame.cov() to compute pairwise covariances among the series in the DataFrame, also excluding NA/null values.
Note
Assuming the missing data are missing at random this results in an estimate for the covariance matrix which is unbiased. However, for many applications this estimate may not be acceptable because the estimated covariance matrix is not guaranteed to be positive semi-definite. This could lead to estimated correlations having absolute values which are greater than one, and/or a non-invertible covariance matrix. See Estimation of covariance matrices for more details.
In [8]: frame = pd.DataFrame(np.random.randn(1000, 5), columns=['a', 'b', 'c', 'd', 'e'])
In [9]: frame.cov()
Out[9]:
a b c d e
a 1.000882 -0.003177 -0.002698 -0.006889 0.031912
b -0.003177 1.024721 0.000191 0.009212 0.000857
c -0.002698 0.000191 0.950735 -0.031743 -0.005087
d -0.006889 0.009212 -0.031743 1.002983 -0.047952
e 0.031912 0.000857 -0.005087 -0.047952 1.042487
DataFrame.cov also supports an optional min_periods keyword that specifies the required minimum number of observations for each column pair in order to have a valid result.
In [10]: frame = pd.DataFrame(np.random.randn(20, 3), columns=['a', 'b', 'c'])
In [11]: frame.loc[frame.index[:5], 'a'] = np.nan
In [12]: frame.loc[frame.index[5:10], 'b'] = np.nan
In [13]: frame.cov()
Out[13]:
a b c
a 1.123670 -0.412851 0.018169
b -0.412851 1.154141 0.305260
c 0.018169 0.305260 1.301149
In [14]: frame.cov(min_periods=12)
Out[14]:
a b c
a 1.123670 NaN 0.018169
b NaN 1.154141 0.305260
c 0.018169 0.305260 1.301149
Correlation¶
Correlation may be computed using the corr() method. Using the method parameter, several methods for computing correlations are provided:
Method name Description
pearson (default) Standard correlation coefficient
kendall Kendall Tau correlation coefficient
spearman Spearman rank correlation coefficient
All of these are currently computed using pairwise complete observations. Wikipedia has articles covering the above correlation coefficients:
Note
Please see the caveats associated with this method of calculating correlation matrices in the covariance section.
In [15]: frame = pd.DataFrame(np.random.randn(1000, 5), columns=['a', 'b', 'c', 'd', 'e'])
In [16]: frame.iloc[::2] = np.nan
# Series with Series
In [17]: frame['a'].corr(frame['b'])
Out[17]: 0.013479040400098794
In [18]: frame['a'].corr(frame['b'], method='spearman')
Out[18]: -0.0072898851595406371
# Pairwise correlation of DataFrame columns
In [19]: frame.corr()
Out[19]:
a b c d e
a 1.000000 0.013479 -0.049269 -0.042239 -0.028525
b 0.013479 1.000000 -0.020433 -0.011139 0.005654
c -0.049269 -0.020433 1.000000 0.018587 -0.054269
d -0.042239 -0.011139 0.018587 1.000000 -0.017060
e -0.028525 0.005654 -0.054269 -0.017060 1.000000
Note that non-numeric columns will be automatically excluded from the correlation calculation.
Like cov, corr also supports the optional min_periods keyword:
In [20]: frame = pd.DataFrame(np.random.randn(20, 3), columns=['a', 'b', 'c'])
In [21]: frame.loc[frame.index[:5], 'a'] = np.nan
In [22]: frame.loc[frame.index[5:10], 'b'] = np.nan
In [23]: frame.corr()
Out[23]:
a b c
a 1.000000 -0.121111 0.069544
b -0.121111 1.000000 0.051742
c 0.069544 0.051742 1.000000
In [24]: frame.corr(min_periods=12)
Out[24]:
a b c
a 1.000000 NaN 0.069544
b NaN 1.000000 0.051742
c 0.069544 0.051742 1.000000
A related method corrwith() is implemented on DataFrame to compute the correlation between like-labeled Series contained in different DataFrame objects.
In [25]: index = ['a', 'b', 'c', 'd', 'e']
In [26]: columns = ['one', 'two', 'three', 'four']
In [27]: df1 = pd.DataFrame(np.random.randn(5, 4), index=index, columns=columns)
In [28]: df2 = pd.DataFrame(np.random.randn(4, 4), index=index[:4], columns=columns)
In [29]: df1.corrwith(df2)
Out[29]:
one -0.125501
two -0.493244
three 0.344056
four 0.004183
dtype: float64
In [30]: df2.corrwith(df1, axis=1)
Out[30]:
a -0.675817
b 0.458296
c 0.190809
d -0.186275
e NaN
dtype: float64
Data ranking¶
The rank() method produces a data ranking with ties being assigned the mean of the ranks (by default) for the group:
In [31]: s = pd.Series(np.random.np.random.randn(5), index=list('abcde'))
In [32]: s['d'] = s['b'] # so there's a tie
In [33]: s.rank()
Out[33]:
a 5.0
b 2.5
c 1.0
d 2.5
e 4.0
dtype: float64
rank() is also a DataFrame method and can rank either the rows (axis=0) or the columns (axis=1). NaN values are excluded from the ranking.
In [34]: df = pd.DataFrame(np.random.np.random.randn(10, 6))
In [35]: df[4] = df[2][:5] # some ties
In [36]: df
Out[36]:
0 1 2 3 4 5
0 -0.904948 -1.163537 -1.457187 0.135463 -1.457187 0.294650
1 -0.976288 -0.244652 -0.748406 -0.999601 -0.748406 -0.800809
2 0.401965 1.460840 1.256057 1.308127 1.256057 0.876004
3 0.205954 0.369552 -0.669304 0.038378 -0.669304 1.140296
4 -0.477586 -0.730705 -1.129149 -0.601463 -1.129149 -0.211196
5 -1.092970 -0.689246 0.908114 0.204848 NaN 0.463347
6 0.376892 0.959292 0.095572 -0.593740 NaN -0.069180
7 -1.002601 1.957794 -0.120708 0.094214 NaN -1.467422
8 -0.547231 0.664402 -0.519424 -0.073254 NaN -1.263544
9 -0.250277 -0.237428 -1.056443 0.419477 NaN 1.375064
In [37]: df.rank(1)
Out[37]:
0 1 2 3 4 5
0 4.0 3.0 1.5 5.0 1.5 6.0
1 2.0 6.0 4.5 1.0 4.5 3.0
2 1.0 6.0 3.5 5.0 3.5 2.0
3 4.0 5.0 1.5 3.0 1.5 6.0
4 5.0 3.0 1.5 4.0 1.5 6.0
5 1.0 2.0 5.0 3.0 NaN 4.0
6 4.0 5.0 3.0 1.0 NaN 2.0
7 2.0 5.0 3.0 4.0 NaN 1.0
8 2.0 5.0 3.0 4.0 NaN 1.0
9 2.0 3.0 1.0 4.0 NaN 5.0
rank optionally takes a parameter ascending which by default is true; when false, data is reverse-ranked, with larger values assigned a smaller rank.
rank supports different tie-breaking methods, specified with the method parameter:
• average : average rank of tied group
• min : lowest rank in the group
• max : highest rank in the group
• first : ranks assigned in the order they appear in the array
Window Functions¶
For working with data, a number of window functions are provided for computing common window or rolling statistics. Among these are count, sum, mean, median, correlation, variance, covariance, standard deviation, skewness, and kurtosis.
The rolling() and expanding() functions can be used directly from DataFrameGroupBy objects, see the groupby docs.
Note
The API for window statistics is quite similar to the way one works with GroupBy objects, see the documentation here.
We work with rolling, expanding and exponentially weighted data through the corresponding objects, Rolling, Expanding and EWM.
In [38]: s = pd.Series(np.random.randn(1000), index=pd.date_range('1/1/2000', periods=1000))
In [39]: s = s.cumsum()
In [40]: s
Out[40]:
2000-01-01 -0.268824
2000-01-02 -1.771855
2000-01-03 -0.818003
2000-01-04 -0.659244
2000-01-05 -1.942133
2000-01-06 -1.869391
2000-01-07 0.563674
...
2002-09-20 -68.233054
2002-09-21 -66.765687
2002-09-22 -67.457323
2002-09-23 -69.253182
2002-09-24 -70.296818
2002-09-25 -70.844674
2002-09-26 -72.475016
Freq: D, Length: 1000, dtype: float64
These are created from methods on Series and DataFrame.
In [41]: r = s.rolling(window=60)
In [42]: r
Out[42]: Rolling [window=60,center=False,axis=0]
These object provide tab-completion of the available methods and properties.
In [14]: r.
r.agg r.apply r.count r.exclusions r.max r.median r.name r.skew r.sum
r.aggregate r.corr r.cov r.kurt r.mean r.min r.quantile r.std r.var
Generally these methods all have the same interface. They all accept the following arguments:
• window: size of moving window
• min_periods: threshold of non-null data points to require (otherwise result is NA)
• center: boolean, whether to set the labels at the center (default is False)
We can then call methods on these rolling objects. These return like-indexed objects:
In [43]: r.mean()
Out[43]:
2000-01-01 NaN
2000-01-02 NaN
2000-01-03 NaN
2000-01-04 NaN
2000-01-05 NaN
2000-01-06 NaN
2000-01-07 NaN
...
2002-09-20 -62.694135
2002-09-21 -62.812190
2002-09-22 -62.914971
2002-09-23 -63.061867
2002-09-24 -63.213876
2002-09-25 -63.375074
2002-09-26 -63.539734
Freq: D, Length: 1000, dtype: float64
In [44]: s.plot(style='k--')
Out[44]: <matplotlib.axes._subplots.AxesSubplot at 0x7f2115c02ba8>
In [45]: r.mean().plot(style='k')
Out[45]: <matplotlib.axes._subplots.AxesSubplot at 0x7f2115c02ba8>
They can also be applied to DataFrame objects. This is really just syntactic sugar for applying the moving window operator to all of the DataFrame’s columns:
In [46]: df = pd.DataFrame(np.random.randn(1000, 4),
....: index=pd.date_range('1/1/2000', periods=1000),
....: columns=['A', 'B', 'C', 'D'])
....:
In [47]: df = df.cumsum()
In [48]: df.rolling(window=60).sum().plot(subplots=True)
Out[48]:
array([<matplotlib.axes._subplots.AxesSubplot object at 0x7f21156c40f0>,
<matplotlib.axes._subplots.AxesSubplot object at 0x7f2115662ef0>,
<matplotlib.axes._subplots.AxesSubplot object at 0x7f21156950f0>,
<matplotlib.axes._subplots.AxesSubplot object at 0x7f211563d2b0>], dtype=object)
Method Summary¶
We provide a number of common statistical functions:
Method Description
count() Number of non-null observations
sum() Sum of values
mean() Mean of values
median() Arithmetic median of values
min() Minimum
max() Maximum
std() Bessel-corrected sample standard deviation
var() Unbiased variance
skew() Sample skewness (3rd moment)
kurt() Sample kurtosis (4th moment)
quantile() Sample quantile (value at %)
apply() Generic apply
cov() Unbiased covariance (binary)
corr() Correlation (binary)
The apply() function takes an extra func argument and performs generic rolling computations. The func argument should be a single function that produces a single value from an ndarray input. Suppose we wanted to compute the mean absolute deviation on a rolling basis:
In [49]: mad = lambda x: np.fabs(x - x.mean()).mean()
Out[50]: <matplotlib.axes._subplots.AxesSubplot at 0x7f21153d6ef0>
Rolling Windows¶
Passing win_type to .rolling generates a generic rolling window computation, that is weighted according the win_type. The following methods are available:
Method Description
sum() Sum of values
mean() Mean of values
The weights used in the window are specified by the win_type keyword. The list of recognized types are the scipy.signal window functions:
• boxcar
• triang
• blackman
• hamming
• bartlett
• parzen
• bohman
• blackmanharris
• nuttall
• barthann
• kaiser (needs beta)
• gaussian (needs std)
• general_gaussian (needs power, width)
• slepian (needs width).
In [51]: ser = pd.Series(np.random.randn(10), index=pd.date_range('1/1/2000', periods=10))
In [52]: ser.rolling(window=5, win_type='triang').mean()
Out[52]:
2000-01-01 NaN
2000-01-02 NaN
2000-01-03 NaN
2000-01-04 NaN
2000-01-05 -1.037870
2000-01-06 -0.767705
2000-01-07 -0.383197
2000-01-08 -0.395513
2000-01-09 -0.558440
2000-01-10 -0.672416
Freq: D, dtype: float64
Note that the boxcar window is equivalent to mean().
In [53]: ser.rolling(window=5, win_type='boxcar').mean()
Out[53]:
2000-01-01 NaN
2000-01-02 NaN
2000-01-03 NaN
2000-01-04 NaN
2000-01-05 -0.841164
2000-01-06 -0.779948
2000-01-07 -0.565487
2000-01-08 -0.502815
2000-01-09 -0.553755
2000-01-10 -0.472211
Freq: D, dtype: float64
In [54]: ser.rolling(window=5).mean()
Out[54]:
2000-01-01 NaN
2000-01-02 NaN
2000-01-03 NaN
2000-01-04 NaN
2000-01-05 -0.841164
2000-01-06 -0.779948
2000-01-07 -0.565487
2000-01-08 -0.502815
2000-01-09 -0.553755
2000-01-10 -0.472211
Freq: D, dtype: float64
For some windowing functions, additional parameters must be specified:
In [55]: ser.rolling(window=5, win_type='gaussian').mean(std=0.1)
Out[55]:
2000-01-01 NaN
2000-01-02 NaN
2000-01-03 NaN
2000-01-04 NaN
2000-01-05 -1.309989
2000-01-06 -1.153000
2000-01-07 0.606382
2000-01-08 -0.681101
2000-01-09 -0.289724
2000-01-10 -0.996632
Freq: D, dtype: float64
Note
For .sum() with a win_type, there is no normalization done to the weights for the window. Passing custom weights of [1, 1, 1] will yield a different result than passing weights of [2, 2, 2], for example. When passing a win_type instead of explicitly specifying the weights, the weights are already normalized so that the largest weight is 1.
In contrast, the nature of the .mean() calculation is such that the weights are normalized with respect to each other. Weights of [1, 1, 1] and [2, 2, 2] yield the same result.
Time-aware Rolling¶
New in version 0.19.0.
New in version 0.19.0 are the ability to pass an offset (or convertible) to a .rolling() method and have it produce variable sized windows based on the passed time window. For each time point, this includes all preceding values occurring within the indicated time delta.
This can be particularly useful for a non-regular time frequency index.
In [56]: dft = pd.DataFrame({'B': [0, 1, 2, np.nan, 4]},
....: index=pd.date_range('20130101 09:00:00', periods=5, freq='s'))
....:
In [57]: dft
Out[57]:
B
2013-01-01 09:00:00 0.0
2013-01-01 09:00:01 1.0
2013-01-01 09:00:02 2.0
2013-01-01 09:00:03 NaN
2013-01-01 09:00:04 4.0
This is a regular frequency index. Using an integer window parameter works to roll along the window frequency.
In [58]: dft.rolling(2).sum()
Out[58]:
B
2013-01-01 09:00:00 NaN
2013-01-01 09:00:01 1.0
2013-01-01 09:00:02 3.0
2013-01-01 09:00:03 NaN
2013-01-01 09:00:04 NaN
In [59]: dft.rolling(2, min_periods=1).sum()
Out[59]:
B
2013-01-01 09:00:00 0.0
2013-01-01 09:00:01 1.0
2013-01-01 09:00:02 3.0
2013-01-01 09:00:03 2.0
2013-01-01 09:00:04 4.0
Specifying an offset allows a more intuitive specification of the rolling frequency.
In [60]: dft.rolling('2s').sum()
Out[60]:
B
2013-01-01 09:00:00 0.0
2013-01-01 09:00:01 1.0
2013-01-01 09:00:02 3.0
2013-01-01 09:00:03 2.0
2013-01-01 09:00:04 4.0
Using a non-regular, but still monotonic index, rolling with an integer window does not impart any special calculation.
In [61]: dft = pd.DataFrame({'B': [0, 1, 2, np.nan, 4]},
....: index = pd.Index([pd.Timestamp('20130101 09:00:00'),
....: pd.Timestamp('20130101 09:00:02'),
....: pd.Timestamp('20130101 09:00:03'),
....: pd.Timestamp('20130101 09:00:05'),
....: pd.Timestamp('20130101 09:00:06')],
....: name='foo'))
....:
In [62]: dft
Out[62]:
B
foo
2013-01-01 09:00:00 0.0
2013-01-01 09:00:02 1.0
2013-01-01 09:00:03 2.0
2013-01-01 09:00:05 NaN
2013-01-01 09:00:06 4.0
In [63]: dft.rolling(2).sum()
Out[63]:
B
foo
2013-01-01 09:00:00 NaN
2013-01-01 09:00:02 1.0
2013-01-01 09:00:03 3.0
2013-01-01 09:00:05 NaN
2013-01-01 09:00:06 NaN
Using the time-specification generates variable windows for this sparse data.
In [64]: dft.rolling('2s').sum()
Out[64]:
B
foo
2013-01-01 09:00:00 0.0
2013-01-01 09:00:02 1.0
2013-01-01 09:00:03 3.0
2013-01-01 09:00:05 NaN
2013-01-01 09:00:06 4.0
Furthermore, we now allow an optional on parameter to specify a column (rather than the default of the index) in a DataFrame.
In [65]: dft = dft.reset_index()
In [66]: dft
Out[66]:
foo B
0 2013-01-01 09:00:00 0.0
1 2013-01-01 09:00:02 1.0
2 2013-01-01 09:00:03 2.0
3 2013-01-01 09:00:05 NaN
4 2013-01-01 09:00:06 4.0
In [67]: dft.rolling('2s', on='foo').sum()
Out[67]:
foo B
0 2013-01-01 09:00:00 0.0
1 2013-01-01 09:00:02 1.0
2 2013-01-01 09:00:03 3.0
3 2013-01-01 09:00:05 NaN
4 2013-01-01 09:00:06 4.0
Rolling Window Endpoints¶
New in version 0.20.0.
The inclusion of the interval endpoints in rolling window calculations can be specified with the closed parameter:
closed Description Default for
right close right endpoint time-based windows
left close left endpoint
both close both endpoints fixed windows
neither open endpoints
For example, having the right endpoint open is useful in many problems that require that there is no contamination from present information back to past information. This allows the rolling window to compute statistics “up to that point in time”, but not including that point in time.
In [68]: df = pd.DataFrame({'x': 1},
....: index = [pd.Timestamp('20130101 09:00:01'),
....: pd.Timestamp('20130101 09:00:02'),
....: pd.Timestamp('20130101 09:00:03'),
....: pd.Timestamp('20130101 09:00:04'),
....: pd.Timestamp('20130101 09:00:06')])
....:
In [69]: df["right"] = df.rolling('2s', closed='right').x.sum() # default
In [70]: df["both"] = df.rolling('2s', closed='both').x.sum()
In [71]: df["left"] = df.rolling('2s', closed='left').x.sum()
In [72]: df["neither"] = df.rolling('2s', closed='neither').x.sum()
In [73]: df
Out[73]:
x right both left neither
2013-01-01 09:00:01 1 1.0 1.0 NaN NaN
2013-01-01 09:00:02 1 2.0 2.0 1.0 1.0
2013-01-01 09:00:03 1 2.0 3.0 2.0 1.0
2013-01-01 09:00:04 1 2.0 3.0 2.0 1.0
2013-01-01 09:00:06 1 1.0 2.0 1.0 NaN
Currently, this feature is only implemented for time-based windows. For fixed windows, the closed parameter cannot be set and the rolling window will always have both endpoints closed.
Time-aware Rolling vs. Resampling¶
Using .rolling() with a time-based index is quite similar to resampling. They both operate and perform reductive operations on time-indexed pandas objects.
When using .rolling() with an offset. The offset is a time-delta. Take a backwards-in-time looking window, and aggregate all of the values in that window (including the end-point, but not the start-point). This is the new value at that point in the result. These are variable sized windows in time-space for each point of the input. You will get a same sized result as the input.
When using .resample() with an offset. Construct a new index that is the frequency of the offset. For each frequency bin, aggregate points from the input within a backwards-in-time looking window that fall in that bin. The result of this aggregation is the output for that frequency point. The windows are fixed size in the frequency space. Your result will have the shape of a regular frequency between the min and the max of the original input object.
To summarize, .rolling() is a time-based window operation, while .resample() is a frequency-based window operation.
Centering Windows¶
By default the labels are set to the right edge of the window, but a center keyword is available so the labels can be set at the center.
In [74]: ser.rolling(window=5).mean()
Out[74]:
2000-01-01 NaN
2000-01-02 NaN
2000-01-03 NaN
2000-01-04 NaN
2000-01-05 -0.841164
2000-01-06 -0.779948
2000-01-07 -0.565487
2000-01-08 -0.502815
2000-01-09 -0.553755
2000-01-10 -0.472211
Freq: D, dtype: float64
In [75]: ser.rolling(window=5, center=True).mean()
Out[75]:
2000-01-01 NaN
2000-01-02 NaN
2000-01-03 -0.841164
2000-01-04 -0.779948
2000-01-05 -0.565487
2000-01-06 -0.502815
2000-01-07 -0.553755
2000-01-08 -0.472211
2000-01-09 NaN
2000-01-10 NaN
Freq: D, dtype: float64
Binary Window Functions¶
cov() and corr() can compute moving window statistics about two Series or any combination of DataFrame/Series or DataFrame/DataFrame. Here is the behavior in each case:
• two Series: compute the statistic for the pairing.
• DataFrame/Series: compute the statistics for each column of the DataFrame with the passed Series, thus returning a DataFrame.
• DataFrame/DataFrame: by default compute the statistic for matching column names, returning a DataFrame. If the keyword argument pairwise=True is passed then computes the statistic for each pair of columns, returning a MultiIndexed DataFrame whose index are the dates in question (see the next section).
For example:
In [76]: df = pd.DataFrame(np.random.randn(1000, 4),
....: index=pd.date_range('1/1/2000', periods=1000),
....: columns=['A', 'B', 'C', 'D'])
....:
In [77]: df = df.cumsum()
In [78]: df2 = df[:20]
In [79]: df2.rolling(window=5).corr(df2['B'])
Out[79]:
A B C D
2000-01-01 NaN NaN NaN NaN
2000-01-02 NaN NaN NaN NaN
2000-01-03 NaN NaN NaN NaN
2000-01-04 NaN NaN NaN NaN
2000-01-05 0.768775 1.0 -0.977990 0.800252
2000-01-06 0.744106 1.0 -0.967912 0.830021
2000-01-07 0.683257 1.0 -0.928969 0.384916
... ... ... ... ...
2000-01-14 -0.392318 1.0 0.570240 -0.591056
2000-01-15 0.017217 1.0 0.649900 -0.896258
2000-01-16 0.691078 1.0 0.807450 -0.939302
2000-01-17 0.274506 1.0 0.582601 -0.902954
2000-01-18 0.330459 1.0 0.515707 -0.545268
2000-01-19 0.046756 1.0 -0.104334 -0.419799
2000-01-20 -0.328241 1.0 -0.650974 -0.777777
[20 rows x 4 columns]
Computing rolling pairwise covariances and correlations¶
Warning
Prior to version 0.20.0 if pairwise=True was passed, a Panel would be returned. This will now return a 2-level MultiIndexed DataFrame, see the whatsnew here.
In financial data analysis and other fields it’s common to compute covariance and correlation matrices for a collection of time series. Often one is also interested in moving-window covariance and correlation matrices. This can be done by passing the pairwise keyword argument, which in the case of DataFrame inputs will yield a MultiIndexed DataFrame whose index are the dates in question. In the case of a single DataFrame argument the pairwise argument can even be omitted:
Note
Missing values are ignored and each entry is computed using the pairwise complete observations. Please see the covariance section for caveats associated with this method of calculating covariance and correlation matrices.
In [80]: covs = df[['B','C','D']].rolling(window=50).cov(df[['A','B','C']], pairwise=True)
In [81]: covs.loc['2002-09-22':]
Out[81]:
B C D
2002-09-22 A 1.367467 8.676734 -8.047366
B 3.067315 0.865946 -1.052533
C 0.865946 7.739761 -4.943924
2002-09-23 A 0.910343 8.669065 -8.443062
B 2.625456 0.565152 -0.907654
C 0.565152 7.825521 -5.367526
2002-09-24 A 0.463332 8.514509 -8.776514
B 2.306695 0.267746 -0.732186
C 0.267746 7.771425 -5.696962
2002-09-25 A 0.467976 8.198236 -9.162599
B 2.307129 0.267287 -0.754080
C 0.267287 7.466559 -5.822650
2002-09-26 A 0.545781 7.899084 -9.326238
B 2.311058 0.322295 -0.844451
C 0.322295 7.038237 -5.684445
In [82]: correls = df.rolling(window=50).corr()
In [83]: correls.loc['2002-09-22':]
Out[83]:
A B C D
2002-09-22 A 1.000000 0.186397 0.744551 -0.769767
B 0.186397 1.000000 0.177725 -0.240802
C 0.744551 0.177725 1.000000 -0.712051
D -0.769767 -0.240802 -0.712051 1.000000
2002-09-23 A 1.000000 0.134723 0.743113 -0.758758
B 0.134723 1.000000 0.124683 -0.209934
C 0.743113 0.124683 1.000000 -0.719088
... ... ... ... ...
2002-09-25 B 0.075157 1.000000 0.064399 -0.164179
C 0.731888 0.064399 1.000000 -0.704686
D -0.739160 -0.164179 -0.704686 1.000000
2002-09-26 A 1.000000 0.087756 0.727792 -0.736562
B 0.087756 1.000000 0.079913 -0.179477
C 0.727792 0.079913 1.000000 -0.692303
D -0.736562 -0.179477 -0.692303 1.000000
[20 rows x 4 columns]
You can efficiently retrieve the time series of correlations between two columns by reshaping and indexing:
In [84]: correls.unstack(1)[('A', 'C')].plot()
Out[84]: <matplotlib.axes._subplots.AxesSubplot at 0x7f210fd6a2b0>
Aggregation¶
Once the Rolling, Expanding or EWM objects have been created, several methods are available to perform multiple computations on the data. These operations are similar to the aggregating API, groupby API, and resample API.
In [85]: dfa = pd.DataFrame(np.random.randn(1000, 3),
....: index=pd.date_range('1/1/2000', periods=1000),
....: columns=['A', 'B', 'C'])
....:
In [86]: r = dfa.rolling(window=60,min_periods=1)
In [87]: r
Out[87]: Rolling [window=60,min_periods=1,center=False,axis=0]
We can aggregate by passing a function to the entire DataFrame, or select a Series (or multiple Series) via standard __getitem__.
In [88]: r.aggregate(np.sum)
Out[88]:
A B C
2000-01-01 -0.289838 -0.370545 -1.284206
2000-01-02 -0.216612 -1.675528 -1.169415
2000-01-03 1.154661 -1.634017 -1.566620
2000-01-04 2.969393 -4.003274 -1.816179
2000-01-05 4.690630 -4.682017 -2.717209
2000-01-06 3.880630 -4.447700 -1.078947
2000-01-07 4.001957 -2.884072 -3.116903
... ... ... ...
2002-09-20 2.652493 -10.528875 9.867805
2002-09-21 0.844497 -9.280944 9.522649
2002-09-22 2.860036 -9.270337 6.415245
2002-09-23 3.510163 -8.151439 5.177219
2002-09-24 6.524983 -10.168078 5.792639
2002-09-25 6.409626 -9.956226 5.704050
2002-09-26 5.093787 -7.074515 6.905823
[1000 rows x 3 columns]
In [89]: r['A'].aggregate(np.sum)
Out[89]:
2000-01-01 -0.289838
2000-01-02 -0.216612
2000-01-03 1.154661
2000-01-04 2.969393
2000-01-05 4.690630
2000-01-06 3.880630
2000-01-07 4.001957
...
2002-09-20 2.652493
2002-09-21 0.844497
2002-09-22 2.860036
2002-09-23 3.510163
2002-09-24 6.524983
2002-09-25 6.409626
2002-09-26 5.093787
Freq: D, Name: A, Length: 1000, dtype: float64
In [90]: r[['A','B']].aggregate(np.sum)
Out[90]:
A B
2000-01-01 -0.289838 -0.370545
2000-01-02 -0.216612 -1.675528
2000-01-03 1.154661 -1.634017
2000-01-04 2.969393 -4.003274
2000-01-05 4.690630 -4.682017
2000-01-06 3.880630 -4.447700
2000-01-07 4.001957 -2.884072
... ... ...
2002-09-20 2.652493 -10.528875
2002-09-21 0.844497 -9.280944
2002-09-22 2.860036 -9.270337
2002-09-23 3.510163 -8.151439
2002-09-24 6.524983 -10.168078
2002-09-25 6.409626 -9.956226
2002-09-26 5.093787 -7.074515
[1000 rows x 2 columns]
As you can see, the result of the aggregation will have the selected columns, or all columns if none are selected.
Applying multiple functions¶
With windowed Series you can also pass a list of functions to do aggregation with, outputting a DataFrame:
In [91]: r['A'].agg([np.sum, np.mean, np.std])
Out[91]:
sum mean std
2000-01-01 -0.289838 -0.289838 NaN
2000-01-02 -0.216612 -0.108306 0.256725
2000-01-03 1.154661 0.384887 0.873311
2000-01-04 2.969393 0.742348 1.009734
2000-01-05 4.690630 0.938126 0.977914
2000-01-06 3.880630 0.646772 1.128883
2000-01-07 4.001957 0.571708 1.049487
... ... ... ...
2002-09-20 2.652493 0.044208 1.164919
2002-09-21 0.844497 0.014075 1.148231
2002-09-22 2.860036 0.047667 1.132051
2002-09-23 3.510163 0.058503 1.134296
2002-09-24 6.524983 0.108750 1.144204
2002-09-25 6.409626 0.106827 1.142913
2002-09-26 5.093787 0.084896 1.151416
[1000 rows x 3 columns]
On a windowed DataFrame, you can pass a list of functions to apply to each column, which produces an aggregated result with a hierarchical index:
In [92]: r.agg([np.sum, np.mean])
Out[92]:
A B C
sum mean sum mean sum mean
2000-01-01 -0.289838 -0.289838 -0.370545 -0.370545 -1.284206 -1.284206
2000-01-02 -0.216612 -0.108306 -1.675528 -0.837764 -1.169415 -0.584708
2000-01-03 1.154661 0.384887 -1.634017 -0.544672 -1.566620 -0.522207
2000-01-04 2.969393 0.742348 -4.003274 -1.000819 -1.816179 -0.454045
2000-01-05 4.690630 0.938126 -4.682017 -0.936403 -2.717209 -0.543442
2000-01-06 3.880630 0.646772 -4.447700 -0.741283 -1.078947 -0.179825
2000-01-07 4.001957 0.571708 -2.884072 -0.412010 -3.116903 -0.445272
... ... ... ... ... ... ...
2002-09-20 2.652493 0.044208 -10.528875 -0.175481 9.867805 0.164463
2002-09-21 0.844497 0.014075 -9.280944 -0.154682 9.522649 0.158711
2002-09-22 2.860036 0.047667 -9.270337 -0.154506 6.415245 0.106921
2002-09-23 3.510163 0.058503 -8.151439 -0.135857 5.177219 0.086287
2002-09-24 6.524983 0.108750 -10.168078 -0.169468 5.792639 0.096544
2002-09-25 6.409626 0.106827 -9.956226 -0.165937 5.704050 0.095068
2002-09-26 5.093787 0.084896 -7.074515 -0.117909 6.905823 0.115097
[1000 rows x 6 columns]
Passing a dict of functions has different behavior by default, see the next section.
Applying different functions to DataFrame columns¶
By passing a dict to aggregate you can apply a different aggregation to the columns of a DataFrame:
In [93]: r.agg({'A' : np.sum,
....: 'B' : lambda x: np.std(x, ddof=1)})
....:
Out[93]:
A B
2000-01-01 -0.289838 NaN
2000-01-02 -0.216612 0.660747
2000-01-03 1.154661 0.689929
2000-01-04 2.969393 1.072199
2000-01-05 4.690630 0.939657
2000-01-06 3.880630 0.966848
2000-01-07 4.001957 1.240137
... ... ...
2002-09-20 2.652493 1.114814
2002-09-21 0.844497 1.113220
2002-09-22 2.860036 1.113208
2002-09-23 3.510163 1.132381
2002-09-24 6.524983 1.080963
2002-09-25 6.409626 1.082911
2002-09-26 5.093787 1.136199
[1000 rows x 2 columns]
The function names can also be strings. In order for a string to be valid it must be implemented on the windowed object
In [94]: r.agg({'A' : 'sum', 'B' : 'std'})
Out[94]:
A B
2000-01-01 -0.289838 NaN
2000-01-02 -0.216612 0.660747
2000-01-03 1.154661 0.689929
2000-01-04 2.969393 1.072199
2000-01-05 4.690630 0.939657
2000-01-06 3.880630 0.966848
2000-01-07 4.001957 1.240137
... ... ...
2002-09-20 2.652493 1.114814
2002-09-21 0.844497 1.113220
2002-09-22 2.860036 1.113208
2002-09-23 3.510163 1.132381
2002-09-24 6.524983 1.080963
2002-09-25 6.409626 1.082911
2002-09-26 5.093787 1.136199
[1000 rows x 2 columns]
Furthermore you can pass a nested dict to indicate different aggregations on different columns.
In [95]: r.agg({'A' : ['sum','std'], 'B' : ['mean','std'] })
Out[95]:
A B
sum std mean std
2000-01-01 -0.289838 NaN -0.370545 NaN
2000-01-02 -0.216612 0.256725 -0.837764 0.660747
2000-01-03 1.154661 0.873311 -0.544672 0.689929
2000-01-04 2.969393 1.009734 -1.000819 1.072199
2000-01-05 4.690630 0.977914 -0.936403 0.939657
2000-01-06 3.880630 1.128883 -0.741283 0.966848
2000-01-07 4.001957 1.049487 -0.412010 1.240137
... ... ... ... ...
2002-09-20 2.652493 1.164919 -0.175481 1.114814
2002-09-21 0.844497 1.148231 -0.154682 1.113220
2002-09-22 2.860036 1.132051 -0.154506 1.113208
2002-09-23 3.510163 1.134296 -0.135857 1.132381
2002-09-24 6.524983 1.144204 -0.169468 1.080963
2002-09-25 6.409626 1.142913 -0.165937 1.082911
2002-09-26 5.093787 1.151416 -0.117909 1.136199
[1000 rows x 4 columns]
Expanding Windows¶
A common alternative to rolling statistics is to use an expanding window, which yields the value of the statistic with all the data available up to that point in time.
These follow a similar interface to .rolling, with the .expanding method returning an Expanding object.
As these calculations are a special case of rolling statistics, they are implemented in pandas such that the following two calls are equivalent:
In [96]: df.rolling(window=len(df), min_periods=1).mean()[:5]
Out[96]:
A B C D
2000-01-01 0.314226 -0.001675 0.071823 0.892566
2000-01-02 0.654522 -0.171495 0.179278 0.853361
2000-01-03 0.708733 -0.064489 -0.238271 1.371111
2000-01-04 0.987613 0.163472 -0.919693 1.566485
2000-01-05 1.426971 0.288267 -1.358877 1.808650
In [97]: df.expanding(min_periods=1).mean()[:5]
Out[97]:
A B C D
2000-01-01 0.314226 -0.001675 0.071823 0.892566
2000-01-02 0.654522 -0.171495 0.179278 0.853361
2000-01-03 0.708733 -0.064489 -0.238271 1.371111
2000-01-04 0.987613 0.163472 -0.919693 1.566485
2000-01-05 1.426971 0.288267 -1.358877 1.808650
These have a similar set of methods to .rolling methods.
Method Summary¶
Function Description
count() Number of non-null observations
sum() Sum of values
mean() Mean of values
median() Arithmetic median of values
min() Minimum
max() Maximum
std() Unbiased standard deviation
var() Unbiased variance
skew() Unbiased skewness (3rd moment)
kurt() Unbiased kurtosis (4th moment)
quantile() Sample quantile (value at %)
apply() Generic apply
cov() Unbiased covariance (binary)
corr() Correlation (binary)
Aside from not having a window parameter, these functions have the same interfaces as their .rolling counterparts. Like above, the parameters they all accept are:
• min_periods: threshold of non-null data points to require. Defaults to minimum needed to compute statistic. No NaNs will be output once min_periods non-null data points have been seen.
• center: boolean, whether to set the labels at the center (default is False).
Note
The output of the .rolling and .expanding methods do not return a NaN if there are at least min_periods non-null values in the current window. For example:
In [98]: sn = pd.Series([1, 2, np.nan, 3, np.nan, 4])
In [99]: sn
Out[99]:
0 1.0
1 2.0
2 NaN
3 3.0
4 NaN
5 4.0
dtype: float64
In [100]: sn.rolling(2).max()
Out[100]:
0 NaN
1 2.0
2 NaN
3 NaN
4 NaN
5 NaN
dtype: float64
In [101]: sn.rolling(2, min_periods=1).max()
Out[101]:
0 1.0
1 2.0
2 2.0
3 3.0
4 3.0
5 4.0
dtype: float64
In case of expanding functions, this differs from cumsum(), cumprod(), cummax(), and cummin(), which return NaN in the output wherever a NaN is encountered in the input. In order to match the output of cumsum with expanding, use fillna():
In [102]: sn.expanding().sum()
Out[102]:
0 1.0
1 3.0
2 3.0
3 6.0
4 6.0
5 10.0
dtype: float64
In [103]: sn.cumsum()
Out[103]:
0 1.0
1 3.0
2 NaN
3 6.0
4 NaN
5 10.0
dtype: float64
In [104]: sn.cumsum().fillna(method='ffill')
Out[104]:
0 1.0
1 3.0
2 3.0
3 6.0
4 6.0
5 10.0
dtype: float64
An expanding window statistic will be more stable (and less responsive) than its rolling window counterpart as the increasing window size decreases the relative impact of an individual data point. As an example, here is the mean() output for the previous time series dataset:
In [105]: s.plot(style='k--')
Out[105]: <matplotlib.axes._subplots.AxesSubplot at 0x7f210fc68518>
In [106]: s.expanding().mean().plot(style='k')
Out[106]: <matplotlib.axes._subplots.AxesSubplot at 0x7f210fc68518>
Exponentially Weighted Windows¶
A related set of functions are exponentially weighted versions of several of the above statistics. A similar interface to .rolling and .expanding is accessed through the .ewm method to receive an EWM object. A number of expanding EW (exponentially weighted) methods are provided:
Function Description
mean() EW moving average
var() EW moving variance
std() EW moving standard deviation
corr() EW moving correlation
cov() EW moving covariance
In general, a weighted moving average is calculated as
$y_t = \frac{\sum_{i=0}^t w_i x_{t-i}}{\sum_{i=0}^t w_i},$
where $$x_t$$ is the input, $$y_t$$ is the result and the $$w_i$$ are the weights.
The EW functions support two variants of exponential weights. The default, adjust=True, uses the weights $$w_i = (1 - \alpha)^i$$ which gives
$y_t = \frac{x_t + (1 - \alpha)x_{t-1} + (1 - \alpha)^2 x_{t-2} + ... + (1 - \alpha)^t x_{0}}{1 + (1 - \alpha) + (1 - \alpha)^2 + ... + (1 - \alpha)^t}$
When adjust=False is specified, moving averages are calculated as
$\begin{split}y_0 &= x_0 \\ y_t &= (1 - \alpha) y_{t-1} + \alpha x_t,\end{split}$
which is equivalent to using weights
$\begin{split}w_i = \begin{cases} \alpha (1 - \alpha)^i & \text{if } i < t \\ (1 - \alpha)^i & \text{if } i = t. \end{cases}\end{split}$
Note
These equations are sometimes written in terms of $$\alpha' = 1 - \alpha$$, e.g.
$y_t = \alpha' y_{t-1} + (1 - \alpha') x_t.$
The difference between the above two variants arises because we are dealing with series which have finite history. Consider a series of infinite history:
$y_t = \frac{x_t + (1 - \alpha)x_{t-1} + (1 - \alpha)^2 x_{t-2} + ...} {1 + (1 - \alpha) + (1 - \alpha)^2 + ...}$
Noting that the denominator is a geometric series with initial term equal to 1 and a ratio of $$1 - \alpha$$ we have
$\begin{split}y_t &= \frac{x_t + (1 - \alpha)x_{t-1} + (1 - \alpha)^2 x_{t-2} + ...} {\frac{1}{1 - (1 - \alpha)}}\\ &= [x_t + (1 - \alpha)x_{t-1} + (1 - \alpha)^2 x_{t-2} + ...] \alpha \\ &= \alpha x_t + [(1-\alpha)x_{t-1} + (1 - \alpha)^2 x_{t-2} + ...]\alpha \\ &= \alpha x_t + (1 - \alpha)[x_{t-1} + (1 - \alpha) x_{t-2} + ...]\alpha\\ &= \alpha x_t + (1 - \alpha) y_{t-1}\end{split}$
which shows the equivalence of the above two variants for infinite series. When adjust=True we have $$y_0 = x_0$$ and from the last representation above we have $$y_t = \alpha x_t + (1 - \alpha) y_{t-1}$$, therefore there is an assumption that $$x_0$$ is not an ordinary value but rather an exponentially weighted moment of the infinite series up to that point.
One must have $$0 < \alpha \leq 1$$, and while since version 0.18.0 it has been possible to pass $$\alpha$$ directly, it’s often easier to think about either the span, center of mass (com) or half-life of an EW moment:
$\begin{split}\alpha = \begin{cases} \frac{2}{s + 1}, & \text{for span}\ s \geq 1\\ \frac{1}{1 + c}, & \text{for center of mass}\ c \geq 0\\ 1 - \exp^{\frac{\log 0.5}{h}}, & \text{for half-life}\ h > 0 \end{cases}\end{split}$
One must specify precisely one of span, center of mass, half-life and alpha to the EW functions:
• Span corresponds to what is commonly called an “N-day EW moving average”.
• Center of mass has a more physical interpretation and can be thought of in terms of span: $$c = (s - 1) / 2$$.
• Half-life is the period of time for the exponential weight to reduce to one half.
• Alpha specifies the smoothing factor directly.
Here is an example for a univariate time series:
In [107]: s.plot(style='k--')
Out[107]: <matplotlib.axes._subplots.AxesSubplot at 0x7f210fbff908>
In [108]: s.ewm(span=20).mean().plot(style='k')
Out[108]: <matplotlib.axes._subplots.AxesSubplot at 0x7f210fbff908>
EWM has a min_periods argument, which has the same meaning it does for all the .expanding and .rolling methods: no output values will be set until at least min_periods non-null values are encountered in the (expanding) window.
EWM also has an ignore_na argument, which determines how intermediate null values affect the calculation of the weights. When ignore_na=False (the default), weights are calculated based on absolute positions, so that intermediate null values affect the result. When ignore_na=True, weights are calculated by ignoring intermediate null values. For example, assuming adjust=True, if ignore_na=False, the weighted average of 3, NaN, 5 would be calculated as
$\frac{(1-\alpha)^2 \cdot 3 + 1 \cdot 5}{(1-\alpha)^2 + 1}.$
Whereas if ignore_na=True, the weighted average would be calculated as
$\frac{(1-\alpha) \cdot 3 + 1 \cdot 5}{(1-\alpha) + 1}.$
The var(), std(), and cov() functions have a bias argument, specifying whether the result should contain biased or unbiased statistics. For example, if bias=True, ewmvar(x) is calculated as ewmvar(x) = ewma(x**2) - ewma(x)**2; whereas if bias=False (the default), the biased variance statistics are scaled by debiasing factors
$\frac{\left(\sum_{i=0}^t w_i\right)^2}{\left(\sum_{i=0}^t w_i\right)^2 - \sum_{i=0}^t w_i^2}.$
(For $$w_i = 1$$, this reduces to the usual $$N / (N - 1)$$ factor, with $$N = t + 1$$.) See Weighted Sample Variance on Wikipedia for further details.
Scroll To Top | 15,272 | 41,901 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-30 | latest | en | 0.599786 |
https://testbook.com/question-answer/if-principal-stresses-in-a-two-dimensional-case-ar--64b107d2f5f81f043868eca8 | 1,726,839,639,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00560.warc.gz | 494,274,766 | 45,681 | # If principal stresses in a two-dimensional case are - 8 MPa and 10 MPa, respectively, then maximum shear stress at the point is:
This question was previously asked in
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1. 14 MPa
2. 20 MPa
3. MPa
4. 18 MPa
## Answer (Detailed Solution Below)
Option 3 : 9 MPa
Free
JSSC JE Full Test 1 (Paper 1)
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## Detailed Solution
Concept:
The maximum shear stress is the half of the difference between the major and minor principal stresses, which can be observed from Mohr’s circle of stress.
σx’ + σy’ = σx + σy
$${{\rm{\sigma }}_{{\rm{avg}}}} = \frac{{{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}}}}{2}{\rm{\;and\;R}} = \sqrt {\left( {\frac{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}{2}} \right){\;^2} + {\rm{\tau }}_{{\rm{xy}}}^2}$$
σmax/min = σavg ± R
$$\tan 2{{\rm{\theta }}_{\rm{p}}} = \frac{{2{{\rm{\tau }}_{{\rm{xy}}}}}}{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}$$
$${{\rm{\tau }}_{{\rm{max}}}} = \frac{1}{2}\left| {{{\rm{\sigma }}_{{\rm{max}}}} - {{\rm{\sigma }}_{{\rm{min}}}}} \right|$$
Calculation:
Major and minor principal stresses on a two dimensional element are 10 MPa and - 8 MPa.
So maximum shear stress = 0.5 × [(10)-(-8)] = 9 MPa
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-> Candidates with a diploma in the concerned engineering branch are eligible for this post. Prepare for the exam with JSSC JE Previous Year Papers. | 655 | 1,968 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.735403 |
https://josmfs.net/2019/01/31/right-triangle-with-roots/ | 1,725,836,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651035.2/warc/CC-MAIN-20240908213138-20240909003138-00409.warc.gz | 308,409,415 | 16,188 | # Right Triangle with Roots
This is an interesting problem from the United Kingdom Mathematics Trust (UKMT) Senior Math Challenge of 2008.
“The length of the hypotenuse of a particular right-angled triangle is given by √(1 + 3 + 5 + … + 23 + 25). The lengths of the other two sides are given by √(1 + 3 + 5 + … + (x – 2) + x) and √ (1 + 3 + 5 + … + (y – 2) + y) where x and y are positive integers. What is the value of x + y?” | 135 | 429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.818733 |
https://sandialabs.github.io/Prove-It/packages/proveit/numbers/multiplication/__pv_it/theorems/c77882abfdeffd2ebb85e1b4d17eb66ec877f6a20/expr.html | 1,716,243,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00696.warc.gz | 442,527,948 | 5,580 | # from the theory of proveit.numbers.multiplication¶
In [1]:
import proveit
# Automation is not needed when building an expression:
proveit.defaults.automation = False # This will speed things up.
proveit.defaults.inline_pngs = False # Makes files smaller.
# import Expression classes needed to build the expression
from proveit import ExprTuple, Function, Q, f
from proveit.core_expr_types import a_1_to_i, b_1_to_j, c_1_to_k
from proveit.numbers import Mult, Sum
from proveit.numbers.summation import summation_b1toj_fQ
In [2]:
# build up the expression from sub-expressions
sub_expr1 = [b_1_to_j]
expr = ExprTuple(Mult(a_1_to_i, summation_b1toj_fQ, c_1_to_k), Sum(index_or_indices = sub_expr1, summand = Mult(a_1_to_i, Function(f, sub_expr1), c_1_to_k), condition = Function(Q, sub_expr1)))
expr:
In [3]:
# check that the built expression is the same as the stored expression
assert expr == stored_expr
assert expr._style_id == stored_expr._style_id
print("Passed sanity check: expr matches stored_expr")
Passed sanity check: expr matches stored_expr
In [4]:
# Show the LaTeX representation of the expression for convenience if you need it.
print(stored_expr.latex())
\left(a_{1} \cdot a_{2} \cdot \ldots \cdot a_{i} \cdot \left[\sum_{b_{1}, b_{2}, \ldots, b_{j}~|~Q\left(b_{1}, b_{2}, \ldots, b_{j}\right)}~f\left(b_{1}, b_{2}, \ldots, b_{j}\right)\right]\cdot c_{1} \cdot c_{2} \cdot \ldots \cdot c_{k}, \sum_{b_{1}, b_{2}, \ldots, b_{j}~|~Q\left(b_{1}, b_{2}, \ldots, b_{j}\right)}~\left(a_{1} \cdot a_{2} \cdot \ldots \cdot a_{i} \cdot f\left(b_{1}, b_{2}, \ldots, b_{j}\right)\cdot c_{1} \cdot c_{2} \cdot \ldots \cdot c_{k}\right)\right)
In [5]:
stored_expr.style_options()
namedescriptiondefaultcurrent valuerelated methods
wrap_positionsposition(s) at which wrapping is to occur; 'n' is after the nth comma.()()('with_wrapping_at',)
justificationif any wrap positions are set, justify to the 'left', 'center', or 'right'leftleft('with_justification',)
In [6]:
# display the expression information
stored_expr.expr_info()
core typesub-expressionsexpression
0ExprTuple1, 2
1Operationoperator: 13
operands: 3
2Operationoperator: 7
operand: 6
3ExprTuple16, 5, 18
4ExprTuple6
5Operationoperator: 7
operand: 10
6Lambdaparameters: 23
body: 9
7Literal
8ExprTuple10
9Conditionalvalue: 11
condition: 15
10Lambdaparameters: 23
body: 12
11Operationoperator: 13
operands: 14
12Conditionalvalue: 17
condition: 15
13Literal
14ExprTuple16, 17, 18
15Operationoperator: 19
operands: 23
16ExprRangelambda_map: 20
start_index: 31
end_index: 21
17Operationoperator: 22
operands: 23
18ExprRangelambda_map: 24
start_index: 31
end_index: 25
19Variable
20Lambdaparameter: 37
body: 26
21Variable
22Variable
23ExprTuple27
24Lambdaparameter: 37
body: 28
25Variable
26IndexedVarvariable: 29
index: 37
27ExprRangelambda_map: 30
start_index: 31
end_index: 32
28IndexedVarvariable: 33
index: 37
29Variable
30Lambdaparameter: 37
body: 34
31Literal
32Variable
33Variable
34IndexedVarvariable: 35
index: 37
35Variable
36ExprTuple37
37Variable | 1,024 | 3,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.461791 |
https://mathematica.stackexchange.com/questions/98445/region-bounded-by-x2y2-1-y-z-x-0-z-0-in-first-octant/98446 | 1,660,003,676,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00456.warc.gz | 373,068,553 | 69,219 | # Region bounded by x^2+y^2=1, y=z, x=0, z=0, in first octant
I need to draw (pencil and paper) the region bounded by $x^2+y^2=1$, $y=z$, $x=0$, and $z=0$ in the first octant. So the first assistance I asked of Mathematica is:
ContourPlot3D[{x^2 + y^2 == 1, y == z, x == 0, z == 0}, {x, 0, 1}, {y,
0, 1}, {z, 0, 1},
ContourStyle -> Opacity[0.5],
AxesLabel -> {"x", "y", "z"},
ViewPoint -> {3, -0.5, 1.5}]
Which gave me this image:
I was then able to draw the image via pencil and paper. Then I thought I'd try RegionFunction.
ContourPlot3D[{x^2 + y^2 == 1, y == z, x == 0, z == 0}, {x, 0, 1}, {y,
0, 1}, {z, 0, 1},
RegionFunction -> Function[{x, y, z}, y <= Sqrt[1 - x^2] && z <= y],
ContourStyle -> Opacity[0.5],
AxesLabel -> {"x", "y", "z"},
ViewPoint -> {3, -0.5, 1.5}]
Which gave me this image.
I was able to repair it by extending my inequalities a bit.
ContourPlot3D[{x^2 + y^2 == 1, y == z, x == 0, z == 0}, {x, 0, 1}, {y,
0, 1}, {z, 0, 1},
RegionFunction ->
Function[{x, y, z}, y <= Sqrt[1 - x^2] + 0.001 && z <= y + 0.001],
ContourStyle -> Opacity[0.5],
AxesLabel -> {"x", "y", "z"},
ViewPoint -> {3, -0.5, 1.5}]
Which gave me this image.
Now, I am aware of RegionPlot3D, but I am not fond of the images it produces, although it is an easy method to get a quick idea of what the image looks like. So, I started trying a little ParametricPlot3D.
Show[
Plot3D[y, {x, 0, 1}, {y, 0, Sqrt[1 - x^2]},
AxesLabel -> {"x", "y", "z"}],
ParametricPlot3D[{x, Sqrt[1 - x^2], z}, {x, 0, 1}, {z, 0,
Sqrt[1 - x^2]},
PlotStyle -> {LightBlue, Opacity[0.8]}],
ViewPoint -> {3, -0.5, 1.5}
]
Which gave me a little bit of strangeness. See the little sudden dipping in the border of the blue side as it approaches the x-axis?
I tried my contour inequality (adding 0.001 here an there) extension approach in several ways, but I could not get it to disappear. Any thoughts?
Update: There is some extremely wonderful work on this page, but I'd also like to add a cylindrical plot based on MichaelE2's suggestion.
Show[ParametricPlot3D[{u Cos[t], u Sin[t], u Sin[t]}, {t, 0, Pi}, {u,
0, 1},
AxesLabel -> {"x", "y", "z"}],
ParametricPlot3D[{Cos[t], Sin[t], u Sin[t]}, {t, 0, Pi/2}, {u, 0, 1},
PlotStyle -> {LightBlue, Opacity[0.8]}],
ViewPoint -> {3, -0.5, 1.5}]
• Seen this? Nov 2, 2015 at 5:57
• Note one can do better in cylindrical coordinates and especially with direct construction of the surfaces. But judging from previous posts, I think the OP would prefer solutions that preserve the equations of the surfaces in cartesian coordinates -- Is that right, David? Nov 2, 2015 at 12:28
• @MichaelE2 Yes, cylindrical would be OK. Nov 2, 2015 at 16:15
A simple alternative is to use Plot3D with both RegionFunction and Filling.
Plot3D[y, {x, 0, 1}, {y, 0, 1},
RegionFunction ->
Function[{x, y, z},
x^2 + y^2 <= 1 && x >= 0 && y >= 0 && z >= 0],
Filling -> 0,
FillingStyle -> Opacity[.75],
PlotStyle -> Opacity[.5],
AxesLabel -> (Style[#, 14, Bold] & /@ {x, y, z}),
BoxRatios -> {1, 1, 1},
ViewPoint -> {3, -1.5, 0.75}]
EDIT: I recommend that you experiment with different settings for PlotTheme to determine which is best for your classroom and smartboard.
Manipulate[
Plot3D[y, {x, 0, 1}, {y, 0, 1},
RegionFunction ->
Function[{x, y, z},
x^2 + y^2 <= 1 && x >= 0 && y >= 0 && z >= 0],
Filling -> 0,
FillingStyle -> Opacity[.75],
PlotStyle -> Opacity[.5],
AxesLabel -> (Style[#, 18, Bold] & /@
{x, y, z}),
BoxRatios -> {1, 1, 1},
ViewPoint -> {3, -1.5, 0.75},
PlotTheme -> pt],
{{pt, "Classic", "Plot Theme"},
"Detailed", "Marketing", "Minimal",
"Monochrome", "Scientific", "Web"}}]
• Very nice, the easiest for the students to understand. I have another question. We use Smartboards in our classrooms to display our Mathematica images. What type of coloring in this example would make this image the easiest to see and interpret for students sitting in the back of the room and viewing the presentation on the Smartboard? Nov 2, 2015 at 16:18
• Really nice going. I am going to give this a test today. Nov 2, 2015 at 21:06
As pointed out by J. M.♦, Simon Woods's approach in #48486 could be used.
sharpregplot[
region_,
{x_, x0_, x1_},
{y_, y0_, y1_},
{z_, z0_, z1_},
opts : OptionsPattern[]
] := Module[
{reg, preds},
reg = LogicalExpand[region && x0 <= x <= x1 && y0 <= y <= y1 && z0 <= z <= z1];
preds = Union@Cases[reg, _Greater | _GreaterEqual | _Less | _LessEqual, -1];
Show @ Table[
ContourPlot3D[
Evaluate[Equal @@ p],
{x, x0, x1},
{y, y0, y1},
{z, z0, z1},
RegionFunction -> Function @@ {{x, y, z}, Refine[reg, p] && Refine[! reg, ! p]},
opts
],
{p, preds}
]
]
Then,
sharpregplot[
y^2 <= 1 - x^2 && z <= y,
{x, 0, 1}, {y, 0, 1}, {z, 0, 1},
AxesLabel -> {"x", "y", "z"},
BoundaryStyle -> None,
ContourStyle -> RandomColor[],
Mesh -> None,
ViewPoint -> 1000 {3, -0.5, 1.5}
]
gives
Here's one approach that uses MeshFunctions to highlight the parts of the bounding surfaces that belong to the region. So many different approaches are possible....
opts = Options[ParametricPlot3D];
SetOptions[ParametricPlot3D,
{Mesh -> {{0}, 15, 15},
MeshStyle -> Opacity[0.], (* ignored -- bug? *)
mfn["y==z"] = Function[{x, y, z, u, v}, z - y];
mfn["x^2+y^2==1"] = Function[{x, y, z, u, v}, x^2 + y^2 - 1];
Show[
ParametricPlot3D[{x, y, y}, {x, 0, 1}, {y, 0, 1},
PlotStyle -> {ColorData[97, 1], Opacity[0.8]},
MeshFunctions -> {mfn["x^2+y^2==1"], #4 &, #5 &}],
ParametricPlot3D[{x, Sqrt[1 - x^2], z}, {x, 0, 1}, {z, 0, 1},
PlotStyle -> {ColorData[97, 2], Opacity[0.8]},
MeshFunctions -> {mfn["y==z"], #4 &, #5 &}],
ParametricPlot3D[{0, y, z}, {y, 0, 1}, {z, 0, 1},
PlotStyle -> {ColorData[97, 3], Opacity[0.8]},
MeshFunctions -> {mfn["y==z"], #4 &, #5 &}],
ParametricPlot3D[{x, y, 0}, {x, 0, 1}, {y, 0, 1},
PlotStyle -> {ColorData[97, 4], Opacity[0.8]},
MeshFunctions -> {mfn["x^2+y^2==1"], #4 &, #5 &}],
ViewPoint -> {3, -0.5, 1.5}]
SetOptions[ParametricPlot3D, opts];
Another bug? This often means Mathematica is about to crash. I think the OP has experienced this one before. (Note: I don't think this is a problem with the uploader. The same happens with Export and if I reevaluate the code. It's a hard to reproduce problem in the FE. I'm on Mac OSX V10.2)
Just to cover more ways of achieving this. We can plot over a Disk and use a PlotTheme.
Plot3D[y, {x, y} ∈ Disk[{0, 0}, 1, {0, π/2}], PlotTheme -> "FilledSurface", | 2,326 | 6,346 | {"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} | 3.265625 | 3 | CC-MAIN-2022-33 | longest | en | 0.875083 |
https://www.doubtnut.com/question-answer/in-delta-abc-a2s-a-b2s-b-c2s-c-642550204 | 1,638,546,295,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00577.warc.gz | 843,713,256 | 82,067 | # In Delta ABC, a^(2)(s-a)+b^(2)(s-b)+c^(2)(s-c)=
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 24-4-2021
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
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Text Solution
4R Delta(cos A+sin B+cos C)4R Delta(sin A+sin B+sin C)4R Delta(1+4sin.(A)/(2)sin.(B)/(2)sin.(C )/(2))none of these
C
Solution :
L.H.S.=(1)/(2)(a^(2)(b+c-a)+b^(2)(c+a-b)+c^(2)(a+b-c)) <br> =(1)/(2)(a(b^(2)+c^(2)-a^(2))+b(c^(2)+a^(2)-b^(2))+c(a^(2)+b^(2)-c^(2))) <br> =(1)/(2)(2abc cos A+2abc cos B + 2abc cos C) <br> =abc(1+4 sin.(A)/(2)sin.(B)/(2)sin.(C )/(2)) <br> =4R Delta (1+4sin.(A)/(2)sin.(B)/(2)sin.(C )/(2))
Transcript
TimeTranscript
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01:00 - 01:59b square c square minus A square right so this will come as b into c square A square minus b square right I rearrange all the form tried so this comes as CC into a square + b square into c square write this as low as we know cosine rule that is a square equal to b b square + c square - 2 BC caused this we can write in this format that this will be as to abc cause a this will come as to abc cosby this will come as to abc costly rate so this will come as a
02:00 - 02:59+ b + c into Cos A + cos b + cos C right now as we know the property of property of trigonometry that ABC we can write cos a cos b as a 2 into Cos A + B by 2 into Cos A minus b to write + cos C this week and right right now as we know that in the triangle angle A + Angle B + angle C is 180 degree angle C is 180 - a minus b Angle B right and a a + b is equal to 180 - 3 so I can write hair
03:00 - 03:59I can write my angle kowsi and Cos equal to 1 minus 2 sin square Sohail right there is angle A + Angle B + angles is 180 degree angle C is one angle A + B that is my angle a + b a + b cos 180 - C survival right here that is a b c into two science Si bi to this will return as their this comes Cos A minus B by 2 and escor Z I can write in this format this is a property 1 minus sin square C by to right now I can write again I can write
04:00 - 04:59this ABC this will come thus to science by 2 cos a minus b by 2 and this will be to this will come as to end this Si I can write it as a a plus bi right so this will be sin square 180 minus b + a bi to write I can write like this this will come ABC this will come science Si bi to I write one science by two common so this will be cause a dynasty by to write writing here we have
05:00 - 05:59so many brackets so this will be cause and this will come as - this will come as costs because this will be cos 90 minus theta + 1 so this will be equal to Cos A + B by to write I am writing this packet here for this right this will become as like this now we have bracket and the cos Cos A minus B by 2 minus Cos A + B by 2 this is a property of trigonometry so this will come ABC 1 - 2 Sin C by to write and I can write this as a to sign by 2 into
06:00 - 06:59sin b by to write so I will have a 20 1 plus 4 sin a by 2 Sin B by 2 into science Si bi to write as we know we can write ABCD in form of this this value at ABC upon 4 is equal to area so this will come as this will be written as 4 into our atah circumradius into area oneplus 4 sin a by 2 Sin B by 2 into sin C by to thank you | 1,166 | 3,843 | {"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} | 4.1875 | 4 | CC-MAIN-2021-49 | latest | en | 0.860486 |
https://docs.quantum.ibm.com/api/qiskit/qiskit.quantum_info.Pauli#insert | 1,718,817,336,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00325.warc.gz | 183,106,984 | 45,536 | # Pauli
class qiskit.quantum_info.Pauli(data=None)
GitHub(opens in a new tab)
Bases: BasePauli
N-qubit Pauli operator.
This class represents an operator $P$ from the full $n$-qubit Pauli group
$P = (-i)^{q} P_{n-1} \otimes ... \otimes P_{0}$
where $q\in \mathbb{Z}_4$ and $P_i \in \{I, X, Y, Z\}$ are single-qubit Pauli matrices:
$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$
Initialization
A Pauli object can be initialized in several ways:
Pauli(obj)
where obj is a Pauli string, Pauli or ScalarOp operator, or a Pauli gate or QuantumCircuit containing only Pauli gates.
Pauli((z, x, phase))
where z and x are boolean numpy.ndarrays and phase is an integer in [0, 1, 2, 3].
Pauli((z, x))
equivalent to Pauli((z, x, 0)) with trivial phase.
String representation
An $n$-qubit Pauli may be represented by a string consisting of $n$ characters from ['I', 'X', 'Y', 'Z'], and optionally phase coefficient in $['', '-i', '-', 'i']$. For example: XYZ or '-iZIZ'.
In the string representation qubit-0 corresponds to the right-most Pauli character, and qubit-$(n-1)$ to the left-most Pauli character. For example 'XYZ' represents $X\otimes Y \otimes Z$ with 'Z' on qubit-0, 'Y' on qubit-1, and 'X' on qubit-2.
The string representation can be converted to a Pauli using the class initialization (Pauli('-iXYZ')). A Pauli object can be converted back to the string representation using the to_label() method or str(pauli).
Note
Using str to convert a Pauli to a string will truncate the returned string for large numbers of qubits while to_label() will return the full string with no truncation. The default truncation length is 50 characters. The default value can be changed by setting the class __truncate__ attribute to an integer value. If set to 0 no truncation will be performed.
Array Representation
The internal data structure of an $n$-qubit Pauli is two length-$n$ boolean vectors $z \in \mathbb{Z}_2^N$, $x \in \mathbb{Z}_2^N$, and an integer $q \in \mathbb{Z}_4$ defining the Pauli operator
$P = (-i)^{q + z\cdot x} Z^z \cdot X^x.$
The $k$-th qubit corresponds to the $k$-th entry in the $z$ and $x$ arrays
\begin{aligned} P &= P_{n-1} \otimes ... \otimes P_{0} \\ P_k &= (-i)^{z[k] * x[k]} Z^{z[k]}\cdot X^{x[k]} \end{aligned}
where z[k] = P.z[k], x[k] = P.x[k] respectively.
The $z$ and $x$ arrays can be accessed and updated using the z and x properties respectively. The phase integer $q$ can be accessed and updated using the phase property.
Matrix Operator Representation
Pauli’s can be converted to $(2^n, 2^n)$ Operator using the to_operator() method, or to a dense or sparse complex matrix using the to_matrix() method.
Data Access
The individual qubit Paulis can be accessed and updated using the [] operator which accepts integer, lists, or slices for selecting subsets of Paulis. Note that selecting subsets of Pauli’s will discard the phase of the current Pauli.
For example
P = Pauli('-iXYZ')
print('P[0] =', repr(P[0]))
print('P[1] =', repr(P[1]))
print('P[2] =', repr(P[2]))
print('P[:] =', repr(P[:]))
print('P[::-1] =', repr(P[::-1]))
Initialize the Pauli.
When using the symplectic array input data both z and x arguments must be provided, however the first (z) argument can be used alone for string label, Pauli operator, or ScalarOp input data.
Parameters
data (str(opens in a new tab) ortuple(opens in a new tab) orPauli orScalarOp) – input data for Pauli. If input is a tuple it must be of the form (z, x) or (z, x, phase) where z and x are boolean Numpy arrays, and phase is an integer from Z_4. If input is a string, it must be a concatenation of a phase and a Pauli string (e.g. ‘XYZ’, ‘-iZIZ’) where a phase string is a combination of at most three characters from [‘+’, ‘-’, ‘’], [‘1’, ‘’], and [‘i’, ‘j’, ‘’] in this order, e.g. ‘’, ‘-1j’ while a Pauli string is 1 or more characters of ‘I’, ‘X’, ‘Y’ or ‘Z’, e.g. ‘Z’, ‘XIYY’.
Raises
QiskitError – if input array is invalid shape.
## Attributes
### dim
Return tuple (input_shape, output_shape).
### name
Unique string identifier for operation type.
### num_clbits
Number of classical bits.
### num_qubits
Return the number of qubits if a N-qubit operator or None otherwise.
### phase
Return the group phase exponent for the Pauli.
### qargs
Return the qargs for the operator.
Return settings.
### x
The x vector for the Pauli.
### z
The z vector for the Pauli.
## Methods
GitHub(opens in a new tab)
Return the adjoint of the Operator.
### anticommutes
anticommutes(other, qargs=None)
GitHub(opens in a new tab)
Return True if other Pauli anticommutes with self.
Parameters
Returns
True if Pauli’s anticommute, False if they commute.
Return type
bool(opens in a new tab)
### apply_layout
apply_layout(layout, num_qubits=None)
GitHub(opens in a new tab)
Apply a transpiler layout to this Pauli
Parameters
• layout (TranspileLayout |list(opens in a new tab)[int(opens in a new tab)] | None) – Either a TranspileLayout, a list of integers or None. If both layout and num_qubits are none, a copy of the operator is returned.
• num_qubits (int(opens in a new tab) | None) – The number of qubits to expand the operator to. If not provided then if layout is a TranspileLayout the number of the transpiler output circuit qubits will be used by default. If layout is a list of integers the permutation specified will be applied without any expansion. If layout is None, the operator will be expanded to the given number of qubits.
Returns
A new Pauli with the provided layout applied
Return type
Pauli
### commutes
commutes(other, qargs=None)
GitHub(opens in a new tab)
Return True if the Pauli commutes with other.
Parameters
Returns
True if Pauli’s commute, False if they anti-commute.
Return type
bool(opens in a new tab)
### compose
compose(other, qargs=None, front=False, inplace=False)
GitHub(opens in a new tab)
Return the operator composition with another Pauli.
Parameters
Returns
The composed Pauli.
Return type
Pauli
Raises
QiskitError – if other cannot be converted to an operator, or has incompatible dimensions for specified subsystems.
Note
Composition (&) by default is defined as left matrix multiplication for matrix operators, while dot() is defined as right matrix multiplication. That is that A & B == A.compose(B) is equivalent to B.dot(A) when A and B are of the same type.
Setting the front=True kwarg changes this to right matrix multiplication and is equivalent to the dot() method A.dot(B) == A.compose(B, front=True).
### conjugate
conjugate()
GitHub(opens in a new tab)
Return the conjugate of each Pauli in the list.
### copy
copy()
GitHub(opens in a new tab)
Make a deep copy of current operator.
### delete
delete(qubits)
GitHub(opens in a new tab)
Return a Pauli with qubits deleted.
Parameters
qubits (int(opens in a new tab) orlist(opens in a new tab)) – qubits to delete from Pauli.
Returns
the resulting Pauli with the specified qubits removed.
Return type
Pauli
Raises
QiskitError – if ind is out of bounds for the array size or number of qubits.
### dot
dot(other, qargs=None, inplace=False)
GitHub(opens in a new tab)
Return the right multiplied operator self * other.
Parameters
Returns
The operator self * other.
Return type
Pauli
### equiv
equiv(other)
GitHub(opens in a new tab)
Return True if Pauli’s are equivalent up to group phase.
Parameters
other (Pauli) – an operator object.
Returns
True if the Pauli’s are equivalent up to group phase.
Return type
bool(opens in a new tab)
### evolve
evolve(other, qargs=None, frame='h')
GitHub(opens in a new tab)
Performs either Heisenberg (default) or Schrödinger picture evolution of the Pauli by a Clifford and returns the evolved Pauli.
Schrödinger picture evolution can be chosen by passing parameter frame='s'. This option yields a faster calculation.
Heisenberg picture evolves the Pauli as $P^\prime = C^\dagger.P.C$.
Schrödinger picture evolves the Pauli as $P^\prime = C.P.C^\dagger$.
Parameters
Returns
the Pauli $C^\dagger.P.C$ (Heisenberg picture) or the Pauli $C.P.C^\dagger$ (Schrödinger picture).
Return type
Pauli
Raises
QiskitError – if the Clifford number of qubits and qargs don’t match.
### expand
expand(other)
GitHub(opens in a new tab)
Return the reverse-order tensor product with another Pauli.
Parameters
other (Pauli) – a Pauli object.
Returns
the tensor product $b \otimes a$, where $a$
is the current Pauli, and $b$ is the other Pauli.
Return type
Pauli
### input_dims
input_dims(qargs=None)
GitHub(opens in a new tab)
Return tuple of input dimension for specified subsystems.
### insert
insert(qubits, value)
GitHub(opens in a new tab)
Insert a Pauli at specific qubit value.
Parameters
Returns
the resulting Pauli with the entries inserted.
Return type
Pauli
Raises
QiskitError – if the insertion qubits are invalid.
### inverse
inverse()
GitHub(opens in a new tab)
Return the inverse of the Pauli.
### output_dims
output_dims(qargs=None)
GitHub(opens in a new tab)
Return tuple of output dimension for specified subsystems.
### power
power(n)
GitHub(opens in a new tab)
Return the compose of a operator with itself n times.
Parameters
n (int(opens in a new tab)) – the number of times to compose with self (n>0).
Returns
the n-times composed operator.
Return type
Clifford
Raises
QiskitError – if the input and output dimensions of the operator are not equal, or the power is not a positive integer.
### reshape
reshape(input_dims=None, output_dims=None, num_qubits=None)
GitHub(opens in a new tab)
Return a shallow copy with reshaped input and output subsystem dimensions.
Parameters
• input_dims (None or tuple(opens in a new tab)) – new subsystem input dimensions. If None the original input dims will be preserved [Default: None].
• output_dims (None or tuple(opens in a new tab)) – new subsystem output dimensions. If None the original output dims will be preserved [Default: None].
• num_qubits (None or int(opens in a new tab)) – reshape to an N-qubit operator [Default: None].
Returns
returns self with reshaped input and output dimensions.
Return type
BaseOperator
Raises
QiskitError – if combined size of all subsystem input dimension or subsystem output dimensions is not constant.
### set_truncation
classmethod set_truncation(val)
GitHub(opens in a new tab)
Set the max number of Pauli characters to display before truncation/
Parameters
val (int(opens in a new tab)) – the number of characters.
Note
Truncation will be disabled if the truncation value is set to 0.
### tensor
tensor(other)
GitHub(opens in a new tab)
Return the tensor product with another Pauli.
Parameters
other (Pauli) – a Pauli object.
Returns
the tensor product $a \otimes b$, where $a$
is the current Pauli, and $b$ is the other Pauli.
Return type
Pauli
Note
The tensor product can be obtained using the ^ binary operator. Hence a.tensor(b) is equivalent to a ^ b.
### to_instruction
to_instruction()
GitHub(opens in a new tab)
Convert to Pauli circuit instruction.
### to_label
to_label()
GitHub(opens in a new tab)
Convert a Pauli to a string label.
Note
The difference between to_label and __str__() is that the later will truncate the output for large numbers of qubits.
Returns
the Pauli string label.
Return type
str(opens in a new tab)
### to_matrix
to_matrix(sparse=False)
GitHub(opens in a new tab)
Convert to a Numpy array or sparse CSR matrix.
Parameters
sparse (bool(opens in a new tab)) – if True return sparse CSR matrices, otherwise return dense Numpy arrays (default: False).
Returns
The Pauli matrix.
Return type
array
### transpose
transpose()
GitHub(opens in a new tab)
Return the transpose of each Pauli in the list. | 3,170 | 11,939 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 36, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-26 | latest | en | 0.634425 |
http://www.transum.org/Software/SW/Starter_of_the_day/students/Basic_Division.asp?Level=5 | 1,582,090,090,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144027.33/warc/CC-MAIN-20200219030731-20200219060731-00349.warc.gz | 248,069,524 | 11,602 | # Basic Division Level 5
## Test your division skills with this self-marking exercise to do without a calculator.
##### Level 1Level 2Level 3Level 4Level 5Level 6Level 7Level 8DescriptionHelpMore Arithmetic
This is level 5: divide a four digit number by a two digit number (most with remainders). You can earn a trophy if you get at least 12 questions correct and you do this activity online.
6782 ÷ 11= remainder 1567 ÷ 13= remainder 2297 ÷ 15= remainder 9302 ÷ 20= remainder 9566 ÷ 18= remainder 3333 ÷ 25= remainder 7535 ÷ 30= remainder 5808 ÷ 25= remainder 3236 ÷ 45= remainder 3490 ÷ 32= remainder 7532 ÷ 36= remainder 1934 ÷ 56= remainder
Check
This is Basic Division Level 5. You can also try:
Level 1 Level 2 Level 3 Level 4 Level 6 Level 7 Level 8
## Instructions
Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.
When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.
## Transum.org
This web site contains over a thousand free mathematical activities for teachers and pupils. Click here to go to the main page which links to all of the resources available.
Please contact me if you have any suggestions or questions.
## More Activities:
Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?
Comment recorded on the 14 October 'Starter of the Day' page by Inger Kisby, Herts and Essex High School:
"Just a quick note to say that we use a lot of your starters. It is lovely to have so many different ideas to start a lesson with. Thank you very much and keep up the good work."
Comment recorded on the 19 October 'Starter of the Day' page by E Pollard, Huddersfield:
"I used this with my bottom set in year 9. To engage them I used their name and favorite football team (or pop group) instead of the school name. For homework, I asked each student to find a definition for the key words they had been given (once they had fun trying to guess the answer) and they presented their findings to the rest of the class the following day. They felt really special because the key words came from their own personal information."
#### Maths Mind Reader
This spectacular magic trick never fails to amaze people of all ages assuming they can add and subtract very simple numbers. The mathematics comes in finding out how the trick works.
There are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer.
A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves.
Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members.
If you would like to enjoy ad-free access to the thousands of Transum resources, receive our monthly newsletter, unlock the printable worksheets and see our Maths Lesson Finishers then sign up for a subscription now:
Subscribe
## Go Maths
Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school.
## Maths Map
Are you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.
## Teachers
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:
Transum,
Wednesday, September 23, 2015
"Here is a good application of division for the inquisitive pupil. Think of a three digit number that has all three digits the same. Divide the number by the sum of the digits. Try some other similar three digit numbers. Interesting isn't it?"
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
For Students:
For All:
## Description of Levels
Close
Level 1 - Divide a two digit number by a single digit number mentally (no remainders)
Level 2 - Divide a two digit number by a single digit number (with remainders)
Level 3 - Divide a three digit number by a single digit number (most with remainders)
Level 4 - Divide a four digit number by a single digit number (most with remainders)
Level 5 - Divide a four digit number by a two digit number (most with remainders)
Level 6 - Divide a four digit number by a one or two digit number where the answer has up to two decimal places
Level 7 - Divide a four digit number by a one or two digit number where the answer has to be rounded to two decimal places
Level 8 - Worded questions requiring interpretation of remainders appropriate for the context
More on this topic including lesson Starters, visual aids, investigations and self-marking exercises.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
## Curriculum Reference
See the National Curriculum page for links to related online activities and resources.
## Division
This video explains how to divide using short division and is from Corbettmaths.
Close | 1,384 | 6,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-10 | latest | en | 0.904876 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/5400/2/f/v/649/1/ | 1,642,893,758,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303917.24/warc/CC-MAIN-20220122224904-20220123014904-00492.warc.gz | 846,457,420 | 75,752 | # Properties
Label 5400.2.f.v.649.1 Level $5400$ Weight $2$ Character 5400.649 Analytic conductor $43.119$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$5400 = 2^{3} \cdot 3^{3} \cdot 5^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 5400.f (of order $$2$$, degree $$1$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$43.1192170915$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(i)$$ Defining polynomial: $$x^{2} + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{13}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 216) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## Embedding invariants
Embedding label 649.1 Root $$-1.00000i$$ of defining polynomial Character $$\chi$$ $$=$$ 5400.649 Dual form 5400.2.f.v.649.2
## $q$-expansion
$$f(q)$$ $$=$$ $$q-3.00000i q^{7} +O(q^{10})$$ $$q-3.00000i q^{7} +4.00000 q^{11} -1.00000i q^{13} -4.00000i q^{17} +1.00000 q^{19} -4.00000i q^{23} -4.00000 q^{31} -9.00000i q^{37} +8.00000i q^{43} -12.0000i q^{47} -2.00000 q^{49} +8.00000i q^{53} -4.00000 q^{59} -5.00000 q^{61} +11.0000i q^{67} +8.00000 q^{71} -1.00000i q^{73} -12.0000i q^{77} +5.00000 q^{79} -8.00000i q^{83} -12.0000 q^{89} -3.00000 q^{91} +5.00000i q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + O(q^{10})$$ $$2 q + 8 q^{11} + 2 q^{19} - 8 q^{31} - 4 q^{49} - 8 q^{59} - 10 q^{61} + 16 q^{71} + 10 q^{79} - 24 q^{89} - 6 q^{91} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/5400\mathbb{Z}\right)^\times$$.
$$n$$ $$1001$$ $$1351$$ $$2377$$ $$2701$$ $$\chi(n)$$ $$1$$ $$1$$ $$-1$$ $$1$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 0 0
$$3$$ 0 0
$$4$$ 0 0
$$5$$ 0 0
$$6$$ 0 0
$$7$$ − 3.00000i − 1.13389i −0.823754 0.566947i $$-0.808125\pi$$
0.823754 0.566947i $$-0.191875\pi$$
$$8$$ 0 0
$$9$$ 0 0
$$10$$ 0 0
$$11$$ 4.00000 1.20605 0.603023 0.797724i $$-0.293963\pi$$
0.603023 + 0.797724i $$0.293963\pi$$
$$12$$ 0 0
$$13$$ − 1.00000i − 0.277350i −0.990338 0.138675i $$-0.955716\pi$$
0.990338 0.138675i $$-0.0442844\pi$$
$$14$$ 0 0
$$15$$ 0 0
$$16$$ 0 0
$$17$$ − 4.00000i − 0.970143i −0.874475 0.485071i $$-0.838794\pi$$
0.874475 0.485071i $$-0.161206\pi$$
$$18$$ 0 0
$$19$$ 1.00000 0.229416 0.114708 0.993399i $$-0.463407\pi$$
0.114708 + 0.993399i $$0.463407\pi$$
$$20$$ 0 0
$$21$$ 0 0
$$22$$ 0 0
$$23$$ − 4.00000i − 0.834058i −0.908893 0.417029i $$-0.863071\pi$$
0.908893 0.417029i $$-0.136929\pi$$
$$24$$ 0 0
$$25$$ 0 0
$$26$$ 0 0
$$27$$ 0 0
$$28$$ 0 0
$$29$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$30$$ 0 0
$$31$$ −4.00000 −0.718421 −0.359211 0.933257i $$-0.616954\pi$$
−0.359211 + 0.933257i $$0.616954\pi$$
$$32$$ 0 0
$$33$$ 0 0
$$34$$ 0 0
$$35$$ 0 0
$$36$$ 0 0
$$37$$ − 9.00000i − 1.47959i −0.672832 0.739795i $$-0.734922\pi$$
0.672832 0.739795i $$-0.265078\pi$$
$$38$$ 0 0
$$39$$ 0 0
$$40$$ 0 0
$$41$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$42$$ 0 0
$$43$$ 8.00000i 1.21999i 0.792406 + 0.609994i $$0.208828\pi$$
−0.792406 + 0.609994i $$0.791172\pi$$
$$44$$ 0 0
$$45$$ 0 0
$$46$$ 0 0
$$47$$ − 12.0000i − 1.75038i −0.483779 0.875190i $$-0.660736\pi$$
0.483779 0.875190i $$-0.339264\pi$$
$$48$$ 0 0
$$49$$ −2.00000 −0.285714
$$50$$ 0 0
$$51$$ 0 0
$$52$$ 0 0
$$53$$ 8.00000i 1.09888i 0.835532 + 0.549442i $$0.185160\pi$$
−0.835532 + 0.549442i $$0.814840\pi$$
$$54$$ 0 0
$$55$$ 0 0
$$56$$ 0 0
$$57$$ 0 0
$$58$$ 0 0
$$59$$ −4.00000 −0.520756 −0.260378 0.965507i $$-0.583847\pi$$
−0.260378 + 0.965507i $$0.583847\pi$$
$$60$$ 0 0
$$61$$ −5.00000 −0.640184 −0.320092 0.947386i $$-0.603714\pi$$
−0.320092 + 0.947386i $$0.603714\pi$$
$$62$$ 0 0
$$63$$ 0 0
$$64$$ 0 0
$$65$$ 0 0
$$66$$ 0 0
$$67$$ 11.0000i 1.34386i 0.740613 + 0.671932i $$0.234535\pi$$
−0.740613 + 0.671932i $$0.765465\pi$$
$$68$$ 0 0
$$69$$ 0 0
$$70$$ 0 0
$$71$$ 8.00000 0.949425 0.474713 0.880141i $$-0.342552\pi$$
0.474713 + 0.880141i $$0.342552\pi$$
$$72$$ 0 0
$$73$$ − 1.00000i − 0.117041i −0.998286 0.0585206i $$-0.981362\pi$$
0.998286 0.0585206i $$-0.0186383\pi$$
$$74$$ 0 0
$$75$$ 0 0
$$76$$ 0 0
$$77$$ − 12.0000i − 1.36753i
$$78$$ 0 0
$$79$$ 5.00000 0.562544 0.281272 0.959628i $$-0.409244\pi$$
0.281272 + 0.959628i $$0.409244\pi$$
$$80$$ 0 0
$$81$$ 0 0
$$82$$ 0 0
$$83$$ − 8.00000i − 0.878114i −0.898459 0.439057i $$-0.855313\pi$$
0.898459 0.439057i $$-0.144687\pi$$
$$84$$ 0 0
$$85$$ 0 0
$$86$$ 0 0
$$87$$ 0 0
$$88$$ 0 0
$$89$$ −12.0000 −1.27200 −0.635999 0.771690i $$-0.719412\pi$$
−0.635999 + 0.771690i $$0.719412\pi$$
$$90$$ 0 0
$$91$$ −3.00000 −0.314485
$$92$$ 0 0
$$93$$ 0 0
$$94$$ 0 0
$$95$$ 0 0
$$96$$ 0 0
$$97$$ 5.00000i 0.507673i 0.967247 + 0.253837i $$0.0816925\pi$$
−0.967247 + 0.253837i $$0.918307\pi$$
$$98$$ 0 0
$$99$$ 0 0
$$100$$ 0 0
$$101$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$102$$ 0 0
$$103$$ − 1.00000i − 0.0985329i −0.998786 0.0492665i $$-0.984312\pi$$
0.998786 0.0492665i $$-0.0156884\pi$$
$$104$$ 0 0
$$105$$ 0 0
$$106$$ 0 0
$$107$$ 12.0000i 1.16008i 0.814587 + 0.580042i $$0.196964\pi$$
−0.814587 + 0.580042i $$0.803036\pi$$
$$108$$ 0 0
$$109$$ 14.0000 1.34096 0.670478 0.741929i $$-0.266089\pi$$
0.670478 + 0.741929i $$0.266089\pi$$
$$110$$ 0 0
$$111$$ 0 0
$$112$$ 0 0
$$113$$ − 12.0000i − 1.12887i −0.825479 0.564433i $$-0.809095\pi$$
0.825479 0.564433i $$-0.190905\pi$$
$$114$$ 0 0
$$115$$ 0 0
$$116$$ 0 0
$$117$$ 0 0
$$118$$ 0 0
$$119$$ −12.0000 −1.10004
$$120$$ 0 0
$$121$$ 5.00000 0.454545
$$122$$ 0 0
$$123$$ 0 0
$$124$$ 0 0
$$125$$ 0 0
$$126$$ 0 0
$$127$$ 4.00000i 0.354943i 0.984126 + 0.177471i $$0.0567917\pi$$
−0.984126 + 0.177471i $$0.943208\pi$$
$$128$$ 0 0
$$129$$ 0 0
$$130$$ 0 0
$$131$$ −16.0000 −1.39793 −0.698963 0.715158i $$-0.746355\pi$$
−0.698963 + 0.715158i $$0.746355\pi$$
$$132$$ 0 0
$$133$$ − 3.00000i − 0.260133i
$$134$$ 0 0
$$135$$ 0 0
$$136$$ 0 0
$$137$$ − 12.0000i − 1.02523i −0.858619 0.512615i $$-0.828677\pi$$
0.858619 0.512615i $$-0.171323\pi$$
$$138$$ 0 0
$$139$$ −9.00000 −0.763370 −0.381685 0.924292i $$-0.624656\pi$$
−0.381685 + 0.924292i $$0.624656\pi$$
$$140$$ 0 0
$$141$$ 0 0
$$142$$ 0 0
$$143$$ − 4.00000i − 0.334497i
$$144$$ 0 0
$$145$$ 0 0
$$146$$ 0 0
$$147$$ 0 0
$$148$$ 0 0
$$149$$ −8.00000 −0.655386 −0.327693 0.944784i $$-0.606271\pi$$
−0.327693 + 0.944784i $$0.606271\pi$$
$$150$$ 0 0
$$151$$ −1.00000 −0.0813788 −0.0406894 0.999172i $$-0.512955\pi$$
−0.0406894 + 0.999172i $$0.512955\pi$$
$$152$$ 0 0
$$153$$ 0 0
$$154$$ 0 0
$$155$$ 0 0
$$156$$ 0 0
$$157$$ − 2.00000i − 0.159617i −0.996810 0.0798087i $$-0.974569\pi$$
0.996810 0.0798087i $$-0.0254309\pi$$
$$158$$ 0 0
$$159$$ 0 0
$$160$$ 0 0
$$161$$ −12.0000 −0.945732
$$162$$ 0 0
$$163$$ 15.0000i 1.17489i 0.809264 + 0.587445i $$0.199866\pi$$
−0.809264 + 0.587445i $$0.800134\pi$$
$$164$$ 0 0
$$165$$ 0 0
$$166$$ 0 0
$$167$$ − 12.0000i − 0.928588i −0.885681 0.464294i $$-0.846308\pi$$
0.885681 0.464294i $$-0.153692\pi$$
$$168$$ 0 0
$$169$$ 12.0000 0.923077
$$170$$ 0 0
$$171$$ 0 0
$$172$$ 0 0
$$173$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$174$$ 0 0
$$175$$ 0 0
$$176$$ 0 0
$$177$$ 0 0
$$178$$ 0 0
$$179$$ −24.0000 −1.79384 −0.896922 0.442189i $$-0.854202\pi$$
−0.896922 + 0.442189i $$0.854202\pi$$
$$180$$ 0 0
$$181$$ 21.0000 1.56092 0.780459 0.625207i $$-0.214986\pi$$
0.780459 + 0.625207i $$0.214986\pi$$
$$182$$ 0 0
$$183$$ 0 0
$$184$$ 0 0
$$185$$ 0 0
$$186$$ 0 0
$$187$$ − 16.0000i − 1.17004i
$$188$$ 0 0
$$189$$ 0 0
$$190$$ 0 0
$$191$$ 12.0000 0.868290 0.434145 0.900843i $$-0.357051\pi$$
0.434145 + 0.900843i $$0.357051\pi$$
$$192$$ 0 0
$$193$$ − 23.0000i − 1.65558i −0.561041 0.827788i $$-0.689599\pi$$
0.561041 0.827788i $$-0.310401\pi$$
$$194$$ 0 0
$$195$$ 0 0
$$196$$ 0 0
$$197$$ − 12.0000i − 0.854965i −0.904024 0.427482i $$-0.859401\pi$$
0.904024 0.427482i $$-0.140599\pi$$
$$198$$ 0 0
$$199$$ −25.0000 −1.77220 −0.886102 0.463491i $$-0.846597\pi$$
−0.886102 + 0.463491i $$0.846597\pi$$
$$200$$ 0 0
$$201$$ 0 0
$$202$$ 0 0
$$203$$ 0 0
$$204$$ 0 0
$$205$$ 0 0
$$206$$ 0 0
$$207$$ 0 0
$$208$$ 0 0
$$209$$ 4.00000 0.276686
$$210$$ 0 0
$$211$$ 13.0000 0.894957 0.447478 0.894295i $$-0.352322\pi$$
0.447478 + 0.894295i $$0.352322\pi$$
$$212$$ 0 0
$$213$$ 0 0
$$214$$ 0 0
$$215$$ 0 0
$$216$$ 0 0
$$217$$ 12.0000i 0.814613i
$$218$$ 0 0
$$219$$ 0 0
$$220$$ 0 0
$$221$$ −4.00000 −0.269069
$$222$$ 0 0
$$223$$ − 4.00000i − 0.267860i −0.990991 0.133930i $$-0.957240\pi$$
0.990991 0.133930i $$-0.0427597\pi$$
$$224$$ 0 0
$$225$$ 0 0
$$226$$ 0 0
$$227$$ − 24.0000i − 1.59294i −0.604681 0.796468i $$-0.706699\pi$$
0.604681 0.796468i $$-0.293301\pi$$
$$228$$ 0 0
$$229$$ −10.0000 −0.660819 −0.330409 0.943838i $$-0.607187\pi$$
−0.330409 + 0.943838i $$0.607187\pi$$
$$230$$ 0 0
$$231$$ 0 0
$$232$$ 0 0
$$233$$ − 16.0000i − 1.04819i −0.851658 0.524097i $$-0.824403\pi$$
0.851658 0.524097i $$-0.175597\pi$$
$$234$$ 0 0
$$235$$ 0 0
$$236$$ 0 0
$$237$$ 0 0
$$238$$ 0 0
$$239$$ −8.00000 −0.517477 −0.258738 0.965947i $$-0.583307\pi$$
−0.258738 + 0.965947i $$0.583307\pi$$
$$240$$ 0 0
$$241$$ −15.0000 −0.966235 −0.483117 0.875556i $$-0.660496\pi$$
−0.483117 + 0.875556i $$0.660496\pi$$
$$242$$ 0 0
$$243$$ 0 0
$$244$$ 0 0
$$245$$ 0 0
$$246$$ 0 0
$$247$$ − 1.00000i − 0.0636285i
$$248$$ 0 0
$$249$$ 0 0
$$250$$ 0 0
$$251$$ −16.0000 −1.00991 −0.504956 0.863145i $$-0.668491\pi$$
−0.504956 + 0.863145i $$0.668491\pi$$
$$252$$ 0 0
$$253$$ − 16.0000i − 1.00591i
$$254$$ 0 0
$$255$$ 0 0
$$256$$ 0 0
$$257$$ 8.00000i 0.499026i 0.968371 + 0.249513i $$0.0802706\pi$$
−0.968371 + 0.249513i $$0.919729\pi$$
$$258$$ 0 0
$$259$$ −27.0000 −1.67770
$$260$$ 0 0
$$261$$ 0 0
$$262$$ 0 0
$$263$$ 8.00000i 0.493301i 0.969104 + 0.246651i $$0.0793300\pi$$
−0.969104 + 0.246651i $$0.920670\pi$$
$$264$$ 0 0
$$265$$ 0 0
$$266$$ 0 0
$$267$$ 0 0
$$268$$ 0 0
$$269$$ 4.00000 0.243884 0.121942 0.992537i $$-0.461088\pi$$
0.121942 + 0.992537i $$0.461088\pi$$
$$270$$ 0 0
$$271$$ −25.0000 −1.51864 −0.759321 0.650716i $$-0.774469\pi$$
−0.759321 + 0.650716i $$0.774469\pi$$
$$272$$ 0 0
$$273$$ 0 0
$$274$$ 0 0
$$275$$ 0 0
$$276$$ 0 0
$$277$$ 10.0000i 0.600842i 0.953807 + 0.300421i $$0.0971271\pi$$
−0.953807 + 0.300421i $$0.902873\pi$$
$$278$$ 0 0
$$279$$ 0 0
$$280$$ 0 0
$$281$$ 32.0000 1.90896 0.954480 0.298275i $$-0.0964112\pi$$
0.954480 + 0.298275i $$0.0964112\pi$$
$$282$$ 0 0
$$283$$ 16.0000i 0.951101i 0.879688 + 0.475551i $$0.157751\pi$$
−0.879688 + 0.475551i $$0.842249\pi$$
$$284$$ 0 0
$$285$$ 0 0
$$286$$ 0 0
$$287$$ 0 0
$$288$$ 0 0
$$289$$ 1.00000 0.0588235
$$290$$ 0 0
$$291$$ 0 0
$$292$$ 0 0
$$293$$ − 12.0000i − 0.701047i −0.936554 0.350524i $$-0.886004\pi$$
0.936554 0.350524i $$-0.113996\pi$$
$$294$$ 0 0
$$295$$ 0 0
$$296$$ 0 0
$$297$$ 0 0
$$298$$ 0 0
$$299$$ −4.00000 −0.231326
$$300$$ 0 0
$$301$$ 24.0000 1.38334
$$302$$ 0 0
$$303$$ 0 0
$$304$$ 0 0
$$305$$ 0 0
$$306$$ 0 0
$$307$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$308$$ 0 0
$$309$$ 0 0
$$310$$ 0 0
$$311$$ 12.0000 0.680458 0.340229 0.940343i $$-0.389495\pi$$
0.340229 + 0.940343i $$0.389495\pi$$
$$312$$ 0 0
$$313$$ − 3.00000i − 0.169570i −0.996399 0.0847850i $$-0.972980\pi$$
0.996399 0.0847850i $$-0.0270203\pi$$
$$314$$ 0 0
$$315$$ 0 0
$$316$$ 0 0
$$317$$ − 24.0000i − 1.34797i −0.738743 0.673987i $$-0.764580\pi$$
0.738743 0.673987i $$-0.235420\pi$$
$$318$$ 0 0
$$319$$ 0 0
$$320$$ 0 0
$$321$$ 0 0
$$322$$ 0 0
$$323$$ − 4.00000i − 0.222566i
$$324$$ 0 0
$$325$$ 0 0
$$326$$ 0 0
$$327$$ 0 0
$$328$$ 0 0
$$329$$ −36.0000 −1.98474
$$330$$ 0 0
$$331$$ 17.0000 0.934405 0.467202 0.884150i $$-0.345262\pi$$
0.467202 + 0.884150i $$0.345262\pi$$
$$332$$ 0 0
$$333$$ 0 0
$$334$$ 0 0
$$335$$ 0 0
$$336$$ 0 0
$$337$$ − 3.00000i − 0.163420i −0.996656 0.0817102i $$-0.973962\pi$$
0.996656 0.0817102i $$-0.0260382\pi$$
$$338$$ 0 0
$$339$$ 0 0
$$340$$ 0 0
$$341$$ −16.0000 −0.866449
$$342$$ 0 0
$$343$$ − 15.0000i − 0.809924i
$$344$$ 0 0
$$345$$ 0 0
$$346$$ 0 0
$$347$$ − 24.0000i − 1.28839i −0.764862 0.644194i $$-0.777193\pi$$
0.764862 0.644194i $$-0.222807\pi$$
$$348$$ 0 0
$$349$$ 33.0000 1.76645 0.883225 0.468950i $$-0.155368\pi$$
0.883225 + 0.468950i $$0.155368\pi$$
$$350$$ 0 0
$$351$$ 0 0
$$352$$ 0 0
$$353$$ − 8.00000i − 0.425797i −0.977074 0.212899i $$-0.931710\pi$$
0.977074 0.212899i $$-0.0682904\pi$$
$$354$$ 0 0
$$355$$ 0 0
$$356$$ 0 0
$$357$$ 0 0
$$358$$ 0 0
$$359$$ −12.0000 −0.633336 −0.316668 0.948536i $$-0.602564\pi$$
−0.316668 + 0.948536i $$0.602564\pi$$
$$360$$ 0 0
$$361$$ −18.0000 −0.947368
$$362$$ 0 0
$$363$$ 0 0
$$364$$ 0 0
$$365$$ 0 0
$$366$$ 0 0
$$367$$ − 23.0000i − 1.20059i −0.799779 0.600295i $$-0.795050\pi$$
0.799779 0.600295i $$-0.204950\pi$$
$$368$$ 0 0
$$369$$ 0 0
$$370$$ 0 0
$$371$$ 24.0000 1.24602
$$372$$ 0 0
$$373$$ 1.00000i 0.0517780i 0.999665 + 0.0258890i $$0.00824165\pi$$
−0.999665 + 0.0258890i $$0.991758\pi$$
$$374$$ 0 0
$$375$$ 0 0
$$376$$ 0 0
$$377$$ 0 0
$$378$$ 0 0
$$379$$ −11.0000 −0.565032 −0.282516 0.959263i $$-0.591169\pi$$
−0.282516 + 0.959263i $$0.591169\pi$$
$$380$$ 0 0
$$381$$ 0 0
$$382$$ 0 0
$$383$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$384$$ 0 0
$$385$$ 0 0
$$386$$ 0 0
$$387$$ 0 0
$$388$$ 0 0
$$389$$ −4.00000 −0.202808 −0.101404 0.994845i $$-0.532333\pi$$
−0.101404 + 0.994845i $$0.532333\pi$$
$$390$$ 0 0
$$391$$ −16.0000 −0.809155
$$392$$ 0 0
$$393$$ 0 0
$$394$$ 0 0
$$395$$ 0 0
$$396$$ 0 0
$$397$$ 14.0000i 0.702640i 0.936255 + 0.351320i $$0.114267\pi$$
−0.936255 + 0.351320i $$0.885733\pi$$
$$398$$ 0 0
$$399$$ 0 0
$$400$$ 0 0
$$401$$ 24.0000 1.19850 0.599251 0.800561i $$-0.295465\pi$$
0.599251 + 0.800561i $$0.295465\pi$$
$$402$$ 0 0
$$403$$ 4.00000i 0.199254i
$$404$$ 0 0
$$405$$ 0 0
$$406$$ 0 0
$$407$$ − 36.0000i − 1.78445i
$$408$$ 0 0
$$409$$ 39.0000 1.92843 0.964213 0.265129i $$-0.0854146\pi$$
0.964213 + 0.265129i $$0.0854146\pi$$
$$410$$ 0 0
$$411$$ 0 0
$$412$$ 0 0
$$413$$ 12.0000i 0.590481i
$$414$$ 0 0
$$415$$ 0 0
$$416$$ 0 0
$$417$$ 0 0
$$418$$ 0 0
$$419$$ 12.0000 0.586238 0.293119 0.956076i $$-0.405307\pi$$
0.293119 + 0.956076i $$0.405307\pi$$
$$420$$ 0 0
$$421$$ 17.0000 0.828529 0.414265 0.910156i $$-0.364039\pi$$
0.414265 + 0.910156i $$0.364039\pi$$
$$422$$ 0 0
$$423$$ 0 0
$$424$$ 0 0
$$425$$ 0 0
$$426$$ 0 0
$$427$$ 15.0000i 0.725901i
$$428$$ 0 0
$$429$$ 0 0
$$430$$ 0 0
$$431$$ −28.0000 −1.34871 −0.674356 0.738406i $$-0.735579\pi$$
−0.674356 + 0.738406i $$0.735579\pi$$
$$432$$ 0 0
$$433$$ 14.0000i 0.672797i 0.941720 + 0.336399i $$0.109209\pi$$
−0.941720 + 0.336399i $$0.890791\pi$$
$$434$$ 0 0
$$435$$ 0 0
$$436$$ 0 0
$$437$$ − 4.00000i − 0.191346i
$$438$$ 0 0
$$439$$ −36.0000 −1.71819 −0.859093 0.511819i $$-0.828972\pi$$
−0.859093 + 0.511819i $$0.828972\pi$$
$$440$$ 0 0
$$441$$ 0 0
$$442$$ 0 0
$$443$$ − 8.00000i − 0.380091i −0.981775 0.190046i $$-0.939136\pi$$
0.981775 0.190046i $$-0.0608636\pi$$
$$444$$ 0 0
$$445$$ 0 0
$$446$$ 0 0
$$447$$ 0 0
$$448$$ 0 0
$$449$$ −4.00000 −0.188772 −0.0943858 0.995536i $$-0.530089\pi$$
−0.0943858 + 0.995536i $$0.530089\pi$$
$$450$$ 0 0
$$451$$ 0 0
$$452$$ 0 0
$$453$$ 0 0
$$454$$ 0 0
$$455$$ 0 0
$$456$$ 0 0
$$457$$ 38.0000i 1.77757i 0.458329 + 0.888783i $$0.348448\pi$$
−0.458329 + 0.888783i $$0.651552\pi$$
$$458$$ 0 0
$$459$$ 0 0
$$460$$ 0 0
$$461$$ 28.0000 1.30409 0.652045 0.758180i $$-0.273911\pi$$
0.652045 + 0.758180i $$0.273911\pi$$
$$462$$ 0 0
$$463$$ 19.0000i 0.883005i 0.897260 + 0.441502i $$0.145554\pi$$
−0.897260 + 0.441502i $$0.854446\pi$$
$$464$$ 0 0
$$465$$ 0 0
$$466$$ 0 0
$$467$$ − 12.0000i − 0.555294i −0.960683 0.277647i $$-0.910445\pi$$
0.960683 0.277647i $$-0.0895545\pi$$
$$468$$ 0 0
$$469$$ 33.0000 1.52380
$$470$$ 0 0
$$471$$ 0 0
$$472$$ 0 0
$$473$$ 32.0000i 1.47136i
$$474$$ 0 0
$$475$$ 0 0
$$476$$ 0 0
$$477$$ 0 0
$$478$$ 0 0
$$479$$ 40.0000 1.82765 0.913823 0.406112i $$-0.133116\pi$$
0.913823 + 0.406112i $$0.133116\pi$$
$$480$$ 0 0
$$481$$ −9.00000 −0.410365
$$482$$ 0 0
$$483$$ 0 0
$$484$$ 0 0
$$485$$ 0 0
$$486$$ 0 0
$$487$$ − 11.0000i − 0.498458i −0.968445 0.249229i $$-0.919823\pi$$
0.968445 0.249229i $$-0.0801771\pi$$
$$488$$ 0 0
$$489$$ 0 0
$$490$$ 0 0
$$491$$ 12.0000 0.541552 0.270776 0.962642i $$-0.412720\pi$$
0.270776 + 0.962642i $$0.412720\pi$$
$$492$$ 0 0
$$493$$ 0 0
$$494$$ 0 0
$$495$$ 0 0
$$496$$ 0 0
$$497$$ − 24.0000i − 1.07655i
$$498$$ 0 0
$$499$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$500$$ 0 0
$$501$$ 0 0
$$502$$ 0 0
$$503$$ − 36.0000i − 1.60516i −0.596544 0.802580i $$-0.703460\pi$$
0.596544 0.802580i $$-0.296540\pi$$
$$504$$ 0 0
$$505$$ 0 0
$$506$$ 0 0
$$507$$ 0 0
$$508$$ 0 0
$$509$$ 12.0000 0.531891 0.265945 0.963988i $$-0.414316\pi$$
0.265945 + 0.963988i $$0.414316\pi$$
$$510$$ 0 0
$$511$$ −3.00000 −0.132712
$$512$$ 0 0
$$513$$ 0 0
$$514$$ 0 0
$$515$$ 0 0
$$516$$ 0 0
$$517$$ − 48.0000i − 2.11104i
$$518$$ 0 0
$$519$$ 0 0
$$520$$ 0 0
$$521$$ −28.0000 −1.22670 −0.613351 0.789810i $$-0.710179\pi$$
−0.613351 + 0.789810i $$0.710179\pi$$
$$522$$ 0 0
$$523$$ 29.0000i 1.26808i 0.773300 + 0.634041i $$0.218605\pi$$
−0.773300 + 0.634041i $$0.781395\pi$$
$$524$$ 0 0
$$525$$ 0 0
$$526$$ 0 0
$$527$$ 16.0000i 0.696971i
$$528$$ 0 0
$$529$$ 7.00000 0.304348
$$530$$ 0 0
$$531$$ 0 0
$$532$$ 0 0
$$533$$ 0 0
$$534$$ 0 0
$$535$$ 0 0
$$536$$ 0 0
$$537$$ 0 0
$$538$$ 0 0
$$539$$ −8.00000 −0.344584
$$540$$ 0 0
$$541$$ 9.00000 0.386940 0.193470 0.981106i $$-0.438026\pi$$
0.193470 + 0.981106i $$0.438026\pi$$
$$542$$ 0 0
$$543$$ 0 0
$$544$$ 0 0
$$545$$ 0 0
$$546$$ 0 0
$$547$$ − 7.00000i − 0.299298i −0.988739 0.149649i $$-0.952186\pi$$
0.988739 0.149649i $$-0.0478144\pi$$
$$548$$ 0 0
$$549$$ 0 0
$$550$$ 0 0
$$551$$ 0 0
$$552$$ 0 0
$$553$$ − 15.0000i − 0.637865i
$$554$$ 0 0
$$555$$ 0 0
$$556$$ 0 0
$$557$$ 28.0000i 1.18640i 0.805056 + 0.593199i $$0.202135\pi$$
−0.805056 + 0.593199i $$0.797865\pi$$
$$558$$ 0 0
$$559$$ 8.00000 0.338364
$$560$$ 0 0
$$561$$ 0 0
$$562$$ 0 0
$$563$$ 32.0000i 1.34864i 0.738440 + 0.674320i $$0.235563\pi$$
−0.738440 + 0.674320i $$0.764437\pi$$
$$564$$ 0 0
$$565$$ 0 0
$$566$$ 0 0
$$567$$ 0 0
$$568$$ 0 0
$$569$$ −28.0000 −1.17382 −0.586911 0.809652i $$-0.699656\pi$$
−0.586911 + 0.809652i $$0.699656\pi$$
$$570$$ 0 0
$$571$$ −33.0000 −1.38101 −0.690504 0.723329i $$-0.742611\pi$$
−0.690504 + 0.723329i $$0.742611\pi$$
$$572$$ 0 0
$$573$$ 0 0
$$574$$ 0 0
$$575$$ 0 0
$$576$$ 0 0
$$577$$ − 13.0000i − 0.541197i −0.962692 0.270599i $$-0.912778\pi$$
0.962692 0.270599i $$-0.0872216\pi$$
$$578$$ 0 0
$$579$$ 0 0
$$580$$ 0 0
$$581$$ −24.0000 −0.995688
$$582$$ 0 0
$$583$$ 32.0000i 1.32530i
$$584$$ 0 0
$$585$$ 0 0
$$586$$ 0 0
$$587$$ 28.0000i 1.15568i 0.816149 + 0.577842i $$0.196105\pi$$
−0.816149 + 0.577842i $$0.803895\pi$$
$$588$$ 0 0
$$589$$ −4.00000 −0.164817
$$590$$ 0 0
$$591$$ 0 0
$$592$$ 0 0
$$593$$ − 40.0000i − 1.64260i −0.570494 0.821302i $$-0.693248\pi$$
0.570494 0.821302i $$-0.306752\pi$$
$$594$$ 0 0
$$595$$ 0 0
$$596$$ 0 0
$$597$$ 0 0
$$598$$ 0 0
$$599$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$600$$ 0 0
$$601$$ 10.0000 0.407909 0.203954 0.978980i $$-0.434621\pi$$
0.203954 + 0.978980i $$0.434621\pi$$
$$602$$ 0 0
$$603$$ 0 0
$$604$$ 0 0
$$605$$ 0 0
$$606$$ 0 0
$$607$$ 5.00000i 0.202944i 0.994838 + 0.101472i $$0.0323552\pi$$
−0.994838 + 0.101472i $$0.967645\pi$$
$$608$$ 0 0
$$609$$ 0 0
$$610$$ 0 0
$$611$$ −12.0000 −0.485468
$$612$$ 0 0
$$613$$ − 35.0000i − 1.41364i −0.707395 0.706818i $$-0.750130\pi$$
0.707395 0.706818i $$-0.249870\pi$$
$$614$$ 0 0
$$615$$ 0 0
$$616$$ 0 0
$$617$$ 36.0000i 1.44931i 0.689114 + 0.724653i $$0.258000\pi$$
−0.689114 + 0.724653i $$0.742000\pi$$
$$618$$ 0 0
$$619$$ 17.0000 0.683288 0.341644 0.939829i $$-0.389016\pi$$
0.341644 + 0.939829i $$0.389016\pi$$
$$620$$ 0 0
$$621$$ 0 0
$$622$$ 0 0
$$623$$ 36.0000i 1.44231i
$$624$$ 0 0
$$625$$ 0 0
$$626$$ 0 0
$$627$$ 0 0
$$628$$ 0 0
$$629$$ −36.0000 −1.43541
$$630$$ 0 0
$$631$$ 7.00000 0.278666 0.139333 0.990246i $$-0.455504\pi$$
0.139333 + 0.990246i $$0.455504\pi$$
$$632$$ 0 0
$$633$$ 0 0
$$634$$ 0 0
$$635$$ 0 0
$$636$$ 0 0
$$637$$ 2.00000i 0.0792429i
$$638$$ 0 0
$$639$$ 0 0
$$640$$ 0 0
$$641$$ 40.0000 1.57991 0.789953 0.613168i $$-0.210105\pi$$
0.789953 + 0.613168i $$0.210105\pi$$
$$642$$ 0 0
$$643$$ 24.0000i 0.946468i 0.880937 + 0.473234i $$0.156913\pi$$
−0.880937 + 0.473234i $$0.843087\pi$$
$$644$$ 0 0
$$645$$ 0 0
$$646$$ 0 0
$$647$$ 32.0000i 1.25805i 0.777385 + 0.629025i $$0.216546\pi$$
−0.777385 + 0.629025i $$0.783454\pi$$
$$648$$ 0 0
$$649$$ −16.0000 −0.628055
$$650$$ 0 0
$$651$$ 0 0
$$652$$ 0 0
$$653$$ − 24.0000i − 0.939193i −0.882881 0.469596i $$-0.844399\pi$$
0.882881 0.469596i $$-0.155601\pi$$
$$654$$ 0 0
$$655$$ 0 0
$$656$$ 0 0
$$657$$ 0 0
$$658$$ 0 0
$$659$$ 24.0000 0.934907 0.467454 0.884018i $$-0.345171\pi$$
0.467454 + 0.884018i $$0.345171\pi$$
$$660$$ 0 0
$$661$$ −13.0000 −0.505641 −0.252821 0.967513i $$-0.581358\pi$$
−0.252821 + 0.967513i $$0.581358\pi$$
$$662$$ 0 0
$$663$$ 0 0
$$664$$ 0 0
$$665$$ 0 0
$$666$$ 0 0
$$667$$ 0 0
$$668$$ 0 0
$$669$$ 0 0
$$670$$ 0 0
$$671$$ −20.0000 −0.772091
$$672$$ 0 0
$$673$$ − 19.0000i − 0.732396i −0.930537 0.366198i $$-0.880659\pi$$
0.930537 0.366198i $$-0.119341\pi$$
$$674$$ 0 0
$$675$$ 0 0
$$676$$ 0 0
$$677$$ − 44.0000i − 1.69106i −0.533930 0.845529i $$-0.679285\pi$$
0.533930 0.845529i $$-0.320715\pi$$
$$678$$ 0 0
$$679$$ 15.0000 0.575647
$$680$$ 0 0
$$681$$ 0 0
$$682$$ 0 0
$$683$$ − 16.0000i − 0.612223i −0.951996 0.306111i $$-0.900972\pi$$
0.951996 0.306111i $$-0.0990280\pi$$
$$684$$ 0 0
$$685$$ 0 0
$$686$$ 0 0
$$687$$ 0 0
$$688$$ 0 0
$$689$$ 8.00000 0.304776
$$690$$ 0 0
$$691$$ −40.0000 −1.52167 −0.760836 0.648944i $$-0.775211\pi$$
−0.760836 + 0.648944i $$0.775211\pi$$
$$692$$ 0 0
$$693$$ 0 0
$$694$$ 0 0
$$695$$ 0 0
$$696$$ 0 0
$$697$$ 0 0
$$698$$ 0 0
$$699$$ 0 0
$$700$$ 0 0
$$701$$ −36.0000 −1.35970 −0.679851 0.733351i $$-0.737955\pi$$
−0.679851 + 0.733351i $$0.737955\pi$$
$$702$$ 0 0
$$703$$ − 9.00000i − 0.339441i
$$704$$ 0 0
$$705$$ 0 0
$$706$$ 0 0
$$707$$ 0 0
$$708$$ 0 0
$$709$$ 7.00000 0.262891 0.131445 0.991323i $$-0.458038\pi$$
0.131445 + 0.991323i $$0.458038\pi$$
$$710$$ 0 0
$$711$$ 0 0
$$712$$ 0 0
$$713$$ 16.0000i 0.599205i
$$714$$ 0 0
$$715$$ 0 0
$$716$$ 0 0
$$717$$ 0 0
$$718$$ 0 0
$$719$$ 8.00000 0.298350 0.149175 0.988811i $$-0.452338\pi$$
0.149175 + 0.988811i $$0.452338\pi$$
$$720$$ 0 0
$$721$$ −3.00000 −0.111726
$$722$$ 0 0
$$723$$ 0 0
$$724$$ 0 0
$$725$$ 0 0
$$726$$ 0 0
$$727$$ 12.0000i 0.445055i 0.974926 + 0.222528i $$0.0714308\pi$$
−0.974926 + 0.222528i $$0.928569\pi$$
$$728$$ 0 0
$$729$$ 0 0
$$730$$ 0 0
$$731$$ 32.0000 1.18356
$$732$$ 0 0
$$733$$ − 2.00000i − 0.0738717i −0.999318 0.0369358i $$-0.988240\pi$$
0.999318 0.0369358i $$-0.0117597\pi$$
$$734$$ 0 0
$$735$$ 0 0
$$736$$ 0 0
$$737$$ 44.0000i 1.62076i
$$738$$ 0 0
$$739$$ −32.0000 −1.17714 −0.588570 0.808447i $$-0.700309\pi$$
−0.588570 + 0.808447i $$0.700309\pi$$
$$740$$ 0 0
$$741$$ 0 0
$$742$$ 0 0
$$743$$ − 4.00000i − 0.146746i −0.997305 0.0733729i $$-0.976624\pi$$
0.997305 0.0733729i $$-0.0233763\pi$$
$$744$$ 0 0
$$745$$ 0 0
$$746$$ 0 0
$$747$$ 0 0
$$748$$ 0 0
$$749$$ 36.0000 1.31541
$$750$$ 0 0
$$751$$ 3.00000 0.109472 0.0547358 0.998501i $$-0.482568\pi$$
0.0547358 + 0.998501i $$0.482568\pi$$
$$752$$ 0 0
$$753$$ 0 0
$$754$$ 0 0
$$755$$ 0 0
$$756$$ 0 0
$$757$$ − 31.0000i − 1.12671i −0.826214 0.563357i $$-0.809510\pi$$
0.826214 0.563357i $$-0.190490\pi$$
$$758$$ 0 0
$$759$$ 0 0
$$760$$ 0 0
$$761$$ −4.00000 −0.145000 −0.0724999 0.997368i $$-0.523098\pi$$
−0.0724999 + 0.997368i $$0.523098\pi$$
$$762$$ 0 0
$$763$$ − 42.0000i − 1.52050i
$$764$$ 0 0
$$765$$ 0 0
$$766$$ 0 0
$$767$$ 4.00000i 0.144432i
$$768$$ 0 0
$$769$$ −47.0000 −1.69486 −0.847432 0.530904i $$-0.821852\pi$$
−0.847432 + 0.530904i $$0.821852\pi$$
$$770$$ 0 0
$$771$$ 0 0
$$772$$ 0 0
$$773$$ 24.0000i 0.863220i 0.902060 + 0.431610i $$0.142054\pi$$
−0.902060 + 0.431610i $$0.857946\pi$$
$$774$$ 0 0
$$775$$ 0 0
$$776$$ 0 0
$$777$$ 0 0
$$778$$ 0 0
$$779$$ 0 0
$$780$$ 0 0
$$781$$ 32.0000 1.14505
$$782$$ 0 0
$$783$$ 0 0
$$784$$ 0 0
$$785$$ 0 0
$$786$$ 0 0
$$787$$ 7.00000i 0.249523i 0.992187 + 0.124762i $$0.0398166\pi$$
−0.992187 + 0.124762i $$0.960183\pi$$
$$788$$ 0 0
$$789$$ 0 0
$$790$$ 0 0
$$791$$ −36.0000 −1.28001
$$792$$ 0 0
$$793$$ 5.00000i 0.177555i
$$794$$ 0 0
$$795$$ 0 0
$$796$$ 0 0
$$797$$ 16.0000i 0.566749i 0.959009 + 0.283375i $$0.0914540\pi$$
−0.959009 + 0.283375i $$0.908546\pi$$
$$798$$ 0 0
$$799$$ −48.0000 −1.69812
$$800$$ 0 0
$$801$$ 0 0
$$802$$ 0 0
$$803$$ − 4.00000i − 0.141157i
$$804$$ 0 0
$$805$$ 0 0
$$806$$ 0 0
$$807$$ 0 0
$$808$$ 0 0
$$809$$ 40.0000 1.40633 0.703163 0.711029i $$-0.251771\pi$$
0.703163 + 0.711029i $$0.251771\pi$$
$$810$$ 0 0
$$811$$ 40.0000 1.40459 0.702295 0.711886i $$-0.252159\pi$$
0.702295 + 0.711886i $$0.252159\pi$$
$$812$$ 0 0
$$813$$ 0 0
$$814$$ 0 0
$$815$$ 0 0
$$816$$ 0 0
$$817$$ 8.00000i 0.279885i
$$818$$ 0 0
$$819$$ 0 0
$$820$$ 0 0
$$821$$ −12.0000 −0.418803 −0.209401 0.977830i $$-0.567152\pi$$
−0.209401 + 0.977830i $$0.567152\pi$$
$$822$$ 0 0
$$823$$ 25.0000i 0.871445i 0.900081 + 0.435723i $$0.143507\pi$$
−0.900081 + 0.435723i $$0.856493\pi$$
$$824$$ 0 0
$$825$$ 0 0
$$826$$ 0 0
$$827$$ 20.0000i 0.695468i 0.937593 + 0.347734i $$0.113049\pi$$
−0.937593 + 0.347734i $$0.886951\pi$$
$$828$$ 0 0
$$829$$ 35.0000 1.21560 0.607800 0.794090i $$-0.292052\pi$$
0.607800 + 0.794090i $$0.292052\pi$$
$$830$$ 0 0
$$831$$ 0 0
$$832$$ 0 0
$$833$$ 8.00000i 0.277184i
$$834$$ 0 0
$$835$$ 0 0
$$836$$ 0 0
$$837$$ 0 0
$$838$$ 0 0
$$839$$ −24.0000 −0.828572 −0.414286 0.910147i $$-0.635969\pi$$
−0.414286 + 0.910147i $$0.635969\pi$$
$$840$$ 0 0
$$841$$ −29.0000 −1.00000
$$842$$ 0 0
$$843$$ 0 0
$$844$$ 0 0
$$845$$ 0 0
$$846$$ 0 0
$$847$$ − 15.0000i − 0.515406i
$$848$$ 0 0
$$849$$ 0 0
$$850$$ 0 0
$$851$$ −36.0000 −1.23406
$$852$$ 0 0
$$853$$ 1.00000i 0.0342393i 0.999853 + 0.0171197i $$0.00544963\pi$$
−0.999853 + 0.0171197i $$0.994550\pi$$
$$854$$ 0 0
$$855$$ 0 0
$$856$$ 0 0
$$857$$ 24.0000i 0.819824i 0.912125 + 0.409912i $$0.134441\pi$$
−0.912125 + 0.409912i $$0.865559\pi$$
$$858$$ 0 0
$$859$$ 27.0000 0.921228 0.460614 0.887601i $$-0.347629\pi$$
0.460614 + 0.887601i $$0.347629\pi$$
$$860$$ 0 0
$$861$$ 0 0
$$862$$ 0 0
$$863$$ 44.0000i 1.49778i 0.662696 + 0.748889i $$0.269412\pi$$
−0.662696 + 0.748889i $$0.730588\pi$$
$$864$$ 0 0
$$865$$ 0 0
$$866$$ 0 0
$$867$$ 0 0
$$868$$ 0 0
$$869$$ 20.0000 0.678454
$$870$$ 0 0
$$871$$ 11.0000 0.372721
$$872$$ 0 0
$$873$$ 0 0
$$874$$ 0 0
$$875$$ 0 0
$$876$$ 0 0
$$877$$ − 33.0000i − 1.11433i −0.830402 0.557165i $$-0.811889\pi$$
0.830402 0.557165i $$-0.188111\pi$$
$$878$$ 0 0
$$879$$ 0 0
$$880$$ 0 0
$$881$$ −28.0000 −0.943344 −0.471672 0.881774i $$-0.656349\pi$$
−0.471672 + 0.881774i $$0.656349\pi$$
$$882$$ 0 0
$$883$$ 7.00000i 0.235569i 0.993039 + 0.117784i $$0.0375792\pi$$
−0.993039 + 0.117784i $$0.962421\pi$$
$$884$$ 0 0
$$885$$ 0 0
$$886$$ 0 0
$$887$$ 56.0000i 1.88030i 0.340766 + 0.940148i $$0.389313\pi$$
−0.340766 + 0.940148i $$0.610687\pi$$
$$888$$ 0 0
$$889$$ 12.0000 0.402467
$$890$$ 0 0
$$891$$ 0 0
$$892$$ 0 0
$$893$$ − 12.0000i − 0.401565i
$$894$$ 0 0
$$895$$ 0 0
$$896$$ 0 0
$$897$$ 0 0
$$898$$ 0 0
$$899$$ 0 0
$$900$$ 0 0
$$901$$ 32.0000 1.06607
$$902$$ 0 0
$$903$$ 0 0
$$904$$ 0 0
$$905$$ 0 0
$$906$$ 0 0
$$907$$ − 5.00000i − 0.166022i −0.996549 0.0830111i $$-0.973546\pi$$
0.996549 0.0830111i $$-0.0264537\pi$$
$$908$$ 0 0
$$909$$ 0 0
$$910$$ 0 0
$$911$$ 16.0000 0.530104 0.265052 0.964234i $$-0.414611\pi$$
0.265052 + 0.964234i $$0.414611\pi$$
$$912$$ 0 0
$$913$$ − 32.0000i − 1.05905i
$$914$$ 0 0
$$915$$ 0 0
$$916$$ 0 0
$$917$$ 48.0000i 1.58510i
$$918$$ 0 0
$$919$$ 20.0000 0.659739 0.329870 0.944027i $$-0.392995\pi$$
0.329870 + 0.944027i $$0.392995\pi$$
$$920$$ 0 0
$$921$$ 0 0
$$922$$ 0 0
$$923$$ − 8.00000i − 0.263323i
$$924$$ 0 0
$$925$$ 0 0
$$926$$ 0 0
$$927$$ 0 0
$$928$$ 0 0
$$929$$ 40.0000 1.31236 0.656179 0.754606i $$-0.272172\pi$$
0.656179 + 0.754606i $$0.272172\pi$$
$$930$$ 0 0
$$931$$ −2.00000 −0.0655474
$$932$$ 0 0
$$933$$ 0 0
$$934$$ 0 0
$$935$$ 0 0
$$936$$ 0 0
$$937$$ 51.0000i 1.66610i 0.553200 + 0.833049i $$0.313407\pi$$
−0.553200 + 0.833049i $$0.686593\pi$$
$$938$$ 0 0
$$939$$ 0 0
$$940$$ 0 0
$$941$$ 12.0000 0.391189 0.195594 0.980685i $$-0.437336\pi$$
0.195594 + 0.980685i $$0.437336\pi$$
$$942$$ 0 0
$$943$$ 0 0
$$944$$ 0 0
$$945$$ 0 0
$$946$$ 0 0
$$947$$ − 36.0000i − 1.16984i −0.811090 0.584921i $$-0.801125\pi$$
0.811090 0.584921i $$-0.198875\pi$$
$$948$$ 0 0
$$949$$ −1.00000 −0.0324614
$$950$$ 0 0
$$951$$ 0 0
$$952$$ 0 0
$$953$$ 36.0000i 1.16615i 0.812417 + 0.583077i $$0.198151\pi$$
−0.812417 + 0.583077i $$0.801849\pi$$
$$954$$ 0 0
$$955$$ 0 0
$$956$$ 0 0
$$957$$ 0 0
$$958$$ 0 0
$$959$$ −36.0000 −1.16250
$$960$$ 0 0
$$961$$ −15.0000 −0.483871
$$962$$ 0 0
$$963$$ 0 0
$$964$$ 0 0
$$965$$ 0 0
$$966$$ 0 0
$$967$$ 59.0000i 1.89731i 0.316310 + 0.948656i $$0.397556\pi$$
−0.316310 + 0.948656i $$0.602444\pi$$
$$968$$ 0 0
$$969$$ 0 0
$$970$$ 0 0
$$971$$ 12.0000 0.385098 0.192549 0.981287i $$-0.438325\pi$$
0.192549 + 0.981287i $$0.438325\pi$$
$$972$$ 0 0
$$973$$ 27.0000i 0.865580i
$$974$$ 0 0
$$975$$ 0 0
$$976$$ 0 0
$$977$$ − 24.0000i − 0.767828i −0.923369 0.383914i $$-0.874576\pi$$
0.923369 0.383914i $$-0.125424\pi$$
$$978$$ 0 0
$$979$$ −48.0000 −1.53409
$$980$$ 0 0
$$981$$ 0 0
$$982$$ 0 0
$$983$$ − 12.0000i − 0.382741i −0.981518 0.191370i $$-0.938707\pi$$
0.981518 0.191370i $$-0.0612931\pi$$
$$984$$ 0 0
$$985$$ 0 0
$$986$$ 0 0
$$987$$ 0 0
$$988$$ 0 0
$$989$$ 32.0000 1.01754
$$990$$ 0 0
$$991$$ 25.0000 0.794151 0.397076 0.917786i $$-0.370025\pi$$
0.397076 + 0.917786i $$0.370025\pi$$
$$992$$ 0 0
$$993$$ 0 0
$$994$$ 0 0
$$995$$ 0 0
$$996$$ 0 0
$$997$$ − 26.0000i − 0.823428i −0.911313 0.411714i $$-0.864930\pi$$
0.911313 0.411714i $$-0.135070\pi$$
$$998$$ 0 0
$$999$$ 0 0
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 5400.2.f.v.649.1 2
3.2 odd 2 5400.2.f.e.649.1 2
5.2 odd 4 5400.2.a.bp.1.1 1
5.3 odd 4 216.2.a.d.1.1 yes 1
5.4 even 2 inner 5400.2.f.v.649.2 2
15.2 even 4 5400.2.a.bn.1.1 1
15.8 even 4 216.2.a.a.1.1 1
15.14 odd 2 5400.2.f.e.649.2 2
20.3 even 4 432.2.a.h.1.1 1
40.3 even 4 1728.2.a.b.1.1 1
40.13 odd 4 1728.2.a.a.1.1 1
45.13 odd 12 648.2.i.a.217.1 2
45.23 even 12 648.2.i.h.217.1 2
45.38 even 12 648.2.i.h.433.1 2
45.43 odd 12 648.2.i.a.433.1 2
60.23 odd 4 432.2.a.a.1.1 1
120.53 even 4 1728.2.a.ba.1.1 1
120.83 odd 4 1728.2.a.bb.1.1 1
180.23 odd 12 1296.2.i.q.865.1 2
180.43 even 12 1296.2.i.a.433.1 2
180.83 odd 12 1296.2.i.q.433.1 2
180.103 even 12 1296.2.i.a.865.1 2
By twisted newform
Twist Min Dim Char Parity Ord Type
216.2.a.a.1.1 1 15.8 even 4
216.2.a.d.1.1 yes 1 5.3 odd 4
432.2.a.a.1.1 1 60.23 odd 4
432.2.a.h.1.1 1 20.3 even 4
648.2.i.a.217.1 2 45.13 odd 12
648.2.i.a.433.1 2 45.43 odd 12
648.2.i.h.217.1 2 45.23 even 12
648.2.i.h.433.1 2 45.38 even 12
1296.2.i.a.433.1 2 180.43 even 12
1296.2.i.a.865.1 2 180.103 even 12
1296.2.i.q.433.1 2 180.83 odd 12
1296.2.i.q.865.1 2 180.23 odd 12
1728.2.a.a.1.1 1 40.13 odd 4
1728.2.a.b.1.1 1 40.3 even 4
1728.2.a.ba.1.1 1 120.53 even 4
1728.2.a.bb.1.1 1 120.83 odd 4
5400.2.a.bn.1.1 1 15.2 even 4
5400.2.a.bp.1.1 1 5.2 odd 4
5400.2.f.e.649.1 2 3.2 odd 2
5400.2.f.e.649.2 2 15.14 odd 2
5400.2.f.v.649.1 2 1.1 even 1 trivial
5400.2.f.v.649.2 2 5.4 even 2 inner | 18,286 | 31,275 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-05 | latest | en | 0.384153 |
https://www.physicsforums.com/threads/g-force-of-wind-on-a-persons-body.619984/ | 1,519,342,472,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814292.75/warc/CC-MAIN-20180222220118-20180223000118-00494.warc.gz | 931,083,429 | 14,535 | # G-force of wind on a persons body.
1. Jul 10, 2012
### cjjjohn2469
After doing some work with a painter friend we were coming home and I put my arm outside of the window as we were going 70 miles an hour. Wondering what the g-forces would be on my arms and hand?
Thank you
2. Jul 11, 2012
### CWatters
There would be no g-force on your arm...
Presumably your arm didn't dissapear behind that car and you were able to hold it out in a more or less stable position? In that case you will feel a force on it due to drag, but it's a constant force not an acceleration so the "g-force" would be zero.
If the airstream was turbulent and your arm flapped up and down then there would be acceleration. Quite hard to quantify though.
3. Jul 11, 2012
### CWatters
Thinking about it some more I suppose you could say that the force due to drag _feels_ like the force due to gravity. I'd estimate it as equivalent to 3-5g.
4. Jul 11, 2012
### cjjjohn2469
Thank you for your reply and explanation you gave. I was just curious and you satisfied that curious thread. Thank you. | 283 | 1,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-09 | longest | en | 0.978385 |
https://dsp.stackexchange.com/questions/3428/time-in-sine-wave-equation/3429 | 1,563,801,171,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528013.81/warc/CC-MAIN-20190722113215-20190722135215-00119.warc.gz | 377,077,005 | 35,230 | time in Sine Wave equation
According to Wikipedia,
The sine wave or sinusoid is a mathematical function that describes a smooth repetitive oscillation.
The formula for the Sine wave is,
A = Amplitude of the Wave ω = the angular frequency, specifies how many oscillations occur in a unit time interval, in radians per second φ, the phase, t = ?
Here ω, is the angular frequency i.e ,
It defines how many cycles of the oscillations are there.
Now let see the frequency,
Frequency is the number of occurrences of a repeating event per unit time
For example, if 100 events occur within 15 seconds the frequency is:
Events = 100;
Time = 15;
FREQ = Events/Time;
Which means there is total 100 cycles of wave in 15 seconds , am i correct ?
Events = 71;
Time = 15;
FREQ = Events/Time;
W = 2*PI*FREQ; % 2*PI for circular path
A = 2 ;
Now what is the use of small 't' ? in the Sine wave equation , what time is it ?
In the context that you described, $t$ is a variable indicating time. That is, the wave will take on a value of $y(t)$ at the time instant $t$. If this is measured in seconds, then $\omega$ specifies the number of radians that the wave passes through in one second. And, as you noted, if the wave has a frequency of $\frac{100}{15}\ \text{Hz}$, then it will pass through 100 periods in 15 seconds. | 337 | 1,321 | {"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} | 3.625 | 4 | CC-MAIN-2019-30 | longest | en | 0.89904 |
https://web2.0calc.com/questions/help-me-with-these-questions | 1,582,502,430,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00366.warc.gz | 585,454,835 | 6,466 | +0
# Help me with these questions
0
318
2
help me with these questions
2)
Nov 27, 2018
#1
+770
0
The volume of a cylinder: $$\pi r^2 \cdot h$$. h = height
The radius is 6 cm, and the height is 4*12 = 48 cm. The volume of the cylinder is $$36\pi \cdot 48 = 1728\pi$$
The volume of a sphere: $$\dfrac{4}{3} \pi r^3$$
The radius is 6 cm, and $$6^3$$ is 216. $$\dfrac{4}{3} \cdot 216 = 288$$. The volume of one sphere is $$288\pi$$. If we multiply by 4, the volume of the four spheres is $$1152\pi$$
We can calculate the empty cylinder space by subtracting 4 VolumeSphere from VolumeCylinder. We have $$1728\pi - 1152\pi$$, which simplifies into $$\boxed{576\pi}$$.
Nov 27, 2018
#2
+107405
+1
For the second one we have
96 = (1/3) (base area) (height)
96 = (1/3) [ x (3x + 1) ] (12)
96 = 4 [x (3x + 1) ] divide both sides by 4
24 = x (3x + 1)
24 = 3x^2 + x rearrange as
3x^2 + x - 24 = 0 factor
(3x - 8) ( x + 3) = 0
Setting each factor to 0 and solve for x and we have that
x = 8/3 cm or x = - 3cm
So...the base area is (x) (3x + 1) = (8/3) (3(8/3) + 1 ) = (8/3)(8 + 1) = (8/3)(9) =
24 cm^2
Nov 27, 2018 | 481 | 1,143 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-10 | latest | en | 0.81866 |
https://bendwavy.org/klitzing/incmats/nd-p.htm | 1,623,742,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487617599.15/warc/CC-MAIN-20210615053457-20210615083457-00482.warc.gz | 134,043,630 | 2,962 | Acronym n/d-p
TOCID symbol (n/d)P
Name n/d-prism,
n-prism with winding number d
` © ©`
Vertex figure [42,n/d]
General of army if d=1: is itself convex
if gcd(n,d)=k: use a (stretched) m-p for its general (with integral m=n/k)
Colonel of regiment (is itself locally convex)
Especially
3/d 4/d 5/d 6/d 7/d 8/d 9/d 10/d 12/d {n/d}-p trip cube pip hip hep op ep dip twip stip 2trip * ship 2cube * step 2pip * 2hip* giship stop 3trip * stiddip 3cube * gistep 2stip * 4trip * stwip
*: Grünbaumian
Confer
special pyramids:
n-p (d=1)
Grünbaumian relatives:
2n/2-p
variations:
(see within files according to individual n/d)
general polytopal classes:
segmentohedra
External
Note that for d odd the 2 n/d-gram layers are aligned in a gyrated way, which then effects that top and bottom vertices are situated on gap. While for d even the 2 n/d-gram layers are still aligned in a gyrated way, but this time it effects in a seeming ungyrated parallel alignment, just like in the mere n/d-p, but now with crossed lacings, so that the according height becomes lesser than 1 here.
Incidence matrix according to Dynkin symbol
```x xn/do (n>2,n/2>d>1)
. . . | 2n | 1 2 | 2 1
--------+----+------+----
x . . | 2 | n * | 2 0
. x . | 2 | * 2n | 1 1
--------+----+------+----
x x . | 4 | 2 2 | n *
. xn/do | n | 0 n | * 2
```
```x2sn/ds (n>2,n/2>d>1)
demi( . . . ) | 2n | 1 2 | 1 2
----------------+----+------+----
demi( x . . ) | 2 | n * | 0 2
sefa( . sn/ds ) | 2 | * 2n | 1 1
----------------+----+------+----
. sn/ds ♦ n | 0 n | 2 *
sefa( x2sn/ds ) | 4 | 2 2 | * n
starting figure: x xn/dx
```
```x2s2n/do (n>2,n/2>d>1)
demi( . . . ) | 2n | 1 2 | 1 2
-----------------+----+------+----
demi( x . . ) | 2 | n * | 0 2
sefa( . s2n/do ) | 2 | * 2n | 1 1
-----------------+----+------+----
. s2n/do ♦ n | 0 n | 2 *
sefa( x2s2n/do ) | 4 | 2 2 | * n
starting figure: x x2n/do
```
```xxn/doo&#x (n>2,n/2>d>1) → height = 1
({n/d} || {n/d})
o.n/do. | n * | 2 1 0 | 1 2 0
.on/d.o | * n | 0 1 2 | 0 2 1
-----------+-----+-------+------
x. .. | 2 0 | n * * | 1 1 0
oon/doo&#x | 1 1 | * n * | 0 2 0
.x .. | 0 2 | * * n | 0 1 1
-----------+-----+-------+------
x.n/do. | n 0 | n 0 0 | 1 * *
xx ..&#x | 2 2 | 1 2 1 | * n *
.xn/d.o | 0 n | 0 0 n | * * 1
``` | 995 | 2,317 | {"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-2021-25 | latest | en | 0.722812 |
https://www.dlsweb.rmit.edu.au/set/Courses/Content/CSIT/oua/cpt160/2014sp4/chapter/01/NumberSystems.html | 1,660,284,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00443.warc.gz | 631,817,674 | 4,536 | # Intro to Computer Systems
## Number Systems
### Binary Numbers
Some quick facts on binary numbers:
• The base for binary numbers is 2
• The symbols are: 0,1
Similarly to decimals, the positional value of each binary digit indicates the corresponding power of 2, as follows:
Digit Power Exponent 1 1 0 0 1 1 0 1 128 64 32 16 8 4 2 1 27 26 25 24 23 22 21 20
Thus, according to this table, the number 11001101 is equivalent to the decimal:
1 * 27 + 1 * 26 + 0 * 25 + 0 * 24 + 1 * 23 + 1 * 22 + 0 * 21 + 1 * 20
= 128+64+8+4+1
= 205
Note: We are calculating the expression above, using the decimal operations that we know.
Find the decimal equivalent of the binary numbers:
%1101 =
%1111111 =
%10000000 =
Some numbers have meaning for different bases, such as 10010 which may be a binary, octal, decimal or hexadecimal number. Hence, it is often necessary to indicate the basis with a little subindex, such as in 100102, 745628 and 7456210. Sometimes a prefix is used, such as % for binary and \$ for hexadecimal.
#### Counting in Binary
Decimal Binary Decimal Binary
0
00000
10
1
00001
11
2
00010
12
3
00011
13
4
14
5
15
6
16
7
17
8
18
9
19
Complete the table.
As we have seen, binary numbers are very easy for computers to store and manipulate, but for humans they are quite uncomfortable to remember and use. The number 010111100110011111000000101 is very hard to remember for a human being, while its equivalent in hexadecimal - as we shall see - is AF33D05, which is much easier to recall.
Quick facts:
• The base for hexadecimal numbers is 16.
• The symbols are: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
Positional Values Decimal Values 164 163 162 161 160 65536 4096 256 16 1
Therefore, the number:
1D7E16
= 1 * 163 + D * 162 + 7 * 161 + E * 160
= 1 * 4096 + 13 * 256 + 7 * 16 + 14 * 1
= 755010
Convert the following numbers to decimal:
\$7E =
\$A8 =
\$13BF =
Decimal Hex Decimal Hex
0
0
10
A
1
1
11
B
2
2
12
3
3
13
4
4
14
5
5
15
6
6
16
7
7
17
8
8
18
9
9
19
Complete the table.
Hexadecimals are inconvenient because they need extra symbols (A-E), but on the other hand they are very useful because each hexa-digit is 4 bits (one nibble), and so a byte is represented neatly by two hexa-digits, as in:
1110 1110 = EE
1111 0110 = F6
1010 0001 = A1
Thus, the most common way to display bytes to human beings is with a two-digit hexadecimal number.
## Conversions Between Number Systems
### Binary to Decimal
Since we know how to operate in decimal, using the positional value of each digit, we can calculate the decimal expression of a binary number:
24 23 22 21 20 1610 810 410 210 110
For example:
101012
= 1 * 16 + 0 * 8 + 1 * 4 + 0 * 2 + 1 * 1
= 16 + 0 + 4 + 0 + 1
= 2110
Convert the following binary values to decimal:
01110 =
11110 =
01101110 =
11100111 =
### Decimal to Binary
Repeatedly divide by 2 until the quotient is zero; for example, to convert 2810 to binary:
2) 28
2) 14 (remainder 0)
2) 7 (remainder 0)
2) 3 (remainder 1)
2) 1 (remainder 1)
2) 0 (remainder 1)
quotient is zero - stop
In the algorithm above, the resulting binary number (from the remainder) should be read upwards.
Therefore, 2810 = 111002
1. Convert the following decimal numbers to binary:
8
17
32
64
127
128
255
2. What is the largest number that may be represented using 4 bits? 8 bits? 16 bits? and 32 bits?
3. How many bits do you need to represent 100010? and 10000010? and 10000000010?
To convert from decimal to hexadecimal the process is the same as above, except now we divide by 16. For example, to convert 34510 to hexadecimal:
16) 345
16) 21 (remainder 9)
16) 1 (remainder 5)
16) 0 (remainder 1)
quotient is zero - stop
Thus 34510 = 15916
16) 74
16) 4 (remainder 10 - i.e. A)
16) 0 (remainder 4)
quotient is zero - stop
Thus 74 10 = 4A16
As we said before, the main advantage of using hexadecimals is that they are convenient for human beings to use, and that it is very easy to convert between them and binary. To convert hexadecimal to binary, write each hexadecimal digit as a 4-bit binary number and put it all together as in the following example:
Hexadecimal Binary F D 6 9 A 1111 1101 0110 1001 1010
The binary expression of \$FD69A is %11111101011010011010. Again, remember that since a byte is 8 bits, each memory location may be represented exactly by two hexadecimal digits.
Convert the following hexadecimal values to binary:
\$FE45 =
\$BBB775 =
\$ABCDEF = | 1,406 | 4,414 | {"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-2022-33 | latest | en | 0.788456 |
https://rpg.stackexchange.com/questions/107220/constant-movement-and-round-based-combat | 1,620,777,721,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00039.warc.gz | 516,526,138 | 43,967 | # Constant Movement and round based combat
This question came up the other day during a session where we have an airship that is basically a hot air balloon attached to a ship and is powered by elementals (Fire elemental for the balloon and air elementals for the propulsion).
While we were on the ship we were attacked by a dragon which landed on the ship. This was fine because we hadn't started moving yet. During the fight one of our PCs initiated the ship bringing it into constant motion. This wouldn't be a problem except that the dragon repeatedly flew off the ship to get a better angle for its breath weapon then landed back on the ship before its turn was over. The reason this was an issue is that the ship moves faster than the dragon. Since it was the dragon's turn it just ignored this as if the ship were standing still even though it was in constant motion.
Is there a good way to solve the problem of moving on and off something in constant motion or are we and the DM just missing something in the rules?
Edit- the dragon has a fly speed of 80ft and never dashed while the ship has a speed of 220ft per round. There was no driver for the ship. The PC started it up and returned to combat. it was traveling through the air in a straight line. Relative motion wouldn’t apply due to the dragons ability to fly against the direction the ship was traveling and still catch back up.
• Related 3.5 question rpg.stackexchange.com/questions/63913/… - about spells, but the key here is framing the encounter or the battlefield. – Wyrmwood Sep 20 '17 at 23:37
• Are you familiar with relative motion? – KorvinStarmast Sep 21 '17 at 2:46
• Relative motion wouldn’t apply because the dragon could fly against the ships direction, breath, then fly back over and land on it. – Chaosweapon Oct 1 '17 at 19:04
## Easiest to treat this as 'Difficult Terrain' for flying movement
Although D&D's round abstraction isn't meant to be completely realistic: the dragon is launching itself from a moving platform, so it could believably even be ahead of the flying ship briefly before landing again (assuming a winged creature the size of a cottage is believably airborne to begin with)... similar to how someone could say, jump 20' forward while on a flying ship without worrying about their jump speed compared to the ship's speed. It's just in the dragon's case: it has some serious wind resistance to contend with (not unlike flying around in strong winds)
Treating this as difficult terrain for flying would be a simple (and supported) method that would halve the distance the Dragon could fly without losing its 'ride' (i.e. something it should probably be able to do briefly). And of course: if it doesn't land on its turn, the ship would move away.
But let's explore a different example: say a melee fighter wants to jump off a very-fast moving wagon, attack someone, then jump back on before the wagon leaves. The DM might decide to use the mounted combat rules as a basis for adjudication: "Once during your move, you can mount a creature that is within 5 feet of you or dismount. Doing so costs an amount of movement equal to half your speed."... which could make the stunt pretty unlikely without further allowances.
• How does this solve the issue? – András Sep 20 '17 at 21:58
• @András By allowing the dragon to jump off a moving ship then quickly land back on it - which a dragon ought to be able to do - while also accounting for the fact that such a maneuver would be more restricted than if the ship were actually stationary so the fact that the ship is moving isn't totally ignored. Sounds good to me. – SirTechSpec Sep 20 '17 at 23:02
• Given the concept of relative motion, the dragon begins at the same speed as the ship. Thus, I'm not sure adding difficult terrain for flying is entirely appropriate. – Wyrmwood Sep 20 '17 at 23:42
• Correct me if I'm wrong, but during a single turn, your melee fighter can either jump off or jump on, not both, no? Per quotes: "Once during your move . . . you can mount . . . or dismount". @Wyrmwood, I think it's about the mentioned wind resistance. – KtX2SkD Sep 21 '17 at 8:59
• @KtX2SkD I'm still pretty certain it would be relative; unless the ship itself is blocking a significant amount while perched. – Wyrmwood Sep 21 '17 at 15:02
## The ship is not moving when the dragon takes its turn.
Sorry, but this is just how the abstraction of turns in the game happens. D&D is not a physics simulator, and never intended to be.
The vehicle moves when it is the driver's turn (just like a mount). On all other turns, the vehicle is considered stationary for all purposes.
So when the vehicle moves, everyone on board of the vehicle move together with the vehicle. Everyone not on board is just left behind and must use their own movement to catch up.
One good strategy is to try and make vehicle move when the dragon is not landed.
The driver can even attempt a maneuver using the vehicle proficiency to shake the dragon off, then move the vehicle. It should be a contested roll, the dragon versus the driver.
On this one the DM should adjucate what ability makes more sense from the roleplaying and description of the maneuver.
Maybe even the PC can use their actions to try and impose advantage / disadvantage on the rolls. Be creative.
This problem also happens during a chase (think a spy vs spy tailing scene in the city alleys). Use contested rolls of stealth, deception versus insight and perception to either tail or shake off the tail.
• If I may, while many times games don't/can't seek 100% realism, it does not necessarily devalue cases where games do/attempt being realistic in certain aspects. In fact, almost any game must fundamentally have a level of realism & correlation to our world, it just boils down most of the time to practicality and/or taste. If I DMed such a situation, I'd investigate all possible alternatives to this... circus of time-freeze... probably not easy, but what's easy about D&D anyways :) – KtX2SkD Sep 21 '17 at 9:17
• @ktx2skd #RantModeOn what baffles me is that people are OK with the HP abstraction: I am a lv 10 barbarian, and I will stand completely still and let my lv 10 fighter friend swing his greataxe at me. For keepers. Green Knight style. There is zero percent chance I'll die. Heck, at lv 20 a barbarian can dive from stratosphere, land on his face, and stand up to fight. but when it comes to the other abstractions the game made to be playable people need to read the book on a headstand to see if they get it. #RantModeOff – Mindwin Sep 21 '17 at 13:29
• Lol, that's news, I'll check the books next time I'm around them for these scenarios. One other example that bugged my taste the most, is the death save system. It's surely not science-based, as I can't fathom a medical injury that without intervention becomes either terminal or not in 30 seconds max., and a corresponding first aid practice that can be "fired & forgotten" after possibly 6 seconds of work. Still pondering it every once in a while... :P – KtX2SkD Sep 21 '17 at 13:53
• @ktx2skd a first aid practice that can be executed without sterelization or implements of any sort with medieval knowledge in 6 seconds. But again, the real world ancient egypt "medics" could predict the gender of a baby from early pregnancy with pinpoint accuracy, so I heard. – Mindwin Sep 21 '17 at 14:06
# The driver should ready an action
If the driver readies an action to move the ship away from the dragon when the dragon tries to land back on the deck, they should be able to escape the dragon. Note that the movement may provoke an attack of opportunity from the dragon, and depending on speed the dragon may be able to hit parts of the ship with its breath weapon the next round.
# If the driver fails to do this, the dragon can land on the ship.
D&D's rules generate a very different metaphysics from the world we live in. Turns are a thing, as much as people would like to pretend that combat rounds happen simultaneously over a 6-second period. You can't remove the effects of the turn-based nature of the combat system without completely replacing the combat system. The ability of the dragon to do this sort of thing is baked into the system and is something you're just going to have to accept, not have the dragon take advantage of, or use a different system.
• Although all of the driver comments are good and make sense the point of the post is that there was no driver, the ship was traveling in a straight line through the air with more fly speed than the dragon but the dragon could just fly off, fly behind the ship in the opposite direction the ship is traveling then use its breath and fly back onto the ship as if it was standing still. This is why i accepted difficult terrain as the answer. – Chaosweapon Oct 1 '17 at 18:53 | 2,037 | 8,835 | {"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-2021-21 | longest | en | 0.978371 |
http://poncelet.math.nthu.edu.tw/d2/curves/lemniscate-tg.html | 1,516,729,402,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00770.warc.gz | 274,637,185 | 1,675 | Tangent to the Lemniscate
Cartesian equation : (x2+y2)2=(x2-y2) Parametric equation : x = a cost / (1+sin2t) , y = a sint cost / (1+sin2t) Polar equation : r = cos 2t | 67 | 170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-05 | latest | en | 0.52477 |
https://www.imathist.com/value-of-log-27-base-3/ | 1,726,141,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00275.warc.gz | 764,950,916 | 50,283 | # Value of log 27 base 3 | Find log_3 27
The value of log 27 base 3 is equal to 3. Here, we learn to find the logarithm of 27 when the base is 3. The formula of log327 is as follows:
log327 = 3.
## Find log 27 base 3
Answer: The value of log 27 with base 3 is equal to 3, that is, log327 = 3.
Explanation:
We know that
27 = 3×3×3
⇒ 27 = 33
Hence, log 27 base 3 will be equal to
log3 27 = log3 33
⇒ log3 27 = 3 log3 3 by the logarithm rule loga bn = n loga b.
⇒ log3 27 = 3 × 1 as we know logaa =1.
⇒ log3 27 = 3.
So the value of log 27 with base 3 is equal to 3, that is, log3 27 = 3.
Read Also: Value of log 8 base 2
Basics of Logarithms
Logarithm rules and properties
Common logarithms and natural logarithms
Solved problems on logarithms
What is log vs ln
## FAQs
### Q1: What is log 27 with base 3?
Answer: Log 27 with base 3 equals 3, that is, log3 27 = 3. Because, 27=33 and take logarithms with base 3. | 338 | 932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.873546 |
https://excel.bigresource.com/copy-Content-variable-in-formula-K3dR0Zcg.html | 1,604,061,344,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00384.warc.gz | 309,645,391 | 13,098 | # Copy Content Variable In Formula
Oct 8, 2013
A
B
1
AAPL
MSFT
2
='C:Documents and SettingsSASTCMy DocumentsStock Data[AAPL.csv]AAPL'!E2
='C:Documents and SettingsSASTCMy DocumentsStock Data[MSFT.csv]MSFT'!E2
How should i COPY automatically "XYL" written as in formula (XYL.csv and XYL') comprehending with the top bar?
Tried "&C1&" but it`s not working.
## Copy The Formula To B3 Based On The Content Of A1?
May 16, 2013
I have created formulas in cells BB2 through BM2 (1-12). I would like to copy the formula to B3 based on the content of A1 (A1 can only be a number from 1-12, representing months).
Example, if A1=3, return the formula in BE2
Formula in BE2 =
=INDEX('1'!\$B\$2:\$AR\$2999,MATCH(INDIRECT("\$A"&ROW()),'1'!\$A\$2:\$A\$2999,0),MATCH(INDIRECT(CHAR(COLUMN()+64)&"1"),'1'!\$B\$1:\$AZ\$1,0))
## Row Variable In Copy Down Formula?
Jun 18, 2014
I would like to do the following in a copied down row where n5 is a cell that contains a number that is added to a row number in order for the range to maintain n5 rows when copied down.
=average(b1:b1+n5)
## VBA - Copy Down Formula With Variable Rows
Apr 13, 2013
Here is my problem. I have a a workbook with 2 sheets.
Sheet 1= Data sheet; Sheet 2 = Table
On sheet 2 I want to copy down an entire row (A8:AH8) but the problem is that the number of copied rows depends on the number of rows contained in sheet 1 column A (-1 row)
So if i have 101 records (100 +label) in column A sheet 1, in my sheet 2, it must copy down the formulas from (A8:AH8) until (A107:AH107)
## Use Range Variable To Copy Formula
Jul 6, 2007
I have below code (option 1) I was hoping would do what I need except instead of copying formula from A1 it copies values... (option 2) copies formula but does not preserve references...any ideas how to copy exact formula so that references are not changed?
Assumptions:
A1 formula = SUM(B2:E2)
in A5 I would like to copy exact formula to keep references to row 2...
I agree I could use \$ in original formula but that would complicate other requirements.
OPTION1:________________
Sub test_var_object()
Dim vRange1 As Range
Set vRange1 = Range("A1")
Range("A5") = vRange1
End Sub
OPTION2:________________
Sub test_var_object()
Dim vRange1 As Range
Set vRange1 = Range("A1")
vRange1.Copy
Range("A5").PasteSpecial
End Sub
## Pattern Search Within A Variable Content
Apr 26, 2007
I have a variable that can contain in no predefined order, a number of letters and one two digit number (code). I would like to extract this two digit number from this variable and based on its the value perform a task. I can test for the specific value of this number as there are only about 10 of these 2 digit codes. Im quite happy to have the routine test for the number by comparing them all one by one as there are only 10 it wount take too long.
The best scenario would be to extract the value into another variable which I then could use to construct a "select case" table from which I could then initiate the required task based on that number.
## Copy Relative Formula & Use Row Reference From Variable
Mar 5, 2008
I have the following code which Clears the content of a cell. The next thing I want to have happen is to have another cell equal a formula but make that formula relative to its place. The formula is the following =IF(ISERROR(VLOOKUP(F56,Routes_All,2,0)),0,VLOOKUP(F56,Routes_All,2,0))
I am not sure how to paste it where the row changes depending on where it is pasted. Here is the code that does not work...
Private Sub CommandButton1_Click()
Worksheets("2008 Log").Select
Dim cRow
cRow = ActiveCell.Row '
Cells(cRow, Range("Column_Type_Of_Ride").Column).ClearContents
Cells(cRow, Range("column_duration").Column).value = "=IF(ISERROR(VLOOKUP(F56,Routes_All,2,0)),0,VLOOKUP(F56,Routes_All,2,0))"
End Sub
As you can see, no matter where I paste it it will always refer to row "F". How can I have it refer to row "cRow"?
## Select Range Based On Variable Cell Content
Apr 30, 2009
Based on if the value in col A contains the characters "TT" I want to select the range starting with this cell and ending at the end at the end of the row I'm using (.End(xlRight) and then merge these cells, change colors etc. And then looping this through a 'range' so that it only occurs where the values occur. I can amend various cells based on this idea, but am unable to identify the range and then merge the cells.
## VBA Macro To Assign HTML Content Of Wikipedia Search To String Variable
Feb 16, 2014
I am doing a Regular Expression search on a string variable assigned to the HTML content of a Wikipedia search. However I am currently manually going to Wikipedia, searching for the term, saving the html page, opening the saved page with Notepad and then copying the content into a cell.
Can the above process be automated with VBA, how to assign the html content of a Wikipedia search to a string variable.
## Insert A Variable Number Of Rows And Copy And Paste From And To Variable Positions
Aug 8, 2009
On the attached Excel file, I have code that will insert a variable number of rows and copy and paste from and to variable positions. That all works fine when run from a command button, but when I try to run it from the Worksheet_Calculate by entering 1 in J1 or K1 (inrange cell is J1+K1 for testing purposes) the CommandButton1_Click sub runs continously until an error occurs.
## Copy Next Cell Content
Feb 10, 2010
Cell A1 needs to contain the contents of A3 without the user having to go and type the entry in each time the next cell along changes.
For example, let's say that last week 1.81 was typed in A2. The user then had to go in to A1 and also type 1.81. This week 1.83 has been entered in A3 so the user will manually have to go in to A1 and type 1.83. Next week when something is entered in A4, the contents of A1 will again need to match the contents of A4 and so on for the next 52 weeks. We'd like a formula in A1 that automatically shows the contents of the next cell along as soon as the content exceeds Zero.
A1 A2 A3 A4 etc
1.81 1.81 1.83 0.00
## VBA To Copy Particular Webpage Content?
Jan 7, 2012
Copy the contents of a text file from a webpage?
For instance, from this page I only want the data in the text box, which can be selected by clicking the Highlight All button.
I've seen code to copy an entire page, but this does not capture the text box contents in this case.
## Copy Content From One Dropdown Box To Another
Feb 22, 2007
I have a combo-box in Sheet Number 1 filled with date. In all my other sheets there a empty combo-boxes. When the workbook opens I automaticly want to copy the content from the combo-box from Sheet 1 into all the other comboboxes in the other sheets. Is there an way to solve this problem with a minimum amount of loops?
## Find Min Then Copy Cell Content
Jul 1, 2009
I have a list of values that are the following:
Column A: City Name
Column B: Distance to destination
Is there a command to find the minimum value in Column B and then copy the corresponding city name into a cell of my choosing.
## Copy Cell Content Between Sheets?
Sep 1, 2013
I want to copy cells content from the sheet1 column A to sheet2,I was tried by the function Sum but problem is some of cells which i tried copy have combination letters and numbers as content.
Maybe, specific things is because i wont copy cell A2 sheet1 to cell A2 sheet2.
## Copy Based On Content To The Left?
May 7, 2014
I would like to paste a formula down a column but it should only paste if there is text in the cell to the right of the column where I am pasting (i.e dragging the formula down).
## Macro - Filter And Copy Associated Content?
Jun 18, 2014
I am having a spot of bother with my spreadsheet, when trying to automate some functionality. Effectively what I am trying to do is...
- With a comprehensive Project Plan press a button that extracts the information of cells that are marked as Critical.
- This information would pull through onto a separate Dashboard sheet, so that those critical items can get flagged to the Project Team.
- The data cannot be copied as a complete table, as there are various columns of data that I do not require copying.
- I have tried recording a macro with me 'filtering' the project plan for critical items and then copying that data across.
- This however only returns the cells originally marked as Critical, it does capture any changes to cells outside of the range in the code.
So,
- In Column C of 'Project Plan' sheet, I have tasks marked as "Critical" or blank.
- I want to copy data of those 'Critical' rows of data, from Columns B,D,F,I
- This data is then to go into the 'Dashboard' sheet, in Columns B,C,E,F.
I embed the code below, from my feeble attempt:
[Code]......
## How Can I Copy Desired Cell Content
Jan 5, 2009
I'm using Excel 2007 and s/s is 325501 rows deep. It consists of series of ranges between 4 and 30 rows deep.
What I want to do is locate the next appearance of a name and copy its accompanying number.
Doing this manully is not feasible, given the large size of the s/s .
I enclose a small attachment showing what I am trying to achieve. For those who don't like opening attachments the wording in it is :
The desired objective is to place in column Q the next appearing number in column L of the name in column C.
The VLOOKUP formula in column Q presents the desired number but (problem!) presents a zero when next appearance = blank.
When this happens I want the formula/code to repeatedly lookup the next appearance until it finds a number.
Examples of where next numbers appear are given here in column R.
## Copy Content To Hidden Sheet
Nov 24, 2006
I have a workbook, wich copies content from an overview sheet to different other sheets. that works fine, but if I try to hide the content-placeholder sheets, I can't copy my content anymore.
here the part where I get the error;
'OldValue contains the name of the "copy to" sheet as a string
Sheets(OldValue).Select
## Copy Email Content To Excel
Dec 10, 2009
I receive about 80-100 emails per week that are computer generated by one of our customers. It contains time entry data that needs to be processed for our internal system.
Today I copy paste the content (it is NOT an attachment!) to Excel and have a macro to decifer and organize the information.
Is there a way to automate the "copy-paste" process, that is, go to the inbox and process each email with a loop like method.
The process needs to be able to go through a Yahoo mail box as well as an Outlook mailbox, Do not know at this time if there is a difference between the two.
## Copy Cells Based On Content
Dec 20, 2006
I have a list of Marketing Product descriptions that I have to match to a list of system codes and was wondering if I can do this in excel.
Marketing Bumph (first Tab):
A B
1 Marketing Code Friendly description
2 XYZ1111 Offer 1
3 ABC111 Offer 2
System Info (real codes not marketing)
A B
1 Marketing Code Real Value
2 XYZ1111 GL321
3 ABC111 FF453
So what I want to end up with is:
A B C
1 Marketing Code Friendly description Real Value
2 XYZ1111 Offer 1 GL321
3 ABC111 Offer 2 FF453
## Copy Content Of Cells From Worksheet To Another
Jul 26, 2007
i would like the code that allows me to copy enything I typing in column a sheet1 to column a in sheet2
May 28, 2008
## Find Cell Content And Copy Row Into New Sheet
Apr 7, 2014
I have tried to these through formulas without success but i think i need VBA also which i am not very experienced.
I want to paste a list in the "InsertList" sheet. This list will only contain the word "Correct" or "False". From then on i need a way to search for the word "Correct" or "False" in the columnS P,Q,R,S,T,U,V.
e.g. If in the column "P" on the "InsertList" sheet the word "Correct" is found, i need that entire row from A to V to be copied onto it's destination, in this case "sheet1".
If the word "Correct" is found on the column "Q" on the "InsertList" sheet, the rows from A to V need to be copied in the Sheet2. And so on..
sheet1.png
Attachment: dropbox.com/s/vgs4kzhoa1pip0a/CopyRows.xltm
## Copy Content Of All Worksheets Onto One Total Sheet
Jan 13, 2009
I have 11 worksheets in one workbook, the last worksheet is a total page.
I'd like to run a macro that copies the information from each sheet and pastes it into the total sheet. My range on every sheet begins at A2, but the end of the range is unknown.
All I have so far is trying to loop to a new empty cell on the total page (coded in a module - is that right?):
## Copy, Insert Row Below And Merge Content Of Two Cels
Feb 9, 2007
I have a list of 12.000 pictures which have to be "connected" with the article number (1000) & color (21). Here starts the problem... Each article usually has more than 2 colors, but is listed only once per row and the colors are stated in columns.
What I would need is a macro or function that would do from such structure:
art description color1 color2 color 3
1000 product A 21 22 23
1001 product B 19 23
this:
art. description picture
1000-21 product A, color 21 1000-21.jpg
1000-22 product A, color 21 1000-21.jpg
1000-23 product A, color 21 1000-21.jpg
1001-19 product B, color 21 1001-19.jpg
1001-23 product B, color 21 1001-19.jpg
1. check if the row in color columns is not null
2. if this is true, then I would like to copy the entire row & paste it below existing
3. then the "art." column would be changed (=A2&"-"&C2), so the article woul get suffix of the color (and column "picture" would be created, maybe something like =A2&".jpg")
4. if any other color column in the same row is greater than null, then proceed to next color & repeat steps 2. & 3., else go to next row
5. until all 5000 rows are converted in about 12.000 rows.
## Macro To Copy Content From Cells Into One Cell
Oct 20, 2008
I have a question.
What code do I need to make the data that is in more cells to be shown as data in one cell ?
ex. A5 D5 to be shown as D2(in a new workbook)
1 1 11
And that for every row?
## VBA To Build Copy Statement From Cell Content
Aug 9, 2006
A single worksheet holds all the values I need to move to various worksheets in the destination workbook. The destination workbook is MASTER.XLS and is already open. The source workbook has various names.
I have the macro walking through each value in column E of the source worksheet. When a match occurs, the corresponding cell in column F has the destination worksheet name, the corresponding cell in column G has the destination cell address and column H has the destination value (string value).
I have dim statements for SheetName, CellAddr and CellVal ; all set for String. I have been playing with "Offset" as well as "Select"ing through the worksheet hierarchy to drive to the desired destination cell. All seem to be more work than necessary and none work properly.
What I am looking for is a set of macro statements that I can use as a "template" within the balance of the macro I have written. I would also happily accept recommendations about books that provide a step-by-step approach to learning the capabilities of Excel's VBA functions. I know from my limited programming background that there are many ways to do the same thing. I'd rather start with the most efficient rather than burn lots of hours experimenting.
## Conditional Copy / Paste Cell Content From One Sheet To Another
May 12, 2014
I need to copy a couple cells from sheet2,3 and 4 to sheet1 depending on value of cell a2 of respective sheets.
I have the basic code here, and what I think I'm missing is the adding row in sheet1.
The below codes can be all wrong by the way, YES, I do not have much knowledge in Macro.
[Code] .....
## VBA / Word Picture Content Control Causing Unreadable Content?
Jan 15, 2013
I am using VBA to create a word document (.docx). This word document contains plain text content controls as well as picture content controls. I then use VBA to automatically select a picture based on the code below
Code:
Set oCC = Word.ActiveDocument.SelectContentControlsByTitle("TabPic").Item(1)
On Error GoTo TabErrorHandler
oCC.Range.InlineShapes.AddPicture Filename:="X:XFERANDREW-TDCD " & LblVL &
[Code].....
After the document has been closed down I try to open it again and I am told "The file cannot be opened because there are problems with the contents."
When I click details it says "Unspecified error" and "Location: Part: /word/document.xml, Line: 2, Column: 0"
If I click ok it says "Word found unreadable content in "". Do you want to recover the contents of this document? If you turst the source of this document, click Yes.
Clicking Yes opens the document with all the contents and it is now renamed to Document 1. If I click no it does not open. | 4,260 | 16,831 | {"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-2020-45 | latest | en | 0.853695 |
http://perplexus.info/show.php?pid=10423&cid=57687 | 1,575,982,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00550.warc.gz | 104,736,620 | 5,031 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Dice Roll Resolution (Posted on 2016-09-08)
Consider an eight-sided die numbered 1 to 8 which is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. For example, a six is three times as probable as a two.
The die is rolled until a run of 8 different faces appears. For example, one might roll the sequence 6565742726486467483472516 with only the last eight rolls all distinct.
(i) What is the expected number of rolls for the above event?
(ii) How will the answer change if the last 8 rolls was required to produce 12345432 (strictly in this order?
No Solution Yet Submitted by K Sengupta No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
simulation of a simpler case | Comment 1 of 4
To test out the possibility that the expected number of rolls is just the reciprocal of the probability of getting the set on the first set of 8, incremented by the 7 rolls before a hit is possible, I started with a simulation of a simpler case, where the digits are equally likely. The expectation in the case this hypothesis is true, in the simplified case, would be 8^8 / 8! + 7 ~= 423. However, presumably because of the non-independence (the overlapping) of successive sets of 8, the simulation shows the average number of rolls to get a hit is around 490. Again, this is the simpler case, where the digits are all equally likely.
Three iterations of 10,000 trials each, show an average number of rolls as:
492.1468 done
488.0111 done
488.6338 done
DefDbl A-Z
Dim crlf\$
Private Sub Command1_Click()
End Sub
Form1.Visible = True
' Text1.Text = ""
crlf = Chr\$(13) + Chr\$(10)
digs\$ = "12345678"
Randomize Timer
tCt = 0: tTr = 0
For trial = 1 To 10000
had\$ = "": good = 0
Do
DoEvents
r = Int(Rnd(1) * 8 + 1)
good = 1
For i = 1 To 8
If InStr(rt\$, Mid(digs, i, 1)) = 0 Then good = 0: Exit For
Next
End If
Loop Until good
tCt = tCt + Len(had): tTr = tTr + 1
Next
Text1.Text = Text1.Text & crlf & tCt / tTr & " done"
End Sub
Posted by Charlie on 2016-09-08 12:10:15
Search: Search body:
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Year 2
# Consolidating and reviewing learning on capacity and volume
## Lesson details
### Key learning points
1. In this lesson, we will review our learning on reading temperature, comparing volumes and solving word problems related to capacity and volume.
### Licence
This content is made available by Oak National Academy Limited and its partners and licensed under Oak’s terms & conditions (Collection 1), except where otherwise stated.
## Video
Share with pupils
## Worksheet
Share with pupils
## Starter quiz
Share with pupils
### 3 Questions
Q1.
A shopkeeper sells water in packs of 4. Each bottle contains 200ml of water. How many millilitres of water are there in each pack?
400ml
600ml
Q2.
Miss Jones has a 1000ml bottle of orange juice. If she drinks 250ml a day. How many days will the orange juice last?
2 days
6 days
Q3.
Sue needs to have a dose of 4ml once a day for five days. How much medicine does the doctor need to make for Sue?
16ml
40ml
## Exit quiz
Share with pupils
### 4 Questions
Q1.
Which amount below is the greatest?
243 ml
325 ml
Q2.
Which amount below is the least amount? | 278 | 1,129 | {"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} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.914847 |
https://strategiccfo.com/articles/accounting/semi-variable-costs/ | 1,718,862,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00235.warc.gz | 477,250,249 | 76,395 | Semi Variable Costs
Semi Variable Costs
Semi Variable Costs Definition
Semi variable costs are costs that include both a fixed and a variable component. They are also called mixed costs.
Semi Variable Cost Example 1
For example, let’s say you subscribe to a phone service that charges \$40 dollars per month, plus \$0.10 per minute for each additional minute beyond 500 minutes per month.
If you talk for less than 500 minutes per month, then the cost is \$40 dollars per month. Beyond 500 minutes, the cost increases. This is an example of a semi variable cost. The flat rate of \$40 dollars for 500 minutes is the fixed cost component. The additional \$0.10 per minute for each additional minute beyond 500 minutes is the variable cost component.
Semi Variable Cost Example 2
Here is an example of a slightly different type of semi variable cost. For example, let’s say a manufacturing company has an electric bill that uses semi variable cost, including a fixed cost component and a variable cost component.
The electric company charges the manufacturing company a flat monthly rate of \$300 dollars per month for basic electricity service. Then they charge \$0.015 per kilowatt hour (kwh). In this example, the flat rate of \$300 dollars per month is the fixed cost component. And the variable cost component is \$0.015 per kwh.
If the manufacturing company uses 50,000 kwhs of electricity in a particular month, then its electric bill would be \$1,050 dollars. (\$1,050 = \$300 + (\$0.015 x 50,000kwhs)). And if the manufacturing company uses 100,000 kwhs of electricity the following month, then its electric bill would be \$1,800 dollars. (\$1,800 = \$300 + (\$0.015 x 100,000kwhs)).
Accounting Treatment
Cost accountants typically separate semi variable costs into their two distinct components – the fixed cost component and the variable cost component – when dealing with semi variable costs. Treat the fixed cost component separately as a fixed cost. Then treat the variable cost component separately as a variable cost. This may cause a differentiation of cost that does not reflect economic reality, but it makes it easier to handle and examine the effects of semi variable costs.
Source:
Barfield, Jesse T., Michael R. Kinney, Cecily A. Raiborn. “Cost Accounting Traditions and Innovations,” West Publishing Company, St. Paul, MN, 1994.
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August 7-10th, 2023 | 584 | 2,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.874411 |
https://socratic.org/questions/how-do-you-rewrite-the-expression-4times-2-6-using-the-distributive-property | 1,638,790,149,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00148.warc.gz | 617,809,591 | 5,757 | # How do you rewrite the expression 4times(2+6) using the distributive property?
Apr 19, 2018
$\left(4 \times 2\right) + \left(4 \times 6\right)$
#### Explanation:
$\text{multiply each of the terms in the parenthesis by } 4$
$\Rightarrow \textcolor{red}{4} \times \left(2 + 6\right) = \textcolor{red}{4} \left(2 + 6\right)$
=(color(red)(4)xx2)+(color(red)(4)xx6))
$= 8 + 24$
$= 32$ | 145 | 389 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-49 | latest | en | 0.510833 |
http://www.aimsuccess.in/2016/11/basic-calculation-quiz-for-ibps-clerk.html | 1,516,531,790,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00001.warc.gz | 388,198,584 | 69,708 | # Basic Calculation Quiz for IBPS Clerk Pre
Directions (Q. 1-10): What will come in place of question mark (?) in the following equations?
1. 34.2 × 17.4 × 1.5 = 2 × ?
1) 432.12
2) 440.62
3) 446.31
4) 448.32
5) 452.4
2. (7776)1.3 × (36)1.25 ÷ (216)² ÷ (1296)–1 = 6?
1) 3
2) 4
3) 5
4) 6
5) 7
3. √1.8225 × √70.56 = ?
1) 11.34
2) 9.72
3) 12.46
4) 8.84
5) None of these
4. 30% of 5/7 of 3/13 of 16/15 of 10920 = ?
1) 448
2) 480
3) 524
4) 576
5) 590
5. 35/7 × 112/3 ÷ 45/42 ÷ 13/5 × 259.5 = ?
1) 920
2) 1050
3) 1130
4) 1280
5) 1520
6. (23.65)² – (48.35)²/0.9 = ?
1) –1976
2) –1864
3) –1724
4) –1684
5) None of these
7. 76% of 960 – 45% of 148 = ?% of 5525
1) 15
2) 20
3) 24
4) 30
5) None of these
8. (4096)3.7 ÷ (256)4.3 × (64)5 ÷ (16)–4 = (4)?
1) 22
2) 24
3) 26
4) 28
5) None of these
9. 32/7 of 45/11 of 3/35 of 3080 = ?
1) 3864
2) 3948
3) 4014
4) 4124
5) None of these
1. 3
2. 5
3. 1
4. 4
5. 2
6. 1
7. 5
8. 4
9. 1
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https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_25&direction=next&oldid=120170 | 1,642,505,211,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00245.warc.gz | 204,189,166 | 11,596 | During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# 2016 AMC 10B Problems/Problem 25
## Problem
Let $f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r$. How many distinct values does $f(x)$ assume for $x \ge 0$?
$\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}$
## Solution
Since $x = \lfloor x \rfloor + \{ x \}$, we have
$f(x) = \product_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)$ (Error compiling LaTeX. ! Undefined control sequence.)
The function can then be simplified into
$$f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)$$
which becomes
$$f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor$$
We can see that for each value of $k$, $\lfloor k \{ x \} \rfloor$ can equal integers from $0$ to $k-1$.
Clearly, the value of $\lfloor k \{ x \} \rfloor$ changes only when $x$ is equal to any of the fractions $\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}$.
So we want to count how many distinct fractions less than $1$ have the form $\frac{m}{n}$ where $n \le 10$. We can find this easily by computing
$$\sum_{k=2}^{10} \phi(k)$$
where $\phi(k)$ is the Euler Totient Function. Basically $\phi(k)$ counts the number of fractions with $k$ as its denominator (after simplification). This comes out to be $31$.
Because the value of $f(x)$ is at least $0$ and can increase $31$ times, there are a total of $\fbox{\textbf{(A)}\ 32}$ different possible values of $f(x)$.
## See Also
2016 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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https://help.scilab.org/docs/6.0.1/ru_RU/taucs_chget.html | 1,620,297,710,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00479.warc.gz | 325,136,653 | 7,863 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
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See the recommended documentation of this function
Справка Scilab >> UMFPACK Interface (sparse) > taucs_chget
# taucs_chget
retrieve the Cholesky factorization at the scilab level
### Syntax
`[Ct,p] = taucs_chget(C_ptr)`
### Arguments
C_ptr
a pointer to the Cholesky factorization (C,p : A(p,p)=CC')
Ct
a scilab sparse matrix (you get the upper triangle i.e. Ct is equal to C')
p
column vector storing the permutation
### Description
This function may be used if you want to plot the sparse pattern of the Cholesky factorization (or if you code something which use the factors). Traditionally, the factorization is written :
`P A P' = C C'`
with P' the permutation matrix associated to the permutation p. As we get the upper triangle Ct (= C'), in scilab syntax we can write :
`A(p,p) = Ct' * Ct`
### Examples
```// Example #1 : a small linear test system
A = sparse( [ 2 -1 0 0 0;
-1 2 -1 0 0;
0 -1 2 -1 0;
0 0 -1 2 -1;
0 0 0 -1 2] );
Cp = taucs_chfact(A);
[Ct, p] = taucs_chget(Cp);
full(A(p,p) - Ct'*Ct) // this must be near the null matrix
taucs_chdel(Cp)```
```// Example #2 a real example
// first load a sparse matrix
// compute the factorization
Cptr = taucs_chfact(A);
// retrieve the factor at scilab level
[Ct, p] = taucs_chget(Cptr);
// plot the initial matrix
scf(0);
clf
PlotSparse(A) ; xtitle("Initial matrix A (bcsstk24.rsa)")
// plot the permuted matrix
B = A(p,p);
scf(1);
clf
PlotSparse(B) ; xtitle("Permuted matrix B = A(p,p)")
// plot the upper triangle Ct
scf(2);
clf
PlotSparse(Ct) ; xtitle("The pattern of Ct (A(p,p) = C*Ct)")
// retrieve cnz
[OK, n, cnz] = taucs_chinfo(Cptr)
// cnz is superior to the realnumber of non zeros elements of C :
cnz_exact = nnz(Ct)
// do not forget to clear memory
taucs_chdel(Cptr)```
• taucs_chfact — cholesky factorization of a sparse s.p.d. matrix
• taucs_chsolve — solves a linear s.p.d. system A*X = B from Cholesky factors of the sparse A
• taucs_chdel — utility function used with taucs_chfact
• taucs_chinfo — get information on Cholesky factors
• taucs_chget — retrieve the Cholesky factorization at the scilab level
• cond2sp — computes an approximation of the 2-norm condition number of a s.p.d. sparse matrix | 742 | 2,364 | {"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-2021-21 | longest | en | 0.559253 |
https://la.mathworks.com/matlabcentral/cody/problems/4-make-a-checkerboard-matrix/solutions/1702243 | 1,606,596,595,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195745.90/warc/CC-MAIN-20201128184858-20201128214858-00447.warc.gz | 368,283,261 | 16,994 | Cody
# Problem 4. Make a checkerboard matrix
Solution 1702243
Submitted on 4 Jan 2019 by Burak Bayram
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 5; a = [1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1]; assert(isequal(a,checkerboard(n)))
2 Pass
n = 4; a = [1 0 1 0; 0 1 0 1; 1 0 1 0; 0 1 0 1]; assert(isequal(a,checkerboard(n)))
3 Pass
n = 7; a = [1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1]; assert(isequal(a,checkerboard(n)))
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Start Hunting! | 345 | 754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.674172 |
http://www.engineeringdrawing.org/2012/05/ellipse-by-arcs-of-circle-method.html | 1,500,684,011,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423839.97/warc/CC-MAIN-20170722002507-20170722022507-00716.warc.gz | 415,448,244 | 10,011 | # Problem 3.6 Engineering Curves
## Procedure:
Step-1 Draw a horizontal major axis of the length 140 mm and give the notations A & B as shown in the figure. And mark a midpoint O on it.
Step-2 Draw a vertical axis, perpendicular to the horizontal axis & passing through the point O; of the length equal to the length of minor axis, which is 100 mm and give the notations C & D as shown in the figure.
Step-3 With the center C or D and length equal to the half of the length of major axis; which is 70 mm; cut the major axis on two sides of the minor axis and give the notations F1& F2 respectively as shown in the figure. These are the focal points of the ellipse.
Step-4 Divide the distance between O & A into five equal divisions and give the notations 1,2,3 etc.
Step-5 Now with the distance equal to A1 and center Fdraw an arc of sufficient length, and with the distance equal to B1 and center Fcut the previously drawn arc on two sides of the major axis and give the notations P1& P1 as shown in the figure.
Step-6 Like in the above stated manner draw arcs with the distances A2 –B2, A3 – B3, A4 – B4 etc. and center F1 – F2 respectively, which should intersect with each other respectively as shown in the figure. And give the notations as P2-P2, P3-P3, P4-P4 etc.
Step-7 Draw a smooth free hand medium dark curve through the points P1, P2, P3 etc. as shown in the figure; in sequence, so the resulting curve is the ellipse.
Step-8 Now mark a point anywhere on the ellipse; i.e., the point M, and connect this point with the focal points F1& F2 with straight lines. Then bisect the angle F1MF2 and draw a line of suitable length and give the notations N – N’ as shown in the figure. This is normal passing through the point M on the ellipse. Then draw a line which is tangent to the previously draw normal and give the notations T-T’, this is tangent passing through the point M on ellipse.
Step-9 Give the dimensions by any one method of dimensions and give the name of the components by leader lines wherever necessary.
## Author: Vasim Machhar
An Engineering Drawing Geek. Working at Noble Group of Institutions, Junagadh as Head of Mechanical Engineering Department. | 545 | 2,188 | {"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} | 3.65625 | 4 | CC-MAIN-2017-30 | latest | en | 0.8983 |
https://practice.geeksforgeeks.org/problems/divisible-by-73224/1/ | 1,638,528,053,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00503.warc.gz | 517,749,130 | 16,735 | Divisible by 7
Easy Accuracy: 65.5% Submissions: 2276 Points: 2
Given an n-digit large number in form of string, check whether it is divisible by 7 or not. Print 1 if divisible by 7, otherwise 0.
Example 1:
Input: num = "8955795758"
Output: 1
Explanation: 8955795758 is divisible
by 7.
Example 2:
Input: num = "1000"
Output: 0
Explanation: 1000 is not divisible
by 7.
You don't need to read input or print anything. Your task is to complete the function isdivisible7​() which takes the string num as inputs and returns the answer.
Expected Time Complexity: O(|num|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |num| ≤ 105
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### Editorial
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Divisible by 7 | 244 | 880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-49 | latest | en | 0.834708 |
https://www.weegy.com/?ConversationId=8VV8ML7E | 1,556,093,325,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578636101.74/warc/CC-MAIN-20190424074540-20190424100540-00343.warc.gz | 873,136,184 | 9,152 | Question and answer
How many yards are in 1 mile?
1 mile = 1,760 yards
Expert answered|Masamune|Points 74520|
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Asked 1/30/2018 9:02:58 AM
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What is the position of 7 in the number 876,543 ?
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The position of 7 in the number 876,543 is: Ten Thousands.
Added 1/25/2018 9:57:20 AM
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What is the sum of 1,015 and 119 ?
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Use the rules of multiplication by powers of 10 to calculate 3×100.
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Which one of the mathematical statement is true? A. 28÷0=0 B.0÷28=0 C. 28+0= 0 D. 28-0=0
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 809 | 2,412 | {"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-2019-18 | latest | en | 0.872819 |
https://acm.wustl.edu/cse232/number-theory.html | 1,550,716,164,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247497858.46/warc/CC-MAIN-20190221010932-20190221032932-00406.warc.gz | 469,127,746 | 40,746 | # Number Theory¶
This page was originally authored by Dr. Brett Olsen, who taught CSE 232 in Fall 2014.
Number theory is the study of the integers. The most basic concept in number theory is divisibility. We say that $b$ divides $a$ (written $b|a$) if $a=bk$ for some integer $k$. We can also say that $a$ is a multiple of $b$, or that $b$ is a divisor of $a$ (if $b \geq 0$). Every positive integer $a$ is divisible by the trivial divisors 1 and $a$, and the nontrivial divisors of $a$ are called the factors of $a$.
##### Greatest Common Divisor¶
The greatest common divisor, or GCD, of two integers $a$ and $b$, is the largest of the common divisors of $a$ and $b$. For example, the factors of 24 are 2, 3, 4, 6, 8, and 12, while the factors of 30 are 2, 3, 5, 6, 10, and 15, so the greatest common divisor of them (or gcd(24,30)) is 6. This has several uses. More prosaically, we can use this to simplify fractions by removing the common factors: e.g., reducing $\frac{24}{30}$ to $\frac{4}{5}$. Euclid's algorithm for calculating the GCD is still the most widely used and simple to program. It exploits the property that:
If $a = bt + r$ for integers $t$ and $r$, then $GCD(a, b) = GCD(b, r)$.
Why? Clearly $a = bt + r$ for some $t$ and $r$ - $r$ is the remainder and $t$ the multiple of $b$. Then $GCD(a, b) = GCD(bt + r, b)$. But any common divisor of $b$ and $bt + r$ must reside entirely in $r$, as $bt$ must necessarily be divisible by any divisor of $b$. So $GCD(a, b) = GCD(r, b)$. So we can write a recursive algorithm to find the GCD of any two positive integers:
In [1]:
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
##### Least Common Multiple¶
The least common multiple (LCM) is a closely related problem, the smallest integer which is divisible by both of a pair of integers. We can calculate it easily using the GCD:
$LCM(a, b) = ab / GCD(a, b)$.
To calculate the GCD and LCM of more than two numbers, we can just nest the calls, e.g., $GCD(a, b, c) = GCD(a, GCD(b, c))$.
#### Primes¶
A prime number is an integer $p > 1$ whose only divisors are 1 and $p$. Primes have a number of useful properties and are essential in number theory. Any number which is not prime is called composite.
##### Testing primality¶
There are an infinite number of primes, but they are not distributed according to any pattern. There are roughly $x / ln(x)$ primes less than or equal to $x$, meaning that roughly 1 out of every $ln(x)$ numbers is prime. The most straightforward way to test whether a number is prime is trial division by candidate divisors. If $N$ is prime, then none of the numbers in $[2, N-1]$ will divide it, so we loop through all of those numbers and test whether any of them are factors of $N$:
In [4]:
def is_prime(N):
for i in xrange(2, N):
if (N % i) == 0:
return False
return True
In [9]:
[p for p in range(2, 50) if is_prime(p)]
Out[9]:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
This is accurate, but we can make a couple of improvements. First, note that any divisor $d$ of $N$ has a paired divisor $d / N$. At least one of $d$ and $d / N$ must be less than or equal to $\sqrt{N}$. That means we only need to test candidate divisors up to a maximum of $\sqrt{N}$ rather than $N-1$. Secondly, we know that 2 is the only even prime, so we can simply skip all other even candidate divisors:
In [18]:
def is_prime_faster(N):
if (N % 2) == 0:
return False
for i in xrange(3, int(sqrt(N)), 2):
if (N % i) == 0:
return False
return True
In [19]:
%timeit [p for p in range(2, 10000) if is_prime(p)]
%timeit [p for p in range(2, 10000) if is_prime_faster(p)]
1 loops, best of 3: 827 ms per loop
100 loops, best of 3: 18.9 ms per loop
Finally, if we already know the primes less than $\sqrt{N}$, we can restrict the candidate divisors to the primes rather than all odd numbers, giving an even larger speed up.
These methods work OK for relatively small numbers, but for large numbers like those primes used for cryptography, faster methods are required. For these, we exploit probabilistic primality testing, which uses a number of trial random numbers to test using methods I won't go into detail about to estimate whether a number is prime. It does have a chance of failure, but that chance can be made arbitrarily low. If you'd like to know more detail, look up the Miller-Rabin primality test algorithm.
##### Generating Primes¶
If we want to generate a list of primes less than some $N$, there's a better way than simply running one of the above is_prime tests on each (odd) number in the range. The algorithm we'll talk about is called the Sieve of Erisothanes.
##### Sieve of Erisothanes¶
The Sieve works by starting with the first prime, 2, and "crossing out" all multiples of 2 between $2^2$ and $N$, marking them as composite. Then it takes the next uncrossed number, 3, and repeats, marking multiples of 3 between $3^2$ and $N$ as composite, leaving the next available prime as 5. We keep doing this, repeating until we've generated all the primes we need. A well-implemented Sieve can generate all the primes less than ~10 million in only a few seconds (which, of course, you can generate before submitting your code and simply load in the list of primes you need from a file or from code). For primes larger than that, you'll want to use some kind of optimized probability primality test.
In [29]:
#Dynamic prime generator using the Sieve
def primes(max=None):
composites = {}
#Yield 2 first, then only loop through odd numbers
yield 2
q = 3
while max is None or q < max:
if q not in composites:
yield q
composites[q * q] = [q]
else:
for p in composites[q]:
try:
composites[p+q].append(p)
except KeyError:
composites[p+q] = [p]
del composites[q]
q += 2
In [32]:
%timeit [p for p in primes(10000000)]
1 loops, best of 3: 7.82 s per loop
#### Prime Factorization¶
The Fundamental Theorem of Arithmetic states that every integer has a unique representation as a multiplication of its prime factors. That is, the prime numbers are the multiplicative building blocks of integers. For example, 1200 can be factored into $2^4 \times 3 \times 5^2$.
A naive algorithm for finding the prime factorization of an integer takes a list of primes (e.g., from a sieve) and simply checks each of them to see which divides the integer. We can do better by dividing the intial integer by each prime factor we find:
In [40]:
def factor(N):
f = {}
for p in primes(sqrt(N)):
if p > N:
break
while (N % p) == 0:
try:
f[p] += 1
except KeyError:
f[p] = 1
N = N // p
if N > 1:
f[N] = 1
return f
In [41]:
print factor(1200)
print factor(136117223861)
print factor(142391208960)
{2: 4, 3: 1, 5: 2}
{104729: 1, 1299709: 1}
{2: 10, 3: 4, 5: 1, 7: 4, 11: 1, 13: 1}
The prime representation of numbers is extremely useful for dealing with very large numbers without integer overflow problems. For example, suppose you were asked how many trailing zeros there are in the decimal representation of 100!. This is a very large number, much too large to fit into a 32-bit integer. Python and Java can do this with their big integer handling, but there's an easier and faster way using the prime representation of the number. A trailing zero at the end of an integer indicates a factor of 10 - the number is divisible by 10. We wish to know how many times we can divide the number by 10, which is equivalent to asking how many prime factors of 2 and 5 are present in the number and reporting the smaller. The factorial function will always add more factors of 2 than 5 to the final integer, so we need only count the number of 5s in the prime factorization.
In [46]:
def count_trailing_zeros(n):
#Count the number of trailing zeros in the factorial of n
count = 0
while n >= 5:
count += n // 5
n = n // 5
return count
In [47]:
print count_trailing_zeros(100)
24
And let's check, just to confirm that our answer is right:
In [44]:
def fact(n):
if n == 1:
return 1
return n * fact(n-1)
print fact(100)
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
#### Modular Arithmetic¶
Sometimes we don't want to work with the exact values of integers - instead we're interested in working with integers modulus some other number. That is, we'll work with the remainder (or residue) of the integer remaining after dividing by the modulus. This could be useful if we want to know, say, what day of the week it will be $N$ days in the future (the $N \; \textrm{mod} \; 7$th day after today), or what the last digit of a large integer is ($N \; \textrm{mod} \; 10$). It's also useful for working with large numbers, where we can more easily work with intermediate values modulo some reasonable number. So let's talk about how modular arithmetic works.
In most languages, there will be a primitive operator (usually %) that gives the remainder of an integer modulus some other number. Let's look at some identities on how to operate on numbers $\; \textrm{mod} \; n$. First, sums and differences modulo $n$ are identical if we first take the modulus of the initial numbers:
$(x + y) \; \textrm{mod} \; n = ((x \; \textrm{mod} \; n) + (y \; \textrm{mod} \; n)) \; \textrm{mod} \; n$
$(x - y) \; \textrm{mod} \; n = ((x \; \textrm{mod} \; n) - (y \; \textrm{mod} \; n)) \; \textrm{mod} \; n$
This suggests we can extend this to multiplication:
$xy \; \textrm{mod} \; n = (x \; \textrm{mod} \; n)(y \; \textrm{mod} \; n) \; \textrm{mod} \; n$
and exponentiation:
$x^y \; \textrm{mod} \; n = (x \; \textrm{mod} \; n)^y \; \textrm{mod} \; n$
Division is tricky - you can't just do this simple method for modular division. In fact, it's complicated enough (and rare enough) that I'm not going to cover it in class, but the short answer is that you need to find the inverse of some $d$ - a number that when multiplied by $d$ is equal to 1 (mod n), and then multiply by that number. But sometimes inverses don't exist, and in general it just gets complicated. Hie thee to a number theory book (e.g. CLRS) if you need more.
Let's look at an example - what is the last digit of $2^{100}$? What we really want to know is $2^{100} \; \textrm{mod} \; 10$. By repeated squaring and taking the remainder mod 10 at each step we can end up with the final result while avoiding ever having to work directly with large numbers:
$2^3 mod 10 = 8$
$2^6 mod 10 = 8 \times 8 mod 10 = 4$
$2^{12} mod 10 = 4 \times 4 mod 10 = 6$
$2^{24} mod 10 = 6 \times 6 mod 10 = 6$
$2^{48} mod 10 = 6 \times 6 mod 10 = 6$
$2^{96} mod 10 = 6 \times 6 mod 10 = 6$
$2^{100} mod 10 = 2^{96} \times 2^3 \times 2 mod 10 = 6 \times 8 \times 2 = 96 mod 10 = 6$
And we can check to make sure that our answer is right:
In [48]:
print pow(2, 100)
1267650600228229401496703205376
##### Chinese Remainder Theorem¶
The Chinese Remainder Theorem deals with finding some $x$ where we know the remainders of $x$ divided by different integers. The theorem states that if the moduli are relatively prime (their GCD is 1) then we are guaranteed a solution (otherwise the different remainders could possibly be inconsistent). Let's look at an example. Suppose we know that
$x \; mod \; 3 = 2$ and
$x \; mod \; 5 = 3$.
We could find the answer(s) by simply listing values of $x$ which hold for the first:
$2, 5, 8, 11, 14, 17, 20, 23, 26, 29\dots$
and second equations:
$3, 8, 13, 18, 23, 28, 33 \dots$
and we can see that $x=8$ and $x=23$ are both valid solutions. But this method could take quite a while if we have large remainders, moduli, or a large number of moduli. | 3,468 | 11,601 | {"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} | 4.8125 | 5 | CC-MAIN-2019-09 | longest | en | 0.904254 |
https://www.coursehero.com/file/6506587/LL2/ | 1,513,160,039,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522343.41/warc/CC-MAIN-20171213084839-20171213104839-00016.warc.gz | 711,501,415 | 32,964 | # LL2 - Part 2 Sampling 1 Statistical Inference 2 Sampling...
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BUS1200-2 1 Part 2 Sampling 1. Statistical Inference 2. Sampling Plans 3. Simple Random Sampling 4. Distributions of Sample Means Section 2.1 Statistical Inference Statistical inference : using data obtained from a sample to make decisions or predictions about the population from which the sample is drawn Parameter : a numerical characteristic of a population (for example, population mean and population variance). The sample results provide estimates of the values of the population parameters. With proper sampling methods, the sample results will provide good estimates of the population parameters. Suppose we want to know the mean income BUS1200-2 2 of all families in a city. The population is therefore the incomes of all families in the city. We then select randomly a family from a city and know its income. In this experiment, the sample space is the city and the sample points are families in the city. The income corresponds to the outcome (a family randomly selected) of the experiment. Therefore we say that the income of a family randomly selected is a random variable. For convenience, we define the distribution of the population as the distribution of this random variable. Section 2.2 Sampling Plans We introduce three different sampling plans. Simple random sampling : all the samples with the same size are equally likely to be chosen. A simple random sample is a sample selected using a simple random sampling plan. BUS1200-2 3 To conduct random sampling, assign a number to each element of the chosen population (or use already given numbers), randomly select the sample numbers using a random number table or a software package. Stratified random sampling
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Ask a homework question - tutors are online | 463 | 2,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-51 | latest | en | 0.844278 |
https://cs.stackexchange.com/questions/7448/is-this-data-structure-a-hypertree-or-they-are-just-isomorphic-trees | 1,713,666,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00735.warc.gz | 165,564,805 | 40,436 | # Is this data structure a hypertree or they are just isomorphic trees?
I have a data structure described as following:
- It's a collection of trees.
- Each tree has the same structure.
- Each tree has information of diferent nature.
A example of this data structure:
4 - - - - - - - 'a' - - - - - - - 3.5
/ \ / \ / \
3 1 'f' 'y' 1.0 3.1
/ \ / \ / \
4 7 'e' 'f' 2.3 7.7
If you see, ignoring the contents of each tree, each of them is just the same tree (the same hierarchy, the same structure). The first contains natural numbers, the second one, characters, and the third one, floating numbers.
The idea of this data structure is that diferent classes (in a C++ program for example) explore diferent "layers" of this tree, in order to each class uses only the information this class needs, ignoring the remaining information. In other words, each class sees only one tree.
Does this data structure match with a hypertree or can it be reformulated/adapted to a hypertree? Or just they are isomorphic trees put together is a same data structure?
• No, this is not a hypertree. You have parallel trees (see parallel arrays). Dec 11, 2012 at 13:43
What you have is in effect only a single tree, e.g. $$T = \left\{\mspace{45mu}\begin{array}{c@{}c@{}c} & (4,a\mspace{-18mu}&,3.5) \\ &/&\backslash \\ &\mspace{-60mu}(3,f,1.0) & (1,y\mspace{-36mu}&,3.1) \\ & & /&\backslash \\ & &\mspace{-60mu}(4,e,2.3) & (7,f,7.7)\mspace{-60mu} \end{array}\mspace{50mu}\right\}$$ and mappings from this tree over tuples, to the trees which are visible by the classes, $$A = \left\{\mspace{-10mu}\begin{array}{c} 4 \\ \;/ \;\;\backslash & \\ \;3 \quad 1 \\ \;~ \quad\; / \;\; \backslash \\ \;~ \quad\;4 \quad7 \end{array}\mspace{-10mu}\right\}, \qquad B = \left\{\mspace{-10mu}\begin{array}{c} a \\ \;/ \;\backslash & \\ \,f \quad y \\ \,~ \quad\; / \;\; \backslash \\ \,~ \quad\;e \quad f \end{array}\mspace{-10mu}\right\}, \qquad C = \left\{\mspace{-20mu}\begin{array}{c} 3.5 \\ \,/ \;\;\backslash & \\ \;1.0 \quad 3.1 \\ \;~ \quad\;\;\;\; / \;\; \backslash \\ \;~ \quad\;\;\;\; 2.3 \quad 7.7 \end{array}\mspace{-10mu}\right\}.$$ The mappings which carry $T \mapsto A$, $T \mapsto B$, and $T \mapsto C$ can be desribed in terms of the projectors \begin{align*} \pi_1 \;\;:\;\; \text{int} \times \text{char} \times \text{float} &\to\;\; \text{int} \;, \\ \pi_2 \;\;:\;\; \text{int} \times \text{char} \times \text{float} &\to\;\; \text{char} \;, \\ \pi_3 \;\;:\;\; \text{int} \times \text{char} \times \text{float} &\to\;\; \text{float} \;. \end{align*} Each of these "projector" maps $\pi_j$ can be extended to a mapping on trees, using the tree functor which carries all functions $f: A \to B$ to functions on trees, $\text{Tree}(f) : \text{Tree}(A) \to \text{Tree}(B)$ in the obvious way, preserving the tree structure for any given tree, and applying the function $f$ to each node in the tree. Thus \begin{align*} A &= \Bigl[\text{Tree}(\pi_1)\Bigr](T), \\ B &= \Bigl[\text{Tree}(\pi_2)\Bigr](T), \\ C &= \Bigl[\text{Tree}(\pi_3)\Bigr](T). \end{align*} So these trees are just the images of $T$ under projection maps in the natural way; they aren't meaningfully independent of one another, and what the classes do which can only access one of these trees is that they can only access information about $T$ through these projections. | 1,145 | 3,427 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-18 | latest | en | 0.78701 |
https://www.gurufocus.com/term/Total%20Equity/WFC/Total%252BEquity/Wells+Fargo+%2526+Co | 1,505,971,825,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687642.30/warc/CC-MAIN-20170921044627-20170921064627-00516.warc.gz | 818,935,621 | 40,543 | Switch to:
Wells Fargo & Co (NYSE:WFC) Total Equity: \$205,230 Mil (As of Jun. 2017)
Wells Fargo & Co's total equity for the quarter that ended in Jun. 2017 was \$205,230 Mil.
Total equity is used to calculate Book Value per Share. Wells Fargo & Co's Book Value per Share for the quarter that ended in Jun. 2017 was \$36.13. The ratio of a company's debt over equity can be used to measure how leveraged this company is. Wells Fargo & Co's Debt-to-Equity for the quarter that ended in Jun. 2017 was 1.63.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Wells Fargo & Co Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Total Equity 157,554.00 170,142.00 184,394.00 192,998.00 199,581.00
Wells Fargo & Co Quarterly Data
Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Total Equity 201,745.00 203,028.00 199,581.00 201,500.00 205,230.00
Calculation
Total Equity refers to the net assets owned by shareholders.
Total Equity and Total Liabilities are the two components for Total Assets.
Wells Fargo & Co's Total Equity for the fiscal year that ended in Dec. 2016 is calculated as
Total Equity = Total Assets(Q: Jun. 2017 ) - Total Liabilities(Q: Jun. 2017 ) = 1930115/td> - 1730534 = 199,581
Wells Fargo & Co's Total Equity for the quarter that ended in Jun. 2017 is calculated as
Total Equity = Total Assets(Q: Jun. 2017 ) - Total Liabilities(Q: Jun. 2017 ) = 1930871 - 1725641 = 205,230
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
1. Total equity is used to calculate book value per share.
Wells Fargo & Co's Book Value per Share for the quarter that ended in Jun. 2017 is
Book Value per Share = (Total Equity - Preferred Stock) / Shares Outstanding (Diluted Average) = (205230 - 25785) / 4966.77 = 36.13
2. The ratio of a company's debt over equity can be used to measure how leveraged this company is.
Wells Fargo & Co's Debt-to-Equity for the quarter that ended in Jun. 2017 is
Debt-to-Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt & Capital Lease Obligation) / Total Equity = (95356 + 238869) / 205230 = 1.63
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
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# p2020chap12 - PHYS-2020: General Physics II Course Lecture...
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Unformatted text preview: PHYS-2020: General Physics II Course Lecture Notes Section XII Dr. Donald G. Luttermoser East Tennessee State University Edition 3.3 Abstract These class notes are designed for use of the instructor and students of the course PHYS-2020: General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. These notes make reference to the College Physics, 9th Edition (2012) textbook by Serway and Vuille. XII. Mirrors and Lenses A. Plane Mirrors. 1. Images formed by plane ( i.e. , flat) mirrors have the following properties: a) The image is as far behind the mirror as the object is in front. b) The image is unmagnified , virtual , and erect . 2. Image orientation: a) Erect : Image is oriented the same as the object. b) Inverted : Image is flipped 180 with respect to the ob- ject. 3. Image classification: a) Real : Image is on the same side of mirror as the object = light rays actually pass through the image point. b) Virtual : Image is on the opposite side of mirror from object = light rays appear to diverge from image point. 4. Image size is determined by the magnification of an object which is given by M image height object height = h prime h . (XII-1) XII1 XII2 PHYS-2020: General Physics II | M | > 1 = Image is bigger than object (magnified). | M | = 1 = Image is unmagnified (like a plane mirror). | M | < 1 = Image is smaller than object (demagnified). M > = Image is erect. M < = Image is inverted. M = 0 = No image is formed. 5. Ray Tracing Rules: a) Images form at the point where rays of light actually in- tersect (for real images) or from which they appear to originate (for virtual images). b) For plane mirrors, p (the object distance from the mirror) = q (the image distance from the mirror) and h = h prime . c) The following diagram shows how images are constructed for a plane mirror. i) One ray runs parallel to the optical axis ( line to the mirror surface at the center of the mirror) from the head of the object ( e.g. , Ray 1 in the figure). ii) One ray travels from the head through the mirror at the point where the optical axis intersects the mirror ( e.g. , Ray 2 in the figure). Mirror OBJECT h IMAGE h Optical Axis (defined wrt object base) p 1 q 2 i r Donald G. Luttermoser, ETSU XII3 B. Spherical Mirrors 1. Spherical mirrors have the shape of a segment of a sphere. a) Concave mirror : Reflecting surface is on the inside of the curved surface. b) Convex mirror : Reflecting surface is on the outside of the curved surface. Concave Mirror (this side) Convex Mirror (this side) 2. Constructing the image. Consider the following concave mirror: O h p Optical Axis (defined wrt mirror center) C Vertex (V) R I h q i r XII4 PHYS-2020: General Physics II a) The line that is normal to the mirror surface at the exact center is called the optical axis of the mirror....
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## p2020chap12 - PHYS-2020: General Physics II Course Lecture...
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Ask a homework question - tutors are online | 898 | 3,626 | {"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-2017-04 | longest | en | 0.888778 |
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### ECE :: Measurements and Instrumentation
1. To increase Q factor of a coil, the wire should be
2. A. long B. thin C. thick D. long and thin
3. An ammeter of 0-25 A range has a guaranteed accuracy of 1% of full scale reading. The current measured is 5 A. The limiting error is
4. A. 2% B. 2.5% C. 4% D. 5%
5. The coil of a moving iron instrument has a resistance of 500 Ω and an inductance of 1 H. It reads 250 V when a 250 V dc is applied. If series resistance is 2000 Ω, its reading when fed by 250 V, 50 Hz ac will be
6. A. 260 V B. 252 V C. 250 V D. 248 V
7. In a CRO which of the following is not a part of electron gun
8. A. cathode B. grid C. accelerating anode D. X - Y plates
9. If reference sound pressure P0 is 2 x 10-5 N/m2, a sound pressure of 90 dB is equal to
10. A. 0.632 N/m2 B. 0.707 N/m2 C. 0.835 N/m2 D. 0.925 N/m2
11. Which of the following voltmeters would you use for measuring voltage across 20 kΩ resistance?
12. A. Voltmeter having a resistance of 5 kΩ B. Voltmeter having a sensitivity of 1 kW/V C. Voltmeter having sensitivity of 10 kW/V D. None of the above
13. A single phase energy meter has the rating 1200 resolutions/ kWh. If a 500 W electric gadget is used for 4 hours, the energy meter will make
14. A. 1200 revolutions B. 1800 revolutions C. 2100 revolutions D. 2400 revolutions
15. Assertion (A): De sauty's bridge is suitable only for pure capacitor.
Reason (R): Capacitors are mostly perfect.
16. A. Both A and R are true and R is correct explanation of A B. Both A and R are true but R is not correct explanation of A C. A is true R is false D. A is false R is true
17. In which of the transformer is the secondary nearly short circuited under normal operating conditions?
18. A. CT B. PT C. Distribution transformer D. Power transformer
19. A moving coil instrument has a resistance of 0.6 Ω and full scale deflection at 0.1 A. To convert it into an ammeter of 0-15 A range, the resistance of shunt should be
20. A. 0.6 Ω B. 0.06 Ω C. 0.1 Ω D. 0.004 Ω | 650 | 2,086 | {"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} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.852495 |
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# STEP maths I, II, III 1991 solutions
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1. (Original post by insparato)
I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.
Exactly.
When i attempted this question, this is what plagued me... I don't think you can assume that the two spheres are going to amount to half the initial height and be half the initial radii.
Form a differential equation for R as a function of time. Solve it for both spheres. Add the answers to get a formula for height as a function of time. Etc...
2. (Original post by insparato)
I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.
Ah, of course! Thanks, this should set me off
3. and
Separating variables
Separating variables on the other one
What do I do with the constants now? They're bugging me.
is the diameter (height) of the upper sphere and is the diameter of the lower sphere. So the height is
So 0.5=h(t) but don't see how this helps. I'm giving up, I have never been able to do rates of change:/
4. I'd do something like:
Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.
So
And
So , so for some constant k.
Subbing in for the initial values, we get .
Now find kt when R_u+R_l = 5R/2, etc...
5. (Original post by DFranklin)
I'd do something like:
Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.
So
And
So , so for some constant k.
Subbing in for the initial values, we get .
Now find kt when R_u+R_l = 5R/2, etc...
This question is number 7 in Advanced Problems in Mathematics
6. II/6:
The inequality is satisfied at x = 0. Differentiating: , which is true. So x increases faster than sin x which increases faster than 0, so the inequality will still be satisfied: so .
The inequality is satisfied at x = 0. Differentiating: , which has been shown previous. So cos x decreases slower than 1 - x^2/2, so the inequality will still be satisfied: so .
is satisfied at x = 0. Differentiating: , which has been previously shown. sin x increases faster than x - x^3/6, so the inequality holds.
Similarly:
7. STEP II Q16
Part 1a
Obviously it needs to rain on both days. The next few arguments assume it does. Note also it is necessary that . If this is not the case, then , and the most specific expression for u is . If then the following applies.
At midnight on the first day the volume will be u below the top.
We require that on the second day it rains at a time t such that ie .
The probability of this being the case is .
So and .
Part 1b
We require where R is a random variable representing the number of times it rains, .
can be approximated by .
where .
We require so we look for a value of k such that .
Now so .
The probability must be strictly less than 0.1 so he should set k=9.
Part 2
The variance in the total amount of rain will be lower, allowing for a lower value of k.
8. On the last part of question 15 of STEP II I got instead of . Did anybody else get this? What are the chances they made a typo? I may type up my solution in a bit, but I just typed up 16 and I'm latexed out.
9. (Original post by justinsh)
I may have done mistakes but here it goes:
In my personal opinion, both solutions to STEP III Q1 part b on here fail to find a general term (although this second one is quite nice with two fairly simple general terms), so I am going to post the way I did it, which is quite similar to the above method except that I actually put a general term, rather than a set of general terms. I don't know whether they expect you to do this, but before looking at these I presumed you had to. Anyway, here goes:
It's just that this question would be too easy without this, I think that what they're testing is your ability to think of a way to merge the xn and x2n terms, and finding a fairly simple function that returns 1 for even n and 0 for odd n.
10. (Original post by Rabite)
I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...
Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors.
I updated the attachment.
Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.
1. There is a typo early on, you put 11 instead of 1.
2. No minimum at (1, -4) of any sort.
3. But there is a minimum at (2, -5), which is also necessary for the graph.
4. In the last part, you are defining f-1(x) so the x we are talking about is a different x to before if that makes sense, ie it is what was on the y-axis before not the x-axis. So your table thing doesn't really make sense.
5. If we ignore point 4, the right-most box is wrong anyway, because the positive root is only valid to the right of 2.
So in the last part you need to consider . The first is undefined, all the others have two possibilities (one of which is the same for all four). It's seriously boring stuff. This question is not particularly challenging, but loads of cases to consider, and loads of room for little errors.
11. STEP III Q16
Part 1
Part 2
Part 3
for even y.
Part 4
for odd y.
Part 5
Woooooooo! This has taken me almost all day and I've finally done it! I'm almost certain the answer is right as it works for a=2 and a=3, and everything follows nicely.
12. This alternative to III/1 part (c) is quite cute.
from the binomial expansion.
13. (Original post by Rabite)
I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...
Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors.
I updated the attachment.
Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.
Hi there just wanted to point out one final thing i noticed when doing this question about the inverse functions is you have to check the functions are still valid in the intervals you gave them:
For instance for -1<x<0
is undefined
Also, in the interval x<-1
is undefined once x<-5 so the interval should be -5<x<-1
and the inverse functions are undefined for the intervals -1<x<0 and x<-5
14. I think these solve 1991 Step 3 Q4 and Q9
Attached Files
15. Step1991Paper3Question4.pdf (56.5 KB, 137 views)
16. Step1991Paper3Question9.pdf (77.4 KB, 119 views)
17. STEP 1991 Paper II question 9
18. An absolute swine to Latex but here goes
19. Attached Thumbnails
20. Attached Thumbnails
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# Two trihybrid plants are mated. What is the expected proportion of progeny plant
wrote...
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6 years ago Edited: 6 years ago, morgan795
Two trihybrid plants are mated. What is the expected proportion of progeny plant Suppose a polygenic system for producing color in kernels of a grain is controlled by three additive genes, G, M, and T. There are two alleles of each gene, G1 and G1, M1 and M2 , and T1 and T2 . The phenotypic effects of the three genotypes of the G gene are G1G1 = 6 units of color, G1G2 = 3 units of color, and G2G2 = 1 unit of color. The phenotypic effects for genes M and T are similar, giving the phenotype of a plant with the genotype G1G1M1M1T1T1 a total of 18 units of color and a plant with the genotype G2G2M2M2T2T2 a total of 3 units of color. Part AHow many units of color are found in trihybrid plants?A) 9 units of colorB) 1 units of colorC) 3 units of colorD) 12 units of colorPart BTwo trihybrid plants are mated. What is the expected proportion of progeny plants displaying 9 units of color?Express your answer as a fraction (example 3/4).Expected proportion: Incorrect Answers: 8/64 & 20/64Part CSuppose that instead of an additive genetic system, kernel-color determination in this organism is a threshold system. The appearance of color in kernels requires 9 or more units of color; otherwise, kernels have no color and appear white. In other words, plants whose phenotypes contain 8 or fewer units of color are white. Based on the threshold model, what proportion of the F2 progeny produced by the trihybrid cross in part (b) will be white?A) 1/8B) 22/64C) 1/4D) 24/64Part DAssuming the threshold model applies to this kernel color system, what proportion of the progeny of the cross G1G2M1M2T2T2 × G1G2M1M2T1T2 do you expect to display colored kernels? A) 14 of the progeny will be white and 34 will be colored.B) 12 of the progeny will be white and 12 will be colored.C) 34 of the progeny will be white and 14 will be colored.D) None of the progeny will be colored.Post Merge: 6 years agoI was thinking that Part B could be 18/64 but I want to be definite before I answer again. Post Merge: 6 years agoPart C is 22/64 Read 6214 times 13 Replies
Replies
jdny96jdny96
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6 years ago
incrediblealex,
### Related Topics
wrote...
6 years ago
Thank You
wrote...
6 years ago
Did you get Part B?So far For Part B: it's NOT 1/64teacher's hint: Keep in mind what the expected frequencies of heterozygous genotypes are at any one locus. (Is it really 1/4??).
wrote...
6 years ago Edited: 6 years ago, morgan795
No, Ive tried 8/6420/64and 18/64and you said its not 1/64. Post Merge: 6 years agoi think it may be 1/8
wrote...
6 years ago
From the textbook (page 42)The probability that offspring of a trihybrid self-fertilization will be trihybrid is (1/2) (1/2) (1/2) = 1/8
wrote...
6 years ago
Yay, thats what I was thinking. Thank You !
wrote...
A year ago
Thank you
wrote...
11 months ago
thank you
wrote...
9 months ago
TY :)
wrote...
7 months ago
Thank you
wrote...
7 months ago
Thank you so much
wrote...
A month ago
Thanks
wrote...
A month ago
TY :)
Explore | 983 | 3,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-49 | latest | en | 0.850712 |
https://www.folkstalk.com/tech/create-list-of-numbers-with-code-examples/ | 1,713,014,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00775.warc.gz | 736,428,047 | 6,208 | # Create List Of Numbers With Code Examples
Create List Of Numbers With Code Examples
In this tutorial, we will try to find the solution to Create List Of Numbers through programming. The following code illustrates this.
```a_list = list(range(1, 5))
print(a_list)
[1,2,3,4,5]
```
The identical problem Create List Of Numbers can be fixed by employing an alternative method, which will be discussed in more detail along with some code samples below.
```import numpy as np
Vi = 0 # Initial value
Vf= 10 # Final Value
## Two ways (A and B)
# A) list (E_l ):
''' Use range function in Python 3.x range is a iterator you need to convert it to a list''';
l = list(range(Vi, Vf))
# B)list incremented by N:
N = 0.5 # Increment by 0.5
l_by = np.arange(Vi, Vf, N).tolist()
# Print
print(f' list = {l}')
print(f' list = {l_by}, by: {N} ')
```
As we have seen, the Create List Of Numbers problemcode was solved by using a number of different instances.
## How do I make a list of numbers?
To start a numbered list, type 1, a period (.), a space, and some text. Word will automatically start a numbered list for you. Type* and a space before your text, and Word will make a bulleted list. To complete your list, press Enter until the bullets or numbering switch off.
## How do you create a list?
Create a new list
• On your Android phone or tablet, open the Google Keep app .
• Next to "Take a note," tap New list .
• When you're done, tap Back .
## How do you make a list of numbers in Python?
Using Python range() Python comes with a direct function range() which creates a sequence of numbers from start to stop values and print each item in the sequence. We use range() with r1 and r2 and then convert the sequence into list.27-Sept-2022
## How do you create a list from 1 to 100 in Python?
To create a list with the numbers from 1 to 100 using Python, we can use the range() function.
• list_1_to_100 = range(1, 101) Python.
• list_1_to_100 = [] for x in range(1,101): list_1_to_100. append(x)
• numbers_1_to_10 = list(range(1,11))
• list_1_to_100 = range(1, 101)
• list_1_to_100 = [] for x in range(1,101): list_1_to_100.
## How do I create a list of numbers in Excel?
Fill a column with a series of numbers
• Select the first cell in the range that you want to fill.
• Type the starting value for the series.
• Type a value in the next cell to establish a pattern.
• Select the cells that contain the starting values.
• Drag the fill handle.
## How do I create a numerical list in Excel?
Click the Home tab in the Ribbon. Click the Bullets and Numbering option in the new group you created. The new group is on the far right side of the Home tab. In the Bullets and Numbering window, select the type of bulleted or numbered list you want to add to the text box and click OK.01-Feb-2021
## How do I make a list in C++?
The C++ fill constructor std::list::list() constructs a new list with n elements and assign zero value to each element of list.
## How do I create a list template?
Open the list that you want to save as a template. Select Settings, and then select List Settings. In the Permissions and Management column, select Save list as template. The Save as Template page appears.
## How do you create a list in Word?
Type * (asterisk) to start a bulleted list or 1. to start a numbered list, and then press Spacebar or the Tab key. Type some text. Press Enter to add the next list item.
## How do you print 1 to 10 in a list Python?
Python: Generate and prints a list of numbers from 1 to 10
• Sample Solution:
• Python Code: nums = range(1,10) print(list(nums)) print(list(map(str, nums)))
• Flowchart:
• Python Code Editor:
• Have another way to solve this solution?
• Previous: Write a Python program to print letters from the English alphabet from a-z and A-Z. | 990 | 3,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-18 | latest | en | 0.79408 |
http://jwilson.coe.uga.edu/EMAT6680Su09/Russell2/SPECPROJ/KIDS/functions/Functions.html | 1,545,067,639,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828697.80/warc/CC-MAIN-20181217161704-20181217183704-00181.warc.gz | 146,234,299 | 2,157 | Valerie Russell
Unit 1 Functions
By Ann, Nidya Javaid, Aaliyah Hale, Yessica Martinez, and Quandra Davis
Our quilt design was based on Unit 1. The design consists of 2 parabolas and 4 square root functions. It also includes 2 cubic functions. We chose our colors because we thought they fit well together. The bright parabola really compliments the dark purple background. The border also fits well with the rest of the design. We chose a light blue color for the parabolas so it would pop out against the dark background. The border is there as a compliment not a prominent figure.
We started out with many complications in the beginning. Before deciding on Unit 1, we were going to do Unit 6 and use the discriminate as the main theme of our design. After much frustration, we decided to settle on capturing what we learned about functions (parabolas, cubic, and square roots) a part of Unit 1. We had to come up with a design that incorporated input from everyone in our group. We asked ourselves, "How can we bring all these functions into one?"
The cloth was and ribbons were purchased separately by three of us. After getting the materials we cut out the fabric from our pattern created on large post-it paper. Getting started was the hardest part because there were so any things to do. Once we divided up our roles it sort of came together.
Having access to the iron and cutting board needed to cut out everything for our quilt wasn't easy. Aaliyah accidentally got burned while we were ironing open the seams. We became impatient while waiting to use the cutting board and decided to use our scissors. Not being master cutters we had uneven slanted edges. Ms. Russell made us re cut everything making sure our measurements were exact on opposite sides.
Creating this quilt gave us a better understanding of the functions we studied. We had to understand the value of translations in creating patterns and what happens to functions when the negative sign appears in different places. We began by creating four quadrants with ribbon for the x and y axes and a red dot for the origin. If you know how to translate one function the others will follow the same pattern. You just need to know what the parent function looks like.
Return | 472 | 2,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-51 | latest | en | 0.978954 |
https://emilhvitfeldt.github.io/rstats-adventofcode/2021.html | 1,716,238,788,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00658.warc.gz | 205,519,693 | 27,687 | # Day 1
## Part 1
``````input <- scan("2021/01-input")
sum(diff(input) > 0)``````
``0.002 sec elapsed``
## Part 2
``````input <- scan("2021/01-input")
sum3 <- slider::slide_dbl(input, mean, .before = 2)
sum(diff(sum3) > 0)
# Simpler solution thanks to https://twitter.com/trang1618
# This works because `lag` happens to use the same direction
# as `.before`
#
# sum(diff(input, lag = 3) > 0)``````
``0.044 sec elapsed``
# Day 2
## Part 1
``````input <- read.delim("2021/02-input", sep = " ", header = FALSE)
horizontal <- 0
depth <- 0
for (i in seq_len(nrow(input))) {
command <- input[i, 1]
value <- input[i, 2]
if (command == "forward") {
horizontal <- horizontal + value
}
if (command == "up") {
depth <- depth - value
}
if (command == "down") {
depth <- depth + value
}
}
depth * horizontal``````
``0.04 sec elapsed``
## Part 2
``````input <- read.delim("2021/02-input", sep = " ", header = FALSE)
horizontal <- 0
depth <- 0
aim <- 0
for (i in seq_len(nrow(input))) {
command <- input[i, 1]
value <- input[i, 2]
if (command == "forward") {
horizontal <- horizontal + value
depth <- depth + aim * value
}
if (command == "up") {
aim <- aim - value
}
if (command == "down") {
aim <- aim + value
}
}
depth * horizontal``````
``0.037 sec elapsed``
# Day 3
## Part 1
``````input <- readLines("2021/03-input") |>
strsplit("") |>
purrr::reduce(rbind)
common_finder <- function(x, fun, even) {
tab <- table(x)
if (length(tab) == 2 & tab[1] == tab[2]) return(even)
names(which(fun(tab) == tab))
}
rate_calculator <- function(mat, fun, even) {
apply(mat, MARGIN = 2, FUN = common_finder, fun, even) |>
paste0(collapse = "") |>
strtoi(base = 2)
}
gamma <- rate_calculator(input, max, "1")
epsilon <- rate_calculator(input, min , "0")
gamma * epsilon``````
``0.088 sec elapsed``
## Part 2
``````input <- readLines("2021/03-input") |>
strsplit("") |>
purrr::reduce(rbind)
common_finder <- function(x, fun, even) {
tab <- table(x)
if (length(tab) == 2 & tab[1] == tab[2]) return(even)
names(which(fun(tab) == tab))
}
rate_calculator <- function(mat, fun, even) {
considered <- !logical(nrow(mat))
res <- c()
for (i in seq_len(ncol(mat))) {
top <- common_finder(mat[considered, i], fun, even)
res <- c(res, top)
considered <- considered & (mat[, i] == top)
if (sum(considered) == 1) break
}
mat[considered, ] |>
paste0(collapse = "") |>
strtoi(base = 2)
}
oxygen <- rate_calculator(input, max, "1")
co2 <- rate_calculator(input, min, "0")
oxygen * co2``````
``0.078 sec elapsed``
# Day 4
The only difference between part 1 and 2 is that part 1 uses `min(win_times)` and part 2 uses `max(win_times)`
## Part 1
``````input <- readLines("2021/04-input")
read_matrix <- function(lines, sep = "", type = identity) {
lines <- stringr::str_trim(lines)
tokens <- strsplit(lines, sep)
token_lengths <- lengths(tokens)
res <- matrix(nrow = length(lines), ncol = max(token_lengths))
for (i in seq_along(lines)) {
res[i, seq_len(token_lengths[i])] <- type(tokens[[i]])
}
res
}
numbers <- strsplit(input[1], ",")[[1]] |> as.integer()
boards <- purrr::map(
0:99,
~ read_matrix(input[3:7 + 6 * .x], "\\s+", type = as.integer)
)
check_board <- function(board) {
for (i in seq_along(numbers)) {
matched <- matrix(board %in% numbers[seq_len(i)], nrow = 5)
row_checks <- apply(matched, MARGIN = 1, prod)
col_checks <- apply(matched, MARGIN = 2, prod)
if (any(c(row_checks, col_checks) == 1)) break
}
i
}
win_times <- purrr::map_int(boards, check_board)
fastest_time <- min(win_times)
fastest_board <- boards[[which(win_times == fastest_time)]]
sum(setdiff(fastest_board, numbers[seq_len(fastest_time)])) *
numbers[fastest_time]``````
``0.503 sec elapsed``
## Part 2
``````input <- readLines("2021/04-input")
read_matrix <- function(lines, sep = "", type = identity) {
lines <- stringr::str_trim(lines)
tokens <- strsplit(lines, sep)
token_lengths <- lengths(tokens)
res <- matrix(nrow = length(lines), ncol = max(token_lengths))
for (i in seq_along(lines)) {
res[i, seq_len(token_lengths[i])] <- type(tokens[[i]])
}
res
}
numbers <- strsplit(input[1], ",")[[1]] |> as.integer()
boards <- purrr::map(
0:99,
~ read_matrix(input[3:7 + 6 * .x], "\\s+", type = as.integer)
)
check_board <- function(board) {
for (i in seq_along(numbers)) {
matched <- matrix(board %in% numbers[seq_len(i)], nrow = 5)
row_checks <- apply(matched, MARGIN = 1, prod)
col_checks <- apply(matched, MARGIN = 2, prod)
if (any(c(row_checks, col_checks) == 1)) break
}
i
}
win_times <- purrr::map_int(boards, check_board)
slowest_time <- max(win_times)
slowest_board <- boards[[which(win_times == slowest_time)]]
sum(setdiff(slowest_board, numbers[seq_len(slowest_time)])) *
numbers[slowest_time]``````
``0.467 sec elapsed``
# Day 5
This is one of the weird days where part 2 solution is simpler than part 1. Simple delete the first `filter()` call.
## Part 1
``````library(tidyverse)
separate(input, into = c("x1", "y1", "x2", "y2"), convert = TRUE) %>%
filter(x1 == x2 | y1 == y2) %>%
group_nest(row_number()) %>%
mutate(crosses = map(data, ~tibble(x = .x\$x1:.x\$x2, y = .x\$y1:.x\$y2))) %>%
unnest(crosses) %>%
count(x, y) %>%
filter(n > 1) %>%
nrow()``````
``2.235 sec elapsed``
## Part 2
``````library(tidyverse)
separate(input, into = c("x1", "y1", "x2", "y2"), convert = TRUE) %>%
group_nest(row_number()) %>%
mutate(crosses = map(data, ~tibble(x = .x\$x1:.x\$x2, y = .x\$y1:.x\$y2))) %>%
unnest(crosses) %>%
count(x, y) %>%
filter(n > 1) %>%
nrow()``````
``2.259 sec elapsed``
# Day 6
## Part 1
``````input <- scan("2021/06-input", sep = ",")
for (i in seq_len(80)) {
input <- input - 1
if (any(input < 0)) {
input <- c(input, rep(8, sum(input < 0)))
input[input < 0] <- 6
}
}
length(input)``````
``0.12 sec elapsed``
## Part 2
``````input <- scan("2021/06-input", sep = ",")
counts <- c(0, tabulate(input, nbins = 8))
for (i in seq_len(256)) {
n0 <- counts[1]
counts[-length(counts)] <- counts[-1]
counts[7] <- counts[7] + n0
counts[9] <- n0
}
options(scipen = 999)
sum(counts)``````
``0.017 sec elapsed``
# Day 7
## Part 1
``````input <- scan("2021/07-input", sep = ",")
values <- seq(min(input), max(input))
fuels <- purrr::map_dbl(values, ~ sum(abs(input - .x)))
min(fuels)
sum(abs(median(input) - input))``````
``0.02 sec elapsed``
## Part 2
``````input <- scan("2021/07-input", sep = ",")
values <- seq(min(input), max(input))
adjust <- function(n) n * (n + 1) / 2
fuels <- purrr::map_dbl(values, ~sum(adjust(abs(input - .x))))
min(fuels)
min(
)``````
``0.067 sec elapsed``
# Day 8
## Part 1
``````library(tidyverse)
str_remove(".*\\| ") %>%
str_split(" ") %>%
map(nchar) %>%
map_int(~length(.x[.x %in% c(2, 4, 3, 7)])) %>%
sum()``````
``0.007 sec elapsed``
## Part 2
``````library(tidyverse)
splitter <- function(x) {
str_split(x, " ") %>%
map(str_split, "") %>%
map(map, sort)
}
setdiff_length <- function(x, y) {
lengths(map(x, ~setdiff(x[[which(y)]], .x)))
}
minus1 <- function(x) x - 1
solver <- function(lights, right) {
x1 <- lengths(lights) == 2
x4 <- lengths(lights) == 4
x7 <- lengths(lights) == 3
x8 <- lengths(lights) == 7
x6 <- lengths(lights) == 6 & setdiff_length(lights, x1) == 1
x0 <- lengths(lights) == 6 & setdiff_length(lights, x4) == 1 & !x6
x9 <- lengths(lights) == 6 & !x6 & !x0
x5 <- lengths(lights) == 5 & setdiff_length(lights, x6) == 1
x3 <- lengths(lights) == 5 & setdiff_length(lights, x9) == 1 & !x5
x2 <- lengths(lights) == 5 & !x5 & !x3
cont <- list(x0, x1, x2, x3, x4, x5, x6, x7, x8, x9) %>%
map(~lights[[which(.x)]]) %>%
map(sort)
right %>%
match(cont) %>%
minus1() %>%
paste(collapse = "") %>%
as.numeric()
}
tibble(input) %>%
separate(input, c("left", "right"), sep = " \\| ") %>%
mutate(across(c("left", "right"), splitter)) %>%
mutate(res = map2_dbl(left, right, solver)) %>%
summarise(total = sum(res))``````
``0.443 sec elapsed``
# Day 9
## Part 1
``````library(purrr)
strsplit("") |>
map(as.integer) |>
reduce(rbind)
row_length <- nrow(input)
col_length <- ncol(input)
mat <- matrix(FALSE, row_length, col_length)
for (row in 1:row_length) {
for (col in 1:col_length) {
row_id <- c(row + 1, row, row - 1, row)
col_id <- c(col, col + 1, col, col - 1)
subset <- !(row_id > row_length | col_id > col_length)
row_id <- row_id[subset]
col_id <- col_id[subset]
if (all(input[cbind(row_id, col_id)] > input[row, col])) {
mat[row, col] <- TRUE
}
}
}
sum(input[mat] + 1)``````
``0.127 sec elapsed``
## Part 2
``````library(purrr)
strsplit("") |>
map(as.integer) |>
reduce(rbind)
row_length <- nrow(input)
col_length <- ncol(input)
mat <- matrix(FALSE, row_length, col_length)
for (row in 1:row_length) {
for (col in 1:col_length) {
row_id <- c(row + 1, row, row - 1, row)
col_id <- c(col, col + 1, col, col - 1)
subset <- !(row_id > row_length | col_id > col_length)
row_id <- row_id[subset]
col_id <- col_id[subset]
if (all(input[cbind(row_id, col_id)] > input[row, col])) {
mat[row, col] <- TRUE
}
}
}
around <- function(row, col) {
list(
list(row = row + 0, col = col + 1),
list(row = row + 1, col = col + 0),
list(row = row + 0, col = col - 1),
list(row = row - 1, col = col + 0)
)
}
range_checker <- function(can) {
!((can\$row == 0) |
(can\$row > row_length) |
(can\$col == 0) |
(can\$col > col_length))
}
value_checker <- function(can, ref) {
value <- input[can\$row, can\$col]
last_value <- input[ref\$row, ref\$col]
if (value == 9) return(FALSE)
value > last_value
}
basin_size <- function(x) {
candidates <- list(
list(row = x[1], col = x[2])
)
saved <- list()
repeat {
new_candidaes <- around(candidates[[1]]\$row, candidates[[1]]\$col)
new_candidaes <- setdiff(new_candidaes, saved)
new_candidaes <- setdiff(new_candidaes, candidates)
new_candidaes <- new_candidaes[map_lgl(new_candidaes, range_checker)]
new_candidaes <- new_candidaes[
map_lgl(new_candidaes, value_checker, candidates[[1]])
]
candidates <- c(candidates, new_candidaes)
saved <- c(saved, candidates[1])
candidates[1] <- NULL
if (length(candidates) == 0) break
}
length(saved)
}
largest_basins <- which(mat, arr.ind = TRUE) |>
apply(1, basin_size)
sort(largest_basins, decreasing = TRUE)[1:3] |>
prod()``````
``3.098 sec elapsed``
# Day 10
## Part 1
``````library(stringr)
library(purrr)
full_pair <- c("\\(\\)" = "", "\\[\\]" = "", "\\{\\}" = "", "<>" = "")
remove_all_pairs <- function(x) {
old <- x
repeat {
new <- str_replace_all(old, full_pair)
if (old == new) break
old <- new
}
old
}
pair_side <- c(
"\\(" = "L", "\\[" = "L", "\\{" = "L", "<" = "L",
"\\)" = "R", "\\]" = "R", "\\}" = "R", ">" = "R"
)
find_corrupted_pair <- function(x) {
value <- str_replace_all(x, pair_side)
loc <- str_locate(value, "LR")
if (is.na(loc[2])) return(NA)
str_sub(x, loc[2], loc[2])
}
cleaned_errors <- map_chr(input, remove_all_pairs)
corrupt <- map_chr(cleaned_errors, find_corrupted_pair)
c(")"= 3, "]" = 57, "}" = 1197, ">" = 25137)[corrupt] |>
sum(na.rm = TRUE)``````
``0.223 sec elapsed``
## Part 2
``````library(stringr)
library(purrr)
full_pair <- c("\\(\\)" = "", "\\[\\]" = "", "\\{\\}" = "", "<>" = "")
remove_all_pairs <- function(x) {
old <- x
repeat {
new <- str_replace_all(old, full_pair)
if (old == new) break
old <- new
}
old
}
pair_side <- c(
"\\(" = "L", "\\[" = "L", "\\{" = "L", "<" = "L",
"\\)" = "R", "\\]" = "R", "\\}" = "R", ">" = "R"
)
find_corrupted_pair <- function(x) {
value <- str_replace_all(x, pair_side)
loc <- str_locate(value, "LR")
if (is.na(loc[2])) return(NA)
str_sub(x, loc[2], loc[2])
}
cleaned_errors <- map_chr(input, remove_all_pairs)
corrupt <- map_chr(cleaned_errors, find_corrupted_pair)
incomplete <- cleaned_errors[is.na(corrupt)]
complete_error <- function(x) {
score <- 0
repeat {
last <- str_sub(x, -1, -1)
pat_com <- c("(" = ")", "[" = "]", "{" = "}", "<" = ">")
score <- score * 5 + match(last, names(pat_com))
x <- paste0(x, pat_com[last])
x <- remove_all_pairs(x)
if (x == "") break
}
score
}
incomplete %>%
map_dbl(complete_error) %>%
median()``````
``0.456 sec elapsed``
# Day 11
## Part 1
``````input <- readLines("2021/11-input") |>
strsplit("") |>
map(as.integer) |>
reduce(rbind)
size <- nrow(input)
around <- function(x) {
row <- x[1]
col <- x[2]
row_id <- c(row - 1, row - 1, row - 1, row, row + 1, row + 1, row + 1, row)
col_id <- c(col - 1, col, col + 1, col + 1, col + 1, col, col - 1, col - 1)
subset <- !(row_id > size | col_id > size)
cbind(row_id[subset], col_id[subset])
}
flashes <- 0
for (i in 1:100) {
flashed <- matrix(FALSE, nrow = size, ncol = size)
input <- input + 1
repeat {
new_flashes <- which((input * !flashed) > 9, arr.ind = TRUE)
if (nrow(new_flashes) == 0) break
flashed <- flashed | (input > 9)
bursts <- map(seq_len(nrow(new_flashes)), ~around(new_flashes[.x, ])) |>
purrr::reduce(rbind)
for (i in seq_len(nrow(bursts))) {
input[bursts[i, 1], bursts[i, 2]] <- input[bursts[i, 1], bursts[i, 2]] + 1
}
}
input[flashed] <- 0
flashes <- flashes + sum(flashed)
}
flashes``````
``0.29 sec elapsed``
## Part 2
``````input <- readLines("2021/11-input") |>
strsplit("") |>
map(as.integer) |>
reduce(rbind)
size <- nrow(input)
around <- function(x) {
row <- x[1]
col <- x[2]
row_id <- c(row - 1, row - 1, row - 1, row, row + 1, row + 1, row + 1, row)
col_id <- c(col - 1, col, col + 1, col + 1, col + 1, col, col - 1, col - 1)
subset <- !(row_id > size | col_id > size)
cbind(row_id[subset], col_id[subset])
}
step <- 0
repeat {
step <- step + 1
flashed <- matrix(FALSE, nrow = size, ncol = size)
input <- input + 1
repeat {
new_flashes <- which((input * !flashed) > 9, arr.ind = TRUE)
if (nrow(new_flashes) == 0) break
flashed <- flashed | (input > 9)
bursts <- map(seq_len(nrow(new_flashes)), ~around(new_flashes[.x, ])) |>
purrr::reduce(rbind)
for (i in seq_len(nrow(bursts))) {
input[bursts[i, 1], bursts[i, 2]] <- input[bursts[i, 1], bursts[i, 2]] + 1
}
}
input[flashed] <- 0
if(all(flashed)) break
}
step``````
``0.919 sec elapsed``
# Day 12
``````library(tidyverse)
input <- tibble(input = readLines("2021/12-input")) %>%
separate(input, into = c("from", "to"))
input <- bind_rows(
input,
input %>% mutate(tmp = from, from = to, to = tmp) %>% select(-tmp)
)
last <- function(x) x[length(x)]
last_x <- last(x)
if (last_x == "end") return(list(x))
new_steps <- input\$to[input\$from == last_x]
map(new_steps, ~c(x, .x))
}
is_correct <- function(x) {
if (length(x) > 1 & last(x) == "start") return(FALSE)
x <- x[!x %in% c("end", "start")]
tab <- table(x)
tab_names <- names(tab)
small <- str_to_lower(tab_names) == tab_names
all(tab[small] == 1)
}
validate_path <- function(x) {
keep(x, is_correct)
}
grow <- function(x) {
x %>%
flatten() %>%
validate_path()
}
old <- list()
old[[1]] <- "start"
repeat {
new <- grow(old)
if (identical(old, new)) break
old <- new
}
length(old)``````
``9.098 sec elapsed``
## Part 2
``````library(tidyverse)
input <- tibble(input = readLines("2021/12-input")) %>%
separate(input, into = c("from", "to"))
input <- bind_rows(
input,
input %>% mutate(tmp = from, from = to, to = tmp) %>% select(-tmp)
)
is_lower <- function(x) tolower(x) == x
paths <- function(current, seen, duplicate) {
if (current == "end") {
return(1)
}
if (current == "start" & !is.null(seen)) {
return(0)
}
if (is_lower(current) & current %in% seen) {
if (is.null(duplicate)) {
duplicate <- current
} else {
return(0)
}
}
seen <- c(seen, current)
out <- 0
out <- out + paths(i, seen, duplicate)
}
out
}
paths(current = "start", seen = NULL, duplicate = NULL)``````
``6.642 sec elapsed``
# Day 13
``````library(tidyverse)
mid <- which(input == "")
p1 <- function(x) x + 1L
folds <- input[seq(mid + 1, length(input))]
points <- input[seq_len(mid - 1)] %>%
str_split(",") %>%
map(as.integer) %>%
map(p1)
mat <- matrix(
FALSE,
nrow = points %>% map_int(~.x[2]) %>% max(),
ncol = points %>% map_int(~.x[1]) %>% max()
)
for (point in points) {
mat[point[[2]], point[[1]]] <- TRUE
}
fold <- folds[1]
axis <- str_extract(fold, "[xy]")
amount <- str_extract(fold, "[0-9]+") %>% as.integer() %>% p1()
if (axis == "y") {
folded <- seq(nrow(mat), amount + 1)
landed <- seq(amount - length(folded), amount - 1)
mat[landed, ] <- mat[landed, ] | mat[folded, ]
mat <- mat[seq_len(amount - 1), ]
} else {
folded <- seq(ncol(mat), amount + 1)
landed <- seq(amount - length(folded), amount - 1)
mat[, landed] <- mat[, landed] | mat[, folded]
mat <- mat[, seq_len(amount - 1)]
}
sum(mat)``````
``0.031 sec elapsed``
## Part 2
``````library(tidyverse)
mid <- which(input == "")
p1 <- function(x) x + 1L
folds <- input[seq(mid + 1, length(input))]
points <- input[seq_len(mid - 1)] %>%
str_split(",") %>%
map(as.integer) %>%
map(p1)
mat <- matrix(
FALSE,
nrow = points %>% map_int(~.x[2]) %>% max(),
ncol = points %>% map_int(~.x[1]) %>% max()
)
for (point in points) {
mat[point[[2]], point[[1]]] <- TRUE
}
for (fold in folds) {
axis <- str_extract(fold, "[xy]")
amount <- str_extract(fold, "[0-9]+") %>% as.integer() %>% p1()
if (axis == "y") {
folded <- seq(nrow(mat), amount + 1)
landed <- seq(amount - length(folded), amount - 1)
mat[landed, ] <- mat[landed, ] | mat[folded, ]
mat <- mat[seq_len(amount - 1), ]
} else {
folded <- seq(ncol(mat), amount + 1)
landed <- seq(amount - length(folded), amount - 1)
mat[, landed] <- mat[, landed] | mat[, folded]
mat <- mat[, seq_len(amount - 1)]
}
}
reshape2::melt(mat) %>%
ggplot(aes(Var2, -Var1, fill = value)) +
geom_raster()``````
``0.076 sec elapsed``
# Day 14
The only difference here between part 1 and part 2 is setting the step range from 10 to 40
``````library(tidyverse)
from <- str_sub(pairs, 1, 2)
adj <- map(pairs, ~ c(paste0(str_sub(.x, 1, 1), str_sub(.x, -1, -1)),
paste0(str_sub(.x, -1, -1), str_sub(.x, 2, 2))))
ref_counts <- counts <- set_names(integer(length(from)), from)
for (i in seq(1, nchar(template) - 1)) {
pair <- str_sub(template, i, i + 1)
counts[pair] <- counts[pair] + 1
}
for (step in 1:10) {
new_counts <- ref_counts
for (pair in names(new_counts)) {
}
counts <- new_counts
}
tibble(
count = c(counts, 1),
char = c(str_sub(names(counts), 1, 1), str_sub(template, -1, -1))
) %>%
count(char, wt = count, sort = TRUE) %>%
summarise(max(n) - min(n))``````
``0.076 sec elapsed``
## Part 2
``````library(tidyverse)
from <- str_sub(pairs, 1, 2)
adj <- map(pairs, ~ c(paste0(str_sub(.x, 1, 1), str_sub(.x, -1, -1)),
paste0(str_sub(.x, -1, -1), str_sub(.x, 2, 2))))
ref_counts <- counts <- set_names(integer(length(from)), from)
for (i in seq(1, nchar(template) - 1)) {
pair <- str_sub(template, i, i + 1)
counts[pair] <- counts[pair] + 1
}
for (step in 1:40) {
new_counts <- ref_counts
for (pair in names(new_counts)) {
}
counts <- new_counts
}
tibble(
count = c(counts, 1),
char = c(str_sub(names(counts), 1, 1), str_sub(template, -1, -1))
) %>%
count(char, wt = count, sort = TRUE) %>%
summarise(max(n) - min(n))``````
``0.079 sec elapsed`` | 6,434 | 18,776 | {"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-2024-22 | latest | en | 0.356216 |
http://pistulka.com/Other/?p=707 | 1,540,000,997,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512501.27/warc/CC-MAIN-20181020013721-20181020035221-00213.warc.gz | 287,476,743 | 14,481 | # Mortgage Pool Price and Average Life One Cell Formulas
See updated formula at:
MBS Math Formula. Servicing, CPR, Payment Delay, Default Rate & Loss Severity
http://pistulka.com/Other/?p=2384
A few posts back I showed the “megaformula” I used back in the day for calculating the price of a mortgage pool with prepayments (CPR). Rather than treating the one cell formula as just an interesting antique, I wanted to show how it might be useful. In order to make it useful, I needed a way to show what happens to the actual term of the mortgage as the CPR changes. I could have used VBA, but I am trying to avoid it. What I needed was an average life formula that would also fit in one cell.
Thanks to Win Smith (The Well-Tempered Spreadsheet) the math had already been done. I just needed to put it together in one cell.
First there is a Price Table with vertical yields and horizontal CPRs. As always, all yellow cells are inputs, so you can make changes.
The second one on the same sheet is an ALM (Asset Liability Management) interest rate shock, plus and minus 400 basis points. The chart shows the negative convexity of the pools value. As the market value increase tappers off when rates are falling, the average life drops quickly.
See what you can come up with one cell formulas
http://www.pistulka.com/Excel_Shared/
1. Richard Vest says:
Hello Don-
I am trying to create an effective spreadsheet for hybrid arms. I have brought most cells in from Bloomberg and am working with Bloomberg API but they can’t figure out how to create book yields with varying cpr rates in the . Would it be possible to collaborate with you on this?
-Richard
1. Don Pistulka says:
Richard,
I am not sure what you mean by varying CPR rates. PSA has varying CPR rates (see my publication “PSA vs. CPR), but it sounds like you what to crate your own vectors. If you are using an amortization schedule, you could create another column that would change the CPR speed.
Can you send my an example what you have so far?
1. Richard Vest says:
Sure Don. What’s the best way to attached the spreadsheet? Do you have a email address I could send it to you as an attachment?
2. Erik says: | 502 | 2,181 | {"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-2018-43 | longest | en | 0.932058 |
https://2021.help.altair.com/2021.2/hwdesktop/engsol/topics/pre_processing/entities/points_r.htm | 1,702,193,213,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00438.warc.gz | 94,553,290 | 11,614 | # Points
A point is a zero-dimensional geometric entity.
A free point is a zero-dimensional geometry entity in space that is not associated with a surface. It is displayed as a small "x", and its color is determined by the component collector to which it belongs. These types of points are typically used for weld locations and connectors.
A fixed point is a zero-dimensional geometry entity that is associated with a surface. It is displayed as a small "o", and its color is determined by the surface to which it is associated. The automesher places a node at each fixed point on the surface being meshed. A fixed point that is placed at the junction of three or more non-suppressed edges is called a vertex or vertex point. Such vertices cannot be suppressed (removed).
## Create Points
### Create Free Points
• xyz - Creates free points by specifying (x,y,z) coordinates (Points panel).
• arc center - Creates at the center of the arc that best approximates the input set of nodes, points or lines (Points panel).
• extract parametric - Creates free points at parametric locations on lines and surfaces (Points panel).
• intersect - Creates free points at the intersection of geometric entities: lines/lines, lines/surfaces, lines/solids, lines/planes, vector/lines, vector/surfaces, vector/solids and vector/plane (Points panel).
• suppressed fixed points - Creates free points at suppressed fixed point locations (Point Edit panel).
• circle center - Creates free points at the center of the circle defined by three free or fixed points (Distance panel).
• duplicate - Creates free points by duplicating existing free or fixed points. This is available in many panels when the "duplicate" advanced entity selector is available on a points collector.
• Misc. API commands that do not have an associated panel.
### Create Fixed Points
• by cursor - Creates fixed points at cursor locations on surfaces and surface edges (Point Edit panel, Quick Edit panel).
• on edge - Creates fixed points at uniform locations on a surface edge (Point Edit panel, Quick Edit panel).
• on surface - Creates fixed points at existing node/free point locations on/near a surface (Point Edit panel).
• project - Creates fixed points on surface edges by projecting existing free or fixed points (Point Edit panel, Quick Edit panel).
• defeature pinholes - When defeaturing pinholes, fixed points are created at the center of the each removed pinhole (Defeature panel).
• Misc. API commands that do not have an associated panel.
## Edit Points
### Edit Free Points
• delete - Deletes free points (Delete panel).
• translate - Moves free points along a vector direction (Translate panel).
• rotate - Rotates free points about a vector axis (Rotate panel).
• scale - Scales the dimensions of free points either proportionally or uniformly (Scale panel).
• reflect - Reflects free points about a plane to create a mirror image (Reflect panel).
• project - Projects free points onto a plane, vector, line/surface edge or surface (Project panel).
• position - Translates and rotate free points into new positions (Position panel).
• permute - Switches the coordinates of free points (Permute panel).
• renumber - Renumbers free points (Renumber panel).
• Misc. API commands that do not have an associated panel.
### Edit Fixed Points
• suppress/remove - Suppresses non-vertex fixed points (Point Edit panel, Quick Edit panel).
• replace - Combines multiple fixed points by moving them to one fixed point location (Point Edit panel, Quick Edit panel).
• release - Releases fixed point vertices such that any shared edges attached to the point become free edges (Point Edit panel, Quick Edit panel).
• renumber - Renumbers fixed points (Renumber panel).
• Misc. API commands that do not have an associated panel.
## Query Points
### Query Free Points
• distance - Finds the distance between two free points (Distance panel).
• shortest distance - Finds the shortest distance between entities (Shortest Distance dialog).
• angle - Finds the angle between three free points (Distance panel).
• organize - Moves free points into different component collectors (Organize panel).
• numbers - Displays the IDs of free points (Numbers panel).
• count - Counts the total or displayed free points (Count panel).
• Misc. API commands that do not have an associated panel.
### Query Fixed Points
• distance - Finds the distance between two fixed points (Distance panel).
• shortest distance - Finds the shortest distance between entities (Shortest Distance dialog).
• angle - Finds the angle between three fixed points (Distance panel).
• numbers - Displays the IDs of fixed points (Numbers panel).
• count - Counts the total or displayed fixed points (Count panel).
• Misc. API commands that do not have an associated panel. | 988 | 4,806 | {"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} | 2.875 | 3 | CC-MAIN-2023-50 | latest | en | 0.87988 |
https://h2maths.com/blog/what-is-functional-maths/ | 1,721,933,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00522.warc.gz | 261,056,359 | 9,584 | # What is functional maths?
Functional maths is an approach to mathematics education that emphasizes the application of mathematical concepts to real-life situations. It focuses on teaching students how to use mathematics to solve problems in practical contexts, rather than just learning mathematical concepts in isolation.
The functional maths approach emphasizes the development of students’ problem-solving and critical thinking skills, as well as their ability to communicate mathematically. It involves the use of real-world examples and scenarios to illustrate mathematical concepts and their applications, and often involves the use of technology such as calculators and computer software.
In functional maths, students learn how to apply mathematical concepts to everyday situations such as calculating taxes, budgeting, and analyzing data. It is particularly relevant for students who may not be planning to pursue advanced mathematical studies but need to have a strong grasp of mathematical concepts and their applications in various fields, including business, economics, and science.
Implementing functional maths requires a combination of resources, teaching strategies, and support. Here are some things that may be needed:
1. A curriculum that emphasizes the application of mathematics to real-life situations, with a focus on problem-solving and critical thinking.
2. Qualified teachers who have a strong understanding of the functional maths approach and are able to teach it effectively.
3. Access to technology and other resources that can be used to support functional maths teaching and learning, such as calculators, computer software, and real-world data sets.
4. Professional development opportunities for teachers to develop their knowledge and skills in the functional maths approach.
5. Support from school administrators and policymakers who can provide funding, resources, and policy guidance to promote the use of functional maths.
6. Collaboration with community partners, such as local businesses and organizations, to provide students with opportunities to apply their functional maths skills in real-world settings.
By providing the necessary resources, support, and training, schools and educational systems can successfully implement functional maths and prepare students for success in a wide range of contexts.
## The Benefits
Implementing functional maths can have several benefits for students, teachers, and the broader community. Here are some of the main benefits:
1. Real-world relevance: Functional maths teaches students how to apply mathematical concepts to real-life situations, making the subject more engaging and relevant to their lives.
2. Problem-solving skills: By emphasizing problem-solving and critical thinking, functional maths helps students develop valuable skills that can be applied in a wide range of contexts.
3. Career preparation: Many careers require a strong understanding of mathematical concepts and their applications. Functional maths can help prepare students for success in these fields.
4. Improved numeracy: Functional maths can help improve students’ numeracy skills, which are essential for success in many areas of life.
5. Increased confidence: By providing students with the skills and knowledge they need to apply mathematical concepts in real-life situations, functional maths can increase their confidence in their abilities.
6. Better understanding of the world: Functional maths can help students better understand and analyze the data that they encounter in their daily lives, from news reports to financial statements.
Overall, implementing functional maths can help students develop important skills and knowledge that will benefit them throughout their lives, while also preparing them for success in a rapidly changing world.
Johann Rizal Alexander is the Perfect Solution Education Group's Digital Marketing Manager. He is passionate about technology, online solutions and digital marketing. He is an advocate for online tuition in Maths and History and encouraging students to tackle subjects by learning each online tuition subject as a language.
Find our more about Johann at https://www.webshopseven.com | 715 | 4,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.951664 |
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