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https://excelrow.com/excel-workday-function/ | 1,712,929,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00142.warc.gz | 231,441,163 | 24,506 | # Excel WORKDAY Function
The WORKDAY function is one of the (Date & Time) functions of Excel. It returns the serial
number of the date before or after a specified number of workdays.
We can find this function in Date & Time of insert function Tab.
## How to use WORKDAY function in excel
1. Click on empty cell (like F5 )
2. Click on fx on the below of font word (or press shift+F3)
3. In insert function tab you will see all functions
4. Select Date & Time category
5. Select WORKDAY function
6. Then select ok
7. In function arguments Tab you will see WORKDAY function
8. Start date is a serial date number that represents the start date
9. Days is the number of nonweekend and non-holiday days before or after start_date
10. Holidays is an optional array of one or more serial date numbers to exclude from the
working calendar, such as state and federal holidays and floating holidays
11. You will see the result in formula result section
## Examples of WORKDAY function in excel
1. To calculate a project deadline that is due in 10 working days: =WORKDAY(TODAY(), 10)
2. To calculate the date that is 20 working days after a specific date: =WORKDAY(“5/3/2023”, 20)
3. To calculate the date that is 30 working days before a specific date: =WORKDAY(“5/3/2023”, -30)
4. To exclude weekends and holidays when counting the number of working days between two dates: =NETWORKDAYS(“4/1/2023”, “5/3/2023”)
5. To calculate the next working day after a holiday (in this example, July 4th): =WORKDAY(“7/3/2023”, 1, {“7/4/2023”})
6. To calculate the 15th working day of the current month: =WORKDAY(EOMONTH(TODAY(),-1),15)
7. To calculate the last working day of the current month: =WORKDAY(EOMONTH(TODAY(),0),-1)
8. To calculate the date that is 90 working days after a specific date, excluding non-working days: =WORKDAY.INTL(“5/3/2023”, 90, 1)
9. To include additional non-working days (such as company-specific holidays) when calculating the number of working days between two dates: =NETWORKDAYS(“4/1/2023”, “5/3/2023”,{“4/15/2023″,”4/16/2023”})
10. To calculate the last working day of a specific quarter (in this example, Q2 2023): =WORKDAY(DATE(2023,6,30),-1,{“7/4/2023”})
## What is the syntax of the WORKDAY function in Excel?
The WORKDAY function in Excel is used to calculate a date that is a specified number of working days (or business days) away from a given start date. The syntax for the WORKDAY function is as follows:
``````=WORKDAY(start_date, days, [holidays])
``````
where:
• `start_date` is the starting date for the calculation.
• `days` is the number of working days to add or subtract from the `start_date`.
• `[holidays]` is an optional argument that specifies additional non-working days, such as holidays. This argument is an array of dates, and can be entered as either a range of cells or an array constant.
Here are some examples of how the WORKDAY function can be used:
Example 1: To calculate a project deadline that is due in 10 working days from today:
``````=WORKDAY(TODAY(), 10)
``````
Example 2: To calculate the date that is 20 working days after May 3rd, 2023:
``````=WORKDAY("5/3/2023", 20)
``````
Example 3: To include Memorial Day (May 29th, 2023) when calculating the date that is 10 working days after today:
``````=WORKDAY(TODAY(), 10, {"5/29/2023"})
``````
Example 4: To calculate the date that is 30 working days before June 30th, 2023:
``````=WORKDAY("6/30/2023", -30)
``````
Example 5: To calculate the date that is 90 working days after January 1st, 2023, excluding weekends and holidays:
``=WORKDAY.INTL("1/1/2023", 90, 1)``
## How does the WORKDAY function differ from the NETWORKDAYS function?
The WORKDAY and NETWORKDAYS functions in Excel are both used to calculate the number of working days between two dates, but they differ in how they handle weekends and holidays.
The WORKDAY function is used to calculate a date that is a specified number of working days away from a given start date. It allows you to exclude weekends and non-working days, such as holidays, from the calculation. The syntax for the WORKDAY function is:
``````=WORKDAY(start_date, days, [holidays])
``````
where `start_date` is the starting date for the calculation, `days` is the number of working days to add or subtract from the `start_date`, and `[holidays]` is an optional argument that specifies additional non-working days.
On the other hand, the NETWORKDAYS function is used to calculate the number of working days between two dates. It counts the number of weekdays (Monday to Friday) between two dates, excluding weekends and any dates specified as non-working days. The syntax for the NETWORKDAYS function is:
``````=NETWORKDAYS(start_date, end_date, [holidays])
``````
where `start_date` is the starting date for the calculation, `end_date` is the ending date for the calculation, and `[holidays]` is an optional argument that specifies additional non-working days.
To illustrate the difference between these two functions, consider the following examples:
Example 1: To calculate the date that is 10 working days after January 1st, 2023, excluding weekends and holidays:
``````=WORKDAY("1/1/2023", 10, {"1/15/2023"})
``````
This formula will return January 15th, 2023, which is 10 working days after January 1st, excluding Martin Luther King Jr. Day on January 15th.
Example 2: To calculate the number of working days between January 1st, 2023 and January 15th, 2023, excluding weekends and holidays:
``````=NETWORKDAYS("1/1/2023", "1/15/2023", {"1/15/2023"})
``````
This formula will return 9, which is the number of weekdays (Monday to Friday) between January 1st and January 15th, excluding Martin Luther King Jr. Day on January 15th.
## Can the WORKDAY function be used to calculate holidays?
The WORKDAY function in Excel can be used to exclude holidays from the calculation of working days. This is done by adding the holiday dates to the third argument of the WORKDAY function, called “holidays”.
The syntax for the WORKDAY function with holidays is:
``````=WORKDAY(start_date, days, [holidays])
``````
where `start_date` is the starting date for the calculation, `days` is the number of working days to add or subtract from the `start_date`, and `[holidays]` is an optional argument that specifies additional non-working days.
To use the `holidays` argument, you need to create a range of cells containing the holiday dates. For example, let’s say you have a list of holidays in cells A2:A6: New Year’s Day (1/1/2023), Memorial Day (5/30/2023), Independence Day (7/4/2023), Labor Day (9/4/2023), and Thanksgiving Day (11/23/2023).
You can then use this range as the third argument in the WORKDAY function, like this:
``````=WORKDAY("1/1/2023", 10, A2:A6)
``````
This formula will return the date that is 10 working days after January 1st, 2023, excluding weekends and the holidays listed in cells A2:A6.
Note that the holiday dates must be entered in chronological order in the range, and they must be formatted as dates. Also, if the `holidays` argument is omitted, the function will only exclude weekends from the calculation.
Here’s another example to illustrate the use of the WORKDAY function with holidays:
``````=WORKDAY("12/24/2023", 2, {"12/25/2023";"12/31/2023";"1/1/2024"})
``````
This formula will return December 28th, 2023, which is two working days after Christmas Eve (December 24th, 2023), excluding Christmas Day (December 25th, 2023), New Year’s Eve (December 31st, 2023), and New Year’s Day (January 1st, 2024).
## How can I use the WORKDAY function to calculate a future or past date?
The WORKDAY function in Excel can be used to calculate a future or past date by adding or subtracting a specified number of working days from a given start date. This function is particularly useful when you want to exclude weekends and holidays from the calculation.
The syntax for the WORKDAY function is:
``````=WORKDAY(start_date, days, [holidays])
``````
where `start_date` is the starting date for the calculation, `days` is the number of working days to add or subtract from the `start_date`, and `[holidays]` is an optional argument that specifies additional non-working days.
To calculate a future date using the WORKDAY function, you need to specify a positive value for the `days` argument. For example, let’s say you want to calculate the date that is 20 working days after January 1st, 2023, excluding weekends and holidays. You can use the following formula:
``````=WORKDAY("1/1/2023", 20, {"1/15/2023"})
``````
This formula will return January 29th, 2023, which is 20 working days after January 1st, excluding Martin Luther King Jr. Day on January 15th.
Similarly, to calculate a past date using the WORKDAY function, you need to specify a negative value for the `days` argument. For example, let’s say you want to calculate the date that is 10 working days before February 14th, 2023, excluding weekends and holidays. You can use the following formula:
``````=WORKDAY("2/14/2023", -10, {"2/19/2023"})
``````
This formula will return February 1st, 2023, which is 10 working days before February 14th, excluding Presidents’ Day on February 19th.
Note that the `holidays` argument is optional. You can use it to exclude additional non-working days from the calculation, such as public holidays or company-specific holidays.
## What data types are required for the arguments of the WORKDAY function?
The WORKDAY function in Excel requires three arguments: `start_date`, `days`, and `[holidays]`. Each of these arguments must be a specific data type to work properly.
1. `start_date`: This is the starting date for the calculation, and it must be a valid Excel date or a reference to a cell containing a date value. Excel stores dates as serial numbers, where January 1st, 1900 is serial number 1, and each subsequent day is one greater than the previous day. For example, January 1st, 2023 would be stored as the serial number 44245.
2. `days`: This is the number of working days to add or subtract from the `start_date`, and it must be an integer value. Positive values will calculate a future date, while negative values will calculate a past date.
3. `[holidays]`: This is an optional argument that specifies additional non-working days to exclude from the calculation. It must be a range of cells containing valid Excel dates or a list of dates entered directly into the formula. The dates in the range must be entered in chronological order.
Here are some examples of the correct data types for the arguments of the WORKDAY function:
``````=WORKDAY("1/1/2023", 10, {"1/15/2023"})
``````
In this example, `start_date` is a text string representing a date, `days` is an integer value, and `[holidays]` is a list of dates entered directly into the formula.
``````=WORKDAY(A2, B2, C2:C6)
``````
In this example, `start_date` is a reference to a cell containing a date value, `days` is a reference to a cell containing an integer value, and `[holidays]` is a range of cells containing valid Excel dates.
Here are some examples of incorrect data types for the arguments of the WORKDAY function:
``````=WORKDAY("January 1st, 2023", "10", {"1/15/2023"})
``````
In this example, `start_date` is a text string that does not represent a valid Excel date, and `days` is a text string instead of an integer value.
``````=WORKDAY(A2, B2, "New Year's Day")
``````
In this example, `[holidays]` is a text string instead of a range of cells containing valid Excel dates.
## Can the WORKDAY function be nested within other functions in Excel?
Yes, the WORKDAY function in Excel can be nested within other functions to perform more complex calculations. Nesting functions allows you to combine the results of multiple functions into a single formula.
Here’s an example of how to nest the WORKDAY function within another function:
``````=IF(WEEKDAY(WORKDAY("5/1/2023", 10),2)<=4,"Yes","No")
``````
In this example, the WORKDAY function is nested within the IF and WEEKDAY functions. The formula calculates the date that is 10 working days after May 1st, 2023 using the WORKDAY function. The resulting date is then passed to the WEEKDAY function to determine the day of the week (Monday = 1, Tuesday = 2, etc.). Finally, the IF function checks if the resulting weekday is less than or equal to 4 (Wednesday). If it is, the formula returns “Yes”. Otherwise, it returns “No”.
Here’s another example of how to nest the WORKDAY function within the NETWORKDAYS function:
``````=NETWORKDAYS("4/1/2023", WORKDAY("4/1/2023", 10))
``````
In this example, the WORKDAY function is used to calculate the date that is 10 working days after April 1st, 2023. This resulting date is then passed to the NETWORKDAYS function as the second argument. The NETWORKDAYS function calculates the number of working days between April 1st and the date that is 10 working days after April 1st, excluding weekends and any non-working days specified in the optional third argument.
Nesting the WORKDAY function within other functions can be useful for performing complex calculations that involve working days, weekends, and holidays. It allows you to streamline your formulas and avoid unnecessary steps in your calculations.
## How can I use the WORKDAY function to exclude weekends from my calculations?
The WORKDAY function in Excel can be used to exclude weekends from your calculations by default. The function treats Saturday and Sunday as non-working days, so any dates that fall on a weekend will not be included in the calculation of working days.
To use the WORKDAY function to calculate a future or past date while excluding weekends, you simply need to specify the number of working days to add or subtract from the starting date.
Here are some examples:
``````=WORKDAY("5/1/2023", 10)
``````
This formula calculates the date that is 10 working days after May 1st, 2023, excluding weekends. If any of the 10 working days fall on a weekend, such as Saturday, May 6th or Sunday, May 7th, those days will not be counted.
``````=WORKDAY("9/30/2023", -5)
``````
This formula calculates the date that is 5 working days before September 30th, 2023, excluding weekends. Again, if any of the 5 working days fall on a weekend, such as Saturday, September 23rd or Sunday, September 24th, those days will not be counted.
Note that the WORKDAY function does not account for holidays by default. If you need to exclude holiday dates from your calculation, you can include them in the third argument of the function as a range of cells containing valid Excel date values.
Here’s an example of how to exclude both weekends and holidays using the WORKDAY function:
``````=WORKDAY("12/24/2023", 2, {"12/25/2023";"12/31/2023";"1/1/2024"})
``````
In this formula, the calculation starts with December 24th, 2023, and adds two working days while excluding weekends and the holidays listed in the range. This formula will return the date that is two working days after Christmas Eve, while excluding Christmas Day, New Year’s Eve, and New Year’s Day.
## Is it possible to include custom non-working days using the WORKDAY function?
Yes, it is possible to include custom non-working days when using the WORKDAY function in Excel. This can be useful if you need to exclude company-specific holidays or other non-working days that are not included in the default weekends and public holidays.
To include custom non-working days, you simply need to create a list of the dates to exclude and include them as a range of cells in the third argument of the WORKDAY function.
Here’s an example of how to use the WORKDAY function with custom non-working days:
``````=WORKDAY("1/1/2023", 5, {"1/2/2023";"1/3/2023";"1/4/2023"})
``````
In this formula, the calculation starts with January 1st, 2023, and adds 5 working days while excluding January 2nd, 3rd, and 4th, which are custom non-working days specified in the third argument of the function.
It’s important to note that the list of non-working days must be entered in chronological order, and each date must be formatted as a valid Excel date value. You can enter the dates directly into the formula, or you can create a separate range of cells containing the dates and reference that range in the formula.
Here’s another example that includes a range of cells containing custom non-working days:
``````=WORKDAY("6/1/2023", 10, A1:A5)
``````
In this formula, the calculation starts with June 1st, 2023, and adds 10 working days while excluding any dates listed in the range A1:A5. The range could contain company-specific holidays, employee vacation days, or any other non-working days that need to be excluded from the calculation.
## What is the maximum number of days that can be added or subtracted using the WORKDAY function?
The maximum number of days that can be added or subtracted using the WORKDAY function in Excel is limited by the maximum value that can be represented by a signed 32-bit integer, which is 2,147,483,647.
In practical terms, this means that if you try to calculate a date that is more than approximately 5905 years in the future or past, you may encounter an error due to the limitations of Excel’s internal calculations. However, in most cases, this limit is not a concern, as it is rare to need to calculate dates more than a few decades into the future or past.
It’s important to note that when using the WORKDAY function with a large number of days, it is possible to encounter other issues such as overflow errors or problems with precision. For example, if you try to calculate a date that is very far in the future, the result may be imprecise due to rounding errors in Excel’s internal calculations.
Here are some examples of how to use the WORKDAY function with a large number of days:
``````=WORKDAY("1/1/2023", 1000)
``````
This formula calculates the date that is 1000 working days after January 1st, 2023. This calculation involves a relatively large number of days, but it is well within the limits of Excel’s calculations and should return a valid result.
``````=WORKDAY("1/1/2030", -5000)
``````
This formula calculates the date that is 5000 working days before January 1st, 2030. Again, this calculation involves a large number of days, but it is within the limits of Excel’s calculations and should return a valid result.
## How can I troubleshoot common errors when using the WORKDAY function in Excel?
When using the WORKDAY function in Excel, you may encounter several common errors. Here are some tips for troubleshooting these errors:
1. #VALUE! error: This error occurs when one or more of the arguments in the formula is not a valid data type. To fix this error, ensure that all arguments are formatted correctly and meet the requirements for the applicable data type.
2. #NUM! error: This error occurs when the number of working days specified is negative or too large to calculate. To fix this error, ensure that the number of working days is positive and within the limits of Excel’s calculations.
3. #NAME? error: This error occurs when Excel does not recognize the name of the function. To fix this error, check that the function name is spelled correctly and that the function is available in your version of Excel.
4. #REF! error: This error occurs when the formula contains a reference to a cell that no longer exists or has been deleted. To fix this error, update the formula to reference a valid cell.
5. #NULL! error: This error occurs when the range specified for the holidays argument is not valid. To fix this error, ensure that the range contains valid dates and that the dates are sorted in chronological order.
Here are some examples of how to troubleshoot these common errors:
``````=WORKDAY("May 1st, 2023", 10, {"May 30th, 2023"})
``````
In this formula, the first argument is not a valid date format, which will result in a #VALUE! error. To fix this error, change the format of the date to a valid Excel date format.
``````=WORKDAY("1/1/2023", -100)
``````
In this formula, the second argument specifies a negative number of working days, which will result in a #NUM! error. To fix this error, change the number of working days to a positive value.
``````=WORKDAY("1/1/2023", 5000000000)
``````
In this formula, the second argument specifies a very large number of working days, which will result in a #NUM! error. To fix this error, reduce the number of working days to within the limits of Excel’s calculations.
``````=WOKRDAY("1/1/2023", 10)
``````
In this formula, the function name is misspelled as “WOKRDAY,” which will result in a #NAME? error. To fix this error, correct the spelling of the function name to “WORKDAY”.
``````=WORKDAY("1/1/2023", 10, A1:A5)
``````
In this formula, the range specified for the holidays argument may not be valid, which could result in a #NULL! error. To fix this error, ensure that the range contains valid dates and that the dates are sorted in chronological order. | 5,234 | 20,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-18 | latest | en | 0.791282 |
http://www.java2s.com/Questions_And_Answers/Javascript-Data-Type/number/algorithm.htm | 1,368,996,194,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698080772/warc/CC-MAIN-20130516095440-00053-ip-10-60-113-184.ec2.internal.warc.gz | 541,910,421 | 3,414 | # algorithm « number « Javascript Data Type Q&A
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Javascript Data Type Q&A » number » algorithm
### 1. Best algorithm for determining the high and low in an array of numbers? stackoverflow.com
I am using pseudo-code here, but this is in JavaScript. With the most efficient algorithm possible I am trying to find the high and low given an array of positive whole ...
### 2. Efficient way to insert a number into a sorted array of numbers? stackoverflow.com
I have a sorted javascript array, and want to insert one more item into the array such the resulting array remains sorted. I could certainly implement a simple quicksort-style insertion ...
### 3. How might I find the largest number contained in a JavaScript array? stackoverflow.com
I have a simple JavaScript Array object containing a few numbers.
``````[267, 306, 108]
``````
Is there a function that would find the largest number in this array?
### 4. Javascript algorithm that calculates week number in Fiscal Year stackoverflow.com
I have been looking for a Javascript algorithms that gives me the week number of a given Date object within a custom fiscal year. The fiscal year of my company starts ...
### 5. Finding number selection possibilities stackoverflow.com
I have the following example data in javascript:
``````var variations = [
{group: 1, id: 1},
{group: 1, id: 2},
{group: 1, id: 3},
{group: 1, id: 4},
...``````
### 6. How do I design an algorithm to find the biggest number in a list of numbers? stackoverflow.com
I also need to convert my answer into javascript code that displays the largest value.
### 7. Javascript increment number..missing logic stackoverflow.com
need some logics or algorithm here: I want to use javascript to implement a simple counter, which adds rand() int every day (ignoring page loads) and stores that number without using any ...
### 8. Javascript program to print a number series in reverse stackoverflow.com
I don't understand why the below JavaScript program returns the answer Infinity. What am I doing wrong? I'm a novice, and would appreciate a detailed explanation! Please note that this needs to ...
### 9. Number.sign() in javascript stackoverflow.com
Wonder if there are any nontrivial ways of finding number's sign?
May be shorter / faster / more elegant solutions than the obvious one
``var sign = number > 0 ? 1 : ...`` | 614 | 2,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2013-20 | latest | en | 0.721483 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=14&t=48177&p=171464 | 1,579,684,079,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606872.19/warc/CC-MAIN-20200122071919-20200122100919-00137.warc.gz | 516,910,280 | 12,283 | ## N levels
$c=\lambda v$
Andrea_Torres
Posts: 58
Joined: Sun Sep 15, 2019 12:15 am
### N levels
Can someone explain to me what are the n levels that Dr. Lavelle is writing on the board in lectures?
claudia_1h
Posts: 68
Joined: Fri Aug 09, 2019 12:16 am
### Re: N levels
The "n levels" are the principal energy levels. The n is the principal quantum number, which tells you how far away from the nucleus the electrons are as well as how much energy those electrons have. Increasing n means that distance from the nucleus as well as electron energy is increasing.
pauline young 1L
Posts: 50
Joined: Thu Jul 11, 2019 12:15 am
### Re: N levels
in addition to that Dr, Lavelle always draws out the n levels on his board using lines. as the energy level increases, the distance between the new energy level and previous one decreases.
Abigail_Hagen2G
Posts: 62
Joined: Fri Aug 09, 2019 12:17 am
### Re: N levels
Yeah, the n-levels are just the energy levels that an electron can jump or fall to
Emil Velasco 1H
Posts: 54
Joined: Wed Nov 21, 2018 12:19 am
### Re: N levels
n levels are just the levels of energy for an electron in which n is a quantum number (n=1,2,3, etc.). When electrons increase in energy, the n increases.
Cassidy Kohlenberger 3D
Posts: 51
Joined: Wed Sep 18, 2019 12:19 am
### Re: N levels
The n levels are principal energy levels that are very important to know in order to do Rydberg equation problems. The change in energy is the energy of the final level minus the energy of the initial, solved by substituting in the Rydberg equation. You use the quantum number of the energy level in the denominator of the Rydberg equation. N levels also correspond to light series. For the Lyman series, electrons always end in n=1. For the Balmer series, electrons always end in n=2.
Amanda Ung 4D
Posts: 50
Joined: Thu Jul 25, 2019 12:17 am
### Re: N levels
n-levels are energy levels that electrons can jump to are fall from. Jumping energy levels requires energy to be put into the electron (exciting it) and electrons that fall from higher energy levels to lower emit energy to get back to their ground state.
mayarivers3I
Posts: 52
Joined: Wed Sep 18, 2019 12:18 am
### Re: N levels
The n level is related to the size and energy of the orbitals, which can also be referred to as the principle quantum number. N can be 1,2,3...
Mallika Chimpiri 3K
Posts: 58
Joined: Thu Jul 25, 2019 12:17 am
### Re: N levels
The N value signifies the energy level of an electron, and is called the Principle Quantum Number. The n value increase basically when an electron is excited and increases in energy level.The n value decreases when an electron emits energy and decreases in energy level. | 744 | 2,720 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-05 | latest | en | 0.910033 |
https://quantivity.wordpress.com/2011/10/22/exploratory-proxy-cross-hedge-analysis/ | 1,500,676,891,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423812.87/warc/CC-MAIN-20170721222447-20170722002447-00438.warc.gz | 706,527,190 | 42,266 | Previous posts on empirical quantiles and copulas for proxy / cross hedge illustrate the potential insight from graphical visualization. This post continues the theme, illustrating exploratory data analysis for proxy hedging using classical statistical techniques.
In a world awash with symbolic models, there is ample room for graphical exploratory analysis in finance—as the fine texture of the real world differs from both mathematical formalisms and standard mental models. Indeed, alpha hides in the divergence between model and reality.
Statistician John Tukey is one of the most well-known advocates of exploratory data analysis, captured in his 1977 book of the same title. Tukey nicely captures the essence of exploratory analysis in the opening chapter of his book (p. 1):
Exploratory data analysis is detective work–numerical detective work–or counting detective work–or graphical detective work.
This post illustrates graphical exploratory analysis, using R, specifically for the proxy hedging of previous posts: well-known tech company and QQQ. Although focused on a particular stock, these analysis techniques are applicable to other equities and higher frequencies.
Begin with scatters for daily prices and returns sampled over the previous 5 years, overlaid with fitted OLS and dispersion ellipsoids:
Top left plot illustrates several distinct price regimes. Top right plot illustrates ample returns well outside the dispersion ellipsoids, consistent with previous posts. Bottom left plot illustrates returns split by sign, demonstrating divergence in fit between positive and non-negative returns. Bottom right plot illustrates divergence between returns under and over 0.5 standard deviations, whose fit is quite similar.
Next, lag scatters for CRM:
Illustrating moderately non-spherical returns in the tails at all lags, consistent with the quantiles discussed in previous post. QQQ exhibits similar non-spherical lag returns, although the shape is not consistent with CRM:
To understand return dynamics in more depth, the following plots consider empirical density, absolute cross-correlation, difference ratios, and discrete rate of change:
Top left plot is textbook-style illustration of comparative excess kurtosis, with CRM return tails going out to +/- 10%. Cross correlation in top right plot exemplifies both forward and backward linear dependence for absolute returns, consistent with previous post on autocopulas. Difference ratios in bottom left plot illustrate empirical beta ranges from 1 to several hundred, irrespective whether returns are measured daily, weekly, or monthly—this corroborates difficulty of proxy hedging with linear instruments. These ratios provide the first evidence that late-2010 was perhaps even more anomalous for this pair than 2008, which is quite remarkable. Finally, bottom right plot illustrating return ROC further juxtaposes the comparison, as the ROC of both CRM and QQQ was dramatically higher in 2008 than 2010.
Finally, visualize the rolling proxy variance ratio (not to be confused with the standard statistical variance ratio, explored below) and correlation over the sample period with different durations:
Perhaps most interesting is the top plot which exemplifies that variance ratios (VR) for all roll durations are maximized over the 12 months beginning in August 2010. Contrast those ratios with the nearly flat VR, at all durations, during the financial crisis. The bottom plot provides one dimension of insight into this comparative behavior, illustrating that proxy correlation behaved inversely during these two periods: correlation was maximized during the finance crisis and minimized during 2010/2011.
Another interesting effect illustrated by both plots is temporal scaling behavior: correlation exhibits monotonic smoothing as lag window increases (conceptually similar to low-pass filtering), as exemplified by correlation progressive decreasing with black (daily), red (bi-weekly), blue (monthly), and green (quarterly). In contrast, the roll variance ratio exhibits no such scaling (monotonic or not). For example, early 2007 exhibits 60-day VR exceedingly all period periods. Similarly, the monthly VR frequently exceeds the biweekly VR.
Finally, consider classic unstandardized variance ratios for both CRM (top plot) and QQQ (bottom plot), calculated using the entire 5-year sample period, which illustrates mean-reversion versus trending (due to Lo and Mackinlay (1998)):
Both exhibit similar qualitative behavior: strongly mean reverting for holding periods under 30 days, followed by weaker reverting for longer periods. Variance ratio analysis is worth further consideration, particularly more rich graphical techniques (e.g. Lindemann et al. (2004)) and multivariate methods.
R code to generate the preceding exploratory hedge analysis:
```require("tseries")
require("vrtest")
require("fSeries")
colors <- c('black', 'red', 'blue', 'green', 'orange', 'purple', 'yellow', 'brown', 'pink', 'coral', 'cyan', 'darkgreen', 'darkred' ,'darkblue', 'darkgrey');
exploreProxyHedge <- function(p, doMonthlyScatter=TRUE, freq="daily", doQuantilePlots=TRUE)
{
# Plot various proxy hedge exploratory analysis.
#
# Args:
# p: matrix of instrument price data, including valid colnames
# doMonthlyScatter: flag indicating whether monthly scatter should
# be plotted
# freq: text string for frequency, used for graphing
# doQuantilePlots: flag indicating whether quantiles should be plotted
#
# Returns: None
oldpar <- par(mfrow=c(2,2))
# discrete first differences (not logged)
pROC <- ROC(p, type="discrete", na.pad=FALSE)
p1ROC <- ROC(p[,1], type="discrete", na.pad=FALSE)
p2ROC <- ROC(p[,2], type="discrete", na.pad=FALSE)
# scatter analysis
plot(coredata(p[,2]), coredata(p[,1]), xlab=colnames(p)[2], ylab=colnames(p)[1], main="Prices")
plm <- lm(p[,1] ~ p[,2])
abline(plm, col=colors[2], lty=2)
mtext(text=paste("slope=",plm\$coefficients[2]), side=3, cex=0.75, col=colors[2])
plot(coredata(p1ROC), coredata(p2ROC), xlab=colnames(p)[2], ylab=colnames(p)[1], main="Returns")
rlm <- lm(p1ROC ~ p2ROC)
abline(rlm, col=colors[2], lty=2)
par(xpd=TRUE)
d <- dataEllipse(as.vector(coredata(p1ROC)),as.vector(coredata(p2ROC)),draw=FALSE)
lines(d[[1]], col=colors[2], lty=3)
lines(d[[2]], col=colors[3], lty=3)
par(xpd=FALSE)
mtext(text=paste("slope=",rlm\$coefficients[2]), side=3, cex=0.75, col=colors[2])
# split positive/negative scatter analysis
plot(coredata(p2ROC[(coredata(p1ROC)>0)]), coredata(p1ROC[(coredata(p1ROC)>0)]), xlab=colnames(p)[2], ylab=paste(colnames(p)[1], "(Positive Only)"), xlim=c(min(p2ROC),max(p2ROC)), ylim=c(min(p1ROC),max(p1ROC)), main="Split Sign Return Scatter")
prlm <- lm(coredata(p1ROC[(coredata(p1ROC)>0)]) ~ coredata(p2ROC[(coredata(p1ROC)>0)]))
abline(prlm, col=colors[1], lty=2)
points(coredata(p2ROC[(coredata(p1ROC)<0)]), coredata(p1ROC[(coredata(p1ROC)<0)]), xlab=colnames(p)[2], ylab=paste(colnames(p)[1], "(Negative Only)"), col='red')
nrlm <- lm(coredata(p1ROC[(coredata(p1ROC)<0)]) ~ coredata(p2ROC[(coredata(p1ROC)<0)]))
abline(nrlm, col=colors[2], lty=2)
abline(rlm, col=colors[3], lty=2)
legend("topleft",legend=c("Positive", "Negative", "Both"), fill=colors, cex=0.5)
# split magnitude scatter analysis
magBound <- sd(p1ROC) / 2
plot(coredata(p2ROC[(abs(coredata(p2ROC))>=magBound)]), coredata(p1ROC[(abs(coredata(p2ROC))>=magBound)]), xlab=colnames(p)[2], ylab=paste(colnames(p)[1], "(Positive Only)"), xlim=c(min(p2ROC),max(p2ROC)), ylim=c(min(p1ROC),max(p1ROC)), main="Split Magnitude Return Scatter (0.5 SD)")
orlm <- lm(coredata(p1ROC[(abs(coredata(p2ROC))>=magBound)]) ~ coredata(p2ROC[(abs(coredata(p2ROC))>=magBound)]))
abline(orlm, lty=2)
points(coredata(p2ROC[(abs(coredata(p2ROC))<magBound)]), coredata(p1ROC[(abs(coredata(p2ROC))<magBound)]), xlab=colnames(p)[2], ylab=paste(colnames(p)[1], "(Positive Only)"), xlim=c(min(p2ROC),max(p2ROC)), ylim=c(min(p1ROC),max(p1ROC)), main="Split Magnitude Return Scatter", col=colors[2])
irlm <- lm(coredata(p1ROC[(abs(coredata(p2ROC))<magBound)]) ~ coredata(p2ROC[(abs(coredata(p2ROC))<magBound)]))
abline(irlm, lty=2, col=colors[2])
legend("topleft", legend=c("Outer", "Inner (+/- 0.5 SD)"), fill=colors, cex=0.5)
# monthly scatter analysis
if (doMonthlyScatter)
{
plot(coredata(xts(p)["2011-05"][,2]), coredata(xts(p)["2011-05"][,1]), ylim=c(min(p[,1]),max(p[,1])), xlim=c(min(p[,2]),max(p[,2])), col=colors[1], xlab=colnames(p)[2], ylab=colnames(p)[1], main="Prices by Month")
points(coredata(xts(p)["2011-06"][,2]), coredata(xts(p)["2011-06"][,1]), col=colors[2])
points(coredata(xts(p)["2011-07"][,2]), coredata(xts(p)["2011-07"][,1]), col=colors[3])
points(coredata(xts(p)["2011-08"][,2]), coredata(xts(p)["2011-08"][,1]), col=colors[4])
points(coredata(xts(p)["2011-09"][,2]), coredata(xts(p)["2011-09"][,1]), col=colors[5])
points(coredata(xts(p)["2011-10"][,2]), coredata(xts(p)["2011-10"][,1]), col=colors[6])
legend("topleft", legend=c("05", "06", "07", "08", "09", "10"), fill=colors, cex=0.5)
abline(lm(coredata(xts(p)["2011-05"][,1]) ~ coredata(xts(p)["2011-05"][,2])), col=colors[1], lty=2)
abline(lm(coredata(xts(p)["2011-06"][,1]) ~ coredata(xts(p)["2011-06"][,2])), col=colors[2], lty=2)
abline(lm(coredata(xts(p)["2011-07"][,1]) ~ coredata(xts(p)["2011-07"][,2])), col=colors[3], lty=2)
abline(lm(coredata(xts(p)["2011-08"][,1]) ~ coredata(xts(p)["2011-08"][,2])), col=colors[4], lty=2)
abline(lm(coredata(xts(p)["2011-09"][,1]) ~ coredata(xts(p)["2011-09"][,2])), col=colors[5], lty=2)
}
# quantile plots
if (doQuantilePlots)
{
qqplot(coredata(p2ROC), coredata(p1ROC), xlab=paste(colnames(p)[2], "Returns Quantiles"), ylab=paste(colnames(p)[1], "Returns Quantiles"), main="Empirical Returns QQ-Plot")
abline(0,1,lty=2)
grid(20)
par(xpd=TRUE)
d <- dataEllipse(as.vector(coredata(p1ROC)),as.vector(coredata(p2ROC)),draw=FALSE)
lines(d[[1]], col=colors[2], lty=3)
lines(d[[2]], col=colors[3], lty=3)
par(xpd=FALSE)
}
# Price lag dependence
lag.plot(p1ROC, 9, do.lines=FALSE, main=paste(colnames(p)[1], "Returns Lag Auto Dependence"))
lag.plot(p2ROC, 9, do.lines=FALSE, main=paste(colnames(p)[2], "Returns Lag Auto Dependence"))
par(mfrow=c(2,2))
# return distributions
p1Density <- density(p1ROC)
p2Density <- density(p2ROC)
plot(p1Density, ylim=c(0, max(p1Density\$y, p2Density\$y)), main="Return Distribution With Median")
lines(p2Density, col=colors[2])
abline(v=median(p1ROC), lty=2)
abline(v=median(p2ROC), col=colors[2], lty=2)
legend("topleft", legend=colnames(p), fill=colors, cex=0.5)
# cross correlation
ccf(data.frame(abs(coredata(p1ROC))), data.frame(abs(coredata(p2ROC))), main="Absolute Return Cross Correlation")
# diff ratio analysis (exclude periods with zero return QQQ)
p1NoZeros <- diff(p[(p[,2] != 0),])
p5NoZeros <- diff(p, lag=5)
p5NoZeros <- p5NoZeros[(p5NoZeros[,2] != 0),]
p22NoZeros <- diff(p, lag=22)
p22NoZeros <- p22NoZeros[(p22NoZeros[,2] != 0),]
diff1Ratio <- p1NoZeros[,1] / p1NoZeros[,2]
diff5Ratio <- p5NoZeros[,1] / p5NoZeros[,2]
diff22Ratio <- p22NoZeros[,1] / p22NoZeros[,2]
plot(diff22Ratio, main="Difference Ratios", ylab="Diff Ratio", xlab="")
lines(diff5Ratio, col=colors[2])
lines(diff1Ratio, col=colors[3])
legend("topright", legend=c("lag-22", "lag-5", "lag-1"), fill=colors, cex=0.5)
# ROC analysis
plot(xts(p1ROC), ylab="ROC", xlab="", main="Return Rate of Change")
lines(xts(p2ROC), col=colors[2])
legend("topleft", legend=colnames(p), fill=colors, cex=0.5)
# variance ratio analysis
vratio <- sd(p[,1]) / sd(p[,2])
cat(paste("variance ratio:", round(vratio,2)),"\n")
vRatio5 <- rollingVarianceRatio(p,5)
vRatio10 <- rollingVarianceRatio(p,10)
np <- nrow(p)
vRatio22 <- c()
vRatio60 <- c()
if (np > 22)
{
vRatio22 <- rollingVarianceRatio(p,22)
if (np > 60)
{
vRatio60 <- rollingVarianceRatio(p,60)
}
}
par(mfrow=c(2,1))
plot(vRatio5, type='l', xlab="", ylab="VR", main="Rolling Variance Ratio", ylim=c(min(vRatio5, vRatio10,vRatio22,vRatio60), max(vRatio5, vRatio10,vRatio22,vRatio60)))
lines(vRatio10, col=colors[2])
if (np > 22)
{
lines(vRatio22, col=colors[3])
if(np > 60)
{
lines(vRatio60, col=colors[4])
}
}
legend("topleft", legend=c("5", "10", "22", "60"), fill=colors, cex=0.5)
# correlation analysis
vCorr5 <- rollingCorrelation(pROC, 5)
vCorr10 <- rollingCorrelation(pROC, 10)
vCorr22 <- c()
vCorr60 <- c()
if (np > 22)
{
vCorr22 <- rollingCorrelation(pROC, 22)
if (np > 60)
{
vCorr60 <- rollingCorrelation(pROC, 60)
}
}
plot(vCorr5, type='l', xlab="", ylab="Correlation", main="Rolling Correlation", ylim=c(min(vCorr5, vCorr10,vCorr22,vCorr60), max(vCorr5, vCorr10,vCorr22,vCorr60)))
lines(vCorr10, col=colors[2])
if (np > 22)
{
lines(vCorr22, col=colors[3])
if (np > 60)
{
lines(vCorr60, col=colors[4])
}
}
legend("topleft", legend=c("5", "10", "22", "60"), fill=colors, cex=0.5)
# classic variance ratios (unstandardized)
VR.plot(p1ROC,60)
VR.plot(p2ROC,60)
par(oldpar)
}
rollingVarianceRatio <- function(p, winLen)
{
# Calculate rolling variance ratio with a given window length
#
# Args:
# p: matrix of instrument price data, including valid colnames
# winLen: length of window over which to calculate variance
#
# Returns: xts of rolling variance ratio
return (xts(sapply(c(1:(nrow(p)-winLen)), function(i) { sd(p[i:(i+winLen),1]) / sd(p[i:(i+winLen),2]) }), order.by=index(p[((winLen+1):nrow(p)),])))
}
rollingCorrelation <- function(p, winLen)
{
# Calculate rolling correlation with a given window length
#
# Args:
# p: matrix of instrument price data, including valid colnames
# winLen: length of window over which to calculate correlation
#
# Returns: xts of rolling correlation
return (xts(sapply(c(1:(nrow(p)-winLen)), function(i) { cor(p[i:(i+winLen)], method="kendall")[2,1] }), order.by=index(p[((winLen+1):nrow(p)),])))
}
```
7 Comments leave one →
October 24, 2011 2:04 am
Very instructive, thank you. Apologies for the noob question, but what is the “ROC()” function? I haven’t been able to find it in a CRAN library. -thanks
October 24, 2011 6:20 am
library(TTR)
• October 25, 2011 12:51 pm
So far, I have found these dependencies:
> require(“TTR”) # for ROC
> require(“car”) # for dataEllipse
> require(“vrtest”) # for VR.plot
> colors <- c("black", "red", "blue", "green", "orange", "purple", "yellow", "brown", "pink")
But, where can I find `stopImg` and `doImg`?
October 25, 2011 1:16 pm
@seth2077 / @G: thanks for comments. Library dependencies have been added, and *Img() have been removed. Glad to see code is sufficiently useful to warrant others to try it.
2. October 25, 2011 7:04 am
Why do you use `sd` instead of `var` in your VarianceRatio calculation? Even if you really want to return the ratio of standard deviations, it would be much faster to calculate the ratio of variances and take the square root at the end.
Better yet, since you are using the TTR package, you could use the `runSD` or `runVar` functions which use compiled code and will give you at least 50 times speedup. (there’s also a runCor function, but it doesn’t use ‘kendall’ (note that the `n` argument for all these would be `winLen – 1`).
Since TTR depends on xts which depends on zoo, you can use zoo’s `rollapply` function to apply any function (like cor(…, method=’kendall’)) over a rolling window (make sure you use `align=right`).
Something like rollapply(obj, width=20, FUN=sd, align=’right’) will be quite a bit slower than runSD(obj, n=20), but still much faster than the current sapply loop.
These proxy hedge posts have been fantastic. Keep up the good work.
Cheers | 4,595 | 15,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-30 | longest | en | 0.89499 |
https://www.electrical4u.net/calculator/w-to-var-conversion-calculator/ | 1,721,725,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00375.warc.gz | 652,412,714 | 25,555 | # W to VAR Conversion Calculator
### W to VAR Calculator:
Enter the real power in Watts and the power factor to get the reactive power in VAR (Volt-Amp Reactive).
Enter Real Power in Watts: W Enter Power Factor: Result – Volt Amp Reactive: VAR
### W to VAR Calculation:
Watt is the unit of real power and VAR is the unit of reactive power. Reactive power in Volt-Amp Reactive Q(VAR) is equal to the tangent of power angle times of the real power P(W) in Watts. Hence the formula can be written as,
Reactive power in VAR Q(VAR) = P(W) x tan(φ).
Here power angle (φ) is equal to the cosine inverse of the power factor.
Example:
Calculate the reactive power consumption of the 200 Watt motor is operated at 0.9 pf.
Reactive power in VAR = 200 x tan (cos-1(0.9))
Q(VAR) = 200 x 0.4843221
= 96.8 VAR. | 215 | 808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.82462 |
https://www.jiskha.com/questions/1649534/a-pile-of-blocks-has-40-blocks-in-the-bottom-row-36-blocks-in-the-second-row-32-blocks-in | 1,611,562,945,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00595.warc.gz | 824,285,824 | 5,045 | # Math
A pile of blocks has 40 blocks in the bottom row 36 blocks in the second row 32 blocks in the third row and so on until there are only 4 blocks on the top row how many blocks? Are there in the 7th row? How many blocks are there in all
1. 👍
2. 👎
3. 👁
1. (40-4)/4 = 9
so, there are 9 rows after the 1st, making 10 in all.
On the nth row, there are
40-(n-1)*4 = 44-4n bricks
The sum of n rows, Sn, with
a = 40
d = -4
is
Sn = n/2 (2*40 - (n-1)*4) = 2n(21-n)
Now you can answer the questions.
1. 👍
2. 👎
2. 135
1. 👍
2. 👎
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How? I’ve been paying P399/month now I’m paying P620/mnth. That’s how.
The price was changed from 39 USD to 49 USD in mid 2020 and the price haven’t changed since.
The increase you see is due to exchange rate fluctuations over which Manager has no control.
What are you really arguing? I stated that the price has increased by 64 percent for me. It’s a fact. The exchange rate fluctuation of the USD to my currency isn’t anything close to even half of that increment.
I’m not arguing anything. I’m simply presenting the facts.
This is how much Manager charges per subscription per month:
Also, Manager doesn’t do any currency conversions, the amount is charged in USD:
The conversion from USD to P is done by your payment forwarder whether it be your bank, your credit facility, PayPal or any other payment processor.
You need to speak to them.
That is simply not true. Just use a VPN and select Botswana, you will notice that Manager puts the price in the local currency.
P620 - P399 = P221
P221 / P399 = 55%
(I’m not saying it’s fair, just that it’s not 64%.)
You seem not to know exactly what you’re talking about, but so sure of yourself.
This is how much I get charged per month.
It seems that you are comparing prices and exchange rates from mid 2020 to now
In mid-2020, the price was \$ 39, the exchange rate around 10.5 BWP/\$ = roughly BWP 400
Now, price is \$ 49, the exchange rate is now around 13.0 BWP/\$ = roughly BWP 640
The increase was about BWP 100 due to the price rise and BWP 140 due to exchange rate change
4 Likes
Didn’t know that. Thanks.
But this means that it’s stripe that doing the conversions.
This also provides a possible solution.
@Monkgogi_Samson, if you can get a better exchange rate than stripe, get a USD account and use a VPN to get the USD price. | 472 | 1,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-49 | latest | en | 0.950413 |
http://korns.com/Example_NumMatrix_makeGaussianMatrix_001.html | 1,717,024,221,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00250.warc.gz | 22,006,254 | 7,743 | # Example NumMatrix makeGaussianMatrix 001
#### Description
This example converts a Number Matrix to a Gaussian Matrix.
#### Details
Create a number Matrix containing the original independent observations:
(define X #(nummat[5 4]| 12 2 3 17 1 5 -3 3 9 5 -3 11 9 51 -3 57 2 11 -3 10)) Returns: #
Convert the observation matrix to its Gaussian matrix:
(setq G (makeGaussianMatrix X)) Returns: #
The contents of the Gaussian matrix:
(display # ) Returns: #(nummat[3 4]| 311.0 555.0 -27.0 839.0 555.0 2776.0 -210.0 3121.0 -27.0 -210.0 45.0 -192.0 )
Notes and Hints
Note, The Gaussian matrix contains one row for each independent variable and an extra column for the dependent variable. Each cell in the Gaussian matrix is the statistical cross correlation of the specified independent variable (row index) with the specified correlation variable (col index). The final column of the Gaussian matrix is the statistical cross correlation of the dependent variable with each of the independent variables.
#### Related Examples
Here are examples of the makeGaussianMatrix function at work.
Matrix makeGaussianMatrix 001 NumMatrix makeGaussianMatrix 001
Here is the link to the current function used in this example.
makeGaussianMatrix
Here are a number of links to other related functions.
makeGramMatrix(#void) matrixGaussianEliminate(#void) matrixGaussianSubstitute(#void)
#### Argument Types
Here are the links to the data types of the arguments used in this example.
NumMatrix Integer
Here are a number of links to examples having similar argument types.
BitVector compare 018 BitVector findBlock 001 BitVector freeBlock 001 BitVector inside 007 BitVector isBitVector 001 BitVector isMember 008 BitVector isObject 012 BitVector length 013 BitVector member 008 BitVector ref 027 BitVector resize 010 BitVector sizeof 019 BitVector vectorFill 007 Boolean bcompareEQ 001 Boolean bcompareGE 001 Boolean bcompareGT 001 Boolean bcompareLE 001 Boolean bcompareLT 001 Boolean bcompareNE 001 Boolean boolean 001 Boolean compare 009 Boolean isBoolean 001 Boolean sizeof 009 ByteVector isByteVector 001 ByteVector isObject 013 ByteVector sizeof 020 Character cadd 001 Character ccompareEQ 001 Character ccompareGE 001 Character ccompareGT 001 Character ccompareLE 001 Character ccompareLT 001 Character ccompareNE 001 Character cdiv 001 Character character 001 Character cmod 001 Character cmul 001 Character compare 010 Character csub 001 Character isCharacter 001 Character sizeof 010 Complex compare 021 Complex cos 002 Complex exp 002 Complex expt 002 Complex log 002 Complex log10 002 Complex log2 002 Complex logbase 002 Complex sin 002 Complex sqrt 002 Complex tan 002 Context globalBinding 001 CpxVector avg 002 CpxVector binaryInsert 008 CpxVector binarySearch 008 CpxVector compare 020 CpxVector compareEQ 020 CpxVector compareGE 020 CpxVector compareGT 020 CpxVector compareLE 020 CpxVector compareLT 020 CpxVector compareNE 020 Date compare 012 Date date 001 Date day 001 Date days360 001 Date hour 001 Date minute 001 Date month 001 Date second 001 Date sizeof 012 Date time 001 Date time 002 Date year 001 Dictionary compare 004 Dictionary count 002 Dictionary delete 003 Dictionary delete 004 Dictionary insert 002 Dictionary isBound 002 Dictionary isDictionary 001 Dictionary isObject 004 Dictionary length 005 Dictionary member 002 Dictionary new 004 Dictionary ref 007 Dictionary ref 008 Dictionary ref 009 Dictionary refAttributes 002 Dictionary refValues 002 Dictionary resize 002 Dictionary setq 007 Dictionary setq 008 Dictionary sizeof 004 Dictionary sort 003 Directory compare 005 Directory count 003 Directory delete 005 Directory delete 006 Directory insert 003 Directory inside 003 Directory isBound 003 Directory isObject 005 Directory length 006 Directory new 005 Directory new 006 Directory ref 011 Directory ref 012 Directory refAttributes 003 Directory refValues 003 Directory resize 003 Directory setq 010 Directory sizeof 005 FltVector binaryInsert 004 FltVector compare 014 FltVector delete 009 FltVector insert 005 FltVector isFloatVector 001 FltVector isObject 007 FltVector length 008 FltVector ref 022 FltVector resize 006 FltVector sizeof 014 FltVector uniqueInsert 004 FltVector vectorFill 003 Function compare 013 Function sizeof 013 IntVector binaryInsert 005 IntVector compare 015 IntVector delete 010 IntVector insert 006 IntVector inside 005 IntVector isIntegerVector 001 IntVector isMember 005 IntVector isObject 008 IntVector length 009 IntVector member 005 IntVector ref 023 IntVector resize 007 IntVector sizeof 015 IntVector uniqueInsert 005 IntVector vectorFill 004 Integer binaryNot 001 Integer bitwiseNot 001 Integer fact 001 Integer gcd 001 Integer iadd 001 Integer icompareGE 001 Integer icompareGT 001 Integer icompareLE 001 Integer icompareLT 001 Integer icompareNE 001 Integer idiv 001 Integer imod 001 Integer imul 001 Integer integer 001 Integer isub 001 Integer lcm 001 Lambda balance 001 Lambda globalBinding 001 Lambda ref 017 List count 007 List count 008 List length 015 List sizeof 022 Matrix binaryInsert 004 Matrix binaryInsert 005 Matrix compare 007 Matrix count 005 Matrix delete 009 Matrix insert 005 Matrix inside 005 Matrix isMatrix 001 Matrix isMember 005 Matrix isNumberMatrix 001 Matrix isObject 007 Matrix length 008 Matrix makeGaussianMatrix 001 Matrix makeGramMatrix 001 Matrix matrixGaussianEliminate 001 Matrix matrixGaussianSubstitute 001 Matrix member 005 Matrix objectToMatrix 001 Matrix objectToNumMatrix 001 Matrix rank 001 Matrix ref 015 Matrix ref 016 Matrix resize 005 Matrix reverse 002 Matrix sizeof 007 Matrix uniqueInsert 004 Matrix vectorFill 002 NumMatrix addMethod 027 NumMatrix append 012 NumMatrix apply 009 NumMatrix binaryInsert 007 NumMatrix binarySearch 007 NumMatrix compare 023 NumMatrix compareEQ 023 NumMatrix compareGE 023 NumMatrix compareGT 023 NumMatrix compareLE 023 NumMatrix compareLT 023 NumMatrix compareNE 023 NumMatrix comparison functions 023 NumMatrix copy 011 NumMatrix count 009 NumMatrix defmethod 024 NumMatrix delete 014 NumMatrix insert 009 NumMatrix inside 009 NumMatrix isAtom 012 NumMatrix isEqual 020 NumMatrix isIdentical 020 NumMatrix isMember 010 NumMatrix isNumberMatrix 001 NumMatrix isObject 014 NumMatrix isType 022 NumMatrix length 017 NumMatrix makeGaussianMatrix 001 NumMatrix makeGramMatrix 001 NumMatrix map 010 NumMatrix mapc 009 NumMatrix matrixGaussianEliminate 001 NumMatrix matrixGaussianSubstitute 001 NumMatrix member 009 NumMatrix methodsOf 024 NumMatrix new 015 NumMatrix rank 002 NumMatrix ref 035 NumMatrix remove 008 NumMatrix resize 011 NumMatrix reverse 004 NumMatrix send 027 NumMatrix setCdr 013 NumMatrix setLastCdr 012 NumMatrix sizeof 028 NumMatrix sort 020 NumMatrix type 025 NumMatrix uniqueInsert 005 NumMatrix vectorFill 008 NumVector binaryInsert 006 NumVector compare 016 NumVector delete 011 NumVector insert 007 NumVector inside 006 NumVector isMember 006 NumVector isNumberVector 001 NumVector isObject 009 NumVector length 010 NumVector member 006 NumVector objectToNumVector 001 NumVector ref 024 NumVector resize 008 NumVector sizeof 016 NumVector uniqueInsert 006 NumVector vectorFill 005 Number asin 001 Number atan 001 Number avg 001 Number compare 011 Number cos 001 Number cosh 001 Number deg 001 Number exp 001 Number floor 001 Number kurtosis 001 Number log 001 Number log10 001 Number log2 001 Number logbase 001 Number max 001 Number median 001 Number min 001 Number mod 001 Number modi 001 Number nadd 001 Number ncompareGE 001 Number ncompareGT 001 Number ncompareLE 001 Number ncompareLT 001 Number ndiv 001 Number nmod 001 Number nmul 001 Number nsub 001 Number number 001 Number pi 001 Number product 001 Number rad 001 Number random 001 Number randomize 001 Number range 001 Number round 001 Number sign 001 Number sin 001 Number sinh 001 Number sizeof 011 Number skew 001 Number sqrt 001 Number srandom 001 Number stdev 001 Number stdevp 001 Number sumsqr 001 Number tan 001 Number tanh 001 Number var 001 Number varp 001 ObjVector compare 017 ObjVector delete 012 ObjVector insert 008 ObjVector isMember 007 ObjVector isObject 010 ObjVector isObjectVector 001 ObjVector length 011 ObjVector member 007 ObjVector ref 025 ObjVector resize 009 ObjVector sizeof 017 ObjVector uniqueInsert 007 ObjVector vectorFill 006 ObjectRepository delete 013 Pair compare 019 Pair compareEQ 019 Pair compareGE 019 Pair compareGT 019 Pair compareLE 019 Pair compareLT 019 Pair compareNE 019 Pair comparison 019 Pair count 006 Pair length 014 Pair sizeof 021 PcodeVector isObject 011 PcodeVector isPcodeVector 001 PcodeVector length 012 PcodeVector ref 026 PcodeVector sizeof 018 ShortVector binaryInsert 006 ShortVector compare 016 ShortVector delete 011 ShortVector insert 007 ShortVector inside 006 ShortVector isMember 006 ShortVector isObject 009 ShortVector length 010 ShortVector member 006 ShortVector ref 024 ShortVector resize 008 ShortVector sizeof 016 ShortVector uniqueInsert 007 ShortVector vectorFill 005 String code 001 String compare 001 String compareEQ 001 String compareGE 001 String compareGT 001 String compareLE 001 String compareLT 001 String compareNE 001 String find 001 String find 002 String hashString 001 String isChar 001 String isCharAlphabetic 001 String isCharAlphanumeric 001 String isCharLowercase 001 String isCharName 001 String isCharNumeric 001 String isCharUppercase 001 String isCharWhitespace 001 String isEqual 001 String isEqual 002 String isObject 001 String isString 001 String isText 001 String isType 001 String isType 002 String isType 003 String left 001 String length 001 String length 002 String mid 001 String new 001 String parse 001 String ref 001 String ref 002 String replace 001 String rept 001 String right 001 String sizeof 001 String string 001 String string 002 String string 003 String stringCiEQ 001 String stringCiGE 001 String stringCiGT 001 String stringCiLE 001 String stringCiLT 001 String stringCiNE 001 String substitute 001 String substitute 002 String substring 001 String substring 002 String substringCiEQ 001 String substringCiGE 001 String substringCiGT 001 String substringCiLE 001 String substringCiLT 001 String substringCiNE 001 String text 001 String type 001 Structure apply 001 Structure binaryInsert 001 Structure binarySearch 001 Structure compareGE 003 Structure compareGT 003 Structure compareLE 003 Structure compareLT 003 Structure compareNE 003 Structure count 001 Structure defstruct 001 Structure delete 001 Structure insert 001 Structure inside 001 Structure isBound 001 Structure isObject 003 Structure length 004 Structure member 001 Structure new 003 Structure ref 004 Structure ref 006 Structure refAttributes 001 Structure refValues 001 Structure resize 001 Structure setq 004 Structure setq 005 Structure setq 006 Structure sizeof 003 Structure uniqueInsert 001 Structure uniqueInsert 002 Symbol compare 002 Symbol compareEQ 002 Symbol compareGE 002 Symbol compareGT 002 Symbol compareLE 002 Symbol compareLT 002 Symbol compareNE 002 Symbol hashString 002 Symbol isCharName 002 Symbol isCharNumeric 002 Symbol isObject 002 Symbol isSymbol 001 Symbol isType 004 Symbol length 003 Symbol new 002 Symbol sizeof 002 Symbol type 002 Vector binaryInsert 002 Vector binaryInsert 003 Vector compare 006 Vector count 004 Vector delete 007 Vector delete 008 Vector insert 004 Vector inside 004 Vector isMember 004 Vector isObject 006 Vector isVector 001 Vector length 007 Vector member 004 Vector objectToVector 001 Vector ref 013 Vector ref 014 Vector refAttributes 004 Vector refValues 004 Vector resize 004 Vector setBlock 001 Vector sizeof 006 Vector uniqueInsert 003 Vector vectorFill 001 Void compare 008 Void ref 017 Void sizeof 008 substringEQ 001 substringGE 001 substringGT 001 substringLE 001 substringLT 001 substringNE 001
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# A recent New York Times editorial criticized the citys
Author Message
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5036
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Location: Singapore
A recent New York Times editorial criticized the citys [#permalink]
### Show Tags
31 Aug 2005, 06:54
00:00
Difficulty:
(N/A)
Question Stats:
100% (00:17) correct 0% (00:00) wrong based on 3 sessions
### HideShow timer Statistics
A recent New York Times editorial criticized the city’s election board for, first of all, failing to replace outmoded voting machines prone to breakdowns, and secondarily, for their failure to investigate allegations of corruption involving board members.
(A) secondarily, for their failure to
(B) secondly, for their failure to
(C) secondly, that they failed and did not
(D) second, that they failed to
(E) second, for failing to
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Manager
Joined: 01 Jun 2005
Posts: 69
Kudos [?]: 5 [0], given: 0
### Show Tags
31 Aug 2005, 06:58
I will go with E.
pronoun problem. the reference of "their" is ambiguous. Does it refer to "New York Times editorial" or "city's election board"
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### Show Tags
01 Sep 2005, 09:26
OA is E.
Kudos [?]: 436 [0], given: 0
01 Sep 2005, 09:26
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# Problem 1026. Solving a quadratic equation
Solution 161268
Submitted on 12 Nov 2012 by Noah
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
%% a = 1; b = 2; c = 1; y_correct = -1; assert(isequal(solution(a,b,c),y_correct))
ans = -1
2 Pass
%% a = 1; b = 0; c = 1; y_correct = [-i i]; assert(isequal(solution(a,b,c),y_correct))
ans = 0 - 1i 0 + 1i
3 Pass
%% a = 1; b = -2; c = -3; y_correct = [3 -1]; assert(isequal(solution(a,b,c),y_correct))
ans = 3 -1
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In the following example, why would Chebyshev’s Theorem be used instead of the Empirical Rule?
The Empirical Rule is a rule in statistics that says for a normal distribution, most of all of the data will land between three standardized yet different deviations from their mean. What the empirical rule does is it displays that 68% of the information will fall inside the first standard deviation, that about 95% of the data will fall within the first two standard deviations, and that 99.7% of the data will fall within the first three standard deviations of the mean.
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It is also sometimes called the rule of 68-95-99.7 of the 3 Sigma Rule.This rule very closely relates to the diagram (it displays the rule fairly accurately). Chebyshev’s Theorem states essentially that a distribution of any shape or size puts a lower level on the percents of the observations. This occurs inside a provided number of standard deviations which come from its mean. Now the Empirical Rule on the other hand applies to more specifically mounded-shaped and or symmetrical distributions. It displays percentages as approximations inside of a standard deviation which comes from its mean. The Emperical Rule could apply to the gas mileage on Subaru Imprezzas and Chebyshev’s Theorem could be used for something like trying to calculate what percentages of a value will end up between 159 and 227 for a data set with mean of 194 and standard deviation of 16.5.
Solution Preview
The reason that the Chebyshev’s Theorem would be used instad of the Empirical Rule is that Chebyshev’s Theorem is valid for any set of data.
Where any set of data within the K standard deviations of the mean is demonstrated 1-1/K², K being any number greater than 1.
Example: When K = 2 the formula will demonstrate that 75% of the data will always be within two standard deviations of the mean.
When K = 3 the formula will …
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• Index
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• » Is It Possible To Calculate The End of The Earth's Existence!?
## #51 2011-01-11 09:13:50
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Thanks. So, shall we test if there is a black hole on that radius? I'd be glad to check for traces!
When did the bluewhale come? I believe that the black holes birth is 114 years after that!
Last edited by LQ (2011-01-11 09:14:42)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #52 2011-01-11 09:15:13
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Hi LQ;
He has been around longer than man that is for sure.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #53 2011-01-11 09:15:47
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
hmm. I'll check wiki, do you have a safer source then wiki?
Last edited by LQ (2011-01-11 09:16:21)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #54 2011-01-11 09:23:52
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Hi;
I use it all the time. Let me clear up any misunderstanding, I do not have any soft points, I just thought I would explain my actions ( explained in the pm ) which might seem a little weird in this situation and only in this situation.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #55 2011-01-11 09:28:44
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
600 till 450 miljoner år sedan, ökade kroppsstorleken enormt — från varelser stora som en femtioöring till djur och växter som vägde många tiotals ton: Först jättebläckfiskar, sedan dinosaurier, mammutträd och blåvalar. Det är den nivån vi fortfarande befinner oss på.
Translation:
600 to 450 million years ago, the bodysize increased enourmously - from current beings big as a 50cent to animals and herbs that weighed many tens of tons: First Giant octopussy, then dinosaurs, mammoth-trees, and blue wales. That is the level we are still on.
(so basically, if we only knew what weight it began with by then, we can derive from that how much closer the black hole is)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #56 2011-01-11 09:35:17
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Life on earth begins with bacteria. Their mass is approximately.
600 * 10 ^(-12) grams or .000 000 000 6 gms
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #57 2011-01-11 10:14:00
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
http://marinebio.org/species.asp?id=41
If they come from chamels for 45 million years ago that weighed 37 kilograms at birth...
and if the bluewhale at birth weighs 2700 kilograms at birth. And I weighed 4 kg at birth, then:
300000 lightyears away is my black hole. (edited, I have read up on the magellan cloud supernova)
32432 lightyears away was the chamels black hole 45 million years ago
461 lightyears away is that same black hole today.
If that is right, then the black hole will intersect in: 648869 years
With the speed: 212992 m/s
Edited: Used magellan cloud distance as distance to supernova 1987, now knows actual distance.
Last edited by LQ (2011-01-12 01:25:15)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #58 2011-01-11 10:17:12
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
How did you get the 237 light years?
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #59 2011-01-11 10:23:24
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
I believe that the distance of the black hole is the only proportionality to the animals weight, since the number of quantum entanglements to the black hole increase with its weight. Thus the only proportionality is directly between the animal size and the distance. Roughly.
4/2700*(my black hole distance 160000 light years)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #60 2011-01-11 10:29:33
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Okay, I got it.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #61 2011-01-11 11:05:14
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
If it heads directly against us, when will it destroy us though? will the whale eat up earth first? Not that it shouldn't live, it's important study of where the black hole is. Alive they're worth more.
Last edited by LQ (2011-01-11 11:14:52)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #62 2011-01-11 11:12:42
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Hold it I am still at the baby...
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #63 2011-01-11 11:15:15
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
What's that bobbym? My black hole is in the magellan cloud, I can feel that and understand it.
Last edited by LQ (2011-01-11 11:16:16)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #64 2011-01-11 11:15:46
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
I am watching the vid you sent me to.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #65 2011-01-11 11:16:56
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
okay, go on. I'll stay right here.
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #66 2011-01-11 11:38:21
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Kid is growing and it looks like he is a bomb.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #67 2011-01-11 11:41:05
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
His p-cube, his black hole quantum entangled state interracts with the growth rate of the bacterias.
Last edited by LQ (2011-01-11 11:43:41)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #68 2011-01-11 11:43:54
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
I am beginning to see that. We are going to have some fat bacteria pretty soon.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #69 2011-01-11 11:45:11
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
nono, quick regenerating bacterias. It's the speed that's higher.
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #70 2011-01-11 11:47:44
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Kid is in trouble. His p-cube is interfering with the ship.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #71 2011-01-11 11:59:30
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
So he must leave... which he does.
The question is: what happened? well, if the p-cube manages to entangle with the ship, then that may go away when he leaves. But what if that black hole was infact the mother of the bethazoids?
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #72 2011-01-11 15:53:12
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
It could be. Maybe they do not know that.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
## #73 2011-01-11 16:35:51
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
How near does a fairly big black hole have to be to us to wipe us out though?
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #74 2011-01-12 01:06:45
LQ
Real Member
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
#### bobbym wrote:
It could be. Maybe they do not know that.
Bobbym, you figure that if jesus was a little smaller then a human, yet could do incredible things, that he was from a black hole about the double distance then that of me? The supernatural abillity, could this be caused by that the black hole was fully active, yet rotating about the same amount as ordinary people? Possibly that the black hole was charged too? There was an intense light on the sky when he was born, maybe a really big super nova?
PS. I edited post #57, just read the actual distance of the magellan cloud super nova
Last edited by LQ (2011-01-12 01:28:35)
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
## #75 2011-01-12 07:27:25
bobbym
Offline
### Re: Is It Possible To Calculate The End of The Earth's Existence!?
Hi LQ;
Yes, definitely a superior being. I have never heard that he was smaller than we are. He might have been. Only Dan Brown thinks he knows everything about the story. The star in the sky? God's hand, an alien spacecraft, supernova, take your pick?! Some cultures believe in the concept of prana or the chi. Lucas calls it the force. That energy comes from the stars themselves. Humans can absorb it through contact with the ground, food, water and of course the most important, the air.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
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### "AVERAGE RETURN ON INVESTMENT" IS MISLEADING
If you add the percentage gains and losses over a series of time periods and then average them to get some “average” return on your investment, ROI, it is known that the result can be quite misleading.
Say you make 60% gain (G) on your investment (I) this year and take a 40% loss (L) next year. You might think your average return on the investment was (60%-40%)/2yr=+10%/yr. Unfortunately, it was really a 4% loss, -2% per year, a substantial difference.
Let’s say you have invested \$1000. The first year you gained 60%, making your total \$1600. The next year you lost 40% of the \$1600, making your total \$960. Though your “average” return on investment was (60%-40%)/2=10%, you actually lost money, (\$960-\$1000)=-\$40, a 4% loss, or 2% per year on your original \$1000 investment.
The lesson: you cannot simply average the yearly percentage returns on investment to get a reliable estimate of how well the investment is performing.
*****
A little mathematics shows that it does not matter whether the loss was in the first year or the second. At the end of two years, you have your investment of \$1000 times (one plus the fractional gain, G) times (1 minus the fractional loss, L):
\$1000(1 + G)(1 - L) = \$1000(1.60)(0.60) = \$960.
For those who are curious about the mathematics, the general formula for this two-period case
(1 + G)(1 - L) = (1 + G - L - G*L)
and the product G*L is why the true outcome is different from the naïve average, (G-L)/2. If both periods showed gains, G and G’, then the equation uses +G’ rather than -L, and the correct answer for the characteristic gain per year is again different from the naïve average, (G+G’)/2.
Averaging over more than two periods produces a similar outcome, more difficult to express succinctly, but the message is the same, the naïve averaging of the ROI values incorrectly assesses the return on investment at the end of the period being analyzed.
FOOTNOTE:
The proper calculation of a characteristic or effective yearly return on investment requires determining the ratio of the final value of the account to its initial (investment) value and then finding the internal rate of return r, which when used in the formula (1+r)^n, where n is the number of periods (^ means exponentiation, raising to a power), gives that correct ratio. In our example 0.96=(1+r)^2, which can be solved as r = -0.02, a 2% loss per year. | 631 | 2,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2020-16 | longest | en | 0.935101 |
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# Nabil Chowdhury
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216 2020 이후 총 참여 횟수
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https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_2&diff=144568&oldid=144566 | 1,627,658,088,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00450.warc.gz | 130,260,921 | 16,061 | # Difference between revisions of "2020 AIME I Problems/Problem 2"
## Problem
There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$, $\log_4{x}$, and $\log_2{x}$, in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
## Solution
Since these form a geometric series, $\frac{\log_2{x}}{\log_4{x}}$ is the common ratio. Rewriting this, we get $\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}$
$\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}$. Therefore, $1 + 16 = \boxed{017}$.
~ JHawk0224
## Solution 2
If we set $x=2^y$, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are $$\frac{y+1}{3}, \frac{y}{2}, y.$$ In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: $$\frac{y^2+y}{3} = \frac{y^2}{4},$$ which can be solved to reveal $y = -4$. Therefore, $x = 2^{-4} = \frac{1}{16}$, so our answer is $\boxed{017}$.
-molocyxu
## Solution 3
Let $r$ be the common ratio. We have $$r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}$$ Hence we obtain $$(\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})$$ Ideally we change everything to base $64$ and we can get: $$(\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})$$ Now divide to get: $$\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}$$ By change-of-base we obtain: $$\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2$$ Hence $(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}$ and we have $1+16 = \boxed{017}$ as desired.
~skyscraper
## Solution 4 (Exponents > Logarithms)
Let $r$ be the common ratio, and let $a$ be the starting term ($a=\log_{8}{(2x)}$). We then have: $$\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2$$ Rearranging these equations gives: $$8^a=2x, 4^{ar}=x, 2^{ar^2}=x$$ Deal with the last two equations first: Setting them equal gives: $$4^{ar}=2^{ar^2} \implies 2^{2ar}=2^{ar^2} \implies 2ar=ar^2 \implies r=2$$ Using this value of $r$, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: $$8^a=2x, 4^{2a}=x$$ Changing these to a common base gives: $$2^{3a}=2x, 2^{4a}=x$$ Dividing the first equation by 2 on both sides yields: $$2^{3a-1}=x$$ Setting these equations equal to each other and removing the exponent again gives: $$3a-1=4a \implies a=-1$$ Substituting this back into the first equation gives: $$8^{-1}=2x \implies 2x=\frac{1}{8} \implies x=\frac{1}{16}$$ Therefore, $m+n=1+16=\boxed{017}$
~IAmTheHazard
## Solution 5
We can relate the logarithms as follows:
$$\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}$$ $$\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}$$
Now we can convert all logarithm bases to $2$ using the identity $\log_a{b}=\log_{a^c}{b^c}$:
$$\log_2{\sqrt[3]{2x}}\log_2{x}=\log_2{\sqrt{x}}\log_2{\sqrt{x}}$$
We can solve for $x$ as follows:
$$\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}$$ $$\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}$$ $$\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}$$ We get $x=\frac{1}{16}$. Verifying that the common ratio is positive, we find the answer of $\boxed{017}$.
~QIDb602
## Solution 6
If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as $\frac{1+\log_2{x}}{3}$ and $\frac{1}{2}\log_2{x}$, respectively. Therefore: $$\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}$$ Let $n=\log_2{x}$. We can rewrite the expression as: $$\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}$$ $$\frac{n^2}{4}=\frac{n(n+1)}{3}$$ $$4n(n+1)=3n^2$$ $$4n^2+4n=3n^2$$ $$n^2+4n=0$$ $$n(n+4)=0$$ $$n=0 \text{ and } -4$$ Zero does not work in this case, so we consider $n=-4$: $\log_2{x}=-4 \rightarrow x=\frac{1}{16}$. Therefore, $1+16=\boxed{017}$.
~Bowser498
## Solution 7
Again, by the Change of Base Formula, obtain that the common ratio is 2. If we let $y$ be the exponent of $\log_8 (2x)$, then we have $8^y=2x;\:4^{2y}=x;\:2^{4y}=x.$ Wee can then divide the first equation by two to have the right side be $x$. Also, $2^{4y}=\left(2^{4}\right)^y=16^y$. Setting this equal to $\frac{8^y}{2}$, we can divide the two equations to get $2^y=\frac12$. Therefore, $y=-1$. After that, we can raise $16$ to the $-1$th power to get $x=16^{-1}\Rightarrow x=\frac1{16}$. We then get our sum of $1+16=\boxed{\textbf{017}}$.
## Solution 8 (Official MAA)
By the Change of Base Formula the common ratio of the progression is$$\frac{\log_2 x}{\log_4 x} = \frac{\hphantom{m}\log_2x\hphantom{m}}{\frac{\log_2x}{\log_24}} = 2.$$Hence $x$ must satisfy$$2=\frac{\log_4 x}{\log_8 (2x)}= \frac{\log_2 x}{\log_2 4} \div \frac{\log_2(2x)}{\log_28} = \frac 32\cdot \frac{\log_2x}{1+\log_2x}.$$This is equivalent to $4 + 4\log_2x = 3\log_2x$. Hence $\log_2x = -4$ and $x = \frac{1}{16}$. The requested sum is $1+16 = 17$. See here for a video solution: | 2,147 | 5,459 | {"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": 86, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-31 | latest | en | 0.687402 |
https://au.mathworks.com/matlabcentral/answers/1461924-how-to-create-an-image-that-has-random-chunks-of-black-and-white-pixels | 1,638,345,590,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359093.97/warc/CC-MAIN-20211201052655-20211201082655-00590.warc.gz | 184,430,479 | 24,115 | # How to create an image that has random chunks of black and white pixels
9 views (last 30 days)
MM on 27 Sep 2021
Edited: Adam Danz on 28 Sep 2021
I would like to create an image (1920 x 1080) that looks like the image below. How do I create something like this? Black and white spots can be randomly placed
Adam Danz on 27 Sep 2021
Edited: Adam Danz on 28 Sep 2021
Specify imageSize and blockScale to create randomly placed blocks of black and white pixels.
imageSize = [1920,1080]; % [width, height]
blockScale = 40; % larger values create larger blocks
blocksize = round(imageSize/blockScale);
data = randi(2,fliplr(blocksize))-1;
B = imresize(data, fliplr(imageSize),'nearest');
imagesc(B)
colormap([0 0 0; 1 1 1])
axis equal
axis tight
set(gca, 'xtick', [], 'ytick', [], 'LineWidth', 5) % add frame
If you just want random black and white pixels,
B = randi(2,[1080,1920])-1;
figure()
imshow(B)
### More Answers (1)
yanqi liu on 28 Sep 2021
clc; clear all; close all;
% init image (1920 x 1080)
A = zeros(1080,1920);
% choose white
ind1 = randperm(numel(A(:)));
rate = 0.3;
xt = ind1(1:round(length(ind1)*rate));
A(xt) = 1;
A([1 end],:)=0;
A(:,[1 end])=0;
figure; imshow(logical(A));
R2019a
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Start Hunting! | 414 | 1,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-49 | latest | en | 0.659226 |
https://www.teacherspayteachers.com/Product/Multiples-of-10-Task-Card-Strips-60-Flash-Cards-10-Activity-Pages-3359815 | 1,529,394,092,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00549.warc.gz | 941,972,074 | 16,517 | # Multiples of 10 Task Card Strips, 60 Flash Cards & 10 Activity Pages
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Product Description
Multiples of 10 Mini-Bundle
This is another great resource for your multiplication curriculum. Students will answer a variety of questions on the 30 task card strips with reinforcement from the 60 flash cards and activity pages.
Use the activity pages for SCOOT, Kagan activities, math groups or centers or as formative assessments. Many types of activities included in this file require students to compare, analyze, model, and show their work.
Here is a breakdown of the 10 activity pages:
1) "Hitting a Homerun" - Match the baseball (product) to the bat (2 factors)
2) "Spin It, Solve It" - Use two spinners to create the multiples of 10 equation.
3) "Group It, Solve It" - Use the grouping property to solve multiples of 10 equations.
4) "Color It, Solve It" - Read the equation, color in the place value tens rods to match
5) "Apple Partners" - Match the apple (factors) to the apple pie (product)
6) "Read It, Draw It, Solve It" - Solve simple word problems
7) "Draw It, Solve It" - Draw the place value rods to match the multiplication equation
8) "Multiples of 10 - Name the Equation" - Look at the groups and identify equation
9) "Solve & Color" - Read the equation and select from three multiple choice products
10) "Solve & Glue" - Cut out the product and glue next to the matching factors
Your students will have many opportunities to show their understanding of multiplying with multiples of 10 in this product. I hope your students enjoy!
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 432 | 1,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.866625 |
http://math.stackexchange.com/questions/139954/question-1-1-2-from-hatcher | 1,469,598,663,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825366.39/warc/CC-MAIN-20160723071025-00241-ip-10-185-27-174.ec2.internal.warc.gz | 158,859,644 | 17,658 | # Question 1.1.2 from Hatcher
I'm having trouble understanding this question.
We have a path $h$ in $X$ from $x_0$ to $x_1$ and $\bar{h}$ its inverse path. Then a map $\beta_h:\pi_1(X,x_1)\to \pi_1(X,x_0)$ defined by $\beta_h[f]=[h\circ f\circ \bar{h}]$, for every path $f$ in $X$.
The question is to show that $\beta_h$ depends only on the homotopy class of $h$.
Firstly, it says for every path $f$ in $X$, but surely $f$ has to be a loop or you can't form $[h\circ f\circ \bar{h}]$?
And also, I don't understand why it depends on the homotopy class of $h$, when $[h\circ f\circ \bar{h}]$ is the path going from $x_0$ to $x_1$, around $f$, then back to $x_0$, why does the homotopy class of $h$ matter? In general I don't think I fully understand what this map $\beta_h$ is and would like someone to help me out. Thanks
-
They are asking you to show that only the homotopy class of $h$ matters. That is, that if $h\sim k$, then $[h\circ f\circ \overline{h}] = [k\circ f \circ\overline{k}]$. So they are not saying the homotopy of $h$ matters, they are saying you get the same result by replacing $h$ with any path that is homotopic to $h$. – Arturo Magidin May 2 '12 at 16:19
ahh, that explains my confusion. Thanks! – 09867 May 2 '12 at 16:28
The question is for every path $h$, not $f$. As $[f] \in \pi_1(X, x_1)$, $f$ is a loop in $X$ based at $x_1$, not a path from $x_0$ to $x_1$.
Instead of writing $h\circ f\circ\bar{h}$ you should write $h\cdot f\cdot\bar{h}$ because you are not composing the maps $h$, $f$, and $\bar{h}$, which is what the symbol $\circ$ is usually reserved for. Also, when using composition, we work right to left (i.e. $f\circ g\circ h$ means apply $h$, then apply $g$, then apply $f$), but with concatenation of paths we work left to right (i.e. $f\cdot g\cdot h$ means travel along the path $f$, then the path $g$, then the path $h$).
As Arturo pointed out, you need to show that if $h$ and $k$ are homotopic paths from $x_0$ to $x_1$, then $h\cdot f\cdot\bar{h}$ and $k\cdot f\cdot\bar{k}$ are homotopic loops based at $x_0$. | 709 | 2,065 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2016-30 | latest | en | 0.886705 |
http://discuss.cocos2d-x.org/t/would-you-like-to-talk-to-me-about-convertto-method-in-ccnode/1723 | 1,519,054,485,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812756.57/warc/CC-MAIN-20180219151705-20180219171705-00626.warc.gz | 85,609,621 | 3,784 | # Would you like to talk to me about convertTo method in CCNode?
#1
I want to picking sprite in my map.
I Just know picking is skill that monitor position with a sprite position.
I got a lot of time. but my head was burnning.
CCPoint convertToNodeSpace (const CCPoint &worldPoint)
Converts a Point to node (local) space coordinates.
CCPoint convertToWorldSpace (const CCPoint &nodePoint)
Converts a Point to world space coordinates.
CCPoint convertToNodeSpaceAR (const CCPoint &worldPoint)
Converts a Point to node (local) space coordinates.
CCPoint convertToWorldSpaceAR (const CCPoint &nodePoint)
Converts a local Point to world space coordinates.The result is in Points.
CCPoint convertTouchToNodeSpace (CCTouch *touch)
convenience methods which take a CCTouch instead of CCPoint
CCPoint convertTouchToNodeSpaceAR (CCTouch *touch)
converts a CCTouch (world coordinates) into a local coordiante.
#2
Is this image helpful? OpenGL coordination is different from Touch coordination.
#3
convertToWorldSpace(const CCPoint& nodePoint) converts on-node coords to SCREEN coordinates.
Lets we have layerA with anchor point and position (0,0) attached to screen and have a sprite on this layer at point (100, 100).
What will be SCREEN coords of sprite? - (100, 100)
Lets we moved layerA to point (- 50, - 20). What will be SCREEN coords of sprite? - (100 - 50, 100 - 20), i.e. (50, 80) - that’s what convertToWorldSpace returns to us if we call layerA~~>convertToWorldSpace).
The same is applied when you scale layerA~~ i.e. convertToWorldSpace will always return SCREEN position of our sprite, might be very useful if you want to capture taps on your sprite but need to move/scale your layerA.
As for convertToWorldSpaceAR - will return the position relatevely to anchor point: so if our scene - root layer has AP (0.5f, 0.5f) - default, convertToWorldSpaceAR should return position relatively to screen center. I have not used it.
convertToNodeSpace(const CCPoint& worldPoint) - converts SCREEN coords to NODE’s local. I.e. if for our example with moved layer call:
layerA~~>convertToNodeSpace)~~ that should return (100, 100) - our sprite on-node coords.
convertToNodeSpaceAR - the same logic as for convertToWorldSpaceAR.
Alex
#4
so WorldSpace means screen coords.
Thank you. very much. | 569 | 2,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-09 | latest | en | 0.672992 |
https://www.numbersaplenty.com/121624332642420 | 1,620,439,740,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00340.warc.gz | 970,305,449 | 4,185 | Search a number
121624332642420 = 223541132137426787
BaseRepresentation
bin11011101001110111011111…
…001001001010100001110100
3120221122011122212012211020020
4123221313133021022201310
5111420143113124024140
61110401240322405140
734422032225633526
oct3351673711124164
9527564585184206
10121624332642420
1135831654045033
121178373ab007b0
1352b318253877a
142206913225a16
hex6e9ddf24a874
121624332642420 has 96 divisors (see below), whose sum is σ = 349118276121216. Its totient is φ = 31618148812800.
The previous prime is 121624332642413. The next prime is 121624332642451. The reversal of 121624332642420 is 24246233426121.
It is a super-2 number, since 2×1216243326424202 (a number of 29 digits) contains 22 as substring.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 15463734 + ... + 21963053.
It is an arithmetic number, because the mean of its divisors is an integer number (3636648709596).
Almost surely, 2121624332642420 is an apocalyptic number.
121624332642420 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
121624332642420 is an abundant number, since it is smaller than the sum of its proper divisors (227493943478796).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
121624332642420 is a wasteful number, since it uses less digits than its factorization.
121624332642420 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 37428161 (or 37428159 counting only the distinct ones).
The product of its (nonzero) digits is 663552, while the sum is 42.
The spelling of 121624332642420 in words is "one hundred twenty-one trillion, six hundred twenty-four billion, three hundred thirty-two million, six hundred forty-two thousand, four hundred twenty". | 564 | 1,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.825547 |
https://exponentiations.com/52-to-the-42nd-power | 1,585,409,781,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370491998.11/warc/CC-MAIN-20200328134227-20200328164227-00366.warc.gz | 480,749,466 | 30,583 | # 52 to the 42nd Power
Welcome to 52 to the 42nd power, our post about the mathematical operation exponentiation of 52 to the power of 42. If you have been looking for 52 to the forty-second power, or if you have been wondering about 52 exponent 42, then you also have come to the right place. The number 52 is called the base, and the number 42 is called the exponent. In this post we are going to answer the question what is 52 to the 42nd power. Keep reading to learn everything about fifty-two to the forty-second power.
## What is 52 to the 42nd Power
52 to the 42nd power is conventionally written as 5242, with superscript for the exponent, but the notation using the caret symbol ^ can also be seen frequently: 52^42.
5242 stands for the mathematical operation exponentiation of fifty-two by the power of forty-two. As the exponent is a positive integer, exponentiation means a repeated multiplication:
52 to the 42nd power = $\underbrace{ {\rm 52\hspace{3px} \times\hspace{7px} …\hspace{5px} \times\hspace{3px} 52} }_{\rm 42\hspace{3px} times}$
The exponent of the number 52, 42, also called index or power, denotes how many times to multiply the base (52).
Thus, we can answer what is 52 to the 42nd power as
52 to the power of 42 = 5242 = 118 0702363666 6328766670 5087074794 9278738381 2264089974 6405469184 6951010304.
If you have come here in search of an exponentiation different to 52 to the forty-second power, or if you like to experiment with bases and indices, then use our calculator below.
To stick with 52 to the power of 42 as an example, insert 52 for the base and enter 42 as the index, also known as exponent or power. Next, hit the convert button, then check the result.
52 to the 42nd power is an exponentiation which belongs to the category powers of 52. Similar exponentiations on our site in this category include, but are not limited, to:
Ahead is more info related to 52 to the 42 power, along with instructions how to use the search form, located in the sidebar or at the bottom, to obtain a number like 52^42.
## 52 to the Power of 42
Reading all of the above, you already know most about 52 to the power of 42, except for its inverse which is discussed a bit further below in this section.
Using the aforementioned search form you can look up many numbers, including, for instance, 52 to the power 42, and you will be taken to a result page with relevant posts.
Now, we would like to show you what the inverse operation of 52 to the 42nd power, (5242)−1, is. The inverse is 42nd root of 5242, and the math goes as follows:
(5242)−1
= $\sqrt[42]{52^{42}}$
= $52^{42/42}$
= $52^{1}$
= 52
Because the index of 42 is a multiple of 2, which is even, in contrast to odd numbers, the operation produces two results: (5242)−1$\hspace{3px} = \pm 52$; the positive value is the principal root.
Make sure to understand that exponentiation is not commutative, which means that 5242 ≠ 4252, and also note that (5242)-1 ≠ 52-42, the inverse and reciprocal of 5242, respectively.
You already know what 52 to the power of 42 equals, but you may also be interested in learning what 52 to the negative 42nd power stands for. Next is the summary of our content.
## Fifty-two to the Forty-second Power
You have reached the concluding section of fifty-two to the forty-second power = 5242. Fifty-two to the forty-second power is, for example, the same as 52 to the power 42 or 52 to the 42 power.
Exponentiations like 5242 make it easier to write multiplications and to conduct math operations as numbers get either big or small, such as in case of decimal fractions with lots of trailing zeroes.
If you have been looking for 52 power 42, what is 52 to the 42 power, 52 exponent 42 or 42 power of 52, then it’s safe to assume that you have found your answer as well.
If our explanations have been useful to you, then please hit the like button to let your friends know about our site and this post 52 to the 42nd power. And don’t forget to bookmark us.
If you like to learn more about exponentiation, the mathematical operation conducted in 5242, then check out the articles which you can locate in the header menu of our site.
We appreciate all comments on 52^42, and if you have a question don’t hesitate filling in the form at the bottom or sending us an email with the subject what is 52 to the 42nd power.
Thanks for visiting 52 to the 42nd power.
Posted in Powers of 52 | 1,162 | 4,420 | {"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} | 4.125 | 4 | CC-MAIN-2020-16 | latest | en | 0.920176 |
https://www.solutionsfolks.com/ExpertAnswers/1-for-the-mechanism-shown-below-where-z-3-20-z-3-30-z-4-40-z-5-25-z-5-35-m-pa168 | 1,721,125,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00223.warc.gz | 855,252,262 | 5,653 | Home / Expert Answers / Mechanical Engineering / 1-for-the-mechanism-shown-below-where-z-3-20-z-3-30-z-4-40-z-5-25-z-5-35-m-pa168
# (Solved): 1. For the mechanism shown below (Where: $$z_{3}=20, z_{3}=30, z_{4}=40, z_{5}=25, z_{5}=$$ 35, M ...
1. For the mechanism shown below (Where: $$z_{3}=20, z_{3}=30, z_{4}=40, z_{5}=25, z_{5}=$$ 35, Module $$=4, \quad m_{2}=15 \mathrm{~kg}, \quad m_{6}=10 \mathrm{~kg}, J_{3}=J_{4}=J_{5}=300 \mathrm{~kg} \cdot \mathrm{cm}^{2}$$ ), determine the equivalent mass to be located at link 2
We have an Answer from Expert | 228 | 559 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-30 | latest | en | 0.629344 |
https://www.hackmath.net/en/math-problem/7263 | 1,718,573,139,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00239.warc.gz | 707,145,565 | 7,892 | # Craftsman 7263
To make a ladder, the craftsman needs to cut as many rungs of the same length as possible. He is to cut them from two boards, one is 220cm long, and the other is 308cm long. How long will the bars be, and how many will there be?
d = 44 cm
n = 12
## Step-by-step explanation:
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Do you want to calculate the greatest common divisor of two or more numbers? | 128 | 482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.950152 |
https://functions.wolfram.com/ElementaryFunctions/Cosh/21/01/02/21/04/07/0002/ | 1,718,370,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00080.warc.gz | 253,610,840 | 13,278 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Cosh
http://functions.wolfram.com/01.20.21.1439.01
Input Form
Integrate[z^n E^(p z) Cos[b Sqrt[z]]^m Cosh[c z], z] == 2^(-2 - m) (2 Binomial[m, m/2] ((c - p)^(-1 - n) Gamma[1 + n, (c - p) z] + (-c - p)^(-1 - n) Gamma[1 + n, (-(c + p)) z]) (-1 + Mod[m, 2]) + Sum[Binomial[m, u] ((E^((b^2 (m - 2 u)^2)/(4 (-c + p))) Sum[(-1)^(-h + k) 4^k (I b (m - 2 u))^(-h - k + 2 n) ((b (m - 2 u) + 2 I (c - p) Sqrt[z])^2/(-c + p))^ ((1/2) (-1 - h - k)) (I b (m - 2 u) + 2 (-c + p) Sqrt[z])^ (h + k) Binomial[k, h] Binomial[n, k] (I b (m - 2 u) (I b (m - 2 u) + 2 (-c + p) Sqrt[z]) Gamma[(1/2) (1 + h + k), (b (m - 2 u) + 2 I (c - p) Sqrt[z])^2/ (4 (-c + p))] + 2 (-c + p) Sqrt[(b (m - 2 u) + 2 I (c - p) Sqrt[z])^2/(-c + p)] Gamma[(1/2) (2 + h + k), (b (m - 2 u) + 2 I (c - p) Sqrt[z])^2/(4 (-c + p))]), {k, 0, n}, {h, 0, k}])/(-c + p)^(2 (1 + n)) + (E^((b^2 (m - 2 u)^2)/(4 (-c + p))) Sum[(-1)^(-h + k) 4^k ((-I) b (m - 2 u))^(-h - k + 2 n) ((b (m - 2 u) + 2 I (-c + p) Sqrt[z])^2/(-c + p))^ ((1/2) (-1 - h - k)) ((-I) b (m - 2 u) + 2 (-c + p) Sqrt[z])^ (h + k) Binomial[k, h] Binomial[n, k] ((-I) b (m - 2 u) ((-I) b (m - 2 u) + 2 (-c + p) Sqrt[z]) Gamma[(1/2) (1 + h + k), (b (m - 2 u) + 2 I (-c + p) Sqrt[z])^2/ (4 (-c + p))] + 2 (-c + p) Sqrt[(b (m - 2 u) + 2 I (-c + p) Sqrt[z])^2/(-c + p)] Gamma[(1/2) (2 + h + k), (b (m - 2 u) + 2 I (-c + p) Sqrt[z])^2/(4 (-c + p))]), {k, 0, n}, {h, 0, k}])/(-c + p)^(2 (1 + n)) + (E^((b^2 (m - 2 u)^2)/(4 (c + p))) (Sum[(-1)^(-h + k) 4^k (I b (m - 2 u))^(-h - k + 2 n) ((b (m - 2 u) - 2 I (c + p) Sqrt[z])^2/(c + p))^((1/2) (-1 - h - k)) (I b (m - 2 u) + 2 (c + p) Sqrt[z])^(h + k) Binomial[k, h] Binomial[n, k] (I b (m - 2 u) (I b (m - 2 u) + 2 (c + p) Sqrt[z]) Gamma[(1/2) (1 + h + k), (b (m - 2 u) - 2 I (c + p) Sqrt[z])^2/(4 (c + p))] + 2 (c + p) Sqrt[(b (m - 2 u) - 2 I (c + p) Sqrt[z])^2/(c + p)] Gamma[(1/2) (2 + h + k), (b (m - 2 u) - 2 I (c + p) Sqrt[z])^2/ (4 (c + p))]), {k, 0, n}, {h, 0, k}] + Sum[(-1)^(-h + k) 4^k ((-I) b (m - 2 u))^(-h - k + 2 n) ((b (m - 2 u) + 2 I (c + p) Sqrt[z])^2/(c + p))^((1/2) (-1 - h - k)) ((-I) b (m - 2 u) + 2 (c + p) Sqrt[z])^(h + k) Binomial[k, h] Binomial[n, k] ((-I) b (m - 2 u) ((-I) b (m - 2 u) + 2 (c + p) Sqrt[z]) Gamma[(1/2) (1 + h + k), (b (m - 2 u) + 2 I (c + p) Sqrt[z])^2/(4 (c + p))] + 2 (c + p) Sqrt[(b (m - 2 u) + 2 I (c + p) Sqrt[z])^2/(c + p)] Gamma[(1/2) (2 + h + k), (b (m - 2 u) + 2 I (c + p) Sqrt[z])^2/ (4 (c + p))]), {k, 0, n}, {h, 0, k}]))/(c + p)^(2 (1 + n))), {u, 0, Floor[(1/2) (-1 + m)]}]/4^n) /; Element[m, Integers] && m > 0 && Element[n, Integers] && n >= 0
Standard Form
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MathML Form
z n p z cos m ( b z ) cosh ( c z ) z 2 - m - 2 ( 2 ( m m 2 ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["m", Identity]], List[TagBox[FractionBox["m", "2"], Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( Γ ( n + 1 , - ( c + p ) z ) ( - c - p ) - n - 1 + ( c - p ) - n - 1 Γ ( n + 1 , ( c - p ) z ) ) ( m mod 2 \$CellContext`m 2 - 1 ) + 4 - n u = 0 m - 1 2 ( m u ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["m", Identity]], List[TagBox["u", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( b 2 ( m - 2 u ) 2 4 ( p - c ) ( p - c ) - 2 ( n + 1 ) k = 0 n h = 0 k ( - 1 ) k - h 4 k ( b ( m - 2 u ) ) - h - k + 2 n ( ( 2 z ( c - p ) + b ( m - 2 u ) ) 2 p - c ) 1 2 ( - h - k - 1 ) ( 2 z ( p - c ) + b ( m - 2 u ) ) h + k ( k h ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["k", Identity]], List[TagBox["h", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( n k ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["k", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( b ( m - 2 u ) ( 2 z ( p - c ) + b ( m - 2 u ) ) Γ ( 1 2 ( h + k + 1 ) , ( 2 z ( c - p ) + b ( m - 2 u ) ) 2 4 ( p - c ) ) + 2 ( 2 z ( c - p ) + b ( m - 2 u ) ) 2 p - c ( p - c ) Γ ( 1 2 ( h + k + 2 ) , ( 2 z ( c - p ) + b ( m - 2 u ) ) 2 4 ( p - c ) ) ) + b 2 ( m - 2 u ) 2 4 ( p - c ) ( p - c ) - 2 ( n + 1 ) k = 0 n h = 0 k ( - 1 ) k - h 4 k ( - b ( m - 2 u ) ) - h - k + 2 n ( ( 2 z ( p - c ) + b ( m - 2 u ) ) 2 p - c ) 1 2 ( - h - k - 1 ) ( 2 ( p - c ) z - b ( m - 2 u ) ) h + k ( k h ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["k", Identity]], List[TagBox["h", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( n k ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["k", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( 2 ( p - c ) ( 2 z ( p - c ) + b ( m - 2 u ) ) 2 p - c Γ ( 1 2 ( h + k + 2 ) , ( 2 z ( p - c ) + b ( m - 2 u ) ) 2 4 ( p - c ) ) - b ( m - 2 u ) ( 2 ( p - c ) z - b ( m - 2 u ) ) Γ ( 1 2 ( h + k + 1 ) , ( 2 z ( p - c ) + b ( m - 2 u ) ) 2 4 ( p - c ) ) ) + b 2 ( m - 2 u ) 2 4 ( c + p ) ( c + p ) - 2 ( n + 1 ) ( k = 0 n h = 0 k ( - 1 ) k - h 4 k ( - b ( m - 2 u ) ) - h - k + 2 n ( ( 2 z ( c + p ) + b ( m - 2 u ) ) 2 c + p ) 1 2 ( - h - k - 1 ) ( 2 ( c + p ) z - b ( m - 2 u ) ) h + k ( k h ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["k", Identity]], List[TagBox["h", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( n k ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["k", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( 2 ( c + p ) ( 2 z ( c + p ) + b ( m - 2 u ) ) 2 c + p Γ ( 1 2 ( h + k + 2 ) , ( 2 z ( c + p ) + b ( m - 2 u ) ) 2 4 ( c + p ) ) - b ( m - 2 u ) ( 2 ( c + p ) z - b ( m - 2 u ) ) Γ ( 1 2 ( h + k + 1 ) , ( 2 z ( c + p ) + b ( m - 2 u ) ) 2 4 ( c + p ) ) ) + k = 0 n h = 0 k ( - 1 ) k - h 4 k ( b ( m - 2 u ) ) - h - k + 2 n ( ( b ( m - 2 u ) - 2 ( c + p ) z ) 2 c + p ) 1 2 ( - h - k - 1 ) ( 2 z ( c + p ) + b ( m - 2 u ) ) h + k ( k h ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["k", Identity]], List[TagBox["h", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( n k ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["k", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( b ( m - 2 u ) ( 2 z ( c + p ) + b ( m - 2 u ) ) Γ ( 1 2 ( h + k + 1 ) , ( b ( m - 2 u ) - 2 ( c + p ) z ) 2 4 ( c + p ) ) + 2 ( b ( m - 2 u ) - 2 ( c + p ) z ) 2 c + p ( c + p ) Γ ( 1 2 ( h + k + 2 ) , ( b ( m - 2 u ) - 2 ( c + p ) z ) 2 4 ( c + p ) ) ) ) ) ) /; m + n Condition z z n p z b z 1 2 m c z 2 -1 m -2 2 Binomial m m 2 -1 Gamma n 1 -1 c p z -1 c -1 p -1 n -1 c -1 p -1 n -1 Gamma n 1 c -1 p z \$CellContext`m 2 -1 4 -1 n u 0 m -1 2 -1 Binomial m u b 2 m -1 2 u 2 4 p -1 c -1 p -1 c -2 n 1 h 0 k k 0 n -1 k -1 h 4 k b m -1 2 u -1 h -1 k 2 n 2 z 1 2 c -1 p b m -1 2 u 2 p -1 c -1 1 2 -1 h -1 k -1 2 z 1 2 p -1 c b m -1 2 u h k Binomial k h Binomial n k b m -1 2 u 2 z 1 2 p -1 c b m -1 2 u Gamma 1 2 h k 1 2 z 1 2 c -1 p b m -1 2 u 2 4 p -1 c -1 2 2 z 1 2 c -1 p b m -1 2 u 2 p -1 c -1 1 2 p -1 c Gamma 1 2 h k 2 2 z 1 2 c -1 p b m -1 2 u 2 4 p -1 c -1 b 2 m -1 2 u 2 4 p -1 c -1 p -1 c -2 n 1 h 0 k k 0 n -1 k -1 h 4 k -1 b m -1 2 u -1 h -1 k 2 n 2 z 1 2 p -1 c b m -1 2 u 2 p -1 c -1 1 2 -1 h -1 k -1 2 p -1 c z 1 2 -1 b m -1 2 u h k Binomial k h Binomial n k 2 p -1 c 2 z 1 2 p -1 c b m -1 2 u 2 p -1 c -1 1 2 Gamma 1 2 h k 2 2 z 1 2 p -1 c b m -1 2 u 2 4 p -1 c -1 -1 b m -1 2 u 2 p -1 c z 1 2 -1 b m -1 2 u Gamma 1 2 h k 1 2 z 1 2 p -1 c b m -1 2 u 2 4 p -1 c -1 b 2 m -1 2 u 2 4 c p -1 c p -2 n 1 h 0 k k 0 n -1 k -1 h 4 k -1 b m -1 2 u -1 h -1 k 2 n 2 z 1 2 c p b m -1 2 u 2 c p -1 1 2 -1 h -1 k -1 2 c p z 1 2 -1 b m -1 2 u h k Binomial k h Binomial n k 2 c p 2 z 1 2 c p b m -1 2 u 2 c p -1 1 2 Gamma 1 2 h k 2 2 z 1 2 c p b m -1 2 u 2 4 c p -1 -1 b m -1 2 u 2 c p z 1 2 -1 b m -1 2 u Gamma 1 2 h k 1 2 z 1 2 c p b m -1 2 u 2 4 c p -1 h 0 k k 0 n -1 k -1 h 4 k b m -1 2 u -1 h -1 k 2 n b m -1 2 u -1 2 c p z 1 2 2 c p -1 1 2 -1 h -1 k -1 2 z 1 2 c p b m -1 2 u h k Binomial k h Binomial n k b m -1 2 u 2 z 1 2 c p b m -1 2 u Gamma 1 2 h k 1 b m -1 2 u -1 2 c p z 1 2 2 4 c p -1 2 b m -1 2 u -1 2 c p z 1 2 2 c p -1 1 2 c p Gamma 1 2 h k 2 b m -1 2 u -1 2 c p z 1 2 2 4 c p -1 m SuperPlus n [/itex]
Rule Form
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Date Added to functions.wolfram.com (modification date)
2002-12-18 | 16,754 | 44,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.198103 |
https://discuss.leetcode.com/topic/70772/o-1-java-solution-short-and-easy | 1,516,124,777,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886476.31/warc/CC-MAIN-20180116164812-20180116184812-00459.warc.gz | 663,742,483 | 8,538 | # O(1) JAVA Solution Short And Easy
• ``````public class TicTacToe {
int anti_diagonal;
int diagonal;
int[] rows;
int[] cols;
int n;
/** Initialize your data structure here. */
public TicTacToe(int n) {
rows = new int[n];
cols = new int[n];
this.n = n;
}
/** Player {player} makes a move at ({row}, {col}).
@param row The row of the board.
@param col The column of the board.
@param player The player, can be either 1 or 2.
@return The current winning condition, can be either:
0: No one wins.
1: Player 1 wins.
2: Player 2 wins. */
public int move(int row, int col, int player) {
if(row == col)
diagonal = player == 1 ? diagonal + 1 : diagonal - 1;
if(row + col == n - 1)
anti_diagonal = player == 1 ? anti_diagonal + 1 : anti_diagonal - 1;
rows[row] = player == 1 ? rows[row] + 1 : rows[row] - 1;
cols[col] = player == 1 ? cols[col] + 1 : cols[col] - 1;
return player == 1 ? ((rows[row] == n || cols[col] == n || diagonal == n || anti_diagonal == n) ? 1 : 0)
: ((rows[row] == -n || cols[col] == -n || diagonal == -n || anti_diagonal == -n) ? 2 : 0);
}
}
/**
* Your TicTacToe object will be instantiated and called as such:
* TicTacToe obj = new TicTacToe(n);
* int param_1 = obj.move(row,col,player);
*/``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 404 | 1,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.730144 |
http://rainbird.com/landscape/resources/calculators/5000plusPRSrotors_5steps2waterSavingsBAR_Ex.htm | 1,441,247,375,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645298065.68/warc/CC-MAIN-20150827031458-00335-ip-10-171-96-226.ec2.internal.warc.gz | 188,922,379 | 7,003 | You Are Here: Home > Landscape Irrigation > Design & Technical Resources > Calculators > 5 Easy Steps to Calculate Water Savings - 5000/5000 Plus PRS Rotors
# 5 Easy Steps to Calculate Water Savings
## Estimated Water Savings Using Rain Bird 5000/5000 Plus PRS Series Rotors
EXAMPLE:
Calculate your estimated cost savings: Step 1: Enter the pressure (bar) at the nozzle (between 3,1 bar and 5,1 bar): (Either enter the average dynamic pressure for the rotor on one zone or enter the average dynamic pressure for all the rotors on the entire system. If the dynamic pressure is not available, enter the system's static pressure. The static pressure will overstate the water savings, but it will still provide an estimate of the potential water savings.) Step 2: Enter the flow (m3/h) @ 3,1 bar: (If you entered the pressure for one zone above, enter the flow (m3/h) for that zone. If you entered the pressure for the entire system above, enter the flow (m3/h) for the entire system). Step 3: Enter Average Watering Time per Day (min.): Step 4: Enter Average Number of Watering Days per Week: Step 5: Enter Average Watering Weeks per Year:
The initial system waste is: flow (m3/h) The yearly water savings is*: m3
* (Due to varying conditions, your actual cost savings may be different. This formula is designed solely as a basis for estimating potential savings.) | 329 | 1,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2015-35 | longest | en | 0.755241 |
https://groupprops.subwiki.org/w/index.php?title=First-order_subgroup_property&oldid=4885 | 1,591,330,120,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492427.71/warc/CC-MAIN-20200605014501-20200605044501-00169.warc.gz | 368,499,864 | 9,208 | # First-order subgroup property
This article defines a subgroup metaproperty in terms of the model-theoretic approach: viz by looking at the language in which the subgroup property can be defined
## Definition
### Symbol-free definition
A subgroup property is said to be a first-order subgroup property if it can be expressed using a first-order formula, viz a formula that allows:
• Logical operations (conjunction, disjunction, negation)
• Quantification over elements of the group and subgroup
• Group operations (multiplication, inversion and the identity element
Things that are not allowed are quantification over other subgroups, quantification over automorphisms, and quantification over supergroups.
## Importance
First-order language is severely constricted, at least when it comes to subgroup properties. Hence, not only are there very few first-order subgroup properties of interest, also, very few of the subgroup property operators preserve the first-order nature.
## Examples
### Normality
Normality is a first-order subgroup property as can be seen from the following definition: a subgroup $N$ of a group $G$ is termed normal if the following holds:
$\forall g \in G,h \in N, ghg^{-1} \in N$
The formula is universal of quantifier rank 1.
### Centrality
A subgroup is a central subgroup if it lies inside the center, or equivalently, if every element in the subgroup commutes with every element in the group.
Clearly, the property of being a central subgroup is first-order.
The formula is universal of quantifier rank 1.
### Central factor
A subgroup is a central factor if every element in the group can be expressed as a product of an element in the subgroup and an element in the centralizer. This can naturally be expressed as a first-order formula of quantifier rank 3 with the outermost layer being universal.
$\forall g \in G (\exists h \in H, k \in G (\forall m \in H, km = mk))$
## Relation with formalisms
### Function restriction formalism
The general question of interest: given a subgroup property with a function restriction formal expression $a \to b$, can we use the expression to give a first-order definition for the subgroup property? It turns out that the following suffice:
• $a$ should be a first-order enumerable function property (this condition is much stronger than just being a first-order function property because we are not allowed to directly quantify over functions.
• $b$ should be a first-order function property in the sense that given any function, it must be possible to give a first-order formula that outputs whether or not the function satisfies $b$.
The primary example of a first-order enumerable function property is the property of being an inner automorphism. Most function properties that we commonly enoucnter are first-order (that is, they can be tested/verified using first-order formulae). | 600 | 2,882 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-24 | latest | en | 0.854322 |
https://myguidedle.cluster003.ovh.net/pm5wa/6c620a-neet-questions-on-thermodynamics-physics | 1,623,776,406,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621450.29/warc/CC-MAIN-20210615145601-20210615175601-00187.warc.gz | 390,389,397 | 10,649 | An objective or MCQ type book is required which will test our knowledge in the subsequent chapters. Thermodynamics NEET/AIPMT Previous Year Questions It takes 100 calories of heat from source and leaves 80 calories of heat to sink. Here, you will discover the NEET Physics MCQ Questions for all Concepts as per the latest syllabus. Candidates preparing for NEET have always been intimidated by the Physics section since it is the toughest out of all three subjects. Further, these chapters will be divided into various subtopics. This online video lecture covers important questions … 4. The work done by the system in going from A → B → C is 30J and 40J heat is given to the system. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. Free Question Bank for NEET Chemistry Thermodynamics. What is a necessary condition for a reaction to be spontaneous at all temperatures? Multiple Choice Questions. NEET Physics : Thermodynamics. The questions in the Physics … Download the largest collection of free MCQs on Physics for NEET. Free Question Bank for NEET Chemistry Thermodynamics. Physics problems of NEET are comparatively easier. Get all questions and answers of Thermodynamics of NEET Physics on TopperLearning. It covers topics from Physics, Chemistry, Botany and Zoology from class 11 and 12. Self Evaluation Test - Thermodynami.. NEET physics – Thermodynamics – Get video solutions of all the question answers, sample & previous year papers. What is the ratio of $$\frac{C_{p}}{C_{v}}$$ for gas if the pressure of the gas is proportional to the … Important topics in Physics are Thermodynamics, Dimensional Analysis, Current Electricity and Heat Transfer, Kinetic Theory of Gases, Geometrical Optics, … NEET Physics : Thermodynamics. NEET 2020 being just around the corner, aspirants must be busy with their revision plans. Multiple Choice Questions. ∆U = ∆Q − P∆V; Efficiency of a heat engine is the ratio of work done by the engine to input heat. 2 3 Q. The coefficient performance of a refrigerator is 5. This will permanently delete All Practiced Questions. NEET Physics syllabus from class 11 and 12 has 6 prime units – Modern Physics, Mechanics, Thermodynamics, Electrodynamics, SHM-Waves and Optics. Find out the relationship between x, y and z considering below three reactions. The foremost thing that they need to have a clear note of is the updated syllabus for NEET 2020 and chalk out their study calendars accordingly. 12. Bond energy. Important notes of Physics for NEET, JEE for Thermodynamics are useful for all aspirants preparing for entrance exams including JEE, NEET. Thermodynamics is a branch of Physics that helps in understanding heat and temperature and their relation with work, energy, radiation, and other properties of matter. NEET Important Questions Thermodynamics The students who are targeting NEET exam should prepare for all the important chapters of NEET syllabus to secure highest marks in NEET 2020. Practice Now ... First Law of Thermodynamics. Many of these NEET Physics questions were based directly on the concepts included in NCERT. What is the ratio of $$\frac{C_{p}}{C_{v}}$$ for gas if the pressure of the gas is proportional to the cube of its temperature and the process is an adiabatic process. Leave a Comment X. 11. Answer. After you have practiced the NEET Physics questions for chapter, you should ideally give a gap of a week or ten days and then take a … The thickness of a metallic plate is 0.4 cm. 147°C. Required fields are marked *. SIZE – 38.3MB. Comment. Practising papers leverage this. 2 5 Q. Important notes are also helpful for revision when you have less time and have to study many topics. 147°C. Important notes are also helpful for revision when you have less time and have to study many topics. A system is taken from state A to state B along two different paths 1 and 2. View important Questions of Thermodynamics: 2. View Test Papers of Thermodynamics: 4. Typology of questions varies in this section every year. Practice Now. NEET 2019 Question Paper with solutions are available in PDF format to download on Vedantu for practising the upcoming exam. 9. Medical or NEET aspirants need to go beyond the traditional NCERT in order to have a solid grip over it. Self Evaluation Test - Thermodynami.. An engine is working. Questions on Thermodynamics, Physics – MCQs on Thermodynamics for NEET and JEE Main, Class 11 and Class 12 To be able to answer any question, a stronghold on concepts is a must. As a matter of fact, a total of 45 questions will be asked from Physics. Refer the diagram and match the following. NEET 2021 - Achiever Batch - Aryan Raj Singh, Strategies for Enhancement in Food Production, Classification of Elements and Periodicity, Organic Chemistry - Some Basic Principles And Techniques, General Principles and Processes of Isolation of Elements, Differentiations & Conversions - Organic Chemistry, Systems of Particles and Rotational Motion, Crack NEET with Online Course - Free Trial, S-15, 2nd floor Uphar Cinema Market, above Red Chilli Restaurant, Green Park Extension, New Delhi, 110016. NEET Physics Syllabus 2021 will be released by NTA on its official website along with the release of an official information brochure for NEET 2021. Try to finish the NEET Physics Chapter Test well before the allotted time. PAGES – 321. AUTHOR – MTG. It has four fundamental laws. Practice Now. An ideal gas changes from state 'a' to state 'b' as shown in figure. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be -, A refrigerator works between 4°C and 30°C. Answer. Toggle navigation 0 . If the heat absorbed and work done by the system along these two paths are Q1, Q2 and W1, W2 respectively, then, A given mass of gas expands from state A to state B by three paths 1,2 and 3 as shown in the figure. Practice Now. The P-V diagram of a system undergoing thermodynamic transformation is shown in figure. Candidates preparing for NEET have always been intimidated by the Physics section since it is the toughest out of all three subjects. Law is based upon the first law of thermodynamics and states that if a chemical change can be made to take place in two or more ways involving one or more steps, the net amount of heat change in the complete process is the same regardless of the method employed. An objective or MCQ type book is required which will test our knowledge in the subsequent chapters. Undoubtedly, if there is any section that requires to be practised the most, it is the Physics section. 2nd and 3rd law of thermodynamics a.. Thermodynamics is the study of energy transformations. 131. 132. A monoatomic gas supplied the heat Q very slowly keeping the pressure constant. Practice Now. Download NEET UG Physics Thermodynamics MCQs in pdf, Physics chapter wise Multiple Choice Questions free, SUMMARY• Thermal equilibrium, teroth law of thermodynamics, Concept of temperature.• Heat, work and internal energy. A gas at state A changes to state B through path I and II as shown in figure. About the author. PAGES – 321. Thermodynamics is a branch of physics that studies the relationship between heat and other kinds of energy. A small sphere of radius 'r' falls from rest in a viscous liquid. NEET Thermodynamics Practice Question / Answers Question 1. This is a complete set of notes of Thermodynamics which is a part of Physics syllabus for NEET, JEE. Toggle navigation 0 . Your email address will not be published. View Video Lectures of Thermodynamics: 3. To prepare for Physics in NEET, candidates must first get hold of good study material.The important topics with maximum number of questions in the last 10 years are Current Electricity (47 questions) followed by Semiconductors (41 questions). The second law of thermodynamics: The entropy of an isolated system that is not in equilibrium will increase over time and … 3 5 Q. Why solve NEET Physics Question Paper? 2nd and 3rd law of thermodynamics a.. The change in internal energy of the gas during the transition is(a) 20 kJ (b) -20 kJ (c) 20 J (d) -12 kJ, The coefficient of performance of a refrigerator is 5. As per the NEET Chapter Wise Weightage a good number of questions have been asked from the chapter Thermodynamics. Practice Now. 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NEET Physics is the scoring paper in the medical entrance examination. The topic of Thermodynamics carries a weightage of around 8-10% in NEET that means 2-3 questions or around 8- 12 marks can be asked from this section. The second law of thermodynamics: The entropy of an isolated system that is not in equilibrium will increase over time and reaches the maximum value at equilibrium. The first law of thermodynamics is based on the principle of conservation of energy. Then, (a) Work done during isothermal compression is more, (b) Work done is independent of the processes used for compression, (c) Work done is more during the adiabatic process, (d) Work done is dependent on the atomicity of the gas, Answer: (c) Work done is more during the adiabatic process, 10. The temperature between its two surfaces is ... Fungi Questions Multiple Choice NEET Physics Practice Question Paper Kingdom Monera Questions and Answers. (b) x = … If the temperature inside freezer is -20°C, the temperature of the surroundings to which it rejects heat is -(a)31°C (b)41°C (c)11°C (d)21°C. 0 ... Physics; Question Bank for NEET Chemistry Thermodynamics ... Free energy and work function. Bond energy. to do mechanical work as burning of fuel in an engine, provide electrical energy as in dry cell, etc. Vishal Arora. Practice Now. TopperLearning’s Experts and Students has answered all of Thermodynamics of NEET Physics questions in detail. Check NEET Physics Syllabus 8. A gas is compressed to half of its initial volume isothermally. A gas at state A changes to state B through path I and II as shown in figure. After you have practiced the NEET Physics questions for chapter, you should ideally give a gap of a week or ten days and then take a mock test of that chapter. NEET Study Material: The National Eligibility cum Entrance Test, more commonly known as NEET, is the only medical entrance exam in India.It is highly competitive as 14 – 16 lakh students appear for this exam for around 1.63 lakh seats.With this level of competition, clearing the NEET cutoff is not enough. Test Series. 2 3 Q. Your email address will not be published. Let’s go through the important sub-topics of this chapter which will help you crack NEET Physics section. 5. Multiple Choice Questions of NEET Physics are prepared adhering to the latest syllabus guidelines. Answer. NEET 2020 Physics Paper With Solutions September 13 Get access to the complete set of questions from the Physics section of the NEET 2020 Question paper. Detailed study material and exercises for practice. NEET Physics Chapter Tests. NEET 2020 being just around the corner, aspirants must be busy with their revision plans. Some of the important sub-topics of Thermodynamics include- Thermal equilibrium, first law of thermodynamics, second law of thermodynamics, etc. Thermodynamics AIIMS Previous Year Questions: 5. To be able to answer any question, a stronghold on concepts is a must. Home » Alternating Current » Chemistry » Physics » Study Notes » Thermodynamics | Physics Notes for IITJEE/NEET Clear Exam Monday, 20 May 2019 Edit this post Thermodynamics | Physics Notes for IITJEE/NEET Clear Exam 4.9 of 5 Our experts have analysed the NEET Physics questions' level and distribution of marks for the purpose of finding out weightage of marks of respective chapters. Physics NEET syllabus will be based on chapters from Class 11 and 12. Also Check: NEET 2020 Question Paper. This action cannot be undone. The heat absorbed by the system in the process AC will be-. Practice more on a regular basis with these NEET Physics objective questions on air pollution and improve your subject knowledge & problem-solving skills along with time management. Following is the diagram of the thermodynamic process. A carnot engine having an efficiency of 110th of heat engine, is used as a refrigerator. The power required is (Take, 1 cal = 4.2 Joules)(a)23.65W(b)236.5W(c)2365W(d)2.365W, Figure below shows two paths that may be taken by a gas to go from a state A to a state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. Heat of reaction. Practice Now. The average difficulty level for questions in Physics section is 3.5 on a scale of 1 to 10. Law is based upon the first law of thermodynamics and states that if a chemical change can be made to take place in two or more ways involving one or more steps, the net amount of heat change in the complete process is the same regardless of the method employed. Check NEET Physics Syllabus Download Solved Question Papers Free for Offline Practice and view Solutions Online. Answer. 3. What is the work done by the system in the cycle. 12. Heat and thermodynamics notes pdf. The important questions for NEET 2020 designed by Vedantu is one of the most reliable study material as it covers all the important chapters which are in the syllabus. One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure. If W1, W2 and W8 respectively be the work done by the gas along the three paths then, The variation of density (p) of gas with its absolute temperature (T) at constant pressure is best represented by the graph. Super Simple Tricks to Solve Thermodynamics Class 11 NEET Physics Questions | NEET 2020 | Vedantu. Physics NEET syllabus will be based on chapters from Class 11 and 12. To view Explanation, Please buy any of the course from below. (B) The first law of thermodynamics is also known as the law of thermal equilibrium (C) The triple point of water is one of the reference points on the thermodynamic scale of temperature (D) At room temperature, the heat of combustion is not found. NEET UG, JEE Main Physics Question Papers | Topic: Thermodynamics, Question Type: MCQs, Test: 02 neet solved physics mcq, Heat and Thermodynamics ,solved 192 mcq for IIT, AIIMS, NEET, MCAT and State board CET examinations gneet TOPIC : Heat and Thermodynamics number of questions … Biology (Botany + Zoology) – 90 questions: ... (MCI) prescribes syllabus for NEET. Thermodynamics (Syllabus for NEET and JEE ): 1 5 Q. 0 ... Physics; Question Bank for NEET Chemistry Thermodynamics ... Free energy and work function. Name * BOOK NAME – 31 YEARS NEET-AIPMT CHAPTER WISE SOLUTIONS PHYSICS. 12 for NEET, JEE for Thermodynamics are useful for all the Question answers, video,. Solve Thermodynamics Class 11 and 12 be divided into various subtopics for all the Question answers, Sample & year. Physics Chapter Test well before the allotted time explaining the thermodynamic process objective Physics NEET will! Paper with solutions are neet questions on thermodynamics physics in PDF format to download on Vedantu for practising the upcoming.. Temperature, the temperature of sink is solutions with PDF and difficulty level NEET Physics MCQs is a set... Y and z considering below three reactions heat to sink the thermodynamic process Thermodynamics MCQ … NEET Physics in... Physics - 100Q Question Bank for NEET diagram of a heat engine is the Physics.... The figure ratio of work done neet questions on thermodynamics physics the engine to input heat questions in! Format to download on Vedantu for practising the upcoming exam a matter of,... Neet Paper were largely based on chapters from Class 11 and 12 NEET... 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Physics Sample Papers from Chapter Thermodynamics, Physics video lecture covers important questions of Thermodynamics: energy neither. Choice objective Practice questions and answers on Physics for NEET laws of Thermodynamics include- Thermal equilibrium with third! Source and leaves 80 calories of heat to sink with BYJU ’ 31! The help of the following is the scoring Paper in the process half through an adiabatic process →. With a third system are in Thermal equilibrium, first law of Thermodynamics all... The ratio of work done by the system Check all Solved important questions of Thermodynamics and they are 2... The important sub-topics of Thermodynamics include- Thermal equilibrium, first law of Thermodynamics which is on! Can neither be created nor destroyed: two systems in Thermal equilibrium, first law of Thermodynamics which is must! Practising the upcoming exam at state a to state B through path I and II shown... For anyone of the next three below: DC Pandey objective Physics NEET notes PDF as dry. - 100Q Question Bank of mutiple Choice objective Practice questions and answers on Physics NEET. Condition for a reaction to be spontaneous at all neet questions on thermodynamics physics compared to other sections and!, heat is given to the surrounding, aiims or any other exam the subsequent chapters... MCI... Help of the graph ABCD subsequent chapters NEET 2019 Physics Sample Papers from Chapter Thermodynamics, Physics video of.: energy can be transformed from one form to another − P∆V ; of! The relationship between heat and other kinds of energy: ( D at. An engine, provide electrical energy as in dry cell, etc format to download on Vedantu for the! Total 180 questions asked in NEET are from the NEET Physics MCQs is a.... Be able to answer any Question, a total of 45 questions out of the space.... Physics ; Question Bank for NEET Physics are prepared adhering to the system gas at state a to B. Physics as the most challenging compared to other sections to state B along a path AB shown... The concepts included in NCERT two different paths 1 and 2 a small sphere of '... Is 127°C, then temperature of sink is the NEET Physics MCQs is part! And Thermodynamics notes PDF heat engine is the nature of the following is the largest of... And leaves 80 calories of heat from source and leaves 80 calories of heat to sink molecule released... Solve Thermodynamics Class 11 and 12 are also helpful for revision when you have time... Further, these chapters will be based on chapters from Class 11 and 12 going from a state... Aspirants preparing for entrance exams including JEE, NEET its initial volume isothermally Question... To B along two different paths 1 and 2 s Experts and Students has answered all of Thermodynamics is...
Scotts Super Turf Builder With Halts Crabgrass Preventer, Wow Show Telugu Latest Episode, Are Nucleic Acids Hydrophobic Or Hydrophilic, How Mobile Commerce Works, Agnus Dei Mass Xviii, Pumpkin Streusel Muffins, Pyranha Fly Spray Aerosol, | 5,243 | 22,874 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-25 | latest | en | 0.875689 |
http://everything2.com/title/E%2521 | 1,386,455,026,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163055855/warc/CC-MAIN-20131204131735-00071-ip-10-33-133-15.ec2.internal.warc.gz | 64,587,534 | 6,934 | what is E! ?
"What drug are you on?"
"E!"
## E! the network
E! Entertainment Television (http://www.eonline.com) is a cable television channel devoted to entertainment industry news and meta-entertainment (metatainment?). If your cable system gets E!, you can watch the following shows:
E!'s logo consists of a red trapezoid and rectangle arranged to form an exclamation point; contained inside the trapezoid is a white capital E. The following is a rough ASCII art approximation of the E! logo:
```
MMMMMMMMMMMMMMMM
MM) MM
MM) MM
MM) MMMMMMMM
MM) MMMMMMMM
MM) MMMMMMMM
(MM (M)
(MM (M)
(MM MMMMMMM)
(MM MMMMMMM)
(MM MMMMMMM)
(MM MMMMMMM)
(MM) MMMMMMM)
MM) MM
MM) MM
MMMMMMMMMMMMMM
MMMMMMMMMMMMMM
MMMMMMMMMMMMMM
MMMMMMMMMMMMMM
MMMMMMMMMMMMMM
MMMMMMMMMMMMMM
MMMMMMMMMMMMMM
```
E!'s parent company (E! Networks) also owns the style. network (http://www.stylenetwork.com/), which appears to be a cross between HGTV and a fashion show.
## E! the integer
As bis mentioned, in the hexadecimal number system, E equals fourteen, and E!, the factorial of E, that is, E*D*C*B*A*9*8*7*6*5*4*3*2*1, equals 144C3B2800.
## E! the real number
It's possible to take the factorial of almost any real number (except negative integers) using Euler's gamma function, a generalization of the factorial. Thus e!, the factorial of Euler's number, becomes Γ(e + 1) = 4.260820474.
Now go back two letters in the alphabet...
Log in or register to write something here or to contact authors. | 433 | 1,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.647023 |
https://www.hackmath.net/en/math-problem/12751 | 1,708,836,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474581.68/warc/CC-MAIN-20240225035809-20240225065809-00279.warc.gz | 813,961,253 | 9,068 | # Students
If the fifth-grade class left 20% of students and in the seventh grade were added two pupils, in the eighth one pupil, the number has not changed in the ninth grade, but it is now tenth students less than it was in the fifth grade.
How many pupils are in the 9th grade now?
d = 27
### Step-by-step explanation:
b=a - 0.20 a
c = 2+b
d = c+1
d = a - 0.10a
b=a - 0.20·a
c = 2+b
d = c+1
d = a - 0.10·a
0.8a-b = 0
b-c = -2
c-d = -1
0.9a-d = 0
Pivot: Row 1 ↔ Row 4
0.9a-d = 0
b-c = -2
c-d = -1
0.8a-b = 0
Row 4 - 0.8/0.9 · Row 1 → Row 4
0.9a-d = 0
b-c = -2
c-d = -1
-b+0.8889d = 0
Row 4 + Row 2 → Row 4
0.9a-d = 0
b-c = -2
c-d = -1
-c+0.8889d = -2
Row 4 + Row 3 → Row 4
0.9a-d = 0
b-c = -2
c-d = -1
-0.1111d = -3
d = -3/-0.11111111 = 27
c = -1+d = -1+27 = 26
b = -2+c = -2+26 = 24
a = 0+d/0.9 = 0+27/0.9 = 30
a = 30
b = 24
c = 26
d = 27
Our linear equations calculator calculates it.
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Showing 1 comment:
Math student
someone do it!
Tips for related online calculators | 482 | 1,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-10 | latest | en | 0.86609 |
https://www.codingninjas.com/studio/library/breadth-first-search | 1,709,611,422,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00153.warc.gz | 694,194,347 | 56,517 | 'Coding has over 700 languages', '67% of programming jobs aren’t in the technology industry', 'Coding is behind almost everything that is powered by electricity'
Last Updated: May 13, 2022
Raksha jain
1 upvote
Data structures & algorithms (Beginner to Intermediate)
Free guided path
13 chapters
99+ problems
## Introduction-
Today, let's learn about a famous and commonly asked Interview Problem, i.e., Breadth-First Search in a graph. It is a critical concept used to solve various significant problems of graphs.
It is very similar to the Level Order Traversal of a Tree.
In a tree, we cannot revisit the parent node and cannot have a self-loop. But there is no such restriction in the Graph.
Example:
Let's consider this Graph and try to perform a Breadth-First Search(BFS) on the graph.
For calculating the Breadth-First Search, we are always given the source vertex i,e. The start vertex. Here, 2 is our source code.
Like level order traversal, we will discover all the nodes at the same level in Breadth-First Search.
So, first, we visit 2 (our source node) and then all its neighboring nodes, i.e., 0 and 3.
Now, we come to the next level and discover all the neighboring nodes of 0, i.e., 2 and 1.
## Importance of Visited Array
It is vital in graphs as a node can visit any other node. As a result, we would keep visiting the node at the previous level and soon be stuck in an infinite loop.
Like in the above example, if we don't keep a visited array, 2 would add 0 as its neighboring node. 0 would add 2, and again 2 would add 0 and so on.
However, keeping in the visited array helps prevent this. Once a node's neighbors are visited, it is marked as visited and no longer added to our queue.
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## Approach
We would maintain a queue data structure and the visited array to store all the Graph nodes.
Principle of Breadth-First Search: remove - mark visited - add node - mark visited (r m* am*)
So, After adding the source node in the queue, mark it as visited.
Then in the while loop (till the queue is not empty) -
1. Remove the front node from the queue.
2. Check if it is visited - if yes: continue as work for the current node is already done, else mark it visited.
3. Run a loop and add all the neighbors of the node in the queue if that neighbor is not visited.
Continue this same process until the queue is empty.
Let's do a dry run on an example:
Let the start vertex be 0. Initially, the visited array and the queue are empty.
Step1: Check if the start Vertex is null. SInce, start vertex is not null, it is added to the queue. So, the queue [0]
Now, we work while the queue is not empty.
Step2: Front element i.e 0 is removed from the queue.
Step3: Since 0 is not visited, it is marked visited.
Step4: 0’s unvisited adjacent vertex i.e. 1, 2 and 3 are added in the queue.
Step5: Front element i.e 1 is removed from the queue.
Step6: Since 1 is not visited, it is marked visited.
Step7: 1’s unvisited adjacent vertices i.e. 2 is added in the queue.=> Queue [2,3,2]
Step8: Front element i.e 2 is removed from the queue.
Step9: Since 2 is not visited, it is marked visited.
Step 10: 2’s unvisited adjacent vertices i.e. 4 are added in the queue.=> Queue [3,2,4]
Step11: Front element i.e 3 is removed from the queue.
Step12: Since 3 is not visited, it is marked visited.
Step13: 3 has no unvisited adjacent vertex. => Queue [2,4]
Step14: Front element i.e 2 is removed from the queue.
Step15: Since 2 is visited, the loop is continued.
Step16: Front element i.e 4 is removed from the queue.
Step17: Since 4 is not visited, it is marked visited.
Step18: 4 has no unvisited adjacent vertex.=> Queue []
Now, the queue is empty and the loop ends.
Implementation-
Let’s have a look at its implementation in Java -
Output:
## Time and Space Complexity-
Time Complexity: O(E+V) as we are traversing all the nodes of the Graph for finding the Breadth-First Search of a graph once.
Space Complexity: O(V) as extra space for storing nodes in a queue as well as the visited array of O(V) size is being used.
Where V is the number of vertices in the graph.
E is the number of edges in the graph.
1. What is a Graph?
Ans. A Graph is a non-linear data structure consisting of nodes and edges. The nodes are sometimes also referred to as vertices and the edges are lines or arcs that connect any two nodes in the graph.
2. What is the Breadth-First Search?
Ans: Breadth-First Search is an algorithm for searching a tree data structure for a node that satisfies a given property. It starts at the tree root and explores all nodes at the present depth prior to moving on to the nodes at the next depth level.
3. What is the best case time complexity for finding the Breadth-First Search in a Graph?
Ans. The best-case time complexity for finding Breadth-First Search in a graph is O(1), i.e. when only a single or no vertix is present in the Graph.
## Key Takeaways-
In this blog, we learned about the Breadth First Search of a graph.
• Maintain a queue data structure and the visited array to store all the Graph nodes.
• Principle of Breadth-First Search: remove - mark visited - add node - mark visited (r m* am*)
• Maintaining a visited array is important to prevent the possible infinite loop.
• The minimum time complexity required is O(V+E) where v = number of vertices and e = number of edges in a Graph as we need to traverse all the nodes of the Graph once.
Check out more blogs on various different traversals and questions of Graphs like Depth First Traversal, Level Order Traversal to read more about these topics in detail. And if you liked this blog, share it with your friends!
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109+ Problems | 1,516 | 6,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-10 | latest | en | 0.934416 |
https://www.studyadda.com/notes/6th-class/science/distance-time-and-speed/time/6448 | 1,623,782,501,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621519.32/warc/CC-MAIN-20210615180356-20210615210356-00625.warc.gz | 842,513,751 | 22,404 | # 6th Class Science Related to Competitive Exam Time
Time
Category : 6th Class
### Introduction
Time, distance and speed are the characteristics of moving objects. Speed of a moving object is determined by the time taken to travel a distance. Earlier, time of a day was determined by the position of the Sun, seen from the Earth in the sky. Later on, a time measuring device known as Sundial and water clock were used for the measurement of time. The standard unit of time is second and other unit of time is its derived unit. Distance is the shortest length between two points, two places, etc. The standard unit of distance is metre and others are its derived unit. The speed of a running object is determined by the division of distance by time. The unit of speed is meter per second and denoted by m/s. We have studied about the time, distance and speed in previous classes. In this chapter, we will learn detail about the time, distance and speed.
Time
The duration of happening an event is measured in time, and its SI unit is second. Hour, minutes and other units are derived unit of time. Sundial, Sand clock and Water clock are the time measuring devices which were used in ancient days.
Time Measuring Devices of Ancient Time
There were various types of time measuring devices in ancient world. The modern time measuring devices are based on the principle of ancient time measuring devices.
Look at the following pictures of ancient time measuring devices:
.
Sundial was one of the most popular devices, used in ancient period. The name of sundial itself indicates that it is associated with the sun rays. Shadow of vertical blade on the division of the sundial shows the time of a particular period. Shadow of vertical bar is formed according to the position of the Sun in the sky. Sand clock consists of two glass bulbs which is joined by a narrow pipe. Upper bulb contains sand which flows downward till it gets empty.
Another ancient time measuring device was water clock. Function of water clock was same as that of sand clock, only water was used instead of sand. Water in the upper part of the clock flow downward and an empty upper part or a filled lower part of the clock was measured as one hour.
Comparison of Different Units of Time
Time is measured in second but the other unit of time is derived unit as discussed above. There is some relation between the units of time with its standard unit. For example, 1 minute is l/60th of one hour and 1 second is 1/ 60th of 1 minute.
Look at the following table of units and comparison of time:
1000 years 100 years 10 years 12 months 30 days 24 hours 1 hour 1 minute 1 microsecond 1 nanosecond 1 millennium 1 century 1 decade 1 year 1 month 1 day 60 minutes 60 seconds 1/1000000 seconds 1/000000000000 seconds
Give the name of the measuring instrument which was used for measuring the time in ancient time.
(a) Odometre
(b) Crystal watch
(c) Sundial
(d) All of these
(e) None of these
Explanation
Sundial was used for the measurement of time in ancient time. Therefore, option (C) is correct and rest of the options is incorrect.
Which one of the following is the standard unit for the measurement of time?
(a) Second
(b) Minute
(c) Hour
(d) All of these
(e) None of these | 728 | 3,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.976625 |
https://stats.stackexchange.com/questions/319124/how-to-find-multi-layer-perceptron-weights | 1,716,325,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00684.warc.gz | 484,469,220 | 39,285 | # How to find multi-layer perceptron weights?
I want to use a multi-layer perceptron to design the following function :
The architecture I want to use is the following one :
What would be $w_i$ weights ? Is there any guide to find them ?
I tried the following, guessing with $\forall i,w_i=1$.
\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline &w_1&w_2&w_3&w_4&w_5&w_6&w_7&y\\ &1&1&1&1&1&1&1&=\\ \hline -1&-1&-1&1&1&0&0&1&1\\ -2&-2&-2&1&1&1&1&1&0\\ -3&-3&-3&1&1&-2&-2&1&-1?\\ \hline \end{array}
As it is it seems everything goes well from there but that was only a guess ...
Why in the plotted function do I have $\sum_iw_ix_i$ ? I don't understant the $x_i$. Don't I have only one $x$ as input ?
The whole point of the Perceptron model is to find the optimal set of weights with respect to your data. You initialize them at some small random number, then with each iteration the Perceptron adjusts the weights in search of a better solution.
How to actually get the weights out of the model depends on your implementation. If you are running this model in Python / R, they should be saved in a matrix; have a look at the code.
• :/ The whole pedagogical point is to find them by hand Dec 20, 2017 at 20:49
What you are missing is the data with ground truth and loss function, i.e., when we change the parameters, how can we measure goodness of the model.
For example, you may define 0-1 loss as the classification accuracy, suppose you have 100 data points, you can try to change the parameters and see in what setting the performance is the best.
So you are monitoring $$\sum_i L(y_i- f(x_i))$$, where $$x_i$$ is the input, and $$y_i$$ is the prediction target. $$w_i$$ will be used when calculate $$f(x_i)$$. | 517 | 1,716 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | latest | en | 0.888101 |
https://www.expertsmind.com/library/what-is-the-amount-of-gain-or-loss-on-disposal-52609424.aspx | 1,701,639,879,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00063.warc.gz | 847,237,534 | 10,861 | ### What is the amount of gain or loss on disposal
Assignment Help Accounting Basics
##### Reference no: EM132609424
Question - A fixed asset with a cost of \$15,000 and accumulated depreciation of \$12,500 is sold for \$1,750. What is the amount of gain or loss on disposal of the fixed asset?
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Dividends from mutual life insurance company on life insurance policy: \$1,000, How much should Betty report as dividend income for 2011 | 547 | 2,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.919449 |
http://www.eagleee.com/detail?newsId=460252 | 1,548,188,887,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583874494.65/warc/CC-MAIN-20190122202547-20190122224547-00514.warc.gz | 289,092,725 | 5,897 | # Can You Solve The Day Of The Week Riddle?
Brilliance | Dec. 15, 2017
Okay, pay attention. This logic puzzle might rattle your brain. Here is the day of the week riddle, as presented by Presh Talwalkar of the MindYourDecisions YouTube channel: When the day after tomorrow is yesterday, then today will be as far from Wednesday as today was from Wednesday when the day before yesterday was tomorrow. What day is today? Check out the video below to learn how to solve this brain teaser, or scroll down to read the answer. Good luck!
In this logic puzzle, today is Wednesday. To see how this problem was solved, watch the video above.
You're the first to comment
Say something. | 158 | 679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-04 | latest | en | 0.96767 |
https://www.humanprogress.org/thanksgiving-dinner-has-become-cheaper-over-time/ | 1,656,618,037,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00578.warc.gz | 887,031,364 | 13,275 | Thanksgiving is almost upon us and time has come for that most sacred of American traditions: bemoaning the rising cost of living. Per this Bloomberg headline on Thursday, “Thanksgiving Meal Costs Most Ever as Bird Flu Hits Turkeys.”
Well, that’s complete and utter nonsense.
The headline grabbing data comes from the American Farm Bureau Federation, which faithfully records the cost of 12 items (e.g., turkey, pumpkin pie mix, sweet potatoes, etc.) that go into a preparation of a Thanksgiving meal for 10 people.
On the face of it, the nominal cost has risen by \$0.70 from \$49.41 in 2014 to \$50.11 in 2015. Using a BLS calculator, I have inflated \$49.41 in 2014 dollars to \$49.64 in 2015 dollars. So, the real increase amounts to mere \$0.47.
Now let us see what happens when we adjust the nominal cost of Thanksgiving dinners by the rise in nominal wages.
In October 2014, FRED tells us, the average hourly wage of production and nonsupervisory employees in the private sector (i.e., blue collar workers) was \$20.72. In October 2015, it was \$21.18.
That means that in 2014, an average worker had to work 2 hours 23 minutes and 5 seconds to procure all the items needed to buy a Thanksgiving dinner for 10 people. In 2015, s/he had to work 2 hours 21 minutes and 57 seconds to do the same. So, in terms of actual work, the price of a Thanksgiving dinner has decreased by 1 minute and 8 seconds between 2014 and 2015.
That may seem like small beans, but consider what happened to the cost of a Thanksgiving dinner since 1986, which was the first year in which the AFBF collected the pertinent data. In 1986, Thanksgiving dinner cost \$28.74. In October 1986, an average worker made \$8.96 an hour. That means that s/he had to work 3 hours 12 minutes and 27 seconds, or 50 minutes and 30 seconds longer than worker today.
So, enjoy your Thanksgiving dinner and rejoice in knowing that you have worked almost an hour less to earn it than would have been the case in 1986.
This first appeared in Cato at Liberty.
Marian L. Tupy is a senior fellow in the Cato Institute’s Center for Global Liberty and Prosperity and editor of HumanProgress.org. | 534 | 2,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.962462 |
https://www.coursehero.com/file/198018/Tut2-Bearing-1/ | 1,529,897,609,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00374.warc.gz | 789,628,102 | 90,410 | Tut2_Bearing_1
# Tut2_Bearing_1 - CIVL3210 Geotechnical Engineering Tutorial...
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CIVL3210 Geotechnical Engineering Tutorial 2 – Bearing capacity of shallow footings CIVL3210 Geotechnical Engineering Tutorial 2 - Bearing capacity of shallow footings 1. Determine a suitable size for a square footing which is to carry a working load of 333 kN and is to be placed a depth of 1 m below the surface in a clay which has undrained shear strength of 120 kN/m 2 and a unit weight of 18 kN/m 3 . ANS: 1.0 m x 1.0 m 2 . A mass concrete pier of circular section is to carry a load, including its own weight, of 2500 kN. The base of the pier is to be located 4.5m below the surface of a deep layer of saturated clay. The average bulk unit weight of the clay is 18 kN/m 3 and it’s average undrained shear strength c u = 50 kN/m 2 . Choosing a suitable value for the bearing capacity factor N c , determine the diameter of the pier to give a factor of safety of three. Assume that a strength of 0.3c u can be mobilised on the cylindrical surface of the pier below ground level.
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http://www.puzzlevilla.com/puzzles/puzzle/112 | 1,579,846,282,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250615407.46/warc/CC-MAIN-20200124040939-20200124065939-00159.warc.gz | 264,128,944 | 10,921 | # PUZZLEVILLA
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Red and Blue Marbles Puzzle
Difficulty Level
You have two jars, 50 red marbles and 50 blue marbles. You need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. When picking, you will first randomly pick a jar, and then randomly pick a marble out of that jar. You can arrange the marbles however you like, but place all 100 marbles in the jars.
Probability of selecting any one jar is 1/2. Now we want to place these 100 marbles in 2 jars to maximize the probability of red marble pick. We can put 1 red marble in one jar and rest all the 99 marbles (50 blue coloured and 49 red coloured) in other jar.
Now, calculating the probability for red marble pick in the first chance goes this way:
In the first jar in which there is only 1 red marble and the probability of picking red is:
(1/2) * (1/1) = 1/2
If the second jar is considered then it's: (1/2) * (49/99) = 49/198.
So, for calculating the total probability considering both the jars can be done in following way:
((1/2) * (1/1)) + ((1/2) * (49/99)) = (1/2) + (49/198) = 148/198 = 74/99 = 0.7474
That is 74.74 %.
Guest Said:Posted On 2017-05-29
nice - what if you want to maximize drawing of n consecutive reds ? --> n reds in first jar, rest in other ??
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# [Noob] Making a plain and simple cloud
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7 replies to this topic
### #1arnoldlayne42 Members - Reputation: 108
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Posted 03 October 2013 - 02:45 AM
I hope such beginner questions aren't looked down upon here.
Anyway, I've *just* begun with OpenGL and I'm trying to make a simple cloud. Something like these:
So my first instinct was to make separate bezier curves for each of the curves in the cloud. I did it, but it's turning out to be too tedious, with recalculating all the points and controls for each curved segment.
What would be a better way to do this? Also, I want to fill it with a color.
Thanks.
### #2tonemgub Members - Reputation: 454
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Posted 03 October 2013 - 04:37 AM
The easiest way to do it is to use textures, and paint them with whatever image editing software you like.
If you still need other ideas:
The cloud from the first image looks like it can be easily accomplished in a pixel shader, with circles/ellipses. Filling is accomplished by using the filled circle/ellipse equation:
r^2 <= (a * x)^2 + (b * y)^2.
In a pixel shader, you test if your pixel coordinates obey this equation for any of the circles/ellipses, and if so, you fill that pixel with the color of the cloud.
For the edges, you do something similar, but in addition to doing the reverse of the test above (to exclude parts of circle edges that lie inside the cloud), you also check the equation a second time for each circle/ellipse, this time, with a radius that is a bit larger. If this test passes, you paint that pixel with the edge color. Ex.: ((r^2 >= (a * x)^2 + (b * y)^2) && ((r + 10)^2 <= (a * x)^2 + (b * y)^2)) would give you a 10-pixel-wide edge.
You can probably simplify these equations to remove a or b. You have to make sure there aren't any spaces between your circles, or your clouds will have holes in them.
Another way to accomplish the above is to use point sprites with some kind of XOR blending mode, but then you'd have to use textures anyway, and figuring out the exact XOR blending equation would be more tedious than pre-rendering the final clouds into textures.
Edited by tonemgub, 03 October 2013 - 05:49 AM.
### #3arnoldlayne42 Members - Reputation: 108
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Posted 03 October 2013 - 01:58 PM
I tried to generate it using multiple beziers. But I'm having a strange problem while trying to fill it.
On the right is the shape generated by GL_LINE_STRIP and on the left is it generated by GL_POLYGON.
There's a function which generates each point 1 by 1 in order.
Why is there such a big difference in the basic shape? Notice how less 'curved' each portion of the cloud is in the GL_POLYGON version.
Any clue how to fix this? Or maybe use something else to fill this polygon I have?
[1]:
### #4tonemgub Members - Reputation: 454
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Posted 04 October 2013 - 12:37 AM
That looks like OpenGL's internal tesselation at work. Do all of your Bezier curves start or end at the left-most point? They seem to be converging towards it, or like they're converging towards the start point of the previous bezier curve? I don't know how you're building the polygons, but it could also be that you're adding that point to each of them somewhere. Myabe you meant to use the last point of the last curve but instead used the first one?
### #5wintertime Members - Reputation: 1371
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Posted 04 October 2013 - 04:37 AM
I think its wrong because you were drawing a single or very few polygon thats concave. You need to cut it up into convex polygons, preferably triangles to avoid having it draw outside the shape.
Though you better just draw it in a program like Inkscape, convert it into a texture and put the texture on a quad.
### #6arnoldlayne42 Members - Reputation: 108
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Posted 04 October 2013 - 05:03 PM
I think its wrong because you were drawing a single or very few polygon thats concave. You need to cut it up into convex polygons, preferably triangles to avoid having it draw outside the shape.
Though you better just draw it in a program like Inkscape, convert it into a texture and put the texture on a quad.
I had to use GL_TRIANGLES after triangulating the points about a center point inside the cloud to get it right.
Although this method is causing some other problems in another shape:
I have a shape which comprises of a 2 joined bezier curves. I'm trying to fill it using GL_TRIANGLES about an internal point. Here are the results:
On the right is the actual shape displayed via GL_LINE_STRIP.
On the left is the result after GL_TRIANGLES after giving the points in a triangular way about a center point..
As you can see, the curvy shape has disappeared and instead it's (almost) a straight edge.
Any idea how can I fix this?
Also, can you elaborate/link me to some place which explains the Inkscape point? How does that work exactly? I didn't quite understand 'putting the texture on a quad'.
Edited by arnoldlayne42, 04 October 2013 - 05:12 PM.
### #7wintertime Members - Reputation: 1371
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Posted 05 October 2013 - 07:37 AM
You can not cut up a concave polygon in any way you like, you have to make sure that you dont have any triangle that includes an area outside the shape. Your picture looks like you just made 4 triangles from that shape where 2 of the triangles include an edge thats outside the shape.
Because thats complicated people often make their life more easy by just using a simple shape like a quadrilateral(or two triangles near each other), where a texture(an image they made outside of their actual program) is applied onto it using some OpenGL function after loading it from your program. The image can contain invisible areas where the alpha value(the opacity) is zero and all kinds of rgb colors they want and its premade inside a paint program usually.
Inkscape is just a free program for making art using vector-graphics ( http://inkscape.org/ , http://en.wikipedia.org/wiki/Inkscape ), where I thought it could be useful for you to know about. These could then be transformed into a raster image, for example using gimp, another free paint program ( http://www.gimp.org/http://en.wikipedia.org/wiki/GIMP ). Such an image could then be used as a texture.
Edited by wintertime, 05 October 2013 - 07:47 AM.
### #8Erik Rufelt Crossbones+ - Reputation: 2449
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Posted 05 October 2013 - 08:16 AM
An idea that might be easier is than bezier curves if you want to do it in code is to just draw many circles.
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
PARTNERS | 1,744 | 6,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.920453 |
https://www.greenlighttestprep.com/comment/3339 | 1,590,515,247,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00273.warc.gz | 747,670,918 | 15,783 | # Lesson: Mean, Median and Mode
## Comment on Mean, Median and Mode
### I'm at cloud nine now! I'm a
I'm at cloud nine now! I'm a potential GRE aspirant. I find this video very useful for dummies like me to learn Maths. In a funny note, I'm also finding all GRE vocabs words to praise the tutorial. It is so useful that I was able to crack many questions easily.
### Thanks! Glad you like the
Thanks! Glad you like the videos,
### Hi, I just want to make sure
Hi, I just want to make sure about one information about the mode, I know that if all numbers in 1 set are different from each others there will be no mode, but you have said in the e.g at 2:20 mins all of the four numbesr we will consider them as mode so we have 4 modes.
### In the case that all of the
In the case that all of the values are different, then we can say that there is a multi-way tie for the most common value.
So, for example, the set {2, 3, 6, 9} has four modes: 2, 3, 6 and 9
### http://greprepclub.com/forum
http://greprepclub.com/forum/course-of-an-experiment-95-measurements-were-recorded-1723.html
plz explain this sum
### You bet.
You bet.
I've posted my solution here: http://greprepclub.com/forum/course-of-an-experiment-95-measurements-wer...
Cheers,
Brent
Thankyou.
### http://greprepclub.com/forum
http://greprepclub.com/forum/frequency-distribution-of-the-values-of-a-variable-y-1757.html
Sometimes i get confused to put answers in 0.1, 0.01 similar like accuracies. Can you please help me in understanding those concepts well.
### Hi
Hi
Can you pl provide the soln to this - A set of 7 integers has a range of 2, an average of 3, and a mode of 3.
### Set A can be EITHER {2, 2, 3,
Set A can be EITHER {2, 2, 3, 3, 3, 4, 4} OR {2, 3, 3, 3, 3, 3, 4}
Both of these sets meet the given conditions.
Can you tell me how the question is worded?
### This is the link to the exact
This is the link to the exact question: http://greprepclub.com/forum/a-set-of-7-integers-has-a-range-of-2-an-average-of-3-and-a-mode-of-2675.html
I was able to come to the answer. But I wasn't sure till the end. Would be nice if you could help me think through.
### You bet.
You bet.
Here's my step-by-step solution: http://greprepclub.com/forum/a-set-of-7-integers-has-a-range-of-2-an-ave...
Cheers,
Brent
### 2 3 3 3 3 3 4 is one possible
2 3 3 3 3 3 4 is one possible set.
Since they told mode 3 expect 1st and 7th, all member can be 3, we met mode restriction. Range has to be 2 so let 1st member be 2 and certainly 7th member will be 4.
### Hi Brent,
Hi Brent,
For the following question, can we say using the logic of weighted average that the qty B is greater ? Qty A has equal weight for x and y. So average has to be in the middle of x and y. However, Qty B has 2 weights of y and 1 weight of x. So average has to be closer to y than x. Since y > x, the average from qty B has to be greater.
http://greprepclub.com/forum/x-y-3608.html
---------------------------------------------------
Also for this question, http://greprepclub.com/forum/graph-above-shows-the-frequency-distribution-of-50-integer-1805.html, 23 values <5 (median), 16 values =5, 11 values >5. More weightage in < median. So average has to be < 5.
Thanks :)
### Hi Vinodhini,
Hi Vinodhini,
Yes, the weighted approach you describe is perfect. GREAT WORK!!
------------------------------------------
Your approach says that, since there are more values BELOW the median than there are values ABOVE the median, the average must be LESS THAN the median.
This approach can get you in trouble, because the mean (average) depends on the MAGNITUDE of the values above and below the median.
For example, if we replace one of the eleven 6's with 1,000,000 then the average of the 50 values will be greater than the median.
Cheers,
Brent
### Hi Brent,
Hi Brent,
Thank you. Got my mistake in 2nd approach..
### Hi Brent ! I wanted to ask if
Hi Brent ! I wanted to ask if there is any way that i can find all the reinforcement problems in one place .Like there are very good problems that i have encountered but i tend to forget the tricks and complexity some problems involve as i have seen them only once .. but i wish to revise and go through all the problems once. Is there any such link available.
### Sorry, but I don't have a
Sorry, but I don't have a list of all reinforcement problems in the course. Perhaps this is something I will add in the future.
### Ok np sir! Thanks for all the
Ok np sir! Thanks for all the help !
### Thank you very much!
Thank you very much!
### https://greprepclub.com/forum
https://greprepclub.com/forum/if-x-y-3281.html
If x=0, and y= -2, then the A seems to be greater than the answer would be D, wouldn't it?
If x = 0, and y = -2, then we get:
QUANTITY A: Average = [0 + 0 + 0 + (-2) + (-2)]/5 = -0.8
QUANTITY B: Average = [0 + 0 + (-2)]/3 = -0.66666...
So, -0.8 < -0.66666..., so Quantity B is greater.
### https://greprepclub.com/forum
https://greprepclub.com/forum/course-of-an-experiment-95-measurements-were-recorded-1723.html
I have a question here with the calculation of mean and determination of median. I calculated the mean by taking the mid-point of the bar chart i.e.: 15(3) + 35(8) + 15(13) + 13(18) + 10(23) + 5(28) + 3(33) = 1223 . To get mean 1223/ 95 = 12.87.
To calculate the median of 95. I divided it by 2 and got 47.5 and then went for cumulative frequencies to determine the range 6-10. As it was lower than the mean I went for A.
Is my calculation correct or I got lucky?
Unfortunately, that strategy won't always work, since we need to consider possible ranges of each value (median and mean)
Cheers,
Brent
When calculating the answer I got c+d = 100. Choice A states the greatest possible NUMBER for d. I did not mention the word integer. On the gre when the question mention number should I not think of it as integer?
I assumed the value for c = 1 when got me 99 for d, therefore C answer which is not correct.
It's a common mistake to assume that numerical values are integers, but we should never make this assumption unless the question specifically states that a number is an INTEGER.
Cheers,
Brent
I don't understand this
### Can you give me an idea of
Can you give me an idea of which part you don't understand?
### This question is not from the
This question is not from the Greenlight bank of questions, it's one that I found online, but am still having difficulty answering it correctly. Can you help me?
"Of 30 theater tickets sold, 20 tickets were sold at prices between \$10 and \$30 each and 10 tickets were sold at prices between \$40 and \$60 each."
Quantity A: the average (arithmetic mean) of the prices of the 30 tickets
Quantity B: \$50
The explanation says the answer is B, however wouldn't the answer be D due to the varying price ranges of the 30 tickets?
### Yes, the average price does
Yes, the average price does vary.
In fact, it varies from \$20 to \$40
However, since Quantity B = \$50, we can see that Quantity B is greater in all possible cases.
Here's my full solution: https://greprepclub.com/forum/of-30-theater-tickets-sold-20-tickets-were...
Cheers,
Brent
### https://greprepclub.com/forum
https://greprepclub.com/forum/qotd-5-the-sum-of-the-annual-salaries-of-the-21-teachers-a-2654.html
I was the reading the explanation of this question answer and it stated that "when the 21 salaries are listed in increasing order, the first 12 salaries in the list are less than \$37,000"
How can we assume that the salaries as listed in increasing order?
In order to find the median salary (Quantity B), we need to arrange the 21 salaries in ASCENDING order.
Let's say the salaries, when listed in ASCENDING order, are:
a, b, c, d, e, f, g, h, i, j, K, l, m, n, o, p, q, r, s, t, u
We can see that the median salary = K
Since we're told that "Twelve of the 21 teachers have an annual salary that is less than \$37,000.", we know that salaries a to l are each less than \$37,000
This tells us that K is less than \$37,000
So, Quantity B is less than \$37,000
Meanwhile, Quantity A = \$781,200/21 = \$37,200
So, we have:
QUANTITY A: Some number less than \$37,000
QUANTITY B: \$37,200
Does that help?
Cheers,
Brent
### https://greprepclub.com/forum
https://greprepclub.com/forum/50-boxes-each-containing-30-machine-parts-were-examined-1869.html
Thanks!
### Here's my full solution:
Hi Brent! In the above question, can't the value of C be 0? That means 'd' could be 100.
Could be please clarify this point.
Thanks for the help!
The question tells us that a, b, c, and d are positive, and c is not positive.
Cheers,
Brent
### https://greprepclub.com/forum
https://greprepclub.com/forum/graph-above-shows-the-frequency-distribution-of-50-integer-1805.html | 2,391 | 8,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-24 | latest | en | 0.905918 |
https://www.physicsforums.com/threads/linear-equation-ax-b.192520/ | 1,481,115,764,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542112.77/warc/CC-MAIN-20161202170902-00077-ip-10-31-129-80.ec2.internal.warc.gz | 996,772,221 | 15,189 | # Linear equation Ax=b
1. Oct 19, 2007
### eyehategod
let A be a mxn matrix.
prove that the system of linear equations Ax=b is consistnet for all column vectors b if and only if the rank of A is m.
I have no idea how to start, can anyone helo me out?
2. Oct 19, 2007
### ice109
what does it mean if the matrix equation is consistent for all vectors b?
3. Oct 19, 2007
### eyehategod
i guess my problem is that i dont quite understand when it says "consistent for all column vectors b."
4. Oct 19, 2007
### eyehategod
also it would mean that b is in the column space of A.
5. Oct 19, 2007
### ice109
yes but any b?
Last edited: Oct 19, 2007
6. Oct 20, 2007
A system of linear equations is consistent if it has a solution. Of course, this solution need not be unique.
7. Oct 25, 2007
### HallsofIvy
Staff Emeritus
A matrix equation, Ax= b, is "consistent" if it has at least one solution. "Ax= b is consistent for all b" means the equation Ax= b is consistent no matter what vector b is.
The OP said earlier, "also it would mean that b is in the column space of A." Okay. And if b is to be any member of A, what must the column space be? | 339 | 1,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2016-50 | longest | en | 0.92091 |
https://calculat.io/en/number/hex-to-dec/1e46 | 1,723,468,300,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00349.warc.gz | 120,254,156 | 24,254 | # 1e46 in decimal
## What is 1e46 in decimal?
Answer: Hexadecimal Number 1E46 (0x1E46) It Is Decimal: 7750
(Seven thousand, seven hundred fifty)
## Hex to Dec Conversion Table
Hexadecimal NumberNumber
1e377735
1e387736
1e397737
1e3a7738
1e3b7739
1e3c7740
1e3d7741
1e3e7742
1e3f7743
1e407744
1e417745
1e427746
1e437747
1e447748
1e457749
7750
1e477751
1e487752
1e497753
1e4a7754
1e4b7755
1e4c7756
1e4d7757
1e4e7758
1e4f7759
7760
1e517761
1e527762
1e537763
1e547764
## About "Hex to Decimal" Calculator
This calculator will help you to convert hexadecimal numbers to decimal. For example, it can help you find out what is 1e46 in decimal? (The answer is: 7750). Enter hexadecimal number (e.g. '1e46') and hit the 'Convert' button.
## FAQ
### What is 1e46 in decimal?
Hexadecimal Number 1E46 (0x1E46) It Is Decimal: 7750 | 335 | 825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.294171 |
https://www.numbersaplenty.com/123201 | 1,674,936,197,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00565.warc.gz | 954,095,593 | 3,697 | Search a number
123201 = 36132
BaseRepresentation
bin11110000101000001
320021000000
4132011001
512420301
62350213
71022121
oct360501
9207000
10123201
1184621
125b369
1344100
1432c81
1526786
hex1e141
123201 has 21 divisors (see below), whose sum is σ = 200019. Its totient is φ = 75816.
The previous prime is 123191. The next prime is 123203. The reversal of 123201 is 102321.
123201 = T350 + T351.
123201 = 13 + 23 + ... + 263.
It is a happy number.
The square root of 123201 is 351.
It is a perfect power (a square), and thus also a powerful number.
It can be written as a sum of positive squares in only one way, i.e., 18225 + 104976 = 135^2 + 324^2 .
It is not a de Polignac number, because 123201 - 25 = 123169 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (9).
It is a Curzon number.
It is a nialpdrome in base 13.
It is not an unprimeable number, because it can be changed into a prime (123203) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 20 ways as a sum of consecutive naturals, for example, 9471 + ... + 9483.
2123201 is an apocalyptic number.
123201 is the 351-st square number.
123201 is the 176-th centered octagonal number.
It is an amenable number.
123201 is a deficient number, since it is larger than the sum of its proper divisors (76818).
123201 is an frugal number, since it uses more digits than its factorization.
123201 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 44 (or 16 counting only the distinct ones).
The product of its (nonzero) digits is 12, while the sum is 9.
The cubic root of 123201 is about 49.7589733231.
Adding to 123201 its reverse (102321), we get a palindrome (225522).
It can be divided in two parts, 12320 and 1, that added together give a palindrome (12321).
The spelling of 123201 in words is "one hundred twenty-three thousand, two hundred one", and thus it is an iban number. | 606 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.878681 |
https://www.convertunits.com/from/moles+(NH4)HCo3/to/grams | 1,566,434,156,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316555.4/warc/CC-MAIN-20190822000659-20190822022659-00416.warc.gz | 761,883,450 | 10,695 | ## ››Convert moles (NH4)HCo3 to gram
moles (NH4)HCo3 grams
How many moles (NH4)HCo3 in 1 grams? The answer is 0.0051060527148882.
We assume you are converting between moles (NH4)HCo3 and gram.
You can view more details on each measurement unit:
molecular weight of (NH4)HCo3 or grams
The SI base unit for amount of substance is the mole.
1 mole is equal to 1 moles (NH4)HCo3, or 195.846 grams.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between moles (NH4)HCo3 and gram.
Type in your own numbers in the form to convert the units!
## ››Similar chemical formulas
Note that all formulas are case-sensitive. Did you mean to convert one of these similar formulas?
moles (NH4)HCO3 to grams
moles (NH4)HCo3 to grams
## ››Convert another chemical substance
Convert moles to grams
## ››Quick conversion chart of moles (NH4)HCo3 to grams
1 moles (NH4)HCo3 to grams = 195.846 grams
2 moles (NH4)HCo3 to grams = 391.692 grams
3 moles (NH4)HCo3 to grams = 587.538 grams
4 moles (NH4)HCo3 to grams = 783.384 grams
5 moles (NH4)HCo3 to grams = 979.23 grams
6 moles (NH4)HCo3 to grams = 1175.076 grams
7 moles (NH4)HCo3 to grams = 1370.922 grams
8 moles (NH4)HCo3 to grams = 1566.768 grams
9 moles (NH4)HCo3 to grams = 1762.614 grams
10 moles (NH4)HCo3 to grams = 1958.46 grams
## ››Want other units?
You can do the reverse unit conversion from grams (NH4)HCo3 to moles, or enter other units to convert below:
## Enter two units to convert
From: To:
## ››Details on molecular weight calculations
In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.
Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.
A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass.
Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance.
The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. We use the most common isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass.
Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.
If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 1,025 | 4,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-35 | longest | en | 0.778792 |
https://www.manualslib.com/manual/908138/Sony-Dcr-Vx2000.html?page=163 | 1,582,199,503,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00216.warc.gz | 811,272,982 | 37,895 | # Lcd System Adjustment; Vco Adjustment (pd-126 Board) - Sony DCR-VX2000 Service Manual
Digital video camera recorder.
1-5.
Note 1: The back light (fluorescent tube) is driven by a high voltage AC
power supply. Therefore, do not touch the back light holder to
avoid electrical shock.
Note 2: When replacing the LCD unit, be careful to prevent damages
caused by static electricity.
Most of the measuring points for adjusting the LCD system are
concentrated in CN2105 of the PD-126 board. The following table
shows the Pin No. and signal name of CN2105.
Pin No.
Signal Name
1
VG
2
COM
3
GND
4
PSIG
5
HSY
Table 5-1-8.
PD-126 board
Fig. 5-1-20.
CN2105
5
1
### 1. VCO Adjustment (PD-126 board)
Set the VCO free-run frequency. If deviated, the LCD screen will
be blurred.
Mode
Signal
Measurement Point
Measuring Instrument
Specified Value
Note1: NTSC: DCR-VX2000
PAL: DCR-VX2000E
1
0
01
2
D
A2
3
D
A2
4
D
A2
5
6
7
8
D
A3
9
0
01
Note2: Refer to "Table 5-4-1. Hexadecimal-decimal Conversion Table".
5-33
VTR stop
No signal
Pin 5 of CN2105 (HSY)
Frequency counter
D
A2, A3
f = 15734 ± 30Hz (NTSC)
f = 15625 ± 30Hz (PAL)
Procedure
01
Set the data.
Change the data and set the
VCO frequency (f) to the
specified value.
Press PAUSE button.
Read the data, and this data is
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. (Note2) | 624 | 1,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.686085 |
https://www.nagwa.com/en/videos/486195810784/ | 1,679,504,479,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943845.78/warc/CC-MAIN-20230322145537-20230322175537-00770.warc.gz | 1,001,642,643 | 23,696 | # Question Video: Determining If a Graphed Function Is Even, Odd, or Neither Mathematics • 12th Grade
Determine whether the function represented by the figure is even, odd, or neither even nor odd.
02:22
### Video Transcript
Determine whether the function represented by the following figure is even, odd, or neither even nor odd.
And then we have a graph of the function shown. So letβs recall how to check the parity of a function, how to check whether itβs even or odd. Well, the first thing we do is ask ourselves, is the domain of this function centered at π₯ equals zero? We might recall that the domain of a function is the set of possible inputs, the set of values of π₯, that we can substitute into the function. And we can read that domain from the graph.
Now we do need to be a little bit careful because the graph doesnβt actually appear to be defined at π₯ equals zero. In fact, the domain is the union of the left-closed right-open interval from negative eight to zero and the left-open right-closed interval from zero to eight. This is centered at π₯ equals zero. Zero is exactly halfway through this domain. And so we can now say yes to this question. And we are able to move on to the next part.
We can say that if π of negative π₯ is equal to π of π₯, the function is even, and itβs odd if π of negative π₯ is equal to negative π of π₯. Well, one way we can establish whether either of these is true is to choose a value of π₯. For instance, letβs use the point π₯ equals five. When π₯ is equal to five, the value of our function, the π¦-value, is negative one. So π of five is negative one. And then this means that negative π₯ must be negative five. And so we need to read the π¦-value when π₯ is negative five. π of negative five is also negative one. So it does look like this might be an even function. But letβs check with another value.
Letβs choose π₯ equals one. π of one is roughly equal to negative 4.1. Then negative π₯ will be equal to negative one. And once again, π of negative one is roughly equal to negative 4.1. And so for the two values weβve tried, π of negative π₯ is equal to π of π₯. But actually, if we look carefully, we see that that function itself has reflectional symmetry about the π¦-axis. And so indeed, every value of π of π₯ must be equal to every value of π of negative π₯. And so we can say that the function itself must be even. | 694 | 2,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-14 | latest | en | 0.88672 |
https://docs.mfem.org/4.3/navier__solver_8hpp_source.html | 1,726,645,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00433.warc.gz | 186,437,459 | 28,188 | MFEM v4.3.0 Finite element discretization library
navier_solver.hpp
Go to the documentation of this file.
1 // Copyright (c) 2010-2021, Lawrence Livermore National Security, LLC. Produced
3 // LICENSE and NOTICE for details. LLNL-CODE-806117.
4 //
5 // This file is part of the MFEM library. For more information and source code
6 // availability visit https://mfem.org.
7 //
8 // MFEM is free software; you can redistribute it and/or modify it under the
9 // terms of the BSD-3 license. We welcome feedback and contributions, see file
10 // CONTRIBUTING.md for details.
11
12 #ifndef MFEM_NAVIER_SOLVER_HPP
13 #define MFEM_NAVIER_SOLVER_HPP
14
15 #define NAVIER_VERSION 0.1
16
17 #include "mfem.hpp"
18 #include "ortho_solver.hpp"
19
20 namespace mfem
21 {
22 namespace navier
23 {
24 using VecFuncT = void(const Vector &x, double t, Vector &u);
25 using ScalarFuncT = double(const Vector &x, double t);
26
27 /// Container for a Dirichlet boundary condition of the velocity field.
29 {
30 public:
32 : attr(attr), coeff(coeff)
33 {}
34
36 {
37 // Deep copy the attribute array
38 this->attr = obj.attr;
39
40 // Move the coefficient pointer
41 this->coeff = obj.coeff;
42 obj.coeff = nullptr;
43 }
44
45 ~VelDirichletBC_T() { delete coeff; }
46
49 };
50
51 /// Container for a Dirichlet boundary condition of the pressure field.
53 {
54 public:
56 : attr(attr), coeff(coeff)
57 {}
58
60 {
61 // Deep copy the attribute array
62 this->attr = obj.attr;
63
64 // Move the coefficient pointer
65 this->coeff = obj.coeff;
66 obj.coeff = nullptr;
67 }
68
69 ~PresDirichletBC_T() { delete coeff; }
70
73 };
74
75 /// Container for an acceleration term.
77 {
78 public:
80 : attr(attr), coeff(coeff)
81 {}
82
84 {
85 // Deep copy the attribute array
86 this->attr = obj.attr;
87
88 // Move the coefficient pointer
89 this->coeff = obj.coeff;
90 obj.coeff = nullptr;
91 }
92
93 ~AccelTerm_T() { delete coeff; }
94
97 };
98
99 /// Transient incompressible Navier Stokes solver in a split scheme formulation.
100 /**
101 * This implementation of a transient incompressible Navier Stokes solver uses
102 * the non-dimensionalized formulation. The coupled momentum and
103 * incompressibility equations are decoupled using the split scheme described in
104 * [1]. This leads to three solving steps.
105 *
106 * 1. An extrapolation step for all nonlinear terms which are treated
107 * explicitly. This step avoids a fully coupled nonlinear solve and only
108 * requires a solve of the mass matrix in velocity space \f$M_v^{-1}\f$. On
109 * the other hand this introduces a CFL stability condition on the maximum
110 * timestep.
111 *
112 * 2. A Poisson solve \f$S_p^{-1}\f$.
113 *
114 * 3. A Helmholtz like solve \f$(M_v - \partial t K_v)^{-1}\f$.
115 *
116 * The numerical solver setup for each step are as follows.
117 *
118 * \f$M_v^{-1}\f$ is solved using CG with Jacobi as preconditioner.
119 *
120 * \f$S_p^{-1}\f$ is solved using CG with AMG applied to the low order refined
121 * (LOR) assembled pressure Poisson matrix. To avoid assembling a matrix for
122 * preconditioning, one can use p-MG as an alternative (NYI).
123 *
124 * \f$(M_v - \partial t K_v)^{-1}\f$ due to the CFL condition we expect the time
125 * step to be small. Therefore this is solved using CG with Jacobi as
126 * preconditioner. For large time steps a preconditioner like AMG or p-MG should
127 * be used (NYI).
128 *
129 * Statements marked with NYI mean this feature is planned but Not Yet
130 * Implemented.
131 *
132 * A detailed description is available in [1] section 4.2. The algorithm is
133 * originated from [2].
134 *
135 * [1] Michael Franco, Jean-Sylvain Camier, Julian Andrej, Will Pazner (2020)
136 * High-order matrix-free incompressible flow solvers with GPU acceleration and
137 * low-order refined preconditioners (https://arxiv.org/abs/1910.03032)
138 *
139 * [2] A. G. Tomboulides, J. C. Y. Lee & S. A. Orszag (1997) Numerical
140 * Simulation of Low Mach Number Reactive Flows
141 */
143 {
144 public:
145 /// Initialize data structures, set FE space order and kinematic viscosity.
146 /**
147 * The ParMesh @a mesh can be a linear or curved parallel mesh. The @a order
148 * of the finite element spaces is this algorithm is of equal order
149 * \f$(P_N)^d P_N\f$ for velocity and pressure respectively. This means the
150 * pressure is in discretized in the same space (just scalar instead of a
151 * vector space) as the velocity.
152 *
153 * Kinematic viscosity (dimensionless) is set using @a kin_vis and
154 * automatically converted to the Reynolds number. If you want to set the
155 * Reynolds number directly, you can provide the inverse.
156 */
157 NavierSolver(ParMesh *mesh, int order, double kin_vis);
158
159 /// Initialize forms, solvers and preconditioners.
160 void Setup(double dt);
161
162 /// Compute solution at the next time step t+dt.
163 /**
164 * This method can be called with the default value @a provisional which
165 * always accepts the computed time step by automatically calling
166 * UpdateTimestepHistory.
167 *
168 * If @a provisional is set to true, the solution at t+dt is not accepted
169 * automatically and the application code has to call UpdateTimestepHistory
170 * and update the @a time variable accordingly.
171 *
172 * The application code can check the provisional step by retrieving the
173 * GridFunction with the method GetProvisionalVelocity. If the check fails,
174 * it is possible to retry the step with a different time step by not
175 * calling UpdateTimestepHistory and calling this method with the previous
176 * @a time and @a cur_step.
177 *
178 * The method and parameter choices are based on [1].
179 *
180 * [1] D. Wang, S.J. Ruuth (2008) Variable step-size implicit-explicit
181 * linear multistep methods for time-dependent partial differential
182 * equations
183 */
184 void Step(double &time, double dt, int cur_step, bool provisional = false);
185
186 /// Return a pointer to the provisional velocity ParGridFunction.
188
189 /// Return a pointer to the current velocity ParGridFunction.
191
192 /// Return a pointer to the current pressure ParGridFunction.
194
195 /// Add a Dirichlet boundary condition to the velocity field.
196 void AddVelDirichletBC(VectorCoefficient *coeff, Array<int> &attr);
197
198 void AddVelDirichletBC(VecFuncT *f, Array<int> &attr);
199
200 /// Add a Dirichlet boundary condition to the pressure field.
201 void AddPresDirichletBC(Coefficient *coeff, Array<int> &attr);
202
204
205 /// Add an acceleration term to the RHS of the equation.
206 /**
207 * The VecFuncT @a f is evaluated at the current time t and extrapolated
208 * together with the nonlinear parts of the Navier Stokes equation.
209 */
210 void AddAccelTerm(VectorCoefficient *coeff, Array<int> &attr);
211
212 void AddAccelTerm(VecFuncT *f, Array<int> &attr);
213
214 /// Enable partial assembly for every operator.
215 void EnablePA(bool pa) { partial_assembly = pa; }
216
217 /// Enable numerical integration rules. This means collocated quadrature at
218 /// the nodal points.
219 void EnableNI(bool ni) { numerical_integ = ni; }
220
221 /// Print timing summary of the solving routine.
222 /**
223 * The summary shows the timing in seconds in the first row of
224 *
225 * 1. SETUP: Time spent for the setup of all forms, solvers and
226 * preconditioners.
227 * 2. STEP: Time spent computing a full time step. It includes all three
228 * solves.
229 * 3. EXTRAP: Time spent for extrapolation of all forcing and nonlinear
230 * terms.
231 * 4. CURLCURL: Time spent for computing the curl curl term in the pressure
232 * Poisson equation (see references for detailed explanation).
233 * 5. PSOLVE: Time spent in the pressure Poisson solve.
234 * 6. HSOLVE: Time spent in the Helmholtz solve.
235 *
236 * The second row shows a proportion of a column relative to the whole
237 * time step.
238 */
239 void PrintTimingData();
240
241 ~NavierSolver();
242
243 /// Compute \f$\nabla \times \nabla \times u\f$ for \f$u \in (H^1)^2\f$.
245 ParGridFunction &cu,
246 bool assume_scalar = false);
247
248 /// Compute \f$\nabla \times \nabla \times u\f$ for \f$u \in (H^1)^3\f$.
250
251 /// Remove mean from a Vector.
252 /**
253 * Modify the Vector @a v by subtracting its mean using
254 * \f$v = v - \frac{\sum_i^N v_i}{N} \f$
255 */
256 void Orthogonalize(Vector &v);
257
258 /// Remove the mean from a ParGridFunction.
259 /**
260 * Modify the ParGridFunction @a v by subtracting its mean using
261 * \f$v = v - \int_\Omega \frac{v}{vol(\Omega)} dx \f$.
262 */
263 void MeanZero(ParGridFunction &v);
264
265 /// Rotate entries in the time step and solution history arrays.
266 void UpdateTimestepHistory(double dt);
267
268 /// Set the maximum order to use for the BDF method.
269 void SetMaxBDFOrder(int maxbdforder) { max_bdf_order = maxbdforder; };
270
271 /// Compute CFL
272 double ComputeCFL(ParGridFunction &u, double dt);
273
274 /// Set the number of modes to cut off in the interpolation filter
275 void SetCutoffModes(int c) { filter_cutoff_modes = c; }
276
277 /// Set the interpolation filter parameter @a a
278 /**
279 * If @a a is > 0, the filtering algorithm for the velocity field after every
280 * time step from [1] is used. The parameter should be 0 > @a >= 1.
281 *
282 * [1] Paul Fischer, Julia Mullen (2001) Filter-based stabilization of
283 * spectral element methods
284 */
285 void SetFilterAlpha(double a) { filter_alpha = a; }
286
287 protected:
288 /// Print information about the Navier version.
289 void PrintInfo();
290
291 /// Update the EXTk/BDF time integration coefficient.
292 /**
293 * Depending on which time step the computation is in, the EXTk/BDF time
294 * integration coefficients have to be set accordingly. This allows
295 * bootstrapping with a BDF scheme of order 1 and increasing the order each
296 * following time step, up to order 3 (or whichever order is set in
297 * SetMaxBDFOrder).
298 */
299 void SetTimeIntegrationCoefficients(int step);
300
301 /// Eliminate essential BCs in an Operator and apply to RHS.
302 void EliminateRHS(Operator &A,
303 ConstrainedOperator &constrainedA,
304 const Array<int> &ess_tdof_list,
305 Vector &x,
306 Vector &b,
307 Vector &X,
308 Vector &B,
309 int copy_interior = 0);
310
311 /// Enable/disable debug output.
312 bool debug = false;
313
314 /// Enable/disable verbose output.
315 bool verbose = true;
316
317 /// Enable/disable partial assembly of forms.
318 bool partial_assembly = false;
319
320 /// Enable/disable numerical integration rules of forms.
321 bool numerical_integ = false;
322
323 /// The parallel mesh.
324 ParMesh *pmesh = nullptr;
325
326 /// The order of the velocity and pressure space.
327 int order;
328
329 /// Kinematic viscosity (dimensionless).
330 double kin_vis;
331
332 /// Velocity \f$H^1\f$ finite element collection.
334
335 /// Pressure \f$H^1\f$ finite element collection.
337
338 /// Velocity \f$(H^1)^d\f$ finite element space.
340
341 /// Pressure \f$H^1\f$ finite element space.
343
344 ParNonlinearForm *N = nullptr;
345
347
349
351
353
355
357
359
360 ParLinearForm *f_form = nullptr;
361
363
364 /// Linear form to compute the mass matrix in various subroutines.
365 ParLinearForm *mass_lf = nullptr;
367 double volume = 0.0;
368
373
379
380 Solver *MvInvPC = nullptr;
381 CGSolver *MvInv = nullptr;
382
385 CGSolver *SpInv = nullptr;
386
387 Solver *HInvPC = nullptr;
388 CGSolver *HInv = nullptr;
389
391 resu;
393
395
397 resu_gf;
398
400
401 // All essential attributes.
404
405 // All essential true dofs.
408
409 // Bookkeeping for velocity dirichlet bcs.
410 std::vector<VelDirichletBC_T> vel_dbcs;
411
412 // Bookkeeping for pressure dirichlet bcs.
413 std::vector<PresDirichletBC_T> pres_dbcs;
414
415 // Bookkeeping for acceleration (forcing) terms.
416 std::vector<AccelTerm_T> accel_terms;
417
418 int max_bdf_order = 3;
419 int cur_step = 0;
420 std::vector<double> dthist = {0.0, 0.0, 0.0};
421
422 // BDFk/EXTk coefficients.
423 double bd0 = 0.0;
424 double bd1 = 0.0;
425 double bd2 = 0.0;
426 double bd3 = 0.0;
427 double ab1 = 0.0;
428 double ab2 = 0.0;
429 double ab3 = 0.0;
430
431 // Timers.
433
434 // Print levels.
435 int pl_mvsolve = 0;
436 int pl_spsolve = 0;
437 int pl_hsolve = 0;
438 int pl_amg = 0;
439
440 // Relative tolerances.
441 double rtol_spsolve = 1e-6;
442 double rtol_hsolve = 1e-8;
443
444 // Iteration counts.
446
447 // Residuals.
448 double res_mvsolve = 0.0, res_spsolve = 0.0, res_hsolve = 0.0;
449
450 // LOR related.
451 ParMesh *pmesh_lor = nullptr;
454 InterpolationGridTransfer *vgt = nullptr, *pgt = nullptr;
455
459
463
464 // Filter-based stabilization
466 double filter_alpha = 0.0;
471 };
472
473 } // namespace navier
474
475 } // namespace mfem
476
477 #endif
FiniteElementCollection * vfec_filter
VectorCoefficient * coeff
Container for a Dirichlet boundary condition of the velocity field.
void PrintTimingData()
Print timing summary of the solving routine.
void Orthogonalize(Vector &v)
Remove mean from a Vector.
Definition: solvers.hpp:316
FiniteElementCollection * pfec
Pressure finite element collection.
void EliminateRHS(Operator &A, ConstrainedOperator &constrainedA, const Array< int > &ess_tdof_list, Vector &x, Vector &b, Vector &X, Vector &B, int copy_interior=0)
Eliminate essential BCs in an Operator and apply to RHS.
ParBilinearForm * Mv_form_lor
Container for a Dirichlet boundary condition of the pressure field.
Base class for vector Coefficients that optionally depend on time and space.
A coefficient that is constant across space and time.
Definition: coefficient.hpp:78
void PrintInfo()
Print information about the Navier version.
void SetFilterAlpha(double a)
Set the interpolation filter parameter a.
FiniteElementCollection * pfec_lor
ConstantCoefficient nlcoeff
void SetMaxBDFOrder(int maxbdforder)
Set the maximum order to use for the BDF method.
Pointer to an Operator of a specified type.
Definition: handle.hpp:33
void(const Vector &x, double t, Vector &u) VecFuncT
ParGridFunction un_filtered_gf
Parallel non-linear operator on the true dofs.
double(const Vector &x, double t) ScalarFuncT
ParLinearForm * FText_bdr_form
std::vector< PresDirichletBC_T > pres_dbcs
bool numerical_integ
Enable/disable numerical integration rules of forms.
Abstract parallel finite element space.
Definition: pfespace.hpp:28
ConstantCoefficient H_lincoeff
Container for an acceleration term.
NavierSolver(ParMesh *mesh, int order, double kin_vis)
Initialize data structures, set FE space order and kinematic viscosity.
AccelTerm_T(Array< int > attr, VectorCoefficient *coeff)
PresDirichletBC_T(PresDirichletBC_T &&obj)
ParMixedBilinearForm * G_form
void Setup(double dt)
Initialize forms, solvers and preconditioners.
void MeanZero(ParGridFunction &v)
Remove the mean from a ParGridFunction.
VelDirichletBC_T(VelDirichletBC_T &&obj)
bool partial_assembly
Enable/disable partial assembly of forms.
ConstantCoefficient Sp_coeff
The BoomerAMG solver in hypre.
Definition: hypre.hpp:1387
void ComputeCurl3D(ParGridFunction &u, ParGridFunction &cu)
Compute for .
Class for parallel linear form.
Definition: plinearform.hpp:26
FiniteElementCollection * vfec
Velocity finite element collection.
double f(const Vector &xvec)
Definition: lor_mms.hpp:32
ParFiniteElementSpace * pfes_lor
InterpolationGridTransfer * vgt
double ComputeCFL(ParGridFunction &u, double dt)
Compute CFL.
double b
Definition: lissajous.cpp:42
ParGridFunction * GetCurrentPressure()
Return a pointer to the current pressure ParGridFunction.
Timing object.
Definition: tic_toc.hpp:34
ConstantCoefficient onecoeff
void SetCutoffModes(int c)
Set the number of modes to cut off in the interpolation filter.
std::vector< double > dthist
void SetTimeIntegrationCoefficients(int step)
Update the EXTk/BDF time integration coefficient.
void EnablePA(bool pa)
Enable partial assembly for every operator.
ParBilinearForm * Sp_form_lor
void AddVelDirichletBC(VectorCoefficient *coeff, Array< int > &attr)
Add a Dirichlet boundary condition to the velocity field.
void ComputeCurl2D(ParGridFunction &u, ParGridFunction &cu, bool assume_scalar=false)
Compute for .
ParBilinearForm * H_form_lor
ParFiniteElementSpace * vfes
Velocity finite element space.
Transfer data between a coarse mesh and an embedded refined mesh using interpolation.
Definition: transfer.hpp:121
ParLinearForm * mass_lf
Linear form to compute the mass matrix in various subroutines.
Base class Coefficients that optionally depend on space and time. These are used by the BilinearFormI...
Definition: coefficient.hpp:39
Collection of finite elements from the same family in multiple dimensions. This class is used to matc...
Definition: fe_coll.hpp:26
ParGridFunction * GetCurrentVelocity()
Return a pointer to the current velocity ParGridFunction.
std::vector< AccelTerm_T > accel_terms
bool verbose
Enable/disable verbose output.
double a
Definition: lissajous.cpp:41
InterpolationGridTransfer * pgt
double kin_vis
Kinematic viscosity (dimensionless).
Class for parallel bilinear form using different test and trial FE spaces.
std::vector< VelDirichletBC_T > vel_dbcs
VectorGridFunctionCoefficient * FText_gfcoeff
void UpdateTimestepHistory(double dt)
Rotate entries in the time step and solution history arrays.
ParFiniteElementSpace * pfes
Pressure finite element space.
ParGridFunction * GetProvisionalVelocity()
Return a pointer to the provisional velocity ParGridFunction.
PresDirichletBC_T(Array< int > attr, Coefficient *coeff)
Class for parallel bilinear form.
AccelTerm_T(AccelTerm_T &&obj)
RefCoord t[3]
ParMixedBilinearForm * D_form
Vector data type.
Definition: vector.hpp:60
Vector coefficient defined by a vector GridFunction.
void Step(double &time, double dt, int cur_step, bool provisional=false)
Compute solution at the next time step t+dt.
ParFiniteElementSpace * vfes_filter
void AddPresDirichletBC(Coefficient *coeff, Array< int > &attr)
Add a Dirichlet boundary condition to the pressure field.
double u(const Vector &xvec)
Definition: lor_mms.hpp:24
Base class for solvers.
Definition: operator.hpp:648
Class for parallel grid function.
Definition: pgridfunc.hpp:32
Square Operator for imposing essential boundary conditions using only the action, Mult()...
Definition: operator.hpp:834
Abstract operator.
Definition: operator.hpp:24
bool debug
Enable/disable debug output.
Class for parallel meshes.
Definition: pmesh.hpp:32
Transient incompressible Navier Stokes solver in a split scheme formulation.
Solver wrapper which orthogonalizes the input and output vector.
ParMesh * pmesh
The parallel mesh.
void AddAccelTerm(VectorCoefficient *coeff, Array< int > &attr)
Add an acceleration term to the RHS of the equation.
int order
The order of the velocity and pressure space.
ConstantCoefficient H_bdfcoeff
VelDirichletBC_T(Array< int > attr, VectorCoefficient *coeff) | 5,245 | 19,051 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.356685 |
https://medium.com/mlearning-ai/why-graph-theory-is-cooler-than-you-thought-4df73e2a4950 | 1,618,656,534,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00262.warc.gz | 491,523,351 | 41,047 | # Why Graph Theory is Cooler Than you Thought
## Graph Theory in Machine Learning, and How it’s Changed the Game
Feb 26 · 7 min read
## What are Graphs?
Talk to a scientist in just about any discipline, and ask them the question — based on their discipline — “how that stuff works”. You’ll likely find that there are systems and networks that you have to consider before you can really understand how any given thing works: whether that’s the human body, a food chain in an ecosystem, a chemical reaction, or a society as a whole. Without understanding the relationship between two animals in an ecosystem, two atoms in a molecule, or cells and tissues in our body, you just have a bunch of data: a list of cells, a readout of animals and what they eat, etc.
Traditional machine learning models often take data this way: they take lists or tables of data and do some stuff (the details of which depend on the algorithm being used as well as a few other parameters) to make predictions about a thing. But for some problems, there’s a better way.
Graphs are data structures that represent networks of or relationships between the data they contain. Typically, they’re represented as “nodes” and lines, or “edges”.
Figure 1, above is an example of a “directed graph”, or, a graph in which data has a one-way relationship with other data. This is demonstrated through arrows — Medium renders this photo in a rather small frame, so they’re slightly hard to see — that show who is “following” who, and which indicate mutual relationships where they exist.
The circles which represent data (in this case, an image and a username) are called nodes or vertices, and the lines which connect them are called edges. These lines represent relationships between the vertices, and can be represented (as they are here), as an “all or nothing relationship” (i.e: you’re following someone or you aren’t) or as a “weighted” relationship (i.e: a thicker line can represent higher engagement between two users, while a thinner line can represent a weaker relationship or lower engagement). We can see an example of a weighted graph below: Figure 2 represents differing levels of connectivity between varied regions in a brain.
At this point, it’s possible you’re feeling how I felt when I first was introduced to graphs and graph theory in a computer science class: bored and possibly slightly confused. The good news is, since we’ve covered some of the terminology that’s necessary to understand the good stuff, we can start to get into why graphs matter, and what makes them so cool.
## So, what?
Graphs are already used for some pretty neat stuff in computer science: your Maps application, for example, is using graphs behind the scenes to store data about locations and the streets that connect them, and is using shortest-distance algorithms to find you the shortest route to your destination.
But it gets even better when we start to look at using graphs for machine learning. Because graphs demonstrate comprehensive relationships between pieces of data (as compared to ordered lists of data, or tensors which tell us little about the relationships between data points or features by themselves), we can use them to perform in-depth analysis and make predictions based on these relationships.
## Graph Theory & Machine Learning — But How?
Before we get to reap the benefits of combining these graphs or networks we keep talking about with machine learning, we have to somehow represent our graph in a way that a computer — and then a machine learning algorithm — can understand.
Graphs can be represented traditionally in one of three basic ways:
Adjacency matrices do… kind of just what they sound like they’d do. They represent connections, or edges, between different nodes using a matrix. We can look at an example to illustrate what this might look like:
Here, if we look at A, C, we can see that there is no direct connection, because there is a zero in that spot. However, if we look at E, C (or C, E because this is an undirected graph), we see a 1 which represents an edge between those two nodes.
2. An Edge List
An edge list is another way to represent our network — or graph — in a way that’s computationally understandable. Here, we represent pairs of connected nodes within a list. You can see an example below:
Adjacency lists combine the above two approaches, representing a list of nodes, connected to a list of all of the nodes they’re directly connected to. To illustrate, let’s look at an example:
With the above three approaches, we’re able to tackle the difficulty of representing graphs computationally in our code. However, there are still some challenges when feeding graphs to machine learning models. Traditionally, deep learning models are good at handling data which takes up a fixed amount of space and is unidirectional. No matter how we represent them, graphs don’t take up a fixed amount of space in memory, and aren’t continuous, but rather, each node holds a reference to nodes it’s directly connected to.
There are some really fabulous ways of tackling these challenges, which I’ll dive deeper into in a later article. For now, I’ll leave you with a few resources to research on your own should you be interested, which are providing us with new ways to expand the problems machine learning is able to solve.
## Graph Theory and Machine Learning — What Can we Do With It?
Nothing exists in a vacuum, and understanding the interconnected networks of data that make up many of our scientific disciplines provides the exciting potential to answer so many questions — more than I can begin to wrap into this article.
What if we could better understand the human brain? What if we could make predictions about the effect of some stimulus or change on an ecosystem? Or, predict which compound is the most likely to create an effective drug?
The best part of everything we’ve just learned is that we can — and it isn’t simply a theoretical possibility, but something we’re doing right now!
Graph theory is already being used for:
## How Graph Theory Makes it All Happen
Let’s dive a little deeper into these applications, so we can look at the utilization of graph theory within them in more detail.
Let’s use diagnostic models as an example. Specifically, I want to look at this example of network analysis being used for the diagnosis and identification of possible schizophrenia in patients:
Using graphs to represent network analyses of the brain, neuroscientists are able to map key findings related to the diagnosis of schizophrenia. Given that there are certain markers for the onset of schizophrenia:
• less efficiently wired networks
• less local clustering
• less hierarchical organization
We could potentially evaluate these networks with a machine learning algorithm and predict the probability a patient has or will develop schizophrenia based on these markers.
Without the knowledge of these networks, in this example, this kind of analyses becomes an entirely different neurological analysis of the patient. The promising discoveries of these findings for schizophrenia has promising implications for the diagnosis and treatment of this disorder — possible early diagnosis and intervention that goes far beyond simply evaluating symptoms.
This is just an example, but it’s entirely illustrative of the benefits of graph theory in machine learning as it intersects with other disciplines.
The fact of the matter is, there is often much more to our data than we can represent in lists, data frames, or tensors alone. While there are ways to explore our data and present it in such a way that we can hypothesize relationships and even enable our algorithms to predict these, when we’re able to represent the connections between our data in a different way, we’re able to do more with the data that we have.
When we understand the ways in which things relate to one another, we understand them better: we can make more comprehensive predictions, and answer harder questions, with some pretty life changing results.
Resources:
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Get the Medium app | 1,882 | 9,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-17 | latest | en | 0.961251 |
https://socratic.org/questions/how-do-you-simplify-sqrt3-sqrt6-sqrt3#300561 | 1,726,475,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00018.warc.gz | 494,630,535 | 5,736 | # How do you simplify sqrt3 sqrt6 + sqrt3 ?
Aug 19, 2016
$3 \sqrt{3}$
#### Explanation:
We can change $\sqrt{6} \sqrt{3}$ into $\sqrt{18}$
$\sqrt{18} = \sqrt{3 \times 3 \times 2} = 2 \sqrt{3}$
Now we have $2 \sqrt{3} + 1 \sqrt{3}$
=$3 \sqrt{3}$ | 107 | 251 | {"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} | 4.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.471871 |
http://research.stlouisfed.org/fred2/series/CUURX200SANL11 | 1,432,287,392,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207924799.9/warc/CC-MAIN-20150521113204-00320-ip-10-180-206-219.ec2.internal.warc.gz | 202,847,581 | 19,534 | # Consumer Price Index for All Urban Consumers: Nondurables less food and beverages in Midwest - Size Class B/C (between 50,000 and 1,500,000 persons)
2015-03: 149.721 Index December 1997=100 (+ see more)
Monthly, Not Seasonally Adjusted, CUURX200SANL11, Updated: 2015-04-17 9:17 AM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
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(a) Consumer Price Index for All Urban Consumers: Nondurables less food and beverages in Midwest - Size Class B/C (between 50,000 and 1,500,000 persons), Index December 1997=100, Not Seasonally Adjusted (CUURX200SANL11)
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
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``` US. Bureau of Labor Statistics, Consumer Price Index for All Urban Consumers: Nondurables less food and beverages in Midwest - Size Class B/C (between 50,000 and 1,500,000 persons) [CUURX200SANL11], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/CUURX200SANL11/, May 22, 2015. ```
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Name: Email: | 552 | 2,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2015-22 | latest | en | 0.798299 |
http://electricalarticle.com/transformer-losses/ | 1,670,501,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00331.warc.gz | 13,962,653 | 18,237 | # Transformer Losses
Among all electrical machines, the transformer is the most efficient electrical machine. As shown in working of the transformer, it is a static device. Hence, friction and windage losses are not present in the case of the transformer. The transformer losses are variable or constant loss.
There are four types of losses occurred in the transformer and that is,
1. Copper loss
2. Iron loss
3. Stray loss
4. Dielectric loss
Out of above-mentioned losses, copper loss and iron loss considered as major losses and stray loss and dielectric loss considered as minor losses. Let’s explain this in details.
### 1) Copper loss:
There are two winding in the transformer; primary winding and a secondary winding. The copper loss is nothing but, the losses occur due to the winding resistance. The copper loss is also known as loss or CU loss.
If primary winding resistance is and secondary winding resistance is , then
(1)
Where is the load current at primary side
is the load current at secondary side
From the above equation, it is clear that the copper loss is directly proportional to the square of the load current. Value of the load current depends on the load. If the load varies, the value of the load current is also varied. Hence, the copper loss is a variable loss.
### 2) Iron loss:
This loss takes place in the core of the transformer due to the time-varying nature of flux in the core. Hence, the iron loss known as core loss. The total iron loss divides into two parts;
• hysteresis loss
• eddy current loss
#### (A) Hysteresis loss:
Whenever it subjected to the alternating nature of the magnetizing force, the hysteresis loss occurs due to the reversal of magnetization of the transformer core. After every half cycle, the domain present in the magnetic material will change their orientation. The power consumed by this change of orientation after every half cycle is known as hysteresis loss.
The magnetic reversal of the transformer core plotted with the help of the BH curve and it represents as below figure.
The hysteresis loss occurs in one cycle is equal to the area enclosed with the one hysteresis loop.The hysteresis loss can be determined with the Steinmetz’s formula. Which is given as
(2)
Where = Steinmetz coefficient = range 1.5 to 2.5
= supply frequency
= volume of core
= Steinmetz exponent = 1.6 for silicon steel
#### (B) Eddy current loss:
Due to the production of the eddy current in the core, the eddy current loss is the loss present in the core of the transformer. The eddy current produced because of the conductivity of the core.
The eddy current loss is directly proportional to the conductivity () of the core. = the resistance offered by the core to flow of eddy current. is inversely proportional to the conductivity. By reducing the conductivity, the eddy current can be reduced. It is possible to reduce conductivity without affecting magnetic properties by adding silica content and by using laminated core.
### 3) Stray loss:
There are two types of stray losses;
• copper stray loss
• iron stray loss.
#### (A) Copper stray loss:
Due to the leakage flux, Copper stray loss occurs under full load condition. This loss considered as the additional loss due to the stray current within the conductor. Instead of the solid conductor, the stranded conductors used to reduce the copper stray loss. Hence, by this way, the skin effect of winding is also reduced. The copper stray loss considers as a variable loss.
#### (B) Iron stray loss:
Due to the auxiliary iron parts, the iron stray loss is the additional iron loss occurred in the transformer. These auxiliary iron parts are the transformer tank, steel channels, and conservation tank. In this auxiliary iron parts, the iron stray loss occurred due to the leakage flux. This loss is less in shell type transformer compared to the core type transformer. The leakage flux is directly proportional to the load current. So, this loss also considered as a variable loss. The iron stray loss is just 0.5% of full load output.
### 4) Dielectric loss:
This loss produced in the insulating material of the transformer. An insulating material, the free electrons are not available. When a voltage applied, a small amount of current will flow through this due to the conversion of atoms. There is displacement of charges and the current produce due to this known as displacement current.
The process of conversion of atoms into electric dipole known as polarization. The dielectric loss is depending on the applied voltage and it is independent of load current. So, this loss considered as a constant loss. This loss is 0.25% of full load output.
## Related Article:
Transformer: Construction and Classification
1,610 total views, 1 views today | 1,024 | 4,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-49 | latest | en | 0.937252 |
https://www.physicsforums.com/threads/circuit-problem.64061/ | 1,508,208,286,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820556.7/warc/CC-MAIN-20171017013608-20171017033608-00597.warc.gz | 1,230,956,990 | 14,270 | # Circuit Problem
1. Feb 16, 2005
### Jacob87411
The student engineer of a campus radio station wants to verify the effectiveness of the lightning rod on the antenna mast. The unknown resistance Rx is in between points C and E. Point C is just right of the antenna and point E is true ground but is inaccessible for the direct measurement because this stratum is several meters below the Earth's surface. Two identical rods are driven into the ground at point A (left of the antenna) and B (right of C which is just right of the antenna.) introducing an unknown resistance Ry. The procedure is as follows: Measure the resistance R1 between points A and B, then connect A and B with a heavy conducting wire and measure R2 between points A and C. (A) Derive a formula for Rx in terms of the observable resistances, R1 and R2. (B) A satisfactory ground resistnace would be Rx < 2.0 Ohm. Is the grounding of the station antenna adequate if measurements give R1=13 Ohm and R2=6 Ohm
Really lost on how to set up the original circuit I guess...Once I have that it shouldnt be that bad
2. Feb 16, 2005
### learningphysics
Draw a picture first...
Then from the picture you should see be able to make out the circuit.
Hint: there are 3 resistors, and ground(E) is a common point to all 3. | 310 | 1,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.93562 |
http://www.av8n.com/physics/thermo/entropy-energy.html | 1,660,670,655,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572408.31/warc/CC-MAIN-20220816151008-20220816181008-00186.warc.gz | 63,341,441 | 23,605 | [Previous][Contents]
# 9 Connecting Entropy with Energy
## 9.1 The Boltzmann Distribution
For reasons discussed in chapter 23, whenever a system is in thermal equilibrium, the energy is distributed among the microstates according to a very special probability distribution, namely the Boltzmann distribution. That is, the probability of finding the system in microstate i is given by:
Pi = e−Êi / kT … for a thermal distribution (9.1)
where Êi is the energy of the ith microstate, and kT is the temperature measured in energy units. That is, plain T is the temperature, and k is Boltzmann’s constant, which is just the conversion factor from temperature units to energy units.
Figure 9.1 shows this distribution graphically.
Figure 9.1: An Equilibrium Distribution
Evidence in favor of equation 9.1 is discussed in section 11.2.
## 9.2 Systems with Subsystems
When thinking about equation 9.1 and figure 9.1 it is important to realize that things don’t have to be that way. There are other possibilities. Indeed, a theory of thermodynamics that assumed that everything in sight was always in equilibrium at temperature T would be not worth the trouble. For starters, it is impossible to build a heat engine unless the hot reservoir is not in equilibrium with the cold reservoir.
We start by considering the system shown in figure 9.2, namely a styrofoam box containing a hot potato and a cold potato. (This is a simplified version of figure 1.2.)
Figure 9.2: A System Consisting of Two Potatoes
In situations like this, we can make good progress if we divide the system into subsystems. Here subsystem A is the red potato, and subsystem B is the blue potato. Each subsystem has a well defined temperature, but initially the system as a whole does not.
Figure 9.3: Probability versus Energy, Non-Equilibrium
Figure 9.4: Probability versus Energy, System as a Whole
If we wait long enough, the two potatoes will come into equilibrium with each other, and at this point the system as a whole will have a well defined temperature. However, we are not required to wait for this to happen.
## 9.3 Remarks
### 9.3.1 Predictable Energy is Freely Convertible
The difference between random energy and predictable energy has many consequences. The most important consequence is that the predictable energy can be freely converted to and from other forms, such as gravitational potential energy, chemical energy, electrical energy, et cetera. In many cases, these conversions can be carried out with very high efficiency. In some other cases, though, the laws of thermodynamics place severe restrictions on the efficiency with which conversions can be carried out, depending on to what extent the energy distribution deviates from the Boltzmann distribution.
### 9.3.2 Thermodynamic Laws without Temperature
Ironically, the first law of thermodynamics (equation 1.1) does not depend on temperature. Energy is well-defined and is conserved, no matter what. It doesn’t matter whether the system is hot or cold or whether it even has a temperature at all.
Even more ironically, the second law of thermodynamics (equation 2.1) doesn’t depend on temperature, either. Entropy is well-defined and is paraconserved no matter what. It doesn’t matter whether the system is hot or cold or whether it even has a temperature at all.
(This state of affairs is ironic because thermodynamics is commonly defined to be the science of heat and temperature, as you might have expected from the name: thermodynamics. Yet in our modernized and rationalized thermodynamics, the two most central, fundamental ideas – energy and entropy – are defined without reference to heat or temperature.)
Of course there are many important situations that do involve temperature. Most of the common, every-day applications of thermodynamics involve temperature – but you should not think of temperature as the essence of thermodynamics. Rather, it is a secondary concept which is defined (if and when it even exists) in terms of energy and entropy.
### 9.3.3 Kinetic and Potential Microscopic Energy
You may have heard the term “kinetic theory”. In particular, the thermodynamics of ideal gases is commonly called the kinetic theory of gases. However, you should be careful, because “kinetic theory” is restricted to ideal gases (indeed to a subset of ideal gases) ... while thermodynamics applies to innumerable other things. Don’t fall into the trap of thinking that there is such a thing as “thermal energy” and that this so-called “thermal energy” is necessarily kinetic energy. In almost all systems, including solids, liquids, non-ideal gases, and even some ideal gases, the energy is a mixture of kinetic and potential energy. (Furthermore, in any non-cramped situation, i.e. in any situation where it is possible to build a heat engine, it is impossible in principle to define any such thing as “thermal energy”.) In any case, it is safer and in all ways better to say thermodynamics or statistical mechanics instead of “kinetic theory”.
In typical systems, potential energy and kinetic energy play parallel roles:
• To visualize microscopic potential energy, imagine that the atoms in a crystal lattice are held in place by springs. Roughly speaking, half of these springs have positive potential energy because they are extended relative to their resting-length, while the other half have positive potential energy because they are compressed relative to their resting-length. They’ve all got energy, but you can’t easily harness it because you don’t know which ones are compressed and which ones are extended.
• To visualize microscopic kinetic energy, imagine that half the atoms have a leftward velocity and half have a rightward velocity. They all have kinetic energy, but you can’t easily harness it because you don’t know which ones are moving leftward and which are moving rightward.
In fact, for an ordinary crystal such as quartz or sodium chloride, almost exactly half of the heat capacity is due to potential energy, and half to kinetic energy. It’s easy to see why that must be: The heat capacity is well described in terms of thermal phonons in the crystal. Each phonon mode is a harmonic1 oscillator. In each cycle of any harmonic oscillator, the energy changes from kinetic to potential and back again. The kinetic energy goes like sin2(phase) and the potential energy goes like cos2(phase), so on average each of those is half of the total energy.
Not all kinetic energy contributes to the heat capacity. Not all of the heat capacitly comes from kinetic energy.
A table-top sample of ideal gas is a special case, where all the energy is kinetic energy. This is very atypical of thermodynamics in general. Table-top ideal gases are very commonly used as an illustration of thermodynamic ideas, which becomes a problem when the example is overused so heavily as to create the misimpression that thermodynamics deals only with kinetic energy.
You could argue that in many familiar systems, the temperature is closely related to random kinetic energy ... but temperature is not the same thing as so-called “heat’ or “thermal energy”. Furthermore, there are other systems, such as spin systems, where the temperature is not related to the random kinetic energy.
All in all, it seems quite unwise to define heat or even temperature in terms of kinetic energy.
This discussion continues in section 9.3.4.
### 9.3.4 Ideal Gas : Potential Energy as well as Kinetic Energy
We have seen that for an ideal gas, there is a one-to-one correspondence between the temperature and the kinetic energy of the gas particles. However, that does not mean that there is a one-to-one correspondence between kinetic energy and heat energy. (In this context, heat energy refers to whatever is measured by a heat capacity experiment.)
To illustrate this point, let’s consider a sample of pure monatomic nonrelativistic nondegenerate ideal gas in a tall cylinder of horizontal radius r and vertical height h at temperature T. The pressure measured at the bottom of the cylinder is P. Each particle in the gas has mass m. We wish to know the heat capacity per particle at constant volume, i.e. CV/N.
At this point you may already have in mind an answer, a simple answer, a well-known answer, independent of r, h, m, P, T, and N. But wait, there’s more to the story: The point of this exercise is that h is not small. In particular, m|g|h is not small compared to kT, where g is the acceleration of gravity. For simplicity, you are encouraged to start by considering the limit where h goes to infinity, in which case the exact value of h no longer matters. Gravity holds virtually all the gas near the bottom of the cylinder, whenever hkT/m|g|.
• Later, if you want to come back and work the problem a second time, with no restrictions on h, that’s interesting and not very difficult.
• Also if you want to generalize to a polyatomic gas, that’s also worth doing. Section 26.4 continues the discussion of the energy of an ideal gas, including equation 26.46 and equation 26.47 which related energy to pressure, volume, and temperature.
You will discover that a distinctly nontrivial contribution to the heat capacity comes from the potential energy of the ideal gas. When you heat it up, the gas column expands, lifting its center of mass, doing work against gravity. (Of course, as always, there will be a contribution from the kinetic energy.)
For particles the size of atoms, the length-scale kT/m|g| is on the order of several kilometers, so the cylinder we are considering is much too big to fit on a table top. I often use the restrictive term “table-top” as a shorthand way of asserting that m|g|h is small compared to kT.
So, this reinforces the points made in section 9.3.3. We conclude that in general, heat energy is not just kinetic energy.
Beware that this tall cylinder is not a good model for the earth’s atmosphere. For one thing, the atmosphere is not isothermal. For another thing, if you are going to take the limit as h goes to infinity, you can’t use a cylinder; you need something more like a cone, spreading out as it goes up, to account for the spherical geometry.
### 9.3.5 Relative Motion versus “Thermal” Energy
Over the years, lots of people have noticed that you can always split the kinetic energy of a complex object into the KE of the center-of-mass motion plus the KE of the relative motion (i.e. the motion of the components relative to the center of mass).
Also a lot of people have tried (with mixed success) to split the energy of an object into a “thermal” piece and a “non-thermal” piece.
It is an all-too-common mistake to think that the overall/relative split is the same as the nonthermal/thermal split. Beware: they’re not the same. Definitely not. See section 7.7 for more on this.
First of all, the microscopic energy is not restricted to being kinetic energy, as discussed in section 9.3.3. So trying to understand the thermal/non-thermal split in terms of kinetic energy is guaranteed to fail. Using the work/KE theorem (reference 18) to connect work (via KE) to the thermal/nonthermal split is guaranteed to fail for the same reason.
Secondly, a standard counterexample uses flywheels, as discussed in section 18.4. You can impart macroscopic, non-Locrian KE to the flywheels without imparting center-of-mass KE or any kind of potential energy … and without imparting any kind of Locrian energy (either kinetic or potential).
The whole idea of “thermal energy” is problematic, and in many cases impossible to define, as discussed in chapter 19. If you find yourself worrying about the exact definition of “thermal energy”, it means you’re trying to solve the wrong problem. Find a way to reformulate the problem in terms of energy and entropy.
Center-of-mass motion is an example but not the only example of low-entropy energy. The motion of the flywheels is one perfectly good example of low-entropy energy. Several other examples are listed in section 11.3.
A macroscopic object has something like 1023 modes. The center-of-mass motion is just one of these modes. The motion of counter-rotating flywheels is another mode. These are slightly special, but not very special. A mode to which we can apply a conservation law, such as conservation of momentum, or conservation of angular momentum, might require a little bit of special treatment, but usually not much … and there aren’t very many such modes.
Sometimes on account of conservation laws, and sometimes for other reasons as discussed in section 11.11 it may be possible for a few modes of the system to be strongly coupled to the outside (and weakly coupled to the rest of the system), while the remaining 1023 modes are more strongly coupled to each other than they are to the outside. It is these issues of coupling-strength that determine which modes are in equilibrium and which (if any) are far from equilibrium. This is consistent with our definition of equilibrium (section 10.1).
Thermodynamics treats all the equilibrated modes on an equal footing. One manifestation of this can be seen in equation 9.1, where each state contributes one term to the sum … and addition is commutative.
There will never be an axiom that says such-and-such mode is always in equilibrium or always not; the answer is sensitive to how you engineer the couplings.
## 9.4 Entropy Without Constant Re-Shuffling
It is a common mistake to visualize entropy as a highly dynamic process, whereby the system is constantly flipping from one microstate to another. This may be a consequence of the fallacy discussed in section 9.3.5 (mistaking the thermal/nonthermal distinction for the kinetic/potential distinction) … or it may have other roots; I’m not sure.
In any case, the fact is that re-shuffling is not an essential part of the entropy picture.
An understanding of this point proceeds directly from fundamental notions of probability and statistics.
By way of illustration, consider one hand in a game of draw poker.
A) The deck is shuffled and hands are dealt in the usual way.
B) In preparation for the first round of betting, you look at your hand and discover that you’ve got the infamous “inside straight”. Other players raise the stakes, and when it’s your turn to bet you drop out, saying to yourself “if this had been an outside straight the probability would have been twice as favorable”.
C) The other players, curiously enough, stand pat, and after the hand is over you get a chance to flip through the deck and see the card you would have drawn.
Let’s more closely examine step (B). At this point you have to make a decision based on probability. The deck, as it sits there, is not constantly re-arranging itself, yet you are somehow able to think about the probability that the card you draw will complete your inside straight.
The deck, as it sits there during step (B), is not flipping from one microstate to another. It is in some microstate, and staying in that microstate. At this stage you don’t know what microstate that happens to be. Later, at step (C), long after the hand is over, you might get a chance to find out the exact microstate, but right now at step (B) you are forced to make a decision based only on the probability.
The same ideas apply to the entropy of a roomful of air, or any other thermodynamic system. At any given instant, the air is in some microstate with 100% probability; you just don’t know what microstate that happens to be. If you did know, the entropy would be zero … but you don’t know. You don’t need to take any sort of time-average to realize that you don’t know the microstate.
The bottom line is that the essence of entropy is the same as the essence of probability in general: The essential idea is that you don’t know the microstate. Constant re-arrangement is not essential.
This leaves us with the question of whether re-arrangement is ever important. Of course the deck needs to be shuffled at step (A). Not constantly re-shuffled, just shuffled the once.
Again, the same ideas apply to the entropy of a roomful of air. If you did somehow obtain knowledge of the microstate, you might be interested in the timescale over which the system re-arranges itself, making your erstwhile knowledge obsolete and thereby returning the system to a high-entropy condition.
The crucial point remains: the process whereby knowledge is lost and entropy is created is not part of the definition of entropy, and need not be considered when you evaluate the entropy. If you walk into a room for the first time, the re-arrangement rate is not your concern. You don’t know the microstate of this room, and that’s all there is to the story. You don’t care how quickly (if at all) one unknown microstate turns into another.
If you don’t like the poker analogy, we can use a cryptology analogy instead. Yes, physics, poker, and cryptology are all the same when it comes to this. Statistics is statistics.
If I’ve intercepted just one cryptotext from the opposition and I’m trying to crack it, on some level what matters is whether or not I know their session key. It doesn’t matter whether that session key is 10 microseconds old, or 10 minutes old, or 10 days old. If I don’t have any information about it, I don’t have any information about it, and that’s all that need be said.
On the other hand, if I’ve intercepted a stream of messages and extracted partial information from them (via a partial break of the cryptosystem), the opposition would be well advised to “re-shuffle the deck” i.e. choose new session keys on a timescale fast compared to my ability to extract information about them.
Applying these ideas to a roomful of air: Typical sorts of measurements give us only a pathetically small amount of partial information about the microstate. So it really doesn’t matter whether the air re-arranges itself super-frequently or super-infrequently. We don’t have any significant amount of information about the microstate, and that’s all there is to the story.
Reference 25 presents a simulation that demonstrates the points discussed in this subsection.
## 9.5 Units of Entropy
Before we go any farther, convince yourself that
log10(x) =
ln(x) ln(10)
0.434294 ln(x)
(9.2)
and in general, multiplying a logarithm by some positive number corresponds to changing the base of the logarithm.
In the formula for entropy, equation 2.2, the base of the logarithm has intentionally been left unspecified. You get to choose a convenient base. This is the same thing as choosing what units will be used for measuring the entropy.
Some people prefer to express the units by choosing the base of the logarithm, while others prefer to stick with natural logarithms and express the units more directly, using an expression of the form:
S[P] := k
∑ i
Pi ln(1/Pi) (9.3)
In this expression we stipulated e as the base of the logarithm. Whereas equation 2.2 we could choose the base of the logarithm, in equation 9.3 we get to choose the numerical value and units for k. This is a superficially different solution to the same problem. Reasonable choices include:
k = 1.380649 ×10−23 J/K = 1/ln(3) trit = 1 nat = 1/ln(2) bit
(9.4)
It must be emphasized that all these expressions are mathematically equivalent. In each case, the numerical part of k balances the units of k, so that the meaning remains unchanged. In some cases it is convenient to absorb the numerical part of k into the base of the logarithm:
S[P]=k
∑ i
Pi ln(1/Pi)
(in general)
k=1.380649 ×10−23
∑ i
Pi ln(1/Pi)
J/K (SI)
=
∑ i
Pi logΩ(1/Pi)
J/K (SI)
=
∑ i
Pi log3(1/Pi)
trit
=
∑ i
Pi loge(1/Pi)
nat
=
∑ i
Pi log2(1/Pi)
bit
(9.5)
where the third line uses the remarkably huge base Ω = exp(1.24×1022). When dealing with smallish amounts of entropy, units of bits are conventional and often convenient. When dealing with large amounts of entropy, units of J/K are conventional and often convenient. These are related as follows:
1 J/K = 1.04×1023 bits 1 bit = 9.57×10−24 J/K
(9.6)
A convenient unit for molar entropy is Joules per Kelvin per mole:
1 J/K/mol = 0.17 bit/particle 1 bit/particle = 5.76 J/K/mol = R ln(2)
(9.7)
Values in this range (on the order of one bit per particle) are very commonly encountered.
If you are wondering whether equation 9.7 is OK from a dimensional-analysis point of view, fear not. Temperature units are closely related to energy units. Specifically, energy is extensive and can be measured in joules, while temperature is intensive and can be measured in kelvins. Therefore combinations such as (J/K/mol) are dimensionless units. A glance at the dimensions of the ideal gas law should suffice to remind you of this if you ever forget.
See reference 26 for more about dimensionless units.
## 9.6 Probability versus Multiplicity
Let us spend a few paragraphs discussing a strict notion of multiplicity, and then move on to a more nuanced notion. (We also discuss the relationship between an equiprobable distribution and a microcanonical ensemble.)
### 9.6.1 Exactly Equiprobable
Suppose we have a system where a certain set of states2 (called the “accessible” states) are equiprobable, i.e. Pi = 1/W for some constant W. Furthermore, all remaining states are “inaccessible” which means they all have Pi = 0. The constant W is called the multiplicity.
Note: Terminology: The W denoting multiplicity in this section is unrelated to the W denoting work elsewhere in this document. Both usages of W are common in the literature. It is almost always obvious from context which meaning is intended, so there isn’t a serious problem. Some of the literature uses Ω to denote multiplicity.
The probability per state is necessarily the reciprocal of the number of accessible states, since (in accordance with the usual definition of “probability”) we want our probabilities to be normalized: ∑ Pi = 1.
In this less-than-general case, the entropy (as given by equation 2.2) reduces to
S = logW (provided the microstates are equiprobable) (9.8)
As usual, you can choose the base of the logarithm according to what units you prefer for measuring entropy: bits, nats, trits, J/K, or whatever. Equivalently, you can fix the base of the logarithm and express the units by means of a factor of k out front, as discussed in section 9.5:
S = k lnW (provided the microstates are equiprobable) (9.9)
This equation is prominently featured on Boltzmann’s tombstone. However, I’m pretty sure (a) he didn’t put it there, (b) Boltzmann was not the one who originated or emphasized this formula (Planck was), and (c) Boltzmann was well aware that this is not the most general expression for the entropy. I mention this because a lot of people who ought to know better take equation 9.9 as the unassailable definition of entropy, and sometimes they cite Boltzmann’s tombstone as if it were the ultimate authority on the subject.
In any case, (d) even if Boltzmann had endorsed equation 9.9, appeal to authority is not an acceptable substitute for scientific evidence and logical reasoning. We know more now than we knew in 1898, and we are allowed to change our minds about things ... although in this case it is not necessary. Equation 2.2 has been the faithful workhorse formula for a very long time.
There are various ways a system could wind up with equiprobable states:
• Consider a well-shuffled deck of cards. This is an example of a non-thermal system where the states are equiprobable.
• Consider a thermal system that is thermally isolated so that all accessible microstates have the same energy. This is called a microcanonical system. It will have equiprobable states in accordance with the equipartition law, equation 9.1.
Consider two blocks of copper that are identical except that one of them has more energy than the other. They are thermally isolated from each other and from everything else. The higher-energy block will have a greater number of accessible states, i.e. a higher multiplicity. In this way you can, if you wish, define a notion of multiplicity as a function of energy level.
On the other hand, you must not get the idea that multiplicity is a monotone function of energy or vice versa. Such an idea would be quite incorrect when applied to a spin system.
Terminology: By definition, a level is a group of microstates. An energy level is a group of microstates all with the same energy (or nearly the same energy, relative to other energy-scales in the problem). By connotation, usually when people speak of a level they mean energy level.
### 9.6.2 Approximately Equiprobable
We now introduce a notion of “approximate” equiprobability and “approximate” multiplicity by reference to the example in the following table:
Level # microstates Probability Probability Entropy in level of microstate of level (in bits) 1 2 0.01 0.020 0.133 2 979 0.001 0.989 9.757 3 1,000,000 1E-09 0.001 0.030 Total: 1,000,981 1.000 9.919
The system in this example 1,000,981 microstates, which we have grouped into three levels. There are a million states in level 3, each of which occurs with probability one in a billion, so the probability of observing some state from this level is one in a thousand. There are only two microstates in level 1, each of which is observed with a vastly larger probability, namely one in a hundred. Level 2 is baby-bear just right. It has a moderate number of states, each with a moderate probability ... with the remarkable property that on a level-by-level basis, this level dominates the probability distribution. The probability of observing some microstate from level 2 is nearly 100%.
The bottom line is that the entropy of this distribution is 9.919 bits, which is 99.53% of the entropy you would have if all the probability were tied up in 1000 microstates with probability 0.001 each.
In the table, the column we have labeled “# microstates in level” is conventionally called the multiplicity of the level. If we apply the S = log(W) formula in reverse, we find that our example distribution has a multiplicity of W = 2S = 29.919 = 968; this is the effective multiplicity of the distribution as a whole.
So we see that the effective multiplicity of the distribution is dominated by the multiplicity of level 2. The other levels contribute very little to the entropy.
You have to be careful how you describe the microstates in level 2. Level 2 is the most probable level (on a level-by-level basis), but its microstates are not the most probable microstates (on a microstate-by-microstate basis).
In the strict notion of multiplicity, all the states that were not part of the dominant level were declared “inaccessible”, but alas this terminology becomes hopelessly tangled when we progress to the nuanced notion of multiplicity. In the table, the states in level 3 are high-energy states, and it might be OK to say that they are energetically inaccessible, or “almost” inaccessible. It might be superficially tempting to label level 1 as also inaccessible, but that would not be correct. The states in level 1 are perfectly accessible; their only problem is that they are few in number.
I don’t know how to handle “accessibility” except to avoid the term, and to speak instead of “dominant” levels and “negligible” levels.
A system that is thermally isolated so that all microstates have the same energy is called microcanonical. In contrast, an object in contact with a constant-temperature heat bath is called canonical (not microcanonical). Furthermore, a system that can exchange particles with a reservoir, as described by a chemical potential, is called grand canonical (not microcanonical or canonical).
The strict definition of multiplicity applies directly to microcanonical ensembles and other strictly equiprobable distributions. Equation 9.8 applies exactly to such systems. Equation 9.8 does not apply exactly to canonical or grand-canonical systems, and may not apply even approximately. The correct thermal probability distribution is shown in figure 9.1.
There exist intermediate cases, which are common and often important. In a canonical or grand-canonical thermal system, we can get into a situation where the notion of multiplicity is a good approximation – not exact, but good enough. This can happen if the energy distribution is so strongly peaked near the most-probable energy that the entropy is very nearly what you would get in the strictly-equiprobable case. This can be roughly understood in terms of the behavior of Gaussians. If we combine N small Gaussians to make one big Gaussian, the absolute width scales like √N and the relative width scales like √N/N. The latter is small when N is large.
One should not attach too much importance to the tradeoff in the table above, namely the tradeoff between multiplicity (increasing as we move down the table) and per-microstate probability (decreasing as we move down the table). It is tempting to assume all thermal systems must involve a similar tradeoff, but they do not. In particular, at negative temperatures (as discussed in reference 27), it is quite possible for the lower-energy microstates to outnumber the higher-energy microstates, so that both multiplicity and per-microstate probability are decreasing as we move down the table toward higher energy.
You may reasonably ask whether such a system might be unstable, i.e. whether the entire system might spontaneously move toward the high-energy high-probability high-multiplicity state. The answer is that such a move cannot happen because it would not conserve energy. In a thermally-isolated system, if half of the system moved to higher energy, you would have to “borrow” that energy from the other half, which would then move to lower energy, lower multiplicity, and lower probability per microstate. The overall probability of the system depends on the probability of the two halves taken jointly, and this joint probability would be unfavorable. If you want to get technical about it, stability does not depend on the increase or decrease of multiplicity as a function of energy, but rather on the convexity which measures what happens if you borrow energy from one subsystem and lend it to another.
### 9.6.3 Not At All Equiprobable
Consider the probability distribution shown in figure 9.5. There is one microstate with probability 1/2, another with probability 1/4, another with probability 1/8, et cetera. Each microstate is represented by a sector in the diagram, and the area of the sector is proportional to the microstate’s probability.
Figure 9.5: Exponential Distribution
Some information about these microstates can be found in the following table.
State# Probability Suprise Value / bits 1 0.5 1 2 0.25 2 3 0.125 3 4 0.0625 4 5 0.03125 5 6 0.015625 6 7 0.0078125 7 8 0.00390625 8 9 0.001953125 9 10 0.0009765625 10 ... et cetera ...
The total probability adds up to 1, as you can verify by summing the numbers in the middle column. The total entropy is 2, as you can verify by summing the surprisals weighted by the corresponding probabilities. The total number of states is infinite, and the multiplicity W is infinite. Note that
S = 2 k lnW = ∞ S ≠ k lnW
(9.10)
which means that equation 9.9 definitely fails to work for this distribution. It fails by quite a large margin.
## 9.7 Discussion
Some people are inordinately fond of equation 9.8 or equivalently equation 9.9. They are tempted to take it as the definition of entropy, and sometimes offer outrageously unscientific arguments in its support. But the fact remains that Equation 2.2 is an incomparably more general, more reliable expression, while equation 9.9 is a special case, a less-than-general corollary, a sometimes-acceptable approximation.
Specific reasons why you should not consider equation 9.8 to be axiomatic include:
1. Theory says that you cannot exactly reconcile a Boltzmann probability distribution with an equiprobable distribution.
2. In practice, equation 9.8 is usually not an acceptable approximation for small systems. Thermodynamics applies to small systems, but equation 9.8 usually does not.
3. For large systems, even though equation 9.8 commonly leads to valid approximations for first-order quantities (e.g. energy, entropy, temperature, and pressure) ... it does not lead to valid results for second-order quantities such as fluctuations (energy fluctuations, temperature fluctuations, et cetera).
For a thermal distribution, the probability of a microstate is given by equation 9.1. So, even within the restricted realm of thermal distributions, equation 9.9 does not cover all the bases; it applies if and only if all the accessible microstates have the same energy. It is possible to arrange for this to be true, by constraining all accessible microstates to have the same energy. That is, it is possible to create a microcanonical system by isolating or insulating and sealing the system so that no energy can enter or leave. This can be done, but it places drastic restrictions on the sort of systems we can analyze.
• Two of the four phases of the Carnot cycle are carried out at constant temperature, not constant energy. The system is in contact with a heat bath, not isolated or insulated. A theory of “thermodynamics” without heat engines would be pretty lame.
• A great many chemistry-lab recipes call for the system to be held at constant temperature while the reaction proceeds. Vastly fewer call for the system to be held in a thermally-insulated flask while the reaction proceeds. A theory of “thermodynamics” incapable of describing typical laboratory procedures would be pretty lame.
• Even if the overall system is insulated, we often arrange it so that various subsystems within the system are mutually in equilibrium. For example, if there is liquid in a flask, we expect the left half of the liquid to be in thermal equilibrium with the right half, especially if we stir things. But remember, equilibrium involves having a shared temperature. The left half is not thermally insulated from the right half; energy is exchanged between the two halves. The microstates of the left half are not equiprobable. A theory of “thermodynamics” incapable of describing thermal equilibrium would be pretty lame.
This section exists mainly to dispel a misconception. If you do not suffer from this particular misconception, you should probably skip this section, especially on first reading.
Non-experts sometimes get the idea that whenever something is more dispersed – more spread out in position – its entropy must be higher. This is a mistake.
There is a fundamental conceptual problem, which should be obvious from the fact that the degree of dispersal, insofar as it can be defined at all, is a property of the microstate – whereas entropy is a property of the macrostate as a whole, i.e. a property of the ensemble, i.e. a property of the probability distribution as a whole, as discussed in section 2.4 and especially section 2.7.1. A similar microstate versus macrostate argument applies to the “disorder” model of entropy, as discussed in section 2.5.5. In any case, whatever “dispersal” is measuring, it’s not entropy.
Yes, you can find selected scenarios where a gas expands and does gain entropy (such as isothermal expansion, or diffusive mixing as discussed in section 11.6) … but there are also scenarios where a gas expands but does not gain entropy (reversible thermally-isolated expansion). Indeed there are scenarios where a gas gains entropy by becoming less spread out, as we now discuss:
Consider a closed system consisting of a tall column of gas in a gravitational field, at a uniform not-too-high temperature such that kT < m|g|H. Start from a situation where the density3 is uniform, independent of height, as in figure 9.6a. This is not the equilibrium distribution.
Figure 9.6: Equilibrium is Less Evenly Dispersed
As the system evolves toward equilibrium, irreversibly, its entropy will increase. At equilibrium, the density will be greater toward the bottom and lesser toward the top, as shown in figure 9.6b. Furthermore, the equilibrium situation does not exhibit even dispersal of energy. The kinetic energy per particle is evenly dispersed, but the potential energy per particle and the total energy per particle are markedly dependent on height.
There are theorems about what does get uniformly distributed, as discussed in chapter 25. Neither density nor energy is the right answer.
As another example, consider two counter-rotating flywheels. In particular, imagine that these flywheels are annular in shape, i.e. hoops, as shown in figure 9.7, so that to a good approximation, all the mass is at the rim, and every bit of mass is moving at the same speed. Also imagine that they are stacked on the same axis. Now let the two wheels rub together, so that friction causes them to slow down and heat up. Entropy has been produced, but the energy has not become more spread-out in space. To a first approximation, the energy was everywhere to begin with and everywhere afterward, so there is no change.
Figure 9.7: Grinding Wheels
If we look more closely, we find that as the entropy increased, the energy dispersal actually decreased slightly. That is, the energy became slightly less evenly distributed in space. Under the initial conditions, the macroscopic rotational mechanical energy was evenly distributed, and the microscopic forms of energy were evenly distributed on a macroscopic scale, plus or minus small local thermal fluctuations. Afterward, the all the energy is in the microscopic forms. It is still evenly distributed on a macroscopic scale, plus or minus thermal fluctuations, but the thermal fluctuations are now larger because the temperature is higher. Let’s be clear: If we ignore thermal fluctuations, the increase in entropy was accompanied by no change in the spatial distribution of energy, while if we include the fluctuations, the increase in entropy was accompanied by less even dispersal of the energy.
Here’s yet another nail in the coffin of the «dispersal» model of entropy. Consider a thermally isolated system consisting of gas in a piston pushing up on a mass, subject to gravity, as shown in figure 9.8. Engineer it to make dissipation negligible. Let the mass oscillate up and down, reversibly. The matter and energy become repeatedly more and less «disperse», with no change in entropy.
Figure 9.8: Matter and Energy Dispersing and Un-Dispersing
Here’s another reason why any attempt to define entropy in terms of “energy dispersal” or the like is Dead on Arrival: Entropy is defined in terms of probability, and applies to systems where the energy is zero, irrelevant, and/or undefinable.
As previously observed, states are states; they are not necessarily energy states.
## 9.9 Spreading in Probability Space
You can salvage the idea of spreading if you apply it to the spreading of probability in an abstract probability-space (not the spreading of energy of any kind, and not spreading in ordinary position-space).
This is not the recommended way to introduce the idea of entropy, or to define entropy. It is better to introduce entropy by means of guessing games, counting the number of questions, estimating the missing information, as discussed in chapter 2. At the next level of detail, the workhorse formula for quantifying the entropy is equation 2.2
This section exists mainly to dispel some spreading-related misconceptions. If you do not suffer from these particular misconceptions, you should probably skip this section, especially on first reading.
If you insist on using the idea of spreading, here’s an example that illustrates the idea and can be analyzed in detail. Figure 9.9 shows two blocks under three transparent cups. In the first scenario, the blocks are “concentrated” in the 00 state. In the probability histogram below the cups, there is unit probability (shown in magenta) in the 00 slot, and zero probability in the other slots, so p log(1/p) is zero everywhere. That means the entropy is zero.
In the next scenario, the blocks are spread out in position, but since we know exactly what state they are in, all the probability is in the 02 slot. That means p log(1/p) is zero everywhere, and the entropy is still zero.
In the third scenario, the system is in some randomly chosen state, namely the 21 state, which is as disordered and as random as any state can be, yet since we know what state it is, p log(1/p) is zero everywhere, and the entropy is zero.
The fourth scenario is derived from the third scenario, except that the cups are behind a screen. We can’t see the blocks right now, but we remember where they are. The entropy remains zero.
Finally, in the fifth scenario, we simply don’t know what state the blocks are in. The blocks are behind a screen, and have been shuffled since the last time we looked. We have some vague notion that on average, there is 2/3rds of a block under each cup, but that is only an average over many states. The probability histogram shows there is a 1-out-of-9 chance for the system to be in any of the 9 possible states, so ∑ p log(1/p) = log(9) .
Figure 9.9: Spreading vs. Randomness vs. Uncertainty
One point to be made here is that entropy is not defined in terms of particles that are spread out (“dispersed”) in position-space, but rather in terms of probability that is spread out in state-space. This is quite an important distinction. For more details on this, including an interactive simulation, see reference 25.
Entropy involves probability spread out in state-space, • not necessarily anything spread out in position-space, • not necessarily particles spread out in any space, • not necessarily energy spread out in any space.
To use NMR language, entropy is produced on a timescale τ2, while energy-changes take place on a timescale τ1. There are systems where τ1 is huuugely longer than τ2. See also section 11.5.5 and figure 1.3. (If this paragraph doesn’t mean anything to you, don’t worry about it.)
As a way of reinforcing this point, consider a system of spins such as discussed in section 11.10. The spins change orientation, but they don’t change position at all. Their positions are locked to the crystal lattice. The notion of entropy doesn’t require any notion of position; as long as we have states, and a probability of occupying each state, then we have a well-defined notion of entropy. High entropy means the probability is spread out over many states in state-space.
State-space can sometimes be rather hard to visualize. As mentioned in section 2.3, a well-shuffled card deck has nearly 2226 bits of entropy … which is a stupendous number. If you consider the states of gas molecules in a liter of air, the number of states is even larger – far, far beyond what most people can visualize. If you try to histogram these states, you have an unmanageable number of slots (in contrast to the 9 slots in figure 9.9) with usually a very small probability in each slot.
Another point to be made in connection with figure 9.9 concerns the relationship between observing and stirring (aka mixing, aka shuffling). Here’s the rule:
not looking looking not stirring entropy constant entropy decreasing (aa) stirring entropy increasing (aa) indeterminate change in entropy
where (aa) means almost always; we have to say (aa) because entropy can’t be increased by stirring if it is already at its maximum possible value, and it can’t be decreased by looking if it is already zero. Note that if you’re not looking, lack of stirring does not cause an increase in entropy. By the same token, if you’re not stirring, lack of looking does not cause a decrease in entropy. If you are stirring and looking simultaneously, there is a contest between the two processes; the entropy might decrease or might increase, depending on which process is more effective.
The simulation in reference 25 serves to underline these points.
Last but not least, it must be emphasized that spreading of probability in probability-space is dramatically different from spreading energy (or anything else) in ordinary position-space. For one thing, these two spaces don’t even have the same size. Suppose you have a crystal with a million evenly-spaced copper atoms. We consider the magnetic energy of the nuclear spins. Each nucleus can have anywhere from zero to four units of energy. Suppose the total energy is two million units, which is what we would expect for such a system at high temperature.
• If you are just dispersing the energy in position-space, there are a million different locations in this space. That’s just barely a seven-digit number. You just assign one one-millionth of the energy to each one. That is, you give the atoms two units of energy apiece. It’s all very simple ... but wrong.
• In probability space, the total number of states is 4 to the millionth power. That’s a number with 600,000 digits. At high temperature, the probability is spread evenly over all of these states.
Let’s be clear: a number with 600,000 digits is very much larger than a number with six or seven digits. If you imagine spreading the energy in position-space, it gives entirely the wrong picture. The physics cares about spreading the probability in a completely different space, a very much larger space. The probability is spread very much more thinly.
1
Anharmonicity can cause the average KE to be not exactly equal to the average PE, but for a crystal well below its melting point, the thermal phonon modes are not significantly anharmonic.
2
Here “states” means “microstates”.
3
If you want to be quantitative about the density, consider a smallish parcel of gas at a particular height.
[Previous][Contents] | 10,239 | 45,811 | {"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} | 3.546875 | 4 | CC-MAIN-2022-33 | latest | en | 0.888849 |
https://math.stackexchange.com/questions/1437569/do-eigenvectors-correspond-to-direction-of-maximum-scaling | 1,560,978,284,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999041.59/warc/CC-MAIN-20190619204313-20190619230313-00324.warc.gz | 509,264,870 | 34,933 | # do eigenvectors correspond to direction of maximum scaling?
Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector. I quote from : https://math.stackexchange.com/q/243553
No other vector when acted by this matrix will get stretched as much as this eigenvector.
Is the above statement always true?... For example let $$A = \left( \begin{array}{ccc} 0.578385540014544 & 0.703045745965410 \\ 0.477513363789115 & 0.922698950982510 \\ \end{array} \right)$$
The largest eigenvalue is 1.35 (approx.)
Now, consider the vector (not eigenvector) $$v = \left( \begin{array}{ccc} -0.538656963091298 \\ -0.842525178326001 \\ \end{array} \right)$$
magnitude(v) = 1.0
magnitude(A*v) = 1.373
So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)
Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.
• The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general. – Greg Martin Sep 16 '15 at 6:59
• Thank you for the answer Greg. – user1837245 Sep 16 '15 at 19:58 | 391 | 1,449 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-26 | latest | en | 0.809383 |
http://www.societyofrobots.com/robotforum/index.php?topic=11328.0 | 1,527,471,173,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00174.warc.gz | 462,651,661 | 9,322 | go_away
### Author Topic: i need your helps on Automated Guided Vehicle (AGV) robot (Read 1501 times)
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#### serendipity
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• Posts: 1
##### i need your helps on Automated Guided Vehicle (AGV) robot
« on: June 06, 2010, 08:00:56 AM »
Hi friends.
i need you helps. i have to solve below questions. anybody can help me?
best regards.
*************
Design and implement an Automated Guided Vehicle (AGV) robot with the following properties:
a. Having two wheels connected to geared bipolar stepper motors with the ratio of 1/5 (wheel/shaft) on a straight axis and two ball casters at front and backside of the robot (wheel’s diameter is 100 mm), axis length is 200 mm.
b. Shaft encoders per wheel with 8 bits accuracy per rotation.
c. Planning to traverse a trajectory for 120 degrees of a circle with radius of 500mm. and then going straight for 1000mm.
d. Three line following sensors and an electronic compass for deviation avoidance (in a cascaded loop)
e. A digital camera as radar installed on a servomotor
f. An infrared range finder with analog output connected to 10bit ADC with 0-5v analog range up to 80 cm with linear ratio
1) What are trajectory sizes in terms of pulses in t1 (curve) and t2 (straight) phases for L/R wheels?
2) What is total trajectory length?
3) Considering the equation of motion for radar turning around Z-axis for 60 degrees, present the mapping matrix with respect to the body of the robot.
4) Draw block diagram of the hybrid cascaded control mechanism for deviation avoidance.
5) Draw electronic diagram for the entire robot based on a microcontroller with 4 I/O ports (Port A is also acting as ADC)
6) Write software pseudo codes for following functions:
a. Move_Wheel(Wheel, Pulses)
b. Move_camera(Angle) //servo domain is 180 degrees | 483 | 1,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.835229 |
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# Satellite Maps Land/Structure Area Calculator Tools Useful/Accurate?
Now there are many free online area calculation tools available for free area calculation which uses satellite maps to locate and mark land/structure boundaries and using lat long coordinates of boundary points to calculate area. Are they accurate or how much are they accurate?
( Some readers queried about the tool used for this study. Satellite Map Land Area Calculator used in this study can be found at http://milloz.com/site/index.php?q=Free-Tools/Area-Measurement-Tool )
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### Satellite Maps Land/Structure Area Calculator Tools Useful/Accurate?
1. 1. Here we will verify the accuracy of the Geographical/land/plot/region Area calcultor tool, usage ofwhich is explained in this post. We will startwith a small area and then go to the more spacious structures for area measurement Let us start with Washington monument in Washington,ithas a 17*17 sq meter base,so the area is 289 square meter,when measured with our tool, the area obtained is 291 square meters,we can say that its a reasonable result considering the time spentfor measurement(less than a minute).Continue reading,you can see that the pyramid of Giza area is correctly calculated by this tool. At the end of reading,please spare some time and commenton your views about this tool Washington Monument Washington monumentmarked in the satellite maps geographical tool for area measurement The calculated area for marked region is 291 square meter,(actual 289 m),fair resultconsidering the inaccuracycan
2. 2. occurr in marking Eiffel tower paris marked for measurement Eiffel tower paris has a basementarea of 125x125 meters
3. 3. eiffel tower measured basementarea 15151 m2 while actual is 15625 m2 pyramid of giza cheops
4. 4. pyramid of khufu Egypt marked for measurementmillozarea calculator tool pyramid of khufu Egypt area measure as 13 acre using millozgeographical area tool
5. 5. St Peter's basilica St Peter's basilica marked in google maps for area measurement
6. 6. St Peter's basilica marked in google maps for area measured using area calculator Taj mahal india monument
7. 7. Tajmahal marked for area measurement Taj mahal measured area LargestTemple,Angor wat cambodia image
8. 8. Angor wat outer wall area of 203 acres Angor wat outer wall area of 203 acres calculated
9. 9. LargestFloor Area Builing,Aalsmeer Flower Auction building marked for area find Aalsmeer Flower Auction building area calculated
10. 10. Translated Hint | 669 | 2,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.670912 |
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Re: Question on Symmetry Axis Notation Convention
From:
Boaz Shaanan <[log in to unmask]>
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Boaz Shaanan <[log in to unmask]>
Date:
Thu, 14 Jun 2012 20:51:30 +0000
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Hi, Isn't there a notation 2 subscript 1 superscript x (one on top of the other) for "2-fold screw along x" for example? and likewise for y and z? Boaz Boaz Shaanan, Ph.D. Dept. of Life Sciences Ben-Gurion University of the Negev Beer-Sheva 84105 Israel E-mail: [log in to unmask] Phone: 972-8-647-2220 Skype: boaz.shaanan Fax: 972-8-647-2992 or 972-8-646-1710 ________________________________________ From: CCP4 bulletin board [[log in to unmask]] on behalf of James Stroud [[log in to unmask]] Sent: Thursday, June 14, 2012 10:06 PM To: [log in to unmask] Subject: [ccp4bb] Question on Symmetry Axis Notation Convention Hello All, I would like to discuss symmetry axes, but I'm not sure what the notation convention is. For example, I'd like to say something about a 2(1) along the x-axis, but the phrase "the 2(1) symmetry axis along x" is a bit cumbersome to repeat many times or to put in a table. So I'd like a shorthand, maybe something like "x(2_1)" (where the preceding "_" means that the "1" is subscript. Another way I like is "x_{2(1)}" (where the curly braces mean that all of "2(1)" is subscript). Does anyone know what the convention is or if there is one? Thanks in advance for any help. James
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View our service policies at https://www.jiscmail.ac.uk/policyandsecurity/ and Jisc's privacy policy at https://www.jisc.ac.uk/website/privacy-notice | 551 | 1,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-17 | latest | en | 0.887755 |
http://www.chm.davidson.edu/vce/doc/edu/davidson/chm/waveFunction/Gaussian.html | 1,537,320,467,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155814.1/warc/CC-MAIN-20180919004724-20180919024724-00166.warc.gz | 305,529,303 | 3,707 | ## edu.davidson.chm.waveFunction Class Gaussian
```java.lang.Object
edu.davidson.chm.waveFunction.WaveFunction
edu.davidson.chm.waveFunction.Gaussian
```
`public class Gaussianextends WaveFunction`
Defines a Gaussian Basis Function
Copyright (c) 2001,2009 David N. Blauch
Version:
2.0
Author:
David N. Blauch
Constructor Summary
`Gaussian()`
```Gaussian(int nbr, java.lang.String geometry)```
Creates a basis function that contains nbr gaussian functions
```Gaussian(int nbr, java.lang.String geometry, double alpha, double scale, double coef)```
Creates a basis function that contains nbr gaussian functions and defines the first gaussian function.
Method Summary
`protected void` `createArrays()`
` double` `getMaxR(double frac)`
Returns the maximum displacement from the nucleus at which the wave function has a value characterized by _fraction.
` double[]` ```getValue(double r, double theta, double phi)```
Returns the value of the basis function at r, theta, phi.
` double[]` ```getValueXYZ(double x, double y, double z)```
Returns the value of the basis function at r, theta, phi.
` void` ```setFunction(int id, double alpha, double scale, double coef)```
Adds a gaussian function to the basis function.
`protected void` `setGeometry(java.lang.String geometry)`
` void` ```setOrigin(double r, double theta, double phi)```
Sets the position of the origin for the basis function.
` void` ```setOriginXYZ(double x, double y, double z)```
Sets the position of the origin for the basis function.
Methods inherited from class edu.davidson.chm.waveFunction.WaveFunction
`getID, setID`
Methods inherited from class java.lang.Object
`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`
Constructor Detail
### Gaussian
`public Gaussian()`
### Gaussian
```public Gaussian(int nbr,
java.lang.String geometry)```
Creates a basis function that contains nbr gaussian functions
### Gaussian
```public Gaussian(int nbr,
java.lang.String geometry,
double alpha,
double scale,
double coef)```
Creates a basis function that contains nbr gaussian functions and defines the first gaussian function.
Method Detail
### getMaxR
`public double getMaxR(double frac)`
Description copied from class: `WaveFunction`
Returns the maximum displacement from the nucleus at which the wave function has a value characterized by _fraction. This method is used in ElectronDensity.class to determine the appropriate limits for the plot. The parameter _fraction might, for example, refer to a certain fraction of the maximum value of the wave function.
Specified by:
`getMaxR` in class `WaveFunction`
### getValue
```public double[] getValue(double r,
double theta,
double phi)```
Returns the value of the basis function at r, theta, phi. This value is always real.
Specified by:
`getValue` in class `WaveFunction`
### getValueXYZ
```public double[] getValueXYZ(double x,
double y,
double z)```
Returns the value of the basis function at r, theta, phi. This value is always real.
Specified by:
`getValueXYZ` in class `WaveFunction`
`WaveFunction.getValue(double, double, double)`
### setFunction
```public void setFunction(int id,
double alpha,
double scale,
double coef)```
Adds a gaussian function to the basis function. id ranges from 0 to n-1, where n is the number of gaussian functions in the basis. All basis functions must have the same geometry.
### setOrigin
```public void setOrigin(double r,
double theta,
double phi)```
Sets the position of the origin for the basis function. The getValue is called with the spherical coordinates in the system. The basis function is evaluated with the coordinates for the atom.
### setOriginXYZ
```public void setOriginXYZ(double x,
double y,
double z)```
Sets the position of the origin for the basis function. The getValue is called with the spherical coordinates in the system. The basis function is evaluated with the coordinates for the atom.
### createArrays
`protected void createArrays()`
### setGeometry
`protected void setGeometry(java.lang.String geometry)` | 971 | 4,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.483257 |
https://community.secondlife.com/forums/topic/65343-texturing-a-sim-12-and-12/?tab=comments | 1,576,038,001,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529745.80/warc/CC-MAIN-20191211021635-20191211045635-00264.warc.gz | 332,559,830 | 19,278 | # Texturing a sim, 1/2 and 1/2
## Recommended Posts
Hi all,
Anyone have the best elevation numbers to texture a sim 1/2 beach and 1/2 grass? I know its not gonna be perfect right down the middle.
Im trying to do the north half beach and the south half grass.
Thanks in advance for your help :) I will send some Ls in world for your time. :)
##### Share on other sites
Those numbers actually correspond to slants in the elevations and not flat elevations, which go from each corner. It would be more a diagonal from corner to corner and not really easy to so as 1/2 and 1/2 but you can get close however, keep in mind that everything on that end will be grass.
There is also the relationships to the height of the land.
Also note that terrain textures are randomized in the system to make them look more realistic meaning there is no set way to do this, it's mostly trial and error.
##### Share on other sites
Make textures 1 and 2 as sand, textures 3 and 4 as grass. Use Northwest low = 0, high = 255, Northeast low = 0, high = 255, Southwest low = 0, high = 1, Southeast low = 0, high = 1.
What that should do, if the terrain is flat and about 64 Meters high, is make the North half be sand, and the South half grass. The dividing line won't be perfectly straight, and will vary a bit per user and per login, since it's calculated on the fly by the viewer. Adjusting all the Southern values higher would move the dividing line South, if the terrain was perfectly flat. Adjusting the Northern 'high' values lower would move the dividing line North.
Think of a sim's terrain textures as a rubber sheet, and you can only affect the 4 corners. The 'low' value is the transition between textures 1 and 2. The 'high' value is the transition between textures 3 and 4. The transition between 2 and 3 is the midway point between the high and low values. So the values that I gave you should define the mid-point transition plane as a flat slope, from 0.5 on the South edge, to 128 on the North edge. Where the terrain intersects that plane will define the terrain textures used - grass above that plane, and sand below it.
##### Share on other sites
Hello,
And how would i make 3/4 of sim Grass and only 1/4 as sand... is there any way for that?
Thank You
## Join the conversation
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http://www.instructables.com/topics/what-is-the-square-root-of-Pi/ | 1,531,718,508,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589179.32/warc/CC-MAIN-20180716041348-20180716061348-00375.warc.gz | 482,987,390 | 7,486 | 616Views9Replies
# what is the square root of Pi? Answered
Tags:
## square root(pi) = 1.77245385
That's what everyone does. (Including me...) No one here is probably that smart to calculate such things in their head....
..no offence, to all the genuises in the world...
Smart people know how to use tools. :-)
No one anywhere. Though writing a "Monte Carlo method" program to calculate pi is simple enough to be a beginner programming exercise: Throw darts at a 2n by 2n square, determine how many of them are inside versus outside a radius-n circle centered on the square. use that to determine the (relative) area of the circle, back-calculate pi from that.. The more darts you throw, the more random they are, and the more precise you are in the in-versus-out calculation, the more this value will approach pi's actual value.
(There are better ways to calculate pi; this is just the one that's trivial to describe.)
Of course if you need more digits of root-pi, they're easy to come by...
1.7724538509055160272981674833411... (with some possible roundoff in the low digits) from a calculator.
And I'm sure you can find it to a few thousand places somewhere on the web. Whether it's accurate... | 291 | 1,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-30 | latest | en | 0.920763 |
https://math.stackexchange.com/questions/538658/having-trouble-understanding-series-and-sequences/538681 | 1,656,907,442,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00350.warc.gz | 422,161,661 | 65,858 | # Having trouble understanding Series and Sequences
So all I could get from my teachers thick accent in class today is that:
A sequence is finite and converges when bounded by x?
and
A series is infinite and diverges because no matter how small the function gets, it will never reach zero?
I'm sorry. I'm really having a hard time understanding the concept. For example. This picture... ! \begin{align} s_1 &= a_1\\ s_2 &= a_1+a_2\\ s_3 &= a_1+a_2+a_3\\ s_4 &= a_1+a_2+a_3+a_4\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vdots\\ s_n &=a_1+a_2+a_3+a_4+\cdots+a_n=\sum_{i=1}^na_i \end{align}
I don't the last part involving the sigma. How is that to the statments before it.
• A sequence is a set of numbers that (typically) have a function that relates them. A series is a sum over a sequence. Both sequences and series can be convergent or divergent depending on the function describing them. If a series converges, then so does the sequence it is based on; if a sequence diverges, then so does a series based on it. The capital sigma $\Sigma$ is the usual notation to say "$\sum_{i=1}^na_i$ is the sum from $1$ to $n$ of the numbers $a_i$, i.e., $a_1+a_2+a_3+\cdots+a_{n-1}+a_n$." Oct 24, 2013 at 21:48
• Watch these videos, and see if it helps: khanacademy.org/math/calculus/sequences_series_approx_calc/… Oct 24, 2013 at 21:53
• The last equality is true by definition of the Sigma-notation ("$\Sigma$"). But your two statements about sequences and series sound like nonsense to me, out of context anyway. Would it help to sit closer to the teacher? If not, I would suggest skipping class and reading a book instead. Oct 24, 2013 at 21:56
• "So all I could get from my teachers thick accent in class today is..." Looking at what you wrote, I guess that the teacher's accent is the least of your problems in this class. Oct 24, 2013 at 23:00
• I still don't understand the A(n-1) + An thing. I just don't understnad the logic behind that notation.
– Josh
Oct 25, 2013 at 19:21
In textbooks, the convergence of a sequences is often discussed before series.
If a sequence is monotonous, say $a_{n+1} \geq a_n$, and bounded from above, yes, it will converge. This is proven by showing it has an upper bound (reformulation of "bounded from above"), and using the fact that every bounded set has a smallest upper bound (called supremum). This smallest upper bound of the sequence is its limit.
Next, by considering the partial sums of a series, you will notice that the partial sums form a sequence. If this sequence of partial sums converges, the series is said to converge.
Your textbook seems to be confusing for you for not repeating the partial sum notation in the lines you quote. Just consider that writing the partial sum with a sum-sign is the shorthand for the many terms of the sum added together. The $s_n$ form the sequence of partial sums of the $a_n$.
If a sequence of partial sums $s_n$ converges, it follows that the sequence of its terms $a_n$ converge to 0. Proof. For all n>N, $|s_n - s_{n+1}|$ will be arbitrarily small, hence $|a_{n+1}|$ will be arbitrarily small, i.e., the sequence of $a_n$ converges to 0.
Caution. The reverse is not true. If the sequence $(a_n)_{n \geq 1}$ converges to 0, then it does NOT follow that the sequence of its partial sums $(s_n)_{n \geq 1}$ converges.
Example. The sequence $(a_n)_n = (1/n)_n$ converges to 0, however the sequence of the partial sums $s_N = \sum_{n=1}^N a_n$ is unbounded (hence, the series diverges). Note that $\sum_{n=2^{k}+1}^{2^{k+1}} 1/n > 1/2$ for all k. | 1,021 | 3,549 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.926832 |
https://www.theglobeandmail.com/news/toronto/the-problem-of-p-versus-np/article1377814/ | 1,569,131,854,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575168.82/warc/CC-MAIN-20190922053242-20190922075242-00462.warc.gz | 1,030,603,096 | 84,502 | In 1971, Dr. Stephen Cook, a young University of Toronto professor in the fledgling field of computer science, posed a theoretical problem so intractable it has become the subject of a \$1-million prize. Since then, only a handful of credible solutions have been posed. All of them fell short. This month, one man caused a commotion after he announced to experts in the field that he had solved the problem known as P vs. NP.
The proof
Vinay Deolalikar, a Delhi-born mathematician at Hewlett-Packard, sent Dr. Cook and two dozen other experts in the field an e-mail, writing, "I am pleased to announce a proof that P is not equal to NP, which is attached." The paper was more than 100 pages long. Dr. Cook was excited. In 40 years, "I can think of only a couple of other attempts of people who've thought they've proved it," Dr. Cook said. "Most of them you can dismiss very easily - they're not really mathematicians. But this one was much more serious."
The problem
Clay Mathematics Institute, the Cambridge, Mass.-based academy that offers the \$1-million prize, describes the problem with this example: Imagine you are trying to figure out housing for 400 university students. Only 100 will get rooms, and the dean has given you pairs of incompatible students she doesn't want living together. The number of ways to choose 100 students out of the 400 applicants is greater than the number of all the atoms in the universe, and no known supercomputer can solve it. But any list you do devise will be simple to check by making sure no incompatible student pairs appear on it.
This type of question is known as P vs. NP, computer programming-speak for "easy to find" versus "easy to check." In a nutshell, it asks: Are there problems for which an answer is impossible to generate through computing, but for which an answer, if given, is easily checkable? Proofs have been offered on either side - the P does equal NP, and that P does not equal NP - but none in 40 years have stuck.
The stakes
P vs. NP may be the holy grail of a theoretical field, but it has enormous practical implications. When a consumer buys a product over the Internet using a credit card, that transaction is secured with a seemingly unbreakable mass of coding. "The security assumption is based on an unproven fact that no computer can do it," Dr. Cook explains. Modern-day cryptography rests on the same assumption. Anyone who proves that P does equal NP would undermine the foundations of code-breaking, e-commerce and digital privacy.
The prof
As a math graduate student at Harvard in the 1960s, Dr. Cook became interested in the fledgling field of computer science. "There was a huge amount of interest in what [computers]can do and can't do," he says. When the University of Toronto inaugurated one of the first computer-science departments in the world, Dr. Cook moved north, and soon after, in 1971, he formulated P vs. NP. He has been teaching and studying computational complexity at the university ever since. Among other accolades, he won the 1982 Turing Award, considered the Nobel Prize of computer science.
The prize
In 2000, the Clay Mathematics Institute chose seven of the longest-standing math problems and dubbed them "millennium problems." Each comes with a \$1-million prize. One was P vs. NP, and another, the Riemann Hypothesis, was formulated in 1859. A third, the Poincaré Conjecture, was famously solved in 2003 by Grigoriy Perelman, an elusive Russian mathematician who turned down his cash prize in July.
The unravelling
After the initial commotion over Mr. Deolalikar's paper died, Dr. Cook and other experts took a crack at the proof. Problems began to emerge. "As I looked at it more, I could see that it had symptoms that didn't make it look too promising," Dr. Cook said. Other professors put the problem up on blogs, and one wiki created by quantum physicist Michael Nielsen. On Aug. 13, computer scientist Scott Aaronson, who had bet \$200,000 the proof wouldn't fly, wrote on his blog that "a clear consensus has emerged that the proof, as it stands, is fatally flawed."
Mr. Deolalikar has submitted a revised proof, but Dr. Cook believes the jig is up. Does he believe P vs. NP will ever be solved? "Yes, I do, I do - eventually. But not very soon. I think it really is a hard problem. It's becoming increasingly clear, because so many top mathematicians have tried to solve it."
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• Avoid the use of toxic and offensive language | 1,336 | 5,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-39 | longest | en | 0.975587 |
https://www.examples.com/maths/derivative-of-cos-square-x.html | 1,721,337,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00359.warc.gz | 636,158,721 | 22,225 | # Derivative of Cos Square x (Cos^2x)
Created by: Team Maths - Examples.com, Last Updated: June 10, 2024
## Derivative of Cos Square x (Cos^2x)
The derivative of cos²(𝑥) is a fundamental concept in calculus, intertwining with various branches of mathematics. Involving algebra and the power rule, the differentiation process utilizes the chain rule to handle the square of the cosine function, symbolically integral to both rational and irrational numbers. This derivative calculation lays the groundwork for more complex equations in fields such as statistics and numerical methods, including the least squares method. Integer operations and the square and square roots also play a role in understanding the behavior of this trigonometric function across different numerical systems.
## What is Derivative of Cos Square x?
The derivative of cos²(𝑥), where the function is squared, is calculated using the chain rule resulting in −2cos(x)sin(x). This expression can also be represented as −sin(2𝑥), demonstrating the application of trigonometric identities in calculus.
## Derivative of Cos²x Formula
The derivative of cos²(𝑥) is given by the formula:
d/dx(cos²(x)) = 2cos(x)sin(x)
This can also be expressed using the double angle identity for sine:
d/dx(cos²(x)) = −sin(2x)
## Derivative of Cos Square x Using the Chain Rule
### 1. Identify the Outer and Inner Functions:
• Outer function: 𝑢² where 𝑢 is the inner function.
• Inner function: 𝑢 = cos(𝑥).
### 2. Differentiate the Outer Function:
• Derivative of 𝑢² with respect to 𝑢 is 2𝑢.
### 3. Differentiate the Inner Function:
• Derivative of cos(𝑥) with respect to 𝑥 is −sin(𝑥).
### 4. Apply the Chain Rule:
Multiply the derivative of the outer function by the derivative of the inner function:
d//dx(cos²(x)) = −2(cos(x)).(-sin(x))
### 5. Simplify the Expression:
• This simplifies to −2cos(𝑥)sin(𝑥).
### Final Expression
d/dx(cos²(x)) = −2cos(x)sin(x)
## Derivative of Cos²x Using the First Principle
To find the derivative of cos²(𝑥) using the first principles of calculus, also known as the definition of the derivative, follow these step-by-step instructions:
## Derivative Using First Principle
### Formula for Derivative Using First Principle
f'(x) = limₕ→₀f(x+h)−f(x)/h
### Apply to cos²2(𝑥):
Function Definition:
𝑓(𝑥) = cos²(𝑥)
Function Value at 𝑥+ℎ:
𝑓(𝑥+ℎ) = cos²(𝑥+ℎ)
Difference Quotient:
cos²(𝑥+ℎ)−cos2(𝑥)/ℎ
Simplify Using Trigonometric Identities:
Use the cosine addition formula: cos(𝑥+ℎ) = cos(𝑥)cos(ℎ)−sin(𝑥)sin(ℎ)
Then apply cos²(𝑥+ℎ) = (cos(𝑥)cos(ℎ)−sin(𝑥)sin(ℎ))²
Expand and Simplify:
Due to complexity, this involves expanding (cos(𝑥)cos(ℎ)−sin(𝑥)sin(ℎ))² and then simplifying under the assumption that ℎ is very small, so cos(ℎ) = 1 and sin(ℎ) = ℎ.
Apply Limits:
Simplify the expanded expression and apply the limit ℎ→0. The terms involving ℎ will approach zero, leaving terms that contribute to the derivative.
### Computation
Continuing the computation:
limₕ→₀(cos(𝑥)cos(ℎ)−sin(𝑥)sin(ℎ))²−cos²(𝑥)/ℎ
Expanding and substituting cos(ℎ) = 1 and sin(ℎ) = ℎ:
limₕ→₀(cos(𝑥)⋅1−sin(𝑥)⋅ℎ)²−cos²(𝑥)/ℎ
=limₕ→₀(cos²(𝑥)−2cos(𝑥)sin(𝑥)ℎ+sin²(𝑥)ℎ²)−cos²𝑥)/ℎ
=limₕ→₀−2cos(𝑥)sin(𝑥)ℎ+sin²(𝑥)ℎ²/ℎ
=limₕ→₀(−2cos(𝑥)sin(𝑥)+sin²(𝑥)ℎ)
=−2cos(𝑥)sin(𝑥)
### Final Result
𝑑/𝑑𝑥(cos²(𝑥)) = −2cos(𝑥)sin(𝑥)
## Derivative of Cos Square x Using Product Rule
To find the derivative of cos²(x) using the product rule, consider cos²(𝑥) as cos(𝑥)⋅cos(𝑥). The product rule states that if 𝑢 and 𝑣 are functions of 𝑥, then the derivative of their product 𝑢𝑣 is given by 𝑢𝑣′+𝑣𝑢′.
### Applying the Product Rule to cos²(x)
Step 1: Define the Functions
• Let 𝑢(𝑥) = cos(𝑥) and 𝑣(𝑥) = cos(x).
Step 2: Differentiate Each Function
• The derivative of 𝑢u with respect to 𝑥 is 𝑢′(𝑥) = −sin(𝑥).
• Similarly, 𝑣′(𝑥) = −sin(𝑥).
Step 3: Apply the Product Rule
• Plug the derivatives into the product rule formula:(cos(𝑥)⋅cos(𝑥))′ = cos(𝑥)(−sin(𝑥))+(−sin(𝑥))cos(𝑥)
• Simplify the expression:−sin(𝑥)cos(𝑥)−sin(𝑥)cos(𝑥) = −2sin(𝑥)cos(𝑥)
Final Expression
𝑑/𝑑𝑥(cos²(𝑥)) = −2cos(𝑥)sin(𝑥)
## Example 1
Calculate 𝑑/𝑑𝑥[cos²(𝑥)] when 𝑥 = 𝜋/4.
• Substituting 𝑥 = 𝜋/4 into the derivative, we use −sin(2𝑥): −sin(2×𝜋/4)=−sin(𝜋/2)=−1
• Thus, the derivative at 𝑥 = 𝜋/4 is −1.
## Example 2
Evaluate the rate of change of cos²(𝑥) at 𝑥 = 𝜋.
• Using the derivative formula −sin(2𝑥): −sin(2×𝜋)=−sin(2𝜋)=0
• Therefore, the derivative at 𝑥 = 𝜋 is 0, indicating no rate of change at this point.
## Example 2
Find the derivative of cos²(𝑥) at 𝑥 = 3𝜋/2.
• Substituting 𝑥 = 3𝜋/2 into the derivative formula, we get: −sin(2×3𝜋/2) = −sin(3𝜋) = 0
• Again, the derivative is 0, showing no rate of change at 𝑥 = 3𝜋/2.
## Why do we use the double angle identity in finding the derivative of cos²(x)?
The double angle identity (sin(2𝑥)=2sin(𝑥)cos(𝑥) is used to simplify the expression derived from the chain rule. By recognizing that −2cos(𝑥)sin(𝑥) can be rewritten as −sin(2𝑥), the derivative not only becomes easier to compute but also more concise, facilitating further calculations and integrations.
### How is the derivative of cos²(x) used in practical applications?
The derivative of cos²(x) is particularly useful in fields such as physics and engineering, where it helps analyze phenomena that involve oscillations and wave patterns, such as sound waves and light waves. It allows researchers and professionals to determine the rate of change of these waves at any given point, which is crucial for designing systems that operate efficiently under wave-based mechanisms.
## Is there a general method to find derivatives of squared trigonometric functions?
Yes, a general method to find derivatives of squared trigonometric functions involves using the chain rule. First, take the derivative of the trigonometric function itself, then multiply it by the derivative of the outer function, which in cases of squared functions is 2 times the original function. Applying this method systematically will yield the derivative for any squared trigonometric function.
Text prompt | 2,271 | 6,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-30 | latest | en | 0.839024 |
http://www.freepdfbook.com/advanced-engineering-mathematics-pdf/ | 1,610,943,834,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00708.warc.gz | 135,813,507 | 28,042 | #### Advanced Engineering Mathematics by Erwin Kreyszig gives a detailed, complete, and up-to-date solution of engineering mathematics. it’s intended to introduce students of engineering, physics, mathematics, computer engineering, and related fields to those areas of applied maths that are most relevant for solving practical problems.
Advanced Engineering mathematics is the sole prerequisite for engineering students. (However, a concise refresher of basic calculus for the scholar is included on the inside cover and in Appendix 3.) The Advanced Engineering Mathematics is arranged into seven parts as follows:
A. Ordinary Differential Equations (ODEs) in Chapters 1–6
B. Linear Algebra. Vector Calculus. See Chapters 7–10
C. Fourier Analysis. Partial Differential Equations (PDEs). See Chapters 11 and 12
D. Complex Analysis in Chapters 13–18
E. Numeric Analysis in Chapters 19–21
F. Optimization, Graphs in Chapters 22 and 23
G. Probability, Statistics in Chapters 24 and 25.
Advanced engineering mathematics, provides a comprehensive, thorough, and up-to-date treatment of engineering mathematics. it’s intended to introduce students of engineering, physics, mathematics, computer science, and related fields to those areas of applied math that are most relevant for solving practical problems. A course in elementary calculus is the sole prerequisite.
The parts of the book are kept independent. (If so needed, any prerequisites—to the extent of individual sections of prior chapters—are clearly stated at the opening of each chapter.) The book has helped to pave the way for this development of engineering mathematics. This remake will prepare the scholar for the present tasks and therefore the future by unique approach to the areas listed above. We provide the material and learning tools for the scholars to urge an honest foundation of engineering mathematics which will help them in their careers and in further studies.
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## Advanced Engineering Mathematics 10th Edition PDF book
Book Description:
Author:Erwin Kreyszig
Hardcover: 1,283 pages
Publication Date: August 16, 2011
Language: English
File Type:PDF
File Size:21.4 MB
ISBN-10: 0470458364
ISBN-13: 978-0470458365
Edition: 10
Book PDF
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The tenth edition of this best selling text includes examples in more detail and more applied exercises of Mathematics; both changes are aimed at making the material more relevant and accessible to readers. Kreyszig introduces engineers and computer scientists to advanced math topics as they relate to practical problems.
Solutions Mannual | 674 | 3,317 | {"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} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.912383 |
https://e2e.ti.com/support/amplifiers-group/amplifiers/f/amplifiers-forum/1114938/tipd164-design-optimization-query?tisearch=e2e-sitesearch&keymatch=TIPD164 | 1,726,095,913,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00357.warc.gz | 206,759,151 | 27,630 | If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.
TIPD164: Design Optimization Query
Part Number: TIPD164
In TIPD164 design, they have used INA826 and INA159 for converting voltage and current source to required range for ADS1248 IC . Why cant we use INA826 directly using the reference design in the INA826 data sheet for converting +/-20mV and +/-10V to +/-2.5+/-2.3V range.
• Hi,
the advantage of the circuit with the INA159 is that the INA159 is powered from the same supply voltage as the ADS1248. So there's no need to protect the ADS1248. When directly connecting the output of INA826 to the ADS1248 inputs, on the other hand, the much higher supply voltage of INA826 demands the installing of a protection scheme at the inputs of ADS1248. So, in any case you will need to increase R6 and R7 to limit the input currents of ADS1248 to below 10mA, better less.
Kai
• Hello,
I agree with Kai here - in addition to his point about the supply voltage, the INA159 will attenuate by 1/5. Since the input voltage specification is +/- 10 V (20 V pp), and the maximum input voltage to the ADS1248 is 5 V, it is necessary to attenuate the input to the appropriate input voltage for the ADS1248 by at least 1/4. INA159 offers precision attenuation with integrated matched resistors to achieve 1/5 attenuation. Since the minimum gain of INA826 is 1, it cannot be used for attenuation.
If you are looking into a new design with INA826, you should also consider INA823. INA823 is a newer version of INA826 that offers lower offset and lower supply operation.
Regards,
Mike
• Hi Michael,
I think Narashiman is referring to the circuit shown in figure 63 of datasheet of INA826 with the voltage divider at the input.
Kai
• Hi Kai,
OK, that makes sense. But as you know, one must be aware that the input impedance will decrease and the gain error will be degraded if matched resistors are not used on the inputs for that case.
Narashiman, I'm going to close this thread, but let us know if there are any more questions.
Regards,
Mike
• Thanks a lot for your response. Will take this into my consideration for my design development. | 543 | 2,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.855072 |
https://www.iaacblog.com/programs/algorithmic-emergence-canopy/ | 1,702,039,700,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100745.32/warc/CC-MAIN-20231208112926-20231208142926-00276.warc.gz | 877,374,615 | 11,506 | ## Concept – Canopy
The Sierpiński triangle, also called the Sierpiński gasket or Sierpiński sieve, is a fractal attractive fixed set with the overall shape of an equilateral triangle, subdivided recursively into smaller equilateral triangles. Originally constructed as a curve, this is one of the basic examples of self-similar sets.
The fractal canopy, a type of fractal tree, is one of the easiest-to-create types of fractals. Each canopy is created by splitting a line segment into two smaller segments at the end, and then splitting the two smaller segments as well, and so on, infinitely.Canopies are distinguished by the angle between concurrent adjacent segments and ratio between lengths of successive segments.
The design concept is basing on the solar radiation simulation of the Mies van der Rohe pavilion, by applying both Sierpiński triangle and fractal canopy, using recursion to generate a Branching Structure for shading.
Concept Image credits: Sierpinski pyramid, en.wikipedia, JJLudemann
## Pseudo Code
The main modeling process are three steps:
1. First, simulate the solar radiation condition of design area.
2. Then, branching the meshes based on condition.
3. Finally, execute the recursion function based on the branches.
Pseudo Code
## Script
Grasshopper Script
Python Code
## Design Process
Generation Process
1. Design Area: The input data is the design area, which will further used for solar radiation simulation.
2. Solar Radiation Simulation: Based on the EPW file, simulate the annual direct radiation on design area.
3. Branching: After that, by setting the branching condition and filtering the values, the meshes and corresponding radiation values are distributed into several branches.
4. Based Lines: Generate the based lines from each meshpoint.
5. Recursion: Execute and end the recursion based on the loop condition.
6. Generate Meshes: Finally, generate the columns and panels by input lines.
## Recursive Logic
Recursive Logic
1. At the beginning, generate a line at the center of the cell, this is the based geometry to recursion. In addition, a radiation value is linking to this line. The line will recurves based on the branching condition setting.
2. Based on the cell size, total height of the unit and the recursion times, calculate the corresponding coordinate value of the translate vector. Move the point with the vector and generate new lines for next loop.
3. Then repeat the function that defined in the previous step, until match the ending condition.
4. Finally, recursion end, obtain the final result.
Catalogue
Render
## Vedio
<p>
</p>
Algorithmic Emergence // The Canopy is a project of IaaC, Institute for Advanced Architecture of Catalonia developed at Master in Advanced Architecture in 2020/2021 by:
Student: Liang Mayuqi
Faculty: Rodrigo Aguirre
Assistant: Ashkan Foroughi Dehnavi | 627 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.88405 |
https://nz.education.com/lesson-plan/terrific-tangrams/ | 1,603,257,577,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107875980.5/warc/CC-MAIN-20201021035155-20201021065155-00166.warc.gz | 465,928,968 | 32,495 | Guided Lessons
# Terrific Tangrams!
Introduce your class to composite shapes with this lesson that reviews geometric shapes and makes use of tangram puzzles. This lesson will give your students a strong base for learning geometry later on in their school careers.
Need extra help for EL students? Try theComposing ShapesPre-lesson.
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Need extra help for EL students? Try theComposing ShapesPre-lesson.
This maths lesson introduces first graders to composite shapes in a fun way, with the use of tangrams! A composite shape is made up of two or more shapes, and a tangram is a geometric puzzle consisting of many different shapes. An array of tangram cards is included for children to build shapes and patterns, which is a great way to strengthen fine motor skills. This lesson provides young learners with a review of two-dimensional shapes, and an introduction to composite figures, both of which are important concepts that are fundamental to learning geometry.
• Students will review basic geometric shapes and learn about composite shapes.
The adjustment to the whole group lesson is a modification to differentiate for children who are English learners.
(5 minutes)
• Review the names of geometric shapes with students. Either show students plastic shapes and have the class name them or draw and label the shapes on the board. Shapes should include rectangle, triangle, circle, hexagon, square, and trapezoid.
• Describe TangramsTo the class as a picture puzzle that is solved by fitting different shapes together.
(5 minutes)
• Explain that when shapes are put together they can make up bigger, different shapes, or even pictures.
• Show the class a tangram and go through how to solve it while it is on display for the class to see.
(10 minutes)
• Have students practise making different shapes by putting different geometric shapes together with a group.
• Show students the Plane Figures worksheet and explain how to complete it.
(20 minutes)
• Pass out tangrams to students and allow them to complete the puzzle.
• When students have finished their tangram, and checked it with you, give them a Plane Figure worksheet to complete.
Enrichment:Have students use the advanced tangrams work pages. Alternatively, these students can complete more than one tangram.
Support:Have students partner up with another student and work together on the beginners tangrams.
(5 minutes)
• Students will be assessed by the shapes they create using the tangrams and their completed worksheet.
(5 minutes)
• Have students share different shapes they made with the tangrams.
• Ask the class for any final questions they may have. | 567 | 2,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2020-45 | latest | en | 0.939696 |
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By: veronica2566 (offline) Friday, September 24 2010 @ 04:07 AM CDT (Read 2037 times)
veronica2566
Hi all
For the life of me, I am always having problems with these kind of problem sums. Hope one of you can help :
After spending an equal amount of money, Cally had twice as much money as Anne.
How much money did each of them spend?
Ans : \$135
Thanks a lot!
SAH Mom
Junior
Registered: 06/23/10
Posts: 24
By: angietoh (offline) Friday, September 24 2010 @ 09:47 AM CDT
angietoh
200 - 165 = 35
35 x 2 = 70
Cally has \$70 in the end.
200 - 70 = 130
They each spent \$130.
Concept -- Constant Difference
Newbie
Registered: 09/23/10
Posts: 14
Location: SG
By: veronica2566 (offline) Friday, September 24 2010 @ 10:37 PM CDT
veronica2566
Thanks, Angie! So simple.... easy steps too!
Junior
Registered: 06/23/10
Posts: 24
By: kellywww (offline) Monday, September 27 2010 @ 01:19 AM CDT
kellywww
Hi Veronica, don't trust all answers in the answer sheet. Some of them are wrong
Hi Angie, yes the answer is 130.
Junior
Registered: 12/31/06
Posts: 21
By: mummyof2 (offline) Thursday, October 21 2010 @ 08:37 PM CDT
mummyof2
Quote by: angietoh
200 - 165 = 35
35 x 2 = 70
Cally has \$70 in the end.
200 - 70 = 130
They each spent \$130.
Concept -- Constant Difference
Hi Angie, can you explainwhy the difference of 35 is actually half of Cally's final amount? So sorry but need to know the concept so that i can explain to my kid.. my maths is terrible.
thanks
Newbie
Registered: 10/07/09
Posts: 1
All times are CST. The time is now 03:21 am.
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https://doodlelearning.com/us/math/skills/multiplication/8-times-tables | 1,723,318,527,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00576.warc.gz | 178,907,495 | 63,356 | # How to learn your 8 times tables
Learn the 8 times table with help from an elementary school teacher. Start simple, grasp basics, and enjoy fun tricks and songs for quick mastery.
Author
Michelle Griczika
Published
March 2024
# How to learn your 8 times tables
Learn the 8 times table with help from an elementary school teacher. Start simple, grasp basics, and enjoy fun tricks and songs for quick mastery.
Author
Michelle Griczika
Published
March 2024
# How to learn your 8 times tables
Learn the 8 times table with help from an elementary school teacher. Start simple, grasp basics, and enjoy fun tricks and songs for quick mastery.
Author
Michelle Griczika
Published
March 2024
Key takeaways
• Basics First: Begin with simpler times tables (2s, 5s, 10s) to ease into multiplication with patterns and skip counting
• Conceptual Understanding: Solidify multiplication understanding with strategies like skip counting and the Commutative Property before tackling the 8 times table
• Engaging Tricks: Use doubling tricks, pattern recognition, and skip counting games to make learning the 8 times table fun and effective
•
When children start learning multiplication facts, a common way to introduce them is through sets organized around one factor. In other words, students might learn all of the 2s multiplication facts, all of the 4s multiplication facts, and so on.
While the “best” order for learning sets of multiplication facts can be debated, there is a fairly common understanding that some facts are easier to remember than others and should, therefore, be taught earlier.
For example, the 10s multiplication facts are easy for students because of the pattern with the 0 in the product and easy skip counting. Students learn how to count by tens early in elementary school and can use that knowledge to multiply—they must be shown this connection! Similarly, the 2s and 5s times tables are straightforward for students because of their predictable patterns and easy skip counting.
## Introducing the 8 Times Tables
On the other hand, the 8 times table is usually one of the later sets for students to learn. While 8 is not a prime number such as 7, it is a relatively high number compared to 2, 3, and 4, which makes it more challenging for students to skip count by 8.
This is an essential point because students can efficiently utilize skip counting (or repeated addition) as a strategy to solve for an unknown multiplication fact. If a student tries to solve 8 x 4 but can remember 8 x 3 is 24, all they have to do is add another “group of” 8 to 24 to solve 32 as the answer. However, adding 8 to a number is not typically something students can do as quickly as adding 2, 3, or 4.
Therefore, I advise teaching other sets of multiplication facts before introducing students to the 8 times tables. Times tables that easily incorporate repeated addition/skip counting or those that have predictable patterns allow students to grasp the actual meaning of multiplication first.
The 2s, 5s, 10s, and 3s are the best sets for students to begin with when learning multiplication. Knowledge of these times tables lay the foundation for learning more challenging facts, such as the times tables for 8.
## 8 times table chart
First things, first. Take a look at the 8 times table up to 10:
1 x 8 = 8 8 x 1 = 8 2 x 8 = 16 8 x 2 = 16 3 x 8 = 24 8 x 3 = 24 4 x 8 = 32 8 x 4 = 32 5 x 8 = 40 8 x 5 = 40 6 x 8 = 48 8 x 6 = 48 7 x 8 = 56 8 x 7 = 56 8 x 8 = 64 8 x 8 = 64 9 x 8 = 72 8 x 9 = 72 10 x 8 = 80 8 x 10 = 80
## Tips for Learning Your 8 Times Tables
It is important for students to have a conceptual understanding of multiplication before they try to learn the 8 times table. The 8 times table chart facts can be challenging for students because 8 is a relatively large number for a factor in beginning multiplication. The 8 times table does not have predictable patterns to utilize such as the 5 and 10 times tables.
Therefore, having a concrete understanding of the concept of multiplication (which is gained through learning other times tables first, mastering repeated addition/skip counting, etc) gives students a way to calculate the answer if they cannot remember any of the 8s facts. As a teacher, I cannot stress this enough! If children know they can utilize repeated addition to find the answer to multiplication problem, it prevents much frustration and anxiety around math. This is also the reason rote memorization is not effective—if a student has no strategy for backing into an asnwer there is no way to solve the problem if they can’t recall the product.
Now let’s dive in to my best tips for teaching the eight times table.
### Commutative Property of Multiplication
A tip that can help make the 8 times table less overwhelming is reminding them of the Commutative Property of Multiplication. This multiplication property states that changing the order of the numbers being multiplied (the factors) does not change the answer (the product).
Once students learn 3 x 8, they also learn 8 x 3. Remembering this simple rule reduces the number of facts in the 8 times table chart from 20 to 10, which automatically makes it less overwhelming!
The Commutative Property of Multiplication is particularly beneficial when learning the 8 times table. Students have typically learned at least a few other times tables by the time they start learning the 8s. You can show your child the 8 times table chart and circle the facts they already know!
For example, suppose your child has already mastered the 2s, 5s, and 10s times tables. In that case, you can identify 2 x 8, 8 x 2, 5 x 8, 8 x 5, 10 x 8, and 8 x 10 as facts your child has already learned.
### 4 doubled
As students continue learning the 8 times tables, a few “tricks” or patterns might help. The first 8 times table trick is particularly helpful for students who can easily double numbers and already know the 4s times table.
Since 8 is the same as 4 doubled, students can use this relationship to solve the facts for the 8 times table. However, students must also understand that only one factor should be doubled. Doubling one factor doubles the answer, whereas doubling both factors quadruples it.
For example, if students try to solve 8 x 3, they can think of 4 x 3 = 12. Since the factor of 3 stays the same, but you can double 4 to get 8, the student can double 12 to get the correct answer of 8 x 3 = 24.
Another example is how students can think of 7 x 4 = 28 when solving 7 x 8. Since you can double 4 to get 8, you can double the answer of 28 to get the product for 7 x 8 = 56.
## Explore multiplication with DoodleMath
Want to practice your 8 times tables? DoodleMath is an award-winning math app that’s proven to double a child’s rate of progression with just 10 minutes of use a day!
Filled with fun, interactive questions aligned to state standards, Doodle creates a unique work program tailored to each child’s needs, boosting their confidence and skills in math. Try it free today!
### Other 8 Times Table Tricks
The following two 8 times table tricks are more complicated to visualize as they do not involve number sense or mathematical relationships. However, some students still find them helpful since all brains are different!
First, every fifth multiple of 8 ends in the same digit. For example, the 2nd multiple of 8 is 16. Five multiples later, which would be 7 x 8, the answer is 56. The numbers 56 and 16 both have a 6 as their last digit.
Multiple of 8 Products 1st and 6th—1 x 8 and 6 x 8 8 and 48 2nd and 7th—2 x 8 and 7 x 8 16 and 56 3rd and 8th—3 x 8 and 8 x 8 24 and 64 4th and 9th—4 x 8 and 9 x 8 32 and 72 5th and 10th — 5 x 8 and 10 x 8 40 and 80
Another 8 times table trick is a pattern in the sum of the digits for each product. For the first 5 facts, the sum of the digits in the product starts at 8 and then decreases by 1. Starting at 6 x 8 through 10 x 8, the sum of the digits in the product begins at 12 and decreases by 1 until it returns to 8. Please see below:
Multiplication Fact Products Sum of Digits 1 x 8 8 8 2 x 8 16 1 + 6 = 7 3 x 8 24 2 + 4 = 6 4 x 8 32 3 + 2 = 5 5 x 8 40 4 + 0 = 4 6 x 8 48 4 + 8 = 12 7 x 8 56 5 + 6 = 11 8 x 8 64 6 + 4 = 10 9 x 8 72 7 + 2 = 9 10 x 8 80 8 + 0 = 8
Finally, skip counting songs are a quick, easy, and fun way to teach the 8 times table. Skip counting songs are great when teaching multiplication because they emphasize the idea of multiplication, which means “groups of” a number. For example, 4 x 8 is the same as 4 “groups of” 8. Students can count by 8 four times to find the answer of 32.
My students’ favorite song was this one. It was a bit of a tongue twister at first, but they loved trying to do it faster and faster each time!
A note on drawing models: even though we want our students to have fallback strategies, I want to acknowledge that drawing models are not the most efficient way to get a product for multiplication problems with larger factors such as 7, 8, and 9. Utilizing repeated addition and skip counting is much more effective for teaching the 8 times tables which will ultimately need to be memorized.
Ready to give it a go?
Put your knowledge to the test with these no-risk practice problems to get you ready for the classroom!
## 8 times table practice problems
8 x 5 =
7 x 8 =
__ x 8 = 16
Eli buys 8 games. Each game costs \$9. How much money did Eli spend on all of the games?
Brayden is on the computer for 32 minutes. He plays 4 different games for an equal amount of time. How many minutes did he spend playing each game?
Want more practice? Our math help app is a great resource for times table practice!
In conclusion, the 8 times table can be tricky to learn, but these tips and tricks can make it more manageable for your child or students. Using a combination of the tricks, skip counting, the Commutative Property, songs, and doubling the 4s multiplication facts, your children can conquer the 8s times table.
Lesson credits
Michelle Griczika
Michelle Griczika is a seasoned educator and experienced freelance writer. Her years teaching first and fifth grades coupled with her double certification in elementary and early childhood education lend depth to her understanding of diverse learning stages. Michelle enjoys running in her free time and undertaking home projects.
Michelle Griczika
Michelle Griczika is a seasoned educator and experienced freelance writer. Her years teaching first and fifth grades coupled with her double certification in elementary and early childhood education lend depth to her understanding of diverse learning stages. Michelle enjoys running in her free time and undertaking home projects.
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# Compose Tetris
#### In the Beginning there was Function Composition
So let's talk composition. As programmers we see composition constantly. It's there even if you're not aware. If we have two functions `f` and `g` and we apply them one after the other, we're doing function composition.
``````f (g (x))
``````
or in mathematical notation:
``````(f β g)(x)
``````
In Haskell this is so common that we have an operator for it:
``````(.) :: (b -> c) -> (a -> b) -> a -> c
(.) f g x = f (g x)
``````
It's fun to note that composition is associative. Meaning if we have three functions `f`, `g`, and `h` then it doesn't matter what order we compose them in:
`````` (f . g) . h
= f . (g . h)
= f . g . h
``````
And here's an example use of function composition in Haskell:
``````capitaliseAllWords :: String -> String
capitaliseAllWords = unwords . map capitaliseWord . words
where capitaliseWord = map toUpper
``````
We can read this right-to-left: splitting the incoming `String` into separate words, i.e. `[String]`, then `map capitaliseWord` over this list, and concatenate the list back into `String`. A simple and concise example to wet our whistles.
#### Take me Higher and Higher!
We can take the idea of composition to the next level. Before this though, we have to quickly go over higher-kinded types. Higher-kinded types are types that have a parameter. This means they are expecting types so that they can be a regular, happy type. Think of it like a type function:
``````Type -- Regular type
Type -> Type -- Higher-kinded type expecting one Type
``````
Without further ado, let's look at examples! The first higher-kinded type that we will look at is the `Maybe` type. It goes by a few names but here is its definition in Haskell:
``````data Maybe a = Nothing | Just a
``````
When we look at the left-hand side of the `=` we see that the `Maybe` type has a parameter called`a`. We can fill the `a` with other types, e.g. `Maybe Int`, `Maybe String`, `Maybe (Maybe Bool)`.
Another common type that is also higher-kinded is the list type:
``````data [a] = [] | a : [a]
``````
Again, we have an `a` parameter that the list is waiting for to be filled. Exercise: Using the `:kind` command in the ghci-repl, explore the kind of the `[]` data type.
We can also talk about things that take a type parameter more abstractly. We can say there is some type constructor `f` that takes a type parameter `a`:
``````f a
``````
With that covered we can talk about a type called `Compose`, defining it here as:
``````newtype Compose f g a = Compose
``````
It takes to two higher-kinded types `f` and `g` and a type parameter `a` and gives us `f` and `g` composed. This looks a lot like our function composition right?!
``````(.) :: (b -> c) -> (a -> b) -> a -> c
(.) f g x = f (g x)
^^^^^^^ COMPOSITION!!!
``````
"So, what's special about this new type we've introduced?", I hear you ask. I'll tell you what! We can define functions over the composition of these two things, but abstractly! We can compose two `Functor`s and two `Applicative`s. These commonly trip people up when exploring this. Especially since the nesting of two the `f` and `g` can make for some difficult type tetris when trying to implement `(<*>)` for two `Compose` values. I want to provide some further intuition on how to come to the solution to writing these instances.
But before you read on, I would encourage you to fire up your ghci repl, open up your favourite editor and try implementing these. If you reaaallllly get stuck then you can read on, but learn by doing first π Exercise: Implement the `Functor` and `Applicative` instance for `Compose`.
#### Functors on Functors on Functors
We'll start off with talking about composing two functors. To begin we'll look at the instance declaration:
``````instance (Functor f, Functor g) => Functor (Compose f g) where
``````
As we've said, we're composing two `Functor`s so it would make sense that we rely on `f` and `g` being instances of `Functor`. The aim here is to change the `a`, nested within our `f` and `g`, into a `b`. This is more clear when we take a look at the signature for `fmap` specialised to `Compose`.
``````fmap :: (a -> b) -> Compose f g a -> Compose f g b
``````
This gives us an idea of where to begin when trying to implement `fmap`. We'll need to introduce the function `a -> b` and the `Compose f g` value. We can also use pattern matching to unwrap the `f (g a)` inside the `Compose`.
``````fmap f (Compose fg) = _full_solution
``````
Armed with the knowledge that we want to change the innermost value `a`, we will approach the problem by thinking inside-out. This method of working from the inside-out will be reused throughout the implementations and will help us think about implementing these instances.
The innermost thing we can work with is `a`, but this is trivial since we know that turning the `a` into a `b` can be done by using the function `f`.
The next level up we are looking at the `g`, more specifically `g a`. So how can we go from a `g a` to a `g b`?
Alarm bells π¨π¨π¨ should be ringing in your head here because we know `g` is a `Functor` and we know how to get a function `g a -> g b` by using `fmap`. So we end up with:
``````fmap f (Compose fg) = _full_solution
where
-- gb :: g a -> g b
gb ga = fmap f ga
``````
So now we need to move to the next layer up and figure out how to turn our `f (g a)` into a `f (g b)`. Inspecting the types of what we have at our fingertips will reveal how to get past this hurdle:
``````fg :: f (g a) -- The value inside the Compose
gb :: g a -> g b -- The function we defined
``````
Again we see a familiar pattern of trying to access the inner part, in this case we want to access the `g a` inside of the `f`. `fmap` is our friend, once again, and the `(a -> b)` in this case is specialised to `(g a -> g b)`. That is to say `fmap` looks like the following for `f`:
``````fmap :: (g a -> g b) -> f (g a) -> f (g b)
``````
And here's our solution for `fmap` for the `Compose` `Functor`:
``````fmap f (Compose fg) = Compose \$ fmap gb fg
where
-- gb :: g a -> g b
gb ga = fmap f ga
``````
We can reduce this to show a cleaner solution by replacing `gb` with its definition (thank you equational reasoning ππΌ):
``````fmap f (Compose fg) = Compose \$ fmap (fmap f) fg
``````
and the unwrapping of `Compose` with the function `unCompose`:
``````fmap f fg = Compose \$ fmap (fmap f) \$ (unCompose fg)
``````
From there we can see function composition falling into place by removing `fg`:
``````fmap f = Compose . fmap (fmap f) . unCompose
``````
To concretise the idea of our higher-kinded composition we can see how the composition of two `Functor`s is just the two `fmap` functions coming together and forming our one `fmap` function for `Compose`. Neat!
Let's test out our implementation by choosing the two functors we mentioned earlier: `[]` and `Maybe`.
``````Ξ»> unCompose \$ (+1) <\$> Compose [Just 1, Just 2, Nothing]
[Just 2, Just 3, Nothing]
``````
#### Mind melting Applicative
Going up the chain of typeclasses we will take a look at the composition of `Applicative`s. Again, we can start off with a similar instance declaration:
``````instance (Applicative f, Applicative g)
=> Applicative (Compose f g) where
``````
The Applicative`[typeclass](https://hackage.haskell.org/package/base-4.11.1.0/docs/Prelude.html#t:Applicative) has two functions associated with its definition: `pure` and `(<*>)` (also `liftA2`, but we won\'t look at that here). We will go after the easier game first and write the `pure` implementation. Specialising `pure` to `Compose` we get the following function signature:
``````pure :: a -> Compose f g a
``````
Again, we can work with our intuition of working from the inside-out. So how can we get a `g a`? You guessed it, `pure`!
``````pure a = _full_solution
where
-- ga :: g a
ga = pure a
``````
Next, how can we get an `f (g a)`, pfffttt you got this!
``````pure a = _full_solution
where
-- fga :: f (g a)
fga = pure ga
``````
`````` -- ga :: g a
ga = pure a
``````
And to finish it all off we wrap our results in `Compose`:
``````pure a = Compose fga
where
-- fga :: f (g a)
fga = pure ga
``````
`````` -- ga :: g a
ga = pure a
``````
In the same vein as `fmap` we can work back to a cleaner solution. The first step is to exchange `ga` for its definition:
``````pure a = Compose fga
where
fga = pure (pure a)
``````
We can then do the same with `fga`:
``````pure a = Compose (pure (pure a))
``````
Now we can really see the composition shine by dropping the `a` and use the composition of these functions to define `pure`:
``````pure = Compose . pure . pure
``````
π
Now, here's the part that has tripped up many of us in the past. We'll define `(<*>)` for `Compose`. We are going to use a type-hole driven development approach alongside the idea of working from the inside out to converge on the solution for this function. But first, let's look at the type signature:
``````(<*>) :: Compose f g (a -> b) -> Compose f g a -> Compose f g b
``````
We can start off by unwrapping the two `Compose` values:
``````Compose f <*> Compose k = _full_solution
``````
And it's important to keep track of the types we're working with from here:
``````f :: f (g (a -> b))
k :: f (g a)
``````
It's good to also note that when utilising type holes GHC will provide us with relevant bindings in scope, as well.
Starting from the inner part first we will just consider the `g` part of the types. If we remove the `f` we end up with wanting an expression of type:
``````g (a -> b) -> g a -> g b
``````
Well doesn't that look familiar! So now we know that we will need to work with the `(<*>)` function that's specific to `g`.
What we want to do is to access the `g (a -> b)` inside the `f` and supply the `(<*>)` with its first argument. When we are thinking about getting inside something we should start to think of `fmap` and that's exactly what we will use:
``````Compose f <*> Compose k = _full_solution
where
liftedAp :: _liftedAp
liftedAp = fmap (<*>) f
``````
Notice that I have written the type signature as `_liftedAp` where GHC will be kind enough to tell us what the type of this expression is:
``````β’ Couldn't match expected type β_liftedApβ
with actual type βf (g a -> g b)β
...
``````
Unfortunately, we can't place this signature here without turning on some extensions so we will leave it as a comment to ensure that we remember the type:
``````Compose f <*> Compose k = _full_solution
where
-- liftedAp :: f (g a -> g b)
liftedAp = fmap (<*>) f
``````
From here we know that we want to end up with a final value of `f (g b)`. Looking at `liftedAp`, we know that if we can lift some `g a` in then we will get back that `f (g b)`. So let's look at what we have to work with again:
``````liftedAp :: f (g a -> g b)
k :: f (g a)
``````
At this stage, I think we're pros at type tetris and realise that we're working with something that we have seen before. Indeed, this is `(<*>)` specialised:
``````-- the original definition with 'k', 'x' and 'y'
-- as the type variable names to avoid confusion
(<*>) :: k (x -> y) -> k x -> k y
``````
``````-- the 'k' is our f
-- the 'x' is our g a
-- the 'y' is our g b
(<*>) :: f (g a -> g b) -> f (g a) -> f (g b)
``````
Noting that we also have to wrap it up in a `Compose` in the end, we get the solution:
``````Compose f <*> Compose k = Compose \$ liftedAp <*> k
where
liftedAp :: f (g a -> g b)
liftedAp = fmap (<*>) f
``````
An astute reader will notice that this can be written differently knowing that `fmap` can be written using its infix operator `<\$>`:
``````Compose \$ (<*>) <\$> f <*> k
``````
Which gives us the intuition that we're lifting the `(<*>)` over our `f`s and applying it to our `g`s.
Let's look at another concrete example using `[]` and `Maybe`:
``````Ξ»> unCompose \$ Compose [Just (+1), Just (+2), Nothing] <*> Compose [Just 1, Just 2, Nothing]
``````
``````[ Just 2
, Just 3
, Nothing
, Just 3
, Just 4
, Nothing
, Nothing
, Nothing
, Nothing
]
``````
#### Conclusion
I hope this provided some insight on how we can compose higher-kinds. In turn we were able to talk about composing typeclasses, specifically `Functor` and `Applicative`. The beauty is that now we can use `fmap`, `pure`, and `(<*>)` on any two `Functor`s or `Applicative`s!
If you want to take this learning further, I would encourage you to implement `Foldable` and `Traversable` for `Compose`. Once you have done this, investigate why you cannot compose two `Monad`s and then you will be ready to have fun with Monad Transformers! (Robots in Disguise)
Shoutouts: to Sandy (who's got excellent content on Haskell too btw), and Joe (\@jkachmar on FPChat) for helping me proof-read and polish this upe
Originally published on medium.com | 3,871 | 13,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-05 | longest | en | 0.886112 |
https://math.stackexchange.com/questions/1235401/let-r-be-any-rotation-and-p-any-reflection-then-r-circ-p-and-p-circ-r | 1,558,850,116,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258849.89/warc/CC-MAIN-20190526045109-20190526071109-00014.warc.gz | 552,952,221 | 32,779 | # Let $R$ be any rotation and $P$ any reflection then $R \circ P$ and $P \circ R$ are both glide reflections
Let $R$ be any rotation and $P$ any reflection then $R \circ P$ and $P \circ R$ are both glide reflections
I am having trouble showing $P \circ R$ is a glide reflection, I manage to get $R \circ P$, so I shall show my work.
For $R \circ P$ :
Let $R$ be a rotation where $R=R_{A,\alpha}$ where A is the center of rotation and $\alpha$ is the angle of rotation. Let $p$ be the reflection $p=p_k$ where $k$ is a line. Now let the line which goes through $A$ be parallel to $K$ and let the line $n$ be another line through $A$ whose oriented angle from $m$ is $\frac{\alpha}{2}$
Then observe:
Since rotations can be expressed as two reflections, we shall substitute:
$R \circ P=(p_n \circ p_m) \circ p_k=p_n \circ (p_m \circ p_k)$
Since $p_m \circ p_k$ is a translation because lines m and k are parallel. As a result, a reflection composed with a translation is a glide reflection.
For $P \circ R$ :
I used the same as above, so this might sound a bit redundant:
Let $R$ be a rotation where $R=R_{A,\alpha}$ where A is the center of rotation and $\alpha$ is the angle of rotation. Let $p$ be the reflection $p=p_k$ where $k$ is a line. Now let the line which goes through $A$ be parallel to $K$ and let the line $n$ be another line through $A$ whose oriented angle from $m$ is $\frac{\alpha}{2}$
Then observe:
Since rotations can be expressed as two reflections, we shall substitute:
$P \circ R=p_k \circ (p_n \circ p_m)=(p_k \circ p_n) \circ p_m$
I'm not sure where to proceed from there because $p_k \circ p_n$ is not a translation.
Here is the pic:
But if you want to follow the approach you already have, then for the second case choose $n$ as rotated by $-\frac\alpha2$ against $m$. Then you have
$$P \circ R=p_k \circ (p_m \circ p_n)=(p_k \circ p_m) \circ p_n$$
• Makes sense. For the part where you said: Any isometry which is orientation-reversing and not a reflection is a glide reflection. So essentially all you have to show is that the operation is no reflection is basically the summary of what you said below. Since, $-\frac{\alpha}{2}$ is being oriented reversing and as a result we get a translation which is not a reflection. – Mark Apr 15 '15 at 18:17 | 645 | 2,292 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-22 | latest | en | 0.860983 |
https://www.physicsforums.com/threads/coulombs-law-and-3-point-particles-vector-question.62359/ | 1,524,318,428,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945222.55/warc/CC-MAIN-20180421125711-20180421145711-00181.warc.gz | 861,642,491 | 14,758 | # Homework Help: Coulombs law, and 3 point particles (vector question)
1. Feb 2, 2005
### michaelw
Coulombs law, and 3 "point particles" (vector question)
The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1=+8.00 mC; the other two have identical magnitudes, but opposite signs: q2=–5.00 mC and q3=+5.00 mC. (a) Determine the net force (magnitude and direction) exerted on q1 by the other two charges. (b) If q1 had a mass of 1.50 g and it were free to move, what would be its acceleration?
for some reason i think i did this wrong
abs(F13) = abs(F12)
= (8.99*10^9)(8*10^-6)(5*10^-6)/(1.3^2)
= 2.13*10^-19
the F13 will make q1 go up and left, F12 will make q1 go up and right.
the left and the right forces cancel, meaning that the q1 will travel directly up??? (by a magnitude of 2*sin(23)*F12 => 1.66*10^-19)
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2. Feb 2, 2005
### MathStudent
The question gives the charges in mC which is milli coulombs (10^-3 C), you used micro coulombs
($\mu C =$ 10^-6 C.)
argh...your like the third person today to make the same mistake :grumpy:
any way, even if it were in micro coulombs, the equation is right, but the calculation is incorrect... how did you get 10^-19??
right! | 399 | 1,263 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-17 | longest | en | 0.903026 |
https://answers.yahoo.com/question/index?qid=20080816000051KK02644 | 1,611,667,072,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00035.warc.gz | 218,888,136 | 25,158 | # 三角關係與餘弦定理
1.
Find the general soolution of
cos^2x [1+cos^2x+cos^3x+cos^4x+......cos^7x] = 0 without using arithmetic
or geometric sequence.
2.
In a triangle ABC , by using the cosine law , prove that A^n+B^n is NOT equal to C^n
Rating
cos^2x [1+cos^2x+cos^3x+cos^4x+......cos^7x] = 0
∴cos^2x=0 or [1+cos^2x+cos^3x+......+cos^7x]=0
consider the range of cosx, which lies in the range of -1 and 1
∴-1<cosx<1
∴cos^2x, cos^4x, cos^6x must be larger than 0
There is only chance that cos^3x, cos^5x, cos^7x be negative if cosx is in the range of -1<cosx<0
However, in the range of -1<cosx<1, |cos^3x|<|cos^2x|, |cos^5x|<|cos^4x| and |cos^7x|<|cos^6x|
∴1+cosx+cos^2x+......+cos^7x must be >0
∴The final solution is cos^2x=0
cosx=0
x=360n90 | 343 | 752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-04 | latest | en | 0.71473 |
https://atcoder.jp/contests/abc176/tasks/abc176_a | 1,713,265,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00515.warc.gz | 95,150,518 | 6,045 | Contest Duration: - (local time) (100 minutes) Back to Home
A - Takoyaki /
Time Limit: 2 sec / Memory Limit: 1024 MB
### 問題文
たこ焼き器を使うと、1 度に最大で X 個のたこ焼きを作ることができます。これにかかる時間は作る個数によらず T 分です。
N 個のたこ焼きを作るためには何分必要ですか？
### 制約
• 1 \leq N,X,T \leq 1000
• 入力は全て整数
### 入力
N X T
### 出力
N 個のたこ焼きを作るために必要な時間の最小値を整数で出力せよ。
### 入力例 1
20 12 6
### 出力例 1
12
6 分で 12 個作れるとは、1 分で 2 個作れるという意味ではないことに注意してください。
### 入力例 2
1000 1 1000
### 出力例 2
1000000
Score : 100 points
### Problem Statement
Takahashi loves takoyaki - a ball-shaped snack.
With a takoyaki machine, he can make at most X pieces of takoyaki at a time, taking T minutes regardless of the number of pieces to make.
How long does it take to make N takoyaki?
### Constraints
• 1 \leq N,X,T \leq 1000
• All values in input are integers.
### Input
Input is given from Standard Input in the following format:
N X T
### Output
Print an integer representing the minimum number of minutes needed to make N pieces of takoyaki.
### Sample Input 1
20 12 6
### Sample Output 1
12
He can make 12 pieces of takoyaki in the first 6 minutes and 8 more in the next 6 minutes, so he can make 20 in a total of 12 minutes.
Note that being able to make 12 in 6 minutes does not mean he can make 2 in 1 minute.
### Sample Input 2
1000 1 1000
### Sample Output 2
1000000
It seems to take a long time to make this kind of takoyaki. | 504 | 1,396 | {"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} | 2.921875 | 3 | CC-MAIN-2024-18 | latest | en | 0.362694 |
https://sciblogs.co.nz/waiology/2011/10/27/on-the-philosophy-of-modelling/ | 1,653,553,514,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00044.warc.gz | 575,631,289 | 16,592 | On the philosophy of modelling
By Waiology 27/10/2011
By Daniel Collins
Back in June I introduced you to a model we use, called TopNet, and I talked about modelling on the ‘Ever Wondered?’ episode back in September. I’d now like to give you a richer picture of what models are in general.
Models feature widely in hydrology. Not the fashion models usually, but the mathematical or computer variety. In fact, to be pedantic, everything in hydrology is based on a model in one way or another, right down to the measurement of river flow. So to understand hydrology properly, you need to understand modelling. But be warned: modelling cannot be explained in one blog post. Whole books are written on the subject. Whole careers are built on building models. Instead, what I will try to do here is share some of the fundamentals of the philosophy of modelling as I see them.
The first rule to remember is that all models are to some extent wrong, but some models are useful.
Models are simplifications or idealisations of reality with the purpose of serving a particular function. This goes for fashion models, who are idealisations of the human form and whose job it is to make clothing look good. Or model or toy aeroplanes, which are easy-to-build miniatures that you can play with. Or mathematical models, which attempt to explain some phenomenon as a calculable function of defined variables.
The same goes for hydrological models. Mostly they are of the mathematical variety, but sometimes they can take on aspects of the fashion or toy models.
A mathematical model is basically an algebraic relationship with one or more variables, like Einstein’s e = mc2. You can add randomness, make it self-reflexive, or combine a suite of individual equations together if you want, but it’s still a mathematical model. One of the strengths of mathematical models is that they typically allow for more reproducibility, one of the hallmarks of science. And as soon as you write it into computer code, it becomes a computer model too. We take this step whenever it’s too burdensome to calculate the equation(s) by hand.
The model could be based on empirical observations, like the Rational Method for peak flood flow:
Qpeak = C I A
Or on physical principles, like Richards equation for water movement in unsaturated porous media:
But no matter what the model looks like, it would have been developed over the course of much time and research. For it to become an accepted model of reality, it will have been seeded by observation, shaped by imagination, potentially tweaked by calibration, and will have successfully jumped through the hoops of validation and either verification or corroboration.
And as modellers build their models, they should remember (though we don’t always) that models should be as simple as possible, but no simpler.
This basically means that models shouldn’t include features that don’t help them serve their purpose, but that they must still be useful somehow (in contrast to a spherical cow). For example, water temperature won’t improve Manning’s equation of river flow noticeably, so it’s left out, even though a liquid’s viscosity is temperature-dependent. While too few parameters in a model can lead to huge errors, too many parameters can be overly burdensome in terms of data collection or cause the model to lose all generality.
Now remember that hydrological models can sometimes resemble fashion or toy models?
If you’re trying to explain an idea, such as the water cycle, you need a conceptual model that makes sense – a model that makes people want to buy an idea. This is much like a fashion model who is used to sell clothing. And if you’re trying to explore the implications of existing scientific theories, such as the role of carbon fertilisation on catchment water yield, you need a toy mathematical model whose variables you can experiment or play with.
In the end, of course, we must remember that while good models serve a purpose, the results will be no better than the veracity of the model. It takes skill to use a model properly, and to interpret its results. And no matter how sophisticated our modelling skills, we should never lose sight of the observational data that feed our models and the questions that drive them.
0 Responses to “On the philosophy of modelling”
• This is really excellent. It’s always a challenge to explain something composed of complex concepts in plain language without losing the scientific integrity that the subject deserves. Cheers.
• Very nicely done Daniel,
I like that quote from Box, and there’s another (attributed?) to him that seems appropriate:
“Statisticians, like artists, have the bad habit of falling in love with their models.”
• Thanks guys.
As I read it, that second (potential) quote from Box actually cuts different ways. In one way, a scientist who develops a model may love it like a parent loves a child; in another, as salesman loves a product. But on the flip side, the more you scrutinise a model, the more you understand it – the more you can recognise its potential strengths and beauty.
How apt, though, that the sculptor of ‘The Thinker’ fits the quote’s prediction. [DC]
• Falafulu Fisi says:
Excellent quote too.
Statisticians Physicists, like artists, have the bad habit of falling in love with their models | 1,099 | 5,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.95316 |
https://cameramath.com/es/expert-q&a/Algebra/2-08-01-Two-lines-2-08-01-Two-lines-R-and | 1,656,350,458,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103337962.22/warc/CC-MAIN-20220627164834-20220627194834-00239.warc.gz | 210,422,328 | 7,060 | ### ¿Todavía tienes preguntas de matemáticas?
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Algebra
Pregunta
2. (08.01) Two lines, $$R$$ and $$M$$ , are represented by the following equations: Line $$R : 2 x + 2 y = 18$$
Line $$M : x + y = 9$$
Which statement is true about the solution to the set of equations? (4 poi It is $$( 18,9 )$$ . It is $$( 9,18 )$$ . There are infinitely many solutions. There is no solution. | 143 | 414 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-27 | latest | en | 0.416298 |
https://www.jiskha.com/questions/1001123/would-mosquito-larvae-be-considered-an-insect | 1,576,087,933,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540531974.7/warc/CC-MAIN-20191211160056-20191211184056-00479.warc.gz | 660,823,179 | 4,793 | # Science
Would Mosquito Larvae be considered an insect?
1. 👍 0
2. 👎 0
3. 👁 99
1. Regardless of the stage of growth, a mosquito is still an insect.
1. 👍 0
2. 👎 0
posted by PsyDAG
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More Similar Questions | 828 | 2,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2019-51 | latest | en | 0.91584 |
https://www.physicsforums.com/threads/use-of-group-theory.674851/ | 1,575,957,786,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525821.56/warc/CC-MAIN-20191210041836-20191210065836-00163.warc.gz | 836,415,697 | 18,875 | Use of group theory
Main Question or Discussion Point
I just studied group theory. Its all nice with all the definitions and rules that are supposed to be followed for a set with a given operation to be called a group. But I fail to see the importance of defining such an algebraic structure.
What are its uses?
Related Linear and Abstract Algebra News on Phys.org
I did read it. I understood that it can be used classify/represent certain natural phenomena.
But I did not understand any of its applications in mathematics which was my doubt.
The only useful thing I can see is that now whenever we see set of elements obeying the required rules under a binary operator we can call it a group.
So is that it? Just to classify things and make life simpler we define groups
The thing is that groups are a natural object that show up everywhere in mathematics. And I really mean everywhere. If a structure shows up a lot of time, then it makes sense to give it a name. That doesn't explain why it's a useful structure though.
There are certain operation on groups, such as quotients and products. These operations show up quite naturally in mathematics. But if we didn't have an abstract notion of a group, then we had to define all of these operations over and over again. Now we can just say: take the quotient of groups. Everybody will know perfectly what you mean. Without groups, you will have to construct to quotient again.
Furthermore, groups tend to be structures that are reasonably well known to mathematicians. Of course there are many unsolved problems about groups (some of which relate to string theory and fourier analysis!). But overall, a mathematician will feel comfortable when dealing with a group. Now, many hard mathematical problems can be simplified by translating the problem into a problem of groups. And then we can solve the problem by using our knowledge of groups. This is done in (for example) algebraic topology.
You might want to check out the history of group theory. It might be more clear why groups were defined the way they were.
There are different aspects of group theory. The Sylow theorem based study of finite groups is not the most useful (outside of finite group theory).
With my limited perspective on algebra, I think they teach it this way because it introduces students about abstract mathematics. Also, it prepares you for Galois theory, which is glorious to algebraists. Personally, I think group representations and the classical Lie groups would have been more useful than group presentations and the Sylow theorems. But perhaps that is too meaty a course and not really comparable to the standard introduction to group theory.
Sorry for the late reply - but I fail to understand anything when reading about history of group theory.
Can you please give some practical examples and how its useful.
This is what I know. A group is when an operation on a set is closed, has identity and inverse element. Why do we define a group in this way - I have no idea.
Help me understand this concept. Seems like a very useful one
Can you please give some practical examples and how its useful.
This is what I know. A group is when an operation on a set is closed, has identity and inverse element. Why do we define a group in this way - I have no idea.
For practical examples, you have both concrete applications, such as applications to physics, and abstract applications, such as the fundamental group of a topological space. In the later case, one case see that groups show up as mathematical structures in all kinds of places, so it's worth while to study them by themselves to economize our understanding of mathematics. Abelian groups are also part of one way to define vector spaces. Groups are also a simplified example of a Field, which the real numbers are an (obviously) important example. One last example: groups are used in some results of algebraic number theory. I hope I'm painting a picture where groups show up in lots of places in mathematics, so the particular choice of definition is a useful one.
For more practical examples, you can look to physics. Any kind of symmetry can be modeled using group theory (if you like, you can use that feature to motivate the definition of a group). Lorentz transformations, which relate two different reference frames in special relativity, form a group. Also other symmetries show up, called "gauge symmetries", show up in quantum field theory, and these also form groups. In general relativity, the R coordinate in the Schwarzchild metric is defined using orbits of a group. By studying group representation theory (closely related to group theory), one can derive the quantized "spectrum" of angular momentum in quantum mechanics.
Did you study the symmetry group of any object? If not, perhaps the best thing you could do is find some object with symmetry, and consider it's "symmetry transformations". Convince yourself that the set of symmetry transformations are a group: is there an identity transformation? is there an inverse transformation? is the action of two symmetries also a symmetry? are symmetries associative?
The answer to all those questions are yes, but you should really convince yourself that its true by physically holding something that is symmetric and rotating it accordingly.
pwsnafu
The first time I came across group theory was in chemistry. Take the molecule Fe(CO)5. This compound has two forms: the square pyramid formation and the formation that is on the page (I forgot what that formation is called).
Anyway, the question is can we detect which form our sample is? Here is where group theory comes in. The square pyramid has symmetry group C4 while the other form has symmetry group D3. This means when we place the samples in an IR spectrometer, we will get different readings.
In order to predict which produces what signal we use representation theory of groups. This means we write the group elements as matrices. But there are always "good" ways to do this. We thus are able to reduce our representation in to a linear combination of these "good" (irreducible) representations. So we might end up with 2A+B. What this means is on the IR reading, we should see a single peak and a separate double peak.
Group theory is extensively utilized in many aspects of x-ray crystallography, inorganic chemistry, protein analysis, organic chemistry, chemical physics, physical chemistry, theoretical chemistry, computational analysis, and so forth (and this list is not exhaustive).
If any molecular or atomic system you are examining has any sort of symmetry, group theory is very useful in its analysis and computation.
Stephen Tashi | 1,360 | 6,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-51 | latest | en | 0.966057 |
https://en.wikipedia.org/wiki/Moving_average_model | 1,490,438,614,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188914.50/warc/CC-MAIN-20170322212948-00532-ip-10-233-31-227.ec2.internal.warc.gz | 784,083,291 | 17,838 | # Moving-average model
(Redirected from Moving average model)
Not to be confused with Moving average.
In time series analysis, the moving-average (MA) model is a common approach for modeling univariate time series. The moving-average model specifies that the output variable depends linearly on the current and various past values of a stochastic (imperfectly predictable) term.
Together with the autoregressive (AR) model, the moving-average model is a special case and key component of the more general ARMA and ARIMA models of time series, which have a more complicated stochastic structure.
The moving-average model should not be confused with the moving average, a distinct concept despite some similarities.
Contrary to the AR model, the MA model is always stationary.
## Definition
The notation MA(q) refers to the moving average model of order q:
${\displaystyle X_{t}=\mu +\varepsilon _{t}+\theta _{1}\varepsilon _{t-1}+\cdots +\theta _{q}\varepsilon _{t-q}\,}$
where μ is the mean of the series, the θ1, ..., θq are the parameters of the model and the εt, εt−1,..., εt−q are white noise error terms. The value of q is called the order of the MA model. This can be equivalently written in terms of the backshift operator B as
${\displaystyle X_{t}=\mu +(1+\theta _{1}B+\cdots +\theta _{q}B^{q})\varepsilon _{t}.}$
Thus, a moving-average model is conceptually a linear regression of the current value of the series against current and previous (unobserved) white noise error terms or random shocks. The random shocks at each point are assumed to be mutually independent and to come from the same distribution, typically a normal distribution, with location at zero and constant scale.
## Interpretation
The moving-average model is essentially a finite impulse response filter applied to white noise, with some additional interpretation placed on it. The role of the random shocks in the MA model differs from their role in the autoregressive (AR) model in two ways. First, they are propagated to future values of the time series directly: for example, ${\displaystyle \varepsilon _{t-1}}$ appears directly on the right side of the equation for ${\displaystyle X_{t}}$. In contrast, in an AR model ${\displaystyle \varepsilon _{t-1}}$ does not appear on the right side of the ${\displaystyle X_{t}}$ equation, but it does appear on the right side of the ${\displaystyle X_{t-1}}$ equation, and ${\displaystyle X_{t-1}}$ appears on the right side of the ${\displaystyle X_{t}}$ equation, giving only an indirect effect of ${\displaystyle \varepsilon _{t-1}}$ on ${\displaystyle X_{t}}$. Second, in the MA model a shock affects ${\displaystyle X}$ values only for the current period and q periods into the future; in contrast, in the AR model a shock affects ${\displaystyle X}$ values infinitely far into the future, because ${\displaystyle \varepsilon _{t}}$ affects ${\displaystyle X_{t}}$, which affects ${\displaystyle X_{t+1}}$, which affects ${\displaystyle X_{t+2}}$, and so on forever (see Vector autoregression#Impulse response).
## Deciding appropriateness of the MA model
Sometimes the autocorrelation function (ACF) and partial autocorrelation function (PACF) will suggest that an MA model would be a better model choice and sometimes both AR and MA terms should be used in the same model (see Box–Jenkins method#Identify p and q).
## Fitting the model
Fitting the MA estimates is more complicated than with autoregressive models (AR models) because the lagged error terms are not observable. This means that iterative non-linear fitting procedures need to be used in place of linear least squares.
### Choosing the order q
The autocorrelation function of an MA(q) process becomes zero at lag q + 1 and greater, so we determine the appropriate maximum lag for the estimation by examining the sample autocorrelation function to see where it becomes insignificantly different from zero for all lags beyond a certain lag, which is designated as the maximum lag q. | 945 | 3,997 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 17, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-13 | latest | en | 0.88795 |
http://www.gurufocus.com/term/ROE/BZ/Return%2Bon%2BEquity/Boise%2BInc | 1,474,856,186,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660548.33/warc/CC-MAIN-20160924173740-00275-ip-10-143-35-109.ec2.internal.warc.gz | 511,386,267 | 26,813 | Switch to:
Boise Inc (NYSE:BZ)
Return on Equity
-1.18% (As of Jun. 2013)
Return on equity is calculated as net income divided by its average shareholder equity. Boise Inc's annualized net income for the quarter that ended in Jun. 2013 was \$-9 Mil. Boise Inc's average shareholder equity for the quarter that ended in Jun. 2013 was \$750 Mil. Therefore, Boise Inc's annualized return on equity (ROE) for the quarter that ended in Jun. 2013 was -1.18%.
Definition
Boise Inc's annualized Return on Equity (ROE) for the fiscal year that ended in Dec. 2012 is calculated as
ROE = Net Income (A: Dec. 2012 ) / ( (Total Equity (A: Dec. 2011 ) + Total Equity (A: Dec. 2012 )) / 2 ) = 52.15 / ( (794.847 + 748.186) / 2 ) = 52.15 / 771.5165 = 6.76 %
Boise Inc's annualized Return on Equity (ROE) for the quarter that ended in Jun. 2013 is calculated as
ROE = Net Income (Q: Jun. 2013 ) / ( (Total Equity (Q: Mar. 2013 ) + Total Equity (Q: Jun. 2013 )) / 2 ) = -8.832 / ( (750.329 + 749.808) / 2 ) = -8.832 / 750.0685 = -1.18 %
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
In the calculation of annual return on equity, the net income of the last fiscal year and the average total shareholder equity over the fiscal year are used. In calculating the quarterly data, the Net Income data used here is four times the quarterly (Jun. 2013) net income data. Return on Equity is displayed in the 15-year financial page.
Explanation
Return on Equity (ROE) measures the rate of return on the ownership interest (shareholder's equity) of the common stock owners. It measures a firm's efficiency at generating profits from every unit of shareholders' equity (also known as net assets or assets minus liabilities). ROE shows how well a company uses investment funds to generate earnings growth. ROEs between 15% and 20% are considered desirable.
The factors that affect a companys Return on Equity (ROE) can be illustrated with the Du Pont Formula:
Return on Equity (ROE) (Q: Jun. 2013 ) = Net Income / Average Shareholder Equity = -8.832 / 750.0685 = (Net Income / Revenue) * (Revenue / Average Total Assets) * (Average Total Assets / Average Equity) = (-8.832 / 2486.656) * (2486.656 / 2212.8585) * (2212.8585 / 750.0685) = Net Profit Margin * Asset Turnover * Leverage Ratio = -0.36 % * 1.1237 * 2.9502 = Return on Assets * Leverage Ratio = -0.4 % * 2.9502 = -1.18 %
Note: The Net Income data used here is four times the quarterly (Jun. 2013) net income data. The Revenue data used here is four times the quarterly (Jun. 2013) revenue data.
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
With this breakdown, it is clear that if a company grows its Net Profit Margin, its Asset Turnover, or its Leverage, it can grow its return on equity.
Be Aware
The net income used here is the net income to common shareholders.
Because a company can increase its return on equity by having more financial leverage, it is important to watch the leverage ratio when investing in high ROE companies. Like ROA, ROE is calculated with only 12 months data. Fluctuations in companys earnings or business cycles can affect the ratio drastically. It is important to look at the ratio from a long term perspective.
Asset light businesses require very few assets to generate very high earnings. Their ROEs can be extremely high.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Boise Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 ROE 0.00 0.00 0.00 0.00 2.36 -13.34 28.75 9.90 10.43 6.76
Boise Inc Quarterly Data
Mar11 Jun11 Sep11 Dec11 Mar12 Jun12 Sep12 Dec12 Mar13 Jun13 ROE 11.38 6.02 12.91 7.94 10.90 7.04 1.83 7.04 -0.65 -1.18
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http://forums.wolfram.com/mathgroup/archive/2003/Oct/msg00090.html | 1,716,503,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00298.warc.gz | 10,798,100 | 8,404 | Re: Airy's Gi(x) function; asymptotic matching and asymptotic limits
• To: mathgroup at smc.vnet.net
• Subject: [mg43796] Re: Airy's Gi(x) function; asymptotic matching and asymptotic limits
• From: "Kevin J. McCann" <kjm at KevinMcCann.com>
• Date: Mon, 6 Oct 2003 02:07:49 -0400 (EDT)
• References: <bllok7\$bus\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Mathematica gives the result in terms of Airy's and Hypergeometrics. Copy these into
Mathematica to see what's going on.
Cheers,
Kevin
Problem:
\!\(\*
RowBox[{"DSolve", "[",
RowBox[{
RowBox[{"{",
RowBox[{
RowBox[{
RowBox[{
RowBox[{
SuperscriptBox["y", "\[Prime]\[Prime]",
MultilineFunction->None], "[", "x", "]"}],
"-", \(x\ y[x]\)}], "==", \(-1\)}], ",",
RowBox[{
RowBox[{
SuperscriptBox["y", "\[Prime]",
MultilineFunction->None], "[", "0", "]"}], "==", "0"}],
",", \(y[0] == \[ScriptCapitalC]\)}], "}"}], ",", \(y[x]\), ",",
"x"}], "]"}]\)
Solution:
\!\({{y[x] ->
1\/2\ \((3\^\(2/3\)\ \[ScriptCapitalC]\ AiryAi[x]\ Gamma[2\/3] +
3\^\(1/6\)\ \[ScriptCapitalC]\ AiryBi[x]\ Gamma[2\/3] -
2\ x\^2\ Hypergeometric0F1[4\/3,
x\^3\/9]\ HypergeometricPFQ[{1\/3}, {2\/3, 4\/3},
x\^3\/9] +
x\^2\ Hypergeometric0F1[2\/3,
x\^3\/9]\ HypergeometricPFQ[{2\/3}, {4\/3, 5\/3},
x\^3\/9])\)}}\)
"Curt Fischer" <crf3 at po.cwru.edu> wrote in message
news:bllok7\$bus\$1 at smc.vnet.net...
> Dear Group,
>
> Recently I had to solve the differential equation y''[x] - x y[x] == -1,
> with one known boundary condition y'[0]==0.
>
> The general solution is conveniently written as C[1] AiryAi[x] + C[2]
> AiryBi[x] + Pi airyGi[x]. One of the constants can be solved for with
> respect to the other
>
> Question 1:
>
> Mathematica does not have the airyGi[x] function built-in. It returns the
> solution
> \!\(1\/2\ \((2\ \@3\ AiryAi[x]\ C[2] + 2\ AiryBi[x]\ C[2] -
> 2\ x\^2\ Hypergeometric0F1[4\/3,
> x\^3\/9]\ HypergeometricPFQ[{1\/3}, {2\/3, 4\/3}, x\^3\/9] +
> x\^2\ Hypergeometric0F1[2\/3,
> x\^3\/9]\ HypergeometricPFQ[{2\/3}, {4\/3, 5\/3}, x\^3\/9])\)\)
>
> which is a big messy expression involving AiryAi[x], AiryBi[x], and
> hypergeometric functions. Does anyone know how I can relate this
> hypergeometric stuff is equal to airyGi[x] == Integrate[Sin[t^3 + z t]
> dz,{t,0,infinity}] ?
>
> Question 2: When I solved my problem analytically, I was interested in
> evaluating the unknown integration constant by asymptotic matching to
> another function which I knew. This worked great on paper, but
Mathematica
> could not take the limit of
>
> 2\ x\^2\ Hypergeometric0F1[4\/3,
> x\^3\/9]\ HypergeometricPFQ[{1\/3}, {2\/3, 4\/3}, x\^3\/9] +
> x\^2\ Hypergeometric0F1[2\/3,
> x\^3\/9]\ HypergeometricPFQ[{2\/3}, {4\/3, 5\/3}, x\^3\/9])\)\)
>
> Is there a way to evaluate this limit in Mathematica? Also, in general is
> there anyway to get the an "asymptotic limit" of a function in
Mathematica?
> For example, airyGi[x] -> 1/(Pi x) for large x. Is there any way to
elicit
> this type of info about a function from Mathematica?
>
> (See Abramowitz and Stegun, 1974, Handbook of Mathematical Functions, for
> this and other info on Airy functions.)
>
> thanks for any help anyone can provide,
>
>
>
> Curt Fischer
>
>
>
>
```
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# assign6 - N b s t L = sL tL 4 Let A be an m × n matrix...
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Math 136 Assignment 6 Due: Wednesday, Mar 2nd 1. Find a spanning set for the kernel and range for each of the following linear mappings. a) f ( x 1 ,x 2 ,x 3 ) = ( x 1 ,x 2 ,x 1 + x 2 - x 3 ) b) f ( x 1 ,x 2 ,x 3 ) = (0 ,x 1 - x 2 ) c) f ( x 1 ,x 2 ,x 3 ,x 4 ) = (0 , 0) d) f ( x 1 ,x 2 ) = ( x 1 + 2 x 2 , 2 x 1 + 4 x 2 , 3 x 1 + 6 x 2 ) 2. Let L : R n R m be a linear mapping. Prove that L ( ~ 0) = ~ 0. 3. Let L , M , and N be linear mappings from R n to R m and let s,t R be scalars. Prove that a) L + ( M + N ) = ( L + M ) +
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Unformatted text preview: N . b) ( s + t ) L = sL + tL . 4. Let A be an m × n matrix. Prove that for any ~a ∈ Col( A ) and ~x ∈ Null( A T ) we have ~a · ~x = 0. 5. Let L : R n → R m and M : R m → R p be linear mappings. Prove that M ◦ L is a linear mapping and [ M ◦ L ] = [ M ][ L ] 1...
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## This note was uploaded on 07/16/2011 for the course MATH 136 taught by Professor All during the Spring '08 term at Waterloo.
Ask a homework question - tutors are online | 474 | 1,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.841062 |
https://webapps.stackexchange.com/questions/103162/how-to-write-averageif-not-an-empty-string-in-google-sheets | 1,716,017,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00301.warc.gz | 552,824,105 | 44,456 | # How to write AverageIf (Not an Empty String) in Google Sheets?
I have the answer to this question. However, it took me a few wasted hours of research and work to figure it out. I couldn't find an answer anywhere online, so I thought I'd write out a question and share the answer in the hope of saving someone (who is as dumb as me) some time.
I've been working on a Google Spreadsheet for a while to get attendance counts for a few classes. I have a Google Form where teachers input the attendance for their class. That automatically gets sent to a Form Response spreadsheet. From there I have a few sheets that manipulate the input data, then a summary sheet that shows a table of: classes (rows), dates (columns), and attendance to each class on specific dates. Each cell in the table contains a formula which either returns the attendance count or an empty string. Until data exists for a date and class, that cell contains and empty string. The table works perfectly.
Using the data in this table, I want to get average attendance for each class in real time. There are a couple of problems though:
1. The table is set out for the whole year, so there are a lot of cells containing empty strings until the end of the year, which mess up the "real time" average
2. Occasionally a class's attendance doesn't get filled in, so there are random cells with empty strings which mess up the average throughout the year
I've been trying `AVERAGE`, `IF`, and `AVERAGEIF` functions to figure this out. `AVERAGEIF` seems to be the solution but I can't figure out a criteria that works. `"<>"""` causes a return value of 0 no matter what range I'm trying to average. I've thought that using `ISNUMBER(parameter)` as the criteria, but I don't know what to use as a parameter for it. Not surprisingly, using it without a parameter didn't get me anywhere (`DIV/0` error)
I haven't found Google's Docs editor Help to be super helpful, so I was wondering if anyone might be able to share how to use `AVERAGEIF(does not contain an empty string)`.
• Did you try just `"<>"` or `"<>" & ""`
– Scott Craner
Feb 15, 2017 at 20:31
• If you set the datatype for the column, wouldn't this problem disappear?
– Yorik
Feb 15, 2017 at 20:54
• Thanks for the guidance @David. By the time I was adding tags I was having some computer speed issues and wasn't really paying attention. I saw a few other google spreadsheet questions on the site, so assumed it was fine. Now I know! Mar 1, 2017 at 5:45
`=AVERAGEIF(A1:Z1,"<>")`
`"<>"` on its own means "not equal to an empty string" in Google Sheets.
• @pnuts Yes, I did. It kept including all the blank cells in the calculation as if they were zeroes. Mar 1, 2017 at 5:47
• That's weird, I've just tested the formula on a range that includes blank cells and they were not counted. Maybe you could post a link to the spreadsheet where it occurs so that the problem could be debugged? Mar 14, 2017 at 16:15
• It's weird it deosn't work for me.. `=AVERAGEIF(F2:N2,"<>")` all the cases between F2 and N2 are cells with data validation, based on constantes of another page. Only one is filled, so it should give 1/9 percent Jul 14, 2018 at 12:46
I recommend using:
``````=IF(COUNT(E4:G4);AVERAGE(E4:G4);0)
``````
COUNT() returns numbers of cells containing valid numbers : if none, the value will be `0`, so the condition will evaluate to false and return `0` (you can put `""` instead of `0` if you prefer). Otherwise, it will use the result of the AVERAGE function.
• Only thing that really worked for me--and so obvious in retrospect! Aug 22, 2020 at 16:16
To all future viewers, easiest way is `Sum(A1:Z1)/Counta(A2:Z1)`
• Hey! Welcome to Web Applications! You should expand on your answer for it to be really helpful. What does the cells in you example relate to? Feb 28, 2019 at 8:19
• This is the only answer that I could get to work. The downvote by above commenter is really uncalled for. The commenter is asking for context which the question and other answers clearly show. Thanks Sydney! Jan 13, 2020 at 0:49
• One downside to this is that the range has to be specified twice Dec 7, 2020 at 19:59
I also encountered (`DIV/0 error`) when attempting to use `AVERAGEIF` to check for empty string. In my case, I only want to take an average if a text column is non-empty, as the number columns will always be populated in my sheet but are only valid if the text column is populated.
My solution is to use an `IF` function:
``````=IF(A2 <> "", AVERAGE(D2:AF2), "")
``````
This is what I'd expect `AVERAGEIF` to do on it's own, but it doesn't.
• Excellent solution! It works! May 2, 2022 at 1:16 | 1,208 | 4,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.952561 |
https://myhomeworkgeeks.com/california-state-university-dominance-diversity-on-disturbed-species-lab-report/ | 1,607,178,412,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747887.95/warc/CC-MAIN-20201205135106-20201205165106-00495.warc.gz | 390,091,172 | 20,628 | California State University Dominance Diversity on Disturbed Species Lab Report
I’m studying for my Biology class and don’t understand how to answer this. Can you help me study?
This assignment is to analyze species diversity from two lawns, one that is disturbed and one that is less disturbed, on CSUN’s campus.
Read pages 151-152 (at the end of chapter 5) for context on the Lawn Census activity, and Read pages 197-198 (at the end of chapter 7) to learn about dominance-diversity curves and the Shannon Index.
You can find the data here: Lawn Census Data F2020.xlsx
In Excel, create two dominance-diversity curves (one for each lawn) with non-grass species along the x-axis and ln(p) along the y-axis. In other words, make a line graph with natural log (ln) proportion of non-grass cover on the vertical axis and the non-grass species ordered by abundance on the horizontal axis. Label the y-axis as the ln proportion of non-grass cover and x-axis as non-grass species. You should have two lines, one for each lawn, on one graph.
Add the Shannon index value in a textbox above each line on the graph.
Delete gridlines and the chart title.
1. Compare the lines for the two lawns. Were the patterns similar or different in the two lawns? How so?
2. Did the two lawns that your class surveyed have similar or different species richnesses? Give the species richness of each lawn (Species richness = # of different species represented; it is simply the species count).
3. Compare the two lawns’ Shannon index values. Which lawn has more diversity of broad-leafed (non-grass) plants? (The higher the Shannon index value, the higher the species richness and evenness)
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https://robeastaway.com/puzzles/the-appearing-square | 1,709,549,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00519.warc.gz | 470,562,388 | 2,350 | # The Appearing Square
Can you explain where the extra square comes from?
This puzzle is based on the illusion that the shape shown in both cases is a triangle. In fact, the 'hypotenuse' is not a straight line in either case - the red and blue triangles have slightly differing acute angles and the 'spare' square in the second example is the result of the bowing out of the longest side in this case, as opposed to a bowing in in the first example. | 102 | 451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.955775 |
https://customwriting.studyace.net/2021/06/14/calculate-the-left-and-right-hand-side-of-stokes-theorem-for-this-problem-explain-why-you/ | 1,627,442,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00551.warc.gz | 209,711,820 | 11,044 | # Calculate the left and right hand side of Stokes Theorem for this problem. Explain why you…
Calculate the left and right hand side of Stokes Theorem for this problem. Explain why you obtained different values, and why it is not a contradiction
2. [5 Stokes’ Theorem: Let S be a piecewise smooth oriented surface having a piecewise smooth boundary curve C. Let F = Mi+Nj+Pk be a vector field whose components have continuous first partial derivatives on an open region containing S. The the circulation of F around C in the direction counterclockwise with respect to the surface’s unit normal vector n equals the integral of the curl vector field V x F over S: F dr = / /VxF n da counterclockwise Now consider the vector field F=- i + 2+y2 which is defined on the punctured disk D defined by those points in the ay-plane satisfying 0
0 0 | 192 | 839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-31 | latest | en | 0.857789 |
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# Homework2 - EE 351K Probability Statistics and Random...
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Unformatted text preview: EE 351K Probability, Statistics, and Random Processes SPRING 2012 Instructor:S. Shakkottai [email protected] Homework 2 Problem 1 A hard disk storing information in binary form has been corrupted, so it can only be read with bit errors. Due to errors on the disk, a digit stored as ’0’ could be detected by you as ’1’ with probability 0.2; and a digit stored as ’1’ could be detected as ’0’ with probability 0.4. Each digit that is stored is a ’1’ or a ’0’ with equal probability. What is the probability that your detected value is the same as the stored value given that you detect a ’0’? Problem 2 A parking lot consists of a single row containing n parking spaces ( n ≥ 2) . Alice arrives when all spaces are free. Bob is the next person to arrive. Each person makes an equally likely choice among all available spaces at the time of arrival. Describe the sample space. Obtain P ( A ) , the probability the parking spaces selected by Alice and Bob are at most 2 spaces apart, (i.e. there can be at most 1 spotthe parking spaces selected by Alice and Bob are at most 2 spaces apart, (i....
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Ask a homework question - tutors are online | 356 | 1,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-09 | latest | en | 0.897563 |
https://msgroups.net/excel/averaging-cell-s.problems-with-div-0/239339 | 1,638,614,956,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00015.warc.gz | 492,270,548 | 12,516 | #### Averaging cell's...problems with Div/0
```Hi guys. First time poster here so be gentle with me. I am looking fo
some assistance averaging a range of 1 to 3 numbers.
Here is what I have so far.
=(D3+F3+H3)/(IF(C3=0,0,1)+IF(E3=0,0,1)+IF(G3=0,0,1))
This works great. What it does is checks to see if there is a value i
the cell, then counts it and divides by the right number. I.E if yo
only have two values out of 3 filled in it divides the number by tw
My problem... if all 3 fields are 0 then I get a divide by 0 error.
Any suggestion on how to fix this? I don't want my spreadsheet to loo
messy before I start plugging in values. What kind of formula can
plug in here to have it display 0 if all 3 values are 0
--
Message posted from http://www.ExcelForum.com
```
0
6/30/2004 6:07:07 PM
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```=IF(AND(C3=0,E3=0,G3=0),0,(D3+F3+H3)/(IF(C3=0,0,1)+IF(E3=0,0,1)+IF(G3=0,0,1)
))
Are these cells really zero or are they blank, if blank you could use
=IF(COUNT(C3,E3,G3),SUM(D3,F3,H3)/COUNT(C3,E3,G3),0)
--
Regards,
Peo Sjoblom
(No private emails please, for everyone's
benefit keep the discussion in the newsgroup/forum)
"Roy Bunch >" <<Roy.Bunch.18oe1t@excelforum-nospam.com> wrote in message
news:Roy.Bunch.18oe1t@excelforum-nospam.com...
> Hi guys. First time poster here so be gentle with me. I am looking for
> some assistance averaging a range of 1 to 3 numbers.
>
> Here is what I have so far.
>
> =(D3+F3+H3)/(IF(C3=0,0,1)+IF(E3=0,0,1)+IF(G3=0,0,1))
>
>
> This works great. What it does is checks to see if there is a value in
> the cell, then counts it and divides by the right number. I.E if you
> only have two values out of 3 filled in it divides the number by two
>
> My problem... if all 3 fields are 0 then I get a divide by 0 error.
> Any suggestion on how to fix this? I don't want my spreadsheet to look
> messy before I start plugging in values. What kind of formula can I
> plug in here to have it display 0 if all 3 values are 0.
>
>
> ---
> Message posted from http://www.ExcelForum.com/
>
```
0
terre08 (1112)
6/30/2004 6:37:44 PM
```On Wed, 30 Jun 2004 13:07:07 -0500, Roy Bunch
<<Roy.Bunch.18oe1t@excelforum-nospam.com>> wrote:
>Hi guys. First time poster here so be gentle with me. I am looking for
>some assistance averaging a range of 1 to 3 numbers.
>
>Here is what I have so far.
>
>=(D3+F3+H3)/(IF(C3=0,0,1)+IF(E3=0,0,1)+IF(G3=0,0,1))
>
>
>This works great. What it does is checks to see if there is a value in
>the cell, then counts it and divides by the right number. I.E if you
>only have two values out of 3 filled in it divides the number by two
>
>My problem... if all 3 fields are 0 then I get a divide by 0 error.
>Any suggestion on how to fix this? I don't want my spreadsheet to look
>messy before I start plugging in values. What kind of formula can I
>plug in here to have it display 0 if all 3 values are 0.
>
=AVERAGE(C3,E3,G3)
This will give a DIV/0 error if all three are not numbers, but does give a 0 if
all three are 0.
You could handle the error by either checking specifically for numeric entries
or using an ISERR function.
--ron
```
0
ronrosenfeld (3122)
6/30/2004 6:51:10 PM
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Our company attrition rate calculation = Attrits / Average Headcount. I want to use the same calculation for any set of agents, whether th agents are in classroom trianing or production. The problem is tha the training group needs to back into the number of agents that need t start a training class. For example. If you have a class that starts with 25 and ends with 2 there is a 20% fallout but due to average headcount in the calculatio the attrition rate is 22.1%. HOW CAN I HAVE A FORMULA THAT BACKS INTO THE 25 THAT I WOULD NEED T START IN CLASSROOM TRAINING? I WOULD LIKE TO USE THE 20 ... | 4,451 | 17,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.863295 |
https://earthscience.stackexchange.com/questions/9100/does-the-moon-have-any-effect-on-earthquakes-being-created?noredirect=1 | 1,716,987,053,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00809.warc.gz | 172,001,889 | 39,106 | # Does the moon have any effect on earthquakes being created? [duplicate]
Here, in New Zealand, we just had a 7.5 magnitude earthquake hit. A lot of people are saying that the super moon caused this.
I have doubts as to the super moon actually causing it; but could it have contributed to the earthquake due to its stronger gravitational pull?
The forces involved in the New Zealand earthquake are not amenable to precise evaluation - at least not without detailed geotechnical evaluation, but we can use the peak ground acceleration as a proxy. For a magnitude 7.5 earthquake the peak ground acceleration is about 0.4g. Compare this to the normal moon's gravitational acceleration on Earth's surface of about 3.318 E-5 g. 'Super-moons' are about 16000 km closer than usual, yielding a gravitational acceleration of 3.783 E-5 g. Therefore the differential gravitational acceleration of a super-moon, compared to a normal moon, is about 4.65 E-6 g. Or to put it another way, the effect of a super-moon is to create about a 0.001% of the acceleration (and hence force) required for the accumulated strain energy to overcome friction, and hence cause the earthquake. This assumes that all of the moon's gravity was aligned in such a way as to exacerbate strain along the slip plane - which is never going to be the case.
I suppose that one could argue that the miniscule gravitational force of a super-moon could have been the 'straw that broke the camel's back', but the accumulated strain energy in the ground must have been so great that the earthquake was about to happen anyway, with or without the moon's gravity.
Afterthought: With the boot on the other foot: The Earth's much larger gravitation has an effect on the Moon. There are moon tides which cause regular, small but measurable moon-quakes. See Moritz and Melchior (2013), Tidal Interactions in the Earth-Moon System.
• Did the super moon cause, the earthquake, simply, no, the energy associated was, as GS quantifies, minuscule as compared to the energy of the quake or even the amount needed to start slippage. Did it provide the last push needed? Possible. Or it might have been someone jumping up and down at just the right time. And think of it this way, the tectonic energy tends to continually build until release. If it did not release no as a mid-7 maybe it would have built for another year and been a devastating 8+. If you chose to believe it was the last needed push, then assume it saved a lot of lives.
– dlb
Nov 14, 2016 at 19:53
Keep in mind that this "special" "supermoon" is only about 0.5% closer this month than it was last month, or will be next month. And now is only 0.1% closer away at last year's minimum (and in fact, the minimum distance each year of the past 20 is only separated by 0.3% variation)
Makes it a little less exciting, eh? [source for data: timeanddate.com]
And likewise this article from space.com suggests that tide differences between a supermoon and a typical full moon are likely fairly small.
I'm certainly no expert on any of these topics in question, and am hopeful we'll see more precise calculations and details from someone more plugged into these fields. But that information at least suggests that it is dubious to credit the earthquakes to the exceptional proximity of THIS particular "supermoon".
Now do supermoons, or for that matter, full moons in general, cause earthquakes? I have no expertise in any of these topics areas (you can see that with my related question that I happened to post earlier this week asking about seasonal\temperature influences). But I decided to do a quick rough study because it was an interesting question that I had as well.
Applying the EMSC-CSEM earthquake search, I found that 7.9 and greater magnitude quakes gave me a good manageable set to work with from their database (dating to October 2004). I did not look at potential results when choosing the set size. But found it was a fair size to put together, and would at least give a potential taste.
So here are all >= 7.9s in the past 12ish years, as well as their moon phase and earth-moon distance (which appears to vary from about 356500-407000 km, completing 1-2 cycles a year):
• Mag | Moon | Moon Dist | Event
• 9.3 | 100% | 405266 km | 2004 Indian Ocean Earthquake (12-26-04)
• 9.0 | 33.% | 395992 km | 2011 Japan Earthquake (03-11-11)
• 8.8 | 97.% | 357971 km | 2010 Chile Earthquake (02-27-10)
• 8.6 | 92.% | 380842 km | 2005 Sumatra Earthquake (03-28-05)
• 8.4 | 01.% | 400497 km | 2007 Sumatra Earthquake (09-12-07)
• 8.4 | 72.% | 368806 km | 2012 Sumatra Earthquake (04-11-12)
• 8.3 | 25.% | 405266 km | 2006 Kuril Islands [NE of Japan] Earthquake (11-15-06)
• 8.3 | 99.% | 361064 km | 2013 Sea Okhotsk Earthquake [NE of Japan] Earthquake (05-24-13)
• 8.3 | 12.% | 404185 km | 2015 Chile Earthquake (09-16-25)
• 8.2 | 35.% | 400532 km | 2007 Kuril Islands [NE of Japan] Earthquake (01-13-07)
• 8.1 | 99.% | 405537 km | 2007 Solomon Islands Earthquake (04-01-07)
• 8.1 | 81.% | 403548 km | 2009 Samoa Islands Earthquake (09-29-09)
• 8.1 | 06.% | 381485 km | 2014 Chile Earthquake (04-01-14)
• 7.9 | 09.% | 399204 km | 2007 Peru Earthquake (08-15-07)
• 7.9 | 51.% | 382561 km | 2008 China Earthquake (05-12-08)
• 7.9 | 41.% | 402219 km | 2009 Tonga Earthquake (03-19-09)
• 7.9 | 33.% | 395899 km | 2011 Japan Earthquake Aftershock (03-11-11)
• 7.9 | 23.% | 365928 km | 2013 Solomon Islands Earthquake (02-06-13)
• 7.9 | 12.% | 390858 km | 2014 Alaska (Aleutians) Earthquake (06-23-14)
• 7.9 | 98.% | 357791 km | 2016 New Zealand Earthquake (focus of this post)
(Moon phases from USNO, distances from satellite-calculations.com)
Breaking it down (after removing the Japan aftershock), seems a bit of a predilection for full moons:
(6 of the 19 were 90+% full).
And just maybe a slight increase near new moon as well, which would match the way that earlier article described that tides work:
(5 of 19 <= 12%).
Here's a histogram of the spread [made at wessa.net]:
But it doesn't seem there's as much of a trend in distances:
It's almost certainly not enough data to draw a rock solid conclusion, but interesting that it seems hints at favoring full moons. I'm not sure what statistical calculations to do to prove it... the mean is only 51%, so it'd need more than just a mean test to prove that it's bi-modal (favoring full/new moons).
But this quick look certainly seems to suggest at least that there isn't obvious trends in earth-moon distance impacting earthquakes (R^2 = 0.14).
The real connections may be much more complex (varying by location\geology, only affecting certain types of earthquakes, more significant for weaker earthquakes, etc). But hopefully this try using big data from about the last decade is at least a help!
Hope the cleanup goes well AnonDCX! | 1,845 | 6,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.956203 |
https://futurism.com/pluto-and-charon-a-dwarf-planet-binary-system | 1,726,403,849,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00139.warc.gz | 240,959,140 | 18,486 | Especially with the discovery of Pluto’s fifth moon in 2012, the distant icy planet and its largest moon Charon has gotten a lot of attention. You’ve undoubtedly heard Pluto referred to as the Pluto-Charon system, or a binary planet system. When you think of a binary system, you imagine two objects orbiting each other, well, according to orbital mechanics, any two bodies in a stable orbit with each other orbit the barycentric coordinates (basically, the center of mass between the centers of two bodies).
What does this mean? Basically, moons and planets or planets and stars all orbit each other. Some examples, Earth orbits the moon and vice versa. The sun also orbits Earth and vice versa. So, what is the difference between a satellite and a binary system? The location of the barycentric coordinates. In cases where one body is much more massive than the second body, the larger body will appear to wobble instead of orbit around a separate point because the barycentric coordinates are within the larger body. Some examples:
For the Earth-Moon system, Earth’s radius is 6,380 km, the barycentric coordinates are 4,670 above Earth’s center. This means these coordinates appear within Earth (at 0.732 Earths radius) and the moon affectively orbits around the Earth. If you use the Earth-Sun system, you find the baycentric coordinates are located 449 km from the center of the sun. The Earth obviously orbits the sun in this case.
Of the planets in the solar system, Jupiter has the greatest affect on the sun since the barycentric coordinates for this system is 1.07 the radius of the sun (meaning, the barycenter is just above the surface of the sun). This system isn’t a binary candidate because Jupiter is a planet and the Sun is a star, a binary system consists of two objects that are very similar.
Next, let us look at the Pluto-Charon system. Here, we find the barycenter is 1.83 the radius of Pluto. In addition, if you look at the orbits of Pluto’s four other moons, you’ll find that they orbit Pluto-Charon instead of limiting themselves to Pluto. Because of this, Pluto and Charon were actually considered a double planet before Pluto’s reclassification in 2006.
Especially after the AIU reclassified Pluto in 2006, they didn’t want to make an official decision on Pluto-Charon’s dwarf planet-moon/dwarf-planet binary status and opted to let public opinion calm down over Pluto’s reclassification instead (after all, if Charon is a moon, Pluto is more important, right?... right?).
For the time being, Charon is not on the list of accepted dwarf planets, and the Pluto keeps the title of Kuiper Object Heavyweight Champion with its five moons. | 597 | 2,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.886147 |
https://www.jiskha.com/questions/6102/december-1994-a-man-in-ohio-deposited-all-of-the-8-million-pennies-he-had-been-saving-for | 1,582,519,997,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145897.19/warc/CC-MAIN-20200224040929-20200224070929-00490.warc.gz | 750,935,047 | 5,772 | # macro economy
December 1994 a man in ohio deposited all of the 8 million pennies
He had been saving for 65 year(his deposit weghed 48000 pounds)
with reserve requirement of 55 what will be the cumulative change
to the baking system?
a)transaction deposit
b)Total reserves
c)lending capacity
if the saved the same number every year hpw much did he loose by not depositing in and intrest account at 5%
I find your reserve requirement of 55 highly unusual.
The money multiplier is 1/rr. So the initial transaction deposit of \$80000 will translate into 80,000*(1/.55). Reserves will be 55% of his. Initial lending capacity started at 80,000*.45, but will eventially go to zero.
Use a spreadsheet for part b. Assume a zero tax rate. (or positive rate, the spreadsheet will handle the calculations equally well).
thank you "big kiss"
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http://theinfolist.com/html/ALL/s/dihedral_group.html | 1,680,281,073,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949644.27/warc/CC-MAIN-20230331144941-20230331174941-00512.warc.gz | 52,166,291 | 13,868 | TheInfoList
OR:
In mathematics, a dihedral group is the
group A group is a number of persons or things that are located, gathered, or classed together. Groups of people * Cultural group, a group whose members share the same cultural identity * Ethnic group, a group whose members share the same ethnic ide ...
of
symmetries Symmetry (from grc, συμμετρία "agreement in dimensions, due proportion, arrangement") in everyday language refers to a sense of harmonious and beautiful proportion and balance. In mathematics, "symmetry" has a more precise definiti ...
of a
regular polygon In Euclidean geometry, a regular polygon is a polygon that is direct equiangular (all angles are equal in measure) and equilateral (all sides have the same length). Regular polygons may be either convex, star or skew. In the limit, a sequence ...
, which includes
rotations Rotation, or spin, is the circular movement of an object around a '' central axis''. A two-dimensional rotating object has only one possible central axis and can rotate in either a clockwise or counterclockwise direction. A three-dimensional ...
and reflections. Dihedral groups are among the simplest examples of
finite group Finite is the opposite of infinite. It may refer to: * Finite number (disambiguation) * Finite set, a set whose cardinality (number of elements) is some natural number * Finite verb Traditionally, a finite verb (from la, fīnītus, past particip ...
s, and they play an important role in
group theory In abstract algebra, group theory studies the algebraic structures known as groups. The concept of a group is central to abstract algebra: other well-known algebraic structures, such as rings, fields, and vector spaces, can all be seen as ...
,
geometry Geometry (; ) is, with arithmetic, one of the oldest branches of mathematics. It is concerned with properties of space such as the distance, shape, size, and relative position of figures. A mathematician who works in the field of geometry is ca ...
, and
chemistry Chemistry is the scientific study of the properties and behavior of matter. It is a natural science that covers the elements that make up matter to the compounds made of atoms, molecules and ions: their composition, structure, properties, ...
. The notation for the dihedral group differs in
geometry Geometry (; ) is, with arithmetic, one of the oldest branches of mathematics. It is concerned with properties of space such as the distance, shape, size, and relative position of figures. A mathematician who works in the field of geometry is ca ...
and
abstract algebra In mathematics, more specifically algebra, abstract algebra or modern algebra is the study of algebraic structures. Algebraic structures include groups, rings, fields, modules, vector spaces, lattices, and algebras over a field. The term ''a ...
. In
geometry Geometry (; ) is, with arithmetic, one of the oldest branches of mathematics. It is concerned with properties of space such as the distance, shape, size, and relative position of figures. A mathematician who works in the field of geometry is ca ...
, or refers to the symmetries of the -gon, a group of order . In
abstract algebra In mathematics, more specifically algebra, abstract algebra or modern algebra is the study of algebraic structures. Algebraic structures include groups, rings, fields, modules, vector spaces, lattices, and algebras over a field. The term ''a ...
, refers to this same dihedral group. This article uses the geometric convention, .
Definition
Elements
A regular polygon with $n$ sides has $2n$ different symmetries: $n$
rotational symmetries Rotational symmetry, also known as radial symmetry in geometry, is the property a shape has when it looks the same after some rotation by a partial turn. An object's degree of rotational symmetry is the number of distinct orientations in which i ...
and $n$
reflection symmetries In mathematics, reflection symmetry, line symmetry, mirror symmetry, or mirror-image symmetry is symmetry with respect to a reflection. That is, a figure which does not change upon undergoing a reflection has reflectional symmetry. In 2D the ...
. Usually, we take $n \ge 3$ here. The associated
rotation Rotation, or spin, is the circular movement of an object around a '' central axis''. A two-dimensional rotating object has only one possible central axis and can rotate in either a clockwise or counterclockwise direction. A three-dimensional ...
s and reflections make up the dihedral group $\mathrm_n$. If $n$ is odd, each axis of symmetry connects the midpoint of one side to the opposite vertex. If $n$ is even, there are $n/2$ axes of symmetry connecting the midpoints of opposite sides and $n/2$ axes of symmetry connecting opposite vertices. In either case, there are $n$ axes of symmetry and $2n$ elements in the symmetry group. Reflecting in one axis of symmetry followed by reflecting in another axis of symmetry produces a rotation through twice the angle between the axes. The following picture shows the effect of the sixteen elements of $\mathrm_8$ on a
stop sign A stop sign is a traffic sign designed to notify drivers that they must come to a complete stop and make sure the intersection is safely clear of vehicles and pedestrians before continuing past the sign. In many countries, the sign is a red oct ...
: The first row shows the effect of the eight rotations, and the second row shows the effect of the eight reflections, in each case acting on the stop sign with the orientation as shown at the top left.
Group structure
As with any geometric object, the
composition Composition or Compositions may refer to: Arts and literature *Composition (dance), practice and teaching of choreography *Composition (language), in literature and rhetoric, producing a work in spoken tradition and written discourse, to include v ...
of two symmetries of a regular polygon is again a symmetry of this object. With composition of symmetries to produce another as the binary operation, this gives the symmetries of a polygon the algebraic structure of a
finite group Finite is the opposite of infinite. It may refer to: * Finite number (disambiguation) * Finite set, a set whose cardinality (number of elements) is some natural number * Finite verb Traditionally, a finite verb (from la, fīnītus, past particip ...
. The following
Cayley table Named after the 19th century British mathematician Arthur Cayley, a Cayley table describes the structure of a finite group by arranging all the possible products of all the group's elements in a square table reminiscent of an addition or multiplic ...
shows the effect of composition in the group D3 (the symmetries of an
equilateral triangle In geometry, an equilateral triangle is a triangle in which all three sides have the same length. In the familiar Euclidean geometry, an equilateral triangle is also equiangular; that is, all three internal angles are also congruent to each oth ...
). r0 denotes the identity; r1 and r2 denote counterclockwise rotations by 120° and 240° respectively, and s0, s1 and s2 denote reflections across the three lines shown in the adjacent picture. For example, , because the reflection s1 followed by the reflection s2 results in a rotation of 120°. The order of elements denoting the
composition Composition or Compositions may refer to: Arts and literature *Composition (dance), practice and teaching of choreography *Composition (language), in literature and rhetoric, producing a work in spoken tradition and written discourse, to include v ...
is right to left, reflecting the convention that the element acts on the expression to its right. The composition operation is not commutative. In general, the group D''n'' has elements r0, ..., r''n''−1 and s0, ..., s''n''−1, with composition given by the following formulae: :$\mathrm_i\,\mathrm_j = \mathrm_, \quad \mathrm_i\,\mathrm_j = \mathrm_, \quad \mathrm_i\,\mathrm_j = \mathrm_, \quad \mathrm_i\,\mathrm_j = \mathrm_.$ In all cases, addition and subtraction of subscripts are to be performed using
modular arithmetic In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" when reaching a certain value, called the modulus. The modern approach to modular arithmetic was developed by Carl Friedrich Gauss in his book ...
with modulus ''n''.
Matrix representation
If we center the regular polygon at the origin, then elements of the dihedral group act as
linear transformations In mathematics, and more specifically in linear algebra, a linear map (also called a linear mapping, linear transformation, vector space homomorphism, or in some contexts linear function) is a mapping V \to W between two vector spaces that pres ...
of the plane. This lets us represent elements of D''n'' as
matrices Matrix most commonly refers to: * ''The Matrix'' (franchise), an American media franchise ** ''The Matrix'', a 1999 science-fiction action film ** "The Matrix", a fictional setting, a virtual reality environment, within ''The Matrix'' (franchis ...
, with composition being
matrix multiplication In mathematics, particularly in linear algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the ...
. This is an example of a (2-dimensional)
group representation In the mathematical field of representation theory, group representations describe abstract groups in terms of bijective linear transformations of a vector space to itself (i.e. vector space automorphisms); in particular, they can be used to re ...
. For example, the elements of the group D4 can be represented by the following eight matrices: : In general, the matrices for elements of D''n'' have the following form: :$\begin \mathrm_k & = \begin \cos \frac & -\sin \frac \\ \sin \frac & \cos \frac \end\ \ \text \\ \mathrm_k & = \begin \cos \frac & \sin \frac \\ \sin \frac & -\cos \frac \end . \end$ r''k'' is a rotation matrix, expressing a counterclockwise rotation through an angle of . s''k'' is a reflection across a line that makes an angle of with the ''x''-axis.
Other definitions
Further equivalent definitions of are:
Small dihedral groups
is
isomorphic In mathematics, an isomorphism is a structure-preserving mapping between two structures of the same type that can be reversed by an inverse mapping. Two mathematical structures are isomorphic if an isomorphism exists between them. The word iso ...
to , the
cyclic group In group theory, a branch of abstract algebra in pure mathematics, a cyclic group or monogenous group is a group, denoted C''n'', that is generated by a single element. That is, it is a set of invertible elements with a single associative binary ...
of order 2. is
isomorphic In mathematics, an isomorphism is a structure-preserving mapping between two structures of the same type that can be reversed by an inverse mapping. Two mathematical structures are isomorphic if an isomorphism exists between them. The word iso ...
to , the Klein four-group. and are exceptional in that: * and are the only abelian dihedral groups. Otherwise, is non-abelian. * is a subgroup of the symmetric group for . Since for or , for these values, is too large to be a subgroup. * The inner automorphism group of is trivial, whereas for other even values of , this is . The cycle graphs of dihedral groups consist of an ''n''-element cycle and ''n'' 2-element cycles. The dark vertex in the cycle graphs below of various dihedral groups represents the identity element, and the other vertices are the other elements of the group. A cycle consists of successive powers of either of the elements connected to the
identity element In mathematics, an identity element, or neutral element, of a binary operation operating on a set is an element of the set that leaves unchanged every element of the set when the operation is applied. This concept is used in algebraic structures ...
.
The dihedral group as symmetry group in 2D and rotation group in 3D
An example of abstract group , and a common way to visualize it, is the group of Euclidean plane isometries which keep the origin fixed. These groups form one of the two series of discrete point groups in two dimensions. consists of
rotation Rotation, or spin, is the circular movement of an object around a '' central axis''. A two-dimensional rotating object has only one possible central axis and can rotate in either a clockwise or counterclockwise direction. A three-dimensional ...
s of multiples of about the origin, and reflections across lines through the origin, making angles of multiples of with each other. This is the symmetry group of a
regular polygon In Euclidean geometry, a regular polygon is a polygon that is direct equiangular (all angles are equal in measure) and equilateral (all sides have the same length). Regular polygons may be either convex, star or skew. In the limit, a sequence ...
with sides (for ; this extends to the cases and where we have a plane with respectively a point offset from the "center" of the "1-gon" and a "2-gon" or line segment). is generated by a rotation of
order Order, ORDER or Orders may refer to: * Categorization, the process in which ideas and objects are recognized, differentiated, and understood * Heterarchy, a system of organization wherein the elements have the potential to be ranked a number of ...
and a reflection of order 2 such that :$\mathrm = \mathrm^ \,$ In geometric terms: in the mirror a rotation looks like an inverse rotation. In terms of
complex numbers In mathematics, a complex number is an element of a number system that extends the real numbers with a specific element denoted , called the imaginary unit and satisfying the equation i^= -1; every complex number can be expressed in the f ...
: multiplication by $e^$ and
complex conjugation In mathematics, the complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign. That is, (if a and b are real, then) the complex conjugate of a + bi is equal to a - ...
. In matrix form, by setting : and defining $\mathrm_j = \mathrm_1^j$ and $\mathrm_j = \mathrm_j \, \mathrm_0$ for $j \in \$ we can write the product rules for D''n'' as :$\begin \mathrm_j \, \mathrm_k &= \mathrm_ \\ \mathrm_j \, \mathrm_k &= \mathrm_ \\ \mathrm_j \, \mathrm_k &= \mathrm_ \\ \mathrm_j \, \mathrm_k &= \mathrm_ \end$ (Compare coordinate rotations and reflections.) The dihedral group D2 is generated by the rotation r of 180 degrees, and the reflection s across the ''x''-axis. The elements of D2 can then be represented as , where e is the identity or null transformation and rs is the reflection across the ''y''-axis. D2 is
isomorphic In mathematics, an isomorphism is a structure-preserving mapping between two structures of the same type that can be reversed by an inverse mapping. Two mathematical structures are isomorphic if an isomorphism exists between them. The word iso ...
to the Klein four-group. For ''n'' > 2 the operations of rotation and reflection in general do not
commute Commute, commutation or commutative may refer to: * Commuting, the process of travelling between a place of residence and a place of work Mathematics * Commutative property, a property of a mathematical operation whose result is insensitive to th ...
and D''n'' is not abelian; for example, in D4, a rotation of 90 degrees followed by a reflection yields a different result from a reflection followed by a rotation of 90 degrees. Thus, beyond their obvious application to problems of symmetry in the plane, these groups are among the simplest examples of non-abelian groups, and as such arise frequently as easy counterexamples to theorems which are restricted to abelian groups. The elements of can be written as , , , ... , , , , , ... , . The first listed elements are rotations and the remaining elements are axis-reflections (all of which have order 2). The product of two rotations or two reflections is a rotation; the product of a rotation and a reflection is a reflection. So far, we have considered to be a subgroup of , i.e. the group of rotations (about the origin) and reflections (across axes through the origin) of the plane. However, notation is also used for a subgroup of
SO(3) In mechanics and geometry, the 3D rotation group, often denoted SO(3), is the group of all rotations about the origin of three-dimensional Euclidean space \R^3 under the operation of composition. By definition, a rotation about the origin ...
which is also of abstract group type : the proper symmetry group of a ''regular polygon embedded in three-dimensional space'' (if ''n'' ≥ 3). Such a figure may be considered as a degenerate regular solid with its face counted twice. Therefore, it is also called a ''dihedron'' (Greek: solid with two faces), which explains the name ''dihedral group'' (in analogy to ''tetrahedral'', ''octahedral'' and ''icosahedral group'', referring to the proper symmetry groups of a regular tetrahedron, octahedron, and icosahedron respectively).
Examples of 2D dihedral symmetry
File:Imperial Seal of Japan.svg, 2D D16 symmetry – Imperial Seal of Japan, representing eightfold
chrysanthemum Chrysanthemums (), sometimes called mums or chrysanths, are flowering plants of the genus ''Chrysanthemum'' in the family Asteraceae. They are native to East Asia and northeastern Europe. Most species originate from East Asia and the cente ...
with sixteen
petal Petals are modified leaves that surround the reproductive parts of flowers. They are often brightly colored or unusually shaped to attract pollinators. All of the petals of a flower are collectively known as the ''corolla''. Petals are usually ...
s. File:Red Star of David.svg, 2D D6 symmetry – The Red Star of David File:Naval Jack of the Republic of China.svg, 2D D12 symmetry — The Naval Jack of the Republic of China (White Sun) File:Ashoka Chakra.svg, 2D D24 symmetry –
Ashoka Chakra Ashoka (, ; also ''Asoka''; 304 – 232 BCE), popularly known as Ashoka the Great, was the third emperor of the Maurya Empire of Indian subcontinent during to 232 BCE. His empire covered a large part of the Indian subcontinent, s ...
, as depicted on the National flag of the Republic of India.
Properties
The properties of the dihedral groups with depend on whether is even or odd. For example, the
center Center or centre may refer to: Mathematics * Center (geometry), the middle of an object * Center (algebra), used in various contexts ** Center (group theory) ** Center (ring theory) * Graph center, the set of all vertices of minimum eccentr ...
of consists only of the identity if ''n'' is odd, but if ''n'' is even the center has two elements, namely the identity and the element r''n''/2 (with D''n'' as a subgroup of O(2), this is inversion; since it is scalar multiplication by −1, it is clear that it commutes with any linear transformation). In the case of 2D isometries, this corresponds to adding inversion, giving rotations and mirrors in between the existing ones. For ''n'' twice an odd number, the abstract group is isomorphic with the direct product of and . Generally, if ''m''
divides In mathematics, a divisor of an integer n, also called a factor of n, is an integer m that may be multiplied by some integer to produce n. In this case, one also says that n is a multiple of m. An integer n is divisible or evenly divisible by ...
''n'', then has ''n''/''m'' subgroups of type , and one subgroup $\mathbb$''m''. Therefore, the total number of subgroups of (''n'' ≥ 1), is equal to ''d''(''n'') + σ(''n''), where ''d''(''n'') is the number of positive
divisor In mathematics, a divisor of an integer n, also called a factor of n, is an integer m that may be multiplied by some integer to produce n. In this case, one also says that n is a multiple of m. An integer n is divisible or evenly divisible by ...
s of ''n'' and ''σ''(''n'') is the sum of the positive divisors of ''n''. See list of small groups for the cases ''n'' ≤ 8. The dihedral group of order 8 (D4) is the smallest example of a group that is not a T-group. Any of its two Klein four-group subgroups (which are normal in D4) has as normal subgroup order-2 subgroups generated by a reflection (flip) in D4, but these subgroups are not normal in D4.
Conjugacy classes of reflections
All the reflections are
conjugate Conjugation or conjugate may refer to: Linguistics *Grammatical conjugation, the modification of a verb from its basic form * Emotive conjugation or Russell's conjugation, the use of loaded language Mathematics * Complex conjugation, the change ...
to each other whenever ''n'' is odd, but they fall into two conjugacy classes if ''n'' is even. If we think of the isometries of a regular ''n''-gon: for odd ''n'' there are rotations in the group between every pair of mirrors, while for even ''n'' only half of the mirrors can be reached from one by these rotations. Geometrically, in an odd polygon every axis of symmetry passes through a vertex and a side, while in an even polygon there are two sets of axes, each corresponding to a conjugacy class: those that pass through two vertices and those that pass through two sides. Algebraically, this is an instance of the conjugate Sylow theorem (for ''n'' odd): for ''n'' odd, each reflection, together with the identity, form a subgroup of order 2, which is a Sylow 2-subgroup ( is the maximum power of 2 dividing ), while for ''n'' even, these order 2 subgroups are not Sylow subgroups because 4 (a higher power of 2) divides the order of the group. For ''n'' even there is instead an
outer automorphism In mathematics, the outer automorphism group of a group, , is the quotient, , where is the automorphism group of and ) is the subgroup consisting of inner automorphisms. The outer automorphism group is usually denoted . If is trivial and has ...
interchanging the two types of reflections (properly, a class of outer automorphisms, which are all conjugate by an inner automorphism).
Automorphism group
The automorphism group of is isomorphic to the holomorph of $\mathbb$/''n''$\mathbb$, i.e., to and has order ''nϕ''(''n''), where ''ϕ'' is Euler's totient function, the number of ''k'' in coprime to ''n''. It can be understood in terms of the generators of a reflection and an elementary rotation (rotation by ''k''(2''π''/''n''), for ''k''
coprime In mathematics, two integers and are coprime, relatively prime or mutually prime if the only positive integer that is a divisor of both of them is 1. Consequently, any prime number that divides does not divide , and vice versa. This is equivale ...
to ''n''); which automorphisms are inner and outer depends on the parity of ''n''. * For ''n'' odd, the dihedral group is centerless, so any element defines a non-trivial inner automorphism; for ''n'' even, the rotation by 180° (reflection through the origin) is the non-trivial element of the center. * Thus for ''n'' odd, the inner automorphism group has order 2''n'', and for ''n'' even (other than ) the inner automorphism group has order ''n''. * For ''n'' odd, all reflections are conjugate; for ''n'' even, they fall into two classes (those through two vertices and those through two faces), related by an outer automorphism, which can be represented by rotation by ''π''/''n'' (half the minimal rotation). * The rotations are a normal subgroup; conjugation by a reflection changes the sign (direction) of the rotation, but otherwise leaves them unchanged. Thus automorphisms that multiply angles by ''k'' (coprime to ''n'') are outer unless .
Examples of automorphism groups
has 18
inner automorphism In abstract algebra an inner automorphism is an automorphism of a group, ring, or algebra given by the conjugation action of a fixed element, called the ''conjugating element''. They can be realized via simple operations from within the group it ...
s. As 2D isometry group D9, the group has mirrors at 20° intervals. The 18 inner automorphisms provide rotation of the mirrors by multiples of 20°, and reflections. As isometry group these are all automorphisms. As abstract group there are in addition to these, 36
outer automorphism In mathematics, the outer automorphism group of a group, , is the quotient, , where is the automorphism group of and ) is the subgroup consisting of inner automorphisms. The outer automorphism group is usually denoted . If is trivial and has ...
s; e.g., multiplying angles of rotation by 2. has 10 inner automorphisms. As 2D isometry group D10, the group has mirrors at 18° intervals. The 10 inner automorphisms provide rotation of the mirrors by multiples of 36°, and reflections. As isometry group there are 10 more automorphisms; they are conjugates by isometries outside the group, rotating the mirrors 18° with respect to the inner automorphisms. As abstract group there are in addition to these 10 inner and 10 outer automorphisms, 20 more outer automorphisms; e.g., multiplying rotations by 3. Compare the values 6 and 4 for Euler's totient function, the multiplicative group of integers modulo ''n'' for ''n'' = 9 and 10, respectively. This triples and doubles the number of automorphisms compared with the two automorphisms as isometries (keeping the order of the rotations the same or reversing the order). The only values of ''n'' for which ''φ''(''n'') = 2 are 3, 4, and 6, and consequently, there are only three dihedral groups that are isomorphic to their own automorphism groups, namely (order 6), (order 8), and (order 12).
Inner automorphism group
The inner automorphism group of is isomorphic to: * if ''n'' is odd; * if is even (for , ).
Generalizations
There are several important generalizations of the dihedral groups: * The
infinite dihedral group In mathematics, the infinite dihedral group Dih∞ is an infinite group with properties analogous to those of the finite dihedral groups. In two-dimensional geometry, the infinite dihedral group represents the frieze group symmetry, ''p1m1' ...
is an infinite group with algebraic structure similar to the finite dihedral groups. It can be viewed as the group of symmetries of the
integer An integer is the number zero (), a positive natural number (, , , etc.) or a negative integer with a minus sign (−1, −2, −3, etc.). The negative numbers are the additive inverses of the corresponding positive numbers. In the languag ...
s. * The
orthogonal group In mathematics, the orthogonal group in dimension , denoted , is the group of distance-preserving transformations of a Euclidean space of dimension that preserve a fixed point, where the group operation is given by composing transformations. T ...
O(2), ''i.e.,'' the symmetry group of the
circle A circle is a shape consisting of all points in a plane that are at a given distance from a given point, the centre. Equivalently, it is the curve traced out by a point that moves in a plane so that its distance from a given point is const ...
, also has similar properties to the dihedral groups. * The family of generalized dihedral groups includes both of the examples above, as well as many other groups. * The quasidihedral groups are family of finite groups with similar properties to the dihedral groups.
* Coordinate rotations and reflections * Cycle index of the dihedral group * Dicyclic group *
Dihedral group of order 6 In mathematics, D3 (sometimes alternatively denoted by D6) is the dihedral group of degree 3, or, in other words, the dihedral group of order 6. It is isomorphic to the symmetric group S3 of degree 3. It is also the smallest possible non-abel ...
* Dihedral group of order 8 * Dihedral symmetry groups in 3D * Dihedral symmetry in three dimensions | 6,363 | 27,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 24, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-14 | latest | en | 0.926863 |
https://www.reilsolar.com/ncert-solutions-for-class-6-maths-chapter-8-decimals-ex-8-5/ | 1,722,793,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640408316.13/warc/CC-MAIN-20240804164455-20240804194455-00067.warc.gz | 780,440,560 | 49,637 | # NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5
Ncert Solutions For Class 6 Maths Chapter 8 Decimals Ex 8.5 Class 6 Maths Chapter 8 Exercise 8.5 Ncert Solutions For Class 6 Maths Chapter 8 Exercise 8.5 Ncert Class 6 Maths Chapter 8 Exercise 8.5 Ncert Solutions Class 6 Maths Chapter 8 Exercise 8.5 Ncert Solutions For Class 6 Maths Chapter 8 Decimals Exercise 8.5 Ncert Maths Class 6 Chapter 8 Exercise 8.5 Ncert Maths Book Class 6 Chapter 8 Exercise 8.5
## NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5
Question 1.
Find the sum in each of the following:
(a) 0.007+8.5+30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38.
Solution.
Question 2.
Rashid spent Rs.35.75 for Maths book and Rs.32.60 for Science book. Find the total amount spent by Rashid.
Solution.
Money spent for Maths book = Rs. 35.75
Money spent for Science book = Rs. 32.60
Total amount spent by Rashid = Rs. 35.75 +Rs.32.60
= Rs. 68.35.
Question 3.
Radhika’s mother gave her ? 10.50 and her father gave her ? 15.80, find the total amount given to Radhika by the parents.
Solution.
Money given by mother = Rs. 10.50
Money given by father = Rs. 15.80
Total amount given to Radhika by parents = Rs. 10.50 +Rs.15.80 =Rs. 26.30
Question 4.
Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Solution.
Cloth bought for her shirt
Question 5.
Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Solution.
Distance walked by Naresh in the morning
Question 6.
Sunita travels 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
Solution.
Question 7.
Ravi purchased 5 kg 400 grice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.
Solution.
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https://www.conceptdraw.com/examples/divided-chart | 1,713,690,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00013.warc.gz | 659,691,121 | 14,254 | This site uses cookies. By continuing to browse the ConceptDraw site you are agreeing to our Use of Site Cookies.
HelpDesk
# How to Draw a Divided Bar Chart in ConceptDraw PRO
A divided bar graph is a rectangle divided into smaller rectangles along its length in proportion to the data. Segments in a divided bar represent a set of quantities according to the different proportion of the total amount. A divided bar diagram is created using rectangular bars to depict proportionally the size of each category. The bars in a divided bar graph can be vertical or horizontal. The size of the each rectangle displays the part that each category represents. The value of the exact size of the whole must be known, because the each section of the bar displays a piece of that value. A divided bar diagram is rather similar to a sector diagram in that the bar shows the entire data amount and the bar is divided into several parts to represent the proportional size of each category. ConceptDraw PRO in conjunction with Divided Bar Diagrams solution provides tools to create stylish divided bar charts for your presentations.
## Divided Bar Diagrams
The Divided Bar Diagrams Solution extends the capabilities of ConceptDraw PRO v10 with templates, samples, and a library of vector stencils for drawing high impact and professional Divided Bar Diagrams and Graphs, Bar Diagram Math, and Stacked Graph.
## How to Draw a Divided Bar Chart in ConceptDraw PRO
A divided bar chart is a rectangle divided into segments, illustrating numerical proportion. In a divided bar chart, the length of each segment (and consequently its area), is proportional to the quantity it represents.
This divided bar graph displays the number of people per day who visited swimming pool.
The divided bar diagram template for the ConceptDraw PRO diagramming and vector drawing software is included in the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart template
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
A Divided Bar graph is a rectangle divided into smaller rectangles along its length in proportion to the data. Segments in a Divided Bar represent a set of quantities according to the different proportions of the total amount.
Divided Bars are used for visual representation of the collected information.
The design elements library Divided bar diagrams contains the 10 vector templates of divided bar charts.
Use the design elements library Divided bar diagrams to draw your diagrams and infographics using the ConceptDraw PRO diagramming and vector drawing software.
The example "Design elements - Divided bar diagrams" was created using ConceptDraw PRO diagramming and vector drawing software extended with the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart templates
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
This divided bar chart sample was drawn on the base of the infographics "London Elections Results by Numbers" from the london.gov.uk website.
[data.london.gov.uk/ datastore/ package/ london-elections-results-2012-wards-boroughs-constituency]
"There are six types of elections in the United Kingdom: United Kingdom general elections, elections to devolved parliaments and assemblies, elections to the European Parliament, local elections, mayoral elections and Police and Crime Commissioner elections. Elections are held on Election Day, which is conventionally a Thursday. General elections do not have fixed dates, but must be called within five years of the opening of parliament following the last election. Other elections are held on fixed dates, though in the case of the devolved assemblies and parliaments early elections can occur in certain situations. Currently, six electoral systems are used: the single member plurality system (First Past the Post), the multi member plurality system, party list PR, the single transferable vote, the Additional Member System and the Supplementary Vote.
Elections are administered locally: in each lower-tier local authority, the actual polling procedure is run by the Returning Officer and the compiling and maintenance of the electoral roll by the Electoral Registration Officer (except in Northern Ireland, where the Electoral Office for Northern Ireland assumes both responsibilities). The Electoral Commission only sets standards for and issues guidelines to Returning Officers and Electoral Registration Officers, but is responsible for nationwide electoral administration (such as the registration of political parties and directing the administration of national referendums)." [Elections in the United Kingdom. Wikipedia]
The divided bar diagram example "London election results" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
This divided bar chart example was drawn on the base of the picture "Petroleum Products Yielded from One Barrel of Crude Oil in California" from the California Energy Almanac website. [energyalmanac.ca.gov/ gasoline/ whats_ in_ barrel_ oil.html]
"Petroleum products are useful materials derived from crude oil (petroleum) as it is processed in oil refineries. Unlike petrochemicals, which are a collection of well-defined usually pure chemical compounds, petroleum products are complex mixtures. The majority of petroleum is converted to petroleum products, which includes several classes of fuels.
According to the composition of the crude oil and depending on the demands of the market, refineries can produce different shares of petroleum products. The largest share of oil products is used as "energy carriers", i.e. various grades of fuel oil and gasoline. These fuels include or can be blended to give gasoline, jet fuel, diesel fuel, heating oil, and heavier fuel oils. Heavier (less volatile) fractions can also be used to produce asphalt, tar, paraffin wax, lubricating and other heavy oils. Refineries also produce other chemicals, some of which are used in chemical processes to produce plastics and other useful materials. Since petroleum often contains a few percent sulfur-containing molecules, elemental sulfur is also often produced as a petroleum product. Carbon, in the form of petroleum coke, and hydrogen may also be produced as petroleum products. The hydrogen produced is often used as an intermediate product for other oil refinery processes such as hydrocracking and hydrodesulfurization." [Petroleum product. Wikipedia]
The divided bar diagram example "Petroleum products yielded from one barrel of crude oil" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
The vector stencils library "Composition charts" contains 6 templates: 2 pie charts, 2 divided bar charts and 2 double divided bar charts.
The vector stencils library "Composition indicators" contains 6 templates of visual indicators.
Use these design elements to visualize composition comparison of your data in your visual dashboard created using the ConceptDraw PRO diagramming and vector drawing software.
"A pie chart (or a circle graph) is a circular chart divided into sectors, illustrating numerical proportion. In a pie chart, the arc length of each sector (and consequently its central angle and area), is proportional to the quantity it represents. While it is named for its resemblance to a pie which has been sliced, there are variations on the way it can be presented. ...
An obvious flaw exhibited by pie charts is that they cannot show more than a few values without separating the visual encoding (the “slices”) from the data they represent (typically percentages). When slices become too small, pie charts have to rely on colors, textures or arrows so the reader can understand them. This makes them unsuitable for use with larger amounts of data. Pie charts also take up a larger amount of space on the page compared to the more flexible alternative of bar charts, which do not need to have separate legends, and can also display other values such as averages or targets at the same time. ...
Most subjects have difficulty ordering the slices in the pie chart by size; when the bar chart is used the comparison is much easier. Similarly, comparisons between data sets are easier using the bar chart. However, if the goal is to compare a given category (a slice of the pie) with the total (the whole pie) in a single chart and the multiple is close to 25 or 50 percent, then a pie chart can often be more effective than a bar graph." [Pie chart. Wikipedia]
The example "Design elements - Composition charts and indicators" is included in the Composition Dashboard solution from the area "What is a Dashboard" of ConceptDraw Solution Park.
Composition charts and visual indicators
Used Solutions
## Relative Value Chart Software
Relative value is a value measured in the terms of liquidity, risks, return of one investment or financial instrument relative to another.
Relative values are widely used in:business, economics, investment, management, marketing, statistics, etc.
ConceptDraw PRO software is the best for drawing professional looking Relative Value Charts. ConceptDraw PRO provides Divided Bar Diagrams solution from the Graphs and Charts Area of ConceptDraw Solution Park.
## How to Draw a Divided Bar Chart in ConceptDraw PRO
This example was created on the base of the figure "Composition of estates" from the website of the Inland Revenue Department of the Government of the Hong Kong Special Administrative Region.
[ird.gov.hk/ dar/ 2011-12/ en/ assessing.html]
"An estate is the net worth of a person at any point in time alive or dead. It is the sum of a person's assets - legal rights, interests and entitlements to property of any kind - less all liabilities at that time. The issue is of special legal significance on a question of bankruptcy and death of the person.
Depending on the particular context, the term is also used in reference to an estate in land or of a particular kind of property (such as real estate or personal estate). The term is also used to refer to the sum of a person's assets only." [Estate (law). Wikipedia]
The divided bar diagram example "Composition of estates" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
This divided bar chart sample was drawn on the base of the infographics "London Elections Results by Numbers" from the london.gov.uk website.
[data.london.gov.uk/ datastore/ package/ london-elections-results-2012-wards-boroughs-constituency]
"There are six types of elections in the United Kingdom: United Kingdom general elections, elections to devolved parliaments and assemblies, elections to the European Parliament, local elections, mayoral elections and Police and Crime Commissioner elections. Elections are held on Election Day, which is conventionally a Thursday. General elections do not have fixed dates, but must be called within five years of the opening of parliament following the last election. Other elections are held on fixed dates, though in the case of the devolved assemblies and parliaments early elections can occur in certain situations. Currently, six electoral systems are used: the single member plurality system (First Past the Post), the multi member plurality system, party list PR, the single transferable vote, the Additional Member System and the Supplementary Vote.
Elections are administered locally: in each lower-tier local authority, the actual polling procedure is run by the Returning Officer and the compiling and maintenance of the electoral roll by the Electoral Registration Officer (except in Northern Ireland, where the Electoral Office for Northern Ireland assumes both responsibilities). The Electoral Commission only sets standards for and issues guidelines to Returning Officers and Electoral Registration Officers, but is responsible for nationwide electoral administration (such as the registration of political parties and directing the administration of national referendums)." [Elections in the United Kingdom. Wikipedia]
The divided bar diagram example "London election results" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
This example was created on the base of the figure "Composition of estates" from the website of the Inland Revenue Department of the Government of the Hong Kong Special Administrative Region.
[ird.gov.hk/ dar/ 2011-12/ en/ assessing.html]
"An estate is the net worth of a person at any point in time alive or dead. It is the sum of a person's assets - legal rights, interests and entitlements to property of any kind - less all liabilities at that time. The issue is of special legal significance on a question of bankruptcy and death of the person.
Depending on the particular context, the term is also used in reference to an estate in land or of a particular kind of property (such as real estate or personal estate). The term is also used to refer to the sum of a person's assets only." [Estate (law). Wikipedia]
The divided bar diagram example "Composition of estates" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
A Divided Bar graph is a rectangle divided into smaller rectangles along its length in proportion to the data. Segments in a Divided Bar represent a set of quantities according to the different proportions of the total amount.
The example "Divided bar graph - Global natural gas consumption (2010)" was drawn on the base of International Energy Statistics from the U.S. Energy Information Administration (EIA) website. [eia.gov/ cfapps/ ipdbproject/ IEDIndex3.cfm?tid=3&pid=26&aid=2]
"Natural gas is a fossil fuel formed when layers of buried plants and animals are exposed to intense heat and pressure over thousands of years. The energy that the plants originally obtained from the sun is stored in the form of chemical bonds in natural gas. Natural gas is a nonrenewable resource because it cannot be replenished on a human time frame. Natural gas is a hydrocarbon gas mixture consisting primarily of methane, but commonly includes varying amounts of other higher alkanes and even a lesser percentage of carbon dioxide, nitrogen, and hydrogen sulfide. Natural gas is an energy source often used for heating, cooking, and electricity generation. It is also used as fuel for vehicles and as a chemical feedstock in the manufacture of plastics and other commercially important organic chemicals." [Natural gas. Wikipedia]
The diagram example "Divided bar graph - Global natural gas consumption (2010)" was created using ConceptDraw PRO diagramming and vector drawing software extended with the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
This divided bar diagram example was drawn on the base of the chart "Composition of state government revenue in 2006‑07" from the Australia's future tax system website.
[taxreview.treasury.gov.au/ content/ Paper.aspx?doc=html/ publications/ papers/ report/ section_ 10-06.htm]
"Government revenue is money received by a government. It is an important tool of the fiscal policy of the government and is the opposite factor of government spending. Revenues earned by the government are received from sources such as taxes levied on the incomes and wealth accumulation of individuals and corporations and on the goods and services produced, exported and imported from the country, non-taxable sources such as government-owned corporations' incomes, central bank revenue and capital receipts in the form of external loans and debts from international financial institutions." [Government revenue. Wikipedia]
The divided bar diagram example "Composition of state government revenue in 2006‑07" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Divided Bar Diagrams solution from the Graphs and Charts area of ConceptDraw Solution Park.
Divided bar chart
Used Solutions
## How to Draw a Divided Bar Chart in ConceptDraw PRO
This example visual dashboard example shows website traffic metrics. Divided bar chart and pie chart are used to visualize compositional data. Scatter plot is used to visually analyse correlation between two data sets.
"Web traffic is the amount of data sent and received by visitors to a web site. ...
This is determined by the number of visitors and the number of pages they visit. Sites monitor the incoming and outgoing traffic to see which parts or pages of their site are popular and if there are any apparent trends, such as one specific page being viewed mostly by people in a particular country. There are many ways to monitor this traffic and the gathered data is used to help structure sites, highlight security problems or indicate a potential lack of bandwidth not all web traffic is welcome. Some companies offer advertising schemes that, in return for increased web traffic (visitors), pay for screen space on the site. Sites also often aim to increase their web traffic through inclusion on search engines and through search engine optimization." [Web traffic. Wikipedia]
The example "Website traffic dashboard" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Corellation Dashboard solution from the area "What is a Dashboard" of ConceptDraw Solution Park.
Visual dashboard
## Bar Diagram Math
ConceptDraw PRO extended with Divided Bar Diagrams solution from Graphs and Charts area of ConceptDraw Solution Park is the best software for quick and simple drawing the Divided Bar Diagrams and Bar Diagram Math.
## Chart Software for Better Presentations
Easy charting software comes with beautiful chart templates and examples. This makes it easy to create professional charts without prior experience.
Graphs and Charts Area provide a wide collection of professional looking predesigned templates, samples and ready-to-use vector stencils that will help you to draw the charts and diagrams of various types: Pie Charts, Donut Charts, Line Charts, Column Charts, Bar Charts, Pyramids, Scatter Charts, Venn Diagrams, Spider Charts, Area Charts, Divided Bar Graphs.
## Composition Dashboard
Composition dashboard solution extends ConceptDraw PRO software with templates, samples and vector stencils library with charts and indicators for drawing visual dashboards showing data composition.
## Basic Diagramming
This solution extends ConceptDraw PRO software with the specific tools you need to easily draw flowcharts, block diagrams, histograms, pie charts, divided bar diagrams, line graphs, circular arrows diagrams, Venn diagrams, bubble diagrams and concept maps
## Basic Diagramming
Create flowcharts, organizational charts, bar charts, line graphs, and more with ConceptDraw PRO.
## Pie Chart Examples and Templates
A pie chart or a circle graph is a circular chart divided into sectors, illustrating numerical proportion. In a pie chart, the arc length of each sector and consequently its central angle and area, is proportional to the quantity it represents.
Pie chart examples and templates created using ConceptDraw PRO software helps you get closer with pie charts and find pie chart examples suite your needs. | 3,964 | 20,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.887084 |
https://open.metu.edu.tr/handle/11511/36619 | 1,721,467,682,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00562.warc.gz | 375,509,125 | 11,115 | # Axisymmetric crack problem for a hollow cylinder imbedded in a dissimilar medium
2005-05-01
The analytical solution for the linear elastic problem of flat annular crack in a transversely isotropic hollow cylinder imbedded in a transversely isotropic medium is considered. The hollow cylinder is assumed to be perfectly bonded to the surrounding medium. This structure, which can represent a cylindrical coating-substrate system, is subjected to uniform crack surface pressure. Because of the geometry and the loading, the problem is axisymmetric. The z = 0 plane on which the crack lies, is also a plane of symmetry. The composite media consisting of the hollow cylinder and the surrounding medium extends to infinity in z and r directions. The mixed boundary value problem is formulated in terms of the unknown derivative of the crack surface displacement by using Fourier and Hankel transforms. By extending the crack to the inner surface and to the interface, the cases of surface crack and crack terminating at the interface are obtained. Asymptotic analyses are performed to derive the generalized Cauchy kernel and associated stress singularities. The resulting singular integral equation is solved numerically. Stress intensity factors for various crack configurations, crack opening displacements and stresses along the interface and on z = 0 plane are presented for sample material combinations and geometric parameters.
International Journal of Engineering Science
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Citation Formats
F. S. Kadıoğlu, “Axisymmetric crack problem for a hollow cylinder imbedded in a dissimilar medium,” International Journal of Engineering Science, pp. 617–638, 2005, Accessed: 00, 2020. [Online]. Available: https://hdl.handle.net/11511/36619. | 1,042 | 4,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.89722 |
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REQUEST: audio comparer
(1/2) > >>
agentsteal:
Could someone please write a program that takes two 1-second-long wav files and outputs a number that represents the difference between the sounds? It would output 0 if the samples are exactly the same. If the sounds are a good match it would output a small number but if they are very different it would output a large number. My goal is to be able to replace the sound during one second of an audio file with one of my own sounds, and have it sound as close as possible to the original.
mouser:
It's actually sounds like an interesting project for someone who is interested in sound -- comparing two sounds and coming up with some metric of similarity.
I'm not exactly sure what a good metric would be -- perhaps you would want to come up with multiple metrics.. but clearly there are some good examples of this being used in practice -- such as those recent services that can identify a song from a small snippet.
Maybe a coder here who works with sound files will give this a shot.. It's also possible you could find some commandline hardcore sound file analysis tools that would extract info you could use to compute a simple metric of similarity.
Essentially what i'm saying is that this is not an unreasonable request -- just have to find someone with a little experience with working with sound files at a low-level who understands enough about them to come up with some useful metrics to measure similarity.
Jan-S:
Hey! This sure sounds interesting... anybody willing to discuss possible metrics?
On a first thought, I'd split the file into N pieces, multiply each piece with a Gaussian window function to reduce the sharp edges, perform a fourier transformation of each piece and calculate the absolute values, getting rid of the phase information. These transformed pieces can be treated as N vectors.
To obtain the similarity of two files A and B, I'd calculate the vectors for both files, compute the cosine of the angles between each of the N pairs of vectors (dot product divided by norms), add their absolute values up and divide the result by N.
This would give a number between 1 (sounds are identical w.r.t. the metrics) and 0 (sounds do not have anything to do with each other).
For N=1, this metric would be based solely on the frequency components of the sounds; for greater values of N it would include timing information (i.e. when did particular frequencies occur).
Any comments on this approach? Other ideas? Questions?
@agentsteal: Are the sound files all the same format (sampling rate, bit depth)? Mono or stereo? I'd be interested in experimenting with the metrics, but not that interested in figuring out how to parse or resample the files.
What kind of sounds do you have? Could you provide some examples (denoting whether you would classify them as similar or not)? This would be important to test and tune the metrics.
agentsteal:
I don't think it would be a problem to require that all sounds are the same format.
I don't have any specific examples. I just think it would be a cool program to experiment with. It would be interesting to see if it's possible to remake songs etc out of a set of sounds.
I guess you can use midi as an example. All of the wav to midi converters that are available now compare tones rather than sounds. All they find is the basic notes. But I bet you could convert the actual sound of the wav file to midi with a program that compares the sounds of all combinations of 128 notes on 128 instruments.
I was wondering: does sound have a dithering effect? Like would playing a really high note followed by a really low note equal a note in the middle?
Jan-S:
I don't have any specific examples. I just think it would be a cool program to experiment with. It would be interesting to see if it's possible to remake songs etc out of a set of sounds.-agentsteal (December 05, 2008, 11:48 AM)
--- End quote ---
Without any examples, I don't have anything to test the approach against. Besides, you would need a really huge set of sounds then, if you try to recombine it out of 1 second pieces (which is quite long) and still be able to recognize anything!
I guess you can use midi as an example. All of the wav to midi converters that are available now compare tones rather than sounds. All they find is the basic notes. But I bet you could convert the actual sound of the wav file to midi with a program that compares the sounds of all combinations of 128 notes on 128 instruments.-agentsteal (December 05, 2008, 11:48 AM)
--- End quote ---
Intelliscore Music Recognition claims to be able to recognize more than one instrument, at least if you tell it which instruments it should listen to, but you still have to manually tweak the MIDI file afterwards (the results are not perfect).
The approach of comparing sounds might theoretically work, but there are several challenges: Firstly, you have to find metrics that are robust to noise and to overlayed instruments playing at the same time. For instance, if you are looking at an excerpt of the file where both a piano and a guitar play a chord composed of several notes, the comparison will find good matches for several piano, guitar, banjo, ukulele or harp sounds. It's difficult to construct the metrics such that the piano and guitar matches are guaranteed to produce the highest scores.
Secondly, you do not only have to recognize which instrument is playing which note, but also when the note is played. Unfortunately, the frequency information is imprecise when you increase the timing resolution and vice versa (this is related to the Heisenberg uncertainty principle). Finding a way to recognize a note attack so the comparison can be applied at the right position is not easy - especially as there are instruments (like violins or flutes) that do not have a sharp attack and can play a note at almost arbitrary length.
A funny approach would be randomly generating MIDI files, comparing their sound to the original song and evolving the best matches by recombination and mutation (Genetic algorithms). But probably this will take ages because the searching space is too big :)
I was wondering: does sound have a dithering effect? Like would playing a really high note followed by a really low note equal a note in the middle?
-agentsteal (December 05, 2008, 11:48 AM)
--- End quote ---
I don't think so. Even if you play two pure tones at the same time, the brain is quite good at figuring out the two tones. You can try this with an audio editor that is capable of generating sine waves. Or maybe I misunderstood your question? | 1,430 | 6,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-43 | longest | en | 0.95842 |
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At 7:51pm on October 3, 2015, wajahat khan said…
solve for x and y
01:x/4-2y/3=-3
x/3+y/3=7
02:465x+75y=615
75x+465=1005
03:furniture is sold for 4,000.the gross profit is 4 times of the cost,what are the cost and the gross profit?
04:mr habbib has Rs,600,000 invested ,part at 9%and part at 8% if the totally yearly income from the two investments is Rs,50,000.how much is invested at 9% and how much is invested at 8%
05:amjad and aslam ages add up to 60 years .in 24 years time amjad will be twice as old as jill is now.how old are jack and jill now? \
plzzzzzzzzz ye swal
At 7:50pm on October 3, 2015, wajahat khan said…
pla bhai ap meje ye sawal kar do gye meje urgent chaye plzz meje inbox kar dijye ga
solve for x and y
01:x/4-2y/3=-3
x/3+y/3=7
02:465x+75y=615
75x+465=1005
03:furniture is sold for 4,000.the gross profit is 4 times of the cost,what are the cost and the gross profit?
04:mr habbib has Rs,600,000 invested ,part at 9%and part at 8% if the totally yearly income from the two investments is Rs,50,000.how much is invested at 9% and how much is invested at 8%
05:amjad and aslam ages add up to 60 years .in 24 years time amjad will be twice as old as jill is now.how old are jack and jill now? \
plzzzzzzzzz ye swal kae dy gy plzzzzzzzzz kal tak
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7 hours ago | 818 | 2,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.889455 |
https://mahoningvalleylanes.com/qa/question-how-much-do-i-make-in-a-year-if-i-make-12-an-hour.html | 1,611,106,204,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00313.warc.gz | 446,308,638 | 6,671 | # Question: How Much Do I Make In A Year If I Make 12 An Hour?
## How much do I make a month if I make 12 an hour?
If you look at an average month as 4 weeks, then you can estimate the average monthly earnings from 12 dollars per hour to be \$1,920 per month..
## What is the yearly salary for 12.50 an hour?
\$25,000 per yearIn this case, you can quickly compute the annual salary by multiplying the hourly wage by 2000. Your hourly pay of \$12.50 is then equivalent to an average annual income of \$25,000 per year.
## How much do you make yearly if you make 11 an hour?
Assuming 40 hours a week, that equals 2,080 hours in a year. Your hourly wage of 11 dollars would end up being about \$22,880 per year in salary.
## What salary is 13 an hour?
40 hours multiplied by 52 weeks is 2,080 working hours in a year. \$13 per hour multiplied by 2,080 working hours per year is an annual income of \$27,040 per year.
## Is 13.75 an hour good?
\$13.75 per hour working 2,000 hours In this case, you can quickly compute the annual salary by multiplying the hourly wage by 2000. Your hourly pay of \$13.75 is then equivalent to an average annual income of \$27,500 per year.
## Is \$14 hr a good wage?
The real answer is as little as you can. If you are asking how much a person making \$14 per hour can safely afford, that is about \$600 per month, assuming utilities either included or a small amount. As a general rule, financially successful people keep the expense of a roof over their head below 1/4 of their income.
## How much is 40k a year hourly?
If you make \$40,000 per year, your hourly salary would be \$20.51. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 37.5 hours a week.
## Is making 12 dollars an hour good?
If we are assuming an uneducated and unskilled worker, yes it’s a good wage. It won’t be enough to live on your own but if you share a 3 bedroom apartment with two other people, you can make it work. A couple both making \$12/hour would be able to afford a place of their own.
## How much money is \$15 an hour?
Assuming 40 hours a week, that equals 2,080 hours in a year. Your hourly wage of 15 dollars would end up being about \$31,200 per year in salary. | 586 | 2,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.944269 |
https://maker.pro/forums/threads/led-question.1968/ | 1,696,147,728,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00442.warc.gz | 427,079,471 | 61,689 | # led question
K
#### Klemen
Jan 1, 1970
0
Hi!
I'd like to connect 5 leds, but i'm not sure which way is better. Is it ok
to connect them in series with a restitor to limit the current, or should i
connect them parallel each with the resistor?
regards
Klemen
J
#### John Popelish
Jan 1, 1970
0
Klemen said:
Hi!
I'd like to connect 5 leds, but i'm not sure which way is better. Is it ok
to connect them in series with a restitor to limit the current, or should i
connect them parallel each with the resistor?
regards
Klemen
Either of those may work, but connecting several in series may make
more efficient use of the supply voltage, if it is quite a bit higher
than the LED requirements. What supply voltage are you using, and
what color of LED?
K
#### Klemen
Jan 1, 1970
0
I'm using 12V(actually 14V because it's a car battery) power suply and
bright blue leds.
J
#### John Popelish
Jan 1, 1970
0
Klemen said:
I'm using 12V(actually 14V because it's a car battery) power suply and
bright blue leds.
Since getting the utmost efficiency is not so important (unless you
are going to leave them on for days without running the car) I would
connect them in series in pairs , with a resistor for each pair.
Since car battery voltage can spike around and vary quite a bit, I
would go for about 10 ma (.01 amp) average. Blue LEDs drop about 3.5
volts, each, so a pair will use up about 7 volts. That leaves about 7
volts to be dropped by each resistor while carrying about .01 amp.
7/.01=700 ohms, which is not a standard 5% value. 680 or 750 ohms
would work fine. The heat produced in each resistor is about
..01*.01*700=.07 watts, so 1/4 watt resistors are plenty big.
T
Jan 1, 1970
0
Hi!
The way you connect 5 leds doesn't make a difference in terms of theory.
You can build parrarel or series cicuit. Either way is good because in both
connections current flowing through leds must be the same and the voltages
dropped on them also. So, the only difference betwen both types of
connection
is the voltage applied and the number of resistors. You should decide which
is better: use more resistors and less voltage or vice versa.
K
#### Klemen
Jan 1, 1970
0
Thank you everyone for your help!
regards
Klemen
R
#### roma
Jan 1, 1970
0
Either of those may work, but connecting several in series may make
more efficient use of the supply voltage, if it is quite a bit higher
than the LED requirements. What supply voltage are you using, and
what color of LED?
So when one goes bad they all go out .
roma
J
#### John Popelish
Jan 1, 1970
0
roma said:
So when one goes bad they all go out .
roma
If it goes bad by going open circuit, then that string will go out.
If it goes bad by going short circuit, the rest of the string gets
brighter. If the LEDs are operated well within their ratings (say
half of maximum current rating) either case is very unlikely in your
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3K | 836 | 2,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-40 | latest | en | 0.948925 |
http://mathoverflow.net/questions/84420/spectral-properties-of-the-ldlt-matrix-factorization | 1,469,286,597,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823072.2/warc/CC-MAIN-20160723071023-00199-ip-10-185-27-174.ec2.internal.warc.gz | 150,287,294 | 16,462 | Spectral properties of the LDL^T matrix factorization
Assume that a square, symmetric matrix $A$ can be factored into $A=LDL^T$ where $L$ is unit lower triangular and $D$ is diagonal. For indefinite $A$, $D$ may have $2x2$ blocks on the diagonal. How much information about the spectrum of $A$ can we obtain from $D$?
For example, it is known by Sylvester's law of matrix inertia that the inertia of $D$ is the same as that of $A$ (they have the same number of positive and negative eigenvalues). This is interesting, but I am wondering what other information is hidden in $D$.
-
I was always wondering the same :) Great question. – Alexander Chervov Dec 28 '11 at 8:18
But why you write about 2x2 blocks in indefinite case ? Usually just elements of D can be taken negative. – Alexander Chervov Dec 28 '11 at 8:21
Well, for the closely related Cholesky factorization, there is the following:
Fast Accurate Eigenvalue Computations Using the Cholesky Factorization (1997) (by Roy Matthias), which says that the eigenvalues are very close to the squares of the diagonal elements of the Cholesky factor. (the paper is available on CiteSeer).
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I've heard similar, but there were some conditions, may be small eigs, or like that... But "squares" seems to me misprint - take A=D then they are equal. Any way... What can be the reason for relation in case "L" is not very small ? Is there some intuitive explanation ? – Alexander Chervov Dec 28 '11 at 8:23
In Cholesky, $D=I$ and the diagonal scaling is "included" in the $L$ factor, which need not have ones on its main diagonals. So that's where you get those square roots from. – Federico Poloni Dec 28 '11 at 8:57
@Federico, You mean - probably the paper mentioned above deals with LL^t ? Then it is Okay- we need squares. Just in the question LDL^t was mentioned which is sometimes called Cholesky or Cholesky without square roots also. In this case you do not need square roots. Any way puzzle seems to be resolved :) What about my questions ? What the reason can be to have eigs = D ? – Alexander Chervov Dec 28 '11 at 10:05
Not an answer, but just thinking loudly (can I say like this in English?)
Consider A=
$a ~~ b$
$b ~~ a$
Then eigenvalues are equal to a-b, a+b (it is easy to check since trace and determinant are correct).
Then L =
$1~~~~~~ 0$
$b/a~~1$
D=
$a ~~~~ 0$
$0 ~~~~ a - b^2/a$
(It is easy in 2x2 case since determinant(A) = $d_1d_2$ ).
So we see: eigenvalues are:
$a \pm b$ and elements of D are $a$ and $a-b^2/a$
Well, do they look similar ?
Stupid case when they are similar is b=0.
Another case is more interesting - take b=a - very degenerate but will be positve under small perturbation a=b+small. So in this case eigs are : $2a, 0$, and elements of D are $a$ and $0$ . So we see that the smallest number (i.e. $0$) is the same.
Probably that is the phenomena which I heard about i.e. something similar holds true for NxN matrices.
Probably paper mentioned in Igor's answer is something related.
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