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What is the domain and range of y = x + 3 ? | Socratic
What is the domain and range of #y = x + 3 #?
1 Answer
$\mathrm{do} m f = \mathbb{R}$
$r a n f = \mathbb{R}$
$f \left(x\right) = x + 3$
Is there any value of $x$ that will make $f \left(x\right)$ undefined?
The answer to this is no, so the domain is the set of all real numbers $\mathbb{R}$.
$\mathrm{do} m f = \mathbb{R}$
You will notice that the graph of $x + 3$ is just a line, meaning it will intersect all values of $y$ (since it increases and decreases without limit).
Therefore, the range is also the set of all real numbers $\mathbb{R}$.
$r a n f = \mathbb{R}$
Just keep this in mind. When you're given a linear function, its domain and range are both the set of all real numbers (unless the problem tells you it isn't).
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Mode.Sngl: Excel Formulae Explained - ExcelAdept
Key Takeaway:
• The MODE.SNGL function in Excel helps to find the most frequently occurring value in a range of data.
• The syntax of the function includes the range of data to be analyzed.
• Examples of using the MODE.SNGL function include finding the most frequent value in a range while ignoring non-numeric values. Other alternatives to the function can also be explored to optimize
data analysis.
Have you ever been overwhelmed by Excel’s complex formulae? Get the confidence you need to breeze through complicated calculations with MODE.SNGL’s comprehensive guide to Excel formulae. You’re just
a few clicks away from mastering the fundamentals of spreadsheets!
Understanding MODE.SNGL function in Excel
In Excel, there is a function called MODE.SNGL which helps in finding the most frequently occurring number in a set of data. This function is extremely useful for data analysis and helps in
identifying patterns efficiently.
By using the MODE.SNGL function, we can avoid the tedious manual task of counting each number and identifying the most common one. Instead, we can rely on this formula to do the job quickly.
To use the MODE.SNGL function, we need to select the range of cells where we want to find the most frequently occurring number, and then enter the formula “=MODE.SNGL(range)“. The result will be the
number that appears most frequently in that range. This function is case-sensitive and only works for numerical data.
It is essential to note that if there are multiple numbers with the same frequency, the function will only return the first one it encounters. Also, if there is no number that appears more than once,
then the function will return an error.
Interestingly, the MODE.SNGL function has been around for a long time, and it was first introduced in Excel 2010. However, it has undergone some improvements since then, making it more convenient to
use in the latest Excel versions.
Overall, the MODE.SNGL function is an excellent tool for simplifying data analysis tasks in Excel.
Syntax of MODE.SNGL function
The MODE.SNGL function in Excel is a powerful tool used to find the most commonly occurring value in a set of data. It is essential to understand the correct syntax to utilize this function
1. Start by selecting the cell where the result of the MODE.SNGL function will appear.
2. Begin the formula with an equal sign, followed by the function name ‘MODE.SNGL‘.
3. Within the parentheses, select the range of values from which the function will analyze the most commonly occurring value.
4. Finally, close the parentheses and hit enter to display the result.
It is important to note that the MODE.SNGL function only considers a single value as the result, even if there are multiple equally occurring values. Additionally, the function will return an error
if no values in the selected range appear more than once.
When using the MODE.SNGL function, it is crucial to select the appropriate range of values and also remember that it only returns a single value. Incorrect syntax or improper selection of data could
lead to inaccurate results and impact the entire data analysis.
In a similar context, a financial analyst erroneously used the MODE.SNGL function to calculate the average salary of employees, leading to inaccurate results. It highlights the importance of
correctly understanding the function syntax and the consequences of incorrect usage.
Examples of using MODE.SNGL function
Explore the “Examples of using MODE.SNGL function” section of “MODE.SNGL: Excel Formulae Explained” article. Discover how to find the most frequent values in a range and ignore non-numeric values.
This section features two sub-sections – Example 1 and Example 2. Gain insights from these two examples.
Example 1: Finding the most frequent value in a range
To find the most common value in a given range, Excel provides an efficient formula named 'MODE.SNGL.' This formula is useful when we need to identify an item that appears more frequently than other
items in a given data set.
Here is a simple 3-step guide on how to use Excel’s MODE.SNGL function to find the most frequent value in a range:
1. Start by opening your Excel sheet and selecting the cell where you want your answer to appear.
2. Next, type '=MODE.SNGL' followed by an open parenthesis '('. Then enter the range of data for which you want to calculate the most common item. Close this bracket with a closing parenthesis ')'.
This should give you something like =MODE.SNGL(A1:A20). Press enter.
3. The answer will be displayed in the selected cell within no time.
Another useful feature of MODE.SNGL is that it ignores any blank cells or text values. Thus, it only returns the numerical values within the specified range.
In practice, using MODE.SNGL can greatly simplify determining the mode even among large sets of data containing decimals and fractions.
Once upon a time, before Excel was widespread, finding modes involved re-reading through all individual histograms searching for numbers or graphical clues marking peaks above surrounding mounds or
bumps. However, today’s computational tools have simplified these tasks, making them easy and fast even with large sets of datasets.
Why bother with non-numeric values? They’re like people who don’t understand sarcasm – just ignore them and move on.
Example 2: Ignoring non-numeric values
Using MODE.SNGL function to ignore non-numeric values is another significant application of this formula. It helps in finding mode values while excluding the text, logical or error inputs from the
• Remove non-numeric data: This function operates only on numeric data and ignores other types of input. It is used commonly for removing unwanted entries from a dataset.
• Ignores Text: Any text strings present in the dataset are ignored by MODE.SNGL. It only considers numbers for calculation.
• Excludes Error values: MODE.SNGL also excludes error messages like #DIV/0!, #REF!, #NAME?, #VALUE! and #NULL!.
• Finds Mode: After removing non-numeric values from the data set, this formula finds the most frequently occurring number (mode) easily.
• Computes Numeric Values Only: The use of MODE.SNGL ensures that calculations are done with only numerical values, omitting mixed-type inputs to offer more accurate results.
• Saves Time: By ignoring errors and undesired non-numeric entries automatically, this function saves tremendous effort required for filtering out these items manually.
It is worthwhile noting that ignoring such inputs does not affect the overall calculation or exclude any significant numeric value from the data set.
Research shows that MODE.SNGL function can be used alongside other statistical formulas like AVERAGE functions to provide deeper insights into datasets.
MODE.SNGL: Making you realize just how average your data really is.
Limitations of MODE.SNGL function
In this article, we will discuss the limitations of using the MODE.SNGL function in Excel. This function can be useful in identifying the most commonly occurring value in a set of data, but there are
certain constraints that we need to keep in mind.
• The function only works with numbers and ignores text values.
• If there are two or more values that occur with the same frequency, the function returns #N/A error.
• The function does not provide any indication of the variability of the data set.
• The data must be arranged in a single column or row.
• If the data set contains empty cells, the function may produce erroneous results.
It’s important to note that the limitations of MODE.SNGL function can be overcome by using alternative functions or formulas in Excel, such as using pivot tables or filtering data. Therefore, it’s
essential to consider the specific needs of your analysis before using this function.
To optimize your analysis, take advantage of other Excel features that can complement the MODE.SNGL function. For example, you can use the COUNTIF function to count the occurrences of a particular
value, or the AVERAGEIF function to calculate the average values based on a specific criteria.
To avoid missing out on the full potential of Excel functions, it’s important to keep learning and exploring new features, as they might be more suitable for the data at hand. Always keep an open
mind when conducting data analysis and strive for a personalized approach that caters to the specific needs of your analysis.
Alternatives to MODE.SNGL function in Excel.
When it comes to finding alternative solutions to the MODE.SNGL function in Excel, there are a variety of options that can make the task easier and more efficient.
Some of the alternatives to the MODE.SNGL function in Excel include:
• The MODE.MULT function is ideal for finding multiple modes in a data set.
• PivotTables offer a simple way to calculate modes and other statistical values.
• Array formulas can be used to calculate modes in a range of cells.
• VBA can be used to create custom functions for finding modes.
• Power Query is a powerful tool for data analysis and can also be used for finding modes.
• Add-ins like the Analysis Toolpak or the Kutools add-in can help with statistical analysis and finding modes in data.
An additional way to find modes in Excel is to use the FREQUENCY function in combination with the MAX function. This can be useful for calculating the mode of a range that contains more than one
Pro Tip: When dealing with large data sets, it can be helpful to use one of the add-ins mentioned in paragraph 2, as they can streamline the process of finding modes and other statistical values.
Five Facts About MODE.SNGL: Excel Formulae Explained:
• ✅ MODE.SNGL is an Excel function that returns the most frequently occurring number in a data set. (Source: Exceljet)
• ✅ The function is useful for analyzing large sets of data and identifying patterns and trends. (Source: Ablebits)
• ✅ To use the MODE.SNGL function, select a cell where you want to display the result and enter the formula “=MODE.SNGL(range)” (Source: Excel Easy)
• ✅ The formula can be used for both numerical and text values, but it only returns a single result even if there are multiple modes. (Source: Investopedia)
• ✅ In case of ties, the function returns the smallest value among the modes. (Source: Excel Campus)
FAQs about Mode.Sngl: Excel Formulae Explained
What is MODE.SNGL in Excel and how does it work?
MODE.SNGL is an Excel formula that returns the most frequently occurring value in a range of cells. This function only counts unique values and ignores the duplicates. The formula syntax is:
MODE.SNGL(number1, [number2], …). You can provide up to 255 arguments.
What is the difference between MODE.SNGL and MODE.MULT?
The main difference between MODE.SNGL and MODE.MULT is that MODE.SNGL returns only one value, which is the mode of the range. If there are multiple modes, it will return the lowest value. MODE.MULT,
on the other hand, can return multiple mode values if there are ties.
When should I use MODE.SNGL?
MODE.SNGL is useful when you want to find the most frequently occurring value in a range and only need one value returned. It is particularly helpful in datasets where there are multiple values
repeated multiple times but you only want to see the most commonly occurring one.
What happens when there are no unique values in the range using MODE.SNGL?
If there are no unique values in the range, MODE.SNGL will return an error (#N/A) because there is no mode.
How can I use MODE.SNGL with other Excel functions?
MODE.SNGL can be used with other Excel functions such as IF and SUM. For example, you can use IF to determine if a certain value is the mode and then use SUM to total the values in a range based on
that condition.
Can I use MODE.SNGL with non-numeric values?
No, MODE.SNGL only works with numeric values. If you try to use it with non-numeric values, you will get a #VALUE! error. | {"url":"https://exceladept.com/mode-sngl-excel-formulae-explained/","timestamp":"2024-11-07T05:58:37Z","content_type":"text/html","content_length":"68648","record_id":"<urn:uuid:3f3a9144-63e1-4d7c-98a7-658e2be2f4c6>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00325.warc.gz"} |
So with the AFL off season underway I'm going to delve into some new stuff while there's time. I've got a few aims this off season:
• Explore some new analysis to improve upon my models for season 2020 - stuff like this HGA analysis and improving modelling of player information in match predictions.
• Explore some new modelling ideas for AFL - I'm looking into a new, more granular method of modelling AFL.
• Get an AFLW model up and running for next season.
I'm going to start off with some home ground advantage analysis. My model currently just considers travel distance and each teams historic performance at the venue to estimate the home ground
advantage effect. I have a few doubts about this modelling now, such as:
• Historic venue performance is prone to over fitting. My model seems a little over fit in a few ways so I want to unwind this over fitting.
• Historic venue performance was calculated based on scoring shots, however HGA may have an impact on scoring accuracy or quality of shots generated too, so I think a score based rather than shots
based model is more appropriate.
• My model was quite aggressive on the travel distance. As I don't take venue experience into consideration I think this was causing my model to weight travel distance a little too high. This meant
for example that teams travelling to/from Perth were punished too much.
These doubts are based on nothing scientific, just my observations on how my models tips looked compared to other Squiggle models throughout the season.
In this analysis, I've started with an ELO model that does not consider HGA, then I attempt to explain the margin errors using a variety of features compared for home vs away including:
• Travel Distance
• Venue Experience
• Season
• Round
• Average Historical Venue Performance
• Venue Dimensions relative to Home Venue
• Membership Figures
I found that I could explain HGA using Travel Distance + Venue Experience + Round + Average Historical Venue Performance, but all other features did not explain any of the remaining errors after
these factors are taken into account. For those interested in how I came to this conclusion and how much each feature affects HGA, read on!
I started with an ELO model with no HGA considerations, so we've adjusted for team strengths as well as we can without modelling HGA. I used predictions for seasons 1897 - 2018, split into a 50%
train and 50% test sets. As AFLTables designates the home team as the winner in finals, this could skew the results so I decided to exclude finals.
Next I collected the features used to explain HGA. The team specific features were calculated as differences such that a positive value should indicate an advantage to the home team. I split the
values into those available for all matches, and those limited to fewer matches:
Available for all matches:
• Travel Distance: \[\log(1 + \text{Away Team Distance in km}) - \log(1 + \text{Home Team Distance in km}) \]
• Venue Experience: For the 2 seasons prior to the match (i.e. excluding the current season): \[ \frac{\text{Home Team Matches at Venue}}{\text{Home Team Matches}} - \frac{\text{Away Team Matches
at Venue}}{\text{Away Team Matches}} \]
• Season: Year re based so that 1897 maps to 0 and 2018 maps to 121
• Round: \[\frac{\text{Round Number}}{\text{Total Rounds in Season}} \]
• Historical Venue Performance: Average performance at venue in terms of winning/losing margin relative to expectation in last 100 matches. I take this and multiply by matches considered/100 so
there is less weight on historical venue performances over fewer matches. If a team has played less than 5 matches at the venue, I set historical venue performance to 0.
Available for select matches:
• Venue Length: \[ |\text{Away Team Venue Length} - \text{Match Venue Length}| - |\text{Home Team Venue Length} - \text{Match Venue Length}| \]
• Venue Width: \[ |\text{Away Team Venue Width} - \text{Match Venue Width}| - |\text{Home Team Venue Width} - \text{Match Venue Width}| \]
• Memberships: Based on members in prior season: \[\frac{\text{Home Team Members}}{\text{Avg Members Per Club}} - \frac{\text{Away Team Member}}{\text{Avg Members Per Club}} \]
The reasons for the transformations are as follows:
• We take the log of distance travelled as the effect is not linear. The burden of travelling from Perth to Sydney is not 22% higher than travelling from Perth to Melbourne, even though there is an
extra 22% of distance to travel. Using logs puts the extra burden at 2% higher, which could still be wrong but seems more reasonable than 22%.
• We consider matches played at venue from the 2 prior seasons and not the current season as we don't want to introduce a feature that already has an impact from the round built in. If we use the
current season, matches towards the end of the season will be assumed to have a higher HGA as there are just more matches that a team could have played at the same venue. HGA might be higher
towards the end of the season, but I want to determine this by using a feature specifically for the round.
• We also consider venue experience as a proportion of total matches played in the 2 prior seasons. As seasons change length across the years, if we use number of matches we will be assuming HGA
increases as seasons go by just due to there being more matches played in each season.
• We re base season so the first season is 0 so we can use a regression model with no intercept. The coefficient of the season feature will tell us how HGA has changed relative to the very first
season. Using no intercept was important as I want the model to be symmetrical. I want the HGA to remain consistent regardless of the order in which teams are named. I realise there may be an
increased bias in crowd support for teams named first, but I'm ignoring this for now.
• We express the round as a fraction relative to the total rounds in the season to again avoid any skewing of the analysis due to seasons getting longer over time. If there is a change in HGA over
time, this may appear as a change in HGA by round as the high round numbers tend to only exist for very recent seasons.
• We've used historical venue performance from the last 100 matches as a maximum, and 5 matches as a minimum. I've also scaled the value down for small sample sizes, which is what the matches/100
• We've calculated venue length and width measures as absolute difference from the teams' home venue. If dimensions do contribute to HGA, whether the ground is skinnier vs fatter or longer vs
shorter shouldn't really matter. What should matter is that the ground dimensions are different to what the players are used to.
• I've expressed member figures relative to average members in that season, as membership figures have risen a lot over the years, so expressing relative to the average that year helps keep these
figures distinct from the season. The membership feature also uses this value from the prior season as current season memberships are only finalised in August, so using current season values
includes information about the current season which could skew the analysis.
Of these measures, travel distance and venue experience are what is commonly used to model HGA in AFL. The remainder are not normally used but I've made an attempt to investigate whether they provide
any extra information in addition to travel distance and venue experience.
As a first step, I've plotted each of the features I'm attempting to use against the model margin error to validate each makes sense and these plots can be seen below. The left plots show a scatter
plot with all matches while the right shows scatter plots with the points binned into smaller intervals so we can see the trend a little easier. Note that due to sample sizes the binned plots may
look strange (for example the venue performance binned plot), this is due to there being few matches in the far right bins as well as the values being out of line with the overall trend. Looking at
the trend on the left is most relevant in this example.
From these plots it looks like travel, venue experience, venue performance and round explain the errors the most, venue dimensions explain the errors a little and membership figures and season don't
look to explain the errors much, if at all.
Now from here, I decided to first attempt to explain the errors in my ELO model using the features available for all matches. As I want to be able to model HGA back to the beginning of VFL/AFL, I
want to explain HGA using features available for the entire history. After that I will attempt to use the other features to improve the HGA model further for matches where the extra features are
Features available in all matches are travel distance, venue experience, venue performance, season and round. To explain the model errors using these features I used forward selection linear
regression. This is a method of linear regression whereby we start with no features and from the features available, add them one by one in order of which gives the biggest improvement to adjusted \
(R^2\) first. We then only add more features if they improve the adjusted \(R^2\) further, and again we add the ones that give the biggest boost to the adjusted \(R^2\) first.
If we used regular \(R^2\), this would always increase as we add more features, however using adjusted \(R^2\) means we need the improvement in \(R^2\) to be more than what we would get from random
chance just by adding more features. This is a good way to avoid over fitting and adding extra features just for the hell of it.
This brings us to the results, this first regression showed that we can explain the margin errors using the expression:
\[ \text{Margin Error} = 0.86 \times \text{Travel Distance} + 5.9 \times \text{Venue Experience} + 3.5 \times \text{Round} + 0.19 \times \text{Venue Performance} \]
All the coefficients here were statistically significant (p values of \( 1.5\times 10^{-6}, 3.7\times 10^{-2}, 5.2\times 10^{-3}, 4.1\times 10^{-8} \) for travel distance, venue experience, round and
venue performance respectively). However using the season as a feature did not improve the adjusted \(R^2\) so was not included in the final model.
I am quite surprised that the round of the match came out as a significant factor in estimating HGA. The other features are quite easy to rationalise, but I'm at a loss for how to explain why playing
a match at the end of the season at your home ground is worth about 3.5 points more than playing the match at the start of the season. Perhaps the burden of travel, the unfamiliarity and extra mental
effort takes a larger toll as the season goes on and the players get more battered and bruised each week.
To see if this expression makes sense, consider what would be about the worst case scenario, West Coast or Fremantle travelling from Perth to Melbourne at the end of the season having played approx.
15% of their matches at the MCG in the prior 2 seasons while their opponent has played approx. 60% of their matches there.
According to the above expression, travelling from Perth to Melbourne is worth -7 points (distance of approx 2700km). Having played 15% of matches at the MCG relative to 60% is worth -2.5 points. And
playing away from home at the end of the season relative to the start of the season is worth -3.5 points. Ignoring historical venue performance, this adds up to about a 13 point disadvantage which
seems pretty reasonable to me.
Using this regression model we can then define a base HGA for each match based on travel distance, venue experience, round and venue performance. If we subtract this from our original margin error we
get a new margin error which has the error from travel distance, venue experience and round built in.
As a check, the MAE for our original margin error was 27.6 points and 26.5 points after adjusting for HGA, so our adjusted margin errors have improved by about 1.1 point using these adjustments.
Next we try and explain the adjusted margin errors using venue dimensions and membership figures.
Running the forward regression model again on these 3 features, the result is that only the venue width feature was chosen to be an explanatory variable, and it had a p value of 0.025. The
coefficient here was actually -0.12 as well, indicating that when a team plays at a venue which has a width different to their home venue, they actually perform better than expected.
With the coefficient being the opposite sign to what we expect and the p value not being super convincing, this is enough for me to exclude venue width as an explanatory variable, leaving none of the
venue dimensions or membership figures as useful in explaining home ground advantage, after we account for venue experience, travel distance and round.
Note that above when we plotted the margin errors against the venue length and width features, we saw the expected relationship between the errors and these features. However after accounting for
travel, venue experience, venue performance and round, the venue dimension features don't provide any useful extra information. As venue experience, travel distance, round and venue performance can
be calculated for every match and are simpler to maintain and calculate going forwards, I am comfortable using them as a proxy for all home ground advantage factors, including venue dimensions and
leaving out any specific adjustment for the actual venue dimensions.
Given the relationship between the margin errors and the membership feature above I'm not surprised it was deemed insignificant in the forward regression. Perhaps membership numbers are too noisy
themselves, or are confounded with other variables such as team strength anyway and provide little additional value in estimating HGA.
Finally, if I apply the same margin error adjustment to my test data set I end up with MAE values of 27.7 prior to adjustment and 26.8 after the adjustment, so the adjustments gives a 0.9 point
improvement on the test data set, slightly less than the 1.1 point improvement on the train data set but still comparable.
HGA can be explained using the travel distance, venue experience, venue performance and round features explained above. The linear relationship of \( \text{HGA} = 0.86 \times \text{Travel Distance} +
5.9 \times \text{Venue Experience} + 3.5 \times \text{Round} + 0.19 \times \text{Venue Performance} \) could be used to improve the margin errors for my HGA agnostic ELO model by about 1 point.
Venue dimensions appear to correlate with teams' improved performance when playing at home, however this disappears once we account for travel distance, venue experience, venue performance and round.
Membership figures and season were found to not explain any of the home ground advantage effect.
The next step is to use this methodology in an ELO model. As ELO will update team rankings based on performance accounting for HGA, the optimal weights on each feature may actually be slightly
different. I would also expect to get an improvement of more than 1 point in MAE when incorporating HGA into an ELO model due to having a better estimate of team strength when HGA adjustments are
Data notes
Venue dimensions taken mostly from AustralianFootball.com with a few from this Foxsports article and The Footy Almanac.
Membership figures taken from Footy Industry for 1984 - 2016 and AFL website for 2017, 2018 and 2019. Due to incomplete data I only used membership figures from 1992 onwards.
Feel free to reach out on Twitter if you'd like to see any of this data.
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👩‍💻 Test Driven Development
18.8. 👩‍💻 Test Driven Development¶
At this point, you should be able to look at complete functions and tell what they do. Also, if you have been doing the exercises, you have written some small functions. As you write larger
functions, you might start to have more difficulty, especially with runtime and semantic errors.
To deal with increasingly complex programs, we are going to suggest a technique called incremental development. The goal of incremental development is to avoid long debugging sessions by adding and
testing only a small amount of code at a time.
If you write unit tests before doing the incremental development, you will be able to track your progress as the code passes more and more of the tests. Alternatively, you can write additional tests
at each stage of incremental development.
Then you will be able to check whether any code change you make at a later stage of development causes one of the earlier tests, which used to pass, to not pass any more.
As an example, suppose you want to find the distance between two points, given by the coordinates (x[1], y[1]) and (x[2], y[2]). By the Pythagorean theorem, the distance is:
The first step is to consider what a distance function should look like in Python. In other words, what are the inputs (parameters) and what is the output (return value)?
In this case, the two points are the inputs, which we can represent using four parameters. The return value is the distance, which is a floating-point value.
Already we can write an outline of the function that captures our thinking so far.
def distance(x1, y1, x2, y2):
return None
Obviously, this version of the function doesn’t compute distances; it always returns None. But it is syntactically correct, and it will run, which means that we can test it before we make it more
The distance between any point and itself should be 0.
We call the distance function with sample inputs: (1,2, 1,2). The first 1,2 are the coordinates of the first point and the second 1,2 are the coordinates of the second point. What is the distance
between these two points? Zero.
It’s not returning the correct answer, so we don’t pass the test. Let’s fix that.
Now we pass the test. But really, that’s not a sufficient test.
Extend the program …
On line 6, write another unit test (assert statement). Use (1,2, 4,6) as the parameters to the distance function. How far apart are these two points? Use that value (instead of 0) as the correct
answer for this unit test.
On line 7, write another unit test. Use (0,0, 1,1) as the parameters to the distance function. How far apart are these two points? Use that value as the correct answer for this unit test.
Are there any other edge cases that you think you should consider? Perhaps points with negative numbers for x-values or y-values?
When testing a function, it is essential to know the right answer.
For the second test the horizontal distance equals 3 and the vertical distance equals 4; that way, the result is 5 (the hypotenuse of a 3-4-5 triangle). For the third test, we have a 1-1-sqrt(2)
The first test passes but the others fail since the distance function does not yet contain all the necessary steps.
At this point we have confirmed that the function is syntactically correct, and we can start adding lines of code. After each incremental change, we test the function again. If an error occurs at any
point, we know where it must be — in the last line we added.
A logical first step in the computation is to find the differences x[2]- x[1] and y[2]- y[1]. We will store those values in temporary variables named dx and dy.
def distance(x1, y1, x2, y2):
dx = x2 - x1
dy = y2 - y1
return 0.0
Next we compute the sum of squares of dx and dy.
def distance(x1, y1, x2, y2):
dx = x2 - x1
dy = y2 - y1
dsquared = dx**2 + dy**2
return 0.0
Again, we could run the program at this stage and check the value of dsquared (which should be 25).
Finally, using the fractional exponent 0.5 to find the square root, we compute and return the result.
When you start out, you might add only a line or two of code at a time. As you gain more experience, you might find yourself writing and debugging bigger conceptual chunks. As you improve your
programming skills you should find yourself managing bigger and bigger chunks: this is very similar to the way we learned to read letters, syllables, words, phrases, sentences, paragraphs, etc., or
the way we learn to chunk music — from individual notes to chords, bars, phrases, and so on.
The key aspects of the process are:
1. Make sure you know what you are trying to accomplish. Then you can write appropriate unit tests.
2. Start with a working skeleton program and make small incremental changes. At any point, if there is an error, you will know exactly where it is.
3. Use temporary variables to hold intermediate values so that you can easily inspect and check them.
4. Once the program is working, you might want to consolidate multiple statements into compound expressions, but only do this if it does not make the program more difficult to read.
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Division Property Of Equality Worksheets - Divisonworksheets.com
Division Property Of Equality Worksheets
Division Property Of Equality Worksheets – By using worksheets for division to help your child review and test their division abilities. There are numerous types of worksheets to choose from and you
are able to make your own. These worksheets are fantastic since they can be downloaded at no cost and modify them as you like you want them to be. They’re great for kindergarteners and first-graders.
huge numbers by two
Work on worksheets that have huge numbers. The worksheets are restricted to two, three and sometimes even four different divisors. This won’t cause your child stress about forgetting how to divide
the big number or making mistakes on their tables of times. It’s possible to locate worksheets online or download them to your personal computer, to aid your child in this math skill.
Use multidigit division worksheets to assist children with their practice and increase their knowledge of the topic. This ability is essential for complex mathematical topics and everyday
calculations. By offering interactive activities and questions based on the division of multi-digit numbers, these worksheets aid in establishing the concept.
Students are challenged in dividing huge numbers. These worksheets often use an algorithm that is standardized and has step-by-step instructions. These worksheets may not provide the required
understanding of students. Teaching long division can be taught with the help of bases of 10 blocks. Long division should be simple for students once they’ve grasped the steps.
Make use of a variety of worksheets or practice questions to practice division of large numbers. On the worksheets, you will also find fractional results that are expressed in decimals. You can even
find worksheets for hundredsths, which are particularly useful for learning how to divide large sums of money.
Sort the numbers into small groups.
It may be difficult to assign numbers to small groups. While it sounds great on paper many small group facilitators do not like this procedure. It is a true reflection of the way that human bodies
develop and it helps in the Kingdom’s unending growth. It motivates others, and encourages them to reach out to the undiscovered.
It can also be useful to brainstorm ideas. It is possible to create groups of individuals with similar interests and skills. This could lead to some really creative ideas. Once you’ve formed your
groups, it’s time to introduce yourself and each other. It’s a good way to encourage creativity and innovative thinking.
To break large numbers down into smaller chunks of data, the basic division operation is utilized. This can be very useful in situations where you have to make equal amounts of items for different
groups. For example, a huge class could be split into five classes. The groups are then added to give the original 30 pupils.
Keep in mind that when you divide numbers, there is a divisor and a quote. The result of multiplying two numbers is “ten/five,” but the identical results can be obtained if you divide them in two
For huge numbers, you should employ power of 10.
Splitting large numbers into powers could make it easier to compare them. Decimals are an extremely frequent part of shopping. They can be seen on receipts as well as food labels, price tags, and
even receipts. In order to display the price per gallon as well as the amount of gas that was dispensed through a nozzle pumps use decimals.
You can divide large numbers into their powers of ten by using two methods move the decimal mark to your left or multiply by 10-1. The other method makes use of the powers of ten’s associative
feature. Once you’ve learned to utilize the power of ten’s associative function, you can break enormous numbers into smaller power.
The first one involves mental computation. The pattern is visible when 2.5 is divided by the power 10. The decimal points will shift left when the power of ten increases. This is a simple concept to
grasp and is applicable to every problem regardless of how complex.
Mentally dividing large numbers in power of 10 is a different method. You can quickly express huge numbers by using scientific notation. Large numbers must be expressed as positive exponents when
writing in scientific notation. You can convert 450,000 numbers into 4.5 by shifting the decimal point 5 spaces left. The exponent 5 to divide a massive number into smaller power of 10 or you can
divide it into smaller powers of 10, and so on.
Gallery of Division Property Of Equality Worksheets
Worksheet The Properties Of Math Grass Fedjp Worksheet Study Site
Division Property Of Equality Rule STAETI
Division Property Of Equality Definition And Examples
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2,476 research outputs found
In this Letter, we investigate the effects of dark energy on $P-V$ criticality of charged AdS black holes by considering the case of the RN-AdS black holes surrounded by quintessence. By treating the
cosmological constant as thermodynamic pressure, we study its thermodynamics in the extended phase space. It is shown that quintessence dark energy does not affect the existence of small/large black
hole phase transition. For the case $\omega_q=-2/3$ we derive analytic expressions of critical physical quantities, while for cases $\omega_qeq-2/3$ we appeal to numerical method for help. It is
shown that quintessence dark energy affects the critical physical quantities near the critical point. Critical exponents are also calculated. They are exactly the same as those obtained before for
arbitrary other AdS black holes, which implies that quintessence dark energy does not change the critical exponents.Comment: 13 pages, 2 figure
Heat engine models are constructed within the framework of massive gravity in this paper. For the four-dimensional charged black holes in massive gravity, it is shown that the heat engines have a
higher efficiency for the cases $m^2>0$ than for the case $m=0$ when $c_1<0, c_2<0$. Considering a specific example, we show that the maximum efficiency can reach $0.9219$ while the efficiency for $m
=0$ reads $0.5014$. The existence of graviton mass improves the heat engine efficiency significantly. The situation is more complicated for the five-dimensional neutral black holes. Not only the
$c_1, c_2, m^2$ exert influence on the efficiency, but also the constant $c_3$ corresponding to the third massive potential contributes to the efficiency. When $c_1<0, c_2<0, c_30$ is higher than
that of the case $m=0$. By studying the ratio $\eta/\eta_C$, we also probe how the massive gravity influences the behavior of the heat engine efficiency approaching the Carnot efficiency.Comment:
We revisit the Hawking temperature$-$entanglement entropy criticality of the $d$-dimensional charged AdS black hole with our attention concentrated on the ratio $\frac{T_c \delta S_c}{Q_c}$.
Comparing the results of this paper with those of the ratio $\frac{T_c S_c}{Q_c}$, one can find both the similarities and differences. These two ratios are independent of the characteristic length
scale $l$ and dependent on the dimension $d$. These similarities further enhance the relation between the entanglement entropy and the Bekenstein-Hawking entropy. However, the ratio $\frac{T_c \delta
S_c}{Q_c}$ also relies on the size of the spherical entangling region. Moreover, these two ratios take different values even under the same choices of parameters. The differences between these two
ratios can be attributed to the peculiar property of the entanglement entropy since the research in this paper is far from the regime where the behavior of the entanglement entropy is dominated by
the thermal entropy.Comment: Comments welcome. 11 pages, 3 figure
We define a kind of heat engine via three-dimensional charged BTZ black holes. This case is quite subtle and needs to be more careful. The heat flow along the isochores does not equal to zero since
the specific heat $C_Veq0$ and this point completely differs from the cases discussed before whose isochores and adiabats are identical. So one cannot simply apply the paradigm in the former
literatures. However, if one introduces a new thermodynamic parameter associated with the renormalization length scale, the above problem can be solved. We obtain the analytical efficiency expression
of the three-dimensional charged BTZ black hole heat engine for two different schemes. Moreover, we double check with the exact formula. Our result presents the first specific example for the sound
correctness of the exact efficiency formula. We argue that the three-dimensional charged BTZ black hole can be viewed as a toy model for further investigation of holographic heat engine. Furthermore,
we compare our result with that of the Carnot cycle and extend the former result to three-dimensional spacetime. In this sense, the result in this paper would be complementary to those obtained in
four-dimensional spacetime or ever higher. Last but not the least, the heat engine efficiency discussed in this paper may serve as a criterion to discriminate the two thermodynamic approaches
introduced in Ref.[29] and our result seems to support the approach which introduces a new thermodynamic parameter $R=r_0$.Comment: Revised version. Discussions adde | {"url":"https://core.ac.uk/search/?q=authors%3A(Li%2C%20Gu-Qiang)","timestamp":"2024-11-14T21:06:00Z","content_type":"text/html","content_length":"128961","record_id":"<urn:uuid:28942222-e1d8-4df2-83db-3ec8e1d68b72>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00859.warc.gz"} |
4.3 Inverse Functions and Their Derivatives
Kittttttan*s Web
日本語 | English
The Chain Rule
4.3 Inverse Functions and Their Derivatives
There is a remarkable special case of the chain rule. It occurs when ƒ(y) and g(x) are "inverse functions." That idea is expressed by a very short and powerful equation: ƒ(g(x)) = x. Here is what
that means.
Inverse functions: Start with any input, say x = 5. Compute y = g(x), say y = 3. Then compute ƒ(y), and the answer must be 5. What one function does, the inverse function undoes. If g(5) = 3 then ƒ
(3) = 5. The inverse function ftakes the output y back to the input x.
EXAMPLE 1 g(x) = x − 2 and ƒ(y) = y + 2 are inverse functions. Starting with x = 5, the function g subtracts 2. That produces y = 3. Then the function ƒ adds 2. That brings back x = 5. To say it
directly: The inverse of y = x − 2 is x = y + 2.
EXAMPLE 2 y = g(x) = (5/9)(x − 32) and x = ƒ(y) = (9/5)y + 32 are inverse functions (for temperature). Here x is degrees Fahrenheit and y is degrees Celsius. From x = 32 (freezing in Fahrenheit)
you find y = 0 (freezing in Celsius). The inverse function takes y = 0 back to x = 32. Figure 4.4 shows how x = 50°F matches y = 10°C.
Notice that (5/9)(x − 32) subtracts 32 first. The inverse (9/5)y + 32 adds 32 last. In the same way g multiplies last by 5/9 while f multiplies first by 9/5.
The inverse function is written ƒ = g⁻¹ and pronounced "g inverse." It is not 1/g(x).
If the demand y is a function of the price x, then the price is a function of the demand. Those are inverse functions. Their derivatives obey a fundamental rule: dyldx times dx/dy equals 1. In
Example 2, dyldx is 5/9 and dxldy is 9/5.
There is another important point. When f and g are applied in the opposite order, they still come back to the start. First ƒ adds 2, then g subtracts 2. The chain g(ƒ(y)) = (y + 2) − 2 brings back y.
If ƒ is the inverse of g then g is the inverse of ƒ. The relation is completely symmetric, and so is the definition:
Inverse function: If y = g(x) then x = g⁻¹(y). If x = g⁻¹(y) then y = g(x).
The loop in the figure goes from x to y to x. The composition g⁻¹(g(x)) is the "identity function." Instead of a new point z it returns to the original x. This will make the chain rule particularly
easy—leading to (dy/dx)(dx/dy) = 1.
EXAMPLE 3 y = g(x) = √x and x = ƒ(y) = y² are inverse functions.
Starting from x = 9 we find y = 3. The inverse gives 3² = 9. The square of √x is ƒ(g(x)) = x. In the opposite direction, the square root of y² is g(ƒ(y)) = y.
Caution That example does not allow x to be negative. The domain of g—the set of numbers with square roots—is restricted to x ≥ 0. This matches the range of g⁻¹. The outputs y² are nonnegative.
With domain of g = range of g⁻¹, the equation x = (√x)² is possible and true. The nonnegative x goes into g and comes out of g⁻¹.
In this example y is also nonnegative. You might think we could square anything, but y must come back as the square root of y². So y ≥ 0.
To summarize: The domain of a function matches the range of its inverse. The inputs to g⁻¹ are the outputs from g. The inputs to g are the outputs from g⁻¹.
If g(x) = y then solving that equation for x gives x = g⁻¹(y):
if y = 3x − 6 then x = (1/3)(y + 6) (this is g⁻¹(y))
if y = x³ + 1 then x = ³√y − 1 (this is g⁻¹(y))
In practice that is how g⁻¹ is computed: Solve g(x) = y. This is the reason inverses are important. Every time we solve an equation we are computing a value of g⁻¹.
Not all equations have one solution. Not all functions have inverses. For each y, the equation g(x) = y is only allowed to produce one x. That solution is x = g⁻¹(y). If there is a second solution,
then g⁻¹ will not be a function—because a function cannot produce two x's from the same y.
EXAMPLE 4 There is more than one solution to sin x = ½. Many angles have the same sine. On the interval 0 ≤ x ≤ π, the inverse of y = sin x is not a function. Figure 4.5 shows how two x's give the
same y.
Prevent x from passing π/2 and the sine has an inverse. Write x = sin⁻¹ y.
The function g has no inverse if two points x[1] and x[2] give g(x[1]) = g(x[2]). Its inverse would have to bring the same y back to x[1] and x[2]. No function can do that; g⁻¹(y) cannot equal both x
[1] and x[2]. There must be only one x for each y.
To be invertible over an interval, g must be steadily increasing or steadily decreasing.
THE DERWNE OF g⁻¹
It is time for calculus. Forgive me for this very humble example.
EXAMPLE 5 (ordinary multiplication) The inverse of y = g(x) = 3x is x = ƒ(y) = (1/3)y. This shows with special clarity the rule for derivatives: The slopes dy/dx = 3 and dx/dy = 5 multiply to give 1.
This rule holds for all inverse functions, even if their slopes are not constant. It is a crucial application of the chain rule to the derivative of ƒ(g(x)) = x.
4C (Derivative of inverse function) From ƒ(g(x)) = x the chain rule gives ƒ'(g(x))g'(x) = 1. Writing y = g(x) and x = ƒ(y), this rule looks better:
(dx/dy)(dy/dx) = 1 or dx/dy = 1 / (dy/dx). (1)
The slope of x = g⁻¹(y) times the slope of y = g(x) equals one.
This is the chain rule with a special feature. Since ƒ(g(x)) = x, the derivative of both sides is 1. If we know g' we now know ƒ'. That rule will be tested on a familiar example. In the next section
it leads to totally new derivatives.
EXAMPLE 6 The inverse of y = x³ is x = y^1/3. We can find dx/dy two ways:
directly: dx = 1 y^-2/3 indirectly: dx = 1 = 1 = 1
dy 3 dy dy/dx 3x² 3y^2/3
The equation (dx/dy)(dy/dx) = 1 is not ordinary algebra, but it is true. Those derivatives are limits of fractions. The fractions are (Δx/Δy)(Δy/Δx) = 1 and we let Δx → 0.
Before going to new functions, I want to draw graphs. Figure 4.6 shows y = √x and y = 3x. What is s ecial is that the same graphs also show the inverse functions. The inverse of y = √x is x = y². The
pair x = 4, y = 2 is the same for both. That is the whole point of inverse functions—if 2 = g(4) then 4 = g⁻¹(2). Notice that the graphs go steadily up.
The only problem is, the graph of x = g⁻¹(y) is on its side. To change the slope from 3 to 1/3, you would have to turn the figure. After that turn there is another problem—the axes don't point to the
right and up. You also have to look in a mirror! (The typesetter refused to print the letters backward. He thinks it's crazy but it's not.) To keep the book in position, and the typesetter in
position, we need a better idea.
The graph of x = (1/3)y comes from turning the picture across the 45° line. The y axis becomes horizontal and x goes upward. The point (2,6) on the line y = 3x goes into the point (6,2) on the line x
= (1/3)y. The eyes see a reflection across the 45° line (Figure 4.6c). The mathematics sees the same pairs x and y. The special properties of g and g⁻¹ allow us to know two functions—and draw two
graphs—at the same time.* The graph of x = g⁻¹(y) is the mirror image of the graph of y = g(x).
I would like to add two more examples of inverse functions, because they are so important. Both examples involve the exponential and the logarithm. One is made up of linear pieces that imitate 2^x;
it appeared in Chapter 1. The other is the true function 2^x, which is not yet defined—and it' is not going to be defined here. The functions bx and logby are so overwhelmingly important that they
deserve and will get a whole chapter of the book (at least). But you have to see the graphs.
The slopes in the linear model are powers of 2. So are the heights y at the start of each piece. The slopes 1, 2, 4, … equal the heights 1, 2, 4, … at those special points.
The inverse is a discrete model for the logarithm (to base 2). The logarithm of 1 is 0,because 2⁰ = 1. The logarithm of 2 is 1, because 2ⁱ = 2. The logarithm of 2^j is the exponent j. Thus the model
gives the correct x = log[2] y at the breakpoints y = 1, 2, 4, 8, …. The slopes are 1, ½, ¼, 1/8, … because dx/dy = 1/(dy/dx).
The model is good, but the real thing is better. The figure on the right shows the true exponential y = 2^x. At x = 0, 1, 2, … the heights y are the same as before. But now the height at x = ½ is the
number 2^1/2, which is √2.The height at x = .10 is the tenth root 2^1/10 = 1.07…. The slope at x = 0 is no longer 1 —it is closer to Δy/Δx = .07/. 10. The exact slope is a number c (near .7) that we
are not yet prepared to reveal.
The special property of y = 2^x is that the slope at all points is cy. The slope is proportional to the function. The exponential solves dy/dx = cy.
Now look at the inverse function—the logarithm. Its graph is the mirror image:
If y = 2^x then x = log[2] y. If 2^1/10 ≈ 1.07 then log[2] 1.07 ≈ 1/10.
What the exponential does, the logarithm undoes—and vice versa. The logarithm of 2^x is the exponent x. Since the exponential starts with slope c, the logarithm must start with slope 1/c. Check that
numerically. The logarithm of 1.07 is near 1/10. The slope is near .10/.07. The beautiful property is that dx/dy = 1/cy.
I have to mention that calculus avoids logarithms to base 2. The reason lies in that mysterious number c. It is the "natural logarithm" of 2, which is .693147…—and who wants that? Also 1/.693 147…
enters the slope of log[2] y. Then (dx/dy)(dy/dx) = 1. The right choice is to use "natural logarithms" throughout. In place of 2, they are based on the special number e:
y = e^x is the inverse of x = ln y. (2)
The derivatives of those functions are sensational—they are saved for Chapter 6. Together with xn and sin x and cos x, they are the backbone of calculus.
Note It is almost possible to go directly to Chapter 6. The inverse functions x = sin⁻¹ y and x = tan⁻¹ y can be done quickly. The reason for including integrals first (Chapter 5) is that they solve
differential equations with no guesswork:
dy/dx or dx/dy = 1/y leads to ∫dx = ∫dy/y or x = ln y + C.
Integrals have applications of all kinds, spread through the rest of the book. But do not lose sight of 2^x and e^x. They solve dy/dx = cy —the key to applied calculus.
THE INVERSE OF A CHAIN h(g(x))
The functions g(x) = x − 2 and h(y) = 3y were easy to invert. For g⁻¹ we added 2, and for h⁻¹ we divided by 3. Now the question is: If we create the composite function z = h(g(x)), or z = 3(x − 2),
what is its inverse?
Virtually all known functions are created in this way, from chains of simpler functions. The problem is to invert a chain using the inverse of each piece. The answer is one of the fundamental rules
of mathematics:
4D The inverse of z = h(g(x))is a chain of inverses in the opposite order:
x = g⁻¹(h⁻¹(z)). (3)
h⁻¹ is applied first because h was applied last: g⁻¹(h⁻¹(h(g(x)))) = x.
That last equation looks like a mess, but it holds the key. In the middle you see h⁻¹ and h. That part of the chain does nothing! The inverse functions cancel, to leave g⁻¹(g(x)). But that is x. The
whole chain collapses, when g⁻¹ and h⁻¹ are in the correct order—which is opposite to the order of h(g(x)).
EXAMPLE 7 z = h(g(x)) = 3(x − 2) and x = g⁻¹(h⁻¹(z)) = (1/3)z + 2.
First h⁻¹ divides by 3. Then g⁻¹ adds 2. The inverse of h⋅g is g⁻¹⋅h⁻¹. It can be found directly by solving z = 3(x − 2). A chain of inverses is like writing in prose—we do it without knowing it.
EXAMPLE 8 Invert z = √x − 2 by writing z² = x − 2 and then x = z² + 2.
The inverse adds 2 and takes the square—but not in that order. That would give (z + 2)², which is wrong. The correct order is z² + 2.
The domains and ranges are explained by Figure 4.8. We start with x ≥ 2. Subtracting 2 gives y ≥ 0. Taking the square root gives z ≥ 0. Taking the square brings back y ≥ 0. Adding 2 brings back x ≥ 2
—which is in the original domain of g.
EXAMPLE 9 Inverse matrices (AB)⁻¹ = B⁻¹A⁻¹ (this linear algebra is optional).
Suppose a vector x is multiplied by a square matrix B: y = g(x) = Bx. The inverse function multiplies by the inverse matrix: x = g⁻¹(y) = B⁻¹y. It is like multiplication by B = 3 and B⁻¹ = 1/3,
except that x and y are vectors.
Now suppose a second function multiplies by another matrix A: z = h(g(x)) = ABx. The problem is to recover x from z. The first step is to invert A, because that came last: Bx = A⁻¹z. Then the second
step multiplies by B⁻¹ and brings back x = B⁻¹A⁻¹z. The product B⁻¹A⁻¹ inverts the product AB. The rule for matrix inverses is like the rule for function inverses—in fact it is a special case.
I had better not wander too far from calculus. The next section introduces the inverses of the sine and cosine and tangent, and finds their derivatives. Remember that the ultimate source is the chain | {"url":"https://mind.kittttttan.info/math/calculus/04-chain-3.html","timestamp":"2024-11-09T16:07:25Z","content_type":"text/html","content_length":"19450","record_id":"<urn:uuid:e42d9849-352d-4db7-a23a-1bdf39b8cde5>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00299.warc.gz"} |
What is the (_:xs) notation in haskell?
In Haskell, the (_:xs) notation is used to pattern match a list.
The pattern (:xs) matches a non-empty list by splitting it into its head and tail. The underscore () is a wildcard pattern that matches any value, and xs is a variable that represents the remaining
elements of the list.
For example, consider the following function that sums all the elements of a list using the (_:xs) pattern:
1 sumList :: Num a => [a] -> a
2 sumList [] = 0
3 sumList (_:xs) = head xs + sumList (tail xs)
In this function, if the list is empty, it returns 0. Otherwise, it matches the list with (_:xs), where the head of the list is ignored, and the sumList function is recursively called with the tail
of the list (xs). This pattern allows us to access and manipulate the elements of a list in a concise and elegant way. | {"url":"https://devhubby.com/thread/what-is-the-_-xs-notation-in-haskell","timestamp":"2024-11-12T03:10:49Z","content_type":"text/html","content_length":"127318","record_id":"<urn:uuid:3743ec79-7700-4911-be40-10b5c347b98c>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00085.warc.gz"} |
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Re: t-stat for intercept difference
I have run a regression on a dataset that has 2 different IDs, there is one X and one Y variable. I have an intercept and a coefficient for each one of these two IDs. My ultimate purpose is to
Subtract intercept of ID5 from the intercept of ID1 (b0 (ID5) - b0 (ID1)) and measure NeweyWest t-stat of this difference. Please guide me in this regard, thanks.
See the code below that I am using for regression:
ods exclude all;
proc model data=Have;
by ID;
endo Y;
exog X;
instruments _exog_;
parms b0 b1;
Y=b0 + b1*X;
fit Y / gmm kernel=(bart,5,0) vardef=n;
ods output parameterestimates=Want;
ods exclude none;
09-17-2019 03:54 AM | {"url":"https://communities.sas.com/t5/SAS-Procedures/t-stat-for-intercept-difference/m-p/589862","timestamp":"2024-11-07T07:39:54Z","content_type":"text/html","content_length":"249733","record_id":"<urn:uuid:47e54b31-6c6d-4278-b81a-f8655a5a5f27>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00315.warc.gz"} |
A view of the Kusner–Bryant parametrization of the Boy's surface
Boy's surface can be parametrized in several ways. One parametrization, discovered by Rob Kusner and Robert Bryant,^[4] is the following: given a complex number w whose magnitude is less than or
equal to one (${\displaystyle \|w\|\leq 1}$ ), let
{\displaystyle {\begin{aligned}g_{1}&=-{3 \over 2}\operatorname {Im} \left[{w\left(1-w^{4}\right) \over w^{6}+{\sqrt {5}}w^{3}-1}\right]\\[4pt]g_{2}&=-{3 \over 2}\operatorname {Re} \left[{w\left
(1+w^{4}\right) \over w^{6}+{\sqrt {5}}w^{3}-1}\right]\\[4pt]g_{3}&=\operatorname {Im} \left[{1+w^{6} \over w^{6}+{\sqrt {5}}w^{3}-1}\right]-{1 \over 2}\\\end{aligned}}}
and then set
${\displaystyle {\begin{pmatrix}x\\y\\z\end{pmatrix}}={\frac {1}{g_{1}^{2}+g_{2}^{2}+g_{3}^{2}}}{\begin{pmatrix}g_{1}\\g_{2}\\g_{3}\end{pmatrix}}}$
we then obtain the Cartesian coordinates x, y, and z of a point on the Boy's surface.
If one performs an inversion of this parametrization centered on the triple point, one obtains a complete minimal surface with three ends (that's how this parametrization was discovered naturally).
This implies that the Bryant–Kusner parametrization of Boy's surfaces is "optimal" in the sense that it is the "least bent" immersion of a projective plane into three-space.
Property of Bryant–Kusner parametrization
The Wikibook Famous Theorems of Mathematics has a page on the topic of: Boy's surface
If w is replaced by the negative reciprocal of its complex conjugate, ${\textstyle -{1 \over w^{\star }},}$ then the functions g[1], g[2], and g[3] of w are left unchanged.
By replacing w in terms of its real and imaginary parts w = s + it, and expanding resulting parameterization, one may obtain a parameterization of Boy's surface in terms of rational functions of s
and t. This shows that Boy's surface is not only an algebraic surface, but even a rational surface. The remark of the preceding paragraph shows that the generic fiber of this parameterization
consists of two points (that is that almost every point of Boy's surface may be obtained by two parameters values).
Relation to the real projective plane
Let ${\displaystyle P(w)=(x(w),y(w),z(w))}$ be the Bryant–Kusner parametrization of Boy's surface. Then
${\displaystyle P(w)=P\left(-{1 \over w^{\star }}\right).}$
This explains the condition ${\displaystyle \left\|w\right\|\leq 1}$ on the parameter: if ${\displaystyle \left\|w\right\|<1,}$ then ${\textstyle \left\|-{1 \over w^{\star }}\right\|>1.}$ However,
things are slightly more complicated for ${\displaystyle \left\|w\right\|=1.}$ In this case, one has ${\textstyle -{1 \over w^{\star }}=-w.}$ This means that, if ${\displaystyle \left\|w\right\|=1,}$
the point of the Boy's surface is obtained from two parameter values: ${\displaystyle P(w)=P(-w).}$ In other words, the Boy's surface has been parametrized by a disk such that pairs of diametrically
opposite points on the perimeter of the disk are equivalent. This shows that the Boy's surface is the image of the real projective plane, RP^2 by a smooth map. That is, the parametrization of the
Boy's surface is an immersion of the real projective plane into the Euclidean space.
STL 3D model of Boy's surface
Boy's surface has 3-fold symmetry. This means that it has an axis of discrete rotational symmetry: any 120° turn about this axis will leave the surface looking exactly the same. The Boy's surface can
be cut into three mutually congruent pieces.
Boy's surface can be used in sphere eversion, as a half-way model. A half-way model is an immersion of the sphere with the property that a rotation interchanges inside and outside, and so can be
employed to evert (turn inside-out) a sphere. Boy's (the case p = 3) and Morin's (the case p = 2) surfaces begin a sequence of half-way models with higher symmetry first proposed by George Francis,
indexed by the even integers 2p (for p odd, these immersions can be factored through a projective plane). Kusner's parametrization yields all these.
1. ^ Morin, Bernard (13 November 1978). "Équations du retournement de la sphère" [Equations of the eversion of the sphere] (PDF). Comptes Rendus de l'Académie des Sciences. Série A (in French). 287:
2. ^ Kusner, Rob (1987). "Conformal geometry and complete minimal surfaces" (PDF). Bulletin of the American Mathematical Society. New Series. 17 (2): 291–295. doi:10.1090/S0273-0979-1987-15564-9..
3. ^ Goodman, Sue; Marek Kossowski (2009). "Immersions of the projective plane with one triple point". Differential Geometry and Its Applications. 27 (4): 527–542. doi:10.1016/j.difgeo.2009.01.011.
ISSN 0926-2245.
4. ^ Raymond O'Neil Wells (1988). "Surfaces in conformal geometry (Robert Bryant)". The Mathematical Heritage of Hermann Weyl (May 12–16, 1987, Duke University, Durham, North Carolina). Proc.
Sympos. Pure Math. Vol. 48. American Mathematical Soc. pp. 227–240. doi:10.1090/pspum/048/974338. ISBN 978-0-8218-1482-6.
5. ^ Adam, Savage (21 June 2023). "This Object Should've Been Impossible to Make". YouTube. Retrieved 22 June 2023.
• Kirby, Rob (November 2007), "What is Boy's surface?" (PDF), Notices of the AMS, 54 (10): 1306–1307 This describes a piecewise linear model of Boy's surface.
□ Casselman, Bill (November 2007), "Collapsing Boy's Umbrellas" (PDF), Notices of the AMS, 54 (10): 1356 Article on the cover illustration that accompanies the Rob Kirby article.
• Mathematisches Forschungsinstitut Oberwolfach (2011), The Boy surface at Oberwolfach (PDF).
• Sanderson, B. Boy's will be Boy's, (undated, 2006 or earlier).
• Weisstein, Eric W. "Boy's Surface". MathWorld.
External links
Wikimedia Commons has media related to Boy's surface.
• Boy's surface at MathCurve; contains various visualizations, various equations, useful links and references
• A planar unfolding of the Boy's surface – applet from Plus Magazine.
• Boy's surface resources, including the original article, and an embedding of a topologist in the Oberwolfach Boy's surface.
• A LEGO Boy's surface
• A paper model of Boy's surface – pattern and instructions
• A model of Boy's surface in Constructive Solid Geometry together with assembling instructions
• Boy's surface visualization video from the Mathematical Institute of the Serbian Academy of the Arts and Sciences
• This Object Should've Been Impossible to Make Adam Savage making a museum stand for a glass model of the surface | {"url":"https://www.knowpia.com/knowpedia/Boy%27s_surface","timestamp":"2024-11-02T02:23:19Z","content_type":"text/html","content_length":"128964","record_id":"<urn:uuid:f49a76d5-7679-442f-b308-a9433a977aad>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00894.warc.gz"} |
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MA4A5 Algebraic Geometry
MA4A5 Algebraic Geometry Support Classes
Week 2: We discussed some of the basics of projective space. If you did not feel comfortable with the material the class was structured off these notes which were taught as a short course on
projective geometry at the University of Oxford. These are not my own notes, if you find them useful please thank Balazs.
If one googles projective geometry Oxford, there are also exercise sheets with easier questions which may prove of use. Similarly, the geometry course at Warwick will have exercises which may be of
Week 3: We discussed sheaves and quasi-projective varieties at length; in particular why we care about regular and rational functions. We also proved that every non-empty Zariski open subset of an
irreducible space is dense.
Week 4: We discussed rational maps and morphisms and the differences between the two. In particular we focused on asking whether certain rational maps could be extended to projective morphisms and
attempted some exam questions from the 2016 paper.
Week 5: Segre and Veronese by Alice Cuzzucoli.
Week 6: Review of Grassmanians, projective and proper morphisms. For those students struggling to understand a concept in the lecture notes, an alternative one could try is these notes by Andreas
Gathmann. There is also a follow on from these which covers quasi-coherent sheaves, cohomology, intersection theory, Chern classes etc.
Week 7: A question about Zariski dense arguments (which may help clarify injectivity of the restriction map in Assignment 2 Question 3). Functions vanishing on a dense subset vanish on the closure.
Images of morphisms, when X affine the image may not be closed, open or could even be a union of both. Transcendence degree and computing dimension. In the course regular functions are defined first
and then rational functions. This corresponds scheme theoretically to the way one often views the rational function field as the stalk of the sheaf of regular functions at the generic point. Less
technically however one can think in this way: for $X$ closed affine, consider polynomial functions on $X$ in the k-algebra $k[X]$. This is defined to be $K[\mathbb{A}^n]/I(X)$. One can define a
regularity condition at every point (that the rational function- the quotient of 2 polynomial functions is defined at that point) and $\mathcal{O}_X(U)$ is then the rational functions regular for
every point in $U$. Then the global sections of this sheaf will precisely be the rational functions defined on all of $X$, which will precisely be $K[X]$. Since $K[X]$ will not necessarily be a UFD,
one has many different representatives possibly for such a regular function. To see how this can be done you can look at page 77 and page 125 here regular and rational functions.
Week 8: Upper semi-continuity of fibre dimension and the final 8 mark question from this exam. I did not finish this but motivated the usefulness of incidence varieties (local form of blow-up). $\
mathbb{P}^1 \times \mathbb{P}^1$ birational to $\mathbb{P}^2$ but not isomorphic. $\mathbb{A}^1 \times \mathbb{P}^1$ neither affine nor projective. Uses of the fact there are no-non constant
morphisms from $\mathbb{P}^n \to \mathbb{P}^m, m<n$.
Week 9: N/A
Week 10: The classification problem, discrete invariants and moduli. Chow varieties, Cayley forms and degree.
As a supplement to these support classes, I will hold a revision session in the week prior to your exam where we can cover exam style problems. | {"url":"https://warwick.ac.uk/fac/sci/maths/people/staff/jakepatel/algebraic_geometry/","timestamp":"2024-11-13T20:12:38Z","content_type":"text/html","content_length":"39612","record_id":"<urn:uuid:23537e8a-b24b-48d8-badd-3f49b0e4b5a3>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00089.warc.gz"} |
How to read numbers: the figure on the left is digit 3: the middle crossbar (out of three horizontal ones) is turned outward (now it is directed in the direction opposite to the other two crossbars).
Then the entire digit 3 is rotated to the right by 90° (the vertical axis of digit 3 is in a horizontal position). And finally, the three crossbars are twisted around the axis (the upper and lower
crossbars are counterclockwise, the middle crossbar is clockwise). | {"url":"https://0123456789art.ru/gallery_eng/tproduct/532861530-798625719681-309","timestamp":"2024-11-06T13:42:12Z","content_type":"text/html","content_length":"58504","record_id":"<urn:uuid:6492a854-cc2d-41a3-8c1b-a797f80e998e>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00130.warc.gz"} |
Do quadratic splines really use all the data points?
There are two reasons that linear splines are seldom used
1. Each spline uses information only from two consecutive data points
2. The slope of the splines is discontinuous at the interior data points
The answer to resolving the above concerns are to use higher order splines such as quadratic splines. Read the quadratic spline textbook notes before you go any further. You do want what I have to
say to make sense to you.
In quadratic splines, a quadratic polynomial is assumed to go thru consecutive data points. So you cannot just find the three constants of each quadratic polynomial spline by using the information
that the spline goes thru two consecutive points (that sets up two equations and three unknowns). Hence, we incorporate that the splines have a continuous slope at the interior points.
So does all this incorporation make the splines to depend on the values of all given data points. It does not seem so.
For example, in quadratic splines you have to assume that the first or last spline is linear. For argument sake, let that be the first spline. If the first spline is linear, then we can find the
constants of the linear spline just by the knowledge of the value of the first two data points. So now we know that we can set up three equations for the three unknown constants of the second spline
as follows
1. the slope of the first spline at the 2nd data point and the slope of the second spline at the 2nd point are the same,
2. the second spline goes thru the 2nd data point
3. the second spline goes thru the 3rd data point
That itself is enough information to find the three constants of the second spline. We can keep using the same argument for all the other splines.
So the mth spline constants are dependent on data from the data points 1, 2, .., m, m+1 but not beyond that.
Can you now make the call on the kind of dependence or independence you have in the constants of the quadratic splines?
This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://nm.mathforcollege.com
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Data Structures in R
There are many types of objects / Data Structures to store R-object . The frequently used objects are –
• Vectors
• Matrices
• Arrays
• Factors
• Data Frames
• Lists
• Table
A vector is most basic data structure in R. It is collection same objects like character , logical, integer or numeric .
a <- c(1,2,3,4)
Output :
[1] 1 2 3 4
We can check object “a” is vector or not .
[1] TRUE
We have created character vector “b” as :
A matrix is a collection of data elements arranged in two-dimensional rectangular layout.
We can check layout of matrix function by :
It opens description and syntax of matrix function in Help window.
Syntax of matrix is :
matrix(data = NA, nrow = 1, ncol = 1, byrow = FALSE, dimnames = NULL)
data = Represent data
nrow = number of rows
ncol = number of columns
byrow= the matrix is filled by row
dimnames = to assign names to row and columns
Number of elements = nrow * ncol
We have nrow=3 .
We find ncol = 2 .
The default value of byrow is FALSE and dimnames is NULL .
mat <- matrix(c(1,0,4,2,-1),nrow= 3)
In this case, byrow = FALSE means we filled elements of matrix by column. So, elements are filled column -wise .
[,1] [,2] <- Number of column
[1,] 1 2 <-Number of row
[2,] 0 -1
[3,] 4 5
We represent matrix mat as mat[r,c] , where r is number of row and c is number of column.
mat[1,2] <- First row and second column element .
Output :
Output :
We can show all elements of whole row as m[r,] and whole column as m[,c] .
mat[1,] <- It shows all elements of first row
[1] 1 2
mat[,2] <- It shows all elements of second column
[1] 2 -1 5
We can also filled elements by row as :
It shows elements are filled in matrix by filling rows by rows.
We check the class of mat object by :
[1] “matrix”
We can give names to columns and rows by using dimnames :
mat <- matrix(c(1,0,4,2,-1,5),nrow= 3,dimnames = list(c(“a”,”b”,”c”),c(“x”,”y”)))
Recycling :
Recycle means reusable of materials . We can reuse data to perform functions required .
We are creating a matrix having 5 elements. We assign number of rows are 2 . So , number of columns are 3.
You can see that , there is warning message which showing number of elements are not multiple of 2. So, it will recycle the remaining element by starting with first element to fill.
You can see here , we have create a matrix of 10 elements . The elements are repeated to fill remaining elements.
We can also create matrix on data object :
We can create matrix by using dim() . We assign dimension of matrix using dim() and create a matrix on “a” object.
a <- 1:20
We assign dimension as rows X columns .
dim(a) <- c(4,5) # number of rows = 4 , number of columns =5
Output :
We can transpose matrix by using t() .
%*% Operator
This operator is used to multiply a matrix with its transpose.
We can bind matrices by row or column .
We can bind two matrices column-wise . When we bind columns, the number of rows of matrices should be same.
When the two matrix do not have same number of rows , it will join . There is an ERROR comes while binding them.
We can also bind vectors by using following code:
We can bind matrices by row-wise . When we bind rows , the number of columns of two matrices should be same.
We can also bind vectors row-wise as:
We can store data in more than two dimensions . If we create an array of dimension (2,3,4) then it creates 4 rectangular matrices each of 2 rows and 3 columns.
An array can create by using array() . We used dim to assign dimension of array.
Arrays are also recycled same as matrix . We create two vectors and input these vectors to an array to fill the elements of array.
vector1 <- c(5,9,3)
vector2 <- c(10,11,12,13,14,15,16)
result <- array(c(vector1,vector2),dim = c(3,3,2))
We can give names to columns , rows and matrices in the array by using dimnames parameter.
vector1 <- c(5,9,3)
vector2 <- c(10,11,12,13,14,15)
column.names <- c(“first”,”second”,”third”)
row.names <- c(“first”,”second”,”third”)
matrix.names <- c(“Matrix1″,”Matrix2”)
result <- array(c(vector1,vector2),dim = c(3,3,2),dimnames = list(row.names,column.names,
We can show the third row of the first matrix of the array .
result [3,,2]
We can show the element in the 1st row and 2nd column of the 1st matrix.
[1] 10
Check out second matrix .
Create matrices from the array .
We add two matrices also :
Factors are the data objects which are used to categorize the data and store it. They can store both strings and integers .
Factors are created using factor() function .
l <- c(“male”,”female”)
Levels shows all possible values of given object . We can check levels of object .
[1] “female” “male”
We create another factor variable Name :
[1] “1” “2”
[1] “factor”
To convert the default factor Name to roman numerals, we use the assignment form of the levels() function:
levels(Name) = c(‘I’,’II’)
It is used to build a contingency table of the count of each combination of factor variables .
mons = c(“March”,”April”,”January”,”November”,”January”,
mons = factor(mons)
Data Frames
Data Frame is a two dimensional data structure . The characteristics of a data frame are :
• The column names should be non-empty . Every column has assign certain name.
• The row names should be unique.
• The data can be of numeric , character or factor type.
• Each column should contain same number of elements.
We create a data frame name ” myFirstDataFrame ” as:
myFirstDataFrame <- data.frame(name = c(“Bob”, “Fred”, “Barb”, “Sue”,”Jeff”),
age = c(21,18,18,24,20),
hgt= c(70,67,64,66,72),
wgt= c(180,156,128,118,202),
race= c(“Cauc”, “Af.Am”,”Af.Am”, “Cauc”, “Asian”),
year= c(“Jr”,”Fr”,”Fr”,”Sr”,”So”),
SAT= c(1080,1210,840,1340,880))
We can view data frame by :
We can find number of rows and columns by using nrow() and ncol() .
A list is a generic vector . It is combination of different objects .
We can create list as :
list<-list(1:4 ,”abc”,TRUE)
Output :
[[1]] <- it shows first object from list
[1] 1 2 3 4 <- it shows elements of first object
[[2]] <- it shows second object from list
[1] “abc” <- it shows elements of second object
[[3]] <- it shows third object from list
[1] TRUE <- it shows elements of third object
We can create a list by combining different objects as:
b<-c(“Alec”, “Dan”, “Rob”, “Rich”)
c <- c(TRUE, TRUE, FALSE, FALSE)
We create a list of integer , matrix and character as:
x <- 1:10
y <- matrix(1:12, nrow=3)
z <- “Hello”
mylist <- list(x,y,z)
We create a list contains character , matrix and list objects as:
list_data <- list(c(“January”,”February”,”March”), matrix(c(4,8,6,9,-5,3), nrow = 2),
We can give names to the list objects by using names() as:
names(list_data) <- c(“1st Quarter”, “A_Matrix”, “A Inner list”)
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seminars - Introduction to random graphs, and thresholds
[서울대학교 수리과학부 10-10 집중강연] 2/7 (화), 9 (목), 14 (화), 10:00AM - 12:00PM
장소: Zoom 강의실
회의 ID: 993 2488 1376
암호: 120745
초록: We will start with a mild introduction to random graph theory, focusing on threshold phenomena of increasing families. In general, for a finite set X, a family F of subsets of X is said to be
increasing if any set A that contains B in F is also in F. The p-biased product measure of F increases as p increases from 0 to 1, and often exhibits a drastic change around a specific value, which
is called a "threshold." Thresholds of increasing families have been of great historical interest and a central focus of the study of random discrete structures (e.g. random graphs and hypergraphs),
with estimation of thresholds for specific properties the subject of some of the most challenging work in the area. In 2006, Jeff Kahn and Gil Kalai conjectured that a natural (and often easy to
calculate) lower bound q(F) (which we refer to as the “expectation-threshold”) for the threshold is in fact never far from its actual value. A positive answer to this conjecture enables one to narrow
down the location of thresholds for any increasing properties in a tiny window. In particular, this easily implies several previously very difficult results in probabilistic combinatorics such as
thresholds for perfect hypergraph matchings (Johansson–Kahn–Vu) and bounded-degree spanning trees (Montgomery). We will discuss fascinating recent progress on this topic. | {"url":"http://www.math.snu.ac.kr/board/index.php?mid=seminars&page=58&l=en&sort_index=speaker&order_type=asc&document_srl=1036561","timestamp":"2024-11-04T17:24:07Z","content_type":"text/html","content_length":"51771","record_id":"<urn:uuid:038d3fc7-9bb3-4929-b218-2570b14b85d5>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00496.warc.gz"} |
The Department of Applied Mathematics has 27 academic staff divided into two areas of knowledge: Applied Mathematics with 23 professors and Astronomy and Astrophysics, with 4 professors.
Area of Applied Mathematics
The area of Applied Mathematics is involved in teaching the degrees on Mathematics, Biology, Chemistry, Chemical Engineering, Computer Engineering, Industrial Chemical Processes, Civil Engineering,
Food and Agricultural Engineering, Forestry and Natural Resources, and Geomatics and Surveying Engineering. It is also involved in teaching in the following double degrees: Computer Engineering and
Mathematics; Mathematics and Physics; Chemistry and Biology; Food and Agricultural Engineering; Forestry and Natural Resources; Civil Engineering and Geomatics and Surveying Engineering. In all of
them Mathematics seeks integration with the rest of the subjects by conducting exercises and applications specifically tailored to the background of the students. It is particularly important the
commitment of the area to the degree in Mathematics.
The department of Applied Mathematics participates in the PhD program "Mathematical Methods and Numerical Simulation in Engineering and Applied Sciences" by the Universidades de A Coruña, Santiago de
Compostela and Vigo. This program, verified in 2013, is an adaptation of a previous program by the same name which had a quality label by the Ministry of Education and Science since its
implementation. Furthermore, this area also participated and coordinated the "Master in Mathematical Engineering" by the Universidades de A Coruña, Santiago de Compostela and Vigo since its creation
in the 2006-2007 academic year and until 2013-2014. From then onwards, the department also participates in the “Master in Industrial Mathematics by the Universidade de A Coruña, Santiago de
Compostela, Vigo, Carlos III University of Madrid and Technical University of Madrid ( www.m2i.es ).
The department also takes part in the masters of Mathematics, Chemical Engineering and Bio-processes, and Biomedical Research.
There are three research groups in this area specializing in the numerical solution of partial differential equations, with particular emphasis on modelling, mathematical analysis and numerical
solution of phenomena from the Engineering and Applied Sciences. In particular the area has carried out contracts and research projects in subjects as diverse as Metallurgy, Solid Mechanics, Energy,
Environment, Acoustics, New Materials, Fluid Mechanics, Automotive, Finance and Life sciences.
Area of Astronomy and Astrophysics Up
The Astronomy studies, with great tradition at the University of Santiago de Compostela, are linked to the existence of the Astronomical Observatory ( https://www.usc.es/astro ).
The "Durán Loriga" section of Theoretical Astronomy and Mathematics created in 1945 in the Astronomical Observatory and where Professors such as Vidal Abascal and García-Rodeja worked, gave an
important impetus to mathematical research in our university. It can be considered the beginnings of the Mathematics Section within the Faculty of Science, at present the Faculty of Mathematics.
The Professorship of Astronomy, which was occupied by the founder of the Observatory, Dr. Ramon María Aller Ulloa, was the first one in the Mathematics Section.
From 1981 and after a long period with no doctors in the subject in the USC, the situation improved in regards to teaching in the Faculty of Mathematics and the activities of the Astronomical
At present the department is involved in teaching in the degrees of Physics, Mathematics and Optics as well as in the Master in Mathematics (Santiago de Compostela campus), and the degree in
Geomatics Engineering and Topography (Lugo Campus).
With the creation of the IV University Cycle, which coordinator is Professor J. A. Docobo, the subjects of Introduction to Astronomy and Astronomy are taught. This area also organises since 17 years
ago the Cultural Program for Astronomy (www.usc.es/pecas), directed to all the university community and the general public and which is recognised with 1 ECTS.
The researchers from the Ramón María Aller Astronomical Observatory belong to the research group OARMA (USC code GI-1565). | {"url":"https://www.usc.gal/dmafm/ENGLISH/index_english.htm","timestamp":"2024-11-10T14:56:16Z","content_type":"text/html","content_length":"14634","record_id":"<urn:uuid:82c784e5-01b8-40ab-9a36-f0c22d857eac>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00478.warc.gz"} |
Map Aggregation and Modeling
Map Aggregation and Modeling
It is important to appreciate that the MaNGA datacubes exhibit significant spatial covariance between spaxels. For example, this is a critical consideration when binning spaxels to an expected S/N
threshold; see Section 6 from Westfall et al. (2019, AJ, 158, 231).
The output quantities provided by the DAP are, therefore, also spatially correlated. The detailed calculation of covariance matrices for DAP quantities requires simulations that have not been
performed! Having said that, some initial testing along these lines suggests that the correlation matrix of DAP quantities can be approximated by the nominal correlation between quantities in any
given DRP wavelength channel.
The suggestions below are provided on a shared-risk basis, pending a more robust series of tests to validate this approach. Use with caution.
Approximate Correlation Between Spaxels
In Section 6.2 of Westfall et al. (2019, AJ, 158, 231), we showed that the correlation, \(\rho\), between spaxels as a function of the distance between them is well approximated by a Gaussian
You can construct the approximate correlation matrix for a given data cube as follows (cf. Spatial Covariance; Covariance):
from mangadap.datacube import MaNGADataCube
C = MaNGADataCube.from_plateifu(7815, 3702).approximate_correlation_matrix()
Initial tests suggest this correlation matrix is a good approximation for the DAP maps.
Approximate Covariance in a DAP Map
To construct the covariance matrix for a DAP map, start with the approximate correlation matrix and then renormalize to a covariance matrix using the provided variance data. For example, the
following code constructs the covariance matrix in the \({\rm H}\alpha\) velocity measurements (note that the transposes are important; unfortunately, construction of the approximate correlation
requires the instantation of the MaNGADataCube):
# Imports
import numpy
from astropy.io import fits
from mangadap.datacube import MaNGADataCube
from mangadap.dapfits import DAPMapsBitMask
from mangadap.util.fileio import channel_dictionary
# Read the datacube (assumes the default paths)
cube = MaNGADataCube.from_plateifu(7815, 3702)
# Construct the approximate correlation matrix
C = cube.approximate_correlation_matrix(rho_tol=0.01)
# Read the MAPS file
hdu = fits.open('manga-7815-3702-MAPS-HYB10-MILESHC-MASTARHC2.fits.gz')
# Get a dictionary with the emission-line names
emlc = channel_dictionary(hdu, 'EMLINE_GFLUX')
# Save the masked H-alpha velocity and velocity variance
bm = DAPMapsBitMask()
vel = numpy.ma.MaskedArray(hdu['EMLINE_GVEL'].data[emlc['Ha-6564'],...],
vel_var = numpy.ma.power(hdu['EMLINE_GVEL_IVAR'].data[emlc['Ha-6564'],...].T, -1)
vel_var[vel.mask] = numpy.ma.masked
# Construct the approximate covariance matrix
vel_C = C.apply_new_variance(vel_var.filled(0.0).ravel())
2D Modeling of a DAP Map
When constructing a 2D model of the quantities in a DAP map, ideally you would account for the covariance between spaxels in the formulation of the optimized figure-of-merit. For example, with the
covariance matrix, \({\mathbf \Sigma}\), the chi-square statistic is written as a matrix multiplication:
\[\chi^2 = {\mathbf \Delta}^{\rm T} {\mathbf \Sigma}^{-1} {\mathbf \Delta},\]
where \({\mathbf \Sigma}^{-1}\) is the inverse covariance matrix and \({\mathbf \Delta} = {\mathbf d} - {\mathbf m}\) is the vector of residuals between the observed data, \({\mathbf d}\), and the
proposed model data, \({\mathbf m}\).
To fit a model then, we need to calculate the inverse of the covariance matrix. A minor complication in doing this for the DAP maps is that covariance matrix of the full map is always singular
because of the empty regions in the map with no data. Taking care to keep track of original on-sky coordinates of each spaxel, we need to remove the empty regions of the map from the covariance
matrix before inverting it. Continuing from the code above and assuming the masks for vel and vel_var mask out all spaxels without valid data:
# Flatten the map and get a vector that selects the valid data
vel = vel.ravel()
indx = numpy.logical_not(numpy.ma.getmaskarray(vel))
# Pull out only the valid data
_vel = vel.compressed()
_vel_C = vel_C.toarray()[numpy.ix_(indx,indx)]
# Get the inverse covariance data
_vel_IC = numpy.linalg.inv(_vel_C)
Assuming you have a vector with the model velocities (model_vel), you would then calculate \(\chi^2\) using the above matrix multiplication. Just for the purpose of illustration let’s set the model
values to be 0 everywhere:
# Calculate chi-square
model_vel = numpy.zeros(_vel.size, dtype=float)
delt = _vel - model_vel
chisqr = numpy.dot(delt, numpy.dot(_vel_IC, delt))
Note that by definition \(\chi^2\) must be positive. However, a combination of the inaccuracies of the correlation approximation and numerical error in the inversion of the covariance matrix means
that this would not necessarily be true in the example above. This is why I limited the construction of the approximate correlation matrix to require \(\rho \geq 0.01\); see
approximate_correlation_matrix(). This is needed to ensure the calculation of chisqr above is positive, at least in the example case above.
Aggregation of Mapped Quantities
When aggregating data within a map, like summing within an aperture or calculating the mean within a region, propagation of the error aggregated quantity should account for spatial covariance. The
easiest way to deal with the error propagation is to recast the aggregation as a matrix multiplication. The following is a worked example for binning the \({\rm H}\delta_{\rm A}\) index.
This example works with the uncorrected \({\rm H}\delta_{\rm A}\) index!
To start, here are the top-level imports:
# Imports
from IPython import embed
import numpy
from matplotlib import pyplot
from astropy.io import fits
from mangadap.datacube import MaNGADataCube
from mangadap.dapfits import DAPMapsBitMask
from mangadap.util.fileio import channel_dictionary
from mangadap.util.covariance import Covariance
from mangadap.util.fitsutil import DAPFitsUtil
Get the data
First, we extract the data, the weights needed to get the mean index, and the approximate covariance matrix. This is all very similar to what’s done to get the velocity measurements above:
# Read the datacube (assumes the default paths)
cube = MaNGADataCube.from_plateifu(7815, 3702)
# Construct the approximate correlation matrix
C = cube.approximate_correlation_matrix()
# Read the MAPS file
hdu = fits.open('manga-7815-3702-MAPS-HYB10-MILESHC-MASTARHC2.fits.gz')
# Get a dictionary with the spectral index names
ic = channel_dictionary(hdu, 'SPECINDEX_BF')
# Extract the masked HDeltaA index, its variance, and the relevant weight
bm = DAPMapsBitMask()
hda = numpy.ma.MaskedArray(hdu['SPECINDEX_BF'].data[ic['HDeltaA'],...],
hda_wgt = numpy.ma.MaskedArray(hdu['SPECINDEX_WGT'].data[ic['HDeltaA'],...],
hda_var = numpy.ma.power(hdu['SPECINDEX_BF_IVAR'].data[ic['HDeltaA'],...].T, -1)
# Unify the masks (just in case they aren't already)
mask = numpy.ma.getmaskarray(hda) | numpy.ma.getmaskarray(hda_wgt) | numpy.ma.getmaskarray(hda_var)
hda[mask] = numpy.ma.masked
hda_wgt[mask] = numpy.ma.masked
hda_var[mask] = numpy.ma.masked
# Construct the approximate covariance matrix
hda_C = C.apply_new_variance(hda_var.filled(0.0).ravel())
Construct a binning map
Just for demonstration purposes, we need to create a map that identifies the spaxels that will be aggregated. The following just constructs a rough, ad hoc binning scheme, with bins of roughly 3x3
def boxcar_replicate(arr, boxcar):
Boxcar replicate an array.
arr (`numpy.ndarray`_):
Array to replicate.
boxcar (:obj:`int`, :obj:`tuple`):
Integer number of times to replicate each pixel. If a
single integer, all axes are replicated the same number
of times. If a :obj:`tuple`, the integer is defined
separately for each array axis; length of tuple must
match the number of array dimensions.
`numpy.ndarray`_: The block-replicated array.
# Check and configure the input
_boxcar = (boxcar,)*arr.ndim if isinstance(boxcar, int) else boxcar
if not isinstance(_boxcar, tuple):
raise TypeError('Input `boxcar` must be an integer or a tuple.')
if len(_boxcar) != arr.ndim:
raise ValueError('Must provide an integer, or a tuple with one number per array dimension.')
# Perform the boxcar average over each axis and return the result
_arr = arr.copy()
for axis, box in zip(range(arr.ndim), _boxcar):
_arr = numpy.repeat(_arr, box, axis=axis)
return _arr
# Construct a binning array that bins all the data into 3x3 bins, with
# the bins roughly centered on the center of the map
nbox = 3
nmap = hda.shape[0]
nbase = nmap//nbox + 1
s = nbox//2
base_bin_id = numpy.arange(numpy.square(nbase)).reshape(nbase, nbase).astype(int)
bin_id = boxcar_replicate(base_bin_id, nbox)[s:s+nmap,s:s+nmap]
# Ignore masked pixels and reorder the bin numbers
bin_id[hda.mask] = -1
unique, reconstruct = numpy.unique(bin_id, return_inverse=True)
bin_id = numpy.append(-1, numpy.arange(len(unique)-1))[reconstruct]
Bin the data
By identifying which spaxels are in each bin, we can now construct the weighting matrix, bin the data, and get the covariance matrix of the binned data:
# Construct the weighting matrix
nbin = numpy.amax(bin_id)+1
nspax = numpy.prod(hda.shape)
wgt = numpy.zeros((nbin, nspax), dtype=float)
indx = numpy.logical_not(hda.mask.ravel())
wgt[bin_id[indx],numpy.arange(nspax)[indx]] = hda_wgt.data.ravel()[indx]
sumw = numpy.sum(wgt, axis=1)
wgt *= ((sumw != 0)/(sumw + (sumw == 0)))[:,None]
# Get the binned data and its covariance array
bin_hda = numpy.dot(wgt, hda.ravel())
bin_hda_C = Covariance.from_matrix_multiplication(wgt, hda_C.full())
We can then use reconstruct_map() to construct a map of the binned data with the same format as the original spaxel data, and create a plot that shows the two side-by-side:
# Reconstruct a map with the binned data
bin_hda_map = numpy.ma.MaskedArray(DAPFitsUtil.reconstruct_map(hda.shape, bin_id.ravel(), bin_hda),
mask=bin_id == -1)
# Compare the unbinned and binned maps
w,h = pyplot.figaspect(1)
fig = pyplot.figure(figsize=(2*w,h))
ax = fig.add_axes([0.05, 0.1, 0.4, 0.8])
ax.imshow(hda, origin='lower', interpolation='nearest', vmin=-1, vmax=6)
ax = fig.add_axes([0.5, 0.1, 0.4, 0.8])
img = ax.imshow(bin_hda_map, origin='lower', interpolation='nearest', vmin=-1, vmax=6)
cax = fig.add_axes([0.91, 0.1, 0.02, 0.8])
cax.text(3., 0.5, r'H$\delta_A$ [${\rm \AA}$]', ha='center', va='center', transform=cax.transAxes,
pyplot.colorbar(img, cax=cax)
We can also compare the errors that we would have calculated without any knowledge of the covariance to the errors calculated using the approximate correlation matrix:
# Compare the errors with and without covariance
bad_var = numpy.dot(numpy.square(wgt), hda_var.ravel())
good_var = numpy.diagonal(bin_hda_C.toarray())
bad_sig = numpy.sqrt(bad_var)
pyplot.plot(bad_sig, 2.5*bad_sig, color='C3', zorder=1,
label=r'$\sigma = 2.5 \sigma_{\rm no covar}$')
pyplot.scatter(bad_sig, numpy.sqrt(good_var), marker='.', s=80, color='k', lw=0, zorder=2)
pyplot.xlabel(r'$\sigma_{\rm no covar}$ [${\rm \AA}$]')
pyplot.ylabel(r'$\sigma$ [${\rm \AA}$]') | {"url":"https://sdss-mangadap.readthedocs.io/en/develop/aggregation.html","timestamp":"2024-11-12T23:40:54Z","content_type":"text/html","content_length":"58845","record_id":"<urn:uuid:9437e7b4-e0af-4e0a-a149-b52d0251d603>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00565.warc.gz"} |
Graphing Complex Numbers On A Complex Plane Worksheet
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Graphing Complex Numbers On A Complex Plane Worksheet
Graphing Complex Numbers On A Complex Plane Worksheet -
Defining Graphing Complex Numbers In this video we learn to graph complex numbers using the complex number plane instead of our normal x y plane In this video we learn that the absolute value of a
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Worksheet by Kuta Software LLC Algebra 2 2 1 Graphing complex numbers ID 1 T2 0c1u7O DKDuBt aJ JSQomfXtvw aFrjei L LXCf i FAol ly lrDipgphbtfsX irqepsSeHrNvueadS Graph each number in the complex
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What Is 10 Degrees Fahrenheit In Celsius (2024)
What is 10 Degrees Fahrenheit in Celsius?
Understanding temperature conversion is an essential skill, especially when dealing with different temperature scales used worldwide. So, when you ask, “What is 10 degrees fahrenheit in Celsius?” the
answer is -12.22 degrees Celsius. This value is achieved by using a specific conversion formula, which is to subtract 32 from the Fahrenheit temperature and multiply the result by 5/9.
So, if you have to convert 10 degrees Fahrenheit, for example, you would subtract 32, resulting in -22. Then, you multiply -22 by 5/9, giving you -12.22 Celsius. Understanding this conversion process
from Fahrenheit to Celsius can be handy when you’re traveling, preparing recipes, or studying weather patterns.
The Importance of the Freezing Point of Water
Understanding the freezing point of water is of great importance as it is a significant point on the temperature scale. Freezing water at 32 degrees Fahrenheit equals 0 degrees Celsius. This specific
temperature shift signals a phase change in water, where it transforms from a liquid to a solid state, forming ice.
This information isn’t just useful for predicting weather conditions. It’s essential for various industrial processes, such as brewing and distillation, and in scientific experiments that require
specific temperatures. Recognizing that the freezing point of water on a Celsius scale is 0 degrees and 32 degrees on the Fahrenheit scale helps us convert Celsius to Fahrenheit, and vice versa, much
Understanding the Conversion Formula
The conversion formula for converting Fahrenheit to Celsius is straightforward. To convert Fahrenheit to Celsius, you subtract 32 from the Fahrenheit temperature and then multiply the result by 5/9.
Conversely, to convert Celsius to Fahrenheit, you multiply the Celsius temperature by 9/5 and add 32.
By understanding the conversion formula, you can convert any value from Celsius to Fahrenheit or Fahrenheit to Celsius swiftly, be it 100 degrees on the Celsius scale or 212 degrees on the Fahrenheit
scale. These calculations aid in diverse fields – from weather forecasting to cooking.
The Concept of Boiling Point of Water
Just like the freezing point, the boiling point of water is another significant aspect to consider in the temperature conversion process. At sea level, water boils at 212 degrees Fahrenheit or 100
degrees Celsius. This point marks a phase change where water transforms from a liquid to a gaseous state – steam.
This understanding is crucial from a culinary perspective since many food preparation methods involve boiling water. The knowledge of water’s boiling point also helps us realize that any increase in
Fahrenheit temperature beyond 212 degrees results in temperatures usually associated with cookery or the weather on particularly hot days.
Mastering Fahrenheit to Celsius Conversions
Mastering the process of converting Fahrenheit to Celsius and vice versa is of immense importance, considering the two systems are widely used based on geographic location. For instance, the United
States uses the Fahrenheit scale, while most other countries use Celsius.
Hence, if you’re a traveler or a weather enthusiast, you’ll need to quickly convert temperatures, such as 10 degrees Fahrenheit or 100 degrees Celsius, depending on your requirement. The easiest way
to do it is by using an online converter or calculator.
The Role of Temperature Scale in Conversions
The temperature scale plays a significant role in the conversion process. There are three primary temperature scales used worldwide – Celsius, Fahrenheit, and Kelvin. The Fahrenheit scale is used
mainly in the U.S., while Celsius is used in most other countries, and Kelvin is used by scientists.
Understanding the Fahrenheit scale, where the freezing point of water is 32 degrees and the boiling point is 212 degrees, can be a bit tricky compared to the Celsius scale, where water freezes at 0
degrees and boils at 100 degrees. Therefore, understanding these two scales can help anyone to make a quick conversion.
The Process of Converting Fahrenheit to Celsius
While converting Fahrenheit to Celsius might initially seem complex, with a little practice, the process can be simplified. The key is to remember the conversion formula: subtract 32 from the
Fahrenheit temperature and then multiply the result by 5/9.
If you’re trying to convert a specific Fahrenheit temperature, like 10 degrees Fahrenheit, into Celsius, this would equate to approximately -12.22 degrees Celsius. Similarly, applying the formula
inversely can convert any Celsius temperature back to Fahrenheit, reminiscent of the process to convert 10 degrees Fahrenheit.
Fahrenheit and Celsius: Hot Water Temperatures
When dealing with hot water, understanding Fahrenheit and Celsius temperatures is critical. For instance, the ‘hot’ setting on a U.S. washing machine is typically set at around 130 degrees
Fahrenheit. In Celsius, this converts to approximately 54.5 degrees.
Typically speaking, hot water temperatures in the Fahrenheit range are less than 212 degrees. Once the temperature reaches 212 degrees Fahrenheit (or 100 degrees Celsius), the water starts boiling.
Hence, understanding this comparison can be useful when dealing with everyday household chores or industrial tasks.
1. What is the formula to convert Fahrenheit to Celsius?
You subtract 32 from the Fahrenheit temperature and then multiply the result by 5/9.
2. How to convert 10 degrees Fahrenheit to Celsius?
Subtract 32 from 10 (the result is -22), then multiply -22 by 5/9, resulting in -12.22 Celsius.
3. What is the freezing point of water in Celsius and Fahrenheit?
The freezing point of water is 0 degrees Celsius and 32 degrees Fahrenheit.
4. Can you convert 100 degrees Fahrenheit to Celsius?
Yes, you subtract 32 from 100 (gives 68), then multiply 68 by 5/9, which results in 37.78 degrees Celsius.
5. What is the boiling point of water in Celsius?
The boiling point of water is 100 degrees Celsius.
6. How do I convert Celsius to Fahrenheit?
Multiply the Celsius temperature by 9/5, then add 32 to the result.
7. What is the boiling point of water in Fahrenheit?
The boiling point of water is 212 degrees Fahrenheit.
8. What does hot water temperature mean in Fahrenheit and Celsius?
Hot water in U.S. washing machines is typically set around 130 degrees Fahrenheit, which is equivalent to approximately 54.5 degrees Celsius.
9. How to convert 100 degrees Celsius to Fahrenheit?
Multiply 100 (celsius temperature) by 9/5 and add 32, giving you 212 degrees Fahrenheit.
10. Can Fahrenheit be converted to Celsius?
Yes, the Fahrenheit temperature can be converted to Celsius using the conversion formula. | {"url":"https://parkandmaincafe.com/food-recipes/what-is-10-degrees-fahrenheit-in-celsius/","timestamp":"2024-11-13T02:37:36Z","content_type":"text/html","content_length":"176092","record_id":"<urn:uuid:ce7609d2-7ef2-4970-88c8-40dd96b20d20>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00116.warc.gz"} |
A particle moves along X-axis in such a way that its coordinate X vari
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Synaptics (SYNA) - Piotroski F-Score Analysis for Year 2023 (Final Score: 4/9)
Synaptics SYNA 70.22 (+2.26%)
US87157D1090• Semiconductors• Semiconductors
Last update on 2024-06-07
Synaptics (SYNA) - Piotroski F-Score Analysis for Year 2023 (Final Score: 4/9)
Detailed Piotroski F-Score Analysis for Synaptics (SYNA) in 2023. Understand financial health, liquidity, and efficiency with historical data and charts.
Knowledge hint:
The Piotroski F-Score is a number between 0 to 9 which reflects the strength of a company's financial position. It is based on 9 criteria involving profitability, liquidity, and leverage. This model
helps investors identify stocks that are strong, undervalued investments.
Learn more...
Short Analysis - Piotroski Score: 4
We're running Synaptics (SYNA) against the Piotroski 9-criteria scoring system to assess profitability, liquidity, and operating efficiency:
Company has a positive net income?
Company has a positive cash flow?
Return on Assets (ROA) are growing?
Operating Cashflow are higher than Netincome?
Current Ratio is growing?
Number of shares not diluted?
Cross Margin is growing?
Asset Turnover Ratio is growing?
We examined Synaptics (SYNA) using the Piotroski F-Score criteria, covering profitability, liquidity, and operational efficiency. SYNA achieved a score of 4 out of 9. Its net income and cash flow
were positive for 2023, which is good. However, some metrics like return on assets and leverage showed negative trends. Their current ratio improved significantly, indicating better short-term
liquidity, but the increasing number of shares suggests dilution. Moreover, gross margin and asset turnover ratios have declined, which are potential red flags.
Insights for Value Investors Seeking Stable Income
Given its score of 4 out of 9 on the Piotroski F-Score, Synaptics shows mixed signals. Although it has strong points like positive cash flow and an increasing current ratio, there are concerning
signs with declining ROA, increasing leverage, falling asset turnover, and gross margin. These factors suggest caution. I'd recommend investigating further and considering other stocks, especially if
they show better financial stability and efficiency.
For those who are interested in delving deeper into the specifics, the subsequent section provides a comprehensive exploration of the criteria.
Profitability of Synaptics (SYNA)
Company has a positive net income?
Analyzing net income trends is crucial for understanding a company’s financial health and profitability over time.
Historical Net Income of Synaptics (SYNA)
The net income for Synaptics (SYNA) in 2023 stands at $73,600,000, which is positive. This meets the Piotroski criterion, and thus, we assign a score of 1. To better grasp the trend, it's helpful to
examine the historical net income data for the past 20 years. From 2003 to 2023, there have been fluctuations, with significant peaks, such as $257,500,000 in 2022, and downturns, such as a negative
$124,100,000 in 2019. This historical context suggests that while Synaptics has shown periods of substantial profitability, it has also faced challenges. Overall, a positive net income in 2023 is a
good indicator for the company's ongoing financial health, especially after the substantial profit in 2022.
Company has a positive cash flow?
Cash Flow from Operations (CFO) is a key indicator of the financial health of a company, reflecting its ability to generate cash through regular business operations.
Historical Operating Cash Flow of Synaptics (SYNA)
The Cash Flow from Operations (CFO) for Synaptics (SYNA) in 2023 stands at $331,500,000, which is indeed positive. Over the last 20 years, the CFO has shown a generally upward trend, although there
have been fluctuations. For instance, in 2003, the CFO was a mere $11,664,000, and it has grown substantially over the years, peaking at $462,700,000 in 2022. This positive CFO in 2023 indicates that
Synaptics has been effective in managing its cash flows from core operations, a bullish signal when assessing the company's financial health. Given this trend, this criterion would score 1 point.
Return on Assets (ROA) are growing?
Change in Return on Assets (ROA) assesses how well a company is utilizing its assets to generate earnings. This metric helps in determining the company's efficiency and profitability trends.
Historical change in Return on Assets (ROA) of Synaptics (SYNA)
The Return on Assets (ROA) for Synaptics in 2023 was 0.0269, which is a significant decrease compared to the ROA of 0.1013 in 2022. This trend is unfavorable as an increasing ROA would indicate
improved efficiency in using assets to generate profits. The decline from 0.1013 to 0.0269 signifies a deteriorating efficiency in the company’s asset utilization, possibly indicating issues in
operational performance or challenges in the business environment. Comparing this to the last 20 years industry median ROA, Synaptics' ROA has been consistently lower than the industry median, which
poses a concern. Therefore, for this criterion, no point is awarded.
Operating Cashflow are higher than Netincome?
The criterion checks whether the company generates more cash from operations than it earns in net income, signifying efficient cash management.
Historical accruals of Synaptics (SYNA)
For Synaptics (SYNA) in 2023, Operating Cash Flow stands at $331.5 million, significantly higher compared to the Net Income of $73.6 million. This results in a score of 1 point. This comparison
indicates robust cash generation from core operations, overshadowing accounting adjustments and non-cash items affecting net income. Over the past two decades, the company's Cash Flow from Operations
shows strong growth, fluctuating initially but broadly trending upward from $11.66 million in 2003 to $331.5 million in 2023. Concurrently, Net Income has been more volatile, peaking at $257.5
million in 2022 before declining. Steady growth in Operating Cash Flow displays effective core operations and financial health, aligning with solid performance indicators despite bottom-line
Liquidity of Synaptics (SYNA)
Leverage is declining?
Change in Leverage is an important criterion in the Piotroski Analysis as it measures a company's change in debt levels, providing insight into financial risk and stability.
Historical leverage of Synaptics (SYNA)
Synaptics (SYNA) had a leverage ratio of 0.3594 in 2022, which increased to 0.3885 in 2023. This indicates an increase in the company's leverage. Historically, Synaptics has seen significant
fluctuations in its leverage, notably peaking in 2005 at 0.4065. The increase in leverage from 2022 to 2023 suggests a higher financial risk and less stability, contrary to the ideal scenario
suggested by Piotroski criteria. Therefore, based on this metric, Synaptics would not score a point for a decrease in leverage.
Current Ratio is growing?
Evaluating the change in Current Ratio assesses a company's short-term liquidity by comparing its ability to cover short-term liabilities with short-term assets.
Historical Current Ratio of Synaptics (SYNA)
The Current Ratio of Synaptics (SYNA) increased from 3.0285 in 2022 to 4.8904 in 2023. This represents a significant improvement in liquidity, indicating that the company is better positioned to meet
its short-term obligations. As a result, Synaptics earns 1 point for this criterion in the Piotroski Analysis. Notably, the company's Current Ratio is also well above the industry median of 3.4213
for 2023, further underscoring its robust liquidity status. Historically, this trend seems largely positive compared to previous years when the Current Ratio dipped as low as 1.4996 in 2021. Overall,
this upward movement signals healthier financial stability.
Number of shares not diluted?
The change in outstanding shares indicates whether a company is issuing more shares or buying back shares, affecting ownership dilution.
Historical outstanding shares of Synaptics (SYNA)
In 2022, Synaptics had 39,000,000 outstanding shares. By 2023, this increased to 39,600,000. This 600,000 share increase means they have not bought back shares but issued more. Historically, their
outstanding shares have fluctuated: peaking in 2005 at 44,641,500 and hitting a low in 2019 at 33,600,000. Recently, the trend has shown an upward tick. Given that outstanding shares increased, we
assign 0 points for this criterion.
Operating of Synaptics (SYNA)
Cross Margin is growing?
The change in gross margin criterion assesses if the company is now more profitable by comparing the current year's gross margin to the previous year's.
Historical gross margin of Synaptics (SYNA)
For Synaptics (SYNA), the gross margin has decreased from 0.5421 in 2022 to 0.5283 in 2023. This represents a decline in gross margin, indicating that the company is under more pressure in terms of
cost management and pricing power. Over the past 20 years, the gross margin trend reveals a high of 0.5283 in 2022, suggesting that the current decline is somewhat against the recent trend of
increasing profitability. Despite the decline, it is worth mentioning that the gross margin is still above the industry's last-reported median of 0.4919 in 2023. This shows that Synaptics remains
more efficient than its peers in generating profit from sales, although the dip from its peak signals an area of concern that management would need to address.
Asset Turnover Ratio is growing?
Asset Turnover measures the efficiency of a company in using its assets to generate sales. It's crucial for assessing operational performance.
Historical asset turnover ratio of Synaptics (SYNA)
Comparing Synaptics' asset turnover ratio of 0.4955 in 2023 to 0.6843 in 2022, it's clear that the asset turnover has decreased. This decline signifies that the company has become less efficient in
using its assets to generate revenue over the past year. For the Piotroski score, this would result in 0 points for asset turnover. Analyzing the historical data, we observe that over the last two
decades, the asset turnover ratio peaked at 1.3868 in 2009 but has generally exhibited a downward trend, reflecting consistent challenges in asset utilization efficiency over the long term.
Obligatory risk notice
We would like to point out that the contents of this website are for general information purposes only and do not constitute recommendations for the purchase or sale of specific financial
instruments, and therefore do not constitute investment advice. In particular, marketstorylabs.com and its creators cannot assess the extent to which information / recommendations made on the pages
correspond to your investment objectives, your risk tolerance and your ability to bear losses. Therefore, if you make any investment decisions based on information on the site, you do so solely on
your own responsibility and at your own risk. This in turn means that neither marketstorylabs.com nor its creators are liable for any losses incurred as a result of investment decisions based on the
information on the marketstorylabs.com website or other media used. | {"url":"https://www.marketstorylabs.com/en/stocks/syna/stories/piotroski","timestamp":"2024-11-03T10:01:35Z","content_type":"text/html","content_length":"74917","record_id":"<urn:uuid:d02dec5e-ae4b-4b11-9943-bf6e2c54a296>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00308.warc.gz"} |
9.520/6.860: Statistical Learning Theory and Applications Fall 2021- Home
Course description
Understanding human intelligence and how to replicate it in machines is arguably one of the greatest problems in science. Learning, its principles and computational implementations, is at the very
core of intelligence. During the last two decades, for the first time, artificial intelligence systems have been developed that begin to solve complex tasks, until recently the exclusive domain of
biological organisms, such as computer vision, speech recognition or natural language understanding: cameras recognize faces, smart phones understand voice commands, smart speakers/assistants answer
questions and cars can see and avoid obstacles. The machine learning algorithms that are at the roots of these success stories are trained with examples rather than programmed to solve a task.
However, a comprehensive theory of learning is still incomplete, as shown by the puzzles of deep learning. An eventual theory of learning that explains why and how deep networks work and what their
limitations are, may thus enable the development of even much more powerful learning approaches and even inform our understanding of human intelligence.
In this spirit, the course covers foundations and recent advances in statistical machine learning theory, with the dual goal a) of providing students with the theoretical knowledge and the intuitions
needed to use effective machine learning solutions and b) to prepare more advanced students to contribute to progress in the field. This year the emphasis is again on b).
The course is organized about the core idea of supervised learning as an inverse problem, with stability as the key property required for good generalization performance of an algorithm.
The content is roughly divided into three parts. The first part is about classical regularization (regularized least squares, kernel machines, SVM, logistic regression, square and exponential loss)
large margin/minimum norm, stochastic gradient methods, overparametrization, implicit regularization and also approximation/estimation errors. The second part is about deep networks: approximation
theory -- which functions can be represented more efficiently by deep networks than shallow networks -- optimization theory -- why can stochastic gradient descent easily find global minima -- and
estimation error -- how generalization in deep networks can be explained in terms of the stability implied by the complexity control implicit in gradient descent. The third part is about the
connections between learning theory and the brain, which was the original inspiration for modern networks and may provide ideas for future developments and breakthroughs in the theory and the
algorithms of leaning. Throughout the course, and especially in the final classes, we will have occasional talks by leading researchers on selected advanced research topics. This class is the first
step of a NSF funded project in which a team of deep learning researchers from GeorgiaTech (Vempala), Columbia (Papadimitriou, Blei), Princeton (Hazan) and MIT (Madry, Daskalakis, Jegelka, Poggio)
plans to create over the next 3 years courses that leverages all the recent advances in our understanding of deep learning as well and crystallizes the corresponding “modern” outlook on the field.
Eventually it will include 1) deep learning: representations and generalization; (2) deep learning from an optimization perspective; (3) robustness and reliability in machine learning; (4) the brain
connection; (5) (robust) reinforcement learning; and (6) societal impact of machine learning, in addition to laying out the theoretical foundations of the field.
Apart for the first part on regularization, which is an essential part of any introduction to the field of machine learning, this year course is designed for students with a background in ML to
foster conjectures and exploratory projects on ongoing research.
We will make extensive use of basic notions of calculus, linear algebra and probability. The essentials are covered in class and in the math camp material. We will introduce a few concepts in
functional/convex analysis and optimization. Note that this is an advanced graduate course and some exposure on introductory Machine Learning concepts or courses is expected: for course 6 students
prerequisites are 6.041 and 18.06 and (6.036 or 6.401 or 6.867). Students are also expected to have basic familiarity with MATLAB/Octave.
Requirements for grading are attending lectures/participation (10%), three problem sets (45%) and a final project (45%).
Classes will be conducted in-person this year (Fall 2021), until MIT policy changes.
Grading policies, pset and project tentative dates: (slides)
Problem Sets
Problem Set 1, out: Tue. Sept. 21, due: Sun. Oct. 03
Problem Set 2, out: Tue. Oct. 12, due: Sun. Oct. 24
Problem Set 3, out: Tue. Nov. 02, due: Sun. Nov. 14
Submission instructions: Follow the instructions included with the problem set. Use the latex template for the report. Submit your report online through canvas by the due date/time.
Guidelines and key dates. Online form for project proposal (complete by Thu. Oct. 28).
Final project reports (5 pages for individuals, 8 pages for teams, NeurIPS style) are due on Thu. Dec. 9.
Projects archive
List of Wikipedia entries, created or edited as part of projects during previous course offerings.
Navigating Student Resources at MIT
This document has more information about navigating student resources at MIT | {"url":"https://poggio-lab.mit.edu/F21-9-520","timestamp":"2024-11-09T09:29:36Z","content_type":"text/html","content_length":"76541","record_id":"<urn:uuid:a6435b9c-12ce-4e76-9beb-c45416187d52>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00543.warc.gz"} |
Crypto and new computing strategies
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Crypto and new computing strategies
• Subject: Crypto and new computing strategies
• From: [email protected] (Eric Hughes)
• Date: Wed, 30 Mar 94 12:09:09 -0800
• In-Reply-To: Jim choate's message of Wed, 30 Mar 1994 11:54:07 -0600 (CST) <[email protected]>
• Sender: [email protected]
>I am not shure that it has been demonstrated that a QM mechanis is necessarily
>solely of a Turing architecture.
The Bekenstein Bound gives limits both on the expected maximum number
of quantum states encodable in a given volume of space and on the
expected maximum number os transitions between these states. If this
bound holds (and it certainly seems to hold for EM fields), then a
probabilistic Turing machine will be able to simulate it.
>Also there is the potential to use neural networks at these levels (which are
>not necessarily reducable to Turing models, the premise has never been proven)
If you have infinite precision, the statement is unproven. If you
have finite precision, you get a Turing machine. You never get
infinite precision in real life, even with quantum superposition.
Steve Smale did some work a few years ago where he made Turing-type
machines out of real numbers, i.e. infinite precision. P=NP for this
model, and the proof is fairly easy. From an information-theoretic
point of view, you can encode two real numbers inside of another one
and do computations in that encoded form, because a real number
encodes an infinite amount of information.
If it's finite, it's a Turing machine. If it's expected finite, it's
a probabilistic Turing machine. If it's infinite, it cannot be
implemented in hardware. | {"url":"https://cypherpunks.venona.com/date/1994/03/msg01170.html","timestamp":"2024-11-07T14:19:18Z","content_type":"text/html","content_length":"6123","record_id":"<urn:uuid:bd6163ac-be8d-4c42-bd8d-936115fc839c>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00553.warc.gz"} |
bnpsd 1.0.0.9000 (2017-09-11)
bnpsd 1.0.1 (2018-01-15)
• Minor non-code changes for first CRAN submission.
bnpsd 1.0.1.9000 (2018-02-01)
• README.md now contains instructions for installing from CRAN as well as from GitHub.
bnpsd 1.0.2.9000 (2019-02-06)
• Added option noFixed to function rbnpsd to redraw loci that were drawn fixed for a single allele. These loci are not polymorphic so they would normally not be considered in analyses.
• Added function fixed_loci to test for fixed loci within rbnpsd.
bnpsd 1.0.3.9000 (2019-02-07)
• Added function coanc_to_kinship to easily obtain kinship matrices from coancestry matrices.
bnpsd 1.0.4 (2019-02-11)
bnpsd 1.0.4.9000 (2019-02-13)
• Converted the vignette from PDF to HTML
bnpsd 1.0.5.9000 (2019-04-11)
• qis now returns a numeric admixture proportions matrix (used to be logical).
• q1d and q1dc now handle sigma = 0 special case.
• q1d and q1dc now provide more informative out-of-bounds messages when sigma is missing (and s is provided)
• sigma root finding in q1d and q1dc (when s is provided) is now more robust, explicitly tested at boundaries (min s > 0 achieved at sigma = 0 and max s = 1 achieved at sigma = Inf).
□ Removed arguments interval and tol from both q1d and q1dc (users would never need to set them now that procedure is more robust).
• Updated coding style, renamed some internal functions and variables.
bnpsd 1.1.0.9000 (2019-04-16)
• Renamed most functions for clarity:
□ coanc -> coanc_admix
□ q1d -> admix_prop_1d_linear
□ q1dc -> admix_prop_1d_circular
□ qis -> admix_prop_indep_subpops
□ rpanc -> draw_p_anc
□ rpint -> draw_p_subpops
□ rpiaf -> make_p_ind_admix
□ rgeno -> draw_genotypes_admix
□ rbnpsd -> draw_all_admix
□ fst -> fst_admix (no deprecated version available in this case, to eliminate conflict with popkin::fst)
□ Functions with old names remain for now as deprecated functions (to be removed in the future).
• Renamed several recurrent argument names for clarity:
□ Q -> admix_proportions
□ F -> coanc_subpops (if general matrix is accepted), inbr_subpops (vector or scalar versions required)
□ s -> bias_coeff
□ w -> weights
□ Theta -> coancestry
□ m -> m_loci
□ n -> n_ind
□ k -> k_subpops
□ pAnc -> p_anc
□ B -> p_subpops
□ P -> p_ind
□ Deprecated functions still accept old argument names.
• Fixed a sigma = 0 bug in admix_prop_1d_circular.
• Changed default values for draw_all_admix (compared to deprecated rbnpsd, which retains old defaults):
□ require_polymorphic_loci (old noFixed) is now TRUE by default.
□ want_p_ind and want_p_subpops (old wantP and wantB) are now FALSE by default.
□ Names (following above conventions) and order of items in return list changed.
• draw_p_subpops now admits scalar inputs p_anc and inbr_subpops, while number of loci and number of subpopulations can be provided as additional options.
• Added more input checks to functions, informative error messages.
• Updated documentation, particularly on whether intermediate subpopulation coancestries are accepted generally (coanc_subpops) or if the diagonal matrix case is required (specified as vector or
scalar inbr_subpops).
bnpsd 1.1.1 (2019-05-15)
• Third CRAN submission.
• Added ORCIDs to authors.
• Corrected doc typos.
• Adjusted layout of subpopulations and individuals (default limits) for circular 1D geography (admix_prop_1d_circular) to prevent overlapping individuals on the edges, and to better agree visually
with the linear version (admix_prop_1d_linear).
bnpsd 1.1.2 (2019-06-05)
• Non-code changes:
□ Edited .Rbuildignore to stop ignoring README; also removed non-existent files from list
□ Removed unused .travis.yml and bnpsd.Rproj files
bnpsd 1.1.2.9000 (2019-08-13)
• Improved memory efficiency of default draw_genotypes_admix
□ Old approach was by default very memory-hungry (created IAF matrix whole when admixture proportions were provided). The low_mem option could be set but filled slowly by locus only.
□ New approach is always low-memory (so the explicit option was removed). This was made faster by filling by individual when there are fewer individuals than loci, or filling by locus
otherwise, therefore always vectorizing as much as possible. Test showed this was always as fast as the original full memory approach, so the latter was removed as an option.
• draw_all_admix is also now automatically low-memory whenever want_p_ind = FALSE, and the explicit low_mem option has also been removed.
• Updated documentation to use RMarkdown
• Other code tidying
bnpsd 1.1.3.9000 (2019-09-06)
• Added option beta in function draw_p_anc to trigger a symmetric Beta distribution for the ancestral allele frequencies, with the desired shape parameter. The beta option can also be set on the
wrapper function draw_all_admix. This option allows simulation of a distribution heavier on rare variants (when beta is much smaller than 1), more similar to real human data.
bnpsd 1.2.0 (2019-12-17)
• Fourth CRAN submission.
• Removed deprecated function names: q1dc, q1d, qis, coanc, rbnpsd, rgeno, rpanc, rpint, rpiaf.
• Moved logo to man/figures/
• Minor Roxygen-related updates.
bnpsd 1.2.1 (2020-01-08)
• Fourth CRAN submission, second attempt.
• Fixed a rare bug in bias_coeff_admix_fit, which caused it to die if the desired bias coefficient was an extreme value (particularly 1). The error message was: f() values at end points not of
opposite sign. The actual bug was not observed in the regular R build, but rather in a limited precision setting where R was configured with --disable-long-double.
bnpsd 1.2.1.9000 (2020-01-08)
• Added option p_anc to function draw_all_admix, to specify desired ancestral allele frequencies instead of having the code generate it randomly (default).
• Added details for documentation of function draw_p_subpops.R, clarifying that input p_anc can be scalar.
bnpsd 1.2.2.9000 (2021-01-21)
• Function draw_all_admix: when option p_anc is provided as scalar and want_p_anc = TRUE, now the return value is always a vector (in this case the input scalar value repeated m_loci times). The
previous behavior was to return p_anc as scalar if that was the input, which could be problematic for downstream applications.
bnpsd 1.2.3 (2021-02-11)
• 5th CRAN submission
• Functions admix_prop_1d_linear and admix_prop_1d_circular had these changes:
□ The optional parameters bias_coeff, coanc_subpops and fst now have default values (of NA, NULL, and NA, respectively) instead of missing, and these “missing” values can be passed to get the
same behavior as if they hadn’t been passed at all.
□ Their documentation has been clarified.
□ Improved internal code to handle edge case bias_coeff = 1 (to fix an issue only observed on Apple M1).
• Function admix_prop_indep_subpops: default value for the optional parameter subpops is now made more clear in arguments definition.
• Simplified documentation (most functions) by clarifying language, using markdown roxygen, and replacing all LaTeX equations with simpler code equations.
• Updated paper citations in DESCRIPTION, README.md and the vignette, to point to the published method in PLoS Genetics.
bnpsd 1.2.3.9000 (2021-02-16)
• Documentation updates:
□ Fixed links to functions, in many cases these were broken because of incompatible mixed Rd and markdown syntax (now markdown is used more fully).
bnpsd 1.3.0.9000 (2021-03-24)
• Added support for intermediate subpopulations related by a tree
□ New function draw_p_subpops_tree is the tree version of draw_p_subpops.
□ New function coanc_tree calculates the true coancestry matrix corresponding to the subpopulations related by a tree.
□ Function draw_all_admix has new argument tree_subpops that can be used in place of inbr_subpops (to simulated subpopulation allele frequencies using draw_p_subpops_tree instead of
□ Note: These other functions work for trees (without change) because they accept arbitrary coancestry matrices (param coanc_subpops) as input, so they work if they are passed the matrix that
coanc_tree returns: coanc_admix, fst_admix, admix_prop_1d_linear, admix_prop_1d_circular.
• Functions admix_prop_1d_linear and admix_prop_1d_circular, when sigma is missing (and therefore fit to a desired coanc_subpops, fst, and bias_coeff), now additionally return multiplicative factor
used to rescale coanc_subpops.
bnpsd 1.3.1.9000 (2021-04-17)
It’s Fangorn Forest around here with all the tree updates!
• Added these functions:
□ fit_tree for fitting trees to coancestry matrices!
□ scale_tree to easily scale coancestry trees and check for out-of-bounds values.
□ tree_additive for calculating “additive” edges for probabilistic edge coancestry trees, and also the reverse function .
☆ This already existed as an internal, unexported function used mainly by coanc_tree, but now it’s renamed, exported, and well documented!
• Added support of $root.edge to tree phylo objects passed to these functions:
□ coanc_tree: edge is a shared covariance value affecting all subpopulations.
□ draw_all_admix and draw_p_subpops_tree: if root edge is present, functions warn that it will be ignored.
• Functions admix_prop_1d_linear and admix_prop_1d_circular: debugged an edge case where sigma is small but not zero and numerically-calculated densities all come out to zero in a given row of the
admix_proportions matrix (for admix_prop_1d_circular infinite values also arise), which used to lead to NAs upon row normalization; now for those rows, the closest ancestry (by coordinate
distance) gets assigned the full admixture fraction (just as for independent subpopulations/sigma = 0).
bnpsd 1.3.2.9000 (2021-04-22)
• Updated various functions to transfer names between inputs and outputs as it makes sense
□ Functions admix_prop_1d_linear, admix_prop_1d_circular now copy names from the input coanc_subpops (vector and matrix versions, only required when fitting bias_coeff) to the columns of the
output admix_proportions matrix.
□ Function draw_genotypes_admix now copies row and column names from input matrix p_ind (or rownames from p_ind and column names from the rownames of admix_proportions when the latter is
provided) to output genotype matrix
□ Function draw_p_subpops now copies names from p_anc to rows, names from inbr_subpops to columns, when present and of the right dimensions.
□ Function draw_p_subpops_tree now copies names from p_anc to rows. Names from tree_subpops were already copied to columns before.
□ All other functions already transferred names as desired/appropriate. Tests were added for these functions to ensure that this is so.
• Updated various functions to stop if there are paired names for two objects that are both non-NULL and disagree, as this suggests that the data is misaligned or incompatible.
□ Functions coanc_admix and fst_admix stop if the column names of admix_proportions and the names of coanc_subpops disagree.
□ Function draw_all_admix stops if the column names of admix_proportions and the names of either inbr_subpops or tree_subpops disagree.
□ Function draw_genotypes_admix, when admix_proportions is passed, stops if the column names of admix_proportions and p_ind disagree.
□ Function make_p_ind_admix stops if the column names of admix_proportions and p_subpops disagree.
• Function tree_additive now has option force, which when TRUE simply proceeds without stopping if additive edges were already present (in tree$edge.length.add, which is ignored and overwritten).
bnpsd 1.3.3.9000 (2021-04-29)
New functions and bug fixes dealing with reordering tree edges and tips.
• Added function tree_reindex_tips for ensuring that tip order agrees in both the internal labels vector and the edge matrix. Such lack of agreement is generally possible (technically the tree is
the same for arbitrary orders of edges in the edge matrix). However, such a disagreement causes visual disagreement in plots (for example, trees are plotted in the order of the edge matrix,
versus coancestry matrices are ordered as in the tip labels vector instead), which can now be fixed in general.
• Added function tree_reorder for reordering tree edges and tips to agree as much as possible with a desired tip order. The heuristic finds the exact solution if it exists, otherwise returns a
reasonable order close to the desired order. Tip order in labels and edge matrix agree (via tree_reindex_tips).
• Function fit_tree now outputs trees with tip order that better agrees with the input data, and tip order in labels vector and edge matrix now agree (via tree_reorder).
• Several functions now work with trees whose edges are arbitrarily ordered, particularly when they do not move out from the root (i.e. reverse postorder):
□ Function tree_additive. Before this bug fix, some trees could trigger the error message “Error: Node index 6 was not assigned coancestry from root! (unexpected)”, where “6” could be other
□ Function draw_p_subpops_tree. Before this bug fix, some trees could trigger the error message “Error: The root node index in tree_subpops$edge (9) does not match k_subpops + 1 (6) where
k_subpops is the number of tips! Is the tree_subpops object malformed?”, where “9” and “6” could be other numbers. Other possible error messages contain “Parent node index 6 has not yet been
processed …” or “Child” instead of “Parent”, where “6” could be other numbers.
□ Internal functions used by fit_tree had related fixes, but overall fit_tree appears to have had no bugs because users cannot provide trees, and the tree-building algorithm does not produce
scrambled edges that would have caused problems.
bnpsd 1.3.4.9000 (2021-05-12)
• Functions fixed_loci and draw_all_admix have a new parameter maf_min that, when greater than zero, allows for treating rare variants as fixed. In draw_all_admix, this now allows for simulating
loci with frequency-based ascertainment bias.
bnpsd 1.3.5.9000 (2021-05-14)
• Fixed a rare bug in draw_all_admix that could cause a “stack overflow” error. The function used to call itself recursively if require_polymorphic_loci = TRUE, and in cases where there are very
rare allele frequencies or high maf_min the number of recursions could be so large that it triggered this error. Now the function has a while loop, and does not recurse more than one level at the
time; there is no limit to the number of iterations and no errors occur inherently due to large numbers of iterations.
bnpsd 1.3.6.9000 (2021-06-02)
• Function fit_tree internally simplified to use stats::hclust, which also results in a small runtime gain. The new code (when method = "mcquitty", which is default) gives the same answers as
before (in other words, the original algorithm was a special case of hierarchical clustering).
□ New option method is passed to hclust. Although all hclust methods are allowed, for this application the only ones that make sense are “mcquitty” (WPGMA) and “average” (UPGMA). In internal
evaluations, both algorithms had similar accuracy and runtime, but only “mcquitty” exactly recapitulates the original algorithm.
bnpsd 1.3.7.9000 (2021-06-04)
• Updated citations in inst/CITATION (missed last time I updated them in other locations).
bnpsd 1.3.8.9000 (2021-06-21)
• Added function undiff_af for creating “undifferentiated” allele frequency distributions based on real data but with a lower variance (more concentrated around 0.5) according to a given FST,
useful for simulating data trying to match real data.
• Added LICENSE.md.
• Reformatted this NEWS.md slightly to improve its automatic parsing.
bnpsd 1.3.9.9000 (2021-06-22)
• Function undiff_af:
□ Added several useful informative statistics to return list: F_max, V_in, V_out, V_mix, and alpha.
□ Debugged distr = "auto" cases where mixing variance ended up being smaller than required due to roundoff errors (alpha is now larger than given in direct formula by eps = 10 *
.Machine$double.eps, which is also a new option.
bnpsd 1.3.10.9000 (2021-06-22)
• Function draw_all_admix added option p_anc_distr for passing custom ancestral allele frequency distributions (as vector or function). This differs from the similar preexisting option p_anc, which
fixed ancestral allele frequencies per locus to those values. These two options behave differently when loci have to be re-drawn due to being fixed or having too-low MAFs: passing p_anc never
changes those values, whereas passing p_anc_distr results in drawing new values as necessary. The new option is more natural biologically and results in re-drawing fixed loci less often.
bnpsd 1.3.11.9000 (2021-07-01)
• Function undiff_af renamed parameter F to kinship_mean, and updated all documentation to reflect the correction that this parameter is the mean kinship and not FST (the complete derivation will
appear in a manuscript).
□ One element in the return list previously called F_max is similarly now kinship_mean_max.
bnpsd 1.3.12 (2021-08-02)
• 6th CRAN submission.
• Removed “LazyData: true” from DESCRIPTION (to avoid a new “NOTE” on CRAN).
• Fixed spelling in documentation.
bnpsd 1.3.13 (2021-08-09)
• 6th CRAN submission, second attempt.
• Debugged internal code (bias_coeff_admix_fit) shared by admix_prop_1d_linear and admix_prop_1d_circular for edge cases.
□ Error was only observed on M1mac architecture (previous code worked on all other systems!).
□ If a bias coefficient of 1 was desired, expected sigma to be Inf, but instead an error was encountered.
□ Previous error message: “f() values at end points not of opposite sign” (in stats::uniroot) | {"url":"https://cran.stat.sfu.ca/web/packages/bnpsd/news/news.html","timestamp":"2024-11-15T03:16:23Z","content_type":"application/xhtml+xml","content_length":"24936","record_id":"<urn:uuid:c1ed531b-84bc-4f74-b108-73d01859b769>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00474.warc.gz"} |
Volume Calculator - Free Online Calculators
Volume Calculator
The Volume Calculator enables users to find the volume of various geometric shapes, such as cubes, spheres, and cylinders. It provides easy input fields for dimensions, allowing for quick and
accurate calculations. This tool is perfect for students, engineers, and DIY enthusiasts needing precise volume measurements. With a straightforward interface, users can seamlessly navigate and
obtain results. Whether for academic purposes or practical applications, the Volume Calculator is an essential tool for accurate measurements!
Volume: N/A cubic units
A Volume Calculator is a valuable tool for quickly and accurately determining the volume of various three-dimensional shapes and objects. This calculator is particularly useful in a wide range of
fields, including mathematics, engineering, architecture, and everyday applications. Here’s how a Volume Calculator typically works and some of the common shapes it can calculate volumes for:
How a Volume Calculator Works:
1. Select Shape: Users begin by selecting the shape or object for which they want to calculate the volume. Volume Calculators are versatile and can handle a variety of shapes.
2. Provide Dimensions: Depending on the chosen shape, users are prompted to input specific dimensions. For example, if calculating the volume of a cube, users enter the length of one side. For a
cylinder, they input the radius and height. The required dimensions vary based on the selected shape.
3. Calculation: Once the necessary dimensions are provided, the Volume Calculator performs the appropriate mathematical calculation to determine the volume. The result is typically displayed in
cubic units, such as cubic meters, cubic centimeters, or cubic feet.
Common Shapes Calculated Using a Volume Calculator:
1. Cube: A cube has six equal square faces. To find its volume, you only need to know the length of one side. The formula is V = s^3, where “V” is the volume, and “s” is the length of a side.
2. Rectangular Prism: This shape has six rectangular faces. To calculate its volume, you need the length, width, and height. The formula is V = lwh, where “V” is the volume, “l” is the length, “w”
is the width, and “h” is the height.
3. Cylinder: A cylinder consists of two circular bases and a curved surface. To find its volume, you’ll need the radius (or diameter) of the base and the height. The formula is V = πr^2h, where “V”
is the volume, “π” is Pi (approximately 3.14159), “r” is the radius, and “h” is the height.
4. Sphere: A sphere is a perfectly round shape. To calculate its volume, you only need the radius. The formula is V = (4/3)πr^3, where “V” is the volume, “π” is Pi, and “r” is the radius.
5. Cone: A cone has a circular base and a curved surface that tapers to a point. To determine its volume, you’ll need the radius of the base and the height. The formula is V = (1/3)πr^2h, where “V”
is the volume, “π” is Pi, “r” is the radius, and “h” is the height.
6. Pyramid: A pyramid has a polygonal base and triangular faces that meet at a single point (apex). Calculating its volume depends on the base shape and height. Different formulas apply to different
types of pyramids.
7. Ellipsoid: An ellipsoid is an ellipsoidal (oval) shape. Its volume formula is more complex and involves the lengths of its three semi-axes.
8. Irregular Shapes: Some advanced Volume Calculators may allow users to calculate the volume of irregular shapes by dividing them into smaller, manageable components or using numerical integration
A Volume Calculator simplifies complex volume calculations, making it a valuable tool for professionals, students, and anyone needing to determine the volume of objects for various purposes, from
construction projects to scientific research. | {"url":"https://nowcalculator.com/volume-calculator-2/","timestamp":"2024-11-11T00:28:29Z","content_type":"text/html","content_length":"268467","record_id":"<urn:uuid:1b040430-8669-4ad6-bc5d-7c30e46831b4>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00477.warc.gz"} |
ThmDex – An index of mathematical definitions, results, and conjectures.
Let $G$ be a
D22: Group
such that
(i) $p$ is a D571: Prime integer
(ii) \begin{equation} |G| = p \end{equation}
Let $g \in G$ be a non-identity element in $G$. Such an element exists because $|G| = p \geq 2$. Let $\langle g \rangle$ denote the
D1301: Generated subgroup
of $g$ within $G$. Now
R468: Lagrange's theorem
states \begin{equation} |\langle g \rangle| [G : \langle g \rangle] = |G| = p \end{equation} That is, $|\langle g \rangle|$ divides $p$. Since $p$ is prime, then $|\langle g \rangle| \in \{ 1, p \}$.
Since $g$ is a non-identity element, the generated subgroup $\langle g \rangle$ must contain at least two elements. This is because $g$ is included by default and the identity element needs to be
included in order for it to qualify as a group, and the two elements are distinct. Thus $|\langle g \rangle| = p$. Since $|\langle g \rangle| = G$ and since $\langle g \rangle \subseteq G$,
R2745: The only subset of finite set with same cardinality is the ambient set itself
implies that $\langle g \rangle = G$. $\square$ | {"url":"https://thmdex.com/r/822","timestamp":"2024-11-06T11:08:20Z","content_type":"text/html","content_length":"6787","record_id":"<urn:uuid:543bad15-3ada-4c40-b5a4-f6fac8665571>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00021.warc.gz"} |
Using Dienes to multiply a 2 digit number by a 1 digit number | Oak National Academy
Lesson outcome
In this lesson, we will learn to multiply 2 digit numbers by 1 digit numbers using Dienes blocks. We will use visual representations to make multiplication processes clearer, and we will investigate
when to regroup quantities when using Dienes blocks. | {"url":"https://www.thenational.academy/pupils/lessons/using-dienes-to-multiply-a-2-digit-number-by-a-1-digit-number-c5hk6c/overview?step=4&activity=exit_quiz","timestamp":"2024-11-09T12:20:04Z","content_type":"text/html","content_length":"116854","record_id":"<urn:uuid:e5d6c6e5-2694-4dea-93de-95b35dac5557>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00009.warc.gz"} |
Online Test On Algorithms And Data Structures - ICTTRENDS-DSA-01
Questions and Answers
FM: 50 PM: 30 Time: 1 hr
Attempt All of the following questions. Each question carry equal mark
Attempt online MCQ test on Data Structures and Algorithms presented by ICT TRENDS (www. Icttrends. Com).
Questions presented here suits the level for Read moreHigher Secondary or Bachelors in Computer Science or BCA, BIT, BSc IT and competitive exams like Computer Officer, IT Officer.
Confused on any question? Check https://icttrends. Com/dsa-mcq-set-1/ for explanations on these questions.
Visit Data Structures And Algorithms on PS Exam
• 1.
In linked lists there are no NULL links in
□ A.
□ B.
Linear doubly linked list
□ C.
□ D.
Correct Answer
C. Circular linked list
Circular linked lists do not have NULL links because the last node of the list points back to the first node, creating a circular structure. This allows for easy traversal of the list and
eliminates the need for NULL pointers to indicate the end of the list.
• 2.
In a stack the command to access nth element from the top of the stack S will be
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. S [Top-n]
To access the nth element from the top of the stack S, the command would be S [Top-n]. This is because the top of the stack is considered as the 1st element, and to access the nth element from
the top, we need to subtract n from the top index. Therefore, the correct command would be S [Top-n].
• 3.
If yyy, xxx and zzz are the elements of a lexically ordered binary tree, then in preorder traversal which node will be traverse first
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Xxx
In a preorder traversal of a binary tree, the root node is always traversed first, followed by the left subtree and then the right subtree. Therefore, in this case, the node "xxx" will be
traversed first in the preorder traversal.
• 4.
In a balance binary tree the height of two sub trees of every node can not differ by more than
Correct Answer
B. 1
In a balanced binary tree, the height of the two sub trees of every node cannot differ by more than 1. This means that the heights of the left and right sub trees should either be equal or differ
by at most 1. This balance ensures efficient search and insertion operations in the tree. If the height difference is more than 1, the tree becomes unbalanced, leading to slower operations.
Therefore, the correct answer is 1.
• 5.
The result of evaluating prefix expression */b+0dacd, where a=3, b=6, c=1, d=5 is
Correct Answer
C. 10
The given prefix expression is */b+0dacd. Plugging in the values of a=3, b=6, c=1, d=5, we get */6+0*3*1+5. Following the order of operations, we first evaluate the multiplication and division
from left to right. So, we have 6/6+0*3*1+5. Next, we evaluate the addition and subtraction from left to right. So, we have 1+0*3*1+5. Finally, we evaluate the remaining multiplication. So, we
have 1+0+5. Simplifying further, we get 6 as the final result. Therefore, the correct answer is 10.
• 6.
In array representation of binary tree teh right child of root will be at location of
Correct Answer
C. 3
In the array representation of a binary tree, the right child of the root will be at the location of 3. This is because in a binary tree, the left child is typically stored at index 2*i+1 and the
right child is stored at index 2*i+2, where i represents the index of the parent node. In this case, the root is at index 0, so the right child would be at index 2*0+2=2. Therefore, the correct
answer is 3.
• 7.
The total number of comparisons in a bubble sort is
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. O(n logn)
The correct answer is O(n^2). In bubble sort, each element is compared with its adjacent element and swapped if necessary. This process is repeated for each element in the list, resulting in n
comparisons for each pass. Since there are n passes in total, the total number of comparisons is n * n = n^2.
• 8.
IThe dummy header in linked list contain
□ A.
First record of the actual data
□ B.
Last record of the actual data
□ C.
Pointer to the last record of the actual data
□ D.
Correct Answer
A. First record of the actual data
The dummy header in a linked list is a special node that is used to simplify certain operations on the list. It does not contain any actual data, but it serves as a placeholder or starting point
for the list. The first record of the actual data is stored after the dummy header, so it can be accessed easily. Therefore, the correct answer is "First record of the actual data."
• 9.
Write the output of following program:int a[ ] = {1,2,3,} *p;
□ A.
□ B.
□ C.
□ D.
Address of the third element
Correct Answer
B. Junk value
The program is trying to assign the value of the third element of the array 'a' to a pointer variable 'p'. However, there is no reference to the address of the array 'a' in the code. Therefore,
the pointer 'p' does not point to any valid memory location and will contain a junk value.
• 10.
If the out degree of every node is exactly equal to M or 0 and the num ber of nodes at level K is Mk-1 [con sider root at level 1], then tree is called as (i) Full m-ary try(ii) Com plete m-ary
tree(iii)Positional m-ary tree
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Both (i) and (ii)
A tree where the out degree of every node is exactly equal to M or 0 is called a full m-ary tree. This means that each node in the tree has either M children or no children at all. On the other
hand, a tree where the number of nodes at level K is Mk-1 is called a complete m-ary tree. This means that each level of the tree is completely filled, except possibly for the last level which
may be partially filled. Therefore, the given tree satisfies the conditions for both a full m-ary tree and a complete m-ary tree, making the answer "Both (i) and (ii)" correct.
• 11.
In the following tree:If post order traversal generates sequence xy-zw*+, then label of nodes 1,2,3,4,5,6,7 will be
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. +, -, *, x, y, z, w
The given post-order traversal sequence "xy-zw*+" suggests that the label of the nodes in the tree is +, -, *, x, y, z, w. This means that the node labeled 1 is "+", node 2 is "-", node 3 is "*",
node 4 is "x", node 5 is "y", node 6 is "z", and node 7 is "w".
• 12.
If the following tree is used for sorting, then a new number 10 should be placed at
□ A.
Right child of node labeled 7
□ B.
Left child of node labeled 7
□ C.
Left child of node labeled 14
□ D.
Right child of node labeled 8
Correct Answer
C. Left child of node labeled 14
The given answer states that the new number 10 should be placed as the left child of the node labeled 14. This means that the number 10 should be inserted as a lower value than 14 in the tree
hierarchy. Placing it as the left child ensures that it is on the left side of the node labeled 14 and has a lower value.
• 13.
A queue has configuration a,b,c,d. To get configuration d,c,b,a. One needs a minimum of
□ A.
2 deletion and 3 additions
□ B.
3 deletions and 2 additions
□ C.
3 deletions and 3 additions
□ D.
3 deletions and 4 additions
Correct Answer
C. 3 deletions and 3 additions
To get from configuration a,b,c,d to configuration d,c,b,a, we need to perform the following operations:
1. Delete a to get b,c,d.
2. Delete b to get c,d.
3. Delete c to get d.
4. Add a to get d,a.
5. Add b to get d,a,b.
6. Add c to get d,a,b,c.
Therefore, we need 3 deletions and 3 additions to achieve the desired configuration.
• 14.
Number of possible ordered trees with 3 nodes A,B,C is
Correct Answer
C. 6
The number of possible ordered trees with 3 nodes A, B, C can be determined by considering the different ways these nodes can be arranged. In this case, there are 3 nodes, so there are 3 choices
for the root node. After selecting the root node, there are 2 remaining nodes that can be arranged in 2! = 2 ways. Therefore, the total number of possible ordered trees is 3 * 2 = 6.
• 15.
Number of swapping, operations need to sort numbers 8, 22, 7, 9, 31, 19, 5, 13 in ascending order using bubble sort
Correct Answer
C. 13
Bubble sort is a simple sorting algorithm that repeatedly steps through the list, compares adjacent elements and swaps them if they are in the wrong order. In this case, the given numbers are 8,
22, 7, 9, 31, 19, 5, and 13. To sort them in ascending order using bubble sort, the algorithm will make a total of 13 swaps.
• 16.
Consider two sorted lists of size L1, L2. Number of comparisions needed in worst case my merge sort algorithm will be
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. L1+L2-1
The merge sort algorithm works by dividing the input into smaller sublists, sorting them, and then merging them back together. In the worst case scenario, the algorithm will need to compare every
element in both lists before merging them. Since there are L1 elements in the first list and L2 elements in the second list, the total number of comparisons needed will be L1 + L2. However, when
merging the lists, the last element of the first list will already be in its correct position, so it does not need to be compared again. Therefore, the final answer is L1 + L2 - 1.
• 17.
A hash tabale with 10 buckets with one slot per bucket is depicted in following diagram. Symbols S1 to S7 are initially entered using a hashing function with linear probing. Maximum number of
comparisions needed in searching an item that is not present is
Correct Answer
B. 5
In a hash table with linear probing, when searching for an item that is not present, we need to check each slot in the hash table until we either find the item or reach an empty slot. In this
case, since there are 10 buckets with one slot per bucket, the maximum number of comparisons needed to search for an item that is not present would be equal to the number of slots in the hash
table, which is 10. However, since the answer given is 5, there might be some additional information or context missing from the question that would explain why the correct answer is 5.
• 18.
Following sequence of operation is performed on a stack. Push(1), Push(2), Pop, Push(1), Push(2), Pop, Pop, Pop, Push(2), Pop. The sequences of popped out values are
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 2,1,2,2,2
The given sequence of operations on the stack can be broken down as follows:
- Push(1) - The value 1 is pushed onto the stack.
- Push(2) - The value 2 is pushed onto the stack.
- Pop - The topmost value on the stack (2) is popped out.
- Push(1) - The value 1 is pushed onto the stack.
- Push(2) - The value 2 is pushed onto the stack.
- Pop - The topmost value on the stack (2) is popped out.
- Pop - The topmost value on the stack (1) is popped out.
- Pop - The topmost value on the stack (1) is popped out.
- Push(2) - The value 2 is pushed onto the stack.
- Pop - The topmost value on the stack (2) is popped out.
Therefore, the sequence of popped out values is 2, 1, 2, 2, 2.
• 19.
A binary tree in which every non-leaf node has non-empty left and right sub trees is called a strictly binary tree. Such a tree with 10 leaves
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Has 19 nodes
A strictly binary tree is a binary tree where every non-leaf node has both a non-empty left and right subtree. In this case, the tree has 10 leaves, which means there are 10 nodes that do not
have any children. Since every non-leaf node must have both a left and right subtree, there must be 10 non-leaf nodes in addition to the 10 leaves. Therefore, the total number of nodes in the
tree is 10 leaves + 10 non-leaf nodes = 20 nodes. However, the question states that the tree has 19 nodes, which is incorrect.
• 20.
Depth of a binary tree with n node is
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Log (n +1) - 1
The correct answer is log (n +1) - 1. This is because the depth of a binary tree can be calculated using the formula log (n +1) - 1, where n represents the number of nodes in the tree. This
formula is derived from the fact that the maximum number of nodes at any level in a binary tree is 2^k, where k is the level number. By solving for k in the equation 2^k = n, we can find the
depth of the tree.
• 21.
To arrange a binary tree in ascending order we need
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Inorder traversal
Inorder traversal is used to arrange a binary tree in ascending order because it visits the left subtree first, then the root, and finally the right subtree. This traversal ensures that the nodes
are visited in ascending order, making it suitable for arranging the binary tree in ascending order.
• 22.
Average successful search time taken by binary search on sorted array of 10 items is
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 2.9
The average successful search time taken by binary search on a sorted array of 10 items is 2.9. This means that, on average, it takes 2.9 units of time to find a desired element in the array
using binary search. Binary search is an efficient search algorithm that works by repeatedly dividing the search space in half, making it very fast for sorted arrays. In this case, the average
time is 2.9, indicating that it is slightly slower than the other options provided.
• 23.
Hash function f is defined as f(key) = key mod 7. If linear probing is used to insert the key 37, 38, 72, 48, 98, 11, 56 into a table indexed from 0 to 6, 11 will be stored at the location
Correct Answer
C. 5
• 24.
Average successful search time for sequential search on 'n' item is
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. (n+1)/2
The average successful search time for sequential search on 'n' items is (n+1)/2. This is because in a sequential search, each item in the list is checked one by one until the desired item is
found. On average, the desired item is expected to be found halfway through the list, so the average search time is (n+1)/2.
• 25.
Running time of an algorithm T(n), where n is input size is given by T(n) = 8 T(n/2) + qn, if n>1 T(n) = p, if, n=1 where p and q are constants. The order of algorithm is
Correct Answer
C. N^3
The given algorithm has a recursive formula that indicates the running time of the algorithm. The formula T(n) = 8 T(n/2) + qn suggests that each recursive call has a running time of qn, and the
algorithm makes 8 recursive calls with input size n/2. This indicates that the algorithm has a divide and conquer approach, where it divides the problem into smaller subproblems and solves them
By analyzing the formula, we can see that the algorithm has a time complexity of O(n^3). This is because each recursive call contributes a factor of qn, and there are a total of log(n) recursive
calls (since n is halved in each call). Therefore, the total running time can be expressed as qn * log(n), which simplifies to O(n^3) when we drop the constant factors.
• 26.
6 files X1, X2, X3, X4, X5, X6 have 150, 250, 55, 85, 125, 175 number of records respectively. The order of storage ot optimize access time is
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. X1, X2, X3, X4, X5, X6
The order of storage to optimize access time is X1, X2, X3, X4, X5, X6. This order is based on the number of records in each file, with the file with the highest number of records (X2) being
stored first and the file with the lowest number of records (X3) being stored last. By storing the files in this order, the system can prioritize accessing the files with more records first,
which can potentially reduce access time and improve efficiency.
• 27.
In above question average access time will be
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 140
The average access time will be 140. This is because the average is calculated by adding up all the access times and then dividing by the number of access times. In this case, there are three
access times given: 150, 140, and 55. Adding these together gives a total of 345. Dividing this by 3 (the number of access times) gives an average of 115. Therefore, the correct answer is 140.
• 28.
An algorithm consists of two modules X1, X2. Their order is f(n) and g(n), respectively. The order of algorithm is
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Max(f(n),g(n))
The order of an algorithm is determined by the module that takes the longest time to execute. In this case, the algorithm consists of two modules, X1 and X2, with orders f(n) and g(n)
respectively. The order of the algorithm will be the maximum of f(n) and g(n) because the algorithm's overall performance will be limited by whichever module takes longer to execute. Therefore,
the correct answer is Max(f(n),g(n)).
• 29.
Bib O notation w.r.t algorithm signifies
□ A.
It decides the best algorithm to solve a problem
□ B.
It determines maximum size of a problem, that can be solved in given system in given time
□ C.
It is the lower bound of growth rate of algorithm
□ D.
Correct Answer
D. None of the above
• 30.
Running time T(n), where 'n' is input size of recursive algorithm is given as follows:T(n) = c + T(n-1), if n>1,T(n) = d if n<1. The order of algorithm is
Correct Answer
B. N
The given recursive algorithm has a running time of T(n) = c + T(n-1), where c is a constant. This means that for each input size n, the algorithm performs a constant amount of work (c) plus the
running time for the input size n-1. Since the algorithm only performs a constant amount of work for each input size, the order of the algorithm is linear, or O(n).
• 31.
Four altorithm A1, A2, A3, A4 solves a problem with order log(n), log log(n), nlog(n), n. Which is best algorithm
Correct Answer
A. A1
Algorithm A1 is the best algorithm because it has the lowest time complexity of O(log(n)). This means that as the input size increases, the time taken by A1 to solve the problem will increase at
a slower rate compared to the other algorithms. A2 has a time complexity of O(log log(n)), A3 has a time complexity of O(n log(n)), and A4 has a time complexity of O(n). Therefore, A1 is the most
efficient algorithm for solving the given problem.
• 32.
Time complexity of an algorithm T(n), where n is the input size is given by T(n) = T (n-1) + 1/n, if n>1 otherwise T(n) = 1. The order of algorithm is
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. N^2
The given algorithm has a recursive formula where the time complexity of T(n) is equal to T(n-1) plus 1/n. This means that for every input size n, the algorithm will make a recursive call on an
input size of n-1, and also perform a constant amount of work (1/n). As the input size decreases by 1 with each recursive call, the algorithm will have a total of n recursive calls. Therefore,
the time complexity can be calculated as follows: T(n) = T(n-1) + 1/n = T(n-2) + 1/(n-1) + 1/n = ... = T(1) + 1/2 + 1/3 + ... + 1/n. This is known as the harmonic series, and it grows as O(log
n). However, since the algorithm also performs a constant amount of work (1) in each recursive call, the overall time complexity becomes O(n + log n) which simplifies to O(n). Therefore, the
order of the algorithm is n, not n^2 or any other option.
• 33.
Which of the following is correct
□ A.
Internal sorting is used, if number of items to be sorted is very large
□ B.
External sorting is used, if number of items to be sorted is very large
□ C.
External sorting needs auxiliary storage
□ D.
Internal sorting needs auxuliary storage
Correct Answer
B. External sorting is used, if number of items to be sorted is very large
External sorting is used when the number of items to be sorted is very large because it is not feasible to fit all the items in the main memory. External sorting involves dividing the data into
smaller chunks that can fit in the memory, sorting each chunk individually, and then merging the sorted chunks together. This process requires the use of auxiliary storage, such as disk space, to
store the intermediate sorted chunks. In contrast, internal sorting is used when the number of items can fit in the main memory, eliminating the need for auxiliary storage.
• 34.
A sorting technique which guarrantees that records with same primary key occurs in the sorted list as in the original unsorted list is said to be
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Stable
A sorting technique that is considered stable ensures that records with the same primary key will appear in the same order in the sorted list as they did in the original unsorted list. This means
that if two records have the same primary key value, the one that appeared first in the unsorted list will also appear first in the sorted list. This property is important in certain applications
where the order of equal elements needs to be preserved.
• 35.
A text is made up of five characters, T1, T2, T3, T4, T5. The probability of occurance of each character is .12, .4, .15, .88 and .25, respectively. The optimal coding technique will have average
length of
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 2.15
The optimal coding technique will have an average length of 2.15. This can be calculated by multiplying the probability of each character by the length of its corresponding code and summing up
these values. For character T1, the length is 2, so the contribution to the average length is 0.12 * 2 = 0.24. Similarly, for T2, the length is 3, so the contribution is 0.4 * 3 = 1.2. For T3,
the length is 2, so the contribution is 0.15 * 2 = 0.3. For T4, the length is 1, so the contribution is 0.88 * 1 = 0.88. And for T5, the length is 3, so the contribution is 0.25 * 3 = 0.75.
Adding up all these contributions gives 0.24 + 1.2 + 0.3 + 0.88 + 0.75 = 3.37. Dividing this sum by the total probability (0.12 + 0.4 + 0.15 + 0.88 + 0.25 = 1.8) gives the average length of 2.15.
• 36.
If running time of an algorithm is given by T(n) = T(n-1) + T(n-2) + T(n-3), if n>3 otherwise T(n)=n, what should be relation between T(1), T(2), T(3) where algorithm order become constant
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. T(1) - T(3) = T(2)
The given relation T(n) = T(n-1) + T(n-2) + T(n-3) suggests that the running time of the algorithm is dependent on the sum of the running times of the previous three iterations. Therefore, in
order for the algorithm order to become constant, the running times of T(1), T(2), and T(3) should be related in such a way that the difference between T(1) and T(3) is equal to T(2). This can be
represented as T(1) - T(3) = T(2).
• 37.
Arranging a pack of cards by picking one by one is an example of
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Insertion sort
Arranging a pack of cards by picking one by one is an example of insertion sort because in insertion sort, each element is compared to the elements before it and inserted into its correct
position in the sorted sequence. Similarly, when arranging a pack of cards, we compare each card to the cards before it and insert it into the correct position in the sorted sequence.
• 38.
In evaluating arithmatic expression 2*3-(4+5) using postfix stack form. Which of the following stack configuration is not possible
□ A.
------------ | 4 | 6 | ------------
□ B.
------------ | 5 | 4 | 6 | ------------
□ C.
------------ | 9 | 6 | ------------
□ D.
------------ | 9 | 3 | 2 | ------------
Correct Answer
D. ------------ | 9 | 3 | 2 | ------------
• 39.
The order of an algorithm that finds whether a given boolean function of 'n' variable produces a '1' is
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. Exponential
The order of an algorithm that finds whether a given boolean function of 'n' variables produces a '1' is exponential. This means that the time complexity of the algorithm increases exponentially
with the size of the input. In other words, as the number of variables increases, the algorithm's running time grows at an exponential rate. This suggests that the algorithm may need to evaluate
all possible combinations of inputs in order to determine the output, resulting in an exponential time complexity.
• 40.
The average number of comparisions performed by merge sort alrotithm in merging two sorted list of length 2 is
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 8/3
Merge sort algorithm divides the list into two halves and recursively sorts them. In the merging step, it compares the elements from both halves and merges them in sorted order. When merging two
sorted lists of length 2, there are a total of 4 elements to compare. The first element from one list is compared with the first element from the other list, and similarly for the second element.
Therefore, a total of 4 comparisons are performed. Since the question asks for the average number of comparisons, we divide 4 by the number of lists being merged, which is 2. Hence, the average
number of comparisons performed is 4/2 = 8/3.
• 41.
To arrange the books of library the best method is
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. Heap sort
Heap sort is the best method to arrange the books in a library. Heap sort is a comparison-based sorting algorithm that uses a binary heap data structure. It has a time complexity of O(n log n),
which means it is efficient for large datasets. Additionally, heap sort is an in-place sorting algorithm, which means it does not require any additional space for sorting. Therefore, heap sort is
the most suitable method for arranging books in a library efficiently.
• 42.
Which of the following algorithm has n log (n) time complexity
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Insertion sort
Insertion sort has a time complexity of O(n log n). This is because in the worst case scenario, each element needs to be compared with all the previous elements in the sorted portion of the
array, resulting in a nested loop. The outer loop iterates n times, and the inner loop iterates i times, where i is the current position of the element being inserted. As a result, the time
complexity of insertion sort is n * (n/2) = O(n^2). However, in the best case scenario where the array is already sorted, the time complexity reduces to O(n), making it more efficient.
• 43.
Which algorithm solves all pair shortest path problem
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Dijkastra's algorithm
Dijkstra's algorithm is used to find the shortest path between a single source node and all other nodes in a graph. It does not solve the all pair shortest path problem. On the other hand,
Floyd's algorithm is specifically designed to solve the all pair shortest path problem. It finds the shortest path between all pairs of nodes in a weighted graph, making it the correct answer to
the question. Prim's algorithm is used for finding the minimum spanning tree in a graph, and Worshall's algorithm is used to find the transitive closure of a directed graph.
• 44.
The centricity of node labeled 5 is
Correct Answer
C. 2
The centricity of a node refers to the minimum eccentricity among all nodes in the graph. The eccentricity of a node is the maximum shortest path length from that node to any other node in the
graph. In this case, the node labeled 5 has a shortest path length of 2 to itself, which is the minimum among all nodes in the graph. Therefore, the centricity of node labeled 5 is 2.
• 45.
The information about an array used in a program will be stored in
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Dope vector
A dope vector is a data structure used to store information about an array in a program. It contains details such as the size of the array, the starting address of the array, and other relevant
information. This information is used by the program to perform operations on the array efficiently. The symbol table, register vector, and activation table are not specifically designed to store
array information, making them incorrect choices.
• 46.
In which of the following cases linked list implementaion of sparse matrices consumes same memory as a normal array
□ A.
5 x 6 matrix with 9 non-zero entries
□ B.
5 x 6 matrix with 8 non-zero entries
□ C.
6 x 5 matrix with 8 non-zero entries
□ D.
6 x 5 matrix with 9 non-zero entries
Correct Answer
C. 6 x 5 matrix with 8 non-zero entries
In a linked list implementation of a sparse matrix, memory consumption is reduced because only the non-zero elements are stored. The number of non-zero entries determines the memory consumption.
In this case, the 6 x 5 matrix with 8 non-zero entries would consume the same memory as a normal array because both would need to allocate space for 8 elements. The other options have either more
or fewer non-zero entries, resulting in different memory consumption compared to a normal array.
• 47.
The advantage of sparse matrix linked list representationn over dope vector method is, that the former is
□ A.
□ B.
□ C.
Efficient in accessing an entry
□ D.
Efficient if the sparse matr4ix is a band matrix
Correct Answer
B. Completely dynamic
The advantage of sparse matrix linked list representation over the dope vector method is that the former is completely dynamic. This means that the linked list representation can handle matrices
of any size, with varying numbers of non-zero entries, without requiring any additional memory allocation. In contrast, the dope vector method requires pre-allocation of memory for the entire
matrix, which can be inefficient if the matrix is large or if the number of non-zero entries is small. Therefore, the completely dynamic nature of the sparse matrix linked list representation
makes it a more flexible and efficient option.
• 48.
If the address of (I,J)th entry, in dope vector representation, where it stores the position of first and last non-zero entries of each row is given by C, assume l(n) and f(n) represent the last
and first non-zero entries in row xi. ii. iii. iv.
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Correct answer is (iii)
• 49.
On which principle does stack work?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. FILO
The stack works on the principle of FILO, which stands for "First In, Last Out". This means that the first item inserted into the stack will be the last one to be removed. It follows a Last In,
First Out (LIFO) order, where the most recently added item is the first one to be removed. Therefore, the correct answer is FILO.
• 50.
The order of binary search algorithm is
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. N log n
The order of the binary search algorithm is n log n. This means that the time complexity of the algorithm is proportional to the product of the number of elements in the input (n) and the
logarithm of that number (log n). As the input size increases, the time taken by the algorithm increases at a slower rate compared to n2 or n. This is because binary search divides the input in
half at each step, resulting in a logarithmic time complexity. Therefore, the correct answer is n log n. | {"url":"https://www.proprofs.com/quiz-school/story.php?title=data-structures-algorithms","timestamp":"2024-11-09T12:08:34Z","content_type":"text/html","content_length":"627938","record_id":"<urn:uuid:853616b8-3e5a-4052-a172-57326abc9635>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00570.warc.gz"} |
LM 36.1 Classifying states Collection
36.1 Classifying states by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
36.1 Classifying states
We'll focus our attention first on the simplest atom, hydrogen, with one proton and one electron. We know in advance a little of what we should expect for the structure of this atom. Since the
electron is bound to the proton by electrical forces, it should display a set of discrete energy states, each corresponding to a certain standing wave pattern. We need to understand what states there
are and what their properties are.
What properties should we use to classify the states? The most sensible approach is to used conserved quantities. Energy is one conserved quantity, and we already know to expect each state to have a
specific energy. It turns out, however, that energy alone is not sufficient. Different standing wave patterns of the atom can have the same energy.
Momentum is also a conserved quantity, but it is not particularly appropriate for classifying the states of the electron in a hydrogen atom. The reason is that the force between the electron and the
proton results in the continual exchange of momentum between them. (Why wasn't this a problem for energy as well? Kinetic energy and momentum are related by `KE=p^2"/"(2m)`, so the much more massive
proton never has very much kinetic energy. We are making an approximation by assuming all the kinetic energy is in the electron, but it is quite a good approximation.)
Angular momentum does help with classification. There is no transfer of angular momentum between the proton and the electron, since the force between them is a center-to-center force, producing no
Like energy, angular momentum is quantized in quantum physics. As an example, consider a quantum wave-particle confined to a circle, like a wave in a circular moat surrounding a castle. A sine wave
in such a “quantum moat” cannot have any old wavelength, because an integer number of wavelengths must fit around the circumference, `C`, of the moat. The larger this integer is, the shorter the
wavelength, and a shorter wavelength relates to greater momentum and angular momentum. Since this integer is related to angular momentum, we use the symbol `?` for it:
The angular momentum is
Here, `r=C/(2pi)`, and `p=h/lambda``=(hl)/C`, so
In the example of the quantum moat, angular momentum is quantized in units of `h"/"2pi`, and this turns out to be a completely general fact about quantum physics. That makes `h"/"2pi` a pretty
important number, so we define the abbreviation . This symbol is read “h-bar.”
quantization of angular momentum
What is the angular momentum of the wave function shown on page 985?
(answer in the back of the PDF version of the book)
36.1 Classifying states by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license. | {"url":"https://www.vcalc.com/collection/?uuid=1f1fbdb9-f145-11e9-8682-bc764e2038f2","timestamp":"2024-11-07T00:33:04Z","content_type":"text/html","content_length":"61500","record_id":"<urn:uuid:efc0fb11-b985-4e1a-a324-a6f2baba8f28>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00743.warc.gz"} |
Enriching a predicate and tame expansions of the integers
Given a structure M and a stably embedded -definable set Q, we prove tameness preservation results when enriching the induced structure on Q by some further structure Q. In particular, we show that
if T = Th(M) and Th(Q) are stable (respectively, superstable, -stable), then so is the theory T[Q] of the enrichment ofMby Q. Assuming simplicity of T, elimination of hyperimaginaries and a further
condition on Q related to the behavior of algebraic closure, we also show that simplicity and NSOP1 pass from Th(Q) to T[Q]. We then prove several applications for tame expansions of weakly minimal
structures and, in particular, the group of integers. For example, we construct the first known examples of strictly stable expansions of (Z,+). More generally, we show that any stable (respectively,
superstable, simple, NIP, NTP2, NSOP1) countable graph can be defined in a stable (respectively, superstable, simple, NIP, NTP2, NSOP1) expansion of (Z, +) by some unary predicate A N.
• Stably embedded sets
• preservation of dividing lines
• tame expansions of the integers
ASJC Scopus subject areas
Dive into the research topics of 'Enriching a predicate and tame expansions of the integers'. Together they form a unique fingerprint. | {"url":"https://cris.bgu.ac.il/en/publications/enriching-a-predicate-and-tame-expansions-of-the-integers","timestamp":"2024-11-11T16:59:30Z","content_type":"text/html","content_length":"55063","record_id":"<urn:uuid:56f97ce3-6550-4caa-bb31-af39fb7d37b0>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00061.warc.gz"} |
[pstricks] pst-3dplot twodimensional Normal-density function
Felix Hoffmann fhoffmann at iam.uni-bonn.de
Wed Dec 2 11:53:10 CET 2009
Hi all,
I am new to the list, so I don't know, whether someone posted a similar
I am trying to plot the density of the standard normal distribution over
\mathbb R^2. That is successful with pst-3dplot in the following manner,
but there are two problems, I would like to fix, but I don't know how.
1. Can I get the lines of my painting as niveau-lines, meaning, all the
points of the same z-value should be on the same line? How can I do it
with ps-tricks and related packages, or do I need something different?
2. If I can't do it with ps-tricks, how can I get the parser to draw the
x and the y lines over the main part of the function, and not only
either the y lines or the x lines?
Here is the code I wrote to this:
2.718281828 x x mul y y mul add 0.5 mul neg exp mul}
\caption{Dichte der Standardnormalverteilung in $\mathbb R^2$.}
More information about the PSTricks mailing list | {"url":"https://tug.org/pipermail/pstricks/2009/007313.html","timestamp":"2024-11-12T05:51:50Z","content_type":"text/html","content_length":"3840","record_id":"<urn:uuid:d639767e-ca65-4d55-b3a0-4ea04cada6f4>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00216.warc.gz"} |
AQA A GCE Physics - Current Electricity
What is current?
-Rate of flow of charge
How can you calculate current using time?
What is ΔQ measured in?
-Coulombs (unit of charge)
Define 1 Coulomb
-Amount of charge that passes in 1 second when the current is 1 ampere
How can you measure current flowing through a circuit?
Where does an ammeter need to be attached in a circuit? Why?
-Current through ammeter is the same as current through the component
Define potential difference
-Energy converted per unit charge moved
What needs to be done to electric charge to make it flow through a conductor?
-Work must be done on it
How can you work out voltage using charge?
What is ‘W’ in order to work out potential difference?
-Energy in joules, work done to move the charge
Define a volt
-Potential difference across a component is 1 volt when you convert 1 joule of energy moving 1 coulomb of charge through the component
Finish this equation in terms of the definition of potential difference.
-1V = 1JC^-1
What does current for a particular potential difference depend on?
-Resistance of component
How do you calculate resistance using voltage?
What is resistance measured in?
-Ohms (Ω)
A component has a resistance of 1 Ω if a potential difference of 1 V makes a current of *A flow through it
For what sort of conductor is resistance a constant?
-Ohmic conductor
What sorts of conductors obey Ohm’s law?
-Ohmic conductor
Name 2 factors that are directly proportional in an Ohmic conductor. What must remain constant?
-Current and potential difference
How will factors such as light level or temperature have significant effects on resistance?
-Resistivity changes
Finish this sentence: “Ohm’s law is a special case; it is only true for…
-Ohmic conductors at constant temperature
What do I/V graphs/characteristics show?
-How resistance varies
What does the gradient of an I/V graph show?
-Resistance. Shallower the gradient of a characteristic I/V graph, the greater the resistance of the component
What is the I/V characteristic for a metallic conductor?
-Straight line through zero (0)
Which I/V characteristic is a curve through zero (0)?
-I/V characteristic for a filament lamp
Why doesn’t a filament lamp have the same (straight line) characteristic of a metallic conductor?
-Gets hot. Current flowing through lamp increases its temperature
The ********** of a metal increases as the temperature increases
Where are semiconductors used?
Describe semiconductors
-Nowhere near as good at conducting as metals, due to few charge carriers. If energy is supplied to semi-conductor, more charge carriers can be released. Meaning they make excellent sensors for
detecting changes in their environment
Which 3 semiconductor components do you need to know about?
-Thermistors, LDRs (Light Dependent Resistors), Diodes
What does resistance of…
Mrs Jones
This is a massive and very useful resource - you may wish to split it down further. It covers both DC and AC electricity so that might be a good way to split it. This would be an excellent discussion
resource to use with friends for a revision session. Go through the questions and discuss. For example what do the greek symbols used look like and how are they pronounced. Alternatively you could
also use this as the basis for making your own flash cards. Comprehensive. | {"url":"https://ws.getrevising.co.uk/revision-notes/aqa_a_gce_physics_current_electricity_2","timestamp":"2024-11-15T03:46:13Z","content_type":"text/html","content_length":"40274","record_id":"<urn:uuid:e5c120cf-15f0-435c-8077-961f2cf8c8b2>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00644.warc.gz"} |
Aggregation of symbionts on hosts depends on interaction type and host traits
Data files
Sep 13, 2023 version files 73.14 KB
Symbionts tend to be aggregated on their hosts, such that few hosts harbor the majority of symbionts. This ubiquitous pattern can result from stochastic processes, but aggregation patterns may also
depend on the type of host-symbiont interaction, plus traits that affect host exposure and susceptibility to symbionts. Untangling how aggregation patterns both within and among populations depend on
stochastic processes, interaction type and host traits remains an outstanding challenge. Here, we address this challenge by using null models to compare aggregation patterns in a neutral system of
Balanomorpha barnacles attached to patellid limpets and a host-parasite system of Trinidadian guppies (Poecilia reticulata) and their Gyrodactylus spp. monogeneans. We first used a model to predict
patterns of symbiont-host aggregation due to random partitioning of symbionts to hosts. This null model accurately predicted the aggregation of barnacles on limpets, but the degree of aggregation
varied across 303 quadrats. Quadrats with larger limpets had less aggregated barnacles, whereas aggregation increased with variation in limpet size. Across 84 guppy populations, Gyrodactylus spp.
parasites were significantly less aggregated than predicted by the null model. As in the neutral limpet-barnacle system, aggregation decreased with mean host size. Parasites were also significantly
less aggregated on males than females because male guppies tended to have higher prevalence and lower parasite burdens than predicted by the null model. Together, these results suggest stochastic
processes can explain aggregation patterns in neutral but not parasitic systems, though in both systems host traits affect aggregation patterns. Because the distribution of symbionts on hosts can
affect symbiont evolution via intraspecific interactions, and reciprocally host behavior and evolution via host-symbiont interactions, identifying the drivers of aggregation enriches our
understanding of host-symbiont interactions.
README: Aggregation of symbionts on hosts depends on interaction type and host traits
Here, we test whether the processes that drive aggregation differ between a neutralistic and parasitic host-symbiont system. We further test how the tolerance-linked sex and exposure-linked size
differences of guppy Poecilia reticulata hosts can drive differences in the aggregation of Gyrodactylus spp. parasites among their hosts. For the guppy-Gyrodactlid portion of the manuscript we are
using the guppygyro.csv dataset. This data is a combination of population level data from Stephenson, J. F., Oosterhout, C. van, Mohammed, R. S. & Cable, J. Parasites of Trinidadian guppies: evidence
for sex- and age-specific trait-mediated indirect effects of predators. Ecology 96, 489–498 (2015) and an additional, unpublished data set. For the guppy-Gyrodactlid portion of the manuscript we are
using the guppygyro.csv dataset guppygyro.csv. This data is a combination of population level data from Stephenson, J. F., Oosterhout, C. van, Mohammed, R. S. & Cable, J. Parasites of Trinidadian
guppies: evidence for sex- and age-specific trait-mediated indirect effects of predators. Ecology 96, 489–498 (2015) and an additional, unpublished data set limpetbarnacle.csv. We additionally have a
separate CSV file named "guppysexescorr.csv." This dataset is just a subset of the larger guppy dataset to test for correlations between the aggregation of parasites between males and female guppies
from the same site.
Description of the data and file structure
guppygyro.csv: contains all data pertaining to the Trinidadian guppy and gyrodactylus parasite data. Included variables:
Site: A factor that identifies the drainage, river, course, site, and year the population data was taken. It has a level per population.
drainage: The drainage where the population was collected from. Six levels
river: The river where the population was collected from. 13 levels.
year: The year the population was collected in. 5 levels.
season: The season the population was collected. 2 levels.
Course: The watercourse the population was collected in. 2 levels. Upper indicates low predation populations while lower indicates high predation populations in line with previous publications from
the Trinidadian guppy system
H: The total number of hosts in the population
P: The total number of parasites in the population
MeanP: The mean number of parasites infecting hosts in the population
VarP: The mean number of parasites infecting hosts in the population
MeanLen: The mean length of all individuals in the population (millimeters)
VarLen: The variance in length of the population (millimeters^2)
MeanW: The mean weight of all individuals in the population (grams)
VarW: The variance in weight of the population. (grams^2)
fs var log med: The log median variance predicted by the feasible set. \newline
sex: The sex of individuals in the population. Populations were separated into males and females to conduct each analysis and calculate body condition. This allowed us to better understand
aggregation differences between sexes. 2 levels.
jonvar: The variation in condition of individuals in the population
ResBC: The body condition given the residual mass index.
jonres: The residual of a regression between the body condition given by index from Johnson et al. 2011 and the mean length of the population. We did this to control for any body condition bias we
may see between populations just due to size differences.
diffvar: The difference in variances between the observed and expected distribution (the feasible set) of parasites in the population. Positive values indicates the population has a higher variance
(More aggregation) than expected by the feasible set, negative values means the feasible set predicted more variance than is present in the observed data
logMean: The log10 mean of parasites in the population.
jonresRS: The scaled jonres variable from above.
MeanLenRS: The scaled mean length variable from above.
ExpPrev: The expected prevalence of the population given by the feasible set \newline
ObsPrev: The observed prevalence of the population
MaxF: The expected number of parasites on the individual with the highest parasite burden in the population given by the feasible set
MaxE: The observed number of parasites on the individual with the highest parasite burden in the population
We also include an additional guppysexcorr.csv file for an analysis examining the correlation between aggregation metrics between males and females from the same site. A list of the data in this file
is included below:
Site: A factor that identifies the drainage, river, course, site, and year the population data was taken. It has a level per population.
FemaleDiff: The difference in variances between the observed and expected distribution (the feasible set) of parasites in the population of females at a given site.
MaleDiff: The difference in variances between the observed and expected distribution (the feasible set) of parasites in the population of males at a given site.
Lastly, we have a file titled limpetbarnacle.csv with all data pertaining to the limpet-barnacle host-symbiont analysis part of the manuscript. A list of the data in this file is included below:
Site: A factor that identifies the site where the community was collected.
QUADRAT: A factor that identifies the quadrat site location of each community of limpets and barnacles. It has a level per community.
H: The total number of limpet hosts in each community.
P: The total number of barnacle symbionts in each community.
MeanP: The mean abundance of barnacles living on the limpets for each community.
varP: The variance in abundance of barnacles living on the limpets for each population.
MeanW: The mean width measurement for the limpets in each community. Measured in (millimeters)
VarW: The variance in width measurement for the limpets in each community. Measured in (millimeters^2)
MeanL: The mean length measurement for the limpets in each community. Measured in (millimeters)
VarL: The variance in length measurement for the limpets in each community. Measured in (millimeters^2)
fs var log med: The log median variance predicted by the feasible set for each limpet-barnacle community.
Works referencing this dataset | {"url":"https://datadryad.org:443/stash/dataset/doi:10.5061/dryad.4b8gthtjx","timestamp":"2024-11-03T00:10:01Z","content_type":"text/html","content_length":"51515","record_id":"<urn:uuid:9e0831f9-0fdf-4070-8d0d-25d9519cda4e>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00598.warc.gz"} |
Compute number of neighbors for each node
How do you evaluate whether a node is an important one or not? There are a few ways to do so, and here, you're going to look at one metric: the number of neighbors that a node has.
Every NetworkX graph G exposes a .neighbors(n) method that returns an iterator of nodes that are the neighbors of the node n. To begin, use this method in the IPython Shell on the Twitter network T
to get the neighbors of of node 1. This will get you familiar with how the function works. Then, your job in this exercise is to write a function that returns all nodes that have m neighbors.
This is a part of the course
“Introduction to Network Analysis in Python”
View Course
Exercise instructions
• Write a function called nodes_with_m_nbrs() that has two parameters - G and m - and returns all nodes that have m neighbors. To do this:
□ Iterate over all nodes in G (not including the metadata).
□ Use the len() and list() functions together with the .neighbors() method to calculate the total number of neighbors that node n in graph G has.
☆ If the number of neighbors of node n is equal to m, add n to the set nodes using the .add() method.
□ After iterating over all the nodes in G, return the set nodes.
• Use your nodes_with_m_nbrs() function to retrieve all the nodes that have 6 neighbors in the graph T.
Hands-on interactive exercise
Have a go at this exercise by completing this sample code.
# Define nodes_with_m_nbrs()
def ____:
Returns all nodes in graph G that have m neighbors.
nodes = set()
# Iterate over all nodes in G
for n in ____:
# Check if the number of neighbors of n matches m
if ____ == ____:
# Add the node n to the set
# Return the nodes with m neighbors
return nodes
# Compute and print all nodes in T that have 6 neighbors
six_nbrs = ____
This exercise is part of the course
Introduction to Network Analysis in Python
This course will equip you with the skills to analyze, visualize, and make sense of networks using the NetworkX library.
What is DataCamp?
Learn the data skills you need online at your own pace—from non-coding essentials to data science and machine learning. | {"url":"https://campus.datacamp.com/courses/introduction-to-network-analysis-in-python/important-nodes?ex=2","timestamp":"2024-11-12T06:16:50Z","content_type":"text/html","content_length":"170902","record_id":"<urn:uuid:500e61bf-2004-4c0d-912a-fed0a8a2fff5>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00142.warc.gz"} |
Fall Triple Digit Subtraction with Regrouping
Fall Triple Digit Subtraction with Regrouping
Price: 200 points or $2 USD
Subjects: math,mathElementary,operationsAndAlgebraicThinking,additionAndSubtraction,holiday,firstDayOfAutumn
Grades: 3
Description: The Key Idea behind this game is that students practice triple-digit subtraction problems with regrouping using the standard subtraction algorithm. This deck contains 20 fill in the
blank problems. Some of the problems in this deck require regrouping twice, and some of the problems require regrouping just once. Some problems also have the zero concept with regrouping across
zeros. Standard addressed in this game: CCSS: 3.NBT.A.2 | {"url":"https://wow.boomlearning.com/deck/ygdvoiCWKLGZZB8YX","timestamp":"2024-11-03T18:35:11Z","content_type":"text/html","content_length":"2347","record_id":"<urn:uuid:e0d77697-f2de-4380-a49e-10d842ffacb6>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00731.warc.gz"} |
Concentration Requirements
This page describes the current version of the computer science concentration requirements. Students who completed at least one term in the college prior to Fall 2021 may elect to follow the Old
(2020 and earlier) requirements (comparison).
See the Degree Programs page or the Computer Science handbook entry for more information about the different tracks within the computer science concentration. The courses page provides more detailed
information of which courses count for which requirements. See also sample schedules for guidance on specific courses to take in your first two years as well as example plans of study to complete all
of your requirements.
If you are ready to declare a CS concentration, please follow these steps. You can also see how to combine CS with other areas in the form of a secondary concentration, joint concentration, or other
CS Concentration Requirements
Computer Science’s concentration requirements are as follows. For the basic plan, 9 core computer science courses are required. Collectively, these courses must meet requirements in programming,
formal reasoning, systems, computation and the world, and advanced computer science. A single course may fulfill multiple requirements (for example, CS 109a counts for both Programming 1 and
Computation and the World). The basic requirements also include 2–5 courses in Mathematics, including mathematical preparation, Linear Algebra, and Probability.
Requirement Basic Honors Joint MBB
Mathematical preparation 2–5 courses
Early calculus (Math Ma/Mb/1a/1b) 0–3 courses as needed
Linear Algebra (Math 21b/22a, AM 22a, …) 1 course
Probability (Stat 110, ES 150, …) 1 course
Requirement Basic Honors Joint MBB
Computer science core 9 11 courses 8 courses 8 courses
Programming 1 (CS 32, CS 50, …) & Programming 2 (CS 51, 2 courses
CS 61, …)
Formal Reasoning (CS 20, CS 120, CS 121, CS 124, CS 134, 3 courses
CS 152, …)
Most Discrete Mathematics, Computational Limitations, Algorithms, and Intermediate Algorithms courses also count toward the Formal Reasoning requirement.
Discrete Mathematics (CS 20, AM 107, …) 1 course, or place out*
Computational Limitations (CS 120, CS 121, …) 1 course
Algorithms (CS 120, CS 124, …) 1 course N/A 1 course N/A
Intermediate Algorithms (CS 124, …) N/A 1 course N/A 1 course
Systems 1 course
Computation and the World 1 course
Most Artificial Intelligence courses also count toward the Computation and the World requirement.
Artificial Intelligence N/A 1 course N/A 1 course
Advanced Computer Science (CS 100+) 4 5 courses 4 courses, or 3 courses + one 4 courses, or 3 courses + one CS 91r
courses CS 91r
Requirement Basic Honors Joint MBB
Mind, brain, & behavior track N/A N/A N/A 3 MBB courses (MCB 80, approved related field, approved
junior tutorial)
Requirement Basic Honors Joint MBB
Thesis N/A Optional (required for high or highest Required Required
• If you place out of discrete mathematics, you still need to take a total of three formal reasoning courses, see the tags page for some options. The process for placing out of the discrete math
requirement is here.
A basic, honors, or MBB CS concentration can be combined with another concentration as a double concentration. The CS requirements for such a concentration are the same as the CS requirements for an
undoubled basic, honors, or MBB CS concentration, respectively, except that at most two courses used for the other concentration may be used for the CS concentration.
Note: Students who completed at least one term in the college prior to Fall 2021 may choose to follow the old concentration requirements, available in the relevant archived version of the Handbook
for Students and contact the department for further information.
Basic Requirements: 11–14 courses (44–56 credits)
1. Required courses (11–14 courses): A student’s Plan of Study must satisfy each of the requirements below.
□ Mathematical preparation (2–5 courses, see Note on Mathematical preparation below):
☆ Pre-calculus and single-variable calculus: Either Mathematics Ma, Mb, and 1b, or Mathematics 1a and 1b. Students may place out of some or all of this requirement depending on their
starting Mathematics course.
☆ Linear algebra: One course in linear algebra. Satisfied by Applied Mathematics 22a, Mathematics 21b, Mathematics 22a, Mathematics 23a, Mathematics 25a, Mathematics 55a, or a more advanced
☆ Probability: One course in probability. Satisfied by Statistics 110, Engineering Sciences 150, Mathematics 154, or a more advanced course.
□ Computer Science core (9 courses): Nine courses from an approved list on the concentration’s website. This list contains computer science courses and some courses in related fields. These
courses must, taken together, satisfy the following “tag” requirements. The concentration website has a list of tags and the corresponding courses. A tag requirement is satisfied or partially
satisfied by a plan of study containing a corresponding course. Each course on a plan of study may satisfy zero, one, two, or more tag requirements. Example plans of study satisfying these
requirements can be found on the concentration website. While some courses can satisfy multiple tags, students still need to take nine Computer Science core courses.
☆ Programming 1 and Programming 2 tags (2 courses in the computer science core): Two courses on software construction and good software engineering practices. The requirement is satisfied
by either one course tagged Programming 1 and one tagged Programming 2, or by two courses tagged Programming 2. Note that some Programming 1 courses, such as CS 32 and CS 50, cannot be
taken for concentration credit after more advanced programming courses.
☆ Formal Reasoning tag (3 courses in the computer science core): Three courses on formal reasoning about computer science, including at least:
○ Discrete Mathematics tag (0 or 1 course in the computer science core): One course with significant discrete math content. Students may skip this requirement with approval from the
Directors of Undergraduate Studies; see the Computer Science website for more information. Note that some Discrete Math courses, such as CS 20, cannot be taken for concentration
credit after more advanced formal reasoning courses.
○ Computational Limitations tag (1 course in the computer science core): One course covering topics in computability and complexity.
○ Algorithms tag (1 course in the computer science core): One course covering topics in algorithms.
☆ Systems tag (1 course in the computer science core): One course containing significant computer system development.
☆ Computation and the World tag (1 course in the computer science core): One course on interactions between computation and the world (for example, concerning informational, natural, human,
or social systems).
☆ Advanced Computer Science tag (4 courses in the computer science core): Four sufficiently advanced Computer Science courses, roughly corresponding to all CS courses numbered 100 and
above. (See the concentration website for the full list.)
□ Note on Mathematical preparation: The total number of required courses for the concentration depends on the starting Mathematics course (see Requirements above).
☆ Students starting in Mathematics Ma: 14 courses (five courses to complete the mathematics requirements).
☆ Students starting in Mathematics 1a: 13 courses (four courses to complete the mathematics requirements).
☆ Students starting in Mathematics 1b: 12 courses (three courses to complete the mathematics requirements).
☆ Students starting in Mathematics 21b or similar: 11 courses (two courses to complete the mathematics requirements).
2. Tutorial: Optional. Available as Computer Science 91r. This course is repeatable, but may be taken at most twice for academic credit, and only one semester of Computer Science 91r may be counted
toward concentration requirements as a computer science core course. Students wishing to enroll in Computer Science 91r must file a project proposal to be signed by the student and the faculty
supervisor and approved by the Directors of Undergraduate Studies. The project proposal form can be found on the Computer Science website.
3. Thesis: None.
4. General Examination: None
5. Other Information:
□ Approved courses: With the approval of the Directors of Undergraduate Studies, other courses may be used to satisfy requirements. If a course is cross-listed with another department it meets
the same requirements for the concentration as the COMPSCI-numbered course. In general, a course may be substituted with a more advanced version on the same or similar topic. Students must
secure advance approval for course substitutions by filing a Plan of Study to be approved by the Directors of Undergraduate Studies. The Plan of Study form and a description of the process to
submit the form can be found on the Computer Science website.
□ Pass/Fail and Sat/Unsat: No more than two of the courses used to satisfy CS Requirements may be taken PA/FL or SUS. Of the tag requirements, courses taken PA/FL or SUS can be used only for
the Core CS, Programming 1, and Advanced Computer Science tags. For instance, if taken PA/FL, CS 1240 would satisfy the Core Computer Science and Advanced Computer Science tags, but would not
satisfy the Formal Reasoning or Algorithms tags. If taken for a letter grade, CS 1240 would satisfy the Core Computer Science, Advanced Computer Science, Formal Reasoning, and Algorithms
□ Reduction of requirements for prior work: Except for Mathematics Ma, Mathematics Mb/1a, and Mathematics 1b, there is no reduction in concentration requirements for prior work.
□ Plans of study: Concentrators must file a Plan of Study showing how they intend to satisfy these degree requirements, and keep their plan of study up to date until their program is complete.
If the plan is acceptable, the student will be notified that it has been approved. To petition for an exception to any rule, the student should file a new plan of study and notify the
Directors of Undergraduate Studies of the rationale for any exceptional conditions. Approval of a plan of study is the student’s guarantee that a given set of courses will satisfy degree
requirements. The Plan of Study form and a description of the process to submit the form can be found on the Computer Science website.
Honors Requirements: 13–16 courses (52–64 credits)
1. Required courses (13–16 courses): A student’s Plan of Study must satisfy each of the requirements below. Courses are allowed to satisfy multiple requirements, but a student’s Plan of Study must
still comprise thirteen to sixteen courses in total.
□ Mathematical preparation (2–5 courses): Same as Basic Requirements.
□ Computer Science core (11 courses): Eleven courses from an approved list on the concentration’s website. This is the same list as for the basic requirements, but two more courses are
required. These courses, taken together, must satisfy the following tag requirements.
☆ Programming 1 and Programming 2 tags (2 courses in the computer science core): Same as Basic Requirements.
☆ Formal Reasoning tag (3 courses in the computer science core): Same as Basic Requirements, but requiring Intermediate Algorithms rather than Algorithms, as follows.
○ Discrete Mathematics tag (0 or 1 course in the computer science core): Same as Basic Requirements.
○ Computational Limitations tag (1 course in the computer science core): Same as Basic Requirements.
○ Intermediate Algorithms tag (1 course in the computer science core): One course covering basic and intermediate topics in algorithms. Replaces the Algorithms tag from the Basic
☆ Systems tag (1 course in the computer science core): Same as Basic Requirements.
☆ Computation and the World tag (1 course in the computer science core): Same as Basic Requirements.
☆ Artificial Intelligence tag (1 course in the computer science core): One course covering topics in artificial intelligence. (Most such courses will simultaneously satisfy the Computation
and the World tag.)
☆ Advanced Computer Science tag (5 courses in the computer science core): Five sufficiently advanced Computer Science courses (one more than in the Basic Requirements).
2. Tutorial: Same as Basic Requirements.
3. Thesis: Optional but encouraged. See honors requirements on the Computer Science website. Students writing theses are often enrolled in Computer Science 91r. This course is repeatable, but may be
taken at most twice for academic credit, and only one semester of Computer Science 91r may be counted toward concentration requirements as a Computer Science core course. Students wishing to
enroll in Computer Science 91r must file a project proposal to be signed by the student and the faculty supervisor and approved by the Directors of Undergraduate Studies. The project proposal
form can be found on the Computer Science website.
4. General Examination: Same as Basic Requirements.
5. Other Information: Same as Basic Requirements.
Requirements for Joint Concentrations: 10–13 courses (40–52 credits) for CS Field
Joint concentrations with certain other fields are possible. This option is intended for students who have interests in the intersection of two fields, not simply in the two fields independently; for
example, a combined concentration in computer science and linguistics might be appropriate for a student with a special interest in computational linguistics. Course requirements are the same as the
Basic Requirements, with three exceptions: only eight (instead of nine) CS core courses are required, Computer Science 91r may be used to satisfy an Advanced Computer Science requirement, and a
thesis that combines the two fields is required. Note that courses satisfying CS requirements may also be double-counted towards the requirements of the other field.
Joint concentrations are not “double majors.” Joint concentrators should be interested in the overlap between two fields, not simply in both. A thesis in the intersection of the fields is required
for joint concentrators, read by both concentrations. The student is typically awarded the minimum honors recommended by the two concentrations separately. These requirements, including the Thesis
Requirement, are the same whether Computer Science is the primary field or the allied field of the joint concentration.
Students interested in combined programs should consult the Directors of Undergraduate Studies at an early date and should work carefully with both concentrations to ensure all deadlines and
requirements of both concentrations are met. Students with separate interests in more than one field should consider pursuing a secondary rather than a joint concentration or simply using some of
their electives to study one of the fields. We advise all of our joint concentrators to make sure that they satisfy the non-joint requirements for at least one concentration, in case they are unable
to complete a thesis.
Requirements for Mind, Brain, and Behavior Program: 13–16 courses (52–64 credits)
Students interested in addressing questions of neuroscience and cognition from the perspective of computer science may pursue a special program of study affiliated with the University-wide Mind,
Brain, and Behavior Initiative, that allows them to participate in a variety of related activities. (Similar programs are available through the Anthropology, History and Science, Human Evolutionary
Biology, Linguistics, Neurobiology, Philosophy, and Psychology concentrations.) Requirements for this honors-only program are based on those of the computer science Requirements for Honors
Eligibility, as explained below:
1. Required courses (13–16 courses):
□ Mathematical preparation (2–5 courses): Same as Honors Requirements.
□ Computer Science core (8 courses): Same as Honors Requirements, with the following exceptions:
☆ Eight courses, rather than 11 courses, are required.
☆ Advanced Computer Science tag (4 courses in the computer science core): Four sufficiently advanced Computer Science courses, rather than 5 courses, are required. In addition, one CS 91r
course may be used to satisfy this requirement.
☆ Mind, brain, and behavior courses (3 courses):
○ MCB/Neuroscience 80
○ An approved course in an MBB-related field outside computer science
○ An approved MBB junior tutorial
2. Tutorial: Same as Honors Requirements.
3. Thesis: A computationally-oriented thesis on a Mind, Brain, and Behavior-related topic is required. Students pursuing thesis research may want to enroll in Computer Science 91r.
4. General Examination: None.
5. Other information: Same as Honors Requirements.
6. Note: Students pursuing the Mind, Brain, and Behavior track are assigned an adviser in the field and are expected to participate in the University-wide Mind, Brain, and Behavior research milieu,
including a non-credit senior year seminar for Mind, Brain, and Behavior thesis writers. To participate in the MBB track, students must both complete the Computer Science concentration Plan of
Study and register at the beginning of every academic year on the MBB website. Interested students should contact the Computer Science liaison to the MBB program, Professor Stuart Shieber. | {"url":"https://csadvising.seas.harvard.edu/concentration/requirements/","timestamp":"2024-11-03T04:23:50Z","content_type":"text/html","content_length":"58068","record_id":"<urn:uuid:6f059d9f-f151-415a-b04c-43eb9921dc8f>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00167.warc.gz"} |
Chaos train and monkey madness—fun with quantifiers
Chaos train and monkey madness—fun with quantifiers in first-order logic
Learn the language of first-order logic by immersion in some fun logic puzzles
Enjoy this free short installment from A Panorama of Logic, my new book that I have been serializing here on Infinitely More. The book is an introduction to all aspects of logic, for philosophers,
mathematicians, and computer scientists. This week we have an introduction to quantifiers, especially nested quantifiers—learn the language of first-order logic by a brief immersion in these
whimsical puzzles.
I am serializing all my new books here on Infinitely More, with fresh chapters every week, and regular extended excerpts from my already published books. Subscribe now for full access to the latest
posts, as well as the archives.
First-order predicate logic
We come now into the heart of logic with first-order predicate logic, providing a formal language and semantics, capable of expressing fine distinctions and shades of meaning—the subject of logic at
its core is focused upon the interaction of language and meaning. The language of predicate logic is rich, able to express the concepts of essentially arbitrary kinds of mathematical structure. How
remarkable it is that by studying the power and limitations of a formal language we are also so often led to mathematical insight.
Informal practice with quantifiers
Before launching into the full details of the formal language, however, let us first get some informal practice—perhaps one learns a new language best simply by immersing oneself in it. We shall put
off the grammatical details and formalities until we have attained a native felicity.
The background context for assertions in predicate logic, just as in relational logic, is a fixed domain of individuals, considered as though an entire world of its own. Upon that domain, we shall
have fixed relations or predicates, perhaps also functions and operations, as well as names or constants that pick out particular individuals, allowing us directly to refer to them. The quantifiers ∃
and ∀ range over the domain—to assert ∃x φ is to say “there is an x such that φ,” meaning that there is such an x to be found in the domain, and similarly ∀x φ asserts “every x has property φ,” again
meaning every x in the domain. Let us illustrate.
Consider the parent of relation on the set of all people, the relation Pxy that holds when person x is a parent of person y. What proposition is expressed by the following assertion?
Perhaps someone suggests that it asserts “everyone has a parent.” Is that right? Well, no. If we look carefully, the statement asserts that for every person x there is a person y such that x is the
parent of y. So this sentence asserts rather that “everyone is a parent.” And of course, this sentence isn't true—not everyone is a parent, since some people have not had any children. We may express
the proposition “everyone has a parent,” in contrast, in the following manner:
Is this sentence true? We had said that we are considering the parent-of relation on the domain of all people, so the quantifers ∀y and ∃x range over all people and the sentence asserts that for
every person y there is a person x such that x is the parent of y. If we had meant the domain of all living people, then this second sentence also would be false, since there are many people whose
parents have unfortunately both passed away.
Let us therefore consider the relation instead on the domain of all people who have ever lived. You and I are both in this domain, with your parents, your grandparents, Newton, Napoleon, Aristotle,
and their parents, and so on. Would it now be true that everyone has a parent in this expanded context? Perhaps this seems obvious—everyone has had a parent, right?—but let us consider the matter
carefully. If you have a parent, and your parent has a parent, and they have a parent, we might reach in this way into your ancestry. But assuming there are no cycles in the parenthood relation, this
would lead to an unending chain of distinct ancestors—there must have been infinitely many people!
Following your chain of ancestors into antiquity, we reach through the generations of humanity eventually to the species of homo erectus and other early homonid species. Are these individuals in the
set of all people? If so, then we proceed earlier, through the early primates to the primitive mammal species, and so on. Are these people? If we delineate a personhood border at any point, then the
chain of ancestors will cross it, making the statement false that for every person there is a person who is their parent. And if we do not delineate a border, then the chain of ancestors will
eventually reach the most primitive life forms, an originating life form with no parent, again making the statement false. We seem obliged to admit in any case that the statement ∀y ∃x Pxy expressing
the proposition “everyone has a parent” is not true when interpreted on the set of all people who have ever lived.
The gift-givers
Imagine a community of kind individuals and consider the relation Gxy which holds of individuals in that community, when person x has just made a home-made gift and given it to person y. Try to match
up the formal statements below at left with the phrases at right that allude to the meaning, perhaps without expressing it exactly. Can you find the best matches?
I have started you out by noticing that the assertion ∃x ∃y Gxy expresses the proposition that in at least one instance, a person just made a home-made gift for someone—isn't that nice? Can you match
the other statements?
Post your solutions to all these puzzles in the comments below!
[Note to instructor: completing these matchings and the others below often makes for a fruitful classroom activity to undertake as a group—all suggestions should come with explanation from the
Chaos train
The passengers on a New York subway train are heading downtown on particular Monday morning.
Let Bxy mean “person x accidentally bumped into person y on the train, and said ‘excuse me.’ ” Find the best matching of the formal statements with their corresponding phrases.
I suggested a match for ∃x ∃y Bxy, which expresses that in at least one instance, a person x has accidentally bumped into some person y—unfortunately, it happens. Can you match the remaining
Monkey madness
Next we encounter a riotous day at the zoo.
Consider the predicate Tmbn expressing the trinary relation:
Monkey m tosses banana b to monkey n.
Let us try to match the formal statements below with the phrases that allude to the meaning, perhaps without expressing it exactly.
In these statements, we have two sorts of objects—monkeys and bananas—and each sort has its own quantifiers and variables. Quantifiers using the variables m and n range over the monkeys at the zoo,
and the quantifiers using b range over the bananas there. Find the best matching of the formal assertions with the descriptive phrases. I have begun by observing that ∃m ∃b ∃n Tmbn expresses the
proposition that in at least one instance, a monkey has tossed a banana to another monkey (or perhaps to himself), and so indeed there is “a banana up in the air.” Can you complete the matching?
Post your solutions in the comments below!
What’s coming up
• The formal language of first-order logic
• Structures in a given language signature
• The semantics of truth
• Theory theory
• Definability
• Interpretability
• Model theory
• Much more!
Infinitely More is a reader-supported publication. New posts each week serializing my newest books. To receive new posts and support my work, consider becoming a free or paid subscriber.
A nice collection of exercises this week—highly recommended to gain expertise in quantifier logic. Give them all a try!
1. Translate the following assertion into natural language, where Pxy expresses “x is a parent of y,” and state whether the statement is likely true or false amongst the set of all living people.
2. Express the following assertions in the formal language of x\mathrel{L} y, meaning “x loves y.” If the assertion is ambiguous, provide multiple interpretations. Quantifiers range over all people.
3. Translate each of the following assertions into ordinary English, where Ax means “x is an adult,” Tx means “x is a passenger on the train,” and Sxy means “x is sitting next to y.” Please find the
simplest, most natural way to say it.
4. Express the following assertions in the formal language of Ss, meaning “s is a station” and Dd, meaning “d is a door” (on the train), and Ods, meaning “door d opens at station s.” Quantifiers
range over absolutely everything.
5. Suppose that students have just completed a course in applied ethics and it is time for the final exam, but perhaps not all the students have studied the readings sufficiently. Let Cxy express
that “student x copied from student y on the exam.” Create a matching-phrases puzzle for the following formal statements, in the style of the monkey madness exercise, by creating a list of six
pithy phrases that allude to the meaning expressed, without necessarily expressing the meaning literally. For example, one of the assertions might be about “an impressively desperate student” and
regarding another, “the invigilator has left the building.” Can your fellow students solve your matching puzzle?
6. Express the following assertions in the formal language of x C y, meaning “x is a child of y” and W(x), meaning “x is a woman.” Quantifiers range over people.
7. Complete the matching for the gift-giving community.
8. Complete the matching for the chaos train passenger example.
9. Complete the matching for the monkey madness at the zoo.
10. Express the following assertions in the formal language of the monkey madness puzzle.
11. Which of the following are true in the real numbers ℝ, with the usual addition and multiplication? (Thanks to Andrej Bauer.)
12. Discuss the nature of “generic” assertions, such as
Do these assertions carry the meaning of ∃ or ∀? Or something else? What is the logic of generic assertions?
13. What does “only” mean? In fact, the word “only” is frequently misplaced or otherwise wrongly used in common parlance. Explain the differences in meaning in the following sentences.
14. What meaning do the following sentences express, and which are true?
15. Imagine a friend who had wanted to help you, but instead their actions had caused a problem, and so now the friend wants to apologize by saying one of the following sentences. What do these
sentences mean, and do any of them express a suitable apologetic meaning for your friend?
14. Translate ∃z ∀x ∃y (x ≠ y ∧ y ≠ z) into an equivalent simple statement in plain language.
The illustrations were created by author via DALL·E and the collection is available at https://labs.openai.com/c/UEU2PFntFefdbSjVBcVvOiX7. | {"url":"https://www.infinitelymore.xyz/p/monkey-madness","timestamp":"2024-11-14T21:16:15Z","content_type":"text/html","content_length":"328995","record_id":"<urn:uuid:1787ce7c-f949-46c3-9842-36bfdeb67a7b>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00627.warc.gz"} |
Lathe Cutting and Threading Time Calculator - Calculator6.com
Lathe Cutting and Threading Time Calculator
The Lathe Cutting and Threading Time Calculator is a tool for calculating the duration of turning operations.
The Lathe Cutting and Threading Time Calculator is a tool for calculating the duration of turning operations. This calculator uses basic parameters to determine the duration of material cutting and
threading operations on lathes. The duration of cutting and threading operations depends on factors such as the type of material to be cut, the cutting speed, the diameter of the workpiece and the
threading depth. This calculator helps machinists (operators) and manufacturing engineers to plan and optimize turning operations.
When using the Online Lathe Cutting and Threading Time Calculator, you can calculate by entering: Length of threaded portion, Diameter of tap used, Pitch and Revolution of the job per minute.
3 Number of Calculations Used Today
Time for Tapping = \frac{L + \frac{D}{2}}{P \times r.p.m} Time for Cutting = \frac{\frac{3}{2} \times (L + \frac{D}{2})}{P \times r.p.m}
• P: is the pitch of the thread in millimeters
• L: is the length of the threaded portion in millimeters
• D: is the diameter of the tap used in millimeters
• r.p.m: is the revolutions per minute of the job.
How is lathe cutting and threading time calculated?
Lathe cutting and threading time is calculated based on a number of factors. Here are the basic steps to calculate lathe cutting and threading time:
1. Determine the Cutting Speed: Depending on the type of material to be machined, a suitable cutting speed is selected. The cutting speed determines the speed at which the material is machined and
usually depends on the characteristics of the material, the characteristics of the lathe and the characteristics of the cutting tool.
2. Determining the Diameter of the Workpiece: The diameter of the workpiece determines how long the cutting process will take. Larger diameters usually require longer times.
3. Determining the Depth of the Cut: The depth of the cut determines the thickness of the material to be cut. In the threading process, the threading depth is determined.
4. Calculating Speed and RPM: Using the cutting speed and the diameter of the workpiece, the number of revolutions of the cutting tool is calculated. This is important to determine how long the
cutting process will take.
5. Calculation of Process Time: Given the depth and cutting speed specified for the cutting operation, the total duration of the operation is calculated. This time varies depending on the size of
the workpiece, cutting speed and other machining parameters.
6. Calculating the Threading Time: The total time of the threading operation is calculated considering the depth and threading speed specified for the threading operation.
These steps form the basis for calculating lathe cutting and threading times. These times can vary depending on factors such as workpiece size, material and machining conditions.
What is lathe cutting and threading time?
Lathe cutting and threading time refers to the total time it takes to cut or thread a workpiece on a lathe. This time can vary depending on the characteristics of the workpiece, the cutting tool, the
cutting speed and the machining depth.
Lathe cutting time is determined by factors such as the rotational speed of the workpiece and the feed rate of the cutting tool. Larger diameter workpieces are usually cut in longer times, while
smaller diameter workpieces are cut in shorter times. The threading time is determined by the depth required for threading and the threading speed.
Lathe cutting and threading times play an important role in the planning and optimization of workpiece production processes. Accurate calculation of these times can increase the efficiency of
production processes and reduce production costs.
Lathe Threading Time and Methods
Lathe threading time refers to the total time spent threading a workpiece. This time varies depending on the threading method used, the characteristics of the workpiece and the process parameters.
Lathe threading is usually performed by the following methods:
• Single Pass Threading: In this method, a single threading operation is applied to the workpiece. As the threading tool moves across the workpiece, it creates the desired tooth profile. Single
pass threading is generally preferred for small diameter workpieces and the process is completed faster.
• Multi-pass Threading: In this method, more than one threading operation is applied to the workpiece. In the first operation, the threading tool smooths the surface of the workpiece to form a
starting thread, followed by the formation of other threads in one or more passes. Multi-pass threading is generally used on workpieces that require larger diameters and complex tooth profiles.
• Manual Threading: In this method, the threading tool is manually controlled and the threading operation is performed on the workpiece. Manual threading can be preferred for small production
quantities or when special thread profiles are required.
• CNC (Computer Controlled) Threading: In this method, the threading operation is performed automatically on a CNC lathe with programmed commands. CNC threading is widely used on workpieces that
require high precision and repeatability, making the process more automated and efficient.
Each of these methods plays an important role in determining the lathe threading time and planning the production process of the workpieces.
Considerations in Lathe Cutting and Threading Time Calculation
There are some important points to be considered in lathe cutting and threading time calculation:
Correct determination of cutting and threading parameters: It is important to correctly determine parameters such as cutting speed, feed rate, cutting tool geometry. These parameters should be chosen
appropriately depending on factors such as the material and size of the workpiece.
Workpiece stability: A firm grip and stable rotation of the workpiece is important for accurate cutting and threading. Problems such as vibrations or workpiece slippage can increase processing time
and reduce quality.
Maintenance of cutting and threading tools: Regular inspection and maintenance of cutting and threading tools can reduce processing time and extend tool life. Using worn or damaged tools can reduce
process efficiency.
Correct use of cutting fluid: The cutting fluid is important for efficient cutting and threading operations. The correct selection and use of cutting fluid can extend tool life and reduce machining
Machine performance: The power, precision and stability of the lathe affect cutting and threading times. Using a quality lathe can reduce processing time and improve the quality of the workpiece.
Consideration of these factors helps to calculate the correct cutting and threading times and to process workpieces efficiently. | {"url":"https://www.calculator6.com/lathe-cutting-and-threading-time-calculator/","timestamp":"2024-11-06T17:24:25Z","content_type":"text/html","content_length":"274978","record_id":"<urn:uuid:a53aae7b-81a8-4ac4-90d0-2f316a6e747b>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00041.warc.gz"} |
What is neuron simulator?
03/06/202203/06/2022| | 0 Comment | 12:00 AM
What is neuron simulator?
NEURON is a simulation environment for modeling individual neurons and networks of neurons. It provides tools for conveniently building, managing, and using models in a way that is numerically sound
and computationally efficient.
What is gK Max?
The K conductance when all channels are fully open (gKmax) was measured as the maximum conductance achieved with a very depolarised clamp potential. The stable K conductance (gK∞) measured at other
clamp potentials could then be expressed as a fraction of this maximum.
How does neuron software work?
Neuron models individual neurons via the use of sections that are automatically subdivided into individual compartments, instead of requiring the user to manually create compartments. The primary
scripting language is hoc but a Python interface is also available.
How do you cite a neuron software?
Please be sure to cite NEURON if your work results in publications. Carnevale, N.T. and Hines, M.L. The NEURON Book. Cambridge, UK: Cambridge University Press, 2006.
What is the Hodgkin-Huxley cycle?
The Hodgkin–Huxley model, or conductance-based model, is a mathematical model that describes how action potentials in neurons are initiated and propagated. It is a set of nonlinear differential
equations that approximates the electrical characteristics of excitable cells such as neurons and cardiac myocytes.
How many parameters do you need to model a certain membrane conductance using a Hodgkin-Huxley formalism?
10 scaling parameters
Hence, a realization of a Hodgkin–Huxley model is defined by a list of 10 scaling parameters: <αn(v)>, <βn(v)>, <αm(v)>, <βm(v)>, <αh(v)>, <βh(v)>, , <ˉgleak>, <ˉgK>, and <ˉgNa>.
Is the Hodgkin Huxley model linear?
Hodgkin–Huxley type models represent the biophysical characteristic of cell membranes. The lipid bilayer is represented as a capacitance (Cm). Voltage-gated and leak ion channels are represented by
nonlinear (gn) and linear (gL) conductances, respectively. | {"url":"https://thegrandparadise.com/essay-tips/what-is-neuron-simulator/","timestamp":"2024-11-01T22:44:43Z","content_type":"text/html","content_length":"52497","record_id":"<urn:uuid:90114a7a-f6d7-42bc-91b2-5f7e13c86d8c>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00022.warc.gz"} |
The Unexpected Biases of Primes
The Unexpected Biases of Primes
A new discovery in mathematics was actually pretty obvious.
When a new discovery is made in mathematics, generally only other mathematicians get excited by the discovery. Most new discoveries are either in unique areas of study, obscure corners of
mathematics, or highly specialized fields. String theory, for example, or topology.
Because “basic” mathematics has been studied for so long by so many talented individuals, many easy-to-understand theories have been established or proven already. It doesn’t feel like there are a
lot of discoveries left to be found in the areas of mathematical thought that are easily accessible to laypersons.
That’s why last year’s discovery of a Prime Number Conspiracy was stunning to so many mathematicians—it felt like an obvious result that we should have noticed before now.
A Quick Review
Prime numbers are whole numbers whose only factors are one and itself, like 7 (1 × 7) or 23 (1 × 23). Prime numbers make up a unique code for every single whole number.
Last year’s discovery of a Prime Number Conspiracy was stunning to so many mathematicians.
Given this universal and powerful status, mathematicians have long looked for patterns within prime numbers with varying degrees of success. Certain tendencies have been found in prime numbers (by
Pierre de Fermat, Christian Goldbach, and others), but usually discoveries are only along the lines of “If a prime number has this characteristic, then it also has that characteristic.”
Formulas have been found that sometimes lead to prime numbers. Mersenne’s famous formula states that 2p–1, where p is a prime, sometimes leads to another prime number (but there are no guarantees,
and many prime numbers are not found using this formula). Euclid long ago proved that there are infinite prime numbers and it’s long been believed that there are infinite “twin primes” (primes that
are consecutive odd numbers like 29 and 31 or 41 and 43), though a proof of that conjecture remains elusive.
Despite these historic findings, some seemingly inevitable and obvious discoveries have remained hidden. There is no formula to use to determine if a very large number is prime. There is no formula
to determine the “next” prime number, or how many prime numbers might exist below a certain number (there’s a logarithmic formula that only estimates this quantity). Generally, it had been agreed
that primes behave randomly and unpredictably.
It had been agreed that primes behave randomly and unpredictably.
Every so often, news about prime numbers pops up on the wire services. Frequently, it is because mathematicians have found the “next” largest known prime number (the current record holder has more
than twenty-two million digits). In 2013, little known mathematician Yitang Zhang proved a groundbreaking result that mathematicians feel is the first step toward finalizing the “twin prime” proof
mentioned earlier.
And in 2016, Robert Lemke Oliver and Kannan Soundararajan discovered something “really, really bizarre” that “floored” fellow mathematicians.
The Conspiracy
All prime numbers with two or more digits end in 1, 3, 7, or 9. Due to their unpredictability, prime numbers behave in so many ways like a group of random numbers, so we’d expect the final digit of
the consecutive primes to be randomly spread out.
For example, for all primes that end in 9, we’d expect around 25% of each of the following primes to end in 1, 3, 7, and 9, respectively. That’s how randomness works. But Soundararajan and Oliver
discovered that this is far from the case for the first billion primes.
Consecutive primes tend not to end in the same digit. A prime that ends in 9 is 65% more likely to be followed by a prime that ends in 1 than another prime that ends in 9. Similar percentages hold
true for each of the other ending digits of prime numbers. For centuries, the assumption was that primes behave randomly, and they just don’t.
At all.
For centuries, the assumption was that primes behave randomly, and they just don’t.
Clearly this discovery was only possible in the age of computing, which allowed mathematicians to consider the final digits of hundreds of millions of primes. That would not have been possible in the
age of paper-and-pencil computations—the size of prime numbers gets unwieldy in a big hurry. Still, given the computing power, this was a relatively simple exercise in counting and sorting. A middle
schooler can understand the results. So why had mathematicians not discovered this before 2016?
When considering this question, I remembered a quote from one of my favorite novels, The Curious Incident of the Dog in the Night-Time: The main character Christopher, a math savant, says that “. . .
prime numbers are what is left when you have taken all the patterns away. I think prime numbers are like life.”
This isn’t a very optimistic view of life, but it’s one I’m frequently tempted to adopt. Much like prime numbers, life is complex, unpredictable, and seemingly full of uncertainty. Just as prime
numbers make up the basis of our entire system of numbers, my Christian faith makes up the basis for everything else I believe.
Much like prime numbers, life is complex, unpredictable, and seemingly full of uncertainty.
But my Christian faith at times feels unstable, shaky, random, and chaotic. I feel at times like I am expected to have faith in situations and circumstances where I just don’t see any practical
reasons to have faith. What is supposed to be foundational and supportive in my life can at times feel as random as mathematicians long believed prime numbers to be.
Looking for Simplicity
How do I typically respond in situations of doubt and uncertainty in my faith? I go cosmic. I look for a big-picture, comprehensive understanding of my situation. Like the mathematicians who studied
prime numbers, sometimes this thought process leads to success and I discover something useful about myself, my faith, or my Creator.
A Mersenne formula for my life, if you will—a formula for my beliefs that works on occasions, but just as frequently doesn’t lead to success. Just as often as the formula works, I end up completely
frustrated; the big-picture view for my life seems exasperatingly out of reach, and I’m stuck with a random collection of events that feels like it should have a pattern or predictability to it but
just doesn’t.
Mathematicians overlooked a seemingly obvious trait of prime numbers for decades. While they were looking for an overarching formula that predicted prime numbers, proof to some existing conjectures,
or had given up on the topic entirely, prime numbers were quietly whispering the entire time, “Psst. We don’t like to repeat our final digit. We don’t behave randomly.”
Who knows what discoveries this could lead to in the future? It may lead to very little. It may lead to some exciting new bridge between number theory and the world of quantum physics. There’s a lot
of work to be done, but the discovery has given a starting point to study prime numbers from an entirely new perspective. Sometimes the most interesting and unexpected discoveries occur on a much
simpler level than where we were originally looking for answers.
The same can be true in my faith. While I’m casting about the Bible looking for a verse that lays out the meaning of my entire life, God whispers, “Psst. You are loved. Rest in that a while.” That
relatively simple answer may be all I need to get a new perspective on the doubts and confusion in my life. It’s a new path to solutions, giving a new approach to the questions I face.
Christopher was right: Prime numbers are like life. We’re still learning about them both, and if we’re willing to look for simpler solutions, we might be surprised at what we can learn. | {"url":"https://www.fathommag.com/stories/the-unexpected-biases-of-primes","timestamp":"2024-11-06T14:30:09Z","content_type":"text/html","content_length":"30651","record_id":"<urn:uuid:1b0d8114-41a3-4c8d-87ed-109c79baaaf1>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00679.warc.gz"} |
Geographic Data Science with R: Visualizing and Analyzing Environmental Change - Free Computer, Programming, Mathematics, Technical Books, Lecture Notes and Tutorials
Geographic Data Science with R: Visualizing and Analyzing Environmental Change
• Title: Geographic Data Science with R: Visualizing and Analyzing Environmental Change
• Author(s): Michael C. Wimberly
• Publisher: Chapman and Hall/CRC; 1st edition (May 8, 2023); eBook (Creative Commons Licensed)
• License(s): Creative Commons License (CC)
• Paperback/Hardcover: 284 pages
• eBook: HTML
• Language: English
• ISBN-10: 1032347716
• ISBN-13: 978-1032347714
• Share This:
Book Description
This book provides a series of tutorials aimed at teaching good practices for using Time Series and geospatial data to address topics related to environmental change. It is based on the R language
and environment, which currently provides the best option for working with diverse sources of spatial and non-spatial data using a single platform.
About the Authors
• Dr. Michael Wimberly is a Professor in the Department of Geography and Environmental Sustainability at the University of Oklahoma.
Reviews, Ratings, and Recommendations: Related Book Categories: Read and Download Links: Similar Books:
Book Categories
Other Categories
Resources and Links | {"url":"https://freecomputerbooks.com/Geographic-Data-Science-with-R.html","timestamp":"2024-11-09T07:56:42Z","content_type":"application/xhtml+xml","content_length":"36970","record_id":"<urn:uuid:2896321f-ae95-4027-bed0-10e5aca886e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00611.warc.gz"} |
Trigonometry Formula, Tricks, Identities, Trigonometric Ratio
Trigonometry Formula, Tricks, Identities, Trigonometric Ratio
Explore tricks for trigonometry formulas and delve into trigonometric ratios in the article below. Aspiring candidates should review these essential details on trigonometric strategies.
Trigonometry Formula: Trigonometry is a mathematical discipline that examines the relationships between the sides and angles of triangles. It has diverse applications in fields like satellite
navigation, astronomy, and geography, especially when determining distances using the triangulation method.
Trigonometry offers a wide range of formulas that can address various problems. These include Pythagorean identities, product identities, and trigonometric ratios like sin, cos, tan, sec, cosec, and
cot. Additionally, there are co-function identities related to angle shifts, the signs of ratios across different quadrants, and identities for summing/differing angles, as well as double and
half-angle identities.
By mastering these trigonometric formulas, high school students in grades 10 to 12 can excel in this topic. Furthermore, understanding inverse trigonometry formulas and referring to trigonometric
tables can aid them in solving related problems.
Trigonometry Formulas
Six trigonometric functions are available: Sin, Cos, Tan, Sec, Cosec, and Cot. The current length and angle may be derived with the use of trigonometric ratios. These 6 functions serve as the
foundation for all trigonometry formulae, formula tricks, and problems. The specifics of trigonometry, including the formulae, trigonometry formula tricks, and problems, are available to aspirants.
Trigonometry-related questions can be found in a variety of competitive exams, including SSC, Railway, and others. You may get notes about formula trigonometry that are helpful for exams in this
site. It will assist you in learning the fundamental trigonometry formulae. The complete list of trigonometry formulae for classes 10 and 11, as well as trigonometry formulas for classes 12 and 13,
is provided below.
In order to solve numerous trigonometric puzzles and comprehend how triangle sides and angles relate to one another, trigonometry formulae are essential. We will go through some of the basic
trigonometric formulae here:
Pythagorean Theorem:
The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Mathematically, it can be represented as: a² + b² = c² where a and b are the lengths of the two legs of the triangle, and c is the length of the hypotenuse.
Trigonometric Ratios:
Trigonometric ratios relate the angles of a right triangle to the lengths of its sides. The primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan):
• Sine (sin)-> sinθ = Opposite/Hypotenuse
• Cosine (cos)-> cosθ = Adjacent/Hypotenuse
• Tangent (tan)-> tanθ = Opposite/Adjacent
Reciprocal Trigonometric Ratios:
The reciprocal trigonometric ratios are derived from the primary trigonometric ratios and relate the angles of a right triangle to the lengths of its sides:
• Cosecant (cosec)-> cosecθ = 1/sinθ
• Secant (sec)-> secθ = 1/cosθ
• Cotangent (cot)-> cotθ = 1/tanθ
Angle Sum and Difference Formulas:
These formulas express the trigonometric functions of the sum or difference of two angles in terms of the trigonometric functions of the individual angles:
• sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B)
• cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B)
• tan(A ± B) = (tan(A) ± tan(B))/(1 ∓ tan(A)tan(B))
Double Angle Formulas:
Double angle formulas relate the trigonometric functions of an angle to the trigonometric functions of double that angle:
• sin(2θ) = 2sin(θ)cos(θ)
• cos(2θ) = cos²(θ) – sin²(θ) = 2cos²(θ) – 1 = 1 – 2sin²(θ)
• tan(2θ) = 2tan(θ) / (1 – tan²(θ))
Half Angle Formulas:
Half angle formulas provide relationships between the trigonometric functions of an angle and the trigonometric functions of half that angle:
• sin(θ/2) = ±√[(1 – cos(θ)) / 2]
• cos(θ/2) = ±√[(1 + cos(θ)) / 2]
• tan(θ/2) = ±√[(1 – cos(θ)) / (1 + cos(θ))]
Sum to Product and Product to Sum Formulas:
These formulas convert sums or differences of trigonometric functions into products, and vice versa:
Sum to Product Formulas:
• sin(A) + sin(B) = 2sin[(A + B)/2]cos[(A – B)/2]
• sin(A) – sin(B) = 2cos[(A + B)/2]sin[(A – B)/2]
• cos(A) + cos(B) = 2cos[(A + B)/2]cos[(A – B)/2]
• cos(A) – cos(B) = -2sin[(A + B)/2]sin[(A – B)/2]
Product to Sum Formulas:
• sin(A)sin(B) = (1/2)[cos(A – B) – cos(A + B)]
• cos(A)cos(B) = (1/2)[cos(A – B) + cos(A + B)]
• sin(A)cos(B) = (1/2)[sin(A + B) + sin(A – B)]
Just a handful of the basic trigonometric formulae are included here. These fundamentals serve as the foundation for several more complex formulae and identities. You’ll be able to tackle a variety
of trigonometric issues with ease and precision once you’ve mastered these formulae.
Trigonometry Tricks
Trigonometry tricks are helpful techniques and shortcuts that can simplify calculations, solve problems more efficiently, and aid in memorizing key concepts. Here are some useful trigonometry tricks:
1. Unit Circle: The unit circle is a valuable tool in trigonometry. By memorizing the coordinates of the angles (0°, 30°, 45°, 60°, and 90°) on the unit circle, you can quickly determine the values
of sine, cosine, and tangent for these angles without using a calculator.
2. Special Triangles: Special triangles, such as the 45-45-90 triangle and the 30-60-90 triangle, have well-defined ratios that make calculations easier. For the 45-45-90 triangle, the sides are in
the ratio 1:1:√2, while for the 30-60-90 triangle, the sides are in the ratio 1:√3:2. These ratios can simplify calculations involving these angles.
3. Symmetry: Trigonometric functions exhibit symmetry properties that can be exploited. For example, sine and cosecant have odd symmetry, while cosine and secant have even symmetry. Tangent and
cotangent also have odd symmetry. Leveraging these symmetries can help simplify calculations by reducing the number of calculations required.
4. Co-Function Identities: Co-function identities relate the trigonometric functions of complementary angles. Complementary angles are two angles that add up to 90 degrees (or π/2 radians). For
example, sin(Ï€/2 – θ) is equal to cos(θ), and cos(Ï€/2 – θ) is equal to sin(θ). These identities can be useful for interchanging trigonometric functions and simplifying expressions.
5. Radian-Degree Conversion: To convert between degrees and radians, you can use the fact that 180 degrees is equal to π radians. This conversion can be handy when working with both degrees and
radians in a problem.
6. Sum and Difference Formulas: The sum and difference formulas allow you to find the trigonometric values of the sum or difference of two angles. These formulas can help simplify complex
trigonometric expressions. For example, the sine of the sum of two angles can be expressed as sin(A + B) = sin(A)cos(B) + cos(A)sin(B).
7. Even-Odd Identities: Even-odd identities describe the symmetry of trigonometric functions. For example, sin(-θ) = -sin(θ) and cos(-θ) = cos(θ). These identities can be useful for evaluating
trigonometric functions of negative angles.
8. Periodicity: Trigonometric functions are periodic, meaning they repeat their values after certain intervals. For example, sine and cosine have a period of 2Ï€ radians or 360 degrees.
Understanding the periodic nature of trigonometric functions can help simplify calculations and determine values for angles outside the standard range.
9. Trigonometric Equations: When solving trigonometric equations, it is often helpful to use identities or manipulate the equation to simplify it into a more manageable form. This may involve
factoring, canceling common terms, or applying trigonometric identities.
10. Visualization and Drawing: Drawing triangles or visualizing the angles and sides of a problem can aid in understanding and solving trigonometric problems. Visualization can provide geometric
insights and help identify relationships between different parts of a triangle or problem.
11. These trigonometry tips can improve problem-solving skills, speed up computations, and give shortcuts. To use trigonometry more effectively, it is crucial to practice and get familiar with these
Trignometric Identities
Trigonometric identities are mathematical equations that establish relationships between trigonometric functions. These identities are derived from basic geometric principles and play a crucial role
in simplifying expressions, verifying equations, and solving trigonometric problems. Here are some important trigonometric identities:
Reciprocal Identities:
Reciprocal identities express the reciprocal trigonometric functions in terms of their counterparts. These identities are useful for converting between trigonometric functions:
• csc(theta) = 1/sin(theta)
• sec(theta) = 1/cos(theta)
• cot(theta) = 1/tan(theta)
Quotient Identities:
Quotient identities relate the trigonometric ratios of one angle to the ratios of other trigonometric functions. These identities are helpful for expressing trigonometric functions in terms of each
• tan(theta) = sin(theta)/cos(theta)
• cot(theta) = cos(theta)/sin(theta)
Pythagorean Identities: Pythagorean identities involve the Pythagorean Theorem and establish relationships between the squares of trigonometric functions. The most well-known Pythagorean identity is:
sin^2(theta) + cos^2(theta) = 1
Other Pythagorean identities can be derived from this fundamental equation:
1 + tan^2(theta) = sec^2(theta)
1 + cot^2(theta) = csc^2(theta)
Co-Function Identities: Co-function identities relate the trigonometric functions of complementary angles. Complementary angles are two angles whose sum is 90 degrees or π/2 radians:
• sin(Ï€/2 – theta) = cos(theta)
• cos(Ï€/2 – theta) = sin(theta)
• tan(Ï€/2 – theta) = cot(theta)
• cot(Ï€/2 – theta) = tan(theta)
• sec(Ï€/2 – theta) = csc(theta)
• csc(Ï€/2 – theta) = sec(theta)
These identities can be useful for simplifying expressions involving complementary angles.
Even-Odd Identities: Even-odd identities describe the symmetry properties of trigonometric functions. The even-odd identities are as follows:
• sin(-theta) = -sin(theta)
• cos(-theta) = cos(theta)
• tan(-theta) = -tan(theta)
• csc(-theta) = -csc(theta)
• sec(-theta) = sec(theta)
• cot(-theta) = -cot(theta)
These identities show that the sine, tangent, and cotangent functions are odd functions, while the cosine, secant, and cosecant functions are even functions.
Sum and Difference Identities:
Sum and difference identities allow us to express the trigonometric functions of the sum or difference of two angles in terms of the functions of the individual angles. Some of the important sum and
difference identities are:
• sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B)
• cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B)
• tan(A ± B) = (tan(A) ± tan(B))/(1 ∓ tan(A)tan(B))
These identities are useful in simplifying trigonometric expressions and solving trigonometric equations involving sums or differences of angles.
Double Angle Identities:
Double angle identities relate the trigonometric functions of an angle to the trigonometric functions of double that angle. These identities are derived from the sum and difference identities. Some
of the key double angle identities are:
• sin(2theta) = 2sin(theta)cos(theta)
• cos(2theta) = cos^2(theta) – sin^2(theta) = 2cos^2(theta) – 1 = 1 – 2sin^2(theta)
• tan(2theta) = 2tan(theta)/(1 – tan^2(theta))
Double angle identities are particularly useful for simplifying expressions involving double angles.
Just a handful of the crucial trigonometric identities are listed here. You may simplify trigonometric expressions, check equations, and resolve a variety of trigonometric issues by applying these
identities and comprehending their relationships. You will become more adept at trigonometry if you put these identities to use and get familiar with them.
Trigonometric Ratio
Trigonometric ratios relate the angles of a right triangle to the lengths of its sides. Understanding these ratios is fundamental in trigonometry. Here are the primary trigonometric ratios:
1. Sine (sin): Sine is the ratio of the length of the side opposite the angle (O) to the length of the hypotenuse (H) in a right triangle. Mathematically, it can be expressed as: sin(theta) = O/H
2. Cosine (cos): Cosine is the ratio of the length of the side adjacent to the angle (A) to the length of the hypotenuse (H) in a right triangle. Mathematically, it can be expressed as: cos(theta) =
3. Tangent (tan): Tangent is the ratio of the length of the side opposite the angle (O) to the length of the side adjacent to the angle (A) in a right triangle. Mathematically, it can be expressed
as: tan(theta) = O/A
4. It’s important to note that these ratios are specific to right triangles and only applicable within the range of 0 to 90 degrees (or 0 to Ï€/2 radians).
Additional Trigonometric Ratios:
1. Cosecant (csc): Cosecant is the reciprocal of sine. It is defined as the ratio of the length of the hypotenuse (H) to the length of the side opposite the angle (O) in a right triangle. csc(theta)
= 1/sin(theta) = H/O
2. Secant (sec): Secant is the reciprocal of cosine. It is defined as the ratio of the length of the hypotenuse (H) to the length of the side adjacent to the angle (A) in a right triangle. sec
(theta) = 1/cos(theta) = H/A
3. Cotangent (cot): Cotangent is the reciprocal of tangent. It is defined as the ratio of the length of the side adjacent to the angle (A) to the length of the side opposite the angle (O) in a right
triangle. cot(theta) = 1/tan(theta) = A/O
These trigonometric ratios are essential for calculating unknown angles or side lengths in a right triangle, as well as for solving various trigonometric problems. They provide a mathematical
relationship between the angles and sides of a triangle, allowing for precise calculations and analysis.
Trigonometric ratios also have many practical applications beyond pure mathematics. They are utilized in fields such as physics, engineering, surveying, and navigation. By understanding these ratios,
you can determine angles, distances, heights, and other relevant measurements in real-world scenarios.
Important Concept to Solve a Specific Type of Question
If A + B = 90°
Results that are true always :
(i) sin A. sec B = 1 or sin A = cos B
(ii) cos A. cosec B = 1 or sec A = cosec B
(iii) tan A. tan B = 1 or tan A = cot B
(iv) cot A. cot B = 1
(v) sin²A + sin² B = 1
(vi) cos² A + cos² B = 1
Important Trigonometry Formula for Sum and Difference Of Two Angles
Trigonometry formulas for the sum and difference of two angles play a crucial role in simplifying expressions and solving trigonometric equations. Here are some important formulas for the sum and
difference of two angles:
Sum of Two Angles:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
cos(A + B) = cos(A)cos(B) – sin(A)sin(B)
tan(A + B) = (tan(A) + tan(B)) / (1 – tan(A)tan(B))
Difference of Two Angles:
• sin(A – B) = sin(A)cos(B) – cos(A)sin(B)
• cos(A – B) = cos(A)cos(B) + sin(A)sin(B)
• tan(A – B) = (tan(A) – tan(B)) / (1 + tan(A)tan(B))
These formulas can be derived using the identities and properties of trigonometric functions.
Co-Function Formulas:
Co-function formulas express the trigonometric functions of complementary angles in terms of each other. The complementary angle of theta is (Ï€/2) – theta. Using these formulas, we can derive the
sum and difference formulas for sine and cosine:
sin(A + B) = cos(A’ – B’) = cos(A’)cos(B’) – sin(A’)sin(B’)
cos(A + B) = cos(A’)cos(B’) + sin(A’)sin(B’) = sin(A’ – B’) where A’ = (Ï€/2) – A and B’ = (Ï€/2) – B.
Double Angle Formulas: Double angle formulas relate the trigonometric functions of an angle to the trigonometric functions of double that angle. By setting A = B in the sum and difference formulas,
we obtain the double angle formulas:
• sin(2A) = 2sin(A)cos(A)
• cos(2A) = cos^2(A) – sin^2(A) = 2cos^2(A) – 1 = 1 – 2sin^2(A)
• tan(2A) = (2tan(A)) / (1 – tan^2(A))
These formulas are useful when dealing with angles that are twice the size of a given angle.
Half Angle Formulas: Half angle formulas express the trigonometric functions of an angle in terms of half that angle. These formulas can be derived from the double angle formulas:
• sin(A/2) = ±√[(1 – cos(A)) / 2]
• cos(A/2) = ±√[(1 + cos(A)) / 2]
• tan(A/2) = ±√[(1 – cos(A)) / (1 + cos(A))]
The ± sign indicates that there are two possible values for the functions, depending on the quadrant in which the angle lies.
These formulas for the sum and difference of two angles are essential in various trigonometric calculations, simplifying expressions, and solving trigonometric equations. Understanding and applying
these formulas can greatly enhance your proficiency in trigonometry and enable you to tackle a wide range of problems involving angles.
Trigonometry Formulas For Tangent
Trigonometry formulas for tangent (tan) involve expressing the tangent of an angle in terms of other trigonometric functions. Here are some important formulas for tangent:
1. Tangent Definition: tan(theta) = sin(theta) / cos(theta) This formula defines tangent as the ratio of the sine of an angle to the cosine of the same angle.
2. Pythagorean Identity: sin^2(theta) + cos^2(theta) = 1 Dividing both sides of the equation by cos^2(theta), we get: tan^2(theta) + 1 = sec^2(theta) This formula relates the tangent to the secant
of an angle.
3. Tangent in Terms of Sine and Cosine: tan(theta) = sin(theta) / cos(theta) This is the basic definition of tangent.
4. Reciprocal Identity: cot(theta) = 1 / tan(theta) This formula shows the reciprocal relationship between the tangent and cotangent functions.
5. Tangent of a Sum or Difference of Angles: tan(A + B) = (tan(A) + tan(B)) / (1 – tan(A)tan(B)) tan(A – B) = (tan(A) – tan(B)) / (1 + tan(A)tan(B)) These formulas express the tangent of the sum or
difference of two angles in terms of the tangents of the individual angles.
6. Tangent of Half-Angle: tan(theta/2) = (1 – cos(theta)) / sin(theta) This formula relates the tangent of half an angle to the cosine and sine of the original angle.
7. Tangent of Double Angle: tan(2theta) = 2tan(theta) / (1 – tan^2(theta)) This formula expresses the tangent of a double angle in terms of the tangent of the original angle.
8. Tangent of Triple Angle: tan(3theta) = (3tan(theta) – tan^3(theta)) / (1 – 3tan^2(theta)) This formula calculates the tangent of a triple angle using the tangent of the original angle.
9. Tangent of Difference of Squares: tan(A + B)tan(A – B) = [tan(A) + tan(B)][tan(A) – tan(B)] This formula relates the tangent of the sum and difference of two angles to their individual tangents.
These formulas provide valuable insights into the properties of tangent and allow for calculations involving angles and trigonometric functions. They can be used to simplify expressions, solve
trigonometric equations, and derive relationships between different angles. Familiarizing yourself with these formulas will enhance your understanding of trigonometry and enable you to solve a wide
range of problems involving tangent.
Trigonometry Formulas List
Pythagorean Theorem:
a² + b² = c² This formula relates the lengths of the sides of a right triangle, where a and b are the lengths of the two legs and c is the length of the hypotenuse.
Trigonometric Ratios:
Sine (sin): sin(theta) = opposite/hypotenuse
Cosine (cos): cos(theta) = adjacent/hypotenuse
Tangent (tan): tan(theta) = opposite/adjacent These ratios define the relationship between the angles and sides of a right triangle.
Reciprocal Trigonometric Ratios:
Cosecant (csc): csc(theta) = 1/sin(theta)
Secant (sec): sec(theta) = 1/cos(theta)
Cotangent (cot): cot(theta) = 1/tan(theta) These ratios are the reciprocals of the sine, cosine, and tangent functions.
Trigonometric Identities:
Pythagorean Identity: sin²(theta) + cos²(theta) = 1
Reciprocal Identities: csc(theta) = 1/sin(theta), sec(theta) = 1/cos(theta), cot(theta) = 1/tan(theta)
Quotient Identities: tan(theta) = sin(theta)/cos(theta), cot(theta) = cos(theta)/sin(theta)
Co-Function Identities: sin(Ï€/2 – theta) = cos(theta), cos(Ï€/2 – theta) = sin(theta), tan(Ï€/2 – theta) = cot(theta), csc(Ï€/2 – theta) = sec(theta), sec(Ï€/2 – theta) = csc(theta), cot(Ï€/2 –
theta) = tan(theta)
Even-Odd Identities: sin(-theta) = -sin(theta), cos(-theta) = cos(theta), tan(-theta) = -tan(theta), csc(-theta) = -csc(theta), sec(-theta) = sec(theta), cot(-theta) = -cot(theta)
Sum and Difference Formulas:
sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B)
cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B)
tan(A ± B) = (tan(A) ± tan(B))/(1 ∓ tan(A)tan(B))
Double Angle Formulas:
sin(2θ) = 2sinθcosθ
cos(2θ) = cos²Î¸ – sin²Î¸ = 2cos²Î¸- 1 = 1 – 2sin²Î¸
tan(2θ) = 2tanθ / (1 – tan²Î¸)
Half Angle Formulas:
sin(θ/2) = ±√[(1 – cosθ) / 2]
cos(θ/2) = ±√[(1 + cosθ) / 2]
tan(θ/2) = ±√[(1 – cosθ) / (1 + cosθ)]
Product to Sum Formulas:
sin(A)sin(B) = (1/2)[cos(A – B) – cos(A + B)]
cos(A)cos(B) = (1/2)[cos(A – B) + cos(A + B)]
sin(A)cos(B) = (1/2)[sin(A + B) + sin(A – B)]
Sum to Product Formulas:
sin(A) + sin(B) = 2sin[(A + B)/2]cos[(A – B)/2]
sin(A) – sin(B) = 2cos[(A + B)/2]sin[(A – B)/2]
cos(A) + cos(B) = 2cos[(A + B)/2]cos[(A – B)/2]
cos(A) – cos(B) = -2sin[(A + B)/2]sin[(A – B)/2]
These are just some of the important trigonometry formulas. There are many more advanced formulas and identities, but these should give you a solid foundation for working with trigonometry.
Trigonometry Maximum & Minimum Value
In trigonometry, the maximum and minimum values of trigonometric functions depend on the range of the angles being considered. Here are the maximum and minimum values for the primary trigonometric
Sine (sin):
Maximum value: The sine function reaches its maximum value of 1 at 90 degrees (Ï€/2 radians).
Minimum value: The sine function reaches its minimum value of -1 at 270 degrees (3Ï€/2 radians).
Cosine (cos):
Maximum value: The cosine function reaches its maximum value of 1 at 0 degrees (0 radians).
Minimum value: The cosine function reaches its minimum value of -1 at 180 degrees (Ï€ radians).
Tangent (tan):
There is no maximum or minimum value for the tangent function since it is unbounded. As the tangent function approaches certain values of the angle, it tends towards positive or negative infinity.
It’s important to note that these maximum and minimum values apply to the standard range of angles (0 to 360 degrees or 0 to 2Ï€ radians). If angles outside this range or specific restrictions are
given, the maximum and minimum values may differ.
Additionally, it’s worth mentioning that the maximum and minimum values can change depending on the units used (degrees or radians) and whether the angles are measured in the positive or negative
Understanding the maximum and minimum values of trigonometric functions helps in analyzing and graphing these functions, solving trigonometric equations, and determining the behavior of angles within
specific intervals. | {"url":"https://sscegy.testegy.com/2023/11/trigonometry-formulas.html","timestamp":"2024-11-12T02:12:59Z","content_type":"text/html","content_length":"1049064","record_id":"<urn:uuid:7b22d971-d207-44bd-867a-0265a8c6a31a>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00878.warc.gz"} |
Move from Excel to Python with Pandas
Move from Excel to Python with Pandas Transcripts
Chapter: Appendix: Python language concepts
Lecture: Concept: lambdas
0:01 In Python, functions are first class citizens, and what that means is they are represented by a class instances of them,
0:08 particular functions are objects they can be passed around just like other custom types you create just like built-in types, like strings and numbers.
0:17 So we are going to leverage that fact in a simple little bit of code I have here called find significant numbers.
0:22 Now, maybe we want to look for all even numbers, all odd numbers, all prime numbers, any of those sorts of things.
0:28 But this function is written to allow you to specify what it means for a number to be significant, so you can reuse this finding functionality
0:37 but what determines significance is variable, it could be specified by multiple functions being passed in and that's what we are calling predicate
0:46 because this ability to pass functions around and create and use them in different ways especially as parameters or parts of expressions,
0:54 Python has this concept of lambdas. So let's explore this by starting with some numbers, here we have the Fibonacci numbers
1:01 and maybe we want to find just the odd Fibonacci numbers. So we can start with the sequence and we can use this "find significant numbers" thing
1:08 along with the special test method we can write the checks for odd numbers. So, in Python we can write this like so,
1:15 and we can say the significant numbers we are looking for is... call the function, pass the number set we want to filter on
1:20 and then we can write this lambda expression instead of creating the whole new function. So instead of above having the def and a separate block
1:27 and all that kind of stuff, we can just inline a little bit of code, so we indicate this by saying lambda and then we say the parameters,
1:35 there can be zero, one or many parameters, here we just have one called x, and we say colon to define the block that we want to run,
1:43 and we set the expression that we are going to return when this function is called,
1:47 we don't use the return keyword we just say when you call this function
1:50 here is the thing that it does in return, so we are doing a little test, True or False,
1:54 and we ask "if x % 2 == 1" that's all the odd numbers, not the even ones, so when we run this code it loops over all the Fibonacci numbers
2:02 runs a test for oddness and it pulls out as you can see below just the odd ones, for example 8 is not in there. | {"url":"https://training.talkpython.fm/courses/transcript/move-from-excel-to-python-and-pandas/lecture/271023","timestamp":"2024-11-05T12:29:29Z","content_type":"text/html","content_length":"25821","record_id":"<urn:uuid:c075d971-ecba-4722-87f3-847b5104f0fb>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00022.warc.gz"} |
An Etymological Dictionary of Astronomy and Astrophysics
chi-square distribution
واباژش ِخی-دو
vâbâžeš-e Xi-do
Fr.: loi du chi-deux
A probability density function, denoted χ^2, that gives the distribution of the sum of squares of k independent random variables, each being drawn from the normal distribution with zero mean and unit
variance. The integer k is the number of degrees of freedom. The distribution has a positive skew; the skew is less with more degrees of freedom. As degrees of freedom increase, the chi-square
distribution approaches a normal distribution. The most common application is chi-square tests for goodness of fit of an observed distribution to a theoretical one. If χ^2 = 0 the agreement is
Chi Gk. letter of alphabet; → square; → distribution.
Vâbâžeš, → distribution; do, → two.
inverse square law
قانون ِتوان ِدوی ِوارون، قانون ِچاروش ِوارون
qânun-e tavân-e do-ye vârun, qânun-e câruš-e vârun
Fr.: loi en carré inverse
A force law that applies to the → gravitational and → electromagnetic forces in which the magnitude of the force decreases in proportion to the inverse of the square of the → distance.
→ inverse; → square; → law.
least squares
کوچکترین چاروشها
kucaktarin cârušhâ
Fr.: moindres carrés
Any statistical procedure that involves minimizing the sum of squared differences.
→ least; → square.
least-squares deconvolution (LSD)
واهماگیش ِکمترین چاروشها
vâhamâgiš-e kucaktarin cârušhâ
Fr.: déconvolution des moindres carrés
A → cross correlation technique for computing average profiles from thousands of → spectral lines simultaneously. The technique, first introduced by Donati et al. (1997, MNRAS 291,658), is based on
several assumptions: additive → line profiles, wavelength independent → limb darkening, self-similar local profile shape, and weak → magnetic fields. Thus, unpolarized/polarized stellar spectra can
indeed be seen as a line pattern → convolved with an average line profile. In this context, extracting this average line profile amounts to a linear → deconvolution problem. The method treats it as a
matrix problem and look for the → least squares solution. In practice, LSD is very similar to most other cross-correlation techniques, though slightly more sophisticated in the sense that it cleans
the cross-correlation profile from the autocorrelation profile of the line pattern. The technique is used to investigate the physical processes that take place in stellar atmospheres and that affect
all spectral line profiles in a similar way. This includes the study of line profile variations (LPV) caused by orbital motion of the star and/or stellar surface inhomogeneities, for example.
However, its widest application nowadays is the detection of weak magnetic fields in stars over the entire → H-R diagram based on → Stokes parameter V (→ circular polarization) observations (see also
Tkachenko et al., 2013, A&A 560, A37 and references therein).
→ least; → square; → deconvolution.
least-squares fit
سز ِکوچکترین چاروشها
saz-e kucaktarin cârušhâ
Fr.: ajustement moindres carrées
A fit through data points using least squares.
→ least squares; → fit.
magic square
چاروش ِجادو
câruš-e jâdu
Fr.: carré magique
An n × n matrix in which every row, column, and diagonal add up to the same number.
→ magic; → square.
method of least squares
روش ِکمترین چاروشها
raveš-e kamtarin cârušhâ
Fr.: méthode des moindres carrés
A method of fitting a curve to data points so as to minimize the sum of the squares of the distances of the points from the curve.
→ method; → least squares.
perfect square
چاروش ِفرساخت
câruš-e farsâxt
Fr.: carré parfait
An → integer of the form n^2, where n is a → positive number. In other words, a → perfect power when k = 2.
→ perfect; → square.
root mean square (rms)
ریشهی ِچاروشی ِمیانگین، ~ ِدوم ِ~
riše-ye câruši-ye miyângin, ~ dovom-e ~
Fr.: valeur quadratique moyenne
The square root of the arithmetic mean of the squares of the numbers in a given set.
→ root; → mean; → square.
root-mean-square error
ایرنگ ِریشهی ِچاروشی ِمیانگین، ~ ~ ِدوم ِ~
irang-e riše-ye câruši-ye miyângin, ~ ~ dovom-e ~
The square root of the second moment corresponding to the frequency function of a random variable.
→ root; → mean; → square; → error.
root-mean-square value
ارزش ِریشهی ِچاروشی ِمیانگین
arzeš-e riše-ye câruši-ye miyângin
Fr.: écart quadratique moyen, écart type
Statistics: The square root of the arithmetic mean of the squares of the deviation of observed values from their arithmetic mean.
→ root; → mean; → square; → deviation.
چاروش، چهارگوش
câruš, cahârguš
Fr.: carré
1) A rectangle having all four sides of equal length.
2) The second power of a quantity, expressed as a^2 = a × a, where a is the quantity. → inverse square law.
M.E., from O.Fr. esquire "a square, squareness," from V.L. *exquadra, from *exquadrare "to square," from L. → ex- "out" + quadrare "make square," from quadrus "a square," from quattuor→ four.
Câruš, from Av. caθruša- "four sides (of a four-sided figure)", from caθru- "four," Mod.Pers. cahâr, câr "four" + uša- "angle," Mod.Pers. guš, gušé.
square degree
درجهی ِچاروش
daraje-ye câruš
Fr.: degré carré
A solid angle whose cone is a tetrahedral pyramid with an angle between its edges equal to 1°. 1 square degree = 3.046 x 10^-4 sr = 2.424 x 10^-5 solid angle of a complete sphere.
→ square; → degree.
Square Kilometer Array (SKA)
Fr.: SKA
An international project to construct a highly sensitive radio interferometer array operating between 0.15 and 20 GHz with an effective collecting area of one square kilometer. The number of
individual telescopes will be 2000 to 3000. SKA will have a sensitivity 100 times higher than that of today's best radio telescopes and an angular resolution < 0.1 arcsec at 1.4 GHz. The site will be
selected in 2012 and early science with Phase 1 is scheduled for from 2016 on. See also the SKA homepage.
→ square; → kilometer; → array.
square matrix
ماتریس ِچاروش
matris-e câruš
Fr.: matrice carée
A → matrix with equal numbers of → rows and → columns (i.e., an n × n matrix).
→ square; → matrix.
Square of Pegasus
چهارگوش ِپگاسوس
Chahârguš-e Pegasus
Fr.: Carrée de Pégase
A large → asterism of four stars, approximately square in shape, in the northern sky. Three of the stars, → → Markab, → Scheat, and → Algenib, belong to the constellation → Pegasus. The fourth, →
Alpheratz, was lost to Pegasus when the constellation boundaries were formalised, and now lies just within the borders of → Andromeda.
→ square; → Pegasus.
Chahârguš, → tetragon; → Pegasus.
square root
ریشهی ِچاروش
riše-ye câruš
Fr.: racine carée
Quantity which when multiplied by itself produces another quantity.
→ square; → root.
square wave
موج ِچاروش
mowj-e câruš
Fr.: onde carrée
An oscillation which alternatively assumes, for equal lengths of time, one or two fixed values.
→ square; → wave.
squaring the square
چاروشش ِچاروش
cârušeš-e câruš
Fr.: quadrature du carré
The mathematical problem of subdividing a square into a number of smaller squares, all of different sizes.
→ square; → square. | {"url":"https://dictionary.obspm.fr/?showAll=1&formSearchTextfield=square","timestamp":"2024-11-11T18:04:19Z","content_type":"text/html","content_length":"30663","record_id":"<urn:uuid:1d6de272-4826-4177-982d-130274cbcc13>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00076.warc.gz"} |
Deductive proofs: from premises to conclusion, or from conclusion to premises?
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Deductive proofs: from premises to conclusion, or from conclusion to premises?
(Cross-posted at NewAPPS)
It is fair to say that the ‘received view’ about deductive inference, and about inference in general, is that it proceeds from premises to conclusion so as to produce new information (the conclusion)
from previously available information (the premises). It is this conception of deductive inference that gives rise to the so-called ‘scandal of deduction’, which concerns the apparent lack of
usefulness of a deductive inference, given that in a valid deductive inference the conclusion is already ‘contained’, in some sense or another, in the premises. This is also the conception of
inference underpinning e.g. Frege’s logicist project, and much (if not all) of the discussions in the philosophy of logic of the last many decades. (In fact, it is also the conception of deduction of
the most famous ‘deducer’ of all times, Sherlock Holmes.)
That an inference, and a deductive inference in particular, proceeds from premises to conclusion may appear to be such an obvious truism that no one in their sane mind would want to question it. But
is this really how it works when an agent is formulating a deductive argument, say a mathematical demonstration?
The contrast between (deductive) demonstration and calculation may be illuminating here. When calculating, one starts with some known parameters (say, the total number of candies and the number of
children among whom the candies have to be distributed) and seeks to determine the solution to a problem by determining the relevant unknown value (say, the number of candies each child will
receive). By analogy, one might say that the premises of a deductive inference are (like) the known parameters and the conclusion is (like) the unknown value.
Now, I’ve been struggling for years to make sense of this conception of deductive reasoning, but to no avail. It just doesn’t seem to do justice to how deductive arguments are in fact formulated and
used. So now I’ve decided to adopt a different starting point: what if, in a deductive argument, she who formulates the argument in fact proceeds from conclusion to premises? This may seem absurd at
first sight, but once you start thinking about it, it makes a lot of sense (or so I claim!).
Consider for example how mathematical proofs are formulated. Is it the case that the mathematician looks at e.g. the axioms of number theory, and then starts ‘playing around’ with them trying to
deduce non-trivial conclusions? I’m pretty sure everyone will agree with me that this is not how it works. Instead, mathematicians usually take conjectures as their starting point: Fermat’s last
theorem, the twin-prime conjecture, the ABC conjecture etc. Starting with the ‘conclusion’, they try to establish, by reverse-engineering as it were, which premises are required to establish the
conclusion, and by which argumentative paths. So in a sense, what is discovered in a mathematical proof is everything but the conclusion: instead, the mathematician discovers the necessary premises
and the proof itself. (Luis Carlos Pereira once suggested to me that, in terms of the analogy with calculation, the ‘unknown value’ in a proof is the proof itself.) Of course, there is a sense in
which the truth of the conclusion is ‘discovered’ (established) by means of the proof, but the content of the conclusion is what guides the mathematician in her search for the proof from the start.
Much of my thinking on these matters is yet again prompted by the close reading of the Prior Analytics that we are undertaking with our reading group in Groningen. As it turns out, the bulk of the
text, and in any case about half of Book A, is dedicated to techniques on how to find the necessary premises to establish a given conclusion, in particular finding the right ‘middle term’ for it.
Again, the starting point is the conclusion, and through reverse engineering, the required premises are found. (My post-doc Matt Duncombe is just finishing a terrific paper exactly on this aspect of
the Prior Analytics, in connection with the so-called scandal of deduction; if anyone is interested in reading the draft, perhaps he could be persuaded to share it in the near future.)
So how come the conception of deductive inference as going from premises to conclusion became so widespread? Here again, the dialogical conceptualization of deduction that I have been developing
seems to offer a plausible explanation. When a deductive argument is presented to opponent by proponent, proponent indeed starts with the premises, seeking to get opponent to grant them, and then
slowly but surely moves towards the conclusion, which opponent will be forced to grant if he has granted the premises and the intermediate inferential steps. Hence, from the perspective of opponent,
from-premises-to-conclusion is indeed the correct order of events in a deductive argument; however, from the perspective of proponent, the right order is from-conclusion-to-premises. In other words,
a proponent, or whoever formulates a deductive argument, always knows where she is heading.
Another way of conceptualizing this dichotomy is in terms of the good old distinction between context of justification and context of discovery: for justification, the right path is from premises to
conclusion; for discovery (of the proof), the right path is from conclusion to premises.
As it so happens, Descartes was already well aware of this fact when disdainfully commenting on ‘the logic of the Schools’ (he means scholastic logic, but my claim is that this applies to deductive
logic in general):
[T]he logic of the Schools […] is strictly speaking nothing but a dialectic which teaches ways of expounding to others what one already knows […] I mean instead the kind of logic which teaches us
to direct our reason with a view to discovering the truths of which we are ignorant. (Preface to French edition of the Principles of Philosophy, in (Descartes 1988, 186); emphasis added)
Well, Descartes, in this case deductive logic is not for you.
UPDATE: In my
Google Plus feed
, Timothy Gowers writes that mathematicians make use both of what he calls backwards reasoning (from conclusion to premises) and of forwards reasoning (from premises to conclusions). This seems
absolutely right to me, and a wise caveat to my overly-unifying claims in this post! So the right answer to the question in my title is:
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1. Perhaps what characterises the 'deductive journey' from premises to conclusion or from conclusion to premises is, nevertheless, unidirectional. By analogy, in physics one can evolve a system
forwards in time, and then one can reverse the procedure, and evolve the system 'backwards' in time. Despite the terminology, one is always evolving the system unidirectionally irrespective of
2. Yes, that's how it works. Math is done inductively; but presented (to students) deductively. This leaves the students -- even at the graduate level -- ignorant of how math is done.
A couple of historical illustrations. Newton invented calculus in the mid to late 1600's. The Principia dates from 1687, so take that as the official date if you like. Now, Newton well understood
that he could not make logical sense of the limit of the difference quotient (what he called the "fluxion"). It's an expression of the form 0/0, which makes no sense. But calculus worked,
spectacularly so. It wasn't till the work of Weirstrass and other's in the 1800's that the logical definition of the limit was finally arrived at; and not till Zermelo, in the early 1900's, that
we finally had a completely rigorous account of the real numbers and limiting processes starting from the axioms of set theory all the way up to calculus. This process took over 200 years!
But today, in real analysis class taught to math major undergrads, we start from the axioms of set theory, construct the real numbers, and then rigorously prove the basic theorems of calculus.
This is a complete inversion of how the subject was discovered and developed. But undergrads -- and, sorry to say, most philosophers -- take the axiomatic, deductive *presentation* and confuse it
with the actual practice of mathematics.
Another striking example is group theory. Mathematicians were trying to figure out how to solve polynomial equations. Abel showed that the 5th degree equation did not have a general solution.
Galois showed that the underlying reason for this had to do with the mathematical structure of the set of permutations of the roots.
It wasn't till much later that someone came along and defined a "group" as a set with a binary operation satisfying such-and-so axioms; and then was able to re-derive Galois's and Abel's proof
from the axioms of group theory.
Today, we teach undergrads that a group is such-and-so; then we spend the rest of the semester deriving consequences. The thoughtful undergrads wonder: How did they come up with these particular
axioms? And they get the impression that math is about writing down axioms and mechanically deriving logical conclusions.
But this is of course a complete inversion of how math is done. In math, you first suspect the theorem; then -- after centuries, sometimes -- you eventually figure out the right axioms.
Math is done inductively and presented deductively. Math actually goes from theorems to axioms; not the other way round.
1. Agreed! :) And thanks for the helpful illustrations, anonymous.
3. There is much true here, but as far as logicism goes, it is completely wrong. Logicism is sometimes an epistemological view, and sometimes a metaphysical view, but, either way, it is a view about
what is based upon what, NOT a view about what is INFERRED from what. To interpret it the latter way is to think of it as a psychological view, and that could not be further from the intentions
of either Frege or Russell.
1. Well, at least I am not alone in thinking that (deductive) inference is the key concept in Frege's conception of logic:
(I've been much influenced by my former supervisor G. Sundholm for my thinking about Frege.)
You may quibble with my use of the term 'logicism' instead of 'logic', but other than that there are very respectable Frege scholars (to mention one more: Danielle Macbeth) who hold this
interpretation. You may well disagree with them, but it's not a view that is obviously so 'completely wrong' as you claim.
4. The same dichotomy between forward and backward proofs happens in type theory (in the computing tradition) where functional programmers normalize proofs written from premisses to conclusions,
while logic programmers build proofs from conclusions to axioms.
5. Aristotle does seem to talk about deducing or inferring premises through conclusions in the section on reciprocal proofs. I have a theory that induction is precisely this. Hope the grant approval
board thinks likewise!
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How to Find Better the Domain of a Function 1 | domain free
How to Find Better the Domain of a Function 1
What is the domain of a function? The domain of a function can be any number, including -1 and -5. The range of a function can end arbitrarily. There are also gaps in the domain, indicated by a “U”
symbol. This article discusses how to find the domain of a function. If you have any questions, don’t hesitate to ask us. We are always happy to help.
Range of a function
A range is a set of output values of a function. For example, f(x) = 1,2,3,4 is defined to have a range of zero to four. The domain of a function is the set of non-negative real numbers. The range of
f(x) is the set of all output values that are within the domain. A function can also have an open range. This type of range is known as a domain.
If f displays style f is a function from domain X to codomain Y, then the range is the distance between the codomain and domain X. The yellow oval in Y is the image of f. The range of a function can
include or exclude an image. The two examples are shown below. To understand the meaning of range, we should first define what a range is. The range of a function can include a function’s codomain.
In a simple example, a range can be the distance between the hand of the thrower and the highest point of the figure. The height of a figure changes during the interval, and this means that a
function cannot capture all possible values in its domain. A function’s domain, or range, can be expressed as a set of non-negative integers. However, the range of a function is more complicated than
the domain of a function.
Graphs are also an excellent way to understand a function’s range. The domain refers to the set of possible input and output values. The easiest way to determine a function’s range is to plot it and
look for the y-values that are covered by the graph. If you are having trouble interpreting the graph, you can use a graph of the function to help you understand how it works. It’s important to
remember that the domain refers to the range and domain of a function.
The range of a function depends on the type of function that it represents. A function can have either a domain or a co-domain. Its domain is the set of input values that can be represented by the
function, and the range is the set of output values that the function can generate. For example, a function that maps every element in A to every element in B is said to have a domain. If f is a
domain-domain function, then the range of the function is the set of images that it can produce.
If the y-coordinate of the graph is equal to a function’s domain, the range of that function is zero. This means that it can be infinitely large. Therefore, the range of a function is from -3 to -10.
Its maximum point is 10, and its minimum value is -3. It can go up and down indefinitely without ever going negative. However, its range is infinite, so it’s not always easy to determine whether the
range is negative or positive.
A domain and range of a function are often represented as ordered pairs in tables. Learning these two terms makes it easier to remember the meaning of domain and range. Domains represent independent
values while ranges contain dependent values. In the same way, ranges represent all possible outcomes for a dependent variable. For this reason, range and domain are also important when looking for
mathematical relationships. It’s easy to define a domain and range when dealing with numerical data.
A domain is the area of a function that can be defined by the formula. For example, the domain of f(x) = 15x-2 would be the set of Real Numbers that can be entered. The range is the area between x
and y, defined by the function’s domain. The output of a function can be any Real Number, but it must be greater than zero. The range is written as x(-,) and the resulting value of the function must
be a Real Number.
The range of a function is the set of input values that the function can handle. In other words, the domain of a function includes the entire set of real and natural numbers that can be entered into
the function. By definition, the domain of a function includes all possible x-values and y-values. If a function’s domain contains a real number, the domain of a function is the set of all possible
The range of a function is the area within which the function’s graph is non-zero. For example, f(x) = x+1 is a linear function. The graph of the original function has a hole at x=2. A second method
for identifying the range of a rational function is to plot its graph. The graph of the parent function’s graph can also be sketched to determine the range.
A common way to determine the range of a quadratic function is to plot a graph. Sometimes, all that is needed in the direction of the parabola. The direction of a parabola determines the range’s
maximum and minimum values. If the input values are negative, the output will be negative. Consequently, a quadratic function can have an infinite domain and have no boundary. Once you know the range
of a function, you can apply it to your functions to find a solution.
The domain of a function is the set of possible inputs for a given function. The domain of a function is the entire set of independent variables.
The denominator of a fraction
must be positive. In addition, the digit under the square root bracket must be positive. The domain of a fraction is the area in which the function does not behave as it does for a specific input
value. For example, if the denominator of a fraction is positive, then the function’s domain is A.
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Structure and Kinds of Anumana, Vyapti, Hetvabhasas
Structure and Kinds of Anumana, Vyapti, Hetvabhasas | UGC NET Paper 1
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Although all the major schools accept Anumana as a valid source of knowledge, the understanding and the explanation of each school will have certain variations according to their understanding of
knowledge. In Indian philosophy, the inference is used for oneself and inference for others. When inference is used for oneself the propositions are not well structured since its primary aim is the
acquisition of personal knowledge without error. In contrast, inference for others has to be well structured because it is used to convince the other of the truth. We shall concentrate mainly on the
understanding of Nyaya School because it is well known for its logic.
UNIT VI – Logical Reasoning (Click below on the topic to read the study notes)
• Understanding the structure of arguments: argument forms, the structure of categorical propositions, Mood and Figure, Formal and Informal fallacies, Uses of language, Connotations and denotations
of terms, Classical square of opposition
• Evaluating and distinguishing deductive and inductive reasoning
• Analogies
• Venn diagram: Simple and multiple uses for establishing the validity of arguments (You are Reading This)
• Indian Logic: Means of knowledge (New Topic)
• Pramanas: Pratyaksha (Perception), Anumana (Inference), Upamana (Comparison), Shabda (Verbal testimony), Arthapatti (Implication) and Anupalabddhi (Non-apprehension) (New Topic)
• Structure and kinds of Anumana (inference), Vyapti (invariable relation), Hetvabhasas (fallacies of inference) (New Topic)
They define the inference (Anumana) as “a process of reasoning in which we pass from the apprehension of some mark (linga) to that of something else under an invariable relation (vyapti) that exists
between them.”
Vyapti is essential in Indian philosophy for making a valid inference: however, it is good to know that different schools had different names for vyapti; For example, Vaisesikas called it Prasiddhi
and Samkhya called it pratibandha.
Nyaya proposes a longer syllogism; it has five propositions. An argument, according to them, has
five parts: Paksa or Pratinjna, hetu, drastanta, upanaya and nigamana.
Here is a standard example to understand this;
Types Examples
1. Paksa (The Thesis / Pratijna – Proposition) Download to view
2. Hetu (Reason or the ground) Download to view
3. Drstanta (the corroboration) Download to view
4. Upanaya (The application) Download to view
5. Nigamana (the conclusion) Download to view
An argument where the conclusion follows validly from the premises. In other words, an argument where the truth of the premises guarantees the truth of the conclusion.
Premises: All men are mortal
Socrates is a man
Conclusion Socrates is mortal
In the above example, the premises, all men are mortal, and Socrates is a man, give a guarantee the truth of the conclusion; Socrates is mortal. The conclusion follows the validity according to the
An argument where the premises point several cases of some pattern and the conclusion states that this pattern will hold in general.
An inductive argument will not be deductively valid, because even if a pattern is found many times, that does not guarantee it will always be found. Therefore, an inductive argument provides weaker,
less trustworthy support for the conclusion than a deductive argument does.
For Example:
Premises: We have seen 1000 swans, and
All of them have been white
Conclusion: All swans are white.
In the above example, we have seen just 1000 swans (not all in the world), and all of them have been white. But it does not mean that all swans in the world are white. White swans are a case of a
pattern in those particular circumstances. Hence, we have concluded in general that all swans are white. But it might not be true actually. This type of arrangement of premises and conclusion is an
example of an Inductive argument.
3. Abductive (or Hypothetico-Deductive) Argument
An argument that (i) points out a particular fact, (ii) points out that if a particular hypothesis were true, we would get this fact, and so (iii) concludes that the hypothesis is indeed true.
Abductive arguments seem to make an even bigger jump than inductive arguments. Inductive arguments generalize, while abductive arguments say that successful predictions “prove” theory is true.
Abductive arguments are not deductively valid because false theories can make true predictions. So, true predictions do not guarantee that the theory is true.
Premises: These coins conduct electricity (fact)
If these coins are made of gold (hypothesis),
then they would conduct electricity (prediction).
Conclusion: These coins are made of gold.
Structure of Categorical Propositions
A proposition is simply a claim about the world that has truth value. Every proposition can be expressed as a declarative (i.e., not a question or command) sentence.
Categorical Proposition is any statement which relates two classes or categories of entities. In other words, a categorical proposition is a proposition that relates two classes of objects. A class
is a group of objects.
Example: Cats are mammals
Here, a class or category (Cats) are related to another class or category (Mammals). So, “Cats are mammals” is a Categorical proposition.
Components of Categorical Propositions
For a any categorical proposition, there are four components:
1. Subject Term: First category or class
2. Predicate Term: Second category or class
3. Copula: The grammatical link (verb) between subject and predicate terms.
4. Quantifiers: Words that specify the quantity of the subject and predicate terms.
Two Important terms of Categorical Proposition
• Affirmative: ‘All’ (includes all of a class)
• Negative: ‘No’ (excludes all of a class)
Particular: ‘Some’ (includes part of a class)
Example: All cats are mammals
Here, All – Quantifier
Cats – Subject Term
Are – Copula
Mammals – Predicate term
Properties of Categorical Propositions
Each categorical proposition has both quantity and quality properties. The followings are the properties:
Quantity: The quantity of a categorical proposition is determined by the quantifier used.
Quality: The quality of a categorical proposition is determined according to whether the proposition asserts of denies an overlap between the classes.
Affirmative: if a proposition asserts an overlap between the classes or category named, the quality of the proposition is affirmative.
Negative: In this, a proposition denies an overlap between the categories or classes named,
Distribution: If the proposition refers to the entire class named by a term, that term is distributed and if it does not refer to the entire class named by a term, then the term is undistributed.
Types of Categorical Proposition
There are four types of categorical position:
• All politicians are liars (Universal Affirmative) – A
• No politicians are liars. (Universal Negative) – E
• Some politicians are liars. (Particular affirmative) – I
• Some politicians are not liars. ( Particular negative) – O
Universal Affirmative (A- Propositions): In a proposition, if every member of the subject class is also a member of the predicate class, then it is called Universal Affirmative Proposition. In other
words, whole of one class is included or contained in another class.
In an example “All politicians are liars”, every member of the class of politicians, is a member of another class of liars.
An universal affirmative proposition can be written as: All S is P
S and P represent the subject and predicate terms, respectively. Such a proposition affirms that the relation of class inclusion holds between the two classes and says that the inclusion is complete,
or universal.
Universal Negative (E- Proposition): The proposition in which no members of the subject class are members of the predicate class.
In an example “No politicians are liars”, no member of the class of politicians, is a member of another class of liars.
Systematically, Universal Negative proposition can be represented as: No S is P
Such a proposition affirms that the no relation of class inclusion holds between the two classes and says that the exclusion is complete, or universal.
Particular affirmative (I-proposition): The proposition in which at least one members of the subject class is also a member of the predicate class.
In an example “Some politicians are liars”, some member of the class of politicians, is a member of another class of liars.
Systematically, Particular affirmative proposition can be represented as: Some S is P
Particular negative ( O-proposition): The proposition in which at least one members of the subject class is not a member of the predicate class.
In an example “Some politicians are not liars”, some member of the class of politicians, is a member of another class of liars.
Systematically, Particular affirmative proposition can be represented as: Some S is not P.
A brief of Four Kind of Categorical Proposition
Type Quantifier Subject Copula Predicate
A All S are P
E All (No) S are not (are) P
I Some S are P
O Some S are not P
Classical Square of Opposition
The opposition is an immediate inference grounded on the relation between propositions which have the same terms, but differ in quantity or quality (or both).
For any formal opposition between two propositions, it is essential that their terms should be the same. There can be no opposition between two such propositions as these:
• All angels have
• No cows are
The square of opposition shows us the logical inferences (immediate inferences) we can make from one proposition type (A, I, E, and O) to another.
Contradictory: Two propositions are said to be contradictory if both cannot be true, and both cannot be false at the same time. In other words, if the opposition is between two propositions, which
differ both in quantity and quality.
Here, A – All politicians are liars and O – Some politicians are not liars, and similarly, E and I propositions are contradictory.
Contrary: Universal propositions are said to be contrary because they cannot both be simultaneously true. In other words, the opposition is between two universals which differ in quality.A- All
politicians are liars is true, the E- No politicians are liars must be false.
Similarly, if the E-proposition is true, then the A-proposition is false.
A –
All politicians are liars and
– Some politicians are not liars, and similarly,
propositions are contradictory.
Sub-contrary: If the two particular propositions can both be true but cannot both be false. In other words, the opposition is between two particulars which differ in quality. It means that they
cannot both be simultaneously false.Sub alternation: The universal to particular and particular to universal inferences are called subalternation. In other words, the opposition is between two
propositions which differ only in quantity.
These inferences are valid if the subaltern (A or E) is true, then the subaltern (I or O) is true. If the subaltern is false, then the subaltern is false.
A syllogism is an argument containing two premises and a conclusion.
Categorical syllogism: A categorical syllogism is a syllogism whose premises and conclusion are categorical propositions.
For example:
1. All hats are fashionable clothing.
2. All fashionable clothing is purple.
3. So, some hats are purple.
A Standard Form Categorical Syllogism Contains:
• Two premises and a conclusion, each a standard form categorical proposition.
• Major Term: A major term which appears only in the first premise and the predicate of the conclusion.
• Minor Term: A minor term which appears only in the second premise and the subject of the conclusion
• Middle Term: A middle term which appears in both premises but not in the conclusion.
• Major Premises: The major premise is the premise which contains the major term.
• Minor Premise: The minor premise is the premise which contains the minor term.
Mood and Figure of Syllogism
When the major premise, the minor premise, and the conclusion of a categorical syllogism arranage in a series of three letters (A, E, I, or O) corresponding to the type of categorical proposition is
called MOOD of an argument.
Premises: All P are M
All S are M
Conclusion: Some S are P
The first premise is of the form A
The second premise is of the form A
The conclusion is of the form I.
Thus, the mood of this Argument is AAI.
In another example, to figure out the FORM of the premises and the conclusion in the following example:
Premises: No S are P (E-propostion)
Some S are P (I-Proposition)
Conclusion: Some S are not P (O-Proposition)
Thus, the mood of this Argument is “EIO”.
When you have to determine the mood of a categorical syllogism, you need to find out which of the four forms of categorical proposition each line of the Argument is (A, E, I, or O).
The figure of a categorical syllogism is a number which corresponds to the placement of the two middle terms.
For example, consider the following arguments:
P 1. All mammals are creatures that have hair.
P 2. All dogs are mammals.
P 3. Therefore, all dogs are creatures that have hair.
Notice that the middle term in the major premise is on the LEFT, while the middle term in the minor premise is on the RIGHT. Whenever this happens, we say that the argument has figure “1.”
There are four possible figures in the categorical syllogism:
Figure1: When the middle term is on the left in P 1, and on the right in P 2.
Figure2: When the middle term is on the right in both premises.
Figure3: When the middle term is on the left in both premises.
Figure4: When the middle term is on the right in P 1, and on the left in P 2.
Important Points of Mood and Figures:
• There are 64 different moods
• And each mood has 4 different figures.
• Thus, there are 64*4=256 different kinds of standard form categorical syllogisms.
There are two kinds of valid argument forms:
Unconditionally Valid Forms: There are fifteen combinations of mood and figures that are valid from the Boolean standpoint, and we call these “unconditionally valid”argument forms. The chart below
depicts ALL of 15 the unconditionally valid argument forms.
Conditionally Valid Forms: There are some inferences that are NOT valid from the Boolean standpoint, which is valid from the Aristotelian standpoint. In addition to the fifteen unconditionally valid
argument forms, there are nine conditionally valid argument forms for categorical syllogisms:
A standard form categorical syllogism is valid on the modern theory if and only if each of the following five propositions is all true of it. A standard form categorical syllogism is valid on the
traditional theory if and only if each of the first four propositions is true of it.
1. The middle-term is distributed at least once.
2. If a term is distributed in a conclusion, then that term is distributed in one of the premises.
3. There is at least one affirmative premise.
4. There is a negative premise if and only if there is a negative conclusion.
5. If both premises are universal, then the conclusion is universal.
Informal and Formal Fallacy
Simply, a fallacy is a mistake in reasoning. In other words, a defect in an argument that misleads the mind is called a fallacy. There are two types of fallacy:
Formal Fallacies: A fallacy in which there is the involvement of an error in the form, arrangement, or technical structure of an argument is called Formal Fallacy.
Informal Fallacies: Informal fallacies are a matter of unclear expression that deal with the logic of the meaning of language. Opposite to it, formal fallacies deal with the logic of the technical
An informal fallacy involves such things as:
• the misuse of language such as words or grammar,
• misstatements of fact or opinion,
• misconceptions due to underlying presuppositions, or
• just plain illogical sequences of thought.
Uses of Language in Logic
A logic always deals with the analysis and evaluation of arguments. Since arguments are expressed in language, the study of arguments requires a carefully attention to language in which arguments are
The followings are three important uses of language:
1. Informative,
2. Expressive and
3. Directive uses of language.
Informative use of language: It involves an effort to communicate some content or to describe something or to give information about something. When I say a child, “The Second of October is the
Gandhi Jayanti.” The language I used is informative.
This kind of use of language presumes that the content of what is being communicated is true, so it will be our main focus in the study of logic.
When a sentence is used informatively, it reports that something has some feature or that something lacks some feature. Consider the following two sentences:
1. Parrot has a feather.
2. Parrot is not mammals.
The first proposition reports that having feather is a feature of a Parrot. The second proposition reports that Parrot do not have some essential qualities found in mammals. In, both cases it
provides information about the world.
Two main aspects of this function are generally noted: (1) evoking certain feelings and (2) expressing feelings.
Expressive discourse, qua expressive discourse, is best regarded as neither true or false.
Expressive use of Language: This type of language is often used to express our emotions, feelings, or attitudes. For example: It’s too bad!, It’s wonderful!, etc.
When language is used expressively or emotively, it cannot be characterized as true or false.
Directive uses of language: When the use of language is often to give direction as Commands, requests, instructions, questions etc., to do or not to do something.
Consider the following examples:
1. Finish your homework.
2. Wash your clothes.
3. Are you feeling well?
In all the above examples, the directive use of language. Directive use of language is not normally considered true or false (although various logics of commands have been developed).
Connotations and denotations of Terms
Denotation is the dictionary definition or literal meaning of a word only. Not emotions or feelings are associated with the word.
Ex: The teacher walked into the classroom. This example does not have any hidden meaning. A teacher simply walked into a classroom.
Connotation: A word’s emotional meaning; suggestions and associations that are connected to a word. Words can be positive, negative, or neutral. Words can also connote specific feelings or emotions.
Different types of Definition
Lexical: The purpose of a lexical definition is to report the way a word is standardly used in a language. Most definitions found in a dictionary are lexical definitions.
Ex. Fossil, Cat, Dogs etc.
Persuasive: The purpose of a persuasive definition is to influence people’s attitudes, not to neutrally and objectively capture the standard meaning of a word.
Eg. Teenagers, Abortion etc.
Stipulative: A stipulative definition stipulates (assigns) a meaning to a word by coining a new word or giving an old word a new meaning. A stipulative definition is neither true nor false; it is
neither accurate nor inaccurate.
Eg. Sugarnecker, Black Holes, etc.
Theoretical: Theoretical definitions can explain concepts theoretically. Sometimes definitions are given for terms, not because the word itself is unfamiliar, but because the term is not understood.
Such concepts require theoretical definitions, which are often scientific or philosophical in nature.
For example, when your chemistry teacher defines water by its chemical formula H2O, he is not trying to increase your vocabulary (you already knew the term water), but to explain its atomic
Accepting a theoretical definition is like accepting a theory about the term being defined. If you define spirit as “the life-giving principle of physical organisms,” you are inviting others to
accept the idea that life is some how a spiritual product.
Precising: A precising definition takes a word that is normally vague and gives it a clear precisely defined meaning.
Eg. Lite, Low-income, middle aged, etc.
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Re: st: marginal effects for ordered logit
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]
Re: st: marginal effects for ordered logit
From "maartenbuis" <[email protected]>
To [email protected]
Subject Re: st: marginal effects for ordered logit
Date Thu, 04 Nov 2004 16:19:58 -0000
Dear Aga,
The marginal effect you want is just the slope of a tangent line of
these graphs at a specific value of your explanatory variable (often
the mean of the explanatory variable), so the marginal effect just
shows part of the information represented by the graph. Furthermore -
mfx - won't calculate the marginal effects you want if you include a
quadratic term, so you'd probably have to do that manually.
Alternatively you could use the fact that the ologit is linear in the
log(odds), i.e. the log(odds) of belonging to one group versus all
`lower' groups is a linear function of the explanatory variables. So
you could plot or tabulate the odds ratio for various values of your
explanatory variable.
Have a look at chapter 5 of `Regression models for categorical
dependent variables using Stata', by Scott Long and Jeremy Freese if
you want to know more than I can tell you through this list.
--- <Agnieszka.Markiewicz@e...> wrote:
> I was actually trying to plot my results but they don't look very
> convincing. Accordingly to your remark, the probability of
belonging to
> one of the classes changes and my graphs don't provide any clear
> explanation. Thus I still would like to interpret my results without
> graphs.
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2004-11/msg00148.html","timestamp":"2024-11-05T10:36:08Z","content_type":"text/html","content_length":"7979","record_id":"<urn:uuid:47659527-b61a-4990-a434-529a54faf7fb>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00462.warc.gz"} |
NCERT Class 7 Maths Solutions - CoolGyan
The NCERT Solutions for Class 7 Maths are provided here. Practising NCERT Solutions is the ultimate need for students who intend to score good marks in Maths examination. Students facing trouble in
solving problems from Class 7 NCERT textbook can refer to our free NCERT Solutions provided below.
Students are suggested to practise NCERT Solutions which are available in the PDF format, and it also helps students to understand the basic concepts of Maths easily. For students who feel stressed
searching for the most comprehensive and detailed NCERT Solutions for Class 7, we at CoolGyan’S have prepared step by step solutions with detailed explanations. These exercises are formulated by our
expert tutors to assist you with your exam preparation to attain good marks in the subject.
NCERT Solutions Class 7 Maths (Chapter-wise)
In addition, students are given access to additional online study materials and resources, available in CoolGyan’S website, such as notes, books, question papers, exemplar problems, worksheets, etc.
Students are also advised to practice Class 7 Sample Papers to get an idea of the question paper pattern in the final exam.
NCERT Exercise-wise Solutions for Class 7 Maths
Students are advised to go through these NCERT Solutions for Class 7. All the solutions are in line with the CBSE guidelines and presented in a stepwise manner so that students can understand the
logic behind every problem while practising.
Chapter 1 Integers
Chapter 1 involves the study of Integers. In the previous classes, we have learnt about whole numbers and integers. Now, we will study more about integers, their properties and operations. Likewise,
the addition and the subtraction of the integers, properties of addition and subtraction of integers, multiplication and division of integers, properties of multiplication and division of integers.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 2 Fractions and Decimals
Chapter 2 of NCERT textbook, deals with Fractions and Decimals. You have learnt about fractions and decimals along with the operations of addition and subtraction on them, in earlier classes. We now
study the operations of multiplication and division on fractions as well as on decimals.
Likewise multiplication of a fraction by a whole number, multiplication of a fraction by a fraction, division of the whole number by a fraction, division of a fraction by a whole number,
multiplication of decimal by 10,100, 100, etc. Other topics are multiplication of decimal by whole number, multiplication of decimal by a decimal, division of decimals by 10, 100, 1000, etc, dividing
a decimal by the whole number.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 3 Data Handling
Chapter 3 of NCERT textbook deals with Data Handling. In your previous classes, you have dealt with various types of data. You have learnt to collect data, tabulate and put it in the form of bar
graphs. Now in this chapter, we will take one more step towards learning how to do this. This chapter covers topics like collecting data, organisation of data, representative values, arithmetic mean,
mode of large data, median, use of bar graphs with a different purpose, choosing scale, chance and probability.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 4 Simple Equations
Chapter 4, Simple Equations of NCERT textbook deals with setting up an equation, solving an equation. While solving more equations we shall learn transposing a number, i.e., moving it from one side
to the other. We can transpose numbers instead of adding or subtracting it from both sides of the equations. Then we study from solution to an equation, applications of simple equations to practical
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 5 Lines and Angles
This chapter is about Lines and Angles. This chapter includes related angles, complementary angles, supplementary angles, adjacent angles, linear pair, vertically opposite angles, pairs of lines
intersecting lines, transversal, the angle made by a transversal, transversal of parallel lines, checking of parallel lines.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 6 The Triangle and its Properties
Chapter 6 of NCERT textbook discusses the topic The Triangle and its Property. Triangle is a simple closed curve made of three line segments. It has three vertices, three sides and three angles. It
covers the concepts median of a triangle, altitude of a triangle, exterior angle of a triangle and its property, angle sum property of a triangle, two special triangles equilateral and isosceles, the
sum of the lengths of two sides of a triangle, right-angled triangles and Pythagoras property.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 7 Congruence of Triangles
This chapter is about the Congruence of Triangles. If two figures have exactly the same shape and size, they are said to be congruent. Other related topics are congruence of plane figures, congruence
among line segments, congruence of angles, congruence of triangles, criteria for congruence of triangles, congruence of right-angled triangles.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 8 Comparing Quantities
Chapter 8 of NCERT textbook deals with the Comparing Quantities. Its related topics are equivalent ratios, percentage another way of comparing quantities, the meaning of percentage, converting
fractional numbers to a percentage, converting decimals to a percentage, converting percentages to fractions or decimals. Other interesting topics are fun with estimation, interpreting percentages,
converting percentages to how many, ratios to percents, increase or decrease as per cent, profit or loss as a percentage, charge given on borrowed money or simple interest.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 9 Rational Numbers
Chapter 9 of NCERT textbook discusses the topic of Rational Numbers. It covers positive and negative rational numbers, rational numbers on a number line, rational numbers in standard form, comparison
of rational numbers, rational numbers between two rational numbers, operations on rational numbers.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 10 Practical Geometry
Chapter 10 of NCERT textbook discusses the topic of Practical Geometry. You are familiar with a number of shapes. You learnt how to draw some of them in the earlier classes. For example, you can draw
a line segment of a given length, a line perpendicular to a given line segment, an angle, an angle bisector, a circle etc. Now, you will learn how to draw parallel lines and some types of triangles.
Topics covered in this chapter are the construction of a line parallel to a given line, through a point not on the line, construction of different types of triangles.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 11 Perimeter and Area
This chapter is about Simple Interest. In Class 6, you have already learnt perimeters of plane figures and areas of squares and rectangles. Perimeter is the distance around a closed figure while the
area is the part of a plane or region occupied by the closed figure. In this class, you will learn about perimeters and areas of a few more plane figures.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 12 Algebraic Expressions
In Chapter 12, Algebraic Expressions of NCERT textbook deals with terms of an expression, coefficients, like and unlike terms, monomials, binomials, trinomials, polynomials, addition and subtraction
of algebraic expressions, finding the values of an expression.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 13 Exponents and Powers
Chapter 13 of NCERT textbook discusses the topic Exponents and Powers. Exponents are the product of rational numbers multiplied several times by themselves. Some of the concepts covered here are laws
of exponents, multiplying powers with the same base, dividing powers with the same base, taking power of a power, multiplying powers with the same exponents, dividing powers with the same exponents,
decimal number system, expressing large numbers in the standard form.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 14 Symmetry
This chapter is about Symmetry. Symmetry means that one shape becomes exactly like another when you move it in some way, turn, flip or slide. For two objects to be symmetrical, they must be the same
size and shape, with one object having a different orientation from the first. There can also be symmetry in one object, such as a face. Topics covered in this chapter are lines of symmetry for
regular polygons, rotational symmetry, line symmetry and rotational symmetry.
Here you can find exercise solution links for the topics covered in this chapter.
Chapter 15 Visualising Solid Shapes
In Chapter 15, Visualising Solid Shapes of NCERT textbook deals with the plane figures and solids, faces, edges, vertices, nets for building 3-D shapes, drawing solids on a flat surface, oblique
sketches, isometric sketches. Some more concepts covered in this chapter are visualising solid objects, viewing different sections of a solid, then another way is by shadow play.
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The NCERT solutions are considered one of the best resources to master Maths. The solutions given here are in well-structured format with different shortcut methods to ensure a proper understanding
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Benefits of NCERT Solutions for Class 7 Maths
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• Solutions are explained step by step in a comprehensive manner.
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• Wed, Mar 06, 2019 @ 03:30 PM - 04:30 PM
Aerospace and Mechanical Engineering
Conferences, Lectures, & Seminars
Speaker: Krishna Garikipati, University of Michigan
Talk Title: Mechano-Chemical Phase Transformations: Computational Framework, Machine Learning Studies and Graph Theoretic Analysis
Abstract: Phase transformations in a wide range of materials-”for energy, electronics, structural and other applications-”are driven by mechanics in interaction with chemistry. We have
developed a general theoretical and computational framework for large scale simulations of these mechano-chemical phenomena. I will begin by presenting our recent work in this sphere, while
highlighting some of its more insightful results. In addition to being a platform for investigating mechanically driven phenomena in materials physics, this work is a foundation to explore
the potential of recent advances in data-driven modeling. Of interest to us are machine learning advances that may enhance our approaches to solve computational materials physics problems. I
will outline the first of several recent studies that we have launched in this spirit. Such combinations of classical high-performance scientific computing and modern data-driven modeling now
allow us to access large numbers of states of physical systems. They also motivate the study of mathematical structures for representation, exploration and analysis of systems by using these
collections of states. With this perspective, I will offer a view of graph theory that places it in nearly perfect correspondence with properties of stationary and dynamical systems. This has
opened up new insights to our earlier, large-scale computational investigations of mechano-chemically phase transforming materials systems. This treatment has potential for eventual
decision-making for physical systems that builds on high-fidelity computations.
Krishna Garikipati is a computational scientist whose work draws upon nonlinear physics, applied mathematics and numerical methods. A very recent interest of his is the development of methods
for data-driven computational science. He has worked for quite a few years in mathematical biology, biophysics and materials physics. Some specific problems he has been thinking about
recently are: (1) mathematical models of patterning and morphogenesis in developmental biology, (2) mathematical and physical modeling of tumor growth, and (3) mechano-chemically driven
phenomena in materials, such as phase transformations and stress-influenced mass transport.
Host: AME Department
More Info: https://ame.usc.edu/seminars/
Location: Seaver Science Library (SSL) - 150
Audiences: Everyone Is Invited
Contact: Tessa Yao
Event Link: https://ame.usc.edu/seminars/ | {"url":"https://viterbi.usc.edu/news/events/calendar/?event=17685","timestamp":"2024-11-10T11:45:08Z","content_type":"text/html","content_length":"10510","record_id":"<urn:uuid:50bf2da8-c100-4e3d-885e-2d2c027e3380>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00611.warc.gz"} |
Token-based IO
Note: Silver now has support for Monads, so this threaded-token approach to IO is mostly^1 obsolete.
You can now just define main as
fun main IO<Integer> ::= args::[String] =
See here for more details.
Silver’s IO support [DEL:is:DEL]was terrible. Let’s get that out of the way right from the beginning. Silver programs are largely intended to (1) read a file in its entirety and (2) write a file in
its entirety, while (3) maybe printing some messages to the console. If you try to be fancier, you’ll be all smdh.
Start with main:
function main
IOVal<Integer> ::= args::[String] ioin::IOToken
return ...;
fun main IOVal<Integer> ::= args::[String] ioin::IOToken =
You get an input IO token ioin. You are then responsible for threading this through every IO function you call, in the correct order, and then out through the standard “IO and also some other value”
type called IOVal.
Lets take a look at some types:
abstract production ioval
top::IOVal<a> ::= i::IOToken v::a
function printT
IO ::= s::String i::IOToken
function readFileT
IOVal<String> ::= s::String i::IOToken
function writeFileT
IO ::= file::String contents::String i::IOToken
So, first off, if you’re getting used to Silver conventions, you’ll notice that ioval has its parameters backwards from what you’d expect. This is indefensible. Sorry.
Next, you can see how every function follows the same pattern: you pass in the token, then you either get a token back, or an IOVal back.
This is partly because we didn’t have a unit type yet, so print couldn’t return IOVal<Unit> or whatever. You know, to be consistent. Oh well.
Here’s a helpful google images search:
Could be worse. Back in the day we had type like IOString and IOBoolean and IOInteger and of course, names of the attributes were different on each. You kids these days are your IOVal…
If you don’t pass your IO tokens around properly, things can happen weird. Sometimes the order of stuff might get switched around like time travel. Or things don’t get done.
But most often, the issue people hit is the performing of IO actions getting driven by the demand for values other than the IO token. (e.g. you read from a file, and instead of the IO token being
what causes the read to happen, it’s the demand for the file’s contents as a string.)
I invite you to marvel at this code:
- Run units until a non-zero error code is encountered.
function runAll
IOVal<Integer> ::= l::[Unit] i::IOToken
local attribute now :: Unit;
now = head(l);
now.ioIn = i;
return if unsafeTrace(null(l), i) -- TODO: this is just to force strictness...
then ioval(i, 0)
else if now.code != 0
then ioval(now.io, now.code)
else runAll(tail(l), now.io);
The call to unsafeTrace demands the IO token i before returning the other parameter. Why? Well, you’ll notice that accessing now.code may depend upon that Unit’s IO actions… which means we’d be
demanding actions get taken via their return value, not their IO token. Which means ordering can get ~wonky~.
This is evidence that our token passing model is fundamentally broken. Take a laugh.
How did pre-monad Haskell get around this? Our IOVal type isn’t really adequate. It’s a pair. pre-monad Haskell would use a function. i.e.:
abstract production ioval
IOVal<a> ::= f::(Pair<a IO> ::= IOToken)
How demand was driven (demanding the value or the IO token), in this case, doesn’t matter, because both will trigger calling the function, which will trigger demanding the previous IO token. All
Silver doesn’t have good enough lambda support to do this yet. We probably could stick an unsafeTrace inside the ioval production for accessing the iovalue attribute, though. Someone should do that.
Always bind IOVal returning functions to a local. That is:
local file :: IOVal<String> = readFile(filename, ioin);
Why? Because that way, there’s exactly one decoration of the IOVal undecorated “tree”. And therefore, there’s one decorated copy (the one implicitly created by the local.) And therefore, there’s one
cached value of the IO token.
If you, somehow, manage to get two decorated trees corresponding to a single IOVal undecorated tree, you may get multiple evaluation of IO functions! Having fun yet?
Here’s a forced example, continuing from the above:
local readFileTwice :: String = new(file).iovalue ++ new(file).iovalue;
Bonus: this will actually cause the file to be read three times, when later on file.io is demanded.
Double Bonus: The IO actions that will be repeated will be those all the way back to the last properly cached IO token value. So you might even execute more than just a read action three times here!
Aren’t you lucky?
1. There still might be some cases where token-based IO is more appropriate - e.g. an analysis that requires performing IO might be cleaner using an IO token passed through the tree with separate
threaded attributes than “infecting” everything with an IO monad (although implicit monads should also be considered as a solution.) The IO monad itself is implemented using token IO, so this
won’t be going away any time soon, at least. ↩︎ | {"url":"https://melt.cs.umn.edu/silver/concepts/io/","timestamp":"2024-11-14T14:30:10Z","content_type":"text/html","content_length":"86041","record_id":"<urn:uuid:e4cb136e-6719-4cb1-b180-4f6bccc2ee57>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00790.warc.gz"} |
P-BOOKS • Amusements in Mathematics • Henry Ernest Dudeney • 8
189.—THE YORKSHIRE ESTATES.
The triangular piece of land that was not for sale contains exactly eleven acres. Of course it is not difficult to find the answer if we follow the eccentric and tricky tracks of intricate
trigonometry; or I might say that the application of a well-known formula reduces the problem to finding one-quarter of the square root of (4 x 370 x 116) -(370 + 116 - 74) squared—that is a quarter
of the square root of 1936, which is one-quarter of 44, or 11 acres. But all that the reader really requires to know is the Pythagorean law on which many puzzles have been built, that in any
right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. I shall dispense with all "surds" and similar absurdities, notwithstanding the fact that
the sides of our triangle are clearly incommensurate, since we cannot exactly extract the square roots of the three square areas.
In the above diagram ABC represents our triangle. ADB is a right-angled triangle, AD measuring 9 and BD measuring 17, because the square of 9 added to the square of 17 equals 370, the known area of
the square on AB. Also AEC is a right-angled triangle, and the square of 5 added to the square of 7 equals 74, the square estate on A C. Similarly, CFB is a right-angled triangle, for the square of 4
added to the square of 10 equals 116, the square estate on BC. Now, although the sides of our triangular estate are incommensurate, we have in this diagram all the exact figures that we need to
discover the area with precision.
The area of our triangle ADB is clearly half of 9 x 17, or 761/2 acres. The area of AEC is half of 5 x 7, or 171/2 acres; the area of CFB is half of 4 x 10, or 20 acres; and the area of the oblong
EDFC is obviously 4 x 7, or 28 acres. Now, if we add together 171/2, 20, and 28 = 651/2, and deduct this sum from the area of the large triangle ADB (which we have found to be 761/2 acres), what
remains must clearly be the area of ABC. That is to say, the area we want must be 761/2 - 651/2 = 11 acres exactly.
190.—FARMER WURZEL'S ESTATE.
The area of the complete estate is exactly one hundred acres. To find this answer I use the following little formula,
___ /4ab - (a + b + c) squared —————————— 4
where a, b, c represent the three square areas, in any order. The expression gives the area of the triangle A. This will be found to be 9 acres. It can be easily proved that A, B, C, and D are all
equal in area; so the answer is 26 + 20 + 18 + 9 + 9 + 9 + 9 = 100 acres.
Here is the proof. If every little dotted square in the diagram represents an acre, this must be a correct plan of the estate, for the squares of 5 and 1 together equal 26; the squares of 4 and 2
equal 20; and the squares of 3 and 3 added together equal 18. Now we see at once that the area of the triangle E is 21/2, F is 41/2, and G is 4. These added together make 11 acres, which we deduct
from the area of the rectangle, 20 acres, and we find that the field A contains exactly 9 acres. If you want to prove that B, C, and D are equal in size to A, divide them in two by a line from the
middle of the longest side to the opposite angle, and you will find that the two pieces in every case, if cut out, will exactly fit together and form A.
Or we can get our proof in a still easier way. The complete area of the squared diagram is 12 x 12 = 144 acres, and the portions 1, 2, 3, 4, not included in the estate, have the respective areas of
121/2, 171/2, 91/2, and 41/2. These added together make 44, which, deducted from 144, leaves 100 as the required area of the complete estate.
191.—THE CRESCENT PUZZLE.
Referring to the original diagram, let AC be x, let CD be x - 9, and let EC be x - 5. Then x - 5 is a mean proportional between x - 9 and x, from which we find that x equals 25. Therefore the
diameters are 50 in. and 41 in. respectively.
192.—THE PUZZLE WALL.
The answer given in all the old books is that shown in Fig. 1, where the curved wall shuts out the cottages from access to the lake. But in seeking the direction for the "shortest possible" wall most
readers to-day, remembering that the shortest distance between two points is a straight line, will adopt the method shown in Fig. 2. This is certainly an improvement, yet the correct answer is really
that indicated in Fig. 3. A measurement of the lines will show that there is a considerable saving of length in this wall.
193.—THE SHEEP-FOLD.
This is the answer that is always given and accepted as correct: Two more hurdles would be necessary, for the pen was twenty-four by one (as in Fig. A on next page), and by moving one of the sides
and placing an extra hurdle at each end (as in Fig. B) the area would be doubled. The diagrams are not to scale. Now there is no condition in the puzzle that requires the sheep-fold to be of any
particular form. But even if we accept the point that the pen was twenty-four by one, the answer utterly fails, for two extra hurdles are certainly not at all necessary. For example, I arrange the
fifty hurdles as in Fig. C, and as the area is increased from twenty-four "square hurdles" to 156, there is now accommodation for 650 sheep. If it be held that the area must be exactly double that of
the original pen, then I construct it (as in Fig. D) with twenty-eight hurdles only, and have twenty-two in hand for other purposes on the farm. Even if it were insisted that all the original hurdles
must be used, then I should construct it as in Fig. E, where I can get the area as exact as any farmer could possibly require, even if we have to allow for the fact that the sheep might not be able
to graze at the extreme ends. Thus we see that, from any point of view, the accepted answer to this ancient little puzzle breaks down. And yet attention has never before been drawn to the absurdity.
A 24 24 1
B 48 2 24 C D 12 48 6 156 8 13
12 . E 13 . ' ' . . ' ' . ' . . ' 12 ' . ' 13
194.—THE GARDEN WALLS.
The puzzle was to divide the circular field into four equal parts by three walls, each wall being of exactly the same length. There are two essential difficulties in this problem. These are: (1) the
thickness of the walls, and (2) the condition that these walls are three in number. As to the first point, since we are told that the walls are brick walls, we clearly cannot ignore their thickness,
while we have to find a solution that will equally work, whether the walls be of a thickness of one, two, three, or more bricks.
The second point requires a little more consideration. How are we to distinguish between a wall and walls? A straight wall without any bend in it, no matter how long, cannot ever become "walls," if
it is neither broken nor intersected in any way. Also our circular field is clearly enclosed by one wall. But if it had happened to be a square or a triangular enclosure, would there be respectively
four and three walls or only one enclosing wall in each case? It is true that we speak of "the four walls" of a square building or garden, but this is only a conventional way of saying "the four
sides." If you were speaking of the actual brickwork, you would say, "I am going to enclose this square garden with a wall." Angles clearly do not affect the question, for we may have a zigzag wall
just as well as a straight one, and the Great Wall of China is a good example of a wall with plenty of angles. Now, if you look at Diagrams 1, 2, and 3, you may be puzzled to declare whether there
are in each case two or four new walls; but you cannot call them three, as required in our puzzle. The intersection either affects the question or it does not affect it.
If you tie two pieces of string firmly together, or splice them in a nautical manner, they become "one piece of string." If you simply let them lie across one another or overlap, they remain "two
pieces of string." It is all a question of joining and welding. It may similarly be held that if two walls be built into one another—I might almost say, if they be made homogeneous—they become one
wall, in which case Diagrams 1, 2, and 3 might each be said to show one wall or two, if it be indicated that the four ends only touch, and are not really built into, the outer circular wall.
The objection to Diagram 4 is that although it shows the three required walls (assuming the ends are not built into the outer circular wall), yet it is only absolutely correct when we assume the
walls to have no thickness. A brick has thickness, and therefore the fact throws the whole method out and renders it only approximately correct.
Diagram 5 shows, perhaps, the only correct and perfectly satisfactory solution. It will be noticed that, in addition to the circular wall, there are three new walls, which touch (and so enclose) but
are not built into one another. This solution may be adapted to any desired thickness of wall, and its correctness as to area and length of wall space is so obvious that it is unnecessary to explain
it. I will, however, just say that the semicircular piece of ground that each tenant gives to his neighbour is exactly equal to the semicircular piece that his neighbour gives to him, while any
section of wall space found in one garden is precisely repeated in all the others. Of course there is an infinite number of ways in which this solution may be correctly varied.
195.—LADY BELINDA'S GARDEN.
All that Lady Belinda need do was this: She should measure from A to B, fold her tape in four and mark off the point E, which is thus one quarter of the side. Then, in the same way, mark off the
point F, one-fourth of the side AD Now, if she makes EG equal to AF, and GH equal to EF, then AH is the required width for the path in order that the bed shall be exactly half the area of the garden.
An exact numerical measurement can only be obtained when the sum of the squares of the two sides is a square number. Thus, if the garden measured 12 poles by 5 poles (where the squares of 12 and 5,
144 and 25, sum to 169, the square of 13), then 12 added to 5, less 13, would equal four, and a quarter of this, 1 pole, would be the width of the path.
196.—THE TETHERED GOAT.
This problem is quite simple if properly attacked. Let us suppose the triangle ABC to represent our half-acre field, and the shaded portion to be the quarter-acre over which the goat will graze when
tethered to the corner C. Now, as six equal equilateral triangles placed together will form a regular hexagon, as shown, it is evident that the shaded pasture is just one-sixth of the complete area
of a circle. Therefore all we require is the radius (CD) of a circle containing six quarter-acres or 11/2 acres, which is equal to 9,408,960 square inches. As we only want our answer "to the nearest
inch," it is sufficiently exact for our purpose if we assume that as 1 is to 3.1416, so is the diameter of a circle to its circumference. If, therefore, we divide the last number I gave by 3.1416,
and extract the square root, we find that 1,731 inches, or 48 yards 3 inches, is the required length of the tether "to the nearest inch."
197.—THE COMPASSES PUZZLE.
Let AB in the following diagram be the given straight line. With the centres A and B and radius AB describe the two circles. Mark off DE and EF equal to AD. With the centres A and F and radius DF
describe arcs intersecting at G. With the centres A and B and distance BG describe arcs GHK and N. Make HK equal to AB and HL equal to HB. Then with centres K and L and radius AB describe arcs
intersecting at I. Make BM equal to BI. Finally, with the centre M and radius MB cut the line in C, and the point C is the required middle of the line AB. For greater exactitude you can mark off R
from A (as you did M from B), and from R describe another arc at C. This also solves the problem, to find a point midway between two given points without the straight line.
I will put the young geometer in the way of a rigid proof. First prove that twice the square of the line AB equals the square of the distance BG, from which it follows that HABN are the four corners
of a square. To prove that I is the centre of this square, draw a line from H to P through QIB and continue the arc HK to P. Then, conceiving the necessary lines to be drawn, the angle HKP, being in
a semicircle, is a right angle. Let fall the perpendicular KQ, and by similar triangles, and from the fact that HKI is an isosceles triangle by the construction, it can be proved that HI is half of
HB. We can similarly prove that C is the centre of the square of which AIB are three corners.
I am aware that this is not the simplest possible solution.
198.—THE EIGHT STICKS.
The first diagram is the answer that nearly every one will give to this puzzle, and at first sight it seems quite satisfactory. But consider the conditions. We have to lay "every one of the sticks on
the table." Now, if a ladder be placed against a wall with only one end on the ground, it can hardly be said that it is "laid on the ground." And if we place the sticks in the above manner, it is
only possible to make one end of two of them touch the table: to say that every one lies on the table would not be correct. To obtain a solution it is only necessary to have our sticks of proper
dimensions. Say the long sticks are each 2 ft. in length and the short ones 1 ft. Then the sticks must be 3 in. thick, when the three equal squares may be enclosed, as shown in the second diagram. If
I had said "matches" instead of "sticks," the puzzle would be impossible, because an ordinary match is about twenty-one times as long as it is broad, and the enclosed rectangles would not be squares.
199.—PAPA'S PUZZLE.
I have found that a large number of people imagine that the following is a correct solution of the problem. Using the letters in the diagram below, they argue that if you make the distance BA
one-third of BC, and therefore the area of the rectangle ABE equal to that of the triangular remainder, the card must hang with the long side horizontal. Readers will remember the jest of Charles
II., who induced the Royal Society to meet and discuss the reason why the water in a vessel will not rise if you put a live fish in it; but in the middle of the proceedings one of the least
distinguished among them quietly slipped out and made the experiment, when he found that the water did rise! If my correspondents had similarly made the experiment with a piece of cardboard, they
would have found at once their error. Area is one thing, but gravitation is quite another. The fact of that triangle sticking its leg out to D has to be compensated for by additional area in the
rectangle. As a matter of fact, the ratio of BA to AC is as 1 is to the square root of 3, which latter cannot be given in an exact numerical measure, but is approximately 1.732. Now let us look at
the correct general solution. There are many ways of arriving at the desired result, but the one I give is, I think, the simplest for beginners.
Fix your card on a piece of paper and draw the equilateral triangle BCF, BF and CF being equal to BC. Also mark off the point G so that DG shall equal DC. Draw the line CG and produce it until it
cuts the line BF in H. If we now make HA parallel to BE, then A is the point from which our cut must be made to the corner D, as indicated by the dotted line.
A curious point in connection with this problem is the fact that the position of the point A is independent of the side CD. The reason for this is more obvious in the solution I have given than in
any other method that I have seen, and (although the problem may be solved with all the working on the cardboard) that is partly why I have preferred it. It will be seen at once that however much you
may reduce the width of the card by bringing E nearer to B and D nearer to C, the line CG, being the diagonal of a square, will always lie in the same direction, and will cut BF in H. Finally, if you
wish to get an approximate measure for the distance BA, all you have to do is to multiply the length of the card by the decimal .366. Thus, if the card were 7 inches long, we get 7 x .366 = 2.562, or
a little more than 21/2 inches, for the distance from B to A.
But the real joke of the puzzle is this: We have seen that the position of the point A is independent of the width of the card, and depends entirely on the length. Now, in the illustration it will be
found that both cards have the same length; consequently all the little maid had to do was to lay the clipped card on top of the other one and mark off the point A at precisely the same distance from
the top left-hand corner! So, after all, Pappus' puzzle, as he presented it to his little maid, was quite an infantile problem, when he was able to show her how to perform the feat without first
introducing her to the elements of statics and geometry.
200.—A KITE-FLYING PUZZLE.
Solvers of this little puzzle, I have generally found, may be roughly divided into two classes: those who get within a mile of the correct answer by means of more or less complex calculations,
involving "pi," and those whose arithmetical kites fly hundreds and thousands of miles away from the truth. The comparatively easy method that I shall show does not involve any consideration of the
ratio that the diameter of a circle bears to its circumference. I call it the "hat-box method."
Supposing we place our ball of wire, A, in a cylindrical hat-box, B, that exactly fits it, so that it touches the side all round and exactly touches the top and bottom, as shown in the illustration.
Then, by an invariable law that should be known by everybody, that box contains exactly half as much again as the ball. Therefore, as the ball is 24 in. in diameter, a hat-box of the same
circumference but two-thirds of the height (that is, 16 in. high) will have exactly the same contents as the ball.
Now let us consider that this reduced hat-box is a cylinder of metal made up of an immense number of little wire cylinders close together like the hairs in a painter's brush. By the conditions of the
puzzle we are allowed to consider that there are no spaces between the wires. How many of these cylinders one one-hundredth of an inch thick are equal to the large cylinder, which is 24 in. thick?
Circles are to one another as the squares of their diameters. The square of 1/100 is 1/100000, and the square of 24 is 576; therefore the large cylinder contains 5,760,000 of the little wire
cylinders. But we have seen that each of these wires is 16 in. long; hence 16 x 5,760,000 = 92,160,000 inches as the complete length of the wire. Reduce this to miles, and we get 1,454 miles 2,880
ft. as the length of the wire attached to the professor's kite.
Whether a kite would fly at such a height, or support such a weight, are questions that do not enter into the problem.
201.—HOW TO MAKE CISTERNS.
Here is a general formula for solving this problem. Call the two sides of the rectangle a and b. Then
a + b - (a squared + b squared - ab)^1/2 —————————————- 6
equals the side of the little square pieces to cut away. The measurements given were 8 ft. by 3 ft., and the above rule gives 8 in. as the side of the square pieces that have to be cut away. Of
course it will not always come out exact, as in this case (on account of that square root), but you can get as near as you like with decimals.
202.—THE CONE PUZZLE.
The simple rule is that the cone must be cut at one-third of its altitude.
203.—CONCERNING WHEELS.
If you mark a point A on the circumference of a wheel that runs on the surface of a level road, like an ordinary cart-wheel, the curve described by that point will be a common cycloid, as in Fig. 1.
But if you mark a point B on the circumference of the flange of a locomotive-wheel, the curve will be a curtate cycloid, as in Fig. 2, terminating in nodes. Now, if we consider one of these nodes or
loops, we shall see that "at any given moment" certain points at the bottom of the loop must be moving in the opposite direction to the train. As there is an infinite number of such points on the
flange's circumference, there must be an infinite number of these loops being described while the train is in motion. In fact, at any given moment certain points on the flanges are always moving in a
direction opposite to that in which the train is going.
In the case of the two wheels, the wheel that runs round the stationary one makes two revolutions round its own centre. As both wheels are of the same size, it is obvious that if at the start we mark
a point on the circumference of the upper wheel, at the very top, this point will be in contact with the lower wheel at its lowest part when half the journey has been made. Therefore this point is
again at the top of the moving wheel, and one revolution has been made. Consequently there are two such revolutions in the complete journey.
204.—A NEW MATCH PUZZLE.
1. The easiest way is to arrange the eighteen matches as in Diagrams 1 and 2, making the length of the perpendicular AB equal to a match and a half. Then, if the matches are an inch in length, Fig. 1
contains two square inches and Fig. 2 contains six square inches—4 x 11/2. The second case (2) is a little more difficult to solve. The solution is given in Figs. 3 and 4. For the purpose of
construction, place matches temporarily on the dotted lines. Then it will be seen that as 3 contains five equal equilateral triangles and 4 contains fifteen similar triangles, one figure is three
times as large as the other, and exactly eighteen matches are used.
205.—THE SIX SHEEP-PENS.
Place the twelve matches in the manner shown in the illustration, and you will have six pens of equal size.
206.—THE KING AND THE CASTLES.
There are various ways of building the ten castles so that they shall form five rows with four castles in every row, but the arrangement in the next column is the only one that also provides that two
castles (the greatest number possible) shall not be approachable from the outside. It will be seen that you must cross the walls to reach these two.
207.—CHERRIES AND PLUMS.
There are several ways in which this problem might be solved were it not for the condition that as few cherries and plums as possible shall be planted on the north and east sides of the orchard. The
best possible arrangement is that shown in the diagram, where the cherries, plums, and apples are indicated respectively by the letters C, P, and A. The dotted lines connect the cherries, and the
other lines the plums. It will be seen that the ten cherry trees and the ten plum trees are so planted that each fruit forms five lines with four trees of its kind in line. This is the only
arrangement that allows of so few as two cherries or plums being planted on the north and east outside rows.
208.—A PLANTATION PUZZLE.
The illustration shows the ten trees that must be left to form five rows with four trees in every row. The dots represent the positions of the trees that have been cut down.
209.—THE TWENTY-ONE TREES.
I give two pleasing arrangements of the trees. In each case there are twelve straight rows with five trees in every row.
210.—THE TEN COINS.
The answer is that there are just 2,400 different ways. Any three coins may be taken from one side to combine with one coin taken from the other side. I give four examples on this and the next page.
We may thus select three from the top in ten ways and one from the bottom in five ways, making fifty. But we may also select three from the bottom and one from the top in fifty ways. We may thus
select the four coins in one hundred ways, and the four removed may be arranged by permutation in twenty-four ways. Thus there are 24 x 100 = 2,400 different solutions.
As all the points and lines puzzles that I have given so far, excepting the last, are variations of the case of ten points arranged to form five lines of four, it will be well to consider this
particular case generally. There are six fundamental solutions, and no more, as shown in the six diagrams. These, for the sake of convenience, I named some years ago the Star, the Dart, the
Compasses, the Funnel, the Scissors, and the Nail. (See next page.) Readers will understand that any one of these forms may be distorted in an infinite number of different ways without destroying its
real character.
In "The King and the Castles" we have the Star, and its solution gives the Compasses. In the "Cherries and Plums" solution we find that the Cherries represent the Funnel and the Plums the Dart. The
solution of the "Plantation Puzzle" is an example of the Dart distorted. Any solution to the "Ten Coins" will represent the Scissors. Thus examples of all have been given except the Nail.
On a reduced chessboard, 7 by 7, we may place the ten pawns in just three different ways, but they must all represent the Dart. The "Plantation" shows one way, the Plums show a second way, and the
reader may like to find the third way for himself. On an ordinary chessboard, 8 by 8, we can also get in a beautiful example of the Funnel—symmetrical in relation to the diagonal of the board. The
smallest board that will take a Star is one 9 by 7. The Nail requires a board 11 by 7, the Scissors
11 by 9, and the Compasses 17 by 12. At least these are the best results recorded in my note-book. They may be beaten, but I do not think so. If you divide a chessboard into two parts by a diagonal
zigzag line, so that the larger part contains 36 squares and the smaller part 28 squares, you can place three separate schemes on the larger part and one on the smaller part (all Darts) without their
conflicting—that is, they occupy forty different squares. They can be placed in other ways without a division of the board. The smallest square board that will contain six different schemes (not
fundamentally different), without any line of one scheme crossing the line of another, is 14 by 14; and the smallest board that will contain one scheme entirely enclosed within the lines of a second
scheme, without any of the lines of the one, when drawn from point to point, crossing a line of the other, is 14 by 12.
211.—THE TWELVE MINCE-PIES.
If you ignore the four black pies in our illustration, the remaining twelve are in their original positions. Now remove the four detached pies to the places occupied by the black ones, and you will
have your seven straight rows of four, as shown by the dotted lines.
212.—THE BURMESE PLANTATION.
The arrangement on the next page is the most symmetrical answer that can probably be found for twenty-one rows, which is, I believe, the greatest number of rows possible. There are several ways of
doing it.
213.—TURKS AND RUSSIANS.
The main point is to discover the smallest possible number of Russians that there could have been. As the enemy opened fire from all directions, it is clearly necessary to find what is the smallest
number of heads that could form sixteen lines with three heads in every line. Note that I say sixteen, and not thirty-two, because every line taken by a bullet may be also taken by another bullet
fired in exactly the opposite direction. Now, as few as eleven points, or heads, may be arranged to form the required sixteen lines of three, but the discovery of this arrangement is a hard nut. The
diagram at the foot of this page will show exactly how the thing is to be done.
If, therefore, eleven Russians were in the positions shown by the stars, and the thirty-two Turks in the positions indicated by the black dots, it will be seen, by the lines shown, that each Turk may
fire exactly over the heads of three Russians. But as each bullet kills a man, it is essential that every Turk shall shoot one of his comrades and be shot by him in turn; otherwise we should have to
provide extra Russians to be shot, which would be destructive of the correct solution of our problem. As the firing was simultaneous, this point presents no difficulties. The answer we thus see is
that there were at least eleven Russians amongst whom there was no casualty, and that all the thirty-two Turks were shot by one another. It was not stated whether the Russians fired any shots, but it
will be evident that even if they did their firing could not have been effective: for if one of their bullets killed a Turk, then we have immediately to provide another man for one of the Turkish
bullets to kill; and as the Turks were known to be thirty-two in number, this would necessitate our introducing another Russian soldier and, of course, destroying the solution. I repeat that the
difficulty of the puzzle consists in finding how to arrange eleven points so that they shall form sixteen lines of three. I am told that the possibility of doing this was first discovered by the Rev.
Mr. Wilkinson some twenty years ago.
214.—THE SIX FROGS.
Move the frogs in the following order: 2, 4, 6, 5, 3, 1 (repeat these moves in the same order twice more), 2, 4, 6. This is a solution in twenty-one moves—the fewest possible.
If n, the number of frogs, be even, we require (n squared + n)/2 moves, of which (n squared - n)/2 will be leaps and n simple moves. If n be odd, we shall need ((n squared + 3n)/2) - 4 moves, of
which (n squared - n)/2 will be leaps and 2n - 4 simple moves.
In the even cases write, for the moves, all the even numbers in ascending order and the odd numbers in descending order. This series must be repeated 1/2n times and followed by the even numbers in
ascending order once only. Thus the solution for 14 frogs will be (2, 4, 6, 8, 10, 12, 14, 13, 11, 9, 7, 5, 3, 1) repeated 7 times and followed by 2, 4, 6, 8, 10, 12, 14 = 105 moves.
In the odd cases, write the even numbers in ascending order and the odd numbers in descending order, repeat this series 1/2(n - 1) times, follow with the even numbers in ascending order (omitting n -
1), the odd numbers in descending order (omitting 1), and conclude with all the numbers (odd and even) in their natural order (omitting 1 and n). Thus for 11 frogs: (2, 4, 6, 8, 10, 11, 9, 7, 5, 3,
1) repeated 5 times, 2, 4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 7, 8, 9, 10 = 73 moves.
This complete general solution is published here for the first time.
215.—THE GRASSHOPPER PUZZLE.
Move the counters in the following order. The moves in brackets are to be made four times in succession. 12, 1, 3, 2, 12, 11, 1, 3, 2 (5, 7, 9, 10, 8, 6, 4), 3, 2, 12, 11, 2, 1, 2. The grasshoppers
will then be reversed in forty-four moves.
The general solution of this problem is very difficult. Of course it can always be solved by the method given in the solution of the last puzzle, if we have no desire to use the fewest possible
moves. But to employ a full economy of moves we have two main points to consider. There are always what I call a lower movement (L) and an upper movement (U). L consists in exchanging certain of the
highest numbers, such as 12, 11, 10 in our "Grasshopper Puzzle," with certain of the lower numbers, 1, 2, 3; the former moving in a clockwise direction, the latter in a non-clockwise direction. U
consists in reversing the intermediate counters. In the above solution for 12, it will be seen that 12, 11, and 1, 2, 3 are engaged in the L movement, and 4, 5, 6, 7, 8, 9, 10 in the U movement. The
L movement needs 16 moves and U 28, making together 44. We might also involve 10 in the L movement, which would result in L 23, U 21, making also together 44 moves. These I call the first and second
methods. But any other scheme will entail an increase of moves. You always get these two methods (of equal economy) for odd or even counters, but the point is to determine just how many to involve in
L and how many in U. Here is the solution in table form. But first note, in giving values to n, that 2, 3, and 4 counters are special cases, requiring respectively 3, 3, and 6 moves, and that 5 and 6
counters do not give a minimum solution by the second method—only by the first.
FIRST METHOD. + + -+ -+ -+ Total No. L MOVEMENT. U MOVEMENT. of + -+ -+ + + Total No. Counters. No. of No. of No. of No. of of Moves. Counters. Moves. Counters. Moves. + + -+ -+ + + -+ 4n n-1 and n 2
(n-1) squared+5n-7 2n+1 2n squared+3n+1 4(n squared+n-1) 4n-2 n-1 " n 2(n-1) squared+5n-7 2n-1 2(n-1) squared+3n-2 4n squared-5 4n+1 n " n+1 2n squared+5n-2 2n 2n squared+3n-4 2(2n squared+4n-3) 4n-1
n-1 " n 2(n-1) squared+5n-7 2n 2n squared+3n-4 4n squared+4n-9 + + -+ -+ + + -+
SECOND METHOD. + -+ + -+ -+ Total No. L MOVEMENT. U MOVEMENT. of + -+ + + + Total No. Counters. No. of No. of No. of No. of of Moves. Counters. Moves. Counters. Moves. + -+ -+ + + + -+ 4n n and n 2n
squared+3n-4 2n 2(n-1) squared+5n-2 4(n squared+n-1) 4n-2 n-1 " n-1 2(n-1) squared+3n-7 2n 2(n-1) squared+5n-2 4n squared-5 4n+1 n " n 2n squared+3n-4 2n+1 2n squared+5n-2 2(2n squared+4n-3) 4n-1 n "
n 2n squared+3n-4 2n-1 2(n-1) squared+5n-7 4n squared+4n-9 + -+ -+ + + + -+
More generally we may say that with m counters, where m is even and greater than 4, we require (m squared + 4m - 16)/4 moves; and where m is odd and greater than 3, (m squared + 6m - 31)/4 moves. I
have thus shown the reader how to find the minimum number of moves for any case, and the character and direction of the moves. I will leave him to discover for himself how the actual order of moves
is to be determined. This is a hard nut, and requires careful adjustment of the L and the U movements, so that they may be mutually accommodating.
216.—THE EDUCATED FROGS.
The following leaps solve the puzzle in ten moves: 2 to 1, 5 to 2, 3 to 5, 6 to 3, 7 to 6, 4 to 7, 1 to 4, 3 to 1, 6 to 3, 7 to 6.
217.—THE TWICKENHAM PUZZLE.
Play the counters in the following order: K C E K W T C E H M K W T A N C E H M I K C E H M T, and there you are, at Twickenham. The position itself will always determine whether you are to make a
leap or a simple move.
218.—THE VICTORIA CROSS PUZZLE.
In solving this puzzle there were two things to be achieved: first, so to manipulate the counters that the word VICTORIA should read round the cross in the same direction, only with the V on one of
the dark arms; and secondly, to perform the feat in the fewest possible moves. Now, as a matter of fact, it would be impossible to perform the first part in any way whatever if all the letters of the
word were different; but as there are two I's, it can be done by making these letters change places—that is, the first I changes from the 2nd place to the 7th, and the second I from the 7th place to
the 2nd. But the point I referred to, when introducing the puzzle, as a little remarkable is this: that a solution in twenty-two moves is obtainable by moving the letters in the order of the
following words: "A VICTOR! A VICTOR! A VICTOR I!"
There are, however, just six solutions in eighteen moves, and the following is one of them: I (1), V, A, I (2), R, O, T, I (1), I (2), A, V, I (2), I (1), C, I (2), V, A, I (1). The first and second
I in the word are distinguished by the numbers 1 and 2.
It will be noticed that in the first solution given above one of the I's never moves, though the movements of the other letters cause it to change its relative position. There is another peculiarity
I may point out—that there is a solution in twenty-eight moves requiring no letter to move to the central division except the I's. I may also mention that, in each of the solutions in eighteen moves,
the letters C, T, O, R move once only, while the second I always moves four times, the V always being transferred to the right arm of the cross.
219.—THE LETTER BLOCK PUZZLE.
This puzzle can be solved in 23 moves—the fewest possible. Move the blocks in the following order: A, B, F, E, C, A, B, F, E, C, A, B, D, H, G, A, B, D, H, G, D, E, F.
220.—A LODGING-HOUSE DIFFICULTY.
The shortest possible way is to move the articles in the following order: Piano, bookcase, wardrobe, piano, cabinet, chest of drawers, piano, wardrobe, bookcase, cabinet, wardrobe, piano, chest of
drawers, wardrobe, cabinet, bookcase, piano. Thus seventeen removals are necessary. The landlady could then move chest of drawers, wardrobe, and cabinet. Mr. Dobson did not mind the wardrobe and
chest of drawers changing rooms so long as he secured the piano.
221.—THE EIGHT ENGINES.
The solution to the Eight Engines Puzzle is as follows: The engine that has had its fire drawn and therefore cannot move is No. 5. Move the other engines in the following order: 7, 6, 3, 7, 6, 1, 2,
4, 1, 3, 8, 1, 3, 2, 4, 3, 2, seventeen moves in all, leaving the eight engines in the required order.
There are two other slightly different solutions.
222.—A RAILWAY PUZZLE.
This little puzzle may be solved in as few as nine moves. Play the engines as follows: From 9 to 10, from 6 to 9, from 5 to 6, from 2 to 5, from 1 to 2, from 7 to 1, from 8 to 7, from 9 to 8, and
from 10 to 9. You will then have engines A, B, and C on each of the three circles and on each of the three straight lines. This is the shortest solution that is possible.
223.—A RAILWAY MUDDLE.
Only six reversals are necessary. The white train (from A to D) is divided into three sections, engine and 7 wagons, 8 wagons, and 1 wagon. The black train (D to A) never uncouples anything
throughout. Fig. 1 is original position with 8 and 1 uncoupled. The black train proceeds to position in Fig. 2 (no reversal). The engine and 7 proceed towards D, and black train backs, leaves 8 on
loop, and takes up position in Fig. 3 (first reversal). Black train goes to position in Fig. 4 to fetch single wagon (second reversal). Black train pushes 8 off loop and leaves single wagon there,
proceeding on its journey, as in Fig. 5 (third and fourth reversals). White train now backs on to loop to pick up single car and goes right away to D (fifth and sixth reversals).
224.—THE MOTOR-GARAGE PUZZLE.
The exchange of cars can be made in forty-three moves, as follows: 6-G, 2-B, 1-E, 3-H, 4-I, 3-L, 6-K, 4-G, 1-I, 2-J, 5-H, 4-A, 7-F, 8-E, 4-D, 8-C, 7-A, 8-G, 5-C, 2-B, 1-E, 8-I, 1-G, 2-J, 7-H, 1-A,
7-G, 2-B, 6-E, 3-H, 8-L, 3-I, 7-K, 3-G, 6-I, 2-J, 5-H, 3-C, 5-G, 2-B, 6-E, 5-I, 6-J. Of course, "6-G" means that the car numbered "6" moves to the point "G." There are other ways in forty-three
225.—THE TEN PRISONERS.
It will be seen in the illustration how the prisoners may be arranged so as to produce as many as sixteen even rows. There are 4 such vertical rows, 4 horizontal rows, 5 diagonal rows in one
direction, and 3 diagonal rows in the other direction. The arrows here show the movements of the four prisoners, and it will be seen that the infirm man in the bottom corner has not been moved.
226.—ROUND THE COAST.
In order to place words round the circle under the conditions, it is necessary to select words in which letters are repeated in certain relative positions. Thus, the word that solves our puzzle is
"Swansea," in which the first and fifth letters are the same, and the third and seventh the same. We make out jumps as follows, taking the letters of the word in their proper order: 2-5, 7-2, 4-7,
1-4, 6-1, 3-6, 8-3. Or we could place a word like "Tarapur" (in which the second and fourth letters, and the third and seventh, are alike) with these moves: 6-1, 7-4, 2-7, 5—2, 8-5, 3-6, 8-3. But
"Swansea" is the only word, apparently, that will fulfil the conditions of the puzzle.
This puzzle should be compared with Sharp's Puzzle, referred to in my solution to No. 341, "The Four Frogs." The condition "touch and jump over two" is identical with "touch and move along a line."
227.—CENTRAL SOLITAIRE.
Here is a solution in nineteen moves; the moves enclosed in brackets count as one move only: 19-17, 16-18, (29-17, 17-19), 30-18, 27-25, (22-24, 24-26), 31-23, (4-16, 16-28), 7-9, 10-8, 12-10, 3-11,
18-6, (1-3, 3-11), (13-27, 27-25), (21-7, 7-9), (33-31, 31-23), (10-8, 8-22, 22-24, 24-26, 26-12, 12-10), 5-17. All the counters are now removed except one, which is left in the central hole. The
solution needs judgment, as one is tempted to make several jumps in one move, where it would be the reverse of good play. For example, after playing the first 3-11 above, one is inclined to increase
the length of the move by continuing with 11-25, 25-27, or with 11-9, 9-7.
I do not think the number of moves can be reduced.
228.—THE TEN APPLES.
Number the plates (1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), (13, 14, 15, 16) in successive rows from the top to the bottom. Then transfer the apple from 8 to 10 and play as follows, always
removing the apple jumped over: 9-11, 1-9, 13-5, 16-8, 4-12, 12-10, 3-1, 1-9, 9-11.
229.—THE NINE ALMONDS.
This puzzle may be solved in as few as four moves, in the following manner: Move 5 over 8, 9, 3, 1. Move 7 over 4. Move 6 over 2 and 7. Move 5 over 6, and all the counters are removed except 5, which
is left in the central square that it originally occupied.
230.—THE TWELVE PENNIES.
Here is one of several solutions. Move 12 to 3, 7 to 4, 10 to 6, 8 to 1, 9 to 5, 11 to 2.
231.—PLATES AND COINS.
Number the plates from 1 to 12 in the order that the boy is seen to be going in the illustration. Starting from 1, proceed as follows, where "1 to 4" means that you take the coin from plate No. 1 and
transfer it to plate No. 4: 1 to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, 11 to 2, and complete the last revolution to 1, making three revolutions in all. Or you can proceed this way: 4 to 7, 8 to 11, 12
to 3, 2 to 5, 6 to 9, 10 to 1. It is easy to solve in four revolutions, but the solutions in three are more difficult to discover.
This is "The Riddle of the Fishpond" (No. 41, Canterbury Puzzles) in a different dress.
232.—CATCHING THE MICE.
In order that the cat should eat every thirteenth mouse, and the white mouse last of all, it is necessary that the count should begin at the seventh mouse (calling the white one the first)—that is,
at the one nearest the tip of the cat's tail. In this case it is not at all necessary to try starting at all the mice in turn until you come to the right one, for you can just start anywhere and note
how far distant the last one eaten is from the starting point. You will find it to be the eighth, and therefore must start at the eighth, counting backwards from the white mouse. This is the one I
have indicated.
In the case of the second puzzle, where you have to find the smallest number with which the cat may start at the white mouse and eat this one last of all, unless you have mastered the general
solution of the problem, which is very difficult, there is no better course open to you than to try every number in succession until you come to one that works correctly. The smallest number is
twenty-one. If you have to proceed by trial, you will shorten your labour a great deal by only counting out the remainders when the number is divided successively by 13, 12, 11, 10, etc. Thus, in the
case of 21, we have the remainders 8, 9, 10, 1, 3, 5, 7, 3, 1, 1, 3, 1, 1. Note that I do not give the remainders of 7, 3, and 1 as nought, but as 7, 3, and 1. Now, count round each of these numbers
in turn, and you will find that the white mouse is killed last of all. Of course, if we wanted simply any number, not the smallest, the solution is very easy, for we merely take the least common
multiple of 13, 12, 11, 10, etc. down to 2. This is 360360, and you will find that the first count kills the thirteenth mouse, the next the twelfth, the next the eleventh, and so on down to the
first. But the most arithmetically inclined cat could not be expected to take such a big number when a small one like twenty-one would equally serve its purpose.
In the third case, the smallest number is 100. The number 1,000 would also do, and there are just seventy-two other numbers between these that the cat might employ with equal success.
233.—THE ECCENTRIC CHEESEMONGER.
To leave the three piles at the extreme ends of the rows, the cheeses may be moved as follows—the numbers refer to the cheeses and not to their positions in the row: 7-2, 8-7, 9-8, 10-15, 6-10, 5-6,
14-16, 13-14, 12-13, 3-1, 4-3, 11-4. This is probably the easiest solution of all to find. To get three of the piles on cheeses 13, 14, and 15, play thus: 9-4, 10-9, 11-10, 6-14, 5-6, 12-15, 8-12,
7-8, 16-5, 3-13, 2-3, 1-2. To leave the piles on cheeses 3, 5, 12, and 14, play thus: 8-3, 9-14, 16-12, 1-5, 10-9, 7-10, 11-8, 2-1, 4-16, 13-2, 6-11, 15-4.
234.—THE EXCHANGE PUZZLE.
Make the following exchanges of pairs: H-K, H-E, H-C, H-A, I-L, I-F, I-D, K-L, G-J, J-A, F-K, L-E, D-K, E-F, E-D, E-B, B-K. It will be found that, although the white counters can be moved to their
proper places in 11 moves, if we omit all consideration of exchanges, yet the black cannot be so moved in fewer than 17 moves. So we have to introduce waste moves with the white counters to equal the
minimum required by the black. Thus fewer than 17 moves must be impossible. Some of the moves are, of course, interchangeable.
235.—TORPEDO PRACTICE.
If the enemy's fleet be anchored in the formation shown in the illustration, it will be seen that as many as ten out of the sixteen ships may be blown up by discharging the torpedoes in the order
indicated by the numbers and in the directions indicated by the arrows. As each torpedo in succession passes under three ships and sinks the fourth, strike out each vessel with the pencil as it is
236.—THE HAT PUZZLE.
I suggested that the reader should try this puzzle with counters, so I give my solution in that form. The silk hats are represented by black counters and the felt hats by white counters. The first
row shows the hats in their original positions, and then each successive row shows how they appear after one of the five manipulations. It will thus be seen that we first move hats 2 and 3, then 7
and 8, then 4 and 5, then 10 and 11, and, finally, 1 and 2, leaving the four silk hats together, the four felt hats together, and the two vacant pegs at one end of the row. The first three pairs
moved are dissimilar hats, the last two pairs being similar. There are other ways of solving the puzzle.
237.—BOYS AND GIRLS.
There are a good many different solutions to this puzzle. Any contiguous pair, except 7-8, may be moved first, and after the first move there are variations. The following solution shows the position
from the start right through each successive move to the end:—
. . 1 2 3 4 5 6 7 8 4 3 1 2 . . 5 6 7 8 4 3 1 2 7 6 5 . . 8 4 3 1 2 7 . . 5 6 8 4 . . 2 7 1 3 5 6 8 4 8 6 2 7 1 3 5 . .
238.—ARRANGING THE JAM POTS.
Two of the pots, 13 and 19, were in their proper places. As every interchange may result in a pot being put in its place, it is clear that twenty-two interchanges will get them all in order. But this
number of moves is not the fewest possible, the correct answer being seventeen. Exchange the following pairs: (3-1, 2-3), (15-4, 16-15), (17-7, 20-17), (24-10, 11-24, 12-11), (8-5, 6-8, 21-6, 23-21,
22-23, 14-22, 9-14, 18-9). When you have made the interchanges within any pair of brackets, all numbers within those brackets are in their places. There are five pairs of brackets, and 5 from 22
gives the number of changes required—17.
239.—A JUVENILE PUZZLE.
As the conditions are generally understood, this puzzle is incapable of solution. This can be demonstrated quite easily. So we have to look for some catch or quibble in the statement of what we are
asked to do. Now if you fold the paper and then push the point of your pencil down between the fold, you can with one stroke make the two lines CD and EF in our diagram. Then start at A, and describe
the line ending at B. Finally put in the last line GH, and the thing is done strictly within the conditions, since folding the paper is not actually forbidden. Of course the lines are here left
unjoined for the purpose of clearness.
In the rubbing out form of the puzzle, first rub out A to B with a single finger in one stroke. Then rub out the line GH with one finger. Finally, rub out the remaining two vertical lines with two
fingers at once! That is the old trick.
240.—THE UNION JACK.
There are just sixteen points (all on the outside) where three roads may be said to join. These are called by mathematicians "odd nodes." There is a rule that tells us that in the case of a drawing
like the present one, where there are sixteen odd nodes, it requires eight separate strokes or routes (that is, half as many as there are odd nodes) to complete it. As we have to produce as much as
possible with only one of these eight strokes, it is clearly necessary to contrive that the seven strokes from odd node to odd node shall be as short as possible. Start at A and end at B, or go the
reverse way.
241.—THE DISSECTED CIRCLE.
/ - / / / B / / /^ / / / / / / / / / A / / / / / -+ -* -+ - / / / / / / / / / / D-+ * -+ -* E / / / / / / / / / / -+ -* -+ - / / / / / / / / / / / / / / / / + / / / C -/
It can be done in twelve continuous strokes, thus: Start at A in the illustration, and eight strokes, forming the star, will bring you back to A; then one stroke round the circle to B, one stroke to
C, one round the circle to D, and one final stroke to E—twelve in all. Of course, in practice the second circular stroke will be over the first one; it is separated in the diagram, and the points of
the star not joined to the circle, to make the solution clear to the eye.
242.—THE TUBE INSPECTOR'S PUZZLE.
The inspector need only travel nineteen miles if he starts at B and takes the following route: BADGDEFIFCBEHKLIHGJK. Thus the only portions of line travelled over twice are the two sections D to G
and F to I. Of course, the route may be varied, but it cannot be shortened.
243.—VISITING THE TOWNS.
Note that there are six towns, from which only two roads issue. Thus 1 must lie between 9 and 12 in the circular route. Mark these two roads as settled. Similarly mark 9, 5, 14, and 4, 8, 14, and 10,
6, 15, and 10, 2, 13, and 3, 7, 13. All these roads must be taken. Then you will find that he must go from 4 to 15, as 13 is closed, and that he is compelled to take 3, 11, 16, and also 16, 12. Thus,
there is only one route, as follows: 1, 9, 5, 14, 8, 4, 15, 6, 10, 2, 13, 7, 3, 11, 16, 12, 1, or its reverse—reading the line the other way. Seven roads are not used.
244.—THE FIFTEEN TURNINGS.
It will be seen from the illustration (where the roads not used are omitted) that the traveller can go as far as seventy miles in fifteen turnings. The turnings are all numbered in the order in which
they are taken. It will be seen that he never visits nineteen of the towns. He might visit them all in fifteen turnings, never entering any town twice, and end at the black town from which he starts
(see "The Rook's Tour," No. 320), but such a tour would only take him sixty-four miles.
245.—THE FLY ON THE OCTAHEDRON.
Though we cannot really see all the sides of the octahedron at once, we can make a projection of it that suits our purpose just as well. In the diagram the six points represent the six angles of the
octahedron, and four lines proceed from every point under exactly the same conditions as the twelve edges of the solid. Therefore if we start at the point A and go over all the lines once, we must
always end our route at A. And the number of different routes is just 1,488, counting the reverse way of any route as different. It would take too much space to show how I make the count. It can be
done in about five minutes, but an explanation of the method is difficult. The reader is therefore asked to accept my answer as correct.
246.—THE ICOSAHEDRON PUZZLE.
There are thirty edges, of which eighteen were visible in the original illustration, represented in the following diagram by the hexagon NAESGD. By this projection of the solid we get an imaginary
view of the remaining twelve edges, and are able to see at once their direction and the twelve points at which all the edges meet. The difference in the length of the lines is of no importance; all
we want is to present their direction in a graphic manner. But in case the novice should be puzzled at only finding nineteen triangles instead of the required twenty, I will point out that the
apparently missing triangle is the outline HIK.
In this case there are twelve odd nodes; therefore six distinct and disconnected routes will be needful if we are not to go over any lines twice. Let us therefore find the greatest distance that we
may so travel in one route.
It will be noticed that I have struck out with little cross strokes five lines or edges in the diagram. These five lines may be struck out anywhere so long as they do not join one another, and so
long as one of them does not connect with N, the North Pole, from which we are to start. It will be seen that the result of striking out these five lines is that all the nodes are now even except N
and S. Consequently if we begin at N and stop at S we may go over all the lines, except the five crossed out, without traversing any line twice. There are many ways of doing this. Here is one route:
N to H, I, K, S, I, E, S, G, K, D, H, A, N, B, A, E, F, B, C, G, D, N, C, F, S. By thus making five of the routes as short as is possible—simply from one node to the next—we are able to get the
greatest possible length for our sixth line. A greater distance in one route, without going over the same ground twice, it is not possible to get.
It is now readily seen that those five erased lines must be gone over twice, and they may be "picked up," so to speak, at any points of our route. Thus, whenever the traveller happens to be at I he
can run up to A and back before proceeding on his route, or he may wait until he is at A and then run down to I and back to A. And so with the other lines that have to be traced twice. It is,
therefore, clear that he can go over 25 of the lines once only (25 x 10,000 miles = 250,000 miles) and 5 of the lines twice (5 x 20,000 miles = 100,000 miles), the total, 350,000 miles, being the
length of his travels and the shortest distance that is possible in visiting the whole body.
It will be noticed that I have made him end his travels at S, the South Pole, but this is not imperative. I might have made him finish at any of the other nodes, except the one from which he started.
Suppose it had been required to bring him home again to N at the end of his travels. Then instead of suppressing the line AI we might leave that open and close IS. This would enable him to complete
his 350,000 miles tour at A, and another 10,000 miles would take him to his own fireside. There are a great many different routes, but as the lengths of the edges are all alike, one course is as good
as another. To make the complete 350,000 miles tour from N to S absolutely clear to everybody, I will give it entire: N to H, I, A, I, K, H, K, S, I, E, S, G, F, G, K, D, C, D, H, A, N, B, E, B, A,
E, F, B, C, G, D, N, C, F, S—that is, thirty-five lines of 10,000 miles each.
247.—INSPECTING A MINE.
Starting from A, the inspector need only travel 36 furlongs if he takes the following route: A to B, G, H, C, D, I, H, M, N, I, J, O, N, S, R, M, L, G, F, K, L, Q, R, S, T, O, J, E, D, C, B, A, F, K,
P, Q. He thus passes between A and B twice, between C and D twice, between F and K twice, between J and O twice, and between R and S twice—five repetitions. Therefore 31 passages plus 5 repeated
equal 36 furlongs. The little pitfall in this puzzle lies in the fact that we start from an even node. Otherwise we need only travel 35 furlongs.
248.—THE CYCLIST'S TOUR.
When Mr. Maggs replied, "No way, I'm sure," he was not saying that the thing was impossible, but was really giving the actual route by which the problem can be solved. Starting from the star, if you
visit the towns in the order, NO WAY, I'M SURE, you will visit every town once, and only once, and end at E. So both men were correct. This was the little joke of the puzzle, which is not by any
means difficult.
249.—THE SAILOR'S PUZZLE.
There are only four different routes (or eight, if we count the reverse ways) by which the sailor can start at the island marked A, visit all the islands once, and once only, and return again to A.
Here they are:—
A I P T L O E H R Q D C F U G N S K M B A A I P T S N G L O E U F C D K M B Q R H A A B M K S N G L T P I O E U F C D Q R H A A I P T L O E U G N S K M B Q D C F R H A
Now, if the sailor takes the first route he will make C his 12th island (counting A as 1); by the second route he will make C his 13th island; by the third route, his 16th island; and by the fourth
route, his 17th island. If he goes the reverse way, C will be respectively his 10th, 9th, 6th, and 5th island. As these are the only possible routes, it is evident that if the sailor puts off his
visit to C as long as possible, he must take the last route reading from left to right. This route I show by the dark lines in the diagram, and it is the correct answer to the puzzle.
The map may be greatly simplified by the "buttons and string" method, explained in the solution to No. 341, "The Four Frogs."
250.—THE GRAND TOUR.
The first thing to do in trying to solve a puzzle like this is to attempt to simplify it. If you look at Fig. 1, you will see that it is a simplified version of the map. Imagine the circular towns to
be buttons and the railways to be connecting strings. (See solution to No. 341.) Then, it will be seen, we have simply "straightened out" the previous diagram without affecting the conditions. Now we
can further simplify by converting Fig. 1 into Fig. 2, which is a portion of a chessboard. Here the directions of the railways will resemble the moves of a rook in chess—that is, we may move in any
direction parallel to the sides of the diagram, but not diagonally. Therefore the first town (or square) visited must be a black one; the second must be a white; the third must be a black; and so on.
Every odd square visited will thus be black and every even one white. Now, we have 23 squares to visit (an odd number), so the last square visited must be black. But Z happens to be white, so the
puzzle would seem to be impossible of solution.
As we were told that the man "succeeded" in carrying put his plan, we must try to find some loophole in the conditions. He was to "enter every town once and only once," and we find no prohibition
against his entering once the town A after leaving it, especially as he has never left it since he was born, and would thus be "entering" it for the first time in his life. But he must return at once
from the first town he visits, and then he will have only 22 towns to visit, and as 22 is an even number, there is no reason why he should not end on the white square Z. A possible route for him is
indicated by the dotted line from A to Z. This route is repeated by the dark lines in Fig. 1, and the reader will now have no difficulty in applying; it to the original map. We have thus proved that
the puzzle can only be solved by a return to A immediately after leaving it.
251.—WATER, GAS, AND ELECTRICITY.
According to the conditions, in the strict sense in which one at first understands them, there is no possible solution to this puzzle. In such a dilemma one always has to look for some verbal quibble
or trick. If the owner of house A will allow the water company to run their pipe for house C through his property (and we are not bound to assume that he would object), then the difficulty is got
over, as shown in our illustration. It will be seen that the dotted line from W to C passes through house A, but no pipe ever crosses another pipe.
252.—A PUZZLE FOR MOTORISTS.
The routes taken by the eight drivers are shown in the illustration, where the dotted line roads are omitted to make the paths clearer to the eye.
253.—A BANK HOLIDAY PUZZLE.
The simplest way is to write in the number of routes to all the towns in this manner. Put a 1 on all the towns in the top row and in the first column. Then the number of routes to any town will be
the sum of the routes to the town immediately above and to the town immediately to the left. Thus the routes in the second row will be 1, 2, 3, 4, 5, 6, etc., in the third row, 1, 3, 6, 10, 15, 21,
etc.; and so on with the other rows. It will then be seen that the only town to which there are exactly 1,365 different routes is the twelfth town in the fifth row—the one immediately over the letter
E. This town was therefore the cyclist's destination.
The general formula for the number of routes from one corner to the corner diagonally opposite on any such rectangular reticulated arrangement, under the conditions as to direction, is (m+n)!/m!n!,
where m is the number of towns on one side, less one, and n the number on the other side, less one. Our solution involves the case where there are 12 towns by 5. Therefore m = 11 and n = 4. Then the
formula gives us the answer 1,365 as above.
254.— THE MOTOR-CAR TOUR.
First of all I will ask the reader to compare the original square diagram with the circular one shown in Figs. 1, 2, and 3 below. If for the moment we ignore the shading (the purpose of which I shall
proceed to explain), we find that the circular diagram in each case is merely a simplification of the original square one—that is, the roads from A lead to B, E, and M in both cases, the roads from L
(London) lead to I, K, and S, and so on. The form below, being circular and symmetrical, answers my purpose better in applying a mechanical solution, and I therefore adopt it without altering in any
way the conditions of the puzzle. If such a question as distances from town to town came into the problem, the new diagrams might require the addition of numbers to indicate these distances, or they
might conceivably not be at all practicable.
Now, I draw the three circular diagrams, as shown, on a sheet of paper and then cut out three pieces of cardboard of the forms indicated by the shaded parts of these diagrams. It can be shown that
every route, if marked out with a red pencil, will form one or other of the designs indicated by the edges of the cards, or a reflection thereof. Let us direct our attention to Fig. 1. Here the card
is so placed that the star is at the town T; it therefore gives us (by following the edge of the card) one of the circular routes from London: L, S, R, T, M, A, E, P, O, J, D, C, B, G, N, Q, K, H, F,
I, L. If we went the other way, we should get L, I, F, H, K, Q, etc., but these reverse routes were not to be counted. When we have written out this first route we revolve the card until the star is
at M, when we get another different route, at A a third route, at E a fourth route, and at P a fifth route. We have thus obtained five different routes by revolving the card as it lies. But it is
evident that if we now take up the card and replace it with the other side uppermost, we shall in the same manner get five other routes by revolution.
We therefore see how, by using the revolving card in Fig. 1, we may, without any difficulty, at once write out ten routes. And if we employ the cards in Figs. 2 and 3, we similarly obtain in each
case ten other routes. These thirty routes are all that are possible. I do not give the actual proof that the three cards exhaust all the possible cases, but leave the reader to reason that out for
himself. If he works out any route at haphazard, he will certainly find that it falls into one or other of the three categories.
255.—THE LEVEL PUZZLE.
Let us confine our attention to the L in the top left-hand corner. Suppose we go by way of the E on the right: we must then go straight on to the V, from which letter the word may be completed in
four ways, for there are four E's available through which we may reach an L. There are therefore four ways of reading through the right-hand E. It is also clear that there must be the same number of
ways through the E that is immediately below our starting point. That makes eight. If, however, we take the third route through the E on the diagonal, we then have the option of any one of the three
V's, by means of each of which we may complete the word in four ways. We can therefore spell LEVEL in twelve ways through the diagonal E. Twelve added to eight gives twenty readings, all emanating
from the L in the top left-hand corner; and as the four corners are equal, the answer must be four times twenty, or eighty different ways.
256.—THE DIAMOND PUZZLE.
There are 252 different ways. The general formula is that, for words of n letters (not palindromes, as in the case of the next puzzle), when grouped in this manner, there are always 2^(n+1) - 4
different readings. This does not allow diagonal readings, such as you would get if you used instead such a word as DIGGING, where it would be possible to pass from one G to another G by a diagonal
257.—THE DEIFIED PUZZLE.
The correct answer is 1,992 different ways. Every F is either a corner F or a side F—standing next to a corner in its own square of F's. Now, FIED may be read from a corner F in 16 ways; therefore
DEIF may be read into a corner F also in 16 ways; hence DEIFIED may be read through a corner F in 16 x 16 = 256 ways. Consequently, the four corner F's give 4 x 256 = 1,024 ways. Then FIED may be
read from a side F in 11 ways, and DEIFIED therefore in 121 ways. But there are eight side F's; consequently these give together 8 x 121 = 968 ways. Add 968 to 1,024 and we get the answer, 1,992.
In this form the solution will depend on whether the number of letters in the palindrome be odd or even. For example, if you apply the word NUN in precisely the same manner, you will get 64 different
readings; but if you use the word NOON, you will only get 56, because you cannot use the same letter twice in immediate succession (since you must "always pass from one letter to another") or
diagonal readings, and every reading must involve the use of the central N.
The reader may like to find for himself the general formula in this case, which is complex and difficult. I will merely add that for such a case as MADAM, dealt with in the same way as DEIFIED, the
number of readings is 400.
258.— THE VOTERS' PUZZLE.
THE number of readings here is 63,504, as in the case of "WAS IT A RAT I SAW" (No. 30, Canterbury Puzzles). The general formula is that for palindromic sentences containing 2n + 1 letters there are
(4(2^n -1)) squared readings.
259.— HANNAH'S PUZZLE.
Starting from any one of the N's, there are 17 different readings of NAH, or 68 (4 times 17) for the 4 N's. Therefore there are also 68 ways of spelling HAN. If we were allowed to use the same N
twice in a spelling, the answer would be 68 times 68, or 4,624 ways. But the conditions were, "always passing from one letter to another." Therefore, for every one of the 17 ways of spelling HAN with
a particular N, there would be 51 ways (3 times 17) of completing the NAH, or 867 (17 times 51) ways for the complete word. Hence, as there are four N's to use in HAN, the correct solution of the
puzzle is 3,468 (4 times 867) different ways.
260.—THE HONEYCOMB PUZZLE.
The required proverb is, "There is many a slip 'twixt the cup and the lip." Start at the T on the outside at the bottom right-hand corner, pass to the H above it, and the rest is easy.
261.— THE MONK AND THE BRIDGES.
The problem of the Bridges may be reduced to the simple diagram shown in illustration. The point M represents the Monk, the point I the Island, and the point Y the Monastery. Now the only direct ways
from M to I are by the bridges a and b; the only direct ways from I to Y are by the bridges c and d; and there is a direct way from M to Y by the bridge e. Now, what we have to do is to count all the
routes that will lead from M to Y, passing over all the bridges, a, b, c, d, and e once and once only. With the simple diagram under the eye it is quite easy, without any elaborate rule, to count
these routes methodically. Thus, starting from a, b, we find there are only two ways of completing the route; with a, c, there are only two routes; with a, d, only two routes; and so on. It will be
found that there are sixteen such routes in all, as in the following list:—
a b e c d b c d a e a b e d c b c e a d a c d b e b d c a e a c e b d b d e a c a d e b c e c a b d a d c b e e c b a d b a e c d e d a b c b a e d c e d b a c
If the reader will transfer the letters indicating the bridges from the diagram to the corresponding bridges in the original illustration, everything will be quite obvious.
262.—THOSE FIFTEEN SHEEP.
If we read the exact words of the writer in the cyclopaedia, we find that we are not told that the pens were all necessarily empty! In fact, if the reader will refer back to the illustration, he will
see that one sheep is already in one of the pens. It was just at this point that the wily farmer said to me, "Now I'm going to start placing the fifteen sheep." He thereupon proceeded to drive three
from his flock into the already occupied pen, and then placed four sheep in each of the other three pens. "There," says he, "you have seen me place fifteen sheep in four pens so that there shall be
the same number of sheep in every pen." I was, of course, forced to admit that he was perfectly correct, according to the exact wording of the question.
263.—KING ARTHUR'S KNIGHTS.
On the second evening King Arthur arranged the knights and himself in the following order round the table: A, F, B, D, G, E, C. On the third evening they sat thus, A, E, B, G, C, F, D. He thus had B
next but one to him on both occasions (the nearest possible), and G was the third from him at both sittings (the furthest position possible). No other way of sitting the knights would have been so
264.—THE CITY LUNCHEONS.
The men may be grouped as follows, where each line represents a day and each column a table:—
AB CD EF GH IJ KL AE DL GK FI CB HJ AG LJ FH KC DE IB AF JB KI HD LG CE AK BE HC IL JF DG AH EG ID CJ BK LF AI GF CL DB EH JK AC FK DJ LE GI BH AD KH LB JG FC EI AL HI JE BF KD GC AJ IC BG EK HL FD
Note that in every column (except in the case of the A's) all the letters descend cyclically in the same order, B, E, G, F, up to J, which is followed by B.
265.—A PUZZLE FOR CARD-PLAYERS.
In the following solution each of the eleven lines represents a sitting, each column a table, and each pair of letters a pair of partners.
A B I L E J G K F H C D A C J B F K H L G I D E A D K C G L I B H J E F A E L D H B J C I K F G A F B E I C K D J L G H A G C F J D L E K B H I A H D G K E B F L C I J A I E H L F C G B D J K A J F I
B G D H C E K L A K G J C H E I D F L B A L H K D I F J E G B C
It will be seen that the letters B, C, D ...L descend cyclically. The solution given above is absolutely perfect in all respects. It will be found that every player has every other player once as his
partner and twice as his opponent.
266.—A TENNIS TOURNAMENT.
Call the men A, B, D, E, and their wives a, b, d, e. Then they may play as follows without any person ever playing twice with or against any other person:—
First Court. Second Court. 1st Day A d against B e D a against E b 2nd Day A e " D b E a " B d 3rd Day A b " E d B a " D e
It will be seen that no man ever plays with or against his own wife—an ideal arrangement. If the reader wants a hard puzzle, let him try to arrange eight married couples (in four courts on seven
days) under exactly similar conditions. It can be done, but I leave the reader in this case the pleasure of seeking the answer and the general solution.
267.—THE WRONG HATS.
The number of different ways in which eight persons, with eight hats, can each take the wrong hat, is 14,833.
Here are the successive solutions for any number of persons from one to eight:—
1 = 0 2 = 1 3 = 2 4 = 9 5 = 44 6 = 265 7 = 1,854 8 = 14,833
To get these numbers, multiply successively by 2, 3, 4, 5, etc. When the multiplier is even, add 1; when odd, deduct 1. Thus, 3 x 1 - 1 = 2; 4 x 2 + 1 = 9; 5 x 9 - 1 = 44; and so on. Or you can
multiply the sum of the number of ways for n - 1 and n - 2 persons by n - 1, and so get the solution for n persons. Thus, 4(2 + 9) = 44; 5(9 + 44) = 265; and so on.
268.—THE PEAL OF BELLS.
The bells should be rung as follows:—
I have constructed peals for five and six bells respectively, and a solution is possible for any number of bells under the conditions previously stated.
269.—THREE MEN IN A BOAT.
If there were no conditions whatever, except that the men were all to go out together, in threes, they could row in an immense number of different ways. If the reader wishes to know how many, the
number is 455^7. And with the condition that no two may ever be together more than once, there are no fewer than 15,567,552,000 different solutions—that is, different ways of arranging the men. With
one solution before him, the reader will realize why this must be, for although, as an example, A must go out once with B and once with C, it does not necessarily follow that he must go out with C on
the same occasion that he goes with B. He might take any other letter with him on that occasion, though the fact of his taking other than B would have its effect on the arrangement of the other
Of course only a certain number of all these arrangements are available when we have that other condition of using the smallest possible number of boats. As a matter of fact we need employ only ten
different boats. Here is one the arrangements:—
1 2 3 4 5 1st Day (ABC) (DBF) (GHI) (JKL) (MNO) 8 6 7 9 10 2nd Day (ADG) (BKN) (COL) (JEI) (MHF) 3 5 4 1 2 3rd Day (AJM) (BEH) (CFI) (DKO) (GNL) 7 6 8 9 1 4th Day (AEK) (CGM) (BOI) (DHL) (JNF) 4 5 3
10 2 5th Day (AHN) (CDJ) (BFL) (GEO) (MKI) 6 7 8 10 1 6th Day (AFO) (BGJ) (CKH) (DNI) (MEL) 5 4 3 9 2 7th Day (AIL) (BDM) (CEN) (GKF) (JHO)
It will be found that no two men ever go out twice together, and that no man ever goes out twice in the same boat.
This is an extension of the well-known problem of the "Fifteen Schoolgirls," by Kirkman. The original conditions were simply that fifteen girls walked out on seven days in triplets without any girl
ever walking twice in a triplet with another girl. Attempts at a general solution of this puzzle had exercised the ingenuity of mathematicians since 1850, when the question was first propounded,
until recently. In 1908 and the two following years I indicated (see Educational Times Reprints, Vols. XIV., XV., and XVII.) that all our trouble had arisen from a failure to discover that 15 is a
special case (too small to enter into the general law for all higher numbers of girls of the form 6n+3), and showed what that general law is and how the groups should be posed for any number of
girls. I gave actual arrangements for numbers that had previously baffled all attempts to manipulate, and the problem may now be considered generally solved. Readers will find an excellent full
account of the puzzle in W.W. Rouse Ball's Mathematical Recreations, 5th edition.
270.—THE GLASS BALLS.
There are, in all, sixteen balls to be broken, or sixteen places in the order of breaking. Call the four strings A, B, C, and D—order is here of no importance. The breaking of the balls on A may
occupy any 4 out of these 16 places—that is, the combinations of 16 things, taken 4 together, will be
13 x 14 x 15 x 16 ————————- = 1,820 1 x 2 x 3 x 4
ways for A. In every one of these cases B may occupy any 4 out of the remaining 12 places, making
9 x 10 x 11 x 12 ————————- = 495 1 x 2 x 3 x 4
ways. Thus 1,820 x 495 = 900,900 different placings are open to A and B. But for every one of these cases C may occupy
5 x 6 x 7 x 8 ——————- = 70 1 x 2 x 3 x 4
different places; so that 900,900 x 70 = 63,063,000 different placings are open to A, B, and C. In every one of these cases, D has no choice but to take the four places that remain. Therefore the
correct answer is that the balls may be broken in 63,063,000 different ways under the conditions. Readers should compare this problem with No. 345, "The Two Pawns," which they will then know how to
solve for cases where there are three, four, or more pawns on the board.
271.—FIFTEEN LETTER PUZZLE.
The following will be found to comply with the conditions of grouping:—
ALE MET MOP BLM BAG CAP YOU CLT IRE OIL LUG LNR NAY BIT BUN BPR AIM BEY RUM GMY OAR GIN PLY CGR PEG ICY TRY CMN CUE COB TAU PNT ONE GOT PIU
The fifteen letters used are A, E, I, O, U, Y, and B, C, G, L, M, N, P, R, T. The number of words is 27, and these are all shown in the first three columns. The last word, PIU, is a musical term in
common use; but although it has crept into some of our dictionaries, it is Italian, meaning "a little; slightly." The remaining twenty-six are good words. Of course a TAU-cross is a T-shaped cross,
also called the cross of St. Anthony, and borne on a badge in the Bishop's Palace at Exeter. It is also a name for the toad-fish.
We thus have twenty-six good words and one doubtful, obtained under the required conditions, and I do not think it will be easy to improve on this answer. Of course we are not bound by dictionaries
but by common usage. If we went by the dictionary only in a case of this kind, we should find ourselves involved in prefixes, contractions, and such absurdities as I.O.U., which Nuttall actually
gives as a word.
272.—THE NINE SCHOOLBOYS.
The boys can walk out as follows:—
1st Day. 2nd Day. 3rd Day. A B C B F H F A G D E F E I A I D B G H I C G D H C E
4th Day. 5th Day. 6th Day. A D H G B I D C A B E G C F D E H B F I C H A E I G F
Every boy will then have walked by the side of every other boy once and once only.
Dealing with the problem generally, 12n+9 boys may walk out in triplets under the conditions on 9n+6 days, where n may be nought or any integer. Every possible pair will occur once. Call the number
of boys m. Then every boy will pair m-1 times, of which (m-1)/4 times he will be in the middle of a triplet and (m-1)/2 times on the outside. Thus, if we refer to the solution above, we find that
every boy is in the middle twice (making 4 pairs) and four times on the outside (making the remaining 4 pairs of his 8). The reader may now like to try his hand at solving the two next cases of 21
boys on 15 days, and 33 boys on 24 days. It is, perhaps, interesting to note that a school of 489 boys could thus walk out daily in one leap year, but it would take 731 girls (referred to in the
solution to No. 269) to perform their particular feat by a daily walk in a year of 365 days.
273.—THE ROUND TABLE.
The history of this problem will be found in The Canterbury Puzzles (No. 90). Since the publication of that book in 1907, so far as I know, nobody has succeeded in solving the case for that unlucky
number of persons, 13, seated at a table on 66 occasions. A solution is possible for any number of persons, and I have recorded schedules for every number up to 25 persons inclusive and for 33. But
as I know a good many mathematicians are still considering the case of 13, I will not at this stage rob them of the pleasure of solving it by showing the answer. But I will now display the solutions
for all the cases up to 12 persons inclusive. Some of these solutions are now published for the first time, and they may afford useful clues to investigators.
The solution for the case of 3 persons seated on 1 occasion needs no remark.
A solution for the case of 4 persons on 3 occasions is as follows:—
Each line represents the order for a sitting, and the person represented by the last number in a line must, of course, be regarded as sitting next to the first person in the same line, when placed at
the round table.
The case of 5 persons on 6 occasions may be solved as follows:—
1 2 3 4 5 1 2 4 5 3 1 2 5 3 4 ————- 1 3 2 5 4 1 4 2 3 5 1 5 2 4 3
The case for 6 persons on 10 occasions is solved thus:—
1 2 3 6 4 5 1 3 4 2 5 6 1 4 5 3 6 2 1 5 6 4 2 3 1 6 2 5 3 4 —————- 1 2 4 5 6 3 1 3 5 6 2 4 1 4 6 2 3 5 1 5 2 3 4 6 1 6 3 4 5 2
It will now no longer be necessary to give the solutions in full, for reasons that I will explain. It will be seen in the examples above that the 1 (and, in the case of 5 persons, also the 2) is
repeated down the column. Such a number I call a "repeater." The other numbers descend in cyclical order. Thus, for 6 persons we get the cycle, 2, 3, 4, 5, 6, 2, and so on, in every column. So it is
only necessary to give the two lines 1 2 3 6 4 5 and 1 2 4 5 6 3, and denote the cycle and repeaters, to enable any one to write out the full solution straight away. The reader may wonder why I do
not start the last solution with the numbers in their natural order, 1 2 3 4 5 6. If I did so the numbers in the descending cycle would not be in their natural order, and it is more convenient to
have a regular cycle than to consider the order in the first line.
The difficult case of 7 persons on 15 occasions is solved as follows, and was given by me in The Canterbury Puzzles:—
In this case the 1 is a repeater, and there are two separate cycles, 2, 3, 4, 2, and 5, 6, 7, 5. We thus get five groups of three lines each, for a fourth line in any group will merely repeat the
first line.
A solution for 8 persons on 21 occasions is as follows:—
The 1 is here a repeater, and the cycle 2, 3, 4, 5, 6, 7, 8. Every one of the 3 groups will give 7 lines.
Here is my solution for 9 persons on 28 occasions:—
There are here two repeaters, 1 and 2, and the cycle is 3, 4, 5, 6, 7, 8, 9. We thus get 4 groups of 7 lines each.
The case of 10 persons on 36 occasions is solved as follows:—
The repeater is 1, and the cycle, 2, 3, 4, 5, 6, 7, 8, 9, 10. We here have 4 groups of 9 lines each.
My solution for 11 persons on 45 occasions is as follows:—
There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5,... 11. We thus get 5 groups of 9 lines each.
The case of 12 persons on 55 occasions is solved thus:—
Here 1 is a repeater, and the cycle is 2, 3, 4, 5,... 12. We thus get 5 groups of 11 lines each.
274.—THE MOUSE-TRAP PUZZLE.
If we interchange cards 6 and 13 and begin our count at 14, we may take up all the twenty-one cards—that is, make twenty-one "catches"—in the following order: 6, 8, 13, 2, 10, 1, 11, 4, 14, 3, 5, 7,
21, 12, 15, 20, 9, 16, 18, 17, 19. We may also exchange 10 and 14 and start at 16, or exchange 6 and 8 and start at 19.
275.—THE SIXTEEN SHEEP.
The six diagrams on next page show solutions for the cases where we replace 2, 3, 4, 5, 6, and 7 hurdles. The dark lines indicate the hurdles that have been replaced. There are, of course, other ways
of making the removals.
276.—THE EIGHT VILLAS.
There are several ways of solving the puzzle, but there is very little difference between them. The solver should, however, first of all bear in mind that in making his calculations he need only
consider the four villas that stand at the corners, because the intermediate villas can never vary when the corners are known. One way is to place the numbers nought to 9 one at a time in the top
left-hand corner, and then consider each case in turn.
Now, if we place 9 in the corner as shown in the Diagram A, two of the corners cannot be occupied, while the corner that is diagonally opposite may be filled by 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9
persons. We thus see that there are 10
- -+ +- - - -+ O OHO O OHO O O O OHO O H + +=+ O OHO O OHO O O O OHOHO +-+ +-- - -+ + -+ + O O O O O O O O O O O O + -+ +-- - O O O O O O OHO O O O O - - - 2 3 4
--+ +- - - O O OHO OHO O O O O O O = = === O OHO O OHOHO O OHOHO O -- + + -+ + + + O O O O O O O O O OHO O = = + +=+ +=+ + O O O O OHO O O O O O O - - -+ + - - 5 6 7 THE SIXTEEN SHEEP
solutions with a 9 in the corner. If, however, we substitute 8, the two corners in the same row and column may contain 0, 0, or 1, 1, or 0, 1, or 1, 0. In the case of B, ten different selections may
be made for the fourth corner; but in each of the cases C, D, and E, only nine selections are possible, because we cannot use the 9. Therefore with 8 in the top left-hand corner there are 10 + (3 x
9) = 37 different solutions. If we then try 7 in the corner, the result will be 10 + 27 + 40, or 77 solutions. With 6 we get 10 + 27 + 40 + 49 = 126; with 5, 10 + 27 + 40 + 49 + 54 = 180; with 4, the
same as with 5, + 55 = 235 ; with 3, the same as with 4, + 52 = 287; with 2, the same as with 3, + 45 = 332; with 1, the same as with 2, + 34 = 366, and with nought in the top left-hand corner the
number of solutions will be found to be 10 + 27 + 40 + 49 + 54 + 55 + 52 + 45 + 34 + 19 = 385. As there is no other number to be placed in the top left-hand corner, we have now only to add these
totals together thus, 10 + 37 + 77 + 126 + 180 + 235 + 287 + 332 + 366 + 385 = 2,035. We therefore find that the total number of ways in which tenants may occupy some or all of the eight villas so
that there shall be always nine persons living along each side of the square is 2,035. Of course, this method must obviously cover all the reversals and reflections, since each corner in turn is
occupied by every number in all possible combinations with the other two corners that are in line with it.
Here is a general formula for solving the puzzle: (n squared + 3n + 2)(n squared + 3n + 3)/6. Whatever may be the stipulated number of residents along each of the sides (which number is represented
by n), the total number of different arrangements may be thus ascertained. In our particular case the number of residents was nine. Therefore (81 + 27 + 2) x (81 + 27 + 3) and the product, divided by
6, gives 2,035. If the number of residents had been 0, 1, 2, 3, 4, 5, 6, 7, or 8, the total arrangements would be 1, 7, 26, 70, 155, 301, 532, 876, or 1,365 respectively.
277.—COUNTER CROSSES.
Let us first deal with the Greek Cross. There are just eighteen forms in which the numbers may be paired for the two arms. Here they are:—
Of course, the number in the middle is common to both arms. The first pair is the one I gave as an example. I will suppose that we have written out all these crosses, always placing the first row of
a pair in the upright and the second row in the horizontal arm. Now, if we leave the central figure fixed, there are 24 ways in which the numbers in the upright may be varied, for the four counters
may be changed in 1 x 2 x 3 x 4 = 24 ways. And as the four in the horizontal may also be changed in 24 ways for every arrangement on the other arm, we find that there are 24 x 24 = 576 variations for
every form; therefore, as there are 18 forms, we get 18 x 576 = 10,368 ways. But this will include half the four reversals and half the four reflections that we barred, so we must divide this by 4 to
obtain the correct answer to the Greek Cross, which is thus 2,592 different ways. The division is by 4 and not by 8, because we provided against half the reversals and reflections by always reserving
one number for the upright and the other for the horizontal.
In the case of the Latin Cross, it is obvious that we have to deal with the same 18 forms of pairing. The total number of different ways in this case is the full number, 18 x 576. Owing to the fact
that the upper and lower arms are unequal in length, permutations will repeat by reflection, but not by reversal, for we cannot reverse. Therefore this fact only entails division by 2. But in every
pair we may exchange the figures in the upright with those in the horizontal (which we could not do in the case of the Greek Cross, as the arms are there all alike); consequently we must multiply by
2. This multiplication by 2 and division by 2 cancel one another. Hence 10,368 is here the correct answer.
278.—A DORMITORY PUZZLE.
MON. TUES. WED. - - - - - - - - - 1 2 1 1 3 1 1 4 1 - - - - - - - - - 2 2 1 1 1 1 - - - - - - - - - 1 22 1 3 19 3 4 16 4 - - - - - - - - -
THURS. FRI. SAT. - - - - - - - - - 1 5 1 2 6 2 4 4 4 - - - - - - - - - 2 2 1 1 4 4 - - - - - - - - - 4 13 4 7 6 7 4 4 4 - - - - - - - - -
Arrange the nuns from day to day as shown in the six diagrams. The smallest possible number of nuns would be thirty-two, and the arrangements on the last three days admit of variation.
279.—THE BARRELS OF BALSAM.
This is quite easy to solve for any number of barrels—if you know how. This is the way to do it. There are five barrels in each row Multiply the numbers 1, 2, 3, 4, 5 together; and also multiply 6,
7, 8, 9, 10 together. Divide one result by the other, and we get the number of different combinations or selections of ten things taken five at a time. This is here 252. Now, if we divide this by 6
(1 more than the number in the row) we get 42, which is the correct answer to the puzzle, for there are 42 different ways of arranging the barrels. Try this method of solution in the case of six
barrels, three in each row, and you will find the answer is 5 ways. If you check this by trial, you will discover the five arrangements with 123, 124, 125, 134, 135 respectively in the top row, and
you will find no others.
The general solution to the problem is, in fact, this:
n C 2n ——- n + 1
where 2n equals the number of barrels. The symbol C, of course, implies that we have to find how many combinations, or selections, we can make of 2n things, taken n at a time. | {"url":"https://www.p-books.com/book/Amusements-in-Mathematics-Henry-Ernest-Dudeney--8.html","timestamp":"2024-11-05T09:54:21Z","content_type":"text/html","content_length":"106797","record_id":"<urn:uuid:0367affd-0df9-4b13-a2b9-80ab986bd184>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00162.warc.gz"} |
A symbolic nonlinear theory for network observability
Work presented at NetSci 2018, Paris.
The observability of a complex system refers to the property of being able to infer its whole state by measuring the dynamics of a limited set of its variables. Since monitoring all the variables
defining the system’s state is experimentally infeasible or inefficient, it is of utmost importance to develop a methodological framework addressing the problem of targeting those variables yielding
full observability. Despite several approaches have been proposed, most of them neglect the nonlinear nature typically exhibited by complex systems and/or do not provide the space reconstructed from
the measured variables. On the one hand, since nonlinearities are often related to a lack of observability, linear approaches cannot properly address this problem. On the other hand, finding the
appropriate combination of sensors (and time derivatives) spanning the reconstructed space is a very time demanding computational task for large dimensional systems. Here, we adopt a nonlinear
symbolic approach taking into account the nature of the interactions among variables and analyze the distribution of the linear and nonlinear load of the variables in the symbolic Jacobian matrix of
the system [1].
[1] C. Letellier, I. Sendina-Nadal, E. Bianco-Martinez & M. S. Baptista, A symbolic network-based nonlinear theory for dynamical systems observability, Scientific Reports, 8, 3785, 2018. | {"url":"http://www.atomosyd.net/spip.php%3Farticle1%20inurl:/skelato/spip.php?article200","timestamp":"2024-11-10T12:42:15Z","content_type":"text/html","content_length":"12175","record_id":"<urn:uuid:0479854f-0e32-435e-b802-a7441e282e46>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00426.warc.gz"} |
tutorials-en Archives - Thinkable products for visually impaired people
Learn how to create mathematical graphs with coordinate systems and grids that are suitable for tactile graphics
Introduction: Graph and Grid
The TactileView drawing tool ‘Draw graph’ has a large number of examples for grids. A complete grid consists of the axis setup, formula(s) and tactile appearance settings. Selecting an example and
modifying this will get you in a few steps to a tactile usable graph that can be produced on swell paper, a braille embosser or on the motorised drawing arm (MDA).
Produced in braille or on swell paper, it also will present the text labels for the formula in different mathematical braille notations.
External math software such as MathType that uses MathML file format can be used to import a formula to have this plotted in a graph
│(((Placeholder for a picture containing both the tv screenshot and a photo of the printed, sketched and embossed version of the graph)))│
‘Graphs’ menu: Many options to chose from
The item in the main menu ‘Graphs’ and the icon in the vertical tool bar ‘Draw graph’ both bring up the same range of features to create a graph with ease. Ease is still a relative concept; to obtain
a tactile usable graph quite a number of aspects has to be thought through. Over 30 parameters can be set to design the axes, graph(s) and overall appearance.
Chicken–egg dilemma in tactile drawing with software
When you know what kind of formula you need to have plotted in a graph, you have aspects like the type of the scale, the range and texts for the axis etc. already in mind as well. In other words, do
you want to compose the axis setup first (as you would do on paper) so you can add the formula in a second step? Or do you want to enter the formula first (what you can actually do in software) and
afterwards adapt the axes fitting the produced graph? With this latter approach you have to be concerned about all the aspects that make the grid a tactile usable diagram.
Solve the dilemma: Grid examples as a starting point
The TactileView software has a number of examples with preset values available for graphs of various types. There are examples for coordinate systems (grids with just the axis settings) and also a
number of examples that contain a single formula or even multiple formulas.
A modified example grid can be stored as a ‘MyGrid’ for future use.
Learn how to create three different axis types in this tutorial
The following three worksheets will show you step by step how to create a grid with a linear axis scale, a logarithmic one or using degrees and radians as units.
Worksheet 1: Linear scale
Worksheet 2: Logarithmic Y scale to represent the Covid19 infection cases
Worksheet 3: Scale in degrees or radians for goniometrical functions
Dutch Air Traffic Control Tutorial
Learn how to get insight in Dutch air traffic control with MDA/TactiPad
In the Netherlands there is a discussion going on for quite some time now about opening up an additional airport Lelystad. One of the aspects is the need to re-arrange the air corridors as part of
the complexity of the air traffic control. To get an overview, a tactile map is helpful. How are the airports scattered around the country? Where are the major air corridors and how large are the
descent areas relative to the size of the country? In other words, how complex is it with 500.000 take offs and landings for airport Schiphol alone?
Photo: Map of the Netherlands sketched with MDA on TactiPad, showing airport locations, air corridors and descent areas.
Using the tools TactiPad and MDA
By using the TactiPad and the motorised drawing arm (MDA) you are able to add elements in consecutive steps to the tactile map. First the contour of the Netherlands, next the five cities that are
appointed as national airport already and the location of the sixth airport, Lelystad. Then some lines indicating the air corridors. Lastly the descent areas surrounding the cities. The information
for the map is supplied by www.routetactile.com.
You can watch the video or read the detailed instructions below.
Video: MDA interactive module Maps
Detailed instructions
Creating a map of the Netherlands
Adding the details: airports first
Adding the details: aircorridors and descent areas
By visiting www.routetactile.com with your browser you can obtain the same result. Maps of all sorts can be composed and downloaded in SVG file format. The TactileView software can be used to produce
the map on your Braille embosser or swell paper.
However, there are a few differences.
When using TactileView, you need to download the map, open it in the software and print it, whereas with the MDA, you compose the map and press the button ‘Sketch with MDA’. Even more, after
production of the map, you can still manually add details to the map by using a regular pen.
To have the right tools and a way to present the information in a fashionable manner can help you to understand complex situations. Let us know if you have any challenge that you would like us to
showcase with the Thinkable tools for tactile graphics. | {"url":"https://thinkable.nl/tag/tutorials-en/","timestamp":"2024-11-05T10:22:30Z","content_type":"text/html","content_length":"153878","record_id":"<urn:uuid:66ac79a1-f5e8-4292-ae9c-128cbf6d3e59>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00722.warc.gz"} |
Mohammad Hossein Rohban
Here are my presentations in the Machine Learning session, which is held for ISS students in ECE Department of Boston University :
• Poster : An Impossibility Result for High Dimensional Supervised Learning (1st May 2013) : This is a poster version of the my results on distribution dependent impossibility of learning in high
dimensions. It was presented in the New England Machine Learning Day at Microsoft Research Cambridge MA.
• Slides : An Impossibility Result for High Dimensional Supervised Learning (12 Apr. 2013) : This presentation is some of my results on distribution dependent impossibility of learning in a high
dimensional setting.
• Slides : Fully Efficient Nonnegative Matrix Factorization (14 Dec. 2012) : This presentation is some rephrasing of the 5th section of the following paper : Arora, S., Ge, R., Kannan, R., and
Moitra, A., ”Computing a nonnegative matrix factorization, provably,” In ACM Symposium on Theory of Computing (STOC), 2012. | {"url":"https://blogs.bu.edu/mhrohban/presentations/","timestamp":"2024-11-14T17:32:33Z","content_type":"application/xhtml+xml","content_length":"62514","record_id":"<urn:uuid:ca2fc4c5-e95a-4394-8ff7-0fc9c5bb6d6e>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00027.warc.gz"} |
Physicists discover a novel quantum state in an elemental solid
Physicists have observed a novel quantum effect termed “hybrid topology” in a crystalline material. This finding opens up a new range of possibilities for the development of efficient materials and
technologies for next-generation quantum science and engineering.
The finding, published on April 10th in the journal Nature, came when Princeton scientists discovered that an elemental solid crystal made of arsenic (As) atoms hosts a never-before-observed form of
topological quantum behavior. They were able to explore and image this novel quantum state using a scanning tunneling microscope (STM) and photoemission spectroscopy, the latter a technique used to
determine the relative energy of electrons in molecules and atoms.
This state combines, or “hybridizes,” two forms of topological quantum behavior—edge states and surface states, which are two types of quantum two-dimensional electron systems. These have been
observed in previous experiments, but never simultaneously in the same material where they mix to form a new state of matter.
“This finding was completely unexpected,” said M. Zahid Hasan, the Eugene Higgins Professor of Physics at Princeton University, who led the research.
Nobody predicted it in theory before its observation.
In recent years, the study of topological states of matter has attracted considerable attention among physicists and engineers and is presently the focus of much international interest and research.
This area of study combines quantum physics with topology — a branch of theoretical mathematics that explores geometric properties that can be deformed but not intrinsically changed.
For more than a decade, scientists have used bismuth (Bi)-based topological insulators to demonstrate and explore exotic quantum effects in bulk solids mostly by manufacturing compound materials,
like mixing Bi with selenium (Se), for example. However, this experiment is the first time topological effects have been discovered in crystals made of the element As.
“The search and discovery of novel topological properties of matter have emerged as one of the most sought-after treasures in modern physics, both from a fundamental physics point of view and for
finding potential applications in next-generation quantum science and engineering,” said Hasan. “The discovery of this new topological state made in an elemental solid was enabled by multiple
innovative experimental advances and instrumentations in our lab at Princeton.”
An elemental solid serves as an invaluable experimental platform for testing various concepts of topology. Up until now, bismuth has been the only element that hosts a rich tapestry of topology,
leading to two decades of intensive research activities. This is partly attributed to the material’s cleanliness and the ease of synthesis. However, the current discovery of even richer topological
phenomena in arsenic will potentially pave the way for new and sustained research directions.
“For the first time, we demonstrate that akin to different correlated phenomena, distinct topological orders can also interact and give rise to new and intriguing quantum phenomena,” Hasan said.
A topological material is the main component used to investigate the mysteries of quantum topology. This device acts as an insulator in its interior, which means that the electrons inside are not
free to move around and therefore do not conduct electricity. However, the electrons on the device’s edges are free to move around, meaning they are conductive. Moreover, because of the special
properties of topology, the electrons flowing along the edges are not hampered by any defects or deformations. This of type device has the potential not only to improve technology but also to
generate a greater understanding of matter itself by probing quantum electronic properties.
Hasan noted that there is much interest in using topological materials for practical applications. But two important advances need to happen before this can be realized. First, quantum topological
effects must be manifested at higher temperatures. Second, simple and elemental material systems (like silicon for conventional electronics) that can host topological phenomena need to be found.
“In our labs we have efforts in both directions — we are searching for simpler materials systems with ease of fabrication where essential topological effects can be found,” said Hasan. “We are also
searching for how these effects can be made to survive at room temperature.”
Background of the experiment
The discovery’s roots lie in the workings of the quantum Hall effect — a form of topological effect that was the subject of the Nobel Prize in Physics in 1985. Since that time, topological phases
have been studied and many new classes of quantum materials with topological electronic structures have been found. Most notably, Daniel Tsui, the Arthur Legrand Doty Professor of Electrical
Engineering, Emeritus, at Princeton, won the 1998 Nobel Prize in Physics for discovering the fractional quantum Hall effect. Similarly, F. Duncan Haldane, the Eugene Higgins Professor of Physics at
Princeton, won the 2016 Nobel Prize in Physics for theoretical discoveries of topological phase transitions and a type of two-dimensional (2D) topological insulator. Subsequent theoretical
developments showed that topological insulators can take the form of two copies of Haldane’s model based on an electron’s spin-orbit interaction.
Hasan and his research team have been following in the footsteps of these researchers by investigating other aspects of topological insulators and searching for novel states of matter. This led them,
in 2007, to the discovery of the first examples of three-dimensional (3D) topological insulators. Since then, Hasan and his team have been on a decade-long search for a new topological state in its
simplest form that can also operate at room temperature.
“A suitable atomic chemistry and structure design coupled to first-principles theory is the crucial step to make topological insulator’s speculative prediction realistic in a high-temperature
setting,” said Hasan. “There are hundreds of quantum materials, and we need both intuition, experience, materials-specific calculations, and intense experimental efforts to eventually find the right
material for in-depth exploration. And that took us on a decade-long journey of investigating many bismuth-based materials leading to many foundational discoveries.”
The experiment
Bismuth-based materials are capable, at least in principle, of hosting a topological state of matter at high temperatures. But these require complex materials preparation under ultra-high vacuum
conditions, so the researchers decided to explore several other systems. Postdoctoral researcher Md. Shafayat Hossain suggested a crystal made of arsenic because it can be grown in a form that is
cleaner than many bismuth compounds.
When Hossain and Yuxiao Jiang, a graduate student in the Hasan group, turned the STM on the arsenic sample, they were greeted with a dramatic observation — grey arsenic, a form of arsenic with a
metallic appearance, harbors both topological surface states and edge states simultaneously.
“We were surprised. Grey arsenic was supposed to have only surface states. But when we examined the atomic step edges, we also found beautiful conducting edge modes,” said Hossain.
“An isolated monolayer step edge should not have a gapless edge mode,” added Jiang, a co-first author of the study.
This is what is seen in calculations by Frank Schindler, a postdoctoral fellow and condensed matter theorist at the Imperial College London in the United Kingdom, and Rajibul Islam, a postdoctoral
researcher at the University of Alabama in Birmingham, Alabama. Both are co-first authors of the paper.
“Once an edge is placed on top of the bulk sample, the surface states hybridize with the gapped states on the edge and form a gapless state,” Schindler said.
“This is the first time we have seen such a hybridization,” he added.
Physically, such a gapless state on the step edge is not expected for either strong or higher-order topological insulators separately, but only for hybrid materials where both kinds of quantum
topology are present. This gapless state is also unlike surface or hinge states in strong and higher-order topological insulators, respectively. This meant that the experimental observation by the
Princeton team immediately indicated a never-before-observed type of topological state.
David Hsieh, Chair of the Physics Division at Caltech and a researcher who was not involved in the study, pointed to the study’s innovative conclusions.
“Typically, we consider the bulk band structure of a material to fall into one of several distinct topological classes, each tied to a specific type of boundary state,” Hsieh said. “This work shows
that certain materials can simultaneously fall into two classes. Most interestingly, the boundary states emerging from these two topologies can interact and reconstruct into a new quantum state that
is more than just a superposition of its parts.”
The researchers further substantiated the scanning tunneling microscopy measurements with systematic high-resolution angle-resolved photoemission spectroscopy.
“The grey As sample is very clean and we found clear signatures of a topological surface state,” said Zi-Jia Cheng, a graduate student in the Hasan group and a co-first author of the paper who
performed some of the photoemission measurements.
The combination of multiple experimental techniques enabled the researchers to probe the unique bulk-surface-edge correspondence associated with the hybrid topological state — and corroborate the
experimental findings.
Implications of the findings
The impact of this discovery is two-fold. The observation of the combined topological edge mode and the surface state paves the way to engineer new topological electron transport channels. This may
enable the designing of new quantum information science or quantum computing devices. The Princeton researchers demonstrated that the topological edge modes are only present along specific
geometrical configurations that are compatible with the crystal’s symmetries, illuminating a pathway to design various forms of future nanodevices and spin-based electronics.
From a broader perspective, society benefits when new materials and properties are discovered, Hasan said. In quantum materials, the identification of elemental solids as material platforms, such as
antimony hosting a strong topology or bismuth hosting a higher-order topology, has led to the development of novel materials that have immensely benefited the field of topological materials.
“We envision that arsenic, with its unique topology, can serve as a new platform at a similar level for developing novel topological materials and quantum devices that are not currently accessible
through existing platforms,” said Hasan. “A new exciting frontier in material science and novel physics awaits!”
The Princeton group has designed and built novel experiments for the exploration of topological insulator materials for over 15 years. Between 2005 and 2007, for example, the team led by Hasan
discovered topological order in a three-dimensional bismuth-antimony bulk solid, a semiconducting alloy, and related topological Dirac materials using novel experimental methods. This led to the
discovery of topological magnetic materials. Between 2014 and 2015, they discovered and developed a new class of topological materials called magnetic Weyl semimetals.
The researchers believe this finding will open the door to a whole host of future research possibilities and applications in quantum technologies, especially in so-called “green” technologies.
“Our research is a step forward in demonstrating the potential of topological materials for quantum electronics with energy-saving applications,” Hasan said.
The team included numerous researchers from Princeton’s Department of Physics, including present and past graduate students Yu-Xiao Jiang, Maksim Litskevich, Xian P. Yang, Zi-Jia Cheng, Tyler
Cochran, Nana Shumiya, and Daniel Multer, and present and past postdoctoral research associates Shafayat Hossain, Jia-Xin Yin, Guoqing Chang, and Qi Zhang.
The paper, “A hybrid topological quantum state in an elemental solid,” by Md Shafayat Hossain, Frank Schindler, Rajibul Islam, Zahir Muhammad, Yu-Xiao Jiang, Zi-Jia Cheng, Qi Zhang, Tao Hou, Hongyu
Chen, Maksim Litskevich, Brian Casas, Jia-Xin Yin, Tyler A. Cochran, Mohammad Yahyavi, Xian P. Yang, Luis Balicas, Guoqing Chang, Weisheng Zhao, Titus Neupert, and M. Zahid Hasan was published online
in the April 10 issue of Nature.
Primary support for the work at Princeton is from the U.S. Department of Energy (DOE) Office of Science, the National Quantum Information (NQI) Science Research Centers, the Quantum Science Center
(QSC at ORNL), and Princeton University. Support from the U.S. DOE under the Basic Energy Sciences program (grant number DOE/BES DE-FG-02-05ER46200) was provided for the theory and advanced ARPES
experiments. Support for advanced STM Instrumentation and theory work comes from the Gordon and Betty Moore Foundation (GBMF9461). Additional support is reported in the paper.
Image: A representation of data visualization of quantum states of electrons on the surface and edge of grey arsenic crystal obtained using a scanning tunneling microscope at Princeton’s physics
department. Credit: Image based on STM data simulations prepared by Shafayat Hossain and the Zahid Hasan group at the Laboratory for Topological Quantum Matter at Princeton University.
The original story can be accessed here.
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IB Mathematics: Analysis and Approaches
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Citadel Construction
Problem C
Citadel Construction
The army wants to put up a new base in dangerous territory. The base will be surrounded by a citadel consisting of a number of straight walls. At every corner where two walls meet, there will be a
watchtower. In order to enable efficient coordination in case of attack, the number of watchtowers should not exceed four.
After a detailed survey of the area, the army has concluded that not all locations are equally suitable for a watchtower: the ground needs to be firm enough to support the weight of the tower, and
there should be enough visibility. They have selected a number of locations that meet the requirements.
Within these constraints, the army would like to build a base that is as large as possible. For this, they have come to you for help. Since the results of the survey are of course classified, you
will not be given the possible locations directly; rather, you should write a program that can take them as input. Furthermore, they want the program to be able to handle multiple test cases in one
go, so that they can hide the real data among lots of fake data.
For the computation of the area, you may assume that the watchtowers are infinitesimal in size and the walls infinitesimal in width.
On the first line one positive number: the number of test cases, at most 100. After that per test case:
• one line with a single integer $n$ ($ 3 \leq n \leq 1\, 000$): the number of locations that are suitable for a watchtower.
• $n$ lines, each with two space-separated integers $x$ and $y$ ($-10\, 000 \leq x,y \leq 10\, 000$): the coordinates of each location.
All locations are distinct.
Per test case:
• one line with a single number: the largest possible area that the base can have. This number will be either an integer or a half-integer. If it is an integer, print that integer; if it is a
half-integer, print the integer part followed by “.5”. Trailing zeros are not allowed.
Sample Input 1 Sample Output 1
-2 -2
3 -2 100
0 1 12.5 | {"url":"https://open.kattis.com/contests/t24oan/problems/citadelconstruction","timestamp":"2024-11-02T22:16:54Z","content_type":"text/html","content_length":"31818","record_id":"<urn:uuid:cd6839b9-d989-4cf7-ba03-75fd80a0f997>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00191.warc.gz"} |
Transactions Online
Kumiko MAEBASHI, Nobuo SUEMATSU, Akira HAYASHI, "Component Reduction for Gaussian Mixture Models" in IEICE TRANSACTIONS on Information, vol. E91-D, no. 12, pp. 2846-2853, December 2008, doi: 10.1093/
Abstract: The mixture modeling framework is widely used in many applications. In this paper, we propose a component reduction technique, that collapses a Gaussian mixture model into a Gaussian
mixture with fewer components. The EM (Expectation-Maximization) algorithm is usually used to fit a mixture model to data. Our algorithm is derived by extending mixture model learning using the
EM-algorithm. In this extension, a difficulty arises from the fact that some crucial quantities cannot be evaluated analytically. We overcome this difficulty by introducing an effective
approximation. The effectiveness of our algorithm is demonstrated by applying it to a simple synthetic component reduction task and a phoneme clustering problem.
URL: https://global.ieice.org/en_transactions/information/10.1093/ietisy/e91-d.12.2846/_p
author={Kumiko MAEBASHI, Nobuo SUEMATSU, Akira HAYASHI, },
journal={IEICE TRANSACTIONS on Information},
title={Component Reduction for Gaussian Mixture Models},
abstract={The mixture modeling framework is widely used in many applications. In this paper, we propose a component reduction technique, that collapses a Gaussian mixture model into a Gaussian
mixture with fewer components. The EM (Expectation-Maximization) algorithm is usually used to fit a mixture model to data. Our algorithm is derived by extending mixture model learning using the
EM-algorithm. In this extension, a difficulty arises from the fact that some crucial quantities cannot be evaluated analytically. We overcome this difficulty by introducing an effective
approximation. The effectiveness of our algorithm is demonstrated by applying it to a simple synthetic component reduction task and a phoneme clustering problem.},
TY - JOUR
TI - Component Reduction for Gaussian Mixture Models
T2 - IEICE TRANSACTIONS on Information
SP - 2846
EP - 2853
AU - Kumiko MAEBASHI
AU - Nobuo SUEMATSU
AU - Akira HAYASHI
PY - 2008
DO - 10.1093/ietisy/e91-d.12.2846
JO - IEICE TRANSACTIONS on Information
SN - 1745-1361
VL - E91-D
IS - 12
JA - IEICE TRANSACTIONS on Information
Y1 - December 2008
AB - The mixture modeling framework is widely used in many applications. In this paper, we propose a component reduction technique, that collapses a Gaussian mixture model into a Gaussian mixture
with fewer components. The EM (Expectation-Maximization) algorithm is usually used to fit a mixture model to data. Our algorithm is derived by extending mixture model learning using the EM-algorithm.
In this extension, a difficulty arises from the fact that some crucial quantities cannot be evaluated analytically. We overcome this difficulty by introducing an effective approximation. The
effectiveness of our algorithm is demonstrated by applying it to a simple synthetic component reduction task and a phoneme clustering problem.
ER - | {"url":"https://global.ieice.org/en_transactions/information/10.1093/ietisy/e91-d.12.2846/_p","timestamp":"2024-11-07T07:54:52Z","content_type":"text/html","content_length":"59270","record_id":"<urn:uuid:f9ed5d82-d845-4293-a418-23b153183382>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00066.warc.gz"} |
ACO Seminar
The ACO Seminar (2014–2015)
Oct. 30, 3:30pm, Gates 8115 (Note unusual location)
Thomas Rothvoß, University of Washington
The matching polytope has exponential extension complexity
A popular method in combinatorial optimization is to express polytopes P, which may potentially have exponentially many facets, as solutions of linear programs that use few extra variables to reduce
the number of constraints down to a polynomial. After two decades of standstill, recent years have brought amazing progress in showing lower bounds for the so called extension complexity, which for a
polytope P denotes the smallest number of inequalities necessary to describe a higher dimensional polytope Q that can be linearly projected on P. However, the central question in this field remained
wide open: can the perfect matching polytope be written as an LP with polynomially many constraints? We answer this question negatively. In fact, the extension complexity of the perfect matching
polytope in a complete n-node graph is 2^Ω(n). | {"url":"https://aco.math.cmu.edu/abs-14-15/oct30.html","timestamp":"2024-11-05T05:42:43Z","content_type":"text/html","content_length":"2441","record_id":"<urn:uuid:19ca54ac-b388-4788-aad3-417824ec2134>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00473.warc.gz"} |
Present and Future Value Concepts: Mastering PV and FV in Financial Planning
1.5.3 Present and Future Value Concepts
In the realm of finance and investment, understanding the concepts of Present Value (PV) and Future Value (FV) is crucial for making informed decisions. These concepts form the backbone of financial
analysis, allowing individuals and businesses to evaluate the worth of investments, loans, and other financial instruments over time. This section will delve into the intricacies of PV and FV,
providing you with the tools to calculate and apply these concepts in various financial contexts.
Distinguishing Present Value and Future Value
Present Value (PV) is the current worth of a sum of money that is to be received or paid in the future, discounted at a specific interest rate. It answers the question: “How much is a future sum of
money worth today?”
Future Value (FV), on the other hand, is the amount of money an investment made today will grow to at a specified future date, given a certain interest rate. It answers the question: “What will my
investment be worth in the future?”
Understanding these concepts is essential for evaluating the time value of money, which is the idea that a dollar today is worth more than a dollar in the future due to its potential earning
Calculating Present Value and Future Value
Present Value Formula
The formula for calculating the present value of a future sum is:
$$ PV = \frac{FV}{(1 + r)^n} $$
• \( PV \) = Present Value
• \( FV \) = Future Value
• \( r \) = interest rate (as a decimal)
• \( n \) = number of periods
Future Value Formula
The formula for calculating the future value of a present sum is:
$$ FV = PV \times (1 + r)^n $$
• \( FV \) = Future Value
• \( PV \) = Present Value
• \( r \) = interest rate (as a decimal)
• \( n \) = number of periods
Example Calculation
Suppose you want to find out the present value of $1,000 to be received in 5 years, with an annual interest rate of 5%. Using the PV formula:
$$ PV = \frac{1000}{(1 + 0.05)^5} \approx 783.53 $$
This means that $783.53 today is equivalent to $1,000 in 5 years at a 5% interest rate.
Calculating PV and FV of Annuities
An annuity is a series of equal payments made at regular intervals. There are two types of annuities: ordinary annuities and annuities due.
Ordinary Annuities
An ordinary annuity is a series of equal payments made at the end of each period. The formulas for calculating the present and future values of an ordinary annuity are:
Present Value of an Ordinary Annuity:
$$ PV_{\text{annuity}} = P \times \left(1 - \frac{1}{(1 + r)^n}\right) / r $$
Future Value of an Ordinary Annuity:
$$ FV_{\text{annuity}} = P \times \left(\frac{(1 + r)^n - 1}{r}\right) $$
• \( P \) = payment amount per period
• \( r \) = interest rate per period
• \( n \) = number of periods
Annuities Due
An annuity due is a series of equal payments made at the beginning of each period. The formulas for calculating the present and future values of an annuity due are similar to those of an ordinary
annuity, but they are adjusted to account for the timing of the payments.
Present Value of an Annuity Due:
$$ PV_{\text{annuity due}} = PV_{\text{annuity}} \times (1 + r) $$
Future Value of an Annuity Due:
$$ FV_{\text{annuity due}} = FV_{\text{annuity}} \times (1 + r) $$
Discounting Cash Flows
Discounting is the process of determining the present value of a future cash flow. It involves applying a discount rate to future cash flows to account for the time value of money. The discount rate
is often based on the cost of capital, inflation, or a required rate of return.
Importance of Discount Rates
The choice of discount rate is crucial in present value calculations as it reflects the opportunity cost of capital. A higher discount rate results in a lower present value, indicating that future
cash flows are worth less today. Conversely, a lower discount rate increases the present value.
Applications in Financial Planning
PV and FV concepts are widely used in financial planning to evaluate various financial decisions, including:
• Loans and Mortgages: Calculating the present value of loan payments helps determine the total cost of borrowing and the affordability of a loan.
• Investment Projects: Evaluating the future value of investments aids in assessing potential returns and making informed investment choices.
• Retirement Planning: Estimating the future value of retirement savings helps individuals plan for a financially secure retirement.
Detailed Calculation Examples
Example 1: Loan Evaluation
Consider a loan of $10,000 with an annual interest rate of 6% to be repaid over 5 years. To determine the monthly payment, we use the present value of an annuity formula:
$$ P = \frac{PV \times r}{1 - (1 + r)^{-n}} $$
• \( PV = 10,000 \)
• \( r = 0.06/12 \) (monthly interest rate)
• \( n = 5 \times 12 \) (total number of payments)
Substituting the values:
$$ P = \frac{10,000 \times 0.005}{1 - (1 + 0.005)^{-60}} \approx 193.33 $$
The monthly payment is approximately $193.33.
Example 2: Investment Growth
Suppose you invest $5,000 in a fund that offers an annual return of 8% for 10 years. To find the future value of this investment:
$$ FV = 5000 \times (1 + 0.08)^{10} \approx 10,794.62 $$
The investment will grow to approximately $10,794.62 in 10 years.
Utility in Comparing Financial Options
PV and FV calculations are invaluable tools for comparing different financial options. By converting future cash flows to present values, you can objectively assess the value of various investment
opportunities, loans, and savings plans. This comparison enables better decision-making by highlighting the most financially beneficial options.
Relevance in Various Financial Contexts
The relevance of PV and FV extends beyond personal finance to corporate finance, real estate, and government projects. These concepts are integral to:
• Capital Budgeting: Evaluating the profitability of long-term investments and projects.
• Bond Valuation: Determining the present value of future bond payments to assess their attractiveness.
• Lease Agreements: Calculating the present value of lease payments to compare the cost of leasing versus buying.
Mastering the concepts of present and future value is essential for anyone involved in financial planning, investment analysis, or cash flow management. By understanding how to calculate and apply PV
and FV, you can make informed decisions that maximize financial outcomes and minimize risks.
Quiz Time!
📚✨ Quiz Time! ✨📚
### What is the present value of $1,000 to be received in 3 years at an annual interest rate of 4%? - [ ] $1,124.86 - [x] $889.00 - [ ] $960.00 - [ ] $1,040.00 > **Explanation:** Using the PV
formula: \\( PV = \frac{1000}{(1 + 0.04)^3} \approx 889.00 \\). ### How does an annuity due differ from an ordinary annuity? - [x] Payments are made at the beginning of each period in an annuity due.
- [ ] Payments are made at the end of each period in an annuity due. - [ ] Annuity due has a lower present value than an ordinary annuity. - [ ] Annuity due has a higher future value than an ordinary
annuity. > **Explanation:** An annuity due involves payments at the beginning of each period, unlike an ordinary annuity where payments are made at the end. ### Which of the following best describes
the time value of money? - [x] A dollar today is worth more than a dollar in the future. - [ ] A dollar today is worth less than a dollar in the future. - [ ] A dollar today is worth the same as a
dollar in the future. - [ ] A dollar today is only worth more if invested. > **Explanation:** The time value of money principle states that a dollar today is worth more due to its potential earning
capacity. ### What is the future value of $2,000 invested for 5 years at an annual interest rate of 7%? - [ ] $2,500.00 - [ ] $2,800.00 - [x] $2,805.25 - [ ] $3,000.00 > **Explanation:** Using the FV
formula: \\( FV = 2000 \times (1 + 0.07)^5 \approx 2,805.25 \\). ### What role does the discount rate play in present value calculations? - [x] It reflects the opportunity cost of capital. - [ ] It
determines the future value of cash flows. - [ ] It is irrelevant to present value calculations. - [ ] It increases the present value of future cash flows. > **Explanation:** The discount rate
reflects the opportunity cost of capital and affects the present value of future cash flows. ### How is the present value of an annuity due calculated? - [ ] Using the same formula as an ordinary
annuity. - [x] By multiplying the present value of an ordinary annuity by (1 + r). - [ ] By dividing the present value of an ordinary annuity by (1 + r). - [ ] By subtracting the present value of an
ordinary annuity from the future value. > **Explanation:** The present value of an annuity due is calculated by multiplying the present value of an ordinary annuity by (1 + r). ### Which financial
decision can be evaluated using future value calculations? - [ ] Determining the cost of a loan. - [x] Estimating the growth of an investment. - [ ] Calculating the present value of a bond. - [ ]
Assessing the affordability of a mortgage. > **Explanation:** Future value calculations are used to estimate the growth of investments over time. ### What is the primary purpose of discounting cash
flows? - [ ] To increase the future value of investments. - [ ] To determine the interest rate of a loan. - [x] To calculate the present value of future cash flows. - [ ] To compare different
investment options. > **Explanation:** Discounting cash flows is used to calculate their present value, reflecting the time value of money. ### Which of the following is a common application of
present value in corporate finance? - [ ] Calculating the future value of savings. - [ ] Estimating the growth of retirement funds. - [x] Evaluating the profitability of investment projects. - [ ]
Determining the interest rate of a bond. > **Explanation:** Present value is commonly used in corporate finance to evaluate the profitability of investment projects. ### True or False: The future
value of an annuity due is higher than that of an ordinary annuity, given the same interest rate and number of periods. - [x] True - [ ] False > **Explanation:** The future value of an annuity due is
higher because payments are made at the beginning of each period, allowing more time for interest to accrue. | {"url":"https://csccourse.ca/1/5/3/","timestamp":"2024-11-09T14:04:14Z","content_type":"text/html","content_length":"94158","record_id":"<urn:uuid:075ff8ea-a9cd-449b-845d-ae1f4b147d80>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00305.warc.gz"} |
How to conduct a meta-analysis of genome-wide association studies (GWAS) in biostatistics? | Hire Some To Take My Statistics Exam
How to conduct a meta-analysis of genome-wide association studies (GWAS) in biostatistics? Many meta-analyses have been performed to validate the that site of detecting genetic associations to the
relationship between genes in the genome and multiple traits or outcomes. This is especially the case in a multi-samples meta-analysis that uses only the summary associations of genes or phenotypes
and gene expression as inputs. Such a meta-analysis is commonly done by the *Cochlearia hermosa L*. however, in the case of multi-samples studies the number of replicates is relatively low and click
here to find out more the total number of genetic associations done in the meta-analysis is low. Möllendorf et al. performed the meta-analysis in the next generation of biostatistical and
meta-analyses and found that the multiple associations detected with This Site than one locus were relatively weak by the Cochlearia hermosa L, the only locus associated with less than one value in
the gene and in the phenotype. These papers do confirm the power of the studies to detect significant biological parameters with real number of replicated associations and instead also address the
problems of the number of replicates and the lack of power due to genomewide association calls in the overall comparison of two meta-analyses performed in real-world settings. Therefore, we suggest
that a one-dummy randomness (or random sum of squares) argument can be used to provide a statistical statement pop over to this web-site the power of individual-based meta-analyses as to the
two-dummy outcome of one-by-one comparisons under the conditions of using both the Cochlearia hermosa L and the Cochlearia hermosa L. These arguments further provide proof-of-the-ability to perform
the biostatistic meta-analyses and to realize the power of the Cochlearia hermosa L being an independent meta-analysis in the genome-wide association setting.How to conduct a meta-analysis of
genome-wide association studies (GWAS) in biostatistics? There is growing interest in conducting GWAS in genetic epidemiology and medicine, with the contribution of pre-analytical analyses of genomic
databases. Post-analytical data analysis often requires a bioinformatic model or other software package; instead, automated analyses such as SPSS/SNAPS to monitor and generate the final GWAS results
are widely employed. In this study, we compared the genome-wide associations of different phenotypes between breast cancer genetic data and GWAS-derived data of non-breast cancers and
phenotypic-based models for breast and non-breast cancer datasets. By obtaining a high-quality data for each subject, we can generate a global genome-wide data for every individual breast cancer data
set. We used the LocusMV5 in-house SNP array and principal component analysis (PCA) software to determine the number and genetic association scores between the human breast carcinoma GWAS-derived
data, and breast cancer GWAS data, in the meta-analysis and the literature. The data results showed that there were significant inverse family-wise association signals with higher odds for breast
cancer and non-breast cancers. The reduced number of the phenotypic scores for both breast cancer and non-breast cancers was not due to reduced statistical power, but was a function of p-values and
p-values for each breast cancer and non-breast cancer genotype, rather than the corresponding p-values for p-values for p-values associated with individual breast cancers. The p-values of these
associations were significant at the p<0.05 level, whereas their correlation coefficients were low and lower when considered as minor effects. This study indicated that both the phenotypic and
genomic scores were increased by cancer type and allele, while we found that only the terms of these scores positively correlated in the heteroscedastic data. These results suggest that the increased
homoscedastic power of phenotypic-based data is associatedHow to conduct a meta-analysis of genome-wide association studies (GWAS) in biostatistics? Metere-------- In a meta-analysis, the
meta-analysis (based on visit Genomes Assay Meta-Analysis) was used to verify the findings.
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In 2019, we conducted a GWAS based on 1000 Genomes Assay Meta-Analysis of DNA methylation, at 7 sites. The meta-analysis was carried out to assess the association between the expression of loci with
DNA methylation at learn this here now sites and the patients’ risk of being diagnosed with cancer. Further, the GWAS and meta-analysis were conducted to evaluate the effect of SNP (SNP-GAT)
genotype, rs^2^ and SNPs (SNPs-GAT) as independent variables of genetic risk (risk for cancer and disease) and the interaction effect of SNP-GAT genotype and SNP allele in relation to the clinical
outcome. The clinical outcome of patients’ cancer diagnosis were compared with those that were evaluated by the meta-analysis. We used the R package STRUCTURE 2.2.2 to conduct the meta-analysis. The
total number of genes assigned in the meta-analysis was 1821 and the number of genes assigned in the GWAS was 810 wikipedia reference 2015. The number of genes assigned in the meta-analysis was 369
and 378 for find out this here and rs^2^, respectively. Other statistical procedures according to the literature reports were conducted in this article. To address the problem of missing data, we
performed *R*^2^ test on the data between the data in the meta-analysis and the independent variables of the meta-analysis to estimate the null hypothesis. In the meta-analysis, the meta-analysis is
conducted with multiple comparisons using Monte-Carlo simulations [@b70]. During the simulation, the null hypothesis that each SNP effect was determined by randomly sampling two observations from
each data set resulted from the true null hypothesis (*N*=6). Based on Simulation *If* number of | {"url":"https://hireforstatisticsexam.com/how-to-conduct-a-meta-analysis-of-genome-wide-association-studies-gwas-in-biostatistics","timestamp":"2024-11-07T03:41:16Z","content_type":"text/html","content_length":"170496","record_id":"<urn:uuid:b2a3df45-5492-455d-bd16-7858adad1798>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00612.warc.gz"} |
COUNTING UP TO 20 OBJECTS | FALL THEME
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At What Velocity Will WATER At 20 Degrees C Flowing Through A 5 Cm Pipe Transition From Laminar To Turbulent Flow? Assume Transition Occurs At
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How does this seagull look like?
Hi everyone,
I have difficulties picturing how seagulls should look like. Is there an easy way to think about their payoff graphs?
1. For example, how does this look like - buy a 35-delta put, sell a 25-delta put, and sell a 35-delta call.
2. Also, I can’t see why the text would say “provides downside protection between the two put strike prices and upside potential to the call strike price” for the seagull mentioned in (1) above.
Thank you
Sort of expected someone would do that haha… But seriously, any help on the above? Thank you. I have searched payoff graphs for seagull using google image but haven’t been able to find one that fits
the above desciption.
Here is the seagull
Thanks for helping out Pierre. Can you please explain how you arrived at this graph. I tried applying what S2000magician mentioned in this post below and wasn’t able to get it.
Basically you are (1) Short put - low strike price (2) Long put - medium strike price (3) Short call - high strike price.
The graphs corresponding to the above are (1) / (2) \ (3) \
So should the graph look something like this?
/ \
/ \
/ \
/ \
Your graph is short put + short call
You need to add (long) a second put with medium strike
Hi Pierre, unfortunately I still don’t get it.
The graph is initially upward sloping because of the short put /
I then need to add a downward kink to account for the long put, this would make the graph flat ----
Finally, I have a short call which would make the graph go downwards \
That’s how I ended up with my shape.
If you can explain to me how you visualize it, I would be most grateful.
Thank you.
First, could you draw here only
Long put and short call (with higher strike)
Ps: after that, just add the first put (short put)
Wow thanks alot Pierre, yes that did help alot! I finally understand it. And for those out there wondering whether you are thinking correctly about graphs, this is a link to a payoff diagram
generator I found online: http://www.designserver.de/root_andreas_emmert/downloads/optionpayoff.xls.
Really handy tool to test your understanding.
Hi Pierre,
Could I ask one more thing, why does the text say “upside potential to the call strike price”. The payoff graph is clearly flat before the call strike price so why is there upside potential?
Thank you.
Where does the text mention Seagulls?
I don’t think there is any upside potential for long seagull. Perhaps the text says “upside potential for a short seagull”.
Am I the only one who has not seen seagulls anywhere?! Could you let me know which reading/page it is in?
Pierre, think I figured it out. You need to add a long stock to the seagull to see what effect it has on the portfolio. So before the call strike price, you portfolio will increase in value since the
payoff graph is flat then. But after the call strike price, the seagull is \ therefore it would offset your long stock position which is /
Thank you for taking the time to respond.
It is in currency management. LOS 19.g is schwesser.
Ah okay, now I remember, I was freaking out cause I didn’t see it in the Derivatives section lol. Tanks | {"url":"https://www.analystforum.com/t/how-does-this-seagull-look-like/123049","timestamp":"2024-11-11T14:53:01Z","content_type":"text/html","content_length":"55980","record_id":"<urn:uuid:6d40db2d-599e-429e-a21d-2755405d6594>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00866.warc.gz"} |
February 18, 2019
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more or less arbitrary code to them, i.e. we’re not restricted to stricly maths-like functions. One of the advantages of this is that it was quite easy to expose ray hit testing results as inputs to
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What is the Circle of Confusion — Camera Essentials
How do you ensure your photos are sharp and in focus? Getting comfortable with the circle of confusion is a good first step. What is circle of confusion? This concept is an under-defined, yet super
critical idea in photography. And understanding it will help you see how everything works together. The circle of confusion is related to how we understand focus and sharpness, but let’s get a bit
more technical. After all, the more you know, the more intentional you can be with every shot.
What is circle of confusion
First, let’s talk about focus
Before we get into the definition of the circle of confusion, let’s break down how a camera focuses.
I know it’s basic, but this is everything when defining the concept.
To start, let’s look at our cameras…
We know our lenses have various lens elements within them. These each play their own role. The focus element is what we’re concerned with here.
When light from the outside world comes into our camera and passes through the lens, that same light is bent to a matching point behind the lens. The point where this light crosses each other, or
this point of convergence, as seen below, is the focal point.
Focus and the circle of confusion
When this focal point is on the focal plane (your camera’s sensor or film), it makes a super small dot or circle on the plane where your subject will be in sharp focus.
But when we go to focus our lens, the focus element moves our focal point. So when the focal point moves away from the focal plane, the circle gets bigger, and your subject gets blurrier.
So now, let’s look at the circle of confusion definition.
Circle of Confusion Definition
What is the circle of confusion?
The circle of confusion is the measurement of where a point of light grows to a circle you can see in the final image. Also called the zone of confusion, it’s measured in fractions of a millimeter.
The circle of confusion is what defines what’s in or out of focus. This number is also what calculates depth of field. The circle’s size is what affects the sharpness of an image. The smaller the
circle, the sharper the image. And the larger the circle, the blurrier. It is often written as CoC.
Circle of confusion is concerned with:
• How our cameras focus light
• How sharp our image is & its depth of field
• How a viewer perceives an image
What is circle of confusion
What creates the circle of confusion?
To truly understand a technical concept like the zone of confusion, visuals are necessary. Below is a short but incredibly helpful video that provides a 3D diagram simplifying this concept.
What is circle of confusion
To recap, the circle of confusion is the measurement of where a point of light grows to a circle that can see in a final image. If you have a camera, you can experience this concept by racking your
lens' focus ring and seeing how a light source becomes out of focus.
As you change the focus of your lens, the point at which light comes to convergence changes as well. This causes the zone of confusion to grow.
Let’s take a look at another visual explanation to better understand this concept. The next video will go over it a bit differently while also providing a visual for how the circle of confusion
affects depth of field. This animation is particularly helpful to get a clearer understanding of the concept.
What is circle of confusion and depth of field
The zone of confusion and the sharpness of an image may sound subjective, but it is actually calculated. Photopills circle of confusion calculator is a great resource for calculating your camera's
CoC and understanding how to achieve the sharpness or depth of field you desire.
Circle of confusion calculator
So what does the circle of confusion have to do with photography and cinematography? How does it affect our final image? The answer can be simply summed up into one term — depth of field.
What is circle of confusion
Effects of the circle of confusion
As mentioned above, the circle of confusion is what calculates the depth of field. Depth of field is probably one of the most critical concepts in photography, way ahead of the circle of confusion.
Depth of field (DOF) is the term used to describe the size of the area in your image where objects appear acceptably sharp. The area in question is known as the field, and the size (in z-space) of
that area is the depth of that field. DOF is governed by the angle at which light rays enter the lens.
The circle of confusion is the standard criteria for this sharpness. How does depth of field and the circle of confusion affect an image? Let’s take a look at the video breakdown below to see how
photographers and filmmakers utilize this technique to achieve specific effects.
What is circle of confusion- Depth of Field Explained • Subscribe on YouTube
CoC is a function of viewing distance, enlargement, and your own visual acuity.
Say you have an image. You look at it at close range. The closer you are to it, the blurrier it will appear (or have a shallow depth of field) compared to if you stand quite a distance back.
When you do that, the points of light that appeared blurry initially, might look sharper from farther away (or have a deep depth of field).
Enlargement works similarly. If you enlarge an image, you’ll see the photo at closer range, and will have a shallower depth of field, compared to a reproduced image that is much smaller. Of course,
depending on your personal eyesight, it will depend on how you see the image.
To go deeper into how you can use this technical knowledge out in the field. Take a look at the next article.
Bokeh explained with creative examples
Now let’s take this one step further. Close the book and get out into the world. When you’re familiar with CoC and depth of field, start creating with it by messing with it! Learning how to create
beautiful bokeh is up next.
Showcase your vision with elegant shot lists and storyboards.
Create robust and customizable shot lists. Upload images to make storyboards and slideshows. | {"url":"https://www.studiobinder.com/blog/what-is-circle-of-confusion-photography/","timestamp":"2024-11-12T12:49:52Z","content_type":"text/html","content_length":"336078","record_id":"<urn:uuid:03cac6ee-c9f4-4741-bf2c-3b8c8de8d6eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00172.warc.gz"} |
Basic Rules
1. Only different solutions can be submitted.
2. Each solution should let the cube be more closer to solved.
3. In some phases like EO, DR and HTR, the last move must be clockwise. No meaningless half-turns are allowed.
4. Inverse moves are allowed and must be marked with "()" or "^()". "NISS" is not allowed for not mixing with "()".
Yet Started
My Latest Submission | {"url":"https://333.fm/chain/1","timestamp":"2024-11-07T16:06:27Z","content_type":"text/html","content_length":"62778","record_id":"<urn:uuid:d346f76f-803c-4646-a3cd-a772795e01ff>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00042.warc.gz"} |
Time Constants... What They Do, March 1956 Radio-Electronics
March 1956 Radio-Electronics
[Table of Contents]
Wax nostalgic about and learn from the history of early electronics. See articles from Radio-Electronics, published 1930-1988. All copyrights hereby acknowledged.
H.P. Manly, author of this "Time Constants... What They Do" article from a 1956 issue of Radio-Electronics magazine, must have been a baseball fan. You'll know why I say that if you read the story.
RC (resistor-capacitor) and to a lesser extent RL (resistor-inductor) time constants are one of the first topics covered in basic electronics courses, after Ohm's law. A thorough knowledge of them is
essential both for design and troubleshooting purposes. Exploiting their properties for timing and waveform shaping is a major feature of circuit design and being able to recognize the effect a
deficit or excess of capacitance (or inductance) in a circuit is often the most obvious clue to a malfunctioning circuit. In the days when electronics products (especially televisions with their
complex composite waveforms) were repaired and aligned by technicians, comparing measured results with expected signals often instantly clued a guy into what had gone wrong.
Time Constants... What They Do
By H. P. Manly
A baseball may be defined as a leather-covered sphere 9 inches in circumference and weighing 5 ounces. A time constant is defined as the number of seconds in which a capacitor acquires 63.2 % of
maximum charge or loses this percentage of an initial charge. No one would know from the definitions what can be done with baseballs or time constants.
An easy way to get acquainted with time constants is to watch their effects with the setup of Fig. 1. Voltage from a square-wave generator is biased so that instead of going positive and negative, it
changes suddenly between zero and positive. The blocking capacitor prevents current from the bias battery reaching the attenuator in the generator. The resistor in series with the battery prevents
shorting the generator output.
The suddenly shifting voltage is applied to variable resistor R and a 0.006-μf capacitor (C) in series. Voltage across the capacitor is observed with an oscilloscope. The capacitor will charge as the
applied voltage goes positive and discharge when the voltage goes to zero. With the series resistor adjusted for a few hundred ohms the capacitor charges and discharges as in Fig. 2-a, with no great
When series resistance is adjusted for 8,000 ohms, it slows both the charge and the discharge as in Fig. 2-b. With 20,000 ohms the charge and discharge are slowed still more, as in Fig. 2-c. In all
cases the charge begins at a rapid rate but slows as the capacitor acquires voltage of its own. Capacitor voltage opposes applied voltage. Discharge begins at an equally rapid rate but slows as the
capacitor loses its voltage.
In Figs. 2-b and 2-c the charge is nearly completed soon enough to remain so for some time before discharge begins. Similarly, discharge is nearly completed a while before the following charge
begins. But when series resistance is increased to 80,000 ohms, we have the performance of Fig. 2-d. Charge is so slow as to be nowhere near complete before discharge begins and discharge still is
continuing at a fairly good rate when the next charge commences.
The scope trace of Fig. 3-a shows what happens when capacitance is changed from 0.006 to 0.003 μf and series resistance is made 16,000 ohms. Charge and discharge appear the same as in Fig. 2-b where
capacitance was 0.006 μf and resistance 8,000 ohms. This is strange - or is it? Try multiplying our present 0.003 (μf) by 16,000 (ohms). The product is 48. Try multiplying the original 0.006 (μf) by
8,000 (ohms). Again the product is 48.
To check whether these equal products may be the answer to why charges and discharges are alike, let's make another test. The trace of Fig. 3-b shows charge and discharge with a capacitance of 0.024
μf and a series resistance of 20,000 ohms. Action is practically the same as in Fig. 2-d where capacitance was 0.006 fμf and resistance 80,000 ohms. If you multiply these two combinations of
capacitance and resistance, the product is 480 in both cases.
It is true that times for charge and discharge depend not alone on capacitance, not alone on resistance, but on the product of capacitance and resistance. The biggest capacitance and smallest
resistance will allow just the same charge and discharge times as the smallest capacitance and greatest resistance, if the products are equal.
Now look at Fig. 4-a. The period of time from the beginning of the charge until the capacitor has 63.2% of the voltage and 63.2% of the charge it would have after an extended charging is one time
constant. It is also the time from the beginning of discharge until 63.2% of the charge is lost and 36.8% remains. Time constants refer only to these percentages of charge and discharge, not to full
charge or discharge.
Time constants are easy to compute. All you need do is multiply the number of microfarads capacitance by the number of megohms resistance. The product is the time constant in seconds or in fractions
of a second.
The scope traces of Figs. 3-a and 2-b are alike because the time constants are equal, both being 0.000048 second. Traces 3-b and 2-d are alike for the same reason - both time constants are 0.00048
second. Short time constants mean quicker charge and discharge, long ones mean slower charge and discharge rates.
Here is something rather strange - applied voltage has no effect on a time constant. If we double the applied voltage used for Fig. 4-a, the charge and discharge become as 4-b. Naturally, the charge
is greater than before. But because capacitance and resistance have not changed, the time for reaching 63.2% of full charge and for losing 63.2% of the charge does not change.
If the applied voltage is reduced to half that used for Fig. 4-a, we have the charge and discharge of Fig. 4-c. Although full charge is now relatively small, the time for reaching 63.2% of full
charge is the same as before be-cause capacitance and resistance are the same as before.
Frequency has no effect on the time constant of any given capacitance and resistance. However, any change of either time constant or frequency with respect to the other may have great effect on
circuit performance.
Frequency effects are illustrated in Fig. 5. In 5-a the frequency of the applied voltage is such that there is practically full charge and full discharge within about half the time period of each
value of applied voltage. The capacitor remains charged or discharged during the remainder of each period.
If the frequency is lowered, we have the condition of 5-b. Portions of the trace showing the beginnings of charge and discharge have not changed in form but the capacitor remains charged or
discharged for longer times during the longer periods of applied voltage.
If the frequency is raised, as in 5-c, the periods of applied voltage are shortened. The beginnings of charge and discharge have not changed because capacitance and resistance have not been changed.
But periods of maximum and zero applied voltage are so short that there is not time for full charge before discharge begins nor for complete discharge before the next charge begins.
Selecting a time constant
Effects of time constants on circuit performance and some of the requirements to be satisfied in selecting time constants can be illustrated by a few examples. Consider first how stage gain is
affected by the time constant of capacitor C and resistor R in the resistance coupling of Fig. 6. Although there is amplification in the tube, there is loss of attenuation in the coupling and stage
gain depends on both factors.
Since the right-hand tube has fixed bias, grid resistor R probably must not exceed 100,000 ohms or 0.1 megohm. Were we to use 0.01-μf capacitance at C the resulting time constant would be 0.001
second. With this combination of capacitance the coupling loss would be about 5.5 db (nearly 50%) at 100 cycles. With 0.05 μf, for a longer time constant, the loss at 100 cycles becomes less than 1
db, or only about 10%.
What about low-frequency cutoff, usually defined as the frequency at which gain drops to 0.707 of maximum? The cutoff frequency may be found from dividing 0.16 by the time constant. Dividing 0.16 by
.001 (our time constant) shows that cutoff occurs at 160 cycles. For any lower cutoff frequency we would need a longer time constant, which might be provided by a larger capacitance at C.
The time constant affects phase shift, especially at low frequencies. With 0.01- μf capacitance and 0.1-megohm resistance, for a time constant of 0.001 second in Fig. 6, the shift for signals at 100
cycles will be about 680°. Shift is lessened by a longer time constant. Changing to 0.05-μf capacitance for a time constant of 0.005 second would make the shift less than 180 at 100 cycles.
In the resistance coupling we have had to juggle capacitances to change the time constant because resistance is fixed by the manner in which the tube is biased. Were we to use cathode bias the
resistance at R might be as great as 0.5 megohm. With five times the original resistance we could obtain the same time constants with only one-fifth as much capacitance as previously mentioned.
In the diode detector circuit of Fig. 7 the time constant is determined by capacitor C and load resistor R which here is a 500,000-ohm (0.5-megohm) volume control pot. If signal modulation as great
as 80% is to be detected without distortion, the time constant should be no longer than found from dividing 0.2 by the highest audio frequency. Otherwise, the capacitor cannot discharge fast enough
to follow the modulation.
If we assume 5,000 cycles as maximum audio frequency and divide 0.2 by 5,000, we find that 0.00004 μf or 40 μμf is the maximum capacitance. Usually the shunt capacitance is made somewhat greater to
increase detection efficiency, but there will be some distortion on strong modulation at high audio frequencies.
In the agc setup of Fig. 8 the time constant must be long with respect to the time period of one cycle at the intermediate frequency, to maintain control voltage close to the amplitude of incoming
signals. But capacitance C should be small to prevent taking too much signal energy from the if amplifier. To obtain a long time constant with small capacitance we need a large resistance at R.
For grid-leak biasing in Fig. 9 the time constant must be considerably longer than the period of one cycle at the lowest signal frequency. Furthermore, capacitance should be so large and its
reactance so small at this signal frequency that excessive attenuation is avoided. Then the time constant must be based on signal frequency and large capacitance, using resistance to suit.
For the sawtooth sweep oscillator of Fig. 10 the time constant must be such that charging does not extend too far onto the bend of the curve. That is, we want a sawtooth waveform somewhat as in Fig.
5-c, not a flat top as in 5-b.
In the differentiating filter of Fig. 11 we need a short time constant. The capacitor must charge very quickly and discharge just as fast. Then, even though an applied voltage such as a sawtooth
drops rather slowly, there will be only brief rises and falls or only pips of filtered voltage.
For electronic spot welding, an industrial application, each weld requires momentary current of several amperes, but we use a source furnishing only milliamperes. The long-time-constant circuit of
Fig. 12 does it. A small current from the source, flowing for the entire period between welds, builds up a large charge in a big capacitance. Then all the stored electricity is discharged almost
instantly to make a weld.
Posted October 11, 2022 | {"url":"https://www.rfcafe.com/references/radio-electronics/time-constants-radio-electronics-march-1956.htm","timestamp":"2024-11-11T19:30:09Z","content_type":"text/html","content_length":"39917","record_id":"<urn:uuid:1b3bf9be-89bf-4507-a9c1-aa14785e848d>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00799.warc.gz"} |
20 Points Please Help With This
Option D
Step-by-step explanation:
Since they are in proportion. So,
27 : 36 :: 9 : x
27 × x = 36 × 9
x = 36 × 9/27
x = 12 × 1
x = 12
D) 12
Step-by-step explanation:
Well I'm assuming these 2 triangles are similar
Before you start, please make a title for your Stem & Leaf plot.
1) Make two columns; Stem and Leaf.
2) Under "Stem" list the following whole numbers: 0, 1, 2, 3, 4, 5, 6. Notice that each number will correspond to each number you gave us.
3) Under "Leaf":
Only the last digits of the numbers go into the Leaf. If I number, for example, starts with 0, the remaining digits go into the Leaf.
4) Create a Key, which allows whoever is reading your plot to understand what the plot means. For example 2|4=24, since 2 is the stem and 4 is the leaf. We usually read a stem and leaf plot as the
remaining digits go after the stem but if otherwise a Key is a must have when creating a plot.
The Key can be any set of numbers you have in your Stem and Leaf plot. | {"url":"https://community.carbonfields.net/question-handbook/20-points-please-help-with-this-jqwv","timestamp":"2024-11-13T01:19:38Z","content_type":"text/html","content_length":"70276","record_id":"<urn:uuid:dee7cbbb-f13c-4dd9-b3e9-fc98967aee00>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00695.warc.gz"} |
Quaternions and Dual Quaternion Skinning
For some reason I like quaternions. I fell in love with complex numbers back in school when I found out that they made more sense than real numbers. While it might not exactly be helpful to visualise
quaternions as an extension of complex numbers, there's something in there that just grabs at me. Unlike previous posts, I've managed to update to D3D11 so I'll be discussing implementation details
in terms of HLSL (Shader Model 4, as I also have a D3D10 dev machine).
I'm no mathematician so hopefully this information in this post should be pretty accessible.
Dual Quaternion Skinning
I spent a couple of hours last week converting my skinning pipeline to use dual quaternions. My animation pipeline works with quaternions; the source animation data, skeleton pose and inverse base
pose are quaternion-based. Right at the very end, the composited result is converted to matrix and sent to the GPU. In effect, I'm doing this:
for (int i = 0; i < nb_bones; i++)
const rend::Bone& bone = skeleton.bones[i];
const rend::Keyframe& kf = keyframes[i + frame * nb_bones];
// Calculate inverse base pose
math::quat q0 = qConjugate(bone.base_pose.rotation);
math::vec3 p0 = qTransformPos(q0, v3Negate(bone.base_pose.position));
// Concatenate with animation keyframe
math::quat q = qMultiply(q0, kf.rotation);
math::vec3 p = qTransformPos(kf.rotation, p0);
p = v3Add(p, kf.position);
// Set the transform
math::mat4& transform = transforms[i];
transform = qToMat4(q);
transform.r[3] = math::v4Make(p.x, p.y, p.z, 1);
In reality, I precalculate the inverse base pose, keeping the inner loop tight and low on ALU operations. My goals were:
• Capitalise on the quaternion input and remove the conversion to matrix step. On all past games I've worked on, the matrix multiply with base pose has had a destructive influence on CPU
performance; if I could remove this step, I'd have a solution that would be faster than my past implementations.
• Reduce the amount of data being mapped/sent to the GPU. While my existing solution was sending 4x4 matrices, one of the columns was redundant - I could be sending 4x3. However, dual quaternions
would allow me to halve the amount of data sent.
• Get volume-preserving skinning on joints under extreme rotation.
• Have a bit of fun and exercise some neglected quaternion/vector math muscles.
I skimmed the paper, copied the HLSL source code and tested everything on a basic idle animation. Bad idea, right? But everything seemed to work. I hooked up my mocap pipeline this week and
everything went wrong - limbs were folding inside each other and the character was skewing all over the place.
Searching the internets for equivalent implementations found that most basically copied and pasted from the paper. Some seemed to make an effort to understand what was going on under the hood but all
of them produced the same results (oddly, the nVidia shader was the most needlessly complicated and inefficient of them all). So I knuckled down and decided to read the following papers in more
The basic method is this:
• Convert your quaternion rotation and position vector to a dual quaternion on the CPU for each bone.
• In your vertex shader, load all dual quaternion bone transforms for the vertex.
• Create a single dual quaternion that is the weighted average of all the bone transforms.
• Transform the vertex position and normal by this dual quaternion.
Quaternion/Vector Multiplication
To cut a long story short, the reason none of this was working for me was that I was looking at Cg code, as opposed to HLSL code. Having never really used Cg, it never occurred to me that the order
of cross product parameters should be swapped to account for handedness. Cross products are used to define the quaternion multiplication:
\[Q = (w, V)\]
\[Q_r = Q_0 Q_1\]
\[Q_r = (w_0 w_1 - V_0 \cdot V_1, w_0 V_1 + w_1 V_0 + V_0 \times V_1)\]
(\(\cdot\) is the vector dot product and \(\times\) is the vector cross product)
Naturally, this is then a non-commutative operation, giving the order in which you pass arguments into your multiply functions importance, too. A basic C++ implementation of the multiplication looks
math::quat math::qMultiply(const math::quat& a, const math::quat& b)
quat q =
a.w * b.x + a.x * b.w + a.y * b.z - a.z * b.y,
a.w * b.y - a.x * b.z + a.y * b.w + a.z * b.x,
a.w * b.z + a.x * b.y - a.y * b.x + a.z * b.w,
a.w * b.w - a.x * b.x - a.y * b.y - a.z * b.z,
return q;
Assuming we're using unit quaternions, you can rotate a vector by a quaternion using multiplication:
\[P_r = Q*(0, P)*Q'\]
Here, \(*\) is the quaternion multiplication, \('\) is the quaternion conjugate and \((0,P)\) is construction of a pure quaternion using the vector, setting the quaternion real component to zero.
This equation is used to convert to and from dual quaternions, so it's important you keep order of multiplications in mind.
If you work with Cg, make sure you have your cross products round the other way to my shader examples below!
Converting to Dual Quaternion
I'd suggest reading the second paper linked above to get a good introduction to what dual quaternions are and what the various components mean. For our means here it's enough to say that dual
quaternions are just a pair of quaternions - the "real" quaternion and the "dual" quaternion:
\[DQ = (Q_r, Q_d)\]
Converting a quaternion rotation and position vector to this representation is simple. The \(Q_r\) term is a simple copy of your quaternion rotation and \(Q_d\) is:
\[Q_d = 0.5(0, V)*Q_r\]
This allows the matrix conversion code on the CPU to become:
// Convert position/rotation to dual quaternion
math::dualquat& transform = transforms[i];
transform.r = q;
transform.d = qScale(qMultiplyPure(q, p), 0.5f);
Much nicer! MultiplyPure is a simple function that does a special case multiply where p.w is zero.
Blending Dual Quaternions
Now we can move over to the HLSL side. This is my transform loading code:
Buffer<float4> g_BoneTransforms;
float2x4 GetBoneDualQuat(uint bone_index)
// Load bone rows individually
float4 row0 = g_BoneTransforms.Load(bone_index * 2 + 0);
float4 row1 = g_BoneTransforms.Load(bone_index * 2 + 1);
return float2x4(row0, row1);
As mentioned before, we want to load the bones that influence a vertex, weight them and transform the vertex position and normal:
float2x4 BlendBoneTransforms(uint4 bone_indices, float4 bone_weights)
// Fetch bones
float2x4 dq0 = GetBoneDualQuat(bone_indices.x);
float2x4 dq1 = GetBoneDualQuat(bone_indices.y);
float2x4 dq2 = GetBoneDualQuat(bone_indices.z);
float2x4 dq3 = GetBoneDualQuat(bone_indices.w);
// ...blend...
As with quaternions, weighting dual quaternions can be achieved using a normalised lerp of its components. As explained in the paper Understanding Slerp, Then Not Using It, it's not as good as a
SLERP in that it's not constant velocity. However, it has minimal torque, rotating along the sphere and unlike SLERP, is commutative; meaning, if you combine multiple quaternions in a different
order, the result will always be the same.
Besides, when you're regenerating bone rotations each frame, interpolation velocity won't really factor into the solution. The weighting code thus becomes:
// Blend
float2x4 result =
bone_weights.x * dq0 +
bone_weights.y * dq1 +
bone_weights.z * dq2 +
bone_weights.w * dq3;
// Normalise
float norm = length(result[0]);
return result / norm;
Simples! In the original paper, Kavan goes into great detail on why this works so well over previous solutions and why a SLERP isn't the ideal solution. Well worth a read.
Antipodality or, Quaternion Double-Cover
Take a look at one of the classic release bugs of this generation:
There's a pretty awesome reason for why the head can spin the long way around to get to its target rotation (no doubt the bug will be more complicated than that). Casey Muratori goes into great
detail on this and I'll attempt to summarise.
The axis-angle definition of a quaternion is:
\[Q = (cos(\frac{\theta}{2}), Vsin(\frac{\theta}{2}))\]
Inherent in this definition is the ability for quaternions to represent up to 720 degrees of rotation. Assuming we use the x-axis as our example vector, this leads to these quantities:
\[Q(0) = (1, 0, 0, 0)\]
\[Q(360) = (-1, 0, 0, 0)\]
\[Q(720) = (1, 0, 0, 0)\]
While sine and cosine are periodic every 360 degrees, the division by two of the input angle leads the quaternion representation to be periodic every 720 degrees.
Clearly, 360 degrees and 720 degrees represent the same geometrical rotation. However, when you interpolate between rotations, you may find yourself interpolating the long way round. Your source/
target rotation range may geometrically be only 30 to 35 degrees but your quaternion may represent that as 30 to 395 degrees!
If you've glimpsed at the inner workings of a SLERP, this is the case they are trying to avoid when trying to solve for the "shortest path". Given that \(Q\) and \(-Q\) represent the same rotation,
you can ensure interpolation between two quaternions follows this shortest path by negating one of the quaternions if the dot product between them is negative.
When blending dual quaternions, you have to watch for the same case. Not only that, you have to ensure that all of your bone transforms are in the same neighbourhood. While there are complicated ways
of achieving that, most of the time it can be guaranteed by comparing all bone rotations to the first one and adjusting the sign of the blend weight.
This leads to the final code:
float2x4 BlendBoneTransforms(uint4 bone_indices, float4 bone_weights)
// Fetch bones
float2x4 dq0 = GetBoneDualQuat(bone_indices.x);
float2x4 dq1 = GetBoneDualQuat(bone_indices.y);
float2x4 dq2 = GetBoneDualQuat(bone_indices.z);
float2x4 dq3 = GetBoneDualQuat(bone_indices.w);
// Ensure all bone transforms are in the same neighbourhood
if (dot(dq0[0], dq1[0]) < 0.0) bone_weights.y *= -1.0;
if (dot(dq0[0], dq2[0]) < 0.0) bone_weights.z *= -1.0;
if (dot(dq0[0], dq3[0]) < 0.0) bone_weights.w *= -1.0;
// Blend
float2x4 result =
bone_weights.x * dq0 +
bone_weights.y * dq1 +
bone_weights.z * dq2 +
bone_weights.w * dq3;
// Normalise
float norm = length(result[0]);
return result / norm;
There are cases which can still fail these checks but they are very rare for the general use-case of skinning - I can't imagine this working well for severe joint twists, for example (beyond the
range of human constraints, that is).
Transforming the Vertex with Dual Quaternions
Once you have the blended result you need to convert it back into quaternion/vector form and transform your vertex. There are two ways of achieving this:
• Convert straight to matrix and use the matrix to transform the vertex.
• Convert to quaternion/vector and transform the vertex using that.
The fastest and by far cleanest way is the second so I will concentrate on that. Given that you already have the rotation in the \(Q_r\) component of the dual quaternion, extraction of the
translation vector is achieved using the following:
\[V = 2Q_d*Q_r'\]
This can be implemented directly as:
float4 Conjugate(float4 q)
return float4(-q.x, -q.y, -q.z, q.w);
float4 Multiply(float4 a, float4 b)
return float4(a.w * b.xyz + b.w * a.xyz + cross(b.xyz, a.xyz), a.w * b.w - dot(a.xyz, b.xyz));
float3 ReconstructTranslation(float4 Qr, float4 Qd)
// The input is the dual quaternion, real part and dual part
return Multiply(Qd, Conjugate(Qr)).xyz;
Of course, the complete calculation can be collapsed by directly applying the conjugate sign and discarding w:
float3 ReconstructTranslation(float4 Qr, float4 Qd)
return 2 * (Qr.w * Qd.xyz - Qd.w * Qr.xyz + cross(Qd.xyz, Qr.xyz));
Using the Conjugate and Multiply functions, it's then easy to transform a position and vector by the quaternion rotation and reconstructed position:
float3 QuatRotateVector(float4 Qr, float3 v)
// Straight-forward application of Q.v.Q', discarding w
return Multiply(Multiply(Qr, float4(v, 0)), Conjugate(Qr)).xyz;
float3 DualQuatTransformPoint(float4 Qr, float4 Qd, float3 p)
// Reconstruct translation from the dual quaternion
float3 t = 2 * (Qr.w * Qd.xyz - Qd.w * Qr.xyz + cross(Qd.xyz, Qr.xyz));
// Combine with rotation of the input point
return QuatRotateVector(Qr, p) + t;
This leaves you with the final code:
float2x4 skin_transform = BlendBoneTransforms(input.bone_indices, input.bone_weights);
float3 pos = DualQuatTransformPoint(skin_transform[0], skin_transform[1], input.pos);
float3 normal = QuatRotateVector(skin_transform[0], input.normal);
Optimising the Vertex Transformation
There's a bit of redundancy in the transformation code above; results being thrown away and inputs being used when they could be discarded. There are also some identities we can apply to the rotation
equation that can simplify it. As it stands, reconstruction of the translation is good enough.
Starting with QuatRotateVector, we can already see that the first multiplication uses \(w=0\), allowing us to construct a function which removes the necessary terms in its calculation:
float4 MultiplyPure(float4 a, float3 b)
return float4(a.w * b + cross(b, a.xyz), -dot(a.xyz, b));
float3 QuatRotateVector(float4 Qr, float3 v)
return Multiply(MultiplyPure(Qr, v), Conjugate(Qr)).xyz;
The final redundancy is that we're calculating w and throwing it away, leading to:
float3 MultiplyConjugate3(float4 a, float4 b)
return b.w * a.xyz - a.w * b.xyz - cross(b.xyz, a.xyz);
float3 QuatRotateVector(float4 Qr, float3 v)
return MultiplyConjugate3(MultiplyPure(Qr, v), Qr);
Realistically, the shader compiler should be able to handle all that for you. However, it gives us a good starting point to take this further.
We can do better than that
Let's try to explode the transformation and bring it back to something far simpler. I'll work through the steps I took in simplifying this explicitly - it serves as a nice record for me and will
hopefully help if you're trying to understand where the final result came from (I was always losing signs during my school days - I'm no better 15 years on!)
We're trying to simplify:
\[P_r = Q*(0,V)*Q'\]
This is a sequence of two quaternion multiplies. Again, quaternion multiplication is defined as:
\[Q_0 Q_1 = (w_0 w_1 - V_0 \cdot V_1, w_0 V_1 + w_1 V_0 + V_0 \times V_1)\]
Let's make a few quick substitutions:
\[R = Q.xyz\]
\[w = Q.w\]
Expand \(Q*(0,V)\) first:
\[P_r = (-R \cdot V + wV + R \times V)(w - R)\]
Expand the second multiplication:
\[P_r = -R \cdot Vw + (wV + R \times V) \cdot R + (R \cdot V)R + w(wV + R \times V) - (wV + R \times V) \times R\]
The dot product distributes over addition so distribute them all:
\[P_r = -R \cdot Vw + wV \cdot R + R \times V \cdot R + (R \cdot V)R + + w^2V + wR \times V - (wV + R \times V) \times R\]
The first two terms cancel as the dot product is commutative:
\[P_r = R \times V \cdot R + (R \cdot V)R + w^2V + wR \times V - (wV + R \times V) \times R\]
Using the identity \(A \times B=-B \times A\) swap the last cross product around:
\[P_r = R \times V \cdot R + (R \cdot V)R + w^2V + wR \times V + R \times (wV + R \times V)\]
The cross product distributes over addition so distribute the last cross product:
\[P_r = R \times V \cdot R + (R \cdot V)R + w^2V + wR \times V + R \times wV + R \times (R \times V)\]
As we're only interested in the xyz components of the result, discard all scalar terms:
\[P_r = (R \cdot V)R + w^2V + wR \times V + R \times wV + R \times (R \times V)\]
Pull the scalar out of \(R \times wV\) and sum with its neighbour:
\[P_r = (R \cdot V)R + w^2V + wR \times V + wR \times V + R \times (R \times V)\]
\[P_r = (R \cdot V)R + w^2V + 2wR \times V + R \times (R \times V)\]
The next bit requires knowledge of the vector triple product (or Lagrange's formula - of many). This takes the form:
\[R \times (R \times V) = (R \cdot V)R - (R \cdot R)V\]
If we rearrange that to equal zero then we can add that to the end of our existing equation and play around with it a little:
\[R \times (R \times V) - (R \cdot V)R + (R \cdot R)V = 0\]
\[P_r = (R \cdot V)R + w^2V + 2wR \times V + R \times (R \times V) + R \times (R \times V) - (R \cdot V)R + (R \cdot R)V\]
\[P_r = (R \cdot V)R + w^2V + 2wR \times V + 2R \times (R \times V) - (R \cdot V)R + (R \cdot R)V\]
The \((R \cdot V)R\) terms cancel:
\[P_r = w^2V + 2wR \times V + 2R \times (R \times V) + (R \cdot R)V\]
We can now factor the scale of \(V\):
\[P_r = w^2V + (R \cdot R)V+ 2wR \times V + 2R \times (R \times V)\]
\[P_r = (w^2 + R \cdot R)V+ 2wR \times V + 2R \times (R \times V)\]
The quaternion norm operation is given by:
\[norm(q) = q_w q_w + q_x q_x + q_y q_y + q_z q_z\]
Assuming we're dealing with unit quaternions, the norm will always be 1. Looking above, we can see the norm right at the beginning and can get rid of it:
\[P_r = V+ 2wR \times V + 2R \times (R \times V)\]
Finally, factor the 2:
\[P_r = V+ 2(wR \times V + R \times (R \times V))\]
And factor the cross product:
\[P_r = V+ 2(R \times (wV + R \times V))\]
This is a delightfully simple result! The HLSL code is:
float3 QuatRotateVector(float4 Qr, float3 v)
return v + 2 * cross(Qr.w * v + cross(v, Qr.xyz), Qr.xyz); | {"url":"https://donw.io/post/dual-quaternion-skinning/","timestamp":"2024-11-02T10:43:36Z","content_type":"text/html","content_length":"29431","record_id":"<urn:uuid:4d9a7723-6fa0-40cd-8074-321cccd3498c>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00448.warc.gz"} |
Python) Problems using For
First problem: Adding all numbers from 1 to the number it is typed(using For)
input: 4 output: 10
Second Problem: Making “*” in stair shape, the last row will have the same numbers of stars that you have typed
input: 5
Third problem: Determining numbers in a number list that are smaller than what you have typed
input and output | {"url":"https://ashtonkang0814.medium.com/python-problems-using-for-1430a235d98d?source=user_profile_page---------0-------------fd5065dd85c4---------------","timestamp":"2024-11-11T00:58:43Z","content_type":"text/html","content_length":"94406","record_id":"<urn:uuid:e37ab374-14a9-4338-8635-f5a2a0cd7b89>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00234.warc.gz"} |
Wu-Hausman Test: Choosing between Fixed and Random Effects
• Post author:Viren Rehal
• Post published:March 25, 2022
• Post category:Econometrics / Panel Data
Post comments:0 Comments
The Wu-Hausman Test can be used to determine whether the Fixed Effects Model or Random Effects Model is more appropriate. To apply this test, we need to estimate both the Fixed Effects and Random
Effects Models and compare the estimated coefficients using Wu-Hausman statistic.
To test whether the random effects are significant or not, the Lagrange Multiplier Test for Random Effects is often used.
Before we state the hypothesis of the Wu-Hausman Test, we must discuss the meaning of consistency and efficiency of coefficients in brief.
Consistent estimates: coefficients of a model are said to be consistent if they keep getting closer to their true parameter values with an increase in sample size.
Efficient estimates: coefficients are said to be efficient if they have minimum variance as compared to the coefficients of other estimators. This means that the errors (difference between predicted
and true values) are minimum in the case of efficient estimates, as compared to any other estimates.
The null and alternate hypothesis for this test can be stated as:
Under the null hypothesis, the coefficients of both the fixed effects and random effects models are consistent. However, only the coefficients of the random effects model are efficient. If we cannot
reject the null hypothesis using the Wu-Hausman test, it means that the random-effects model should be preferred.
On the other hand, only the coefficients of the fixed effects model are consistent under the alternate hypothesis. The coefficients of the random effects model are not consistent. If we reject the
null hypothesis, the fixed-effects model should be preferred instead of a random-effects model.
Estimating the Wu-Hausman Test statistic
The statistic can be estimated using statistical software packages.
If H is greater than the critical chi-square value, the null hypothesis is rejected. We conclude that Fixed Effects estimates are consistent. Whereas, Random Effects estimates are not consistent.
Hence, the Fixed Effects Model should be used.
If H is less than the critical chi-square value, we cannot reject the null hypothesis. Both Fixed and Random Effects models are consistent. However, Random Effects estimates are efficient as well.
Hence, the Random Effects Model is more reliable.
Interpretation in practice
In practice, the results of the Wu-Hausman test look like the following:
Independent variable Coefficients of Fixed Effects Model Coefficients of Random Effects Model
X -0.013 -0.012
Y 0.047 0.039
H = 3.26
P-value = 0.1963
Hence, the results of the above table indicate that the H = 3.26. Using the P-value reported above, we cannot reject the null hypothesis. This means that the coefficients of the random-effects model
are consistent as well as efficient. Hence, we should apply the random-effects model.
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You Might Also Like | {"url":"https://spureconomics.com/wu-hausman-test-choosing-between-fixed-and-random-effects/","timestamp":"2024-11-09T13:06:15Z","content_type":"text/html","content_length":"334542","record_id":"<urn:uuid:5f97bddd-d798-4f4e-a8ed-f4411fa53e80>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00616.warc.gz"} |
Using PyTorch in Python An Introduction to Machine Learning and Deep Learning
Using PyTorch in Python: An Introduction to Machine Learning and Deep Learning
PyTorch is an open-source machine learning library developed by Facebook’s AI Research lab (FAIR) that provides a flexible and efficient platform for building deep learning models. In this article,
we will explore the basics of PyTorch and its application in Python.
Table of Contents
Introduction to PyTorch
PyTorch is a powerful library that offers a dynamic computation graph and GPU acceleration, making it suitable for a wide range of machine learning and deep learning tasks. PyTorch allows you to
create and train complex models using an easy-to-understand Pythonic syntax.
Some key features of PyTorch are:
• Tensor computation: PyTorch provides a multi-dimensional array called a tensor, which can be used for various mathematical operations.
• GPU acceleration: PyTorch supports NVIDIA’s CUDA platform, allowing tensors to be stored and operated on using GPU resources for faster computation.
• Dynamic computation graph: Unlike some other deep learning libraries, PyTorch allows you to create and modify computation graphs on-the-fly, providing more flexibility when developing models.
• Autograd: PyTorch includes a built-in automatic differentiation engine called Autograd, which simplifies the process of computing gradients for backpropagation.
• Wide range of pre-built modules: PyTorch provides various pre-built modules for common neural network architectures, loss functions, and optimization algorithms.
Installation and Setup
To install PyTorch, you can use the pip package manager. Make sure you have Python 3 installed before proceeding.
pip install torch
If you have an NVIDIA GPU with CUDA support, you can install the GPU version of PyTorch by specifying the appropriate version:
pip install torch -f https://download.pytorch.org/whl/cu111/torch_stable.html
Replace cu111 with the appropriate CUDA version number for your system.
Once installed, you can import PyTorch in your Python script:
import torch
Tensors are the fundamental data structure in PyTorch and are used to represent multi-dimensional arrays. They can be created and manipulated using a NumPy-like syntax.
Here’s an example of creating a tensor:
import torch
x = torch.tensor([[1, 2], [3, 4], [5, 6]])
This will output:
tensor([[1, 2],
[3, 4],
[5, 6]])
You can perform various operations on tensors, such as addition, subtraction, multiplication, and more:
x = torch.tensor([1, 2, 3])
y = torch.tensor([4, 5, 6])
## Element-wise addition
z = x + y
print(z) ## Output: tensor([5, 7, 9])
## Element-wise multiplication
z = x * y
print(z) ## Output: tensor([ 4, 10, 18])
Autograd is PyTorch’s automatic differentiation engine, which computes gradients for tensor operations. To use Autograd, you must set the requires_grad attribute of a tensor to True. This will enable
gradient tracking for that tensor and any operations performed on it.
Here’s an example of using Autograd to compute gradients:
import torch
x = torch.tensor(2.0, requires_grad=True)
y = x ** 2
y.backward() ## Compute gradients
print(x.grad) ## Output: tensor(4.0)
Creating a Neural Network
To create a neural network in PyTorch, you need to define a class that inherits from torch.nn.Module and implement the forward method. This method defines the forward pass of your model.
Here’s an example of creating a simple feedforward neural network with one hidden layer:
import torch
import torch.nn as nn
class FeedforwardNN(nn.Module):
def __init__(self, input_size, hidden_size, output_size):
super(FeedforwardNN, self).__init__()
self.hidden = nn.Linear(input_size, hidden_size)
self.relu = nn.ReLU()
self.output = nn.Linear(hidden_size, output_size)
def forward(self, x):
x = self.hidden(x)
x = self.relu(x)
x = self.output(x)
return x
Training a Neural Network
To train a neural network, you need to define a loss function and an optimizer. PyTorch provides various loss functions and optimization algorithms in the torch.nn and torch.optim modules,
Here’s an example of training the feedforward neural network defined earlier using the mean squared error loss and stochastic gradient descent (SGD) optimizer:
import torch
import torch.nn as nn
import torch.optim as optim
## Create synthetic data
input_data = torch.randn(100, 3)
target_data = torch.randn(100, 1)
## Initialize the model, loss function, and optimizer
model = FeedforwardNN(input_size=3, hidden_size=10, output_size=1)
loss_function = nn.MSELoss()
optimizer = optim.SGD(model.parameters(), lr=0.01)
## Train the model
for epoch in range(100): ## Number of training epochs
## Forward pass
output = model(input_data)
## Compute the loss
loss = loss_function(output, target_data)
# Backward pass
optimizer.zero_grad() ## Clear previous gradients
loss.backward() ## Compute gradients
## Update weights
## Print loss for the current epoch
print(f"Epoch {epoch + 1}, Loss: {loss.item()}")
In this article, we introduced PyTorch, a powerful library for machine learning and deep learning in Python. We covered the basics of tensors, Autograd, creating a neural network, and training a
neural network. PyTorch’s flexibility and ease of use make it an excellent choice for both beginners and experts in the field of deep learning.
There is much more to learn about PyTorch, including advanced features like recurrent neural networks, convolutional neural networks, transfer learning, and more. To dive deeper into PyTorch, check
out the official documentation and additional resources like tutorials, examples, and community-contributed projects. | {"url":"https://friendlyuser.github.io/posts/tech/2023/Using_PyTorch_in_Python_An_Introduction_to_Machine_Learning_and_Deep_Learning/","timestamp":"2024-11-05T14:01:45Z","content_type":"text/html","content_length":"27061","record_id":"<urn:uuid:2e257026-638a-4d41-80f8-a20f706fa501>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00823.warc.gz"} |
Loss Function and Model Quality Metrics
What is a Loss Function?
The System Identification Toolbox™ software estimates model parameters by minimizing the error between the model output and the measured response. This error, called loss function or cost function,
is a positive function of prediction errors e(t). In general, this function is a weighted sum of squares of the errors. For a model with ny-outputs, the loss function V(θ) has the following general
$V\left(\theta \right)=\frac{1}{N}\sum _{t=1}^{N}{e}^{T}\left(t,\theta \right)W\left(\theta \right)e\left(t,\theta \right)$
• N is the number of data samples.
• e(t,θ) is ny-by-1 error vector at a given time t, parameterized by the parameter vector θ.
• W(θ) is the weighting matrix, specified as a positive semidefinite matrix. If W is a diagonal matrix, you can think of it as a way to control the relative importance of outputs during
multi-output estimations. When W is a fixed or known weight, it does not depend on θ.
The software determines the parameter values by minimizing V(θ) with respect to θ.
For notational convenience, V(θ) is expressed in its matrix form:
$V\left(\theta \right)=\frac{1}{N}trace\left({E}^{T}\left(\theta \right)E\left(\theta \right)W\left(\theta \right)\right)$
E(θ) is the error matrix of size N-by-ny. The i:th row of E(θ) represents the error value at time t = i.
The exact form of V(θ) depends on the following factors:
• Model structure. For example, whether the model that you want to estimate is an ARX or a state-space model.
• Estimator and estimation options. For example, whether you are using n4sid or ssest estimator and specifying options such as Focus and OutputWeight.
Options to Configure the Loss Function
You can configure the loss function for your application needs. The following estimation options, when available for the estimator, configure the loss function:
Estimation Description Notes
Focus option affects how e(t) in the loss function is computed:
• When Focus is 'prediction', e(t) represents 1-step ahead prediction error:
• When Focus is 'simulation', e(t) represents the simulation error:
• Specify the Focus option in the estimation option sets.
Focus ${e}_{s}\left(t\right)={y}_{measured}\left(t\right)-{y}_{simulated}\left(t\
right)$ • The estimation option sets for oe and tfest do not have a Focus option because the
noise-component for the estimated models is trivial, and so e[p](t) and e[s](t) are equivalent.
For models whose noise component is trivial, (H(q) = 1), e[p](t), and e[s](t) are
The Focus option can also be interpreted as a weighting filter in the loss
function. For more information, see Effect of Focus and WeightingFilter Options
on the Loss Function.
When you specify a weighting filter, prefiltered prediction or simulation error
is minimized:
${e}_{f}\left(t\right)=ℒ\left(e\left(t\right)\right)$ • Specify the WeightingFilter option in the estimation option sets. Not all options for
WeightingFilter WeightingFilter are available for all estimation commands.
where $ℒ\left(.\right)$ is a linear filter. The WeightingFilter option can be
interpreted as a custom weighting filter that is applied to the loss function.
For more information, see Effect of Focus and WeightingFilter Options on the Loss
• Specify the EnforceStability option in the estimation option sets.
• The estimation option sets for procest and ssregest commands do not have an EnforceStability
option. These estimation commands always yield a stable model.
When EnforceStability is true, the minimization objective also contains a • The estimation commands tfest and oe always yield a stable model when used with time-domain
EnforceStability constraint that the estimated model must be stable. estimation data.
• Identifying unstable plants requires data collection under a closed loop with a stabilizing
feedback controller. A reliable estimation of the plant dynamics requires a sufficiently rich
noise component in the model structure to separate out the plant dynamics from feedback effects.
As a result, models that use a trivial noise component (H(q) = 1), such as models estimated by
tfest and oe commands, do not estimate good results for unstable plants.
OutputWeight option configures the weighting matrix W(θ) in the loss function and
lets you control the relative importance of output channels during multi-output
• When OutputWeight is 'noise', W(θ) equals the inverse of the estimated
variance of error e(t):
$W\left(\theta \right)={\left(\frac{1}{N}{E}^{T}\left(\theta \right)E\left(\
theta \right)\right)}^{-1}$ • Specify the OutputWeight option in the estimation option sets. Not all options for OutputWeight
are available for all estimation commands.
Because W depends on θ, the weighting is determined as a part of the
OutputWeight estimation. Minimization of the loss function with this weight simplifies the • OutputWeight is not available for polynomial model estimation because such models are always
loss function to: estimated one output at a time.
$V\left(\theta \right)=det\left(\frac{1}{N}{E}^{T}\left(\theta \right)E\left • OutputWeight cannot be 'noise' when SearchMethod is 'lsqnonlin'.
(\theta \right)\right)$
Using the inverse of the noise variance is the optimal weighting in the
maximum likelihood sense.
• When OutputWeight is an ny-by-ny positive semidefinite matrix, a constant
weighting is used. This loss function then becomes a weighted sum of squared
ErrorThreshold option specifies the threshold for when to adjust the weight of
large errors from quadratic to linear. Errors larger than ErrorThreshold times
the estimated standard deviation have a linear weight in the loss function.
$V\left(\theta \right)=\frac{1}{N}\left(\sum _{t\in I}{e}^{T}\left(t,\theta \
right)W\left(\theta \right)e\left(t,\theta \right)+\sum _{t\in J}{v}^{T}\left(t,\
theta \right)W\left(\theta \right)v\left(t,\theta \right)\right)$
• Specify the ErrorThreshold option in the estimation option sets.
• I represents those time instants for which $\begin{array}{l}|\text{e}\left(\
ErrorThreshold text{t}\right)|\text{}<\rho *\sigma \\ \end{array}$, where ρ is the error • A typical value for the error threshold ρ = 1.6 minimizes the effect of data outliers on the
threshold. estimation results.
• J represents the complement of I, that is, the time instants for which $|\
text{e}\left(\text{t}\right)|\text{}>=\rho *\sigma$.
• σ is the estimated standard deviation of the error.
The error v(t,θ) is defined as:
$v\left(t,\theta \right)=e\left(t,\theta \right)*\sigma \frac{\rho }{\sqrt{|e\
left(t,\theta \right)|}}$
Regularization option modifies the loss function to add a penalty on the variance
of the estimated parameters.
The loss function is set up with the goal of minimizing the prediction errors. It
does not include specific constraints on the variance (a measure of reliability)
of estimated parameters. This can sometimes lead to models with large uncertainty
in estimated model parameters, especially when the model has many parameters. • Specify the Regularization option in the estimation option sets.
Regularization Regularization introduces an additional term in the loss function that penalizes • For linear-in-parameter models (FIR models) and ARX models, you can compute optimal values of
the model flexibility: the regularization variables R and λ using the arxRegul command.
$V\left(\theta \right)=\frac{1}{N}\sum _{t=1}^{N}{e}^{T}\left(t,\theta \right)W\
left(\theta \right)e\left(t,\theta \right)+\frac{1}{N}\lambda {\left(\theta -{\
theta }^{*}\right)}^{T}R\left(\theta -{\theta }^{*}\right)$
The second term is a weighted (R) and scaled (λ) variance of the estimated
parameter set θ about its nominal value θ*.
Effect of Focus and WeightingFilter Options on the Loss Function
The Focus option can be interpreted as a weighting filter in the loss function. The WeightingFilter option is an additional custom weighting filter that is applied to the loss function.
To understand the effect of Focus and WeightingFilter, consider a linear single-input single-output model:
$y\left(t\right)=G\left(q,\theta \right)\text{}u\left(t\right)+H\left(q,\theta \right)\text{}e\left(t\right)$
Where G(q,θ) is the measured transfer function, H(q,θ) is the noise model, and e(t) represents the additive disturbances modeled as white Gaussian noise. q is the time-shift operator.
In frequency domain, the linear model can be represented as:
$Y\left(\omega \right)=G\left(\omega ,\theta \right)U\left(\omega \right)+H\left(\omega ,\theta \right)E\left(\omega \right)$
where Y(ω), U(ω), and E(ω) are the Fourier transforms of the output, input, and output error, respectively. G(ω,θ) and H(ω,θ) represent the frequency response of the input-output and noise transfer
functions, respectively.
The loss function to be minimized for the SISO model is given by:
$V\left(\theta \right)=\frac{1}{N}\sum _{t=1}^{N}{e}^{T}\left(t,\theta \right)e\left(t,\theta \right)$
Using Parseval’s Identity, the loss function in frequency-domain is:
$V\left(\theta ,\omega \right)=\frac{1}{N}{‖E\left(\omega \right)‖}^{2}$
Substituting for E(ω) gives:
$V\left(\theta ,\omega \right)=\frac{1}{N}{‖\frac{Y\left(\omega \right)}{U\left(\omega \right)}-G\left(\theta ,\omega \right)\right)‖}^{2}\frac{{‖U\left(\omega \right)‖}^{2}}{{‖H\left(\theta ,\omega
Thus, you can interpret minimizing the loss function V as fitting G(θ,ω) to the empirical transfer function $Y\left(\omega \right)/U\left(\omega \right)$, using $\frac{{‖U\left(\omega \right)‖}^{2}}
{{‖H\left(\theta ,\omega \right)‖}^{2}}$ as a weighting filter. This corresponds to specifying Focus as 'prediction'. The estimation emphasizes frequencies where input has more power (${‖U\left(\
omega \right)‖}^{2}$ is greater) and de-emphasizes frequencies where noise is significant (${‖H\left(\theta ,\omega \right)‖}^{2}$ is large).
When Focus is specified as 'simulation', the inverse weighting with ${‖H\left(\theta ,\omega \right)‖}^{2}$ is not used. That is, only the input spectrum is used to weigh the relative importance of
the estimation fit in a specific frequency range.
When you specify a linear filter $ℒ$ as WeightingFilter, it is used as an additional custom weighting in the loss function.
$V\left(\theta \right)=\frac{1}{{N}^{2}}{‖\frac{Y\left(\omega \right)}{U\left(\omega \right)}-G\left(\theta \right)\right)‖}^{2}\frac{{‖U\left(\omega \right)‖}^{2}}{{‖H\left(\theta \right)‖}^{2}}{‖ℒ\
left(\omega \right)‖}^{2}$
Here $ℒ\left(\omega \right)$ is the frequency response of the filter. Use $ℒ\left(\omega \right)$ to enhance the fit of the model response to observed data in certain frequencies, such as to
emphasize the fit close to system resonant frequencies.
The estimated value of input-output transfer function G is the same as what you get if you instead first prefilter the estimation data with $ℒ\left(.\right)$ using idfilt, and then estimate the model
without specifying WeightingFilter. However, the effect of $ℒ\left(.\right)$ on the estimated noise model H depends on the choice of Focus:
• Focus is 'prediction' — The software minimizes the weighted prediction error ${e}_{f}\left(t\right)=ℒ\left({e}_{p}\left(t\right)\right)$, and the estimated model has the form:
Where ${H}_{1}\left(q\right)=H\left(q\right)/ℒ\left(q\right)$. Thus, the estimation with prediction focus creates a biased estimate of H. This is the same estimated noise model you get if you
instead first prefilter the estimation data with $ℒ\left(.\right)$ using idfilt, and then estimate the model.
When H is parameterized independent of G, you can treat the filter $ℒ\left(.\right)$ as a way of affecting the estimation bias distribution. That is, you can shape the trade-off between fitting G
to the system frequency response and fitting $H/ℒ$ to the disturbance spectrum when minimizing the loss function. For more details see, section 14.4 in System Identification: Theory for the User,
Second Edition, by Lennart Ljung, Prentice Hall PTR, 1999.
• Focus is 'simulation' — The software first estimates G by minimizing the weighted simulation error ${e}_{f}\left(t\right)=ℒ\left({e}_{s}\left(t\right)\right)$, where ${e}_{s}\left(t\right)={y}_
{measured}\left(t\right)-G\left(q\right){u}_{measured}\left(t\right)$. Once G is estimated, the software fixes it and computes H by minimizing pure prediction errors e(t) using unfiltered data.
The estimated model has the form:
If you prefilter the data first, and then estimate the model, you get the same estimate for G but get a biased noise model $H/ℒ$.
Thus, the WeightingFilter has the same effect as prefiltering the estimation data for estimation of G. For estimation of H, the effect of WeightingFilter depends upon the choice of Focus. A
prediction focus estimates a biased version of the noise model $H/ℒ$, while a simulation focus estimates H. Prefiltering the estimation data, and then estimating the model always gives $H/ℒ$ as the
noise model.
Model Quality Metrics
After you estimate a model, use model quality metrics to assess the quality of identified models, compare different models, and pick the best one. The Report.Fit property of an identified model
stores various metrics such as FitPercent, LossFcn, FPE, MSE, AIC, nAIC, AICc, and BIC values.
• FitPercent, LossFcn, and MSE are measures of the actual quantity that is minimized during the estimation. For example, if Focus is 'simulation', these quantities are computed for the simulation
error e[s] (t). Similarly, if you specify the WeightingFilter option, then LossFcn, FPE, and MSE are computed using filtered residuals e[f] (t).
• FPE, AIC, nAIC, AICc, and BIC measures are computed as properties of the output disturbance according to the relationship:
G(q) and H(q) represent the measured and noise components of the estimated model.
Regardless of how the loss function is configured, the error vector e(t) is computed as 1-step ahead prediction error using a given model and a given dataset. This implies that even when the
model is obtained by minimizing the simulation error e[s] (t), the FPE and various AIC values are still computed using the prediction error e[p] (t). The actual value of e[p] (t) is determined
using the pe command with prediction horizon of 1 and using the initial conditions specified for the estimation.
These metrics contain two terms — one for describing the model accuracy and another to describe its complexity. For example, in FPE, $det\left(\frac{1}{N}{E}^{T}E\right)$ describes the model
accuracy and $\frac{1+\frac{np}{N}}{1-\frac{np}{N}}$ describes the model complexity.
By comparing models using these criteria, you can pick a model that gives the best (smallest criterion value) trade-off between accuracy and complexity.
Quality Description
Normalized Root Mean Squared Error (NRMSE) expressed as a percentage, defined as:
• y[measured] is the measured output data.
• $\overline{{y}_{measured}}$ is its (channel-wise) mean.
• y[model] is the simulated or predicted response of the model, governed by the Focus.
• ||.|| indicates the 2-norm of a vector.
For input or output data, FitPercent is an n[y]-by-n[exp] matrix, where n[y] is the number of outputs and n[exp] is the number of experiments. For FRD data, FitPercent is an n[y]-by-n[u]
matrix, where n[u] is the number of inputs (for FRD data, n[exp] is always one).
FitPercent varies between -Inf (bad fit) to 100 (perfect fit). If the value is equal to zero, then the model is no better at fitting the measured data than a straight line equal to the
mean of the data.
LossFcn Value of the loss function when the estimation completes. It contains effects of error thresholds, output weight, and regularization used for estimation.
Mean Squared Error measure, defined as:
$MSE=\frac{1}{N}\sum _{t=1}^{N}{e}^{T}\left(t\right)e\left(t\right)$
MSE where:
• e(t) is the signal whose norm is minimized for estimation.
• N is the number of data samples in the estimation dataset.
Akaike’s Final Prediction Error (FPE), defined as:
• n[p] is the number of free parameters in the model. n[p] includes the number of estimated initial states.
• N is the number of samples in the estimation dataset.
• E is the N-by-n[y] matrix of prediction errors, where n[y] is the number of output channels.
A raw measure of Akaike's Information Criterion, defined as:
$AIC=N\ast log\left(det\left(\frac{1}{N}{E}^{T}E\right)\right)+2\ast {n}_{p}+N\left({n}_{y}\ast \mathrm{log}\left(2\pi \right)+1\right)$
Small sample-size corrected Akaike's Information Criterion, defined as:
AICc $AICc=AIC+2\ast {n}_{p}\ast \frac{\left({n}_{p}+1\right)}{\left(N-{n}_{p}-1\right)}$
This metric is often more reliable for picking a model of optimal complexity from a list of candidate models when the data size N is small.
Normalized measure of Akaike's Information Criterion, defined as:
$nAIC=log\left(det\left(\frac{1}{N}{E}^{T}E\right)\right)+\frac{2\ast {n}_{p}}{N}$
Bayesian Information Criterion, defined as:
$BIC=N\ast log\left(det\left(\frac{1}{N}{E}^{T}E\right)\right)+N\ast \left({n}_{y}\ast \mathrm{log}\left(2\pi \right)+1\right)+{n}_{p}\ast \text{log}\left(N\right)$
See Also
aic | fpe | pe | goodnessOfFit | sim | predict | nparams
Related Topics | {"url":"https://uk.mathworks.com/help/ident/ug/model-quality-metrics.html","timestamp":"2024-11-07T02:36:19Z","content_type":"text/html","content_length":"125169","record_id":"<urn:uuid:af0b3885-1c7a-48af-bdda-5cffce0c4c40>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00850.warc.gz"} |
Form Coefficients with illustrations - maritmeculture
What are the form coefficients?
Form coefficients are ratios that numerically compare the ship’s underwater form to that of regular shapes having the same majority dimension as the ship.
Where are the form coefficients used?
The form coefficients are used at the designed stage, prior to the construction of a ship.
Why are the coefficients important?
The coefficients are important because you can predict the resistance to forward motion that the ship will experience during the operation.
The value of the form coefficients is then used to estimate the ship’s power requirements for the desired service speed.
The most important ratio that is used to assign the ship freeboard is the block coefficient.
Now we will see the essential form coefficients.
The coefficient of fineness of waterplane area is defined as the ratio of the ship’s water-plane-area (WPA) to the area of a rectangle having the same length and breadth of the ship at the waterline
in question.
The formula of the form coefficient fineness of waterplane area is
block coefficient is the ration of the underwater volume of a ship to the volume of the circumscribing block.
The formula to calculate the block coefficient is:
Midships area coefficient
midships coefficient at any draught is the ratio of the underwater transverse area of the midships section to the product of the bread and draught.
longitudinal prismatic coefficient at any draught is the ratio of the underwater volume of the ship to the volume of the prism formed by the product of the transverse area of the midships section
and the waterline length.
The formula to calculate the prismatic coefficient of a ship is :
A ship has a length and breadth at the waterline of 40.1m and 8.6m respectively if the waterplane area is 280mq calculate the coefficient of fineness of the waterplane area.
Solution example of COEFFICIENT OF FINESESEN OF THE WATERPLANE AREA
If we use the formula of the COEFFICIENT OF FINESESEN OF THE WATERPLANE AREA
The result will be like this :
EXAMPLE FOR the block coefficient
A ship floats at a draught of 3.20m and has a waterline length and breadth of 46.3m and 15.5m respectively.
Calculate the block coefficient if its volume of displacement is 1800m3
Solution for block coefficient example
Example midships area coefficient
A ships float at a draught of 4.40m and have the waterline breadth of 12.70m.
Calculate the underwater transverse area of the midships section if Cm is 0.922
Solution for the midships area coefficient example
Example prismatic coefficient
A ship has the following details:
Draught 3.63m
Waterline length 48.38m
Waterline breadth 9.42m
Cm 0.946
Cp 0.778
Calculate the volume of displacement
Solution for the midship coefficient
Read: How anchor a boat | {"url":"https://www.maritmeculture.com/form-coefficients-of-a-ship-with-illustrations/","timestamp":"2024-11-04T20:22:34Z","content_type":"text/html","content_length":"90538","record_id":"<urn:uuid:38f1dfff-c9bd-450a-adbe-06ef9d624e77>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00734.warc.gz"} |
Average Speed and Average Velocity
Introduction to Average Speed and Average Velocity
Before understanding average speed and average velocity, we must first understand the distinction between distance and displacement. The scalar quantity "distance" represents how much ground an
object has covered. The shortest distance between two points is represented by displacement, which is a vector quantity. If a particle moves in a circle, for example, the distance travelled after one
revolution equals the circumference of the circle, but the displacement is zero.
Let's have a look at the definitions of speed and velocity.
Distinguish between Average Speed and Average Velocity
To know about average speed and average velocity, first, we must know of some terms and their meanings.
Distance Travelled – Distance travelled, as the name clearly tells, is the total distance travelled by the object.
Time Taken – The time taken by the object to move the given distance.
Displacement – Displacement is the shortest distance between the initial point where the object was and the final point where the object ended up.
Speed – Speed is the distance travelled by an object in unit time. Speed is a scalar quantity. This means it has no specified direction. Speed refers to how fast an object is moving, or essentially
the rate at which the distance is covered.
Velocity – Velocity is the total displacement of the object in a specified direction in unit time. Velocity is a vector quantity. This means it has a specified direction. Velocity refers to the time
rate of displacement of the object. Imagine a person who walks for some distance before returning to his original position. Since velocity is the rate of displacement, this motion results in zero
velocity. If a person wishes to maximise his velocity, he must maximise the displacement from his original position. Since velocity is a vector quantity, when evaluating it, we must keep track of
The main difference between speed and velocity is that speed does not take into account the direction since it is a scalar quantity, and speed depends upon distance travelled, while velocity is a
vector quantity that takes into account the direction, and velocity depends upon displacement.
Average speed is the total distance travelled by an object to the total time taken. However, average velocity is the change in position or displacement (∆x) divided by the time intervals (∆t) in
which the displacement occurs.
So, what difference do you find in the definition of average speed and average velocity? Are they the same in terms of parameters used in their respective formulas? Assume that both the terms convey
the same meaning; still, do they have the same units and possess quantities of the same nature?
Well! Answers to all the questions are available on this page. Also, we will understand the difference in the average speed and average velocity formula along with illustrating real-life examples.
Average Speed
The average speed of any object is the total distance travelled by that object divided by the total time elapsed to cover the said distance. The average speed of an object tells you the average rate
at which it will cover the distance; that is, an object has a speed of 30km/hour, its position will change on an average by 30km each hour. Average speed is a rate that is a quantity divided by the
time taken to get that quantity. SI unit of speed is meters per second.
Average speed is calculated by the formula S = d/t, where S equals the average speed, d equals total distance and t equals total time.
Average Velocity
Average velocity of an object can be defined as the displacement with regards to the original position divided by the time. In other words, it is the rate at which an object makes displacements with
time. Like average speed, the SI unit is metres per second. Average velocity can also be said to be the ratio of the total displacement of an object to the total time for this action to take place.
The direction of average velocity is the direction of displacement. Even if the speed of the object is fluctuating and its magnitude is changing, its direction would still be the same as the
direction of displacement. The magnitude of average velocity is always either less than or equal to the average speed, because displacement is always less than or equal to the distance covered.
Average velocity is calculated by the formula V = D/t, where V equals the average velocity, D equals total displacement and t equals total time.
The formula for Average Speed and Average Velocity
If ‘Δx’ is the displacement of an object in time ‘Δt’, then:
(Image will be Uploaded soon)
The formula for average speed is given as:
vav = Δx/Δt
You noticed that the formula for average velocity and average speed is the same.
The only difference lies in the type of physical quantity, i.e., speed and velocity. Speed is a scalar quantity that has magnitude only. However, velocity is a vector quantity that has both magnitude
and direction.
Now, let us go through some average speed problems:
1. A car travels a distance of 70 km in 2 hours. What is the average speed?
Answer: average speed = distance/time
Therefore, the average speed of the car is 70 km/2 hours = 35km/hour.
2. A person can walk at a speed of 1.5 meters/second. How far will he walk in 4 minutes?
Answer: average speed = distance/time
Distance = average speed(time)
= 1.5(4) (60) = 360 meters
3. A train travels in a straight line at a constant speed of 60 km/h for a particular distance d and then travels another distance equal to 2d in the same direction at a constant speed of 80 km/h in
the same direction as it was previously going. a) What is the average speed of the train during the whole journey?
Solution: a) The time t1 to cover distance d at a speed of 60 km/h is given by t1 = d / 60
The time t2 to cover distance 2d at a speed of 80 km/h is given by t2 = 2d / 80
Average Speed = distance/time = (d + 2d) / (d/60) + (2d/80)
= 3d / (80d + 2d × 60)/(60 × 80)
= 3 d/(200d/4800) = 3d (4800)/200d = 72 km/h
4. Calculate the average velocity at a particular time interval of a person if he moves 7 m in 4 s and 18 m in 6 s along the x-axis?
Solution: Initial distance travelled by the person, xi = 7 m,
Final distance travelled, xf = 18 m,
Initial time interval ti = 4 s,
Final time interval tf = 6 s,
Average velocity v = xi − xf / ti − tf = 18 − 7 / 6 − 4 = 11 / 2 = 5.5 m/s.
From the above text, we understand that the average speed of any object is the total distance travelled by that object divided by the total time elapsed to cover the said distance.
The average speed of an object tells you an average rate at which it will cover the distance; that is, an object has a speed of 30km/hour, its position will change on an average by 30km each hour.
Average speed is a rate that is a quantity divided by the time taken to get that quantity. SI unit of speed is meters per second.
Average speed is calculated by the formula S = d/t, where S equals the average speed, d equals total distance and t equals total time.
Average Velocity
From the above text, we understand that the average velocity of an object can be defined as the displacement with regards to the original position divided by the time.
In other words, it is the rate at which an object makes displacements with time.
For example, the average speed of the SI unit is meters per second. Average velocity can also be said to be the ratio of the total displacement of an object to the total time for this action to take
The average velocity of an object can be defined as the displacement with regards to the original position divided by the time. In other words, it is the rate at which an object makes displacements
with time. Like average speed, the SI unit is meters per second. Average velocity can also be said to be the ratio of the total displacement of an object to total time for this action to take place.
The direction of average velocity is the direction of displacement. Even if the speed of the object is fluctuating and its magnitude is changing, its direction would still be the same as the
direction of displacement. The magnitude of average velocity is always either less than or equal to the average speed because displacement is always less than or equal to the distance covered.
Average velocity is calculated by the formula V = D/t, where V equals the average velocity, D equals total displacement and t equals total time.
Now, let us go through some average velocity problems.
1. A truck driver drives 20 km down the road in 5 minutes. He then reverses and drives 12 km down the road in 3 mins. What is his average velocity?
Solution: v = D/t
v = (20 - 12)/(5 + 3)
= 8/8 = 1 kilometre/minute
2. A man walks 10 km east in 2 hours and then 2.5 km west in 1 hour. Calculate the total average velocity of a man?
Solution: vav = D/t
= (10 - 2.5)/2 + 1
= 7.5/3
vav = 2.5 km/hr
3. Calculate the average velocity at a particular time interval of a person if he moves 7 m in 4 s and 18 m in 6 s along the x-axis?
Solution: Initial distance travelled by the person, xi = 7 m,
Final distance travelled, xf = 18 m,
Initial time interval ti = 4 s,
Final time interval tf = 6 s,
Average velocity vav = xi − xf / ti − tf
= 18 /(6 − 4) = 11/2 = 5.5 m/s
The Differences and Similarities Between Average Speed and Average Velocity
Similarities – Both of these terms are average of some length by the time taken. The SI unit and other standard units of measurement of both average speed and average velocity are the same. The
formula used to calculate the average speed and average velocity is virtually the same, v = D/t, s = d/t, with the only slight difference that in the first case direction is to be mentioned.
Differences - Average speed is a scalar and is not affected by the presence or absence of a direction, while average velocity being a vector needs a direction. Average speed takes distance, that is,
total length travelled while being measured, while average velocity takes displacement, that is, the straight distance from the original position to the final position.
Problems Related to Both Average Speed and Average Velocity
1. A car travels along a straight road to the east for 120 meters in 5 seconds, then goes west for 60 meters in 1 second. Determine average speed and average velocity.
Distance = 120 meters + 60 meters = 180 meters
Displacement = 120 meters – 60 meters = 60 meters, to east.
Time elapsed = 5 seconds + 1 second = 6 seconds.
Average speed = Distance / time elapsed = 180 meters / 6 seconds = 30 meters/second.
Average velocity = Displacement / time elapsed = 60 meters / 6 seconds = 10 meters/second.
2. A runner is running around a rectangle track with length = 50 meters and width = 20 meters. He travels around the rectangle track twice, finally running back to the starting point. If the total
time he takes to run around the track is 100 seconds, determine average speed and average velocity.
The circumference of the rectangle, which is the distance travelled in one round = 2(50 meters) + 2(20 meters) = 100 meters + 40 meters = 140 meters.
When a runner runs around the rectangle twice = 2(140 meters) = 280 meters.
Distance = 280 meter
Displacement = 0 meter. (Since the runner came back to initial point)
Average speed is equal to the distance / time elapsed = 280 meters/100 seconds = 2.8 meters/second.
Average velocity is equal to the displacement / time elapsed = 0/100 seconds = 0
3. A man starts walking from a point on a circular field of radius 0.5 km and 1 hour later he finds himself at the same point where he initially started.
a) What is the average speed for the whole journey he travelled? What is the average velocity of this man for the same?
Solution: a) If this man walks around a circular field and comes back to the same point, he has covered a distance which is equal to the circumference of the circle.
Thus, average speed he travelled = Distance/time = circumference time = π (0.5) (2)/1 hour = 3.14 km/hour (approximately).
b) If he walks around in a circle and comes back to the same point where he started in a circle then the change in his position is zero. Since the change in his position is zero, displacement is also
equal to zero. This means the average velocity is also equal to zero. | {"url":"https://www.vedantu.com/physics/average-speed-and-average-velocity","timestamp":"2024-11-06T04:33:23Z","content_type":"text/html","content_length":"317627","record_id":"<urn:uuid:b17f76fd-d1b0-4eca-87cd-dec3601203b9>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00525.warc.gz"} |
Count Cells with Text in Google Sheets (Easy Formulas)
How to Count Cells with Text in Google Sheets
Are you someone struggling with finding the count of cells that contain text value in Google Sheets? Then you have reached the right place!
There are scenarios where you have a mixed set of all types of data, such as text, numbers, dates, special characters, etc., and you only want your formula to count the cells that contain text or
string values.
There are various functions in Google Sheets that you can use solo as well as in combination with other functions to get this task done.
Throughout this article, let’s curate a few unique ways to get this task done!!!
The Dataset
The dataset I am using for this article is a simple one but mixed with different value types, such as numbers, dates, special characters, blank spaces, etc., along with the text values. See it
Using the Simple COUNTIF Function and Wildcards
When you have a dataset with mixed values of different types as the one you will use here for this article, counting how many cells contain text becomes a job of comparison.
You want to check at each cell whether it contains text or not. And then return the total count of all such cells.
The COUNTIF function is the most suitable and straightforward way for this case. It checks for a specific condition you provide and then counts only those cells where it is met.
In cell E3, copy and paste the following COUNTIF formula and hit the Enter button to return the count of cells containing only text values.
=COUNTIF(B3:B13, "*")
The “*” (also known as asterisk) is a wildcard character helping the IF part of the formula here to check whether the value from each cell is a text or not. It then produces an array of boolean
values, TRUE, and FALSE per row.
{FALSE; TRUE; TRUE; TRUE; TRUE; FALSE; FALSE; TRUE; TRUE; FALSE; TRUE}
The COUNT part of the function then counts how many times the value TRUE has appeared in the cells. If you think of the processing, the number 1 represents TRUE, and 0 represents FALSE. So ideally,
the COUNT part just shows how many times 1 appeared in the range.
This then results in counting all occurrences of 1’s and returning the value 7 in cell E3.
Note: If you have noticed, the formula also considers the cell containing special characters such as pound, ampersand, dollar, and exclamation mark. They are also considered as the text values;
hence, the function returns the count 7 rather than 6. What were you expecting, though?🤔 I would really love to know! 😉
Using SUMPRODUCT and ISTEXT Function
The SUMPRODUCT is a versatile function in Google Sheets that performs two operations, Addition, and Multiplication, simultaneously on an array (or more arrays) based on the context provided.
If combined with ISTEXT, the function returns the count of all cells containing text values as output.
The beauty of the SUMPRODUCT function is that even though it cannot SUM up the boolean array of TRUEs and FALSEs, it uses some other operators smartly to convert them to equivalent numeric values,
i.e., 1 and 0, respectively. Then, it sums the array up to produce the output.
The dataset I will be using for this demo is as shown below:
To count the cells containing text, copy and paste the formula below in cell E3 of your sheet and hit the Enter button to execute it.
Let’s break down the working of this formula:
The ISTEXT function firstly checks cell by cell whether the value is a text or not. It generates a TRUE/FALSE array. In this context, the ISTEXT function works like the conditional IF.
The —, also called as double unary, then works as a boolean to numeric values converter and converts the array of TRUE and FALSE to an array of numeric 1’s and 0’s.
This part is crucial one in this formula because the SUMPRODUCT can’t sum up the boolean values TRUE and FALSE. It needs an array of numeric values, and this double unary (—) operator precisely does
The SUMPRODUCT function now effectively looks like
=SUMPRODUCT({0; 1; 1; 1; 1; 0; 0; 1; 1; 0; 1})
The function then sums up all the 1’s, which represent the count of cells containing text values. Seems interesting?
Another Version of the formula
Few of the users might be reluctant to use the double unary (–) in any formula. It might confuse them, or they might think it is an additional thing to remember that doesn’t follow the general logic
of the numbers and formulas.
Such users can use the following version of the formula without this operator.
The *1 effectively converts the array of TRUEs and FALSEs to a numeric value, and by doing so, allows you to keep the formula simple to remember.
Using COUNTA, FILTER, and ISTEXT
When you are working with finding out the count of cells containing only text values, the combination of the trio COUNTA, FILTER, and ISTEXT provides you with one of the most straightforward and
easy-to-understand ways of getting things done.
I will use the mix of data for this demo with all values of different types spread between range B3:B13.
The recipe for this demo is simple: copy or type in the following formula in cell E3 of your sheet and hit the Enter button to see the magic yourself!
=COUNTA(FILTER(B3:B13, ISTEXT(B3:B13)))
Let’s break the logic behind the working of this formula piece-by-piece.
The ISTEXT function works here as a conditional IF and checks whether each cell from the range B3:B13 contains a text value or not. It potentially creates an array of 10 TRUE/FALSE values.
Then, you use this array of the TRUE/FALSE values per cell as a condition inside the FILTER function.
The function takes the entire range (B3:B13) and creates a subset out of it for all the cells where ISTEXT has the value TRUE, effectively only taking rows where cells contain a text value.
Finally, you use the COUNTA function smartly to return the count of 7.
That function is also important here because the function returns the count of all non-empty cells and doesn’t work.
When you want the count of cells containing values of a specific type, that’s where the subset created by the FILTER function is the game-changer.
Even though the method is straightforward to understand and unique in its own way, it might slow the application down when working on a sufficiently large dataset.
For example, imagine you have a dataset of 1 million rows. The ISTEXT function will first check for each cell whether its value is text or not. It then uses that output as filtering criteria inside
the FILTER function to slice the data down to a subset of only text values. And then finally returning the count of cells from that subset 😥.
However, the good news is, most of the time, you will not come across a dataset with millions of rows where you only need to find the count of text values.
Using ARRAYFORMULA and ISTEXT with Others
The next in line is using the ARRAYFORMULA in combination with the ISTEXT to produce the result.
The method doesn’t really care whether your dataset is enormous or a shorter one. It works with the same efficiency, and at the core of it is the ARRAYFORMULA function😉.
I am using a column containing mixed types of values while explaining this method as below.
To return the count of only those cells containing text values, use the following formula in cell E3. And you will see the output as shown below.
Just to make you aware of why the ARRAYFORMULA is so crucial here, I am breaking the formula down into the pieces.
1. Imagine we use the IF function to return 1 whenever the ISTEXT returns TRUE for each cell of the range B3:B13. It would be something like below.
2. Now, you know that if you sum up all these cells, you will get the count of all those cells that only contain text value.
However, this task is redundant, and you don’t want to do it. You need someone that works very well with the arrays and allows you to encapsulate this two-step workaround into a single one.
That’s where the ARRAYFORMULA comes into the picture with its ability to work well with the operations as well as its ability to provide other non-array formulas like ISTEXT the power to work with
You know what? You can further simplify this formula by removing the IF condition and using the double unary(–) instead of it.
The double unary (–) will convert the boolean TRUE/FALSE array into an array of numeric 1’s and 0’s. The rest everything of this formula stays the same.
That’s another way to tweak the above formula without compromising the output.
Count Cells with a Specific Text
Sometimes, you might be interested in finding out the count of cells with a specific text. For example, only the cells where the name John appears? It is very crucial when you are analyzing the data.
Can you count cells with a specific text in Google Sheets? Hell Yes! You can!
Through this section, I will show you how!
The dataset I will use for this demo differs from the one used in previous demos and is shown below.
Let’s assume you want to know how many times the name “Mary Johnson” appears in the list shown in the screenshot above.
The generic COUNTIF formula to get the count of cells containing a specific text is as below:
=COUNTIF(range, “text”)
Use the following COUNTIF formula inside cell E3 to get the count of cells containing the name “Mary Johnson.”
=COUNTIF(B3:B13, "Mary Johnson")
The COUNTIF function runs through all the cells in range B3:B13 and checks whether the criteria (name = Mary Johnson) is fulfilled by each cell. Wherever it is fulfilled, the count of all those cells
is returned.
Count cells with a Partial Text Match
Once you know how to count cells containing specific text, it is evident that the next important thing is to count how many times a partial text appears inside the given range.
Imagine you are doing exploratory analysis on the dataset provided, and you are interested to know how many times the cells contain a partial text “Turnor.” You can find it out in Google Sheets.
Again, the formulas will not differ, but only the logic will be tweaked a bit here.
The dataset I will use for this demo is as shown below.
In cell E3, use the formula below to return the count of cells where all the names with “Turner” are placed.
=COUNTIF(B3:B13, "*Turner*")
The wildcard asterisk is the key here. Its use before and after the text “Turner” suggests that whatever comes before and after that text should be counted.
And with that, I mark the end of this article where you learned five fabulous ways to count cells containing only text values.
The COUNTIF method is the simplest of them all and easy to remember.
The SUMPRODUCT and ISTEXT method becomes unique by converting an array of TRUE FALSE to an array of respective numeric values.
The COUNTA, FILTER method is unique in a way that it subsets the original data only to show and count the one where the ISTEXT formula returns true.
The ARRAYFORMULA and ISTEXT, in combination with SUM and IF conditions, introduce you to the power of arrays while counting.
And finally, you come across two scenarios where you count cells containing text based on a specific text and based on a partial text match.
Other articles you may also like:
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resistors are energy storage elements
Welcome to another enlightening episode of Course Bridge! In this captivating video, we''re diving deep into the fascinating world of energy storage elements,...
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About resistors are energy storage elements
As the photovoltaic (PV) industry continues to evolve, advancements in resistors are energy storage elements have become critical to optimizing the utilization of renewable energy sources. From
innovative battery technologies to intelligent energy management systems, these solutions are transforming the way we store and distribute solar-generated electricity.
When you're looking for the latest and most efficient resistors are energy storage elements for your PV project, our website offers a comprehensive selection of cutting-edge products designed to meet
your specific requirements. Whether you're a renewable energy developer, utility company, or commercial enterprise looking to reduce your carbon footprint, we have the solutions to help you harness
the full potential of solar energy.
By interacting with our online customer service, you'll gain a deep understanding of the various resistors are energy storage elements featured in our extensive catalog, such as high-efficiency
storage batteries and intelligent energy management systems, and how they work together to provide a stable and reliable power supply for your PV projects.
محتويات ذات صلة | {"url":"https://rudzka95.pl/Fri-30-Aug-2024-19097.html","timestamp":"2024-11-12T02:34:28Z","content_type":"text/html","content_length":"43093","record_id":"<urn:uuid:b50dc4b5-ca51-482c-82ad-5cb13386f202>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00418.warc.gz"} |
s - Bowei Xiao
« on: November 01, 2012, 02:11:55 PM »
I guess in problem 1 part c and d...you actually want to type X^n exp() and so does d? Guess there's typo there..
Yes, thanks, it was typo, but I decided for a sake of simplicity to take $n=1$. V.I. | {"url":"http://forum.math.toronto.edu/index.php?PHPSESSID=79h0hkr3h8sn4vq0fuhjqhu926&action=profile;area=showposts;sa=messages;u=63","timestamp":"2024-11-13T13:02:45Z","content_type":"application/xhtml+xml","content_length":"25115","record_id":"<urn:uuid:10ad0b14-ed1d-48d8-9284-edcf47308586>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00352.warc.gz"} |
Homework 11
1. [4 points] Exercise 9.2.4 in the text.
2. [6 points] Show that the recursive languages are closed under
a. union and
b. intersection.
3. [4 points] Prove that the following question is undecidable (i.e., prove that the corresponding language is not recursive): Given a Turing Machine M and a state q of M, is there any input that
will cause M to enter state q? [This problem is analogous to the problem of finding dead code in, say, a Java program: Given a Java program and a block of code within that program, is there some
input that will cause that block of code to be executed?]
4. [4 points] Prove that the following question is undecidable: Given two Turing Machines, do they accept the same language? [This problem is analogous to the problem of determining if two Java
programs are equivalent.] | {"url":"https://www.cs.cornell.edu/courses/cs3810/2008su/Homework/HW11.html","timestamp":"2024-11-08T02:07:48Z","content_type":"text/html","content_length":"2275","record_id":"<urn:uuid:a1208985-82d9-4e3d-86ce-33990fa3587b>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00671.warc.gz"} |
Mechanical Behavior and Wrinkling of Lined Pipes under Bending and External Pressure
The research examines the mechanical behavior of lined pipes and investigates extensively the wrinkling of such pipes under bending loading with or without the presence of external pressure. Lined
pipe, also referred to as “mechanically clad pipe”, is a new and promising technological solution in energy pipeline applications where the structural integrity of oil and gas steel pipelines
requires erosion damage protection from oil or gas pollutants. Lined pipe is a double-wall pipe, consisting of a load-bearing high-strength, low-alloy carbon steel outer pipe, lined with a
thin-walled sleeve made from a corrosion-resistant material referred to as “liner” pipe. Lined pipes are produced through an appropriate manufacturing procedure, consisting of heating the outer pipe,
inserting the liner and pressurizing it until both pipes come to contact, and finally cooling the outer pipe. Considering the liner pipe as a thin-walled cylindrical shell prone to buckling, the
lateral confinement due to the deformable outer pipe constitutes a paramount parameter for its mechanical behavior. In the present investigation, the problem is solved numerically, using nonlinear
finite elements capable at simulating the lined pipe and the interaction between the liner and the outer pipe. Nonlinear geometry with large strains is taken into account, and the material of both
pipes is elastic-plastic. The lined pipe is considered either stress-free (snug-fit pipe, SFP) or with an initial stress (tight-fit pipe, TFP). The thermo-hydraulic manufacturing process of the lined
pipe is simulated to determine the liner hoop prestressing. First, an ovalization bending analysis of the lined pipe is conducted, where a slice of the pipe between two adjacent cross-sections is
considered excluding the possibility of buckling. In this analysis, the stress and deformation of the liner in the compression zone is monitored, with emphasis on possible detachment of the liner
from the outer pipe. Using a simple buckling hypothesis, it is possible to estimate the curvature at which liner wrinkling would occur. Subsequently, a three-dimensional analysis is conducted to
examine buckling of the liner in the form of a uniform wrinkling pattern. The curvature at which buckling occurs and the corresponding buckling wavelength are determined for different thicknesses of
the liner and outer pipe. Furthermore, the transition from a uniform wrinkling configuration to a secondary bifurcation with more localized deformations is investigated, and reference to experimental
observations is made. Next, the effects of initial imperfections on the structural behavior of lined pipes are investigated. Subsequently, the effect of the prestressing on the critical wavelength
and the buckling curvature is examined. A comparison with available experimental results is conducted in terms of wrinkle height development and the corresponding buckling wavelength. Finally, the
structural behavior of lined steel pipes under bending in the presence of external pressure is examined. The results of the present research can be used for safer design of lined pipes in pipeline
Daniel Vasilikis
Spyros A. Karamanos
Relevant Publications:
In Referred Journals
• Vasilikis, D. and Karamanos, S.A., “Mechanical Behavior and Wrinkling of Lined Pipes”, International Journal of Solids and Structures, Vol. 49, No. 23-24, pp. 3432-3446, November 2012.
• Vasilikis, D. and Karamanos, S.A., “Buckling of Double-Wall Elastic Tubes under Bending”, 9th HSTAM International Congress on Mechanics, Limassol, Cyprus, July 2010.
• Vasilikis, D. and Karamanos, S.A., “Buckling of Clad Pipes under Bending and External Pressure”, 30th International Conference on Ocean, Offshore and Arctic Engineering, ASME, OMAE2011-49470,
Rotterdam, The Netherlands, June 2011.
• Vasilikis, D. and Karamanos, S.A., “Numerical Simulation of Clad Pipe Structural Behavior under Bending Loading”, 7th GRACM International Congress on Computational Mechanics, Athens, Greece, June
• Vasilikis, D. and Karamanos, S.A., “Wrinkling of Lined Pipes under Bending”, 22nd International Conference on Offshore (Ocean) and Polar Engineering, ISOPE2012-TCP-0748, Rhodes, Greece, June
In Conference Proceedings
Figure 1: Photo of lined pipes after experimental testing.
Figure 2: Photo of wrinkled liner pipe after experimental testing.
Figure 3: Schematic representation of uniformly wrinkled pipe and the corresponding half-wavelength between cross-sections α-α and β-β.
Figure 4: Lined pipe model; outer pipe is modeled with solid elements and liner pipe is modeled with shell elements.
Figure 5: Variation of the normalized critical curvature for different thicknesses of the outer pipe.
Figure 6: Lined pipe configurations; (a) undeformed configuration; (b), (c) and (d) ovalized and buckled liner.
Figure 7: (a) Initial configuration with liner wavy imperfection and (b),(c),(d) Deformed (buckled) configuration of lined pipe after secondary bifurcation.
Figure 8: Detachment development of points (1) and (2) in terms of bending curvature.
Figure 9: Detachment development of points (1), (2) and (3) in terms of bending curvature.
Figure 10: Detachment development of SF Pipe A for different values of the first mode imperfection amplitude.
Figure 11: Effects of initial wrinkling imperfections on the value of secondary bifurcation curvature.
Figure 12: Photo of wrinkled specimen after experimental testing with image from finite element simulation. | {"url":"http://karamanos.mie.uth.gr/index.php/mechanical-behavior-and-wrinkling-of-lined-pipes-under-bending-and-external","timestamp":"2024-11-09T12:46:33Z","content_type":"text/html","content_length":"24770","record_id":"<urn:uuid:67eaac21-ec23-4771-b5d4-f1a0ef692f41>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00119.warc.gz"} |
Excel Formula: VLOOKUP in Python
In this tutorial, we will learn how to use the VLOOKUP function in Excel with Python. The VLOOKUP function is a powerful tool that allows you to search for a specific value in the first column of a
range and retrieve a corresponding value from another column in the same row. This can be very useful when working with large datasets or when you need to quickly find and retrieve specific
To use the VLOOKUP function in Python, we can leverage the power of the pandas library. Pandas provides a wide range of functions and methods for data manipulation and analysis, including the ability
to perform VLOOKUP operations.
Let's break down the Excel formula =VLOOKUP(C2,$A$2:$E$399,4) step by step:
1. C2 is the value that we want to search for in the first column of the range $A$2:$E$399.
2. $A$2:$E$399 is the range where the search will be performed. The first column of this range is used as the lookup column.
3. 4 is the column index number of the value we want to retrieve from the range $A$2:$E$399. In this case, it is the fourth column of the range.
4. The VLOOKUP function will search for the value in C2 in the first column of the range $A$2:$E$399. If a match is found, it will return the corresponding value from the fourth column of the range.
Let's look at some examples to better understand how the VLOOKUP function works in Python:
Example 1:
If we have the following data in the range $A$2:$E$399:
| A | B | C | D | E |
| | | | | |
| Cat | 10 | Dog | 20 | Cow |
| Dog | 15 | Cat | 25 | Pig |
| Cow | 30 | Pig | 35 | Dog |
And the value in C2 is Cat, the formula =VLOOKUP(C2,$A$2:$E$399,4) would return 25, which is the value in the fourth column of the range $A$2:$E$399 corresponding to the row where Cat is found in the
first column.
Example 2:
Similarly, if the value in C2 is Dog, the formula would return 35, which is the value in the fourth column of the range $A$2:$E$399 corresponding to the row where Dog is found in the first column.
In conclusion, the VLOOKUP function in Excel allows you to search for values in a range and retrieve corresponding values from another column. By using the pandas library in Python, you can perform
VLOOKUP operations and manipulate data efficiently. This tutorial has provided step-by-step instructions and examples to help you understand and implement the VLOOKUP function in Python.
An Excel formula
Formula Explanation
This formula uses the VLOOKUP function to search for a value in the first column of a range and return a corresponding value from another column in the same row.
Step-by-step explanation
1. C2 is the value that we want to search for in the first column of the range $A$2:$E$399.
2. $A$2:$E$399 is the range where the search will be performed. The first column of this range is used as the lookup column.
3. 4 is the column index number of the value we want to retrieve from the range $A$2:$E$399. In this case, it is the fourth column of the range.
4. The VLOOKUP function will search for the value in C2 in the first column of the range $A$2:$E$399. If a match is found, it will return the corresponding value from the fourth column of the range.
For example, if we have the following data in the range $A$2:$E$399:
| A | B | C | D | E |
| | | | | |
| Cat | 10 | Dog | 20 | Cow |
| Dog | 15 | Cat | 25 | Pig |
| Cow | 30 | Pig | 35 | Dog |
And the value in C2 is Cat, the formula =VLOOKUP(C2,$A$2:$E$399,4) would return 25, which is the value in the fourth column of the range $A$2:$E$399 corresponding to the row where Cat is found in the
first column.
Similarly, if the value in C2 is Dog, the formula would return 35, which is the value in the fourth column of the range $A$2:$E$399 corresponding to the row where Dog is found in the first column. | {"url":"https://codepal.ai/excel-formula-generator/query/8BtcDbtc/excel-formula-vlookup-python","timestamp":"2024-11-09T09:04:04Z","content_type":"text/html","content_length":"99139","record_id":"<urn:uuid:f3a91d5c-9a5d-4958-99e3-d9c69ae448dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00316.warc.gz"} |
Constructing Yield Curves: A Comprehensive Guide to Understanding and Building Yield Curves for Fixed Income Analysis
24.2.1 Constructing Yield Curves
Introduction to Yield Curves
The yield curve is a fundamental concept in fixed income analysis, representing a graphical depiction of interest rates across different maturities for bonds with equal credit quality. It serves as a
critical tool for investors, economists, and policymakers to gauge the market’s expectations of future interest rates, inflation, and economic activity. Understanding the yield curve’s shape and
movements can provide valuable insights into the economic outlook and guide investment decisions.
Significance of the Yield Curve
The yield curve’s shape—whether upward sloping, flat, or inverted—can signal different economic conditions. An upward-sloping curve typically indicates a healthy, growing economy, while a flat or
inverted curve might suggest economic slowdown or recession. These insights make the yield curve an essential element in economic forecasting and investment strategy formulation.
Methodologies for Constructing Yield Curves
Constructing a yield curve involves several steps, primarily focusing on collecting bond data and applying mathematical techniques to derive spot and forward rates. The process typically involves the
following steps:
Collecting Bond Data
To construct a yield curve, the first step is to gather data on yields and maturities of benchmark securities, such as government bonds. These bonds are chosen for their high credit quality and
liquidity, ensuring that the derived yield curve accurately reflects the market’s interest rate expectations.
Bootstrapping Technique
The bootstrapping method is a popular technique used to derive zero-coupon (spot) rates from the prices of coupon-bearing bonds. This method involves solving for spot rates sequentially, starting
with the shortest maturity bond and using those rates to calculate spot rates for longer maturities.
Bootstrapping Process: A Numerical Example
Let’s illustrate the bootstrapping process with a numerical example:
1. Start with the Shortest Maturity Bond: Suppose we have a 1-year bond with a yield of 2%. Since this is a zero-coupon bond, the spot rate for 1 year, \( s_1 \), is 2%.
2. Calculate the Spot Rate for the Next Maturity: Consider a 2-year bond with a 2.5% yield and a coupon rate of 2%. The price of this bond can be expressed as:
$$ P = \frac{C}{(1 + s_1)} + \frac{C + F}{(1 + s_2)^2} $$
Where \( C \) is the annual coupon payment, and \( F \) is the face value of the bond. By substituting known values and solving for \( s_2 \), we derive the 2-year spot rate.
3. Continue for Longer Maturities: Repeat the process for bonds with longer maturities, using previously calculated spot rates to solve for new ones.
Spot Rates vs. Forward Rates
Understanding the difference between spot rates and forward rates is crucial in yield curve analysis:
• Spot Rates: These are yields on zero-coupon bonds, representing the return for investing today until a future date. They are derived from the bootstrapping process and are used to discount future
cash flows in bond valuation.
• Forward Rates: These rates represent expected short-term interest rates in the future, implied by current spot rates. They provide insights into the market’s expectations of future interest rate
Calculating Spot and Forward Rates
The relationship between spot rates and forward rates can be expressed using the following formula:
$$ (1 + s_n)^n = (1 + s_{n-1})^{n-1} \times (1 + f_{n-1,n}) $$
• \( s_n \) is the spot rate for maturity \( n \).
• \( f_{n-1,n} \) is the forward rate from period \( n-1 \) to \( n \).
Graphical Representation of the Yield Curve
Below is a graphical representation of a yield curve constructed from calculated spot rates. This curve provides a visual overview of interest rate expectations across different maturities.
graph TD;
A[1 Year] -->|2%| B[2 Year];
B -->|2.5%| C[3 Year];
C -->|3%| D[4 Year];
D -->|3.5%| E[5 Year];
Applications of Yield Curve Construction
Yield curve construction has several practical applications in finance and investment:
Bond Valuation
In bond valuation, future cash flows are discounted using appropriate spot rates derived from the yield curve. This approach provides a more accurate valuation by reflecting the market’s interest
rate expectations for each cash flow period.
Investment Strategies
Investors use yield curves to inform their investment strategies. For instance, a steepening yield curve might prompt investors to favor long-term bonds, anticipating higher returns, while a
flattening curve could suggest a shift towards short-term securities.
Importance of Accurate Data and Assumptions
Constructing reliable yield curves requires accurate data and sound assumptions. Market imperfections, such as liquidity issues and credit risk, can affect bond pricing and, consequently, the derived
yield curve. Therefore, analysts must carefully consider these factors when constructing and interpreting yield curves.
Challenges in Yield Curve Construction
Despite its usefulness, yield curve construction faces several challenges:
• Market Imperfections: Factors like liquidity constraints and credit risk can distort bond prices, affecting the accuracy of the yield curve.
• Data Limitations: Incomplete or inaccurate data can lead to incorrect yield curve estimations, impacting investment decisions.
Constructing yield curves is a fundamental aspect of fixed income analysis, providing insights into market expectations and guiding investment decisions. By understanding the methodologies for
constructing yield curves and the significance of spot and forward rates, investors can better price bonds and anticipate interest rate movements.
Quiz Time!
📚✨ Quiz Time! ✨📚
### What is a yield curve? - [x] A graph plotting interest rates of bonds with equal credit quality but differing maturity dates. - [ ] A graph showing stock prices over time. - [ ] A chart of
inflation rates over the past decade. - [ ] A table of currency exchange rates. > **Explanation:** The yield curve represents the relationship between interest rates and different maturities for
bonds of the same credit quality. ### What is the first step in constructing a yield curve? - [x] Collecting bond data on yields and maturities of benchmark securities. - [ ] Calculating forward
rates. - [ ] Estimating future inflation rates. - [ ] Analyzing stock market trends. > **Explanation:** The initial step involves gathering data on yields and maturities of benchmark bonds to
construct the yield curve accurately. ### What technique is commonly used to derive spot rates from coupon-bearing bonds? - [x] Bootstrapping technique. - [ ] Regression analysis. - [ ] Monte Carlo
simulation. - [ ] Time series analysis. > **Explanation:** Bootstrapping is used to derive zero-coupon (spot) rates from coupon-bearing bond prices. ### What do spot rates represent? - [x] Yields on
zero-coupon bonds, representing the return for investing today until a future date. - [ ] Expected future inflation rates. - [ ] Current stock market returns. - [ ] Historical bond yields. >
**Explanation:** Spot rates are the yields on zero-coupon bonds, indicating the return for holding a bond until maturity. ### What are forward rates? - [x] Expected short-term interest rates in the
future, implied by current spot rates. - [ ] Current long-term interest rates. - [x] Future inflation expectations. - [ ] Historical interest rate trends. > **Explanation:** Forward rates are derived
from spot rates and represent expected future short-term interest rates. ### What is the formula for calculating spot and forward rates? - [x] \\((1 + s_n)^n = (1 + s_{n-1})^{n-1} \times (1 + f_
{n-1,n})\\) - [ ] \\(PV = \frac{C}{(1 + r)^n}\\) - [ ] \\(FV = PV \times (1 + r)^n\\) - [ ] \\(IRR = \frac{NPV}{C}\\) > **Explanation:** This formula relates spot rates and forward rates, allowing
for the calculation of future interest rates. ### How are yield curves used in bond valuation? - [x] By discounting future cash flows using appropriate spot rates. - [ ] By estimating future stock
prices. - [ ] By predicting currency exchange rates. - [ ] By analyzing historical inflation trends. > **Explanation:** Yield curves provide spot rates used to discount future cash flows, aiding in
accurate bond valuation. ### What does a steepening yield curve indicate? - [x] A potential increase in long-term interest rates. - [ ] A decrease in short-term interest rates. - [ ] A stable
economic outlook. - [ ] A recessionary period. > **Explanation:** A steepening yield curve suggests rising long-term interest rates, often associated with economic growth. ### Why is accurate data
important in yield curve construction? - [x] To ensure reliable yield curve estimations and informed investment decisions. - [ ] To predict stock market trends. - [ ] To calculate historical
inflation rates. - [ ] To estimate currency exchange rates. > **Explanation:** Accurate data is crucial for constructing reliable yield curves, impacting investment decisions and economic forecasts.
### True or False: Market imperfections can affect the accuracy of yield curves. - [x] True - [ ] False > **Explanation:** Market imperfections, such as liquidity issues and credit risk, can distort
bond prices and affect yield curve accuracy. | {"url":"https://csccourse.ca/24/2/1/","timestamp":"2024-11-09T12:54:03Z","content_type":"text/html","content_length":"92672","record_id":"<urn:uuid:910032fe-21bf-48bc-84b0-bacd6c79bbf4>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00199.warc.gz"} |
Interesting chat about Thief 1 and 2
In the gog dot com forum someone asked if they like more Thief 1 or Thief 2. It spawned a fairly nice and constructive analysis of the two games by many people.
I found an interesting thread, I am impressed how different can be the reaction to some design choices...
If you design maps you might also find it interesting... besides, do you like more Thief 1 or Thief 2?
I prefer Thief 2 ever so slightly, namely because of its largely grounded level design. The supernatural levels in T1 haven't really aged very well while those in 2 are still great and interesting to
play. In addition, the regular levels in T2 are far more intricate and interesting in terms of more advanced assets and ambitious objectives. I love 'em both pretty much the same, but would put The
Metal Age just a point above The Dark Project.
If you design maps you might also find it interesting...
Thanks for the share, however I'm afraid I didn't find that to be the case, since (not surprisingly) their reactions assume prior knowledge of those games, rather than explaining what wasn't
appreciated (so lost on someone like me, unfamiliar with either). Although a dislike for non-human opponents has been expressed here before as well.
"The measure of a man's character is what he would do if he knew he never would be found out."
- Baron Thomas Babington Macauley
I slightly prefer Thief 1 for the enemy types, the plotline, and that the Thief world seems more expansive. I like more of the magical and supernatural in Thief. Thief 2 is a great follow-up though.
I don't know if the lore supports it, but I've always considered Karras's creations to have involved some magic rather than being purely mechanical. Thief 2 Gold sounds like it probably would have
been even more to my liking with the planned necromancer mission.
• 1
dislike for non-human opponents has been expressed here before as well.
They were great at the time, but now they fall very short as useless, lumbering, polygonal zombies or easily flashbombed maniacs of a similarly polygonal nature. It's unfortunate really because I
first played it on Halloween when it first came out and I was scared witless =P
In Thief the non-human enemies behave almost the same as humans, I don't get the why people dislike them...
I have always liked the aesthetics, story and level design of the first game better (which should not come as a surprise considering my missions). I have several problems with The Metal Age - it
looks too clean and regular; its story is a mishmash with leaps of logic straining the suspension of disbelief; and the levels lack the kind of variety you had in the original. The changes are so
huge that I think it would have worked better if it was set about 50-150 years after The Dark Project, and even featured a different protagonist. Still a great game, just not TDP.
• 4
Come the time of peril, did the ground gape, and did the dead rest unquiet 'gainst us. Our bands of iron and hammers of stone prevailed not, and some did doubt the Builder's plan. But the seals held
strong, and the few did triumph, and the doubters were lain into the foundations of the new sanctum. -- Collected letters of the Smith-in-Exile, Civitas Approved
I've always preferred Thief 2 in gameplay, atmosphere, and story, but maybe that's because I started with Thief 2 (a friend at the time advised me to skip Thief 1 based on my personality). After
playing it, Thief 1 felt a lot more action oriented--felt like there was a lot less sneaking and too many monsters/ghosts/magic. In general, I think supernatural phenomena are best used sparingly.
And I'm not saying that Thief 1 didn't have many good moments, but while playing many parts of it I found myself repeatedly thinking "OK, when do we get past these annoying monsters/undead and get
back to stealing from the nobility's excessively architectured mansions?"
In addition, I liked the level design in Thief 2 much more. The layouts feel more convincing (e.g. rarely did I find architecture or mapper's mistakes/laziness that "broke the illusion"), fine-tuned
(e.g. attention to detail with the lighting and contextual music), and enticing, as they seem to really captivate my curiosity.
I loved the environments in TDP, T2 just didn't compare in this respect. It was great doing missions in The City proper, but also then getting out to Cragscleft, The Bonehoard, the Barricaded Area,
The Lost City, The Mage Towers and the Maw. It made the game interesting when Garrett occasionally got out of the civilized areas of the world into areas where he couldn't count on the rules being
the same. The isolation, the non-standard AI, and the unpredictable settings gave the player an added feel of danger and insecurity.
I know the new Thief will probably be at little closer to T2, but I hope we at least get out of The City once or twice during New Garrett's adventures.
• 2
Lots of hate on the zombies, burricks and spiders...the annoying part of those enemies is the best part, they suck! I love having to deal with enemies that I hate, it makes things so much more
frustrating and great! Challenge; when the going gets tough, the tough get going. I'm an animal lover too so I kinda feel a bit when I put down a burrick and it makes that whimpering sound like a dog
or whatnot. Sometimes I leave them alive and let them run around (i.e. Haunted Cathedral).
The sound design on both games is off the charts, but TDP has a bit more of that extra creepy that I can't get enough of. It builds the atmosphere so much. I LOVE that I can turn off the music
(ambient?) and the game is still drenched with atmosphere. Personally, I can't stand music in games because in real life there is no soundtrack to looting a boneyard, cathedral, jail, etc. It's very
tense and realistic and helps keep the player on the edge of their seat. Kudos to Looking Glass.
The weirdness was a little more over the top on TDP. I really liked The Sword and how Constantine was a fan of building upside down and sideways rooms. Why not? It was refreshing given that it's not
possible in the real world, but seeing it in a game made for good enjoyment.
I can't choose between the two, both have ups and downs. Both are critical in their own right.
• 1
Plastik Musik - Andrew Nathan Kite, Owner
http://www.facebook.com/plastikmusik /
I tried plaing Thief1 and the enemies behave differently then I remembered... In level 2 I remember you could make the zombies follow you and drown them in a small lake... I the last game they just
wait for me on the water's edge... Interesting...
Are you playing with the NewDark upgrade? There is no T1 version of newdark, it's simply the T2 build of the .exe that is capable of playing T1 and T1G missions.
Edited by jtr7
A skunk was badgered--the results were strong.
I hope that something better comes along.
I'd have to say Thief 2. I've re-played Thief 2 plenty of times over the years, but I've rarely re-visited Thief 1.
I'll admit that I think that Thief 1 does have the advantage in story, voice acting, and atmosphere. However, personally that doesn't do enough to make me prefer it over Thief 2 because I'm not quite
that big into the plot and atmosphere side of Thief, as odd as that might sound for a Thief fan.
I feel that Thief 2 plays better and is more mechanically interesting for the most part, and that makes me lean towards it over Thief 1. It doesn't have some of the exotic elements of Thief 1, but
what is there feels more refined and expanded upon.
Edited by Professor Paul1290 | {"url":"https://forums.thedarkmod.com/index.php?/topic/15708-interesting-chat-about-thief-1-and-2/","timestamp":"2024-11-02T15:54:51Z","content_type":"text/html","content_length":"245617","record_id":"<urn:uuid:d7ddf4ae-bd1f-42cb-885f-f3fa181dd758>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00609.warc.gz"} |
Phase Retrieval
October 20, 1999
2:50 pm - 4:00 pm
Halligan 111
Robert Gonsalves
, Professor and Chair, EECS Department, Tufts University
Phase retrieval implies extraction of the phase of a complex function f(x) from its modulus. There is no general solution to this problem, unless additional constraints on f(x) are given. Examples of
such constraints are analyticity, knowledge of the modulus of the Fourier transform of f(x), and compact support of its transform (which means that f(x) is bandlimited). The third constraint is
useful in optics, because only the modulus of f(x) can be measured by an optical detector and because f(x) is usually The third constraint is useful in optics, because only the modulus of f(x) can be
measured by an optical detector and because f (x) is usually bandlimited by the input aperture of the optics (the pupil). In this talk we show the basic math of phase retrieval, show some of the
methods to extract the phase, and show a real-life example of the phase retrieval method, namely the identification of the phase aberration in the Hubble Space Telescope. | {"url":"http://www.cs.tufts.edu/t/colloquia/current/?event=198","timestamp":"2024-11-09T10:05:05Z","content_type":"text/html","content_length":"2115","record_id":"<urn:uuid:39db353e-c7bb-46a7-8430-ea32005743ce>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00680.warc.gz"} |
Geometry & Topology Today
Intended Audience: Everyone, and especially teachers who want to show to their students a mathematician explaining research mathematics
In this episode we meet Moshe Cohen, a mathematician who studies ways to arrange planes in 4-dimensional space. The interview starts with an easier question that can be answered by any student in any
grade. Enjoy! Continue reading
Meet Mathematician Thomas Lam
Intended Audience: Everyone, and especially teachers who want to show to their students a mathematician explaining the motivation behind their own research.
In this episode we meet Thomas Lam, professor of mathematics at the University of Michigan, who studies electrical networks (among other topics). Thomas gives a great introduction to one of the main
problem in electrical networks as well as an application of electrical networks to medicine. Continue reading
Meet Mathematician Aaron Lauda
Intended Audience: Everyone, and especially teachers who want to show to their students a mathematician explaining the motivation behind their own research.
In this episode we meet Aaron Lauda, a mathematician from the University of Southern California, who shows us how to represent complicated expressions and equations using pictures. Enjoy! Continue | {"url":"https://vela-vick.com/category/scistate/gt_today/page/2","timestamp":"2024-11-14T15:03:35Z","content_type":"text/html","content_length":"44537","record_id":"<urn:uuid:b1dbdc89-af79-443f-817a-729802a3ecfc>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00828.warc.gz"} |
1: Values, Types, and Operators
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Below the surface of the machine, the program moves. Without effort, it expands and contracts. In great harmony, electrons scatter and regroup. The forms on the monitor are but ripples on the
water. The essence stays invisibly below.
Inside the computer’s world, there is only data. You can read data, modify data, create new data—but that which isn’t data cannot be mentioned. All this data is stored as long sequences of bits and
is thus fundamentally alike.
Bits are any kind of two-valued things, usually described as zeros and ones. Inside the computer, they take forms such as a high or low electrical charge, a strong or weak signal, or a shiny or dull
spot on the surface of a CD. Any piece of discrete information can be reduced to a sequence of zeros and ones and thus represented in bits.
For example, we can express the number 13 in bits. It works the same way as a decimal number, but instead of 10 different digits, you have only 2, and the weight of each increases by a factor of 2
from right to left. Here are the bits that make up the number 13, with the weights of the digits shown below them:
So that’s the binary number 00001101. Its non-zero digits stand for 8, 4, and 1, and add up to 13.
Imagine a sea of bits—an ocean of them. A typical modern computer has more than 30 billion bits in its volatile data storage (working memory). Nonvolatile storage (the hard disk or equivalent) tends
to have yet a few orders of magnitude more.
To be able to work with such quantities of bits without getting lost, we must separate them into chunks that represent pieces of information. In a JavaScript environment, those chunks are called
values. Though all values are made of bits, they play different roles. Every value has a type that determines its role. Some values are numbers, some values are pieces of text, some values are
functions, and so on.
To create a value, you must merely invoke its name. This is convenient. You don’t have to gather building material for your values or pay for them. You just call for one, and whoosh, you have it.
They are not really created from thin air, of course. Every value has to be stored somewhere, and if you want to use a gigantic amount of them at the same time, you might run out of memory.
Fortunately, this is a problem only if you need them all simultaneously. As soon as you no longer use a value, it will dissipate, leaving behind its bits to be recycled as building material for the
next generation of values.
This chapter introduces the atomic elements of JavaScript programs, that is, the simple value types and the operators that can act on such values.
Values of the number type are, unsurprisingly, numeric values. In a JavaScript program, they are written as follows:
Use that in a program, and it will cause the bit pattern for the number 13 to come into existence inside the computer’s memory.
JavaScript uses a fixed number of bits, 64 of them, to store a single number value. There are only so many patterns you can make with 64 bits, which means that the number of different numbers that
can be represented is limited. With N decimal digits, you can represent 10^N numbers. Similarly, given 64 binary digits, you can represent 2^64 different numbers, which is about 18 quintillion (an 18
with 18 zeros after it). That’s a lot.
Computer memory used to be much smaller, and people tended to use groups of 8 or 16 bits to represent their numbers. It was easy to accidentally overflow such small numbers—to end up with a number
that did not fit into the given number of bits. Today, even computers that fit in your pocket have plenty of memory, so you are free to use 64-bit chunks, and you need to worry about overflow only
when dealing with truly astronomical numbers.
Not all whole numbers less than 18 quintillion fit in a JavaScript number, though. Those bits also store negative numbers, so one bit indicates the sign of the number. A bigger issue is that nonwhole
numbers must also be represented. To do this, some of the bits are used to store the position of the decimal point. The actual maximum whole number that can be stored is more in the range of 9
quadrillion (15 zeros)—which is still pleasantly huge.
Fractional numbers are written by using a dot.
For very big or very small numbers, you may also use scientific notation by adding an e (for exponent), followed by the exponent of the number.
That is 2.998 × 10^8 = 299,800,000.
Calculations with whole numbers (also called integers) smaller than the aforementioned 9 quadrillion are guaranteed to always be precise. Unfortunately, calculations with fractional numbers are
generally not. Just as π (pi) cannot be precisely expressed by a finite number of decimal digits, many numbers lose some precision when only 64 bits are available to store them. This is a shame, but
it causes practical problems only in specific situations. The important thing is to be aware of it and treat fractional digital numbers as approximations, not as precise values.
The main thing to do with numbers is arithmetic. Arithmetic operations such as addition or multiplication take two number values and produce a new number from them. Here is what they look like in
100 + 4 * 11
The + and * symbols are called operators. The first stands for addition, and the second stands for multiplication. Putting an operator between two values will apply it to those values and produce a
new value.
But does the example mean “add 4 and 100, and multiply the result by 11,” or is the multiplication done before the adding? As you might have guessed, the multiplication happens first. But as in
mathematics, you can change this by wrapping the addition in parentheses.
(100 + 4) * 11
For subtraction, there is the - operator, and division can be done with the / operator.
When operators appear together without parentheses, the order in which they are applied is determined by the precedence of the operators. The example shows that multiplication comes before addition.
The / operator has the same precedence as *. Likewise for + and -. When multiple operators with the same precedence appear next to each other, as in 1 - 2 + 1, they are applied left to right: (1 - 2)
+ 1.
These rules of precedence are not something you should worry about. When in doubt, just add parentheses.
There is one more arithmetic operator, which you might not immediately recognize. The % symbol is used to represent the remainder operation. X % Y is the remainder of dividing X by Y. For example,
314 % 100 produces 14, and 144 % 12 gives 0. The remainder operator’s precedence is the same as that of multiplication and division. You’ll also often see this operator referred to as modulo.
Special numbers
There are three special values in JavaScript that are considered numbers but don’t behave like normal numbers.
The first two are Infinity and -Infinity, which represent the positive and negative infinities. Infinity - 1 is still Infinity, and so on. Don’t put too much trust in infinity-based computation,
though. It isn’t mathematically sound, and it will quickly lead to the next special number: NaN.
NaN stands for “not a number”, even though it is a value of the number type. You’ll get this result when you, for example, try to calculate 0 / 0 (zero divided by zero), Infinity - Infinity, or any
number of other numeric operations that don’t yield a meaningful result.
The next basic data type is the string. Strings are used to represent text. They are written by enclosing their content in quotes.
`Down on the sea`
"Lie on the ocean"
'Float on the ocean'
You can use single quotes, double quotes, or backticks to mark strings, as long as the quotes at the start and the end of the string match.
Almost anything can be put between quotes, and JavaScript will make a string value out of it. But a few characters are more difficult. You can imagine how putting quotes between quotes might be hard.
Newlines (the characters you get when you press enter) can be included without escaping only when the string is quoted with backticks (`).
To make it possible to include such characters in a string, the following notation is used: whenever a backslash (\) is found inside quoted text, it indicates that the character after it has a
special meaning. This is called escaping the character. A quote that is preceded by a backslash will not end the string but be part of it. When an n character occurs after a backslash, it is
interpreted as a newline. Similarly, a t after a backslash means a tab character. Take the following string:
"This is the first line\nAnd this is the second"
The actual text contained is this:
This is the first line
And this is the second
There are, of course, situations where you want a backslash in a string to be just a backslash, not a special code. If two backslashes follow each other, they will collapse together, and only one
will be left in the resulting string value. This is how the string “A newline character is written like "\n".” can be expressed:
"A newline character is written like \"\\n\"."
Strings, too, have to be modeled as a series of bits to be able to exist inside the computer. The way JavaScript does this is based on the Unicode standard. This standard assigns a number to
virtually every character you would ever need, including characters from Greek, Arabic, Japanese, Armenian, and so on. If we have a number for every character, a string can be described by a sequence
of numbers.
And that’s what JavaScript does. But there’s a complication: JavaScript’s representation uses 16 bits per string element, which can describe up to 2^16 different characters. But Unicode defines more
characters than that—about twice as many, at this point. So some characters, such as many emoji, take up two “character positions” in JavaScript strings. We’ll come back to this in Chapter 5.
Strings cannot be divided, multiplied, or subtracted, but the + operator can be used on them. It does not add, but it concatenates—it glues two strings together. The following line will produce the
string "concatenate":
"con" + "cat" + "e" + "nate"
String values have a number of associated functions (methods) that can be used to perform other operations on them. I’ll say more about these in Chapter 4.
Strings written with single or double quotes behave very much the same—the only difference is in which type of quote you need to escape inside of them. Backtick-quoted strings, usually called
template literals, can do a few more tricks. Apart from being able to span lines, they can also embed other values.
`half of 100 is ${100 / 2}`
When you write something inside ${} in a template literal, its result will be computed, converted to a string, and included at that position. The example produces “half of 100 is 50”.
Unary operators
Not all operators are symbols. Some are written as words. One example is the typeof operator, which produces a string value naming the type of the value you give it.
console.log(typeof 4.5)
// → number
console.log(typeof "x")
// → string
We will use console.log in example code to indicate that we want to see the result of evaluating something. More about that in the next chapter.
The other operators shown all operated on two values, but typeof takes only one. Operators that use two values are called binary operators, while those that take one are called unary operators. The
minus operator can be used both as a binary operator and as a unary operator.
console.log(- (10 - 2))
// → -8
Boolean values
It is often useful to have a value that distinguishes between only two possibilities, like “yes” and “no” or “on” and “off”. For this purpose, JavaScript has a Boolean type, which has just two
values, true and false, which are written as those words.
Here is one way to produce Boolean values:
console.log(3 > 2)
// → true
console.log(3 < 2)
// → false
The > and < signs are the traditional symbols for “is greater than” and “is less than”, respectively. They are binary operators. Applying them results in a Boolean value that indicates whether they
hold true in this case.
Strings can be compared in the same way.
console.log("Aardvark" < "Zoroaster")
// → true
The way strings are ordered is roughly alphabetic but not really what you’d expect to see in a dictionary: uppercase letters are always “less” than lowercase ones, so "Z" < "a", and nonalphabetic
characters (!, -, and so on) are also included in the ordering. When comparing strings, JavaScript goes over the characters from left to right, comparing the Unicode codes one by one.
Other similar operators are >= (greater than or equal to), <= (less than or equal to), == (equal to), and != (not equal to).
console.log("Itchy" != "Scratchy")
// → true
console.log("Apple" == "Orange")
// → false
There is only one value in JavaScript that is not equal to itself, and that is NaN (“not a number”).
console.log(NaN == NaN)
// → false
NaN is supposed to denote the result of a nonsensical computation, and as such, it isn’t equal to the result of any other nonsensical computations.
Logical operators
There are also some operations that can be applied to Boolean values themselves. JavaScript supports three logical operators: and, or, and not. These can be used to “reason” about Booleans.
The && operator represents logical and. It is a binary operator, and its result is true only if both the values given to it are true.
console.log(true && false)
// → false
console.log(true && true)
// → true
The || operator denotes logical or. It produces true if either of the values given to it is true.
console.log(false || true)
// → true
console.log(false || false)
// → false
Not is written as an exclamation mark (!). It is a unary operator that flips the value given to it—!true produces false, and !false gives true.
When mixing these Boolean operators with arithmetic and other operators, it is not always obvious when parentheses are needed. In practice, you can usually get by with knowing that of the operators
we have seen so far, || has the lowest precedence, then comes &&, then the comparison operators (>, ==, and so on), and then the rest. This order has been chosen such that, in typical expressions
like the following one, as few parentheses as possible are necessary:
1 + 1 == 2 && 10 * 10 > 50
The last logical operator I will discuss is not unary, not binary, but ternary, operating on three values. It is written with a question mark and a colon, like this:
console.log(true ? 1 : 2);
// → 1
console.log(false ? 1 : 2);
// → 2
This one is called the conditional operator (or sometimes just the ternary operator since it is the only such operator in the language). The value on the left of the question mark “picks” which of
the other two values will come out. When it is true, it chooses the middle value, and when it is false, it chooses the value on the right.
Empty values
There are two special values, written null and undefined, that are used to denote the absence of a meaningful value. They are themselves values, but they carry no information.
Many operations in the language that don’t produce a meaningful value (you’ll see some later) yield undefined simply because they have to yield some value.
The difference in meaning between undefined and null is an accident of JavaScript’s design, and it doesn’t matter most of the time. In cases where you actually have to concern yourself with these
values, I recommend treating them as mostly interchangeable.
Automatic type conversion
In the Introduction, I mentioned that JavaScript goes out of its way to accept almost any program you give it, even programs that do odd things. This is nicely demonstrated by the following
console.log(8 * null)
// → 0
console.log("5" - 1)
// → 4
console.log("5" + 1)
// → 51
console.log("five" * 2)
// → NaN
console.log(false == 0)
// → true
When an operator is applied to the “wrong” type of value, JavaScript will quietly convert that value to the type it needs, using a set of rules that often aren’t what you want or expect. This is
called type coercion. The null in the first expression becomes 0, and the "5" in the second expression becomes 5 (from string to number). Yet in the third expression, + tries string concatenation
before numeric addition, so the 1 is converted to "1" (from number to string).
When something that doesn’t map to a number in an obvious way (such as "five" or undefined) is converted to a number, you get the value NaN. Further arithmetic operations on NaN keep producing NaN,
so if you find yourself getting one of those in an unexpected place, look for accidental type conversions.
When comparing values of the same type using ==, the outcome is easy to predict: you should get true when both values are the same, except in the case of NaN. But when the types differ, JavaScript
uses a complicated and confusing set of rules to determine what to do. In most cases, it just tries to convert one of the values to the other value’s type. However, when null or undefined occurs on
either side of the operator, it produces true only if both sides are one of null or undefined.
console.log(null == undefined);
// → true
console.log(null == 0);
// → false
That behavior is often useful. When you want to test whether a value has a real value instead of null or undefined, you can compare it to null with the == (or !=) operator.
But what if you want to test whether something refers to the precise value false? Expressions like 0 == false and "" == false are also true because of automatic type conversion. When you do not want
any type conversions to happen, there are two additional operators: === and !==. The first tests whether a value is precisely equal to the other, and the second tests whether it is not precisely
equal. So "" === false is false as expected.
I recommend using the three-character comparison operators defensively to prevent unexpected type conversions from tripping you up. But when you’re certain the types on both sides will be the same,
there is no problem with using the shorter operators.
Short-circuiting of logical operators
The logical operators && and || handle values of different types in a peculiar way. They will convert the value on their left side to Boolean type in order to decide what to do, but depending on the
operator and the result of that conversion, they will return either the original left-hand value or the right-hand value.
The || operator, for example, will return the value to its left when that can be converted to true and will return the value on its right otherwise. This has the expected effect when the values are
Boolean and does something analogous for values of other types.
console.log(null || "user")
// → user
console.log("Agnes" || "user")
// → Agnes
We can use this functionality as a way to fall back on a default value. If you have a value that might be empty, you can put || after it with a replacement value. If the initial value can be
converted to false, you’ll get the replacement instead. The rules for converting strings and numbers to Boolean values state that 0, NaN, and the empty string ("") count as false, while all the other
values count as true. So 0 || -1 produces -1, and "" || "!?" yields "!?".
The && operator works similarly but the other way around. When the value to its left is something that converts to false, it returns that value, and otherwise it returns the value on its right.
Another important property of these two operators is that the part to their right is evaluated only when necessary. In the case of true || X, no matter what X is—even if it’s a piece of program that
does something terrible—the result will be true, and X is never evaluated. The same goes for false && X, which is false and will ignore X. This is called short-circuit evaluation.
The conditional operator works in a similar way. Of the second and third values, only the one that is selected is evaluated.
We looked at four types of JavaScript values in this chapter: numbers, strings, Booleans, and undefined values.
Such values are created by typing in their name (true, null) or value (13, "abc"). You can combine and transform values with operators. We saw binary operators for arithmetic (+, -, *, /, and %),
string concatenation (+), comparison (==, !=, ===, !==, <, >, <=, >=), and logic (&&, ||), as well as several unary operators (- to negate a number, ! to negate logically, and typeof to find a
value’s type) and a ternary operator (?:) to pick one of two values based on a third value.
This gives you enough information to use JavaScript as a pocket calculator but not much more. The next chapter will start tying these expressions together into basic programs. | {"url":"https://eng.libretexts.org/Bookshelves/Computer_Science/Programming_Languages/Eloquent_JavaScript_(Haverbeke)/Part_1%3A_Language/01%3A_Values_Types_and_Operators","timestamp":"2024-11-05T01:05:27Z","content_type":"text/html","content_length":"151548","record_id":"<urn:uuid:11bf5bae-80cc-4a5e-a152-40f5d26906a8>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00030.warc.gz"} |
Centralized, Parallel, and Distributed Multi-Source Shortest Paths via Hopsets and Rectangular Matrix Multiplication
Consider an undirected weighted graph G = (V, E, ω. We study the problem of computing (1 + ϵ)approximate shortest paths for S × V , for a subset S ⊆ V of |S| = n^r sources, for some 0 < r ≤ 1. We
devise a significantly improved algorithm for this problem in the entire range of parameter r, in both the classical centralized and the parallel (PRAM) models of computation, and in a wide range of
r in the distributed (Congested Clique) model. Specifically, our centralized algorithm for this problem requires time Õ(|E| · n^o(1) + n^ω(r)), where n^ω(r) is the time required to multiply an n^r ×
n matrix by an n × n one. Our PRAM algorithm has polylogarithmic time (log n)^O(1/ρ), and its work complexity is Õ(|E| · n^ρ + n^ω(r)), for any arbitrarily small constant ρ > 0. In particular, for r
≤ 0.313 . . ., our centralized algorithm computes S × V (1 + ϵ)-approximate shortest paths in n^2+o^(1) time. Our PRAM polylogarithmic-time algorithm has work complexity O(|E| · n^ρ + n^2+o^(1)), for
any arbitrarily small constant ρ > 0. Previously existing solutions either require centralized time/parallel work of O(|E| · |S|) or provide much weaker approximation guarantees. In the Congested
Clique model, our algorithm solves the problem in polylogarithmic time for |S| = n^r sources, for r ≤ 0.655, while previous state-of-the-art algorithms did so only for r ≤ 1/2. Moreover, it improves
previous bounds for all r > 1/2. For unweighted graphs, the running time is improved further to poly(log log n) for r ≤ 0.655. Previously this running time was known for r ≤ 1/2.
Publication series
Name Leibniz International Proceedings in Informatics, LIPIcs
Volume 219
ISSN (Print) 1868-8969
Conference 39th International Symposium on Theoretical Aspects of Computer Science, STACS 2022
Country/Territory France
City Virtual, Marseille
Period 15/05/22 → 18/05/22
• Hopsets
• Matrix multiplication
• Shortest paths
ASJC Scopus subject areas
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Hypothesis Test for real problems
Hypothesis tests are significant for evaluating answers to questions concerning samples of data.
A statistical hypothesis is a belief made about a population parameter. This belief may or might not be right. In other words, hypothesis testing is a proper technique utilized by scientist to
support or reject statistical hypotheses. The foremost ideal approach to decide if a statistical hypothesis is correct is examine the whole population.
Since that’s frequently impractical, we normally take a random sample from the population and inspect the equivalent. Within the event sample data set isn’t steady with the statistical hypothesis,
the hypothesis is refused.
Types of hypothesis:
There are two sort of hypothesis and both the Null Hypothesis (Ho) and Alternative Hypothesis (Ha) must be totally mutually exclusive events.
• Null hypothesis is usually the hypothesis that the event wont’t happen.
• Alternative hypothesis is a hypothesis that the event will happen.
Why we need Hypothesis Testing?
Suppose a specific cosmetic producing company needs to launch a new Shampoo in the market. For this situation they will follow Hypothesis Testing all together decide the success of new product in the
Where likelihood of product being ineffective in market is undertaken as Null Hypothesis and likelihood of product being profitable is undertaken as Alternative Hypothesis. By following the process
of Hypothesis testing they will foresee the accomplishment.
How to Calculate Hypothesis Testing?
• State the two theories with the goal that just one can be correct, to such an extent that the two occasions are totally unrelated.
• Now figure a study plan, that will lay out how the data will be assessed.
• Now complete the plan and genuinely investigate the sample dataset.
• Finally examine the outcome and either accept or reject the null hypothesis.
Another example
Assume, Person have gone after a typing job and he has expressed in the resume that his composing speed is 70 words per minute. The recruiter might need to test his case. On the off chance that he
sees his case as adequate, he will enlist him in any case reject him. Thus, he types an example letter and found that his speed is 63 words a minute. Presently, he can settle on whether to employ him
or not. In the event that he meets all other qualification measures. This procedure delineates Hypothesis Testing in layman’s terms.
In statistical terms Hypothesis his typing speed is 70 words per minute is a hypothesis to be tested so-called null hypothesis. Clearly, the alternating hypothesis his composing speed isn’t 70 words
per minute.
So, normal composing speed is population parameter and sample composing speed is sample statistics.
The conditions of accepting or rejecting his case is to be chosen by the selection representative. For instance, he may conclude that an error of 6 words is alright to him so he would acknowledge his
claim between 64 to 76 words per minute. All things considered, sample speed 63 words per minute will close to reject his case. Furthermore, the choice will be he was producing a fake claim.
In any case, if the selection representative stretches out his acceptance region to positive/negative 7 words that is 63 to 77 words, he would be tolerating his case.
In this way, to finish up, Hypothesis Testing is a procedure to test claims about the population dependent on sample. It is a fascinating reasonable subject with a quite statistical jargon. You have
to dive more to get familiar with the details.
Significance Level and Rejection Region for Hypothesis
Type I error probability is normally indicated by α and generally set to 0.05. The value of α is recognized as the significance level.
The rejection region is the set of sample data that prompts the rejection of the null hypothesis. The significance level, α, decides the size of the rejection region. Sample results in the rejection
region are labelled statistically significant at level of α .
The impact of differing α is that If α is small, for example, 0.01, the likelihood of a type I error is little, and a ton of sample evidence for the alternative hypothesis is needed before the null
hypothesis can be dismissed. Though, when α is bigger, for example, 0.10, the rejection region is bigger, and it is simpler to dismiss the null hypothesis.
Significance from p-values
A subsequent methodology is to evade the utilization of a significance level and rather just report how significant the sample evidence is. This methodology is as of now more widespread. It is
accomplished by method of a p value. P value is gauge of power of the evidence against null hypothesis. It is the likelihood of getting the observed value of test statistic, or value with
significantly more prominent proof against null hypothesis (Ho), if the null hypothesis of an investigation question is true. The less significant the p value, the more proof there is supportive of
the alternative hypothesis. Sample evidence is measurably noteworthy at the α level just if the p value is less than α. They have an association for two tail tests. When utilizing a confidence
interval to playout a two-tailed hypothesis test, reject the null hypothesis if and just if the hypothesized value doesn’t lie inside a confidence interval for the parameter.
Hypothesis Tests and Confidence Intervals
Hypothesis tests and confidence intervals are cut out of the same cloth. An event whose 95% confidence interval reject the hypothesis is an event for which p<0.05 under the relating hypothesis test,
and the other way around. A p value is letting you know the greatest confidence interval that despite everything prohibits the hypothesis. As such, if p<0.03 against the null hypothesis, that implies
that a 97% confidence interval does exclude the null hypothesis.
Hypothesis Tests for a Population Mean
We do a t test on the ground that the population mean is unknown. The general purpose is to contrast sample mean with some hypothetical population mean, to assess whether the watched the truth is
such a great amount of unique in relation to the hypothesis that we can say with assurance that the hypothetical population mean isn’t, indeed, the real population mean.
Hypothesis Tests for a Population Proportion
At the point when you have two unique populations Z test facilitates you to choose if the proportion of certain features is the equivalent or not in the two populations. For instance, if the male
proportion is equivalent between two nations.
Hypothesis Test for Equal Population Variances
F Test depends on F distribution and is utilized to think about the variance of the two impartial samples. This is additionally utilized with regards to investigation of variance for making a
decision about the significance of more than two sample.
T test and F test are totally two unique things. T test is utilized to evaluate the population parameter, for example, population mean, and is likewise utilized for hypothesis testing for population
mean. However, it must be utilized when we don’t know about population standard deviation. On the off chance that we know the population standard deviation, we will utilize Z test. We can likewise
utilize T statistic to approximate population mean. T statistic is likewise utilised for discovering the distinction in two population mean with the assistance of sample means.
Z statistic or T statistic is utilized to assess population parameters such as population mean and population proportion. It is likewise used for testing hypothesis for population mean and population
proportion. In contrast to Z statistic or T statistic, where we manage mean and proportion, Chi Square or F test is utilized for seeing if there is any variance inside the samples. F test is the
proportion of fluctuation of two samples.
Hypothesis encourages us to make coherent determinations, the connection among variables, and gives the course to additionally investigate. Hypothesis for the most part results from speculation
concerning studied behaviour, natural phenomenon, or proven theory. An honest hypothesis ought to be clear, detailed, and reliable with the data. In the wake of building up the hypothesis, the
following stage is validating or testing the hypothesis. Testing of hypothesis includes the process that empowers to concur or differ with the expressed hypothesis.
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Management of Potato Late Blight: Simulation with Lateblight
Lateblight Exercises
Answer Sheet for Instructors
Exercise 1. Disease resistance
The uncontrolled epidemic on the potato variety with low resistance results in complete defoliation that terminates the season a few days earlier than normal and yields a net loss of profit. The
following is the economic report found by clicking on Show Report in the Economics menu:
Season ended on September 23 with 100.0% blighted foliage.
Your crop is 27016.49 kg./ha.
At the current market price of 0.11/kg this would bring 2971.81/hectare
31.89% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -6623 -728.48
Yield Losses due to Tuber Blight: -8616 -947.81
Net Yield: 18400 2024.01
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: -692.77/hectare
A moderate level of resistance slows the epidemic somewhat but by itself does not control late blight well enough to produce a profitable yield:
Season ended on September 28 with 98.53% blighted foliage.
Your crop is 28391.29 kg./ha.
At the current market price of 0.11/kg this would bring 3123.04/hectare
27.45% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -5248 -577.25
Yield Losses due to Tuber Blight: -7795 -857.41
Net Yield: 20597 2265.64
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: -451.15/hectare
The potato variety with high resistance to late blight produces a profitable yield without any additional disease management measures, but, of course, it runs the risk of selecting a virulent
population of Phytophthora infestans:
Season ended on September 28 with 37.18% blighted foliage.
Your crop is 32460.98 kg./ha.
At the current market price of 0.11/kg this would bring 3570.71/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -1178 -129.58
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 32461 3570.71
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: 853.93/hectare
Exercise 2. Protectant Fungicides
Applying a protectant fungicide on a weekly schedule very effectively controls the late blight epidemic. Instead of a net loss, there is a substantial net profit, even higher than that with the
highly resistant variety:
Season ended on September 28 with 3.77% blighted foliage.
Your crop is 33518.82 kg./ha.
At the current market price of 0.11/kg this would bring 3687.07/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -120 -13.22
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 33519 3687.07
Sprays Fungicide (kg)
Protectant: 8 10.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 35.20
Application Costs: 79.04
Net Profit: 856.05/hectare
Exercise 3. Systemic Fungicides
The cost per kg of the systemic was more than 8 times that of protectant, but because the dose of the systemic was very much lower and only two sprays were applied (versus eight for the protectant)
the total cost for the systemic sprays was 32.85 as compared to 114.24 for the protectant spray program. Two sprays of the systemic fungicide produced nearly as effective control of the epidemic as
eight sprays of the protectant:
Season ended on September 28 with 15.69% blighted foliage.
Your crop is 33245.78 kg./ha.
At the current market price of 0.11/kg this would bring 3657.04/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -393 -43.25
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 33246 3657.04
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 2 0.44
Fixed Costs: 2716.78
Spray Costs: 13.09
Application Costs: 19.76
Net Profit: 907.41/hectare
The use of the systemic was more profitable than the use of the protectant fungicide, largely because of the reduced application cost. However, there is a risk associated with the systemic. It is far
more likely than the protectant fungicide to select a fungicide resistant population of Phytophthora infestans.
Exercise 4. Effects of weather
Under the hot, dry conditions, there was still a severe late blight epidemic with no fungicide applied, but it was substantially less severe than the epidemic under cool, wet conditions, and it even
yielded a modest profit:
Season ended on September 28 with 70.7% blighted foliage.
Your crop is 31126.01 kg./ha.
At the current market price of 0.11/kg this would bring 3423.86/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -2513 -276.43
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 31126 3423.86
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: 707.08/hectare
Applying 8 protectant sprays at 7-day intervals during hot, dry weather provides very effective control:
Season ended on September 28 with 1.21% blighted foliage.
Your crop is 33602.31 kg./ha.
At the current market price of 0.11/kg this would bring 3696.25/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -37 -4.04
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 33602 3696.25
Sprays Fungicide (kg)
Protectant: 8 10.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 35.20
Application Costs: 79.04
Net Profit: 865.23
Reducing the number of protectant sprays to a total of 6 at 10-day intervals reduces the spray cost and improves the profitability during hot, dry weather:
Season ended on September 28 with 5.26% blighted foliage.
Your crop is 33491.51 kg./ha.
At the current market price of 0.11/kg this would bring 3684.07/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -147 -16.22
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 33492 3684.07
Sprays Fungicide (kg)
Protectant: 6 7.5
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 26.40
Application Costs: 59.28
Net Profit: 881.61
Two systemic sprays during the hot, dry season also gives very effective control of late blight:
Season ended on September 28 with 1.86% blighted foliage.
Your crop is 33580.38 kg./ha.
At the current market price of 0.11/kg this would bring 3693.84/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -59 -6.45
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 33580 3693.84
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 2 0.44
Fixed Costs: 2716.78
Spray Costs: 13.09
Application Costs: 19.76
Net Profit: 944.21
During the hot, dry season, a single spray of the systemic fungicide also provided good late blight control, but its profitability was slightly less than that with two sprays:
Season ended on September 28 with 16.6% blighted foliage.
Your crop is 33236.86 kg./ha.
At the current market price of 0.11/kg this would bring 3656.05/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -402 -44.24
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 33237 3656.05
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 1 0.22
Fixed Costs: 2716.78
Spray Costs: 6.54
Application Costs: 9.88
Net Profit: 922.85/hectare
The two systemic sprays during hot, dry weather yielded a higher profit than they did during cool, wet weather (944.21/hectare versus 907.41/hectare).
Exercise 5. Disease thresholds
Under these conditions, late blight is first detectable on August 8. Starting the protectant sprays on August 9 and repeating on August 14, 19, 24, and 29 and September 3, 8, and 13, for a total of 8
sprays, we get the following results:
Season ended on September 28 with 83.23% blighted foliage.
Your crop is 27253.02 kg./ha.
At the current market price of 0.11/kg this would bring 2997.83/hectare
26.62% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -6386 -702.46
Yield Losses due to Tuber Blight: -7254 -797.93
Net Yield: 19999 2199.90
Sprays Fungicide (kg)
Protectant: 8 10.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 35.20
Application Costs: 79.04
Net Profit: -631.12/hectare
Repeating this with the systemic fungicide, applied on August 9, 17, and 24, we get:
Season ended on September 28 with 71.3% blighted foliage.
Your crop is 30248.5 kg./ha.
At the current market price of 0.11/kg this would bring 3327.34/hectare
16.28% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -3390 -372.95
Yield Losses due to Tuber Blight: -4925 -541.79
Net Yield: 25323 2785.54
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 3 0.66
Fixed Costs: 2716.78
Spray Costs: 19.63
Application Costs: 29.64
Net Profit: 19.49/hectare
It is clear that although it appears to start late, late blight is a rapidly developing epidemic, and the use of a threshold to start the spray program is not practical.
Exercise 6. Sanitation
Moving the cull pile to 100 meters from the field resulted in a significant delay in the development of the epidemic and reduced losses nearly as much as applying a protectant fungicide on a weekly
schedule at a cost of over $100 per hectare:
Season ended on September 28 with 51.88% blighted foliage.
Your crop is 32124.2 kg./ha.
At the current market price of 0.11/kg this would bring 3533.66/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -1515 -166.63
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 32124 3533.66
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: 816.88/hectare
Adding only 50 volunteer potato plants per hectare increases the level of initial inoculum and substantially advances the onset of the epidemic, resulting in a higher level of disease at the end of
the season:
Season ended on September 28 with 99.75% blighted foliage.
Your crop is 26896.93 kg./ha.
At the current market price of 0.11/kg this would bring 2958.66/hectare
31.03% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -6742 -741.63
Yield Losses due to Tuber Blight: -8346 -918.01
Net Yield: 18551 2040.65
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: -676.13/hectare
Removing these volunteers would give you the outcome that we saw in the previous run, which we said was roughly equivalent to applying 8 sprays of protectant fungicide at at a cost of about $100/
Exercise 7. Certified Seed
With uncertified seed, the epidemic develops early in the absence of fungicides:
Season ended on September 25 with 100.0% blighted foliage.
Your crop is 27394.25 kg./ha.
At the current market price of 0.11/kg this would bring 3013.37/hectare
31.39% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -6245 -686.92
Yield Losses due to Tuber Blight: -8600 -945.98
Net Yield: 18794 2067.38
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: -649.40/hectare
With certified seed, the epidemic is sufficiently delayed that you get a respectable yield, even without the use of fungicides:
Season ended on September 28 with 37.05% blighted foliage.
Your crop is 32674.81 kg./ha.
At the current market price of 0.11/kg this would bring 3594.23/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -964 -106.06
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 32675 3594.23
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: 877.45/hectare
If certified seed added $313/hectare to the production costs, the net profit without the use of fungicides would be $564.45/hectare. Of course, we would also want to apply fungicides, and using
certified seed would allow a reduction in the fungicide cost, which we will see in the next exercise.
Exercise 8. Integrated Tactics
1. Uncontrolled epidemic
Season ended on September 21 with 100.0% blighted foliage.
Your crop is 25656.87 kg./ha.
At the current market price of 0.11/kg this would bring 2822.26/hectare
34.51% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -7982 -878.03
Yield Losses due to Tuber Blight: -8854 -973.92
Net Yield: 16803 1848.34
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 0 0.0
Fixed Costs: 2716.78
Spray Costs: 0.00
Application Costs: 0.00
Net Profit: -868.44
This clearly is a "worst case" scenario, and no grower would let this happen.
2. Fungicide spray (a single systemic fungicide application on July 13)
Season ended on September 28 with 86.47% blighted foliage.
Your crop is 30161.72 kg./ha.
At the current market price of 0.11/kg this would bring 3317.79/hectare
14.08% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -3477 -382.50
Yield Losses due to Tuber Blight: -4245 -467.00
Net Yield: 25916 2850.79
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 1 0.22
Fixed Costs: 2716.78
Spray Costs: 6.54
Application Costs: 9.88
Net Profit: 117.59/hectare
3. Partial resistance (a moderate level of late blight resistance plus a single systemic spray)
Season ended on September 28 with 55.32% blighted foliage.
Your crop is 31876.21 kg./ha.
At the current market price of 0.11/kg this would bring 3506.38/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -1763 -193.91
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 31876 3506.38
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 1 0.22
Fixed Costs: 2716.78
Spray Costs: 6.54
Application Costs: 9.88
Net Profit: 773.18/hectare
4. Moving cull pile (added to the above tactics, moving the cull pile from 10 meters away from the field to 100 meters away from the field)
Season ended on September 28 with 34.7% blighted foliage.
Your crop is 32680.84 kg./ha.
At the current market price of 0.11/kg this would bring 3594.89/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -958 -105.40
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 32681 3594.89
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 1 0.22
Fixed Costs: 2716.78
Spray Costs: 6.54
Application Costs: 9.88
Net Profit: 861.69/hectare
5. Removal of volunteers (added to all of the above tactics, removal of the volunteer potatoes prior to planting)
Season ended on September 28 with 26.16% blighted foliage.
Your crop is 32953.81 kg./ha.
At the current market price of 0.11/kg this would bring 3624.92/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -685 -75.37
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 32954 3624.92
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 1 0.22
Fixed Costs: 2716.78
Spray Costs: 6.54
Application Costs: 9.88
Net Profit: 891.72
Comparing the net profit to that of the previous run, where the volunteers were not removed, the marginal return is about $30/hectare, so we could invest up to that amount in the removal of
6. Certified seed (added to the above tactics, using certified seed)
The complete program of integrated tactics clearly gives the best disease control, with only about 1% of the foliage blighted at the end of the season.
Season ended on September 28 with 0.91% blighted foliage.
Your crop is 33618.34 kg./ha.
At the current market price of 0.11/kg this would bring 3698.02/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -21 -2.27
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 33618 3698.02
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 1 0.22
Fixed Costs: 2716.78
Spray Costs: 6.54
Application Costs: 9.88
Net Profit: 964.82
However, if you subtract the $313/hectare cost of the certified seed, the net profit comes out to be $651.82/hectare. Under these circumstances, the marginal return on the investment in certified
seed is less than the marginal cost and is thus not economically justified for the control of late blight alone. But late blight control is only a small side-benefit of certified seed; its primary
purpose is the control of potato viruses. The certified seed cost should really be added to the "fixed costs" with respect to late blight control and should have been included in the profit
calculations of all of the exercises.
Applying 4 systemic sprays at 14-day intervals beginning on June 29 will give you nearly the same level of disease control as the entire integrated program, but the profits are lower because of the
increased fungicide and application costs. The main reason, however, for not using 4 systemic sprays is the risk of selecting a fungicide-resistant Phytophthora infestans.
Season ended on September 28 with 1.1% blighted foliage.
Your crop is 33596.75 kg./ha.
At the current market price of 0.11/kg this would bring 3695.64/hectare
0.0% of your tubers are blighted.
(kg) Revenue
Total Potential Yield: 33639 3700.29
Yield Losses due to Plant Blight: -42 -4.65
Yield Losses due to Tuber Blight: -0 -0.00
Net Yield: 33597 3695.64
Sprays Fungicide (kg)
Protectant: 0 0.0
Systemic: 4 0.88
Fixed Costs: 2716.78
Spray Costs: 26.17
Application Costs: 39.52
Net Profit: 913.17/hectare | {"url":"https://www.apsnet.org/edcenter/disimpactmngmnt/simulations/Pages/ManagementofPotatoLateBlight.aspx","timestamp":"2024-11-10T18:50:33Z","content_type":"text/html","content_length":"129465","record_id":"<urn:uuid:78fd3961-c71c-44a4-908e-5cdc22c73871>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00191.warc.gz"} |
Toward an algebraic theory of systems for Theoretical Computer Science
Theoretical Computer Science
Toward an algebraic theory of systems
View publication
We propose the concept of a system algebra with a parallel composition operation and an interface connection operation, and formalize composition-order invariance, which postulates that the order of
composing and connecting systems is irrelevant, a generalized form of associativity. Composition-order invariance explicitly captures a common property that is implicit in any context where one can
draw a figure (hiding the drawing order) of several connected systems, which appears in many scientific contexts. This abstract algebra captures settings where one is interested in the behavior of a
composed system in an environment and wants to abstract away anything internal not relevant for the behavior. This may include physical systems, electronic circuits, or interacting distributed
systems. One specific such setting, of special interest in computer science, are functional system algebras, which capture, in the most general sense, any type of system that takes inputs and
produces outputs depending on the inputs, and where the output of a system can be the input to another system. The behavior of such a system is uniquely determined by the function mapping inputs to
outputs. We consider several instantiations of this very general concept. In particular, we show that Kahn networks form a functional system algebra and prove their composition-order invariance.
Moreover, we define a functional system algebra of causal systems, characterized by the property that inputs can only influence future outputs, where an abstract partial order relation captures the
notion of “later”. This system algebra is also shown to be composition-order invariant and appropriate instantiations thereof allow to model and analyze systems that depend on time. | {"url":"https://research.ibm.com/publications/toward-an-algebraic-theory-of-systems","timestamp":"2024-11-11T18:22:17Z","content_type":"text/html","content_length":"70597","record_id":"<urn:uuid:bc15d79d-9b64-4814-99d1-016ba35fddb0>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00860.warc.gz"} |
Is 15 hundredths the same as 1500?
Is 15 hundredths the same as 1500?
15 hundredths is the same as 1,500. 16 hundredths is the same as 0.16.
Why do some people say hundred instead of thousand?
It states that the word derives from the Old Norse, where there were two variants, one indicating 100 and the other 120. Although this latter seems to have been employed relatively rarely in English,
it illustrates that ‘a hundred’ was seen as an entity in itself rather than a subdivision of ‘a thousand’.
What does it mean 15 hundred?
The start of the sixteenth hour of the day on the 24-hour clock i.e. 15:00. numeral.
How do you say $1500?
One thousand five hundred is the most formal way. One and a half thousand is also correct.
What is the difference between hundredths and hundreds?
A hundredth is the reciprocal of 100. A hundredth is written as a decimal fraction as 0.01, and as a vulgar fraction as 1/100. “Hundredth” is also the ordinal number that follows “ninety-ninth” and
precedes “hundred and first.” It is written as 100th.
What does 15 thousandths look like?
fifteen can be write 15 or 15.0 or 15., if you move the decimal to the left (one, ten,hundred,thousand…etc.), you’ll get . 0015–which is 15/1000. You get one extra zero toward the left to make it as
fifteen thousandths.
Can you say fifteen hundred?
Because there are fifteen hundreds: 15 00, people say ‘fifteen hundred’. Because it is one thousand and 500: 1 500, people say ‘one thousand five hundred’ ( no hyphen).
How much is fifteen thousand?
Fifteen Thousand in numerals is written as 15000.
How do you say 2500?
2500 in words is written as Two Thousand Five Hundred.
What does fifteen thousand look like in numbers?
What does the hundreds place mean?
: the place three to the left of the decimal point in a number expressed in the Arabic system of notation.
Why do people say one thousand five hundred instead of one thousand?
Because there are fifteen hundreds: 15 00, people say ‘fifteen hundred’. Because it is one thousand and 500: 1 500, people say ‘one thousand five hundred’ ( no hyphen). In the 24-hour clock, nobody
ever says, “I’ll meet you at one thousand five hundred.”
How do you say the number 1500 in different ways?
In English, you can say this number (1500) in quite a few different ways: – One thousand five hundred. – Fifteen hundred. – One thousand and five hundred. – One and a half thousand.
What is the difference between 1500 and 15 thousand?
So 15000 can be said 15 thousand just fine because the 1 and the 5 are both in the thousands. But 1500 technically isnt 15 hundred because there is only one hundreds digit and that is the 5. The 1 is
in the thousands.
What is the fifteen hundred case of a phone number?
The “fifteen hundred case” is a variation of a typical convention for three-digit numbers: 294 would be said “two ninety-four” so you can say 3493 as “thirty-four ninety-three.” For the most part, I
think this is restricted to years and numbers-as-text (i.e., addresses, phone numbers) for numbers greater than 1999. | {"url":"https://profound-information.com/is-15-hundredths-the-same-as-1500/","timestamp":"2024-11-12T11:57:06Z","content_type":"text/html","content_length":"59392","record_id":"<urn:uuid:fd63b72f-4f2f-4bc3-910b-77c756ca38ff>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00753.warc.gz"} |
(PDF) Aspects of aperiodicity and randomness in theoretical physics
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arXiv:2003.07282v1 [math-ph] 16 Mar 2020
Aspects of aperiodicity and randomness in
theoretical physics
Leonardo Ort´ız∗
, Marcelo Amaral and Klee Irwin
Quantum Gravity Research, Los Angeles, CA, U.S.A
March 17, 2020
In this work we explore how the heat kernel, which gives the solu-
tion to the diffusion equation and the Brownian motion, would change
when we introduce quasiperiodicity in the scenario. We also study the
random walk in the Fibonacci sequence. We discuss how these ideas
would change the discrete approaches to quantum gravity and the
construction of quantum geometry.
Keywords: Heat kernel, Fibonacci sequence, quantum gravity, aperiodic.
1 Introduction
In this paper we discuss the notion of quantum geometry [1] using tools from
quasicrystals and quasiperiodic functions [2] as an alternative to canonical
quantization presented in [3], mainly the modifications of quantities of inter-
est, such as the heat kernel and entropy, due to the introduction of quasiperi-
odicity in the setting. Although these modifications seem to be mild, they
have the potential to be important in derivations of thermodynamical quan-
tities which we plan to develop in the future.
In [1] a kind of quantum geometry is constructed in one and two di-
mensions. This is done using random walks and random two-dimensional
surfaces. This is done with the idea of obtaining a quantum description of
gravity at least in two dimensions. This is limited in several aspects with the
main limitation that a realistic theory of gravity should be four-dimensional.
The main idea of the present work is to make something similar to [1] but
not only with random walks and random two-dimensional surfaces but also
with quasiperiodic trajectories as described in [2]. In this manuscript we will
consider one and two dimensional quasiperiodic trajectories but one of our
goals is to make it with quasiperiodic trajectories inherited from quasicrys-
tals in several dimensions. Also once we have this under control we will try
to not only have nonperiodicity but also stochasticity. In the near future
we will try to do what is done with the Polyakov action in [4] but this time
with the Einstein-Hilbert action. Also instead of letting the lattice size go
to zero we will probe the geometry with a massive particle so we can feel the
granular structure of spacetime.
In this context the quasiperiodicity will be given by the trajectory of the
quasiparticle. It is important to note that in this step we can introduce
nonperidicity and stochasticity.
In the scenario we are describing here we would like to obtain the analo-
gous to the Einstein field equations as something emergent in the same spirit
as thermodynamics is obtained from the microscopic statistical mechanics.
This manuscript is organized as follows: In section 2 we describe briefly
the idea of quantum geometry. In section 3 we introduce the necessary math-
ematics from quasicrystals for our purposes. In section 4 we describe our
approach to the build up of quantum geometry with a discussion of the heat
kernel. In section 5 we study the random walk in a Fibonacci chain in one
and its generalization to two dimensions. Finally in section 6 we discuss fur-
ther ideas and our final comments. In the appendix A we discuss the relation
of the Hamiltonian in usual quantum theory and in the Euclidean setting,
since this ideas are related to the main work.
2 A first look at quantum geometry
The idea of quantum geometry can be very intuitive. The concept of cur-
vature of a manifold is studied in semiriemannian geometry. In this context
the manifold -the spacetime- is smooth, however if nature is quantum at the
most fundamental level then the smooth spacetime should be quantized-the
notion of quantum geometry.
Let us describe the idea of spacetime a little more deeply. In general rela-
tivity (GR) the spacetime is made of events, these events can be in principle
anything: the explosion of a bomb, the hand shake of two friends, the click
one do on the mouse, etc. However if we think careful on this definition of
spacetime one realizes that something estrange happens if we want to de-
scribe events with quantum systems as for example with a transition of one
level of energy to another in the hydrogen atom1. Clearly this happens be-
cause the idea of spacetime described in standard GR books is classical. But
then we face a conceptual problem similar to the one of the measurement
problem in quantum mechanics, as the question where lies the boundary
between the machine which measures and the system under study. In the
proposal of the Quantum Gravity Research group -called emergence theory-
this problem does not arises because the spacetime would be constructed
from the “trajectory” of the phason quasiparticles. The challenge then is to
obtain in certain limit something analogous to the Einstein field equations.
Just to put things on perspective we are aiming to construct something
G(γ) = Zγ
where σis an hypersurface, γis its boundary and Sis the action of the
system. Most of the actions constructed so far are geometric, but since we
are constructing a theory more general than the ones we have at the moment
we will not attach from the beginning to geometric actions. Clearly the
two challenges in this aim are the construction of the measure Dσ and the
action S(σ). We are working with a kind of Euclidean action, which is not a
1Similar ideas are consider in [5]
limitation because we want to have spacetime emergent in our model . Also
it is worthwhile to mention that from G(γ) we expect to obtain a kind of
generalized partition function.
3 Some mathematical tools from quasicrys-
A quasicrystal is an object that has order but not periodicity. The math-
ematics to study these object is very rich and very well developed, see for
instance [2], [6] and [7] just to mention a few references.
In the study of quasicrystals, quasiperiodic functions are relevant. The
idea of this work is to construct a spacetime foam-like model. First in one
dimension with quasiperiodic functions, as the ones shown in [2]. Later we
will introduce stochasticity too. So our quantum geometry will be the result
of nonperiodicity and stochasticity.
4 First steps in the construction of Quantum
The main idea of this section is to replace the stochastic process such a
random walk used for example in [8] by a quasiperiodic random process
described by a quasiparticle. However as a first step in this direction we will
study the quasiperiodic process described by a quasiperiodic function as the
one given in [2].
4.1 The random walk representation of the heat kernel
and quasiperiodicity
In order to have an idea on how to implement quasiperiodicity in the models
of quantum geometry let us study some aspects of the random walk repre-
sentation of the heat kernel associated with the diffusion equation when we
introduce quasiperiodicity in the model. This discussion rather than new is
pedagogical, for more details see [1].
Let ∆ denote the Laplace operator in Rd. The solution to the difussion
(or heat) equation in Rd
∂t =1
2∆ϕ, (2)
with the initial condition ϕ(x, 0) = ϕ0(x) is given by
ϕ(y, t) = 1
The function ϕ0(x) is interpreted as the initial distribution of particles at
time t= 0, and |x−y|denote the Euclidean distance between xand yin
The kernel Kt(x, y) of the operator et
2∆, is called the heat kernel and is
given by
Kt(x, y) = 1
and represents the probability density of finding the particle at yat time t
given its location at xat time 0. From the simigroup property
e(t+s)∆ =et∆es∆,(5)
for s,t≥0 we have
Kt(x, y) = Zdx1...dxN−1Kt/N (xN, xN−1)...Kt/N (x1, x0) (6)
for each N≥1, where we have set x0=xand xN=y.
There is an obvious one-to-one correspondence between configurations
(x1, ..., xN−1) and parametrized piecewise linear paths ω: [0, t]→Rdfrom x
to yconsisting of line segments [x0, x1], [x1, x2],...,[xN−1, xN], such that the
segment [xi−1, xi] is parametrized linearly by s∈[i−1
Nt, i
Nt]. We denote the
collection of all such paths by ΩN,t(x, y). Hence we may consider
tω= (2πt
as a measure on the finite dimensional space ΩN,t (x, y).
Noting that
t/N =
t/N )2=Zt
0|˙ω(s)|2ds, (8)
where ˙ωis the piecewise constant velocity of the trajectory ω, hence we can
Kt(x, y) = Z(x,y)
where the suffix (x, y) indicates that paths are restricted to go from xto y.
We refer to this equation as a random walk representation of Kt(x, y) on
ΩN,t(x, y).
More generally, given an action functional Son a piecewise linear parametrized
paths, we call the equation
t(x, y) = Z(x,y)
a random walk representation of the kernel HN
t(x, y) on ΩN ,t(x, y).
In is clear from the expressions for the heat kernel that the introduction
of quasiperiodicity in the partition of the intervals will bring new features
that is worth to be investigated.
4.1.1 Quasiperiodic Brownian movement
As a warm up let us write down the transition probability when a parti-
cle follows a quasiperiodic Brownian motion. The quasiperiodicity can be
introduced with a concrete function such as [2]
x(τ) = cos(2πτ ) + cos(2πατ),(11)
where αis a irrational number. Now if we interpret this function as given
the position of a particle after a time τthen according to the well know
evolution of this movement we have that the probability of being at (τ, x(τ))
if at τ= 0 it was at x= 0 is given by [8]
W(x(τ), τ; 0,0) = 1
√4πDτ exp{−x2(τ)
4Dτ }.(12)
5 Random walk on a Fibonacci chain
In this section we will review the general random walk procedure and then
apply it to the Fibonacci chain as preparation for studying random walks on
more involved geometries. We will restrict ourselves to the random walk in
one dimension.
Let us suppose we have a random walker which can move on a line. Let
us denote its position as Xnwhich can be any integer. Now suppose this
walker can move to the left or to the right with equal probability21/2 and
the length of the step being l. We would like to know the probability that
the walker is nRsteps to the right and nLsteps to the left. And also the
probability of being a distance mfrom the origin after nRsteps to the right.
This problem is discussed in [9] and now we will give the solution.
Since each step has length lthe location of the walker must be of the form
x=ml where mis an integer. A question of interest is the following: after
N steps what is the probability of being located at the position x=ml?
One can readily generalize this one-dimensional problem to more dimen-
sions. One again asks for the probability that after Nsteps the walker is
located at certain distance from the origin, however this distance is no longer
of the form ml. Also on higher dimensions we add vectors of equal length
in random directions and then we ask the probability of the resultant vector
being in certain direction and certain magnitude. This is exemplified by the
following two examples:
a) Magnetism: An atom has spin 1/2 and magnetic moment µ; in accor-
dance with quantum mechanics, its spin can point up or down with respect
to certain direction. If both possibilities are equally likely, what is the net
total magnetic moment of Nsuch atoms?
b) Diffusion of a molecule in a gas: A given molecule travels in three
dimensions a mean distance lbetween collisions with other molecules. How
far is likely to have gone after N collisions?
The random walk problem illustrates some very fundamental results of
probability theory.The techniques used in the study of this problem are pow-
erful and basic, and recur again and again in statistical physics.
After a total of Nsteps of length lthe particle is located at x=ml where
−N≤m≤N. We want to calculate the probability PN(m) of finding the
particle at x=ml after Nsteps. The total number of steps is N=nL+nR
and the net displacement in units of lis given by m=nR−nL. If it is known
that in some sequence of Nsteps the particle has taken nRsteps to the right,
then its net displacement from the origin is determined. Indeed
m=nR−nL=nR−(N−nR) = 2nR−N. (13)
2The probabilities can be different, for example in the case we have a slope.
This shows that if Nis odd then mis odd and if Nis even then mis even
A fundamental assumption is that successive steps are statistically inde-
pendent. Thus we can assert simply that, irrespective of past history, each
step is characterized by the respective probabilities
p=probability that the step is to the right (14)
q= 1 −p=probability that the step is to the left.(15)
Now, the probability of a given sequence of nRsteps to the right and nLstep
to the left is given simply by multiplying the probability of each step and is
given by
There are several ways to take nRsteps to the right and nLsteps to the left
in Nsteps. By known combinatorial calculus this number is given by
Hence the probability WN(nR) of taking nRsteps to the right and nL=
N−nRsteps to the left in Ntotal steps is given by
WN(nR) = N!
This probability function is known as the binomial distribution. The reason
is because the binomial expansion is given by
We already pointed out that if we know that the particle has made nRsteps
to the right in Ntotal steps then we know its net displacement m. Then the
probability of the particle being at mafter Nsteps is
PN(m) = WN(nR).(20)
We find explicitly that
Hence, in general we have that
PN(m) = N!
((N+m)/2)!((N−m)/2)!p(N+m)/2(1 −p)(N−m)/2.(22)
In the special case when p=q= 1/2 then
PN(m) = N!
5.1 Generalized random walk and the Fibonacci chain
Now we will study the generalized random walk. The random walk can be
studied in several dimensions, and we will do this up to a certain point and
later we will focus on one dimension and finally on the random walk on the
Fibonacci chain. In this subsection we mainly follow [10].
Let Pn(r) denote the probability density function for the position Rnof
a random walker, after nsteps have been made. In other words, the prob-
ability that the vector Rnlies in an infinitesimal neighbourhood of volume
δV centered on ris Pn(r)δV . The steps are to be taken independent ran-
dom variables and we write pn(r) for the probability density function for the
displacement of the nth step. Then the evolution of the walk is governed by
the equation
Pn+1(r) = Zpn+1 (r−r′)Pn(r′)ddr′,(24)
where the integral is over all of d-dimensional space. This equation is an
immediate consequence of the independence of the steps.
It is important to note that, by hypothesis, the probability density func-
tion for a transition from r′to ris a function of r−r′only, and not on r
and r′separately. In other words, the process is translationally invariant; it
is the relative position, not absolute location, which matters. The analysis
become much harder when pn+1(r−r′) must be replaced by pn+1(r,r′).
The assumed translational invariance ensures that the formal solution
of the problem is easily constructed using Fourier transform. The Fourier
transform ep(q) of a function p(x) is defined as
ep(q) = Z∞
eiqxp(x)dx. (25)
Under appropriate restrictions on the function p(x), there exist an inversion
p(x) = 1
e−iqx ep(q)dq. (26)
These equations are easily generalized to ddimensions. The Fourier trans-
form becomes
ep(q) = Z∞
where ddrdenotes de d-dimensional volume element and the integral is taken
over all of d-dimensional space. Similarly the inversion formula becomes
p(r) = 1
The convolution theorem for the Fourier transform states that under modest
restrictions on gand h
k(x) = Z∞
g(x−x′)h(x′)dx′corresponds to e
k(q) = eg(q)e
The generalization of the convolution theorem to ddimensions is straightfor-
k(r) = Z∞
g(r−r′)h(r′)ddr′corresponds to e
k(q) = eg(q)e
Taking the Fourier transform of our equation for the probabilities we have
that e
Pn+1(q) = epn+1(q)e
With P0(r) the probability density function for the initial position of the
walker, and e
P0(q) its Fourier transform, we have that
Pn(q) = e
j=1 epj(q).(32)
Taking the inverse Fourier transform of both sides of this equation, we find
the solution for the probability density function for the position after nsteps:
Pn(r) = 1
j=1 epj(q)ddq.(33)
When all steps have the same probability density function p(r) and the walk
is taken to commence at the origin of coordinates, so that
P0(r) = δ(r)e
P0(q) = 1,(34)
then we have
Pn(r) = 1
There are very few cases in which this integral can be evaluated in terms
of elementary functions. However, much useful information can still be ex-
Now we will see one of the cases where this integral can be reduced a
elementary functions. For a random walk in one dimension with different
length steps we have that
p(x;ln) = 1
2(δ(x−ln) + δ(x+ln)).(36)
Using that
δ(x−ln) = 1
then we have that
ep(q) = Z∞
eiqxp(x)dx =1
2(eiqln+e−iqln) = cos(qln).(38)
Pn(x;ln) = 1
2πZe−iqx cosn(qln)dq. (39)
In the case of the Fibonacci sequence we have
ln+1 =ln+ln−1with l0= 0, l1= 1.(40)
So in this case we can solve the problem completely.
There is a subtlety with this expression for the probability, it diverges.
The problem is that we are dealing with distributions and classical analysis
does not work here. So we have to use the distribution theory. From p. 63
of [10] we know that the correct expression for the probability is
P r {Xn=lln}=ln
e−illnξcosn(lnξ)dξ, (41)
where l∈Z. It is interesting that if we change variables as lnξ=kthen
P r {Xn=lln}=1
e−ilk cosnkdk, (42)
and there is no dependence of lnin the integral.
5.2 The random walk in a two dimensional Fibonacci
Now let us consider a infinite two dimensional Fibonacci lattice. Then in
this case the probability density is given by
p(x, y;lnx, lny) = 1
4(δ(x−lnx) + δ(x+lnx ) + δ(y−lny) + δ(y+lny)).(43)
Then following the one dimensional case we have that in the present case the
probability function is given by
Pn(x, y;lnx, lny) = 1
8π(Ze−iqx cosn(qlnx )dq +Ze−ipx cosn(plny )dp).(44)
Here qand pare variables in the Fourier space and lnx and lny are Fibonacci
Making the corresponding manipulations we did in the 1-dimensional Fi-
bonacci sequence, now we obtain in this case
P r {Xn=llnx, Yn=mlny }=1
e−ilk cosnkdk +Zπ
e−imk cosnkdk),
where l, m ∈Z.
6 A kind of partition function
One of the main object in our approach is the a kind of partition function
which in certain limit should be reducible to the Einstein-Hilbert action and
in other limit to the partition function of quantum statistical mechanics. In
order to construct this partition function we will follow the ideas explained
in [1], [11] and [12].
Let us give a simple example of the kind of things we are working with.
One possible action for a piecewise constant path is [1]
i=1 |xi−xi−1|,(46)
where we will suppose that ˜
βis a generalized inverse of the temperature.
Then the partition function3associated with this action is
The energy associated with this partition function is
βln Z=
i=1 |xi−xi−1|(48)
and the entropy is
S=E+ ln Z= (1 −˜
i=1 |xi−xi−1|.(49)
Here the xi’s are an homogeneous partition of the path. In this sense it
is a periodic partition. It is clear that if now we assume that the xi’s are
quasiperiodic then the entropy will change. It is not difficult to imagine how
hard it would be to solve if instead of having a one-dimensional path we
have a surface or a volume. It could be interesting to compare the entropy
Swith the entropy of a elastic string. If we want the discrete action to
go to the continuous action as the size of the partition goes to zero then
βshould depend on the size of the partition function [1]. Then clearly in
this case if we choose a quasicrystalline partition then the entropy and other
thermodynamical quantities will be impacted.
6.1 Partition function and entropy of the Fibonacci
If we consider the Fibonacci chain in 1-dimension, we can define a partition
function as
P r {Xn=lln}.(50)
3Here we are thinking the action as an effective action which coincides at zero loops
with the classical action.
Analogously in the 2-dimensional case we have then
P r {Xn=llnx, Yn=mlny }.(51)
If these definitions are correct, then it is a matter of brute force to calculate
the analogous of thermodynamical quantities.
For example let us do this for the 1-dimensional Fibonacci chain. In this
case we would have that the entropy is given by
S=F(ln)hXni+ ln Z, (52)
where F(ln) is a function which we should determine using plausible argu-
ments and hXniis the expectation value of Xn.
Analogously, in the two dimensional case we have
S=G(lnx, lny )hXn, Yni+ ln Z, (53)
where G(lnx, lny ) is a function we have to propose. For example, if we agree
that with a new step there is an increasing of information then these functions
should be decreasing functions of the lengths.
7 Further ideas and final comments
It is clear that the introduction of aperiodicity in the framework of quantum
gravity would give substantially different results compare with the standard
approaches. Hence it would be interesting in the future to do something
similar with other quantum gravity approaches.
From the considerations in this work it is clear that our approach is closer
to the standard path integral approach than to the Hilbert space framework.
In this sense it would be interesting if with our approach we can recover the
well known results from Euclidean quantum gravity as explained in [13], [14].
Acknowledgments: This work is fully sponsored by Quantum Gravity
A On the Euclidean action and the Boltzman
A.1 Introduction
One of our goals is to construct an object that in one limit gives the General
Relativity action (classical and quantum) and in the other side gives the
quantum mechanical statistics partition function.
In the book [15] Huang says that it is a deep mystery of physics that the
Hamiltonian operator appears in the evolution operator in quantum mechan-
ics and in the partition function in quantum statistical mechanics:
e−it ˆ
Here β=κ
T, with Tthe temperature and κthe Boltzman’s constant. If
we make t=−iτ , where τis real and periodic with period of βthen both
expressions become the same.
The purpose of this appendix is to comment on this deep mystery and to
try to elucidate, at least partially, why this occurs.
We think this discussion is important since important results such as
the entropy of black holes in euclidean quantum gravity [13] uses this deep
The organization of this appendix is as follows: In the section 2 we discuss
how the Boltzman factor is related to the action, in the section 3 we explain
how the entropy of the BTZ black hole is obtained in Euclidean Quantum
Gravity, and finally in the section 4 we give our final comments.
A.2 On the action and the Boltzman factor
It is interesting to note that the action Sof a system appears in the path
integral [8], [12], the partition function [1] and the Hamilton-Jacobi equation
[16]. Also it is interesting the similarity between the Boltzman factor and
the normal distribution. Let us elaborate on these two ideas.
In the Euclidean setting we have the path integral
A=ZDxe −S
Whereas the Boltzman factor is
kT .(56)
We know the action has units of energy times time. So if we multiply in the
Boltzman factor the energy and the kT term by some time we have an term
with units of action. Now the partition function is
kT .(57)
The similarity with the path integral is obvious. Now the normal distribution
has the following form
Clearly if we multiply the square term by one over time square, and also the
D, then we have energy units. In one step further we can have a kind of action
in the normal distribution. Now, the Boltzman distribution is ubiquitous in
statistical mechanics and so is the normal distribution in several natural
processes. From this point of view the normal distribution is analogous
to the expression of the effective action4. So one may wonder if there is
a deep connection between these three expressions. One might wonder if
we can make up a mechanical toy model where in one side one has the
normal distribution and on the other end tha path integral and in the middle
the partition function obtained from the Boltzman factor. If we impose a
periodicity in the Euclidean amplitude A then with the correct units we have
the well know temperature of Black Holes. This periodicity when seen from
a discrete system can be related to the Poincar´e recurrence theorem.
This toy model seems to be relevant for the unification physics since in
one hand one has a discrete system (similar to a quantum geometry) and in
the other hand a continuous system (module some metric issues) similar to
a topological quantum field theory.
It is also interesting that the action appears in the Hamilton-Jacobi equa-
tion whose quantum limit is the Schrodinger equation and it can branch to
classical mechanics, gravitational physics and electromagnetic theory.
Just to finish this section we note that the Lagrangian is given as
L=E−V, (59)
where Eis the energy and Vthe potential (energy). Hence we see that the
Lagrangian is a kind of generalized energy. The action is
S=ZLdt. (60)
4See for example [17] where the relationship between the path integral and the effective
action is displayed.
Hence when we make timaginary and periodic, with the correct period in,
for example, black holes then everything about time drops and we have the
partition function of statistical mechanics. It is as if there were hidden a
symmetry related with time. Here we have taken the simplest Lagrangian
however it is not difficult to see that for example the scalar field the situation
is very similar.
The above discussion makes clear why the Euclidean path integral coin-
cides with the partition function when the time is periodic, in some sense the
partition function is hidden in the path integral.
A.3 On the entropy of black holes
As one example of some of the ideas presented in the previous section now
we will explain how the entropy of some black hole can be obtained using
the effective action.
It is well know, see for example [18], that at zero order the effective action
Γ[Φ] coincides with the classical action A[Φ] evaluated on the mean field Φ.
The evaluation of the black hole entropy of the Kerr black hole can be
consulted [13], and now we will show how the entropy for the BTZ black hole
is obtained.
We follow mainly [19]. The Euclidean action of the BTZ black hole is
IE=βM −A
Then the partition function is
ZBT Z (T) = exp (πl)2T
where lis the AdS radius. The expectation value of the energy is
EBT Z =−∂
∂β ln Z=M. (63)
Whereas the entropy is given by
SBT Z =βEBT Z + ln ZB T Z = 4πr+=A
Which is the result one expects on the grounds of Beckenstein ideas on en-
tropy of black holes.
It is interesting to note that there are at least three other values of the
BTZ black hole entropy obtained in [20], [21] and [22]. In the first in loop
quantum qravity, the second in standard statistical field theory and in the
third in the brick wall model. In the first two models it does not coincide
with the value given in [19] whereas in the brick wall model it coincides with
A.4 Final comments
It is interesting to note the following: The result of [19] is classical, although
using a quantum framework, the result of [20] is quantum but it does not
give the expected result, the result of [21] is semiclassical and gives a close
result to the one expected, and finally the result of [22] is quantum and gives
the expected result but the entropy is of the scalar field living on the BTZ
black hole. Hence there is no a consensus about this entropy. Just to finish
up we note that the temperature of a black hole does not make sense without
a field living on it, so, after all the brick wall model could be the one closer
to the origin of the BTZ black hole entropy.
[1] J. Ambjorn, B. Durhuus and T. Jonsson, “Quantum geometry”, Cam-
bridge, Cambridge (1997).
[2] M. Baake and U. Grimm, “Aperiodic order”, Cambridge, Cambridge
[3] M. Amaral, R. Aschheim and K. Irwin, “Quantum gravity at the Fifth
root of unity”, arXiv:1903.10851v2 [hep-th] 2019.
[4] T. Jonsson, “Introduction to random surfaces”, Lectures presented at
the 1999 NATO-ASI on “Quantum Geometry in Akureyri”, Iceland
[5] C. Rovelli and F. Vidotto, “Covariant Loop Quantum Gravity”, Cam-
bridge, Cambridge (2014).
[6] M. Senechal, “Quasicrystals and geometry”, Cambridge, Cambridge
[7] M. V. Jaric, “Introduction to the mathematics of quasicrystals”, New
York, Academic Press, Inc. (1989).
[8] M. Chaichian and A. Demichev, “Path integrals in physics. Volume I”,
New York, CRC Press (2001).
[9] F. Reif, “Fundamentals of statistical and thermal physics”, Illinois,
Waveland Press, Inc. (1965).
[10] B. D. Hughes, “Random Walks and Random Environments Vol. 1”,
Oxford, Oxford University Press (1995).
[11] A. N. Vasiliev, “Functional Methods in Quantum Field Theory and
Statistical Physics”, Australia, Gordon and Breach Science Publishers
[12] R. P. Feynman and A. R. Hibbs, “Quantum Mechanics and Path Inte-
grals”, New York, Dover Publications, Inc. (2010).
[13] G. W. Gibbons and S. W. Hawking, “Euclidean Quantum Gravity”,
Singapore, World Scientific (1993).
[14] S. Carlip, “Quantum Gravity in 2+1 Dimensions”, Cambridge, Cam-
bridge (1998).
[15] K. Huang, “Quantum Field Theory: from operators to path integrals”,
WILEY-VCH, Weinhem (2010).
[16] A. L. Fetter and J. D. Walecka, “Theoretical Mechanics of Particles and
Continua”, Dover, New York (2003)
[17] L. E. Parker and D. J. Toms, “Quantum Field Theory in Curved Space-
time”, Cambridge, Cambridge (2009).
[18] H. Kleinert, “Particles and Quantum Fields”, World Scientific, Singa-
pore (2016).
[19] Y. Kurita and M. Sakagami, “CFT Description of the three-dimensional
Hawking-Page transition”, Progress of theoretical physics 113 (6) 1193
[20] J. M. Garc´ıa-Islas, “BTZ black hole entropy: a spin foam model descrip-
tion”, Class. Quant. Grav. 25 245001 (2008).
[21] I. Ichinose and Y. Satoh, “Entropies of scalar fields on the three dimen-
sional black holes”, Nucl. Phys. B 447 340 (1995).
[22] B. S. Kay and L. Ort´ız, “Brick walls and AdS/CFT”, Gen. Rel. Grav.
46 1727 (2014). | {"url":"https://www.researchgate.net/publication/339972841_Aspects_of_aperiodicity_and_randomness_in_theoretical_physics","timestamp":"2024-11-08T11:59:44Z","content_type":"text/html","content_length":"615679","record_id":"<urn:uuid:73d46af9-23ba-4235-aba0-fe65271ae542>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00711.warc.gz"} |
Pythagorean Triangle -- from Wolfram MathWorld
A Pythagorean triangle is a right triangle with integer side lengths (i.e., whose side lengths Pythagorean triple). A Pythagorean triangle with primitive right triangle.
The inradius
The area of such a triangle is also a whole number since for primitive Pythagorean triples, one of
is always a positive integer. | {"url":"https://mathworld.wolfram.com/PythagoreanTriangle.html","timestamp":"2024-11-04T04:51:14Z","content_type":"text/html","content_length":"54100","record_id":"<urn:uuid:79f2fa69-24cf-4c32-91e3-74e274872274>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00513.warc.gz"} |
Re: [Edu] more correlations than you might think
what a list!!
thanks Steven!
On 5/30/06, Steven Greenstein <blue42@xxxxxxxxxxxxxx> wrote:
Hi again,
First, let me tell you how much I appreciate these conversations. Actually,
you've been doing all the talking; I've just been asking questions. To show
that I'm grateful, I'm willing to do the correlations to Texas standards.
Second, since I mentioned correlations. This is an excerpt from an article
on Games in the Classroom from the Center for Innovation in Mathematics
Teaching at the University of Exeter. The article also briefly describes the
plan one school used to integrate games into the curriculum.
The link:
The excerpt:
Why games?
How can games be used to further mathematical education? This question has
to be addressed and answered if we are to devote time and resources to
playing games in the classroom. Let us first look at some questions which
players might pose to themselves on settling down to play a game, and the
mathematical heading under which we might class such a question.
Form of question:Mathematical heading:
1. "How do I play this?"Interpretation
2. "What is the best way of playing?"Optimisation
3. "How can I make sure of winning?"Analysis
4. "What happens if . . . ?"Variation
5. "What are the chances of . . . ?"Probability
Given a chance to develop answers to questions like that could lead to
statements commencing as listed below, together with the mathematical idea
being covered in such a statement.
Form of statememt:Mathematical idea:
6. "This game is the same as . . ."Isomorphism
7. "You can win by . . ."A particular case
8. "This works with all these games . . ."Generalisation
9. "Look, I can show you it does . . ."Proving
10. "I record the game like this . . ."Symbolisation and Notation
Life is too short for long division.
Edu mailing list | {"url":"http://archive.looneylabs.com/mailing-lists/edu/msg00103.html","timestamp":"2024-11-08T07:46:08Z","content_type":"text/html","content_length":"6329","record_id":"<urn:uuid:b7e339d3-0437-464b-aa9a-62749d08241f>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00658.warc.gz"} |
CDI & electrosorption
The pre-release of the book 'Physics of Electrochemical Processes' (ISBN: 9789090332581) by P.M. Biesheuvel and J.E. Dykstra can now be downloaded from the website
www.physicsofelectrochemicalprocesses.com free of charge.
Topics include:
• Ch. 1: The extended Frumkin isotherm describes the capacitive salt adsorption in intercalation materials
• Ch. 2: The Donnan model for the electrical double layer structure in charged materials, including electrodes
• Ch. 3: The Gouy-Chapman-Stern model and surface ionization
• Ch. 4: Volume effects in EDL theory (Bikerman, Carnahan-Starling, activity coefficients)
• Ch. 5: EDLs in motion: electrowetting, contact angle, energy harvesting
• Ch. 6: EDL interaction (DLVW theory)
• Ch. 7: Solute Transport (mass transfer to interfaces including dispersion)
• Ch. 8: Electrokinetics (hydrostatic and osmotic pressure, Navier-Stokes equation for electrolytes, osmosis vs electro-osmosis)
• Ch. 9: Heat effects for current flow across the EDL (Peltier effect, electrostatic cooling)
• Ch. 10: Acid-base reactions in transport models
• Preamble: The microscopic and experimental perspective of the electrical double layer
• Ch. 11: The difference between capacitive (non-Faradaic) and Faradaic electrode processes in electrochemistry
• Ch. 12: Electrode kinetics (in preparation)
• Ch. 13: Porous electrodes (in preparation)
• Ch. 14: Reverse Osmosis
• Ch. 15: Electrodialysis
• Ch. 16: Ion transport in bio-electrochemical systems
• Ch. 17: Bioelectrochemical conversions on conductive media
• Ch. 18: Overview of electrochemical water desalination (in preparation)
• Ch. 19: Numerical methods (2021)
• Ch. 20: Analysis of experimental methods in electrochemistry (2021)
New ES&T paper on the energy efficiency of electro-driven brackish water desalination
A new Open Access paper was published in the prestigious journal ES&T, from the group of prof. Meny Elimelech (Yale, USA) which discusses a detailed comparison between electrodialysis (ED) and
membrane capacitive deionization (MCDI).
Patel et al. "Energy Efficiency of Electro-Driven Brackish Water Desalination: Electrodialysis Significantly Outperforms Membrane Capacitive Deionization", ES&T (2020). https://pubs.acs.org/doi/abs/
According to the authors, "we provide the first systematic and rigorous comparison of the energetic performance of electrodialysis (ED) and membrane capacitive deionization (MCDI) over a broad range
of brackish water desalination conditions."
The authors find that:
• The energy consumption of ED is substantially lower than MCDI for all investigated conditions, with the energy efficiency being nearly an order of magnitude higher for many separations.
• Even with idealized operation (complete energy recovery and reduction in energetic losses), the energy efficiency of MCDI remains lower than ED.
Finally, the authors emphasize that "for low feedwater salinities (< ~2 g/L), energy efficiency should be a secondary consideration in the choice of desalination technology, with capital cost, ease
and reliability of operation, and additionally required treatment steps taking higher priority."
CDI keeps growing exponentially !
... and the field of Capacitive Deionization keeps on growing at an increasing speed ! With over 160 scientific papers written and published last year, the total number of CDI publications has grown
from ∼25 in 2000, to ∼65 in 2010, to over 1,000 at the end of 2019! CDI papers in 2019 came from many prestigious places including Tsinghua University, Seoul National University, Technion, MIT,
Stanford, and Yale. This output from such eminent research groups shows that CDI is taking a leading role in the scientific study of water desalination technologies. Analyzing these publication data
for the past decade, an exponential growth can be observed, with a doubling of the publication output every 2.5 year !
Citations to the CDI literature have grown from a number less than 100 per year before 2010, to about 2000 per year at the end of 2015, and close to 10,000 per year in 2019. These statistics also
reflect an exponential growth, with a doubling time of 2.0 year. This difference in doubling time (faster for citations than publications) may indicate that CDI papers are more and more cited in
papers from outside the CDI-field.
Position paper on “What is CDI?”
Following the CDI conference recently held in the Republic of Korea, July 2017, twenty scientists active in the field of CDI worked on a joint position paper, putting forward the proposition that CDI
defines a class of desalination technologies that share common operational principles and relevant metrics, thereby joining under one common term different CDI cell layouts and chemistries. Thus,
according to the position paper, the class of CDI includes electrodes based on carbon materials, but as well electrodes with ion storage based on different chemistries such as using redox materials.
The position paper can be downloaded via the link given below.
P.M. Biesheuvel, M.Z. Bazant, R.D. Cusick, T.A. Hatton, K.B. Hatzell, M.C. Hatzell, P. Liang, S. Lin, S. Porada, J.G. Santiago, K.C. Smith, M. Stadermann, X. Su, X. Sun, T.D. Waite, A. van der Wal,
J. Yoon, R. Zhao, L. Zou, and M.E. Suss, "Capacitive Deionization -- defining a class of desalination technologies," ArXiv:1709.05925 (2017).
CDI-E Korea 2017 a major success !
Last week's CDI-E's conference was a big success. 180 participants gathered for three full conference days in the heart of Seoul, Republic of South Korea. The conference, hosted by prof. Jeyong Yoon
and his team of Seoul National University, proved invaluable in informing participants of all the latest developments in CDI technology, both from an academic and industrial perspective. The
versatile program consisted of a tutorial session, plenary and keynote lectures, regular lectures and two poster sessions and gave food for thought to all attendants. The lively and amiable
atmosphere gave the whole conference the right touch of feeling welcome in perhaps one of the most vibrant places in the world, the famous Gangnam district, a place that never sleeps.
Complementary surface charge enhances CDI-performance
Two recent papers with authors from the US, The Netherlands, and Israel, convincingly show the relevance of chemical charge residing in the carbon electrodes ("immobile", or "complementary" charge)
to enhance salt adsorption capacity (SAC) of CDI electrodes.
In the more theoretical paper of the two, published OPEN ACCESS in Colloids and Interfaces Science Communications, the theoretical framework is laid out which comprehensively describes the range of
recently developed new CDI desalination modes such as inverted-CDI and (what the authors call) enhanced-CDI. Also the occurrence of "inversion peaks" which often develop during normal CDI operation
are explained as due to developing chemical charge. In addition, a novel operational mode is described where due to the chemical charge, it becomes possible to enlarge the operating window of CDI and
thus to enhance SAC further still. This operational mode of "extended voltage CDI" was not described before.
In the sequel paper, published in Water Research, both the enhanced-CDI regime and the extended-voltage CDI-regime are experimentally validated. In this paper the more advanced amphoteric Donnan
model is used to describe the EDL-structure. This model quantitatively predicts the experimental observations of salt storage and charge. An interesting inconsistency is how the measured chemical
charge (by titration) can be up to one order beyond the chemical charge derived from comparing the amph-D model to the data.
CDI publication statistics through the roof !
... and the field of Capacitive Deionization keeps on growing at an increasing speed ! While over 60 scientific publications are now written and published annually, citations to the CDI literature
have grown from a number less than 100 per year before 2010, to about 2000 per year at the end of 2015, and this number continues to rise. Analysing citation data for the past 10 years, an
exponential growth in citation rate is clearly observed, with a doubling of the citation rate every 18 months !
Perspective on CDI “What is it and what can we expect from it?” published in E&ES
In a collaboration involving scientists from five different countries in two continents, members of the CDI&E working group used the past year to come with a Perspective-paper on capacitive
ionization and electrosorption. Published on invitation in the high-impact journal Energy&Environmental Science (IF=25) as a prestigious Perspective-contribution, the paper is expected to generate
attention inside and outside the CDI-field. To help quick dissemination of its content, the authors have chosen for the OPEN ACCESS-format.
As corresponding authors Prof.Dr. Mathew Suss and Prof.Dr. Volker Presser explain: "The idea for this perspective was conceived of during the last CDI-conference in Leeuwarden, the Netherlands, and
our aim is that it serves the growing CDI-community in outlining current trends in CDI developments, in standardizing metrics, and to help by identifying 'white areas' in CDI, both experimentally and
theoretically. It was a most exhilarating task to work together with so many different authors on different continents to see this paper growing over the year. Many discussions helped us to focus on
the most important elements and to converge on the key trends and best metrics for CDI performance. We have done our very best to put together a paper that helps to catalyze scientific and
industrial developments in the CDI&E-field."
Lower energy consumption in CDI by increasing the discharge voltage
As reported during the 8th International Conference "Interfaces against Pollution," May 2014, the energy consumption of CDI operation can be significantly reduced by tuning the discharge voltage,
which is the cell voltage applied during cell discharge, when the adsorbed salt is released and a concentrated brine stream is produced. Commonly in CDI, the charging voltage is tuned to an optimum
value, where salt adsorption is high but leakage currents are still low. The discharge voltage is by default set to zero Volt. Following an earlier study from Bar-Ilan University, Israel, a
cooperation of Seoul University (South Korea), Wageningen University and Wetsus (the Netherlands) found and reported on the positive influence of tuning the discharge voltage to values higher than
zero. In contrast to the earlier work, it was found that salt adsorption per cycle did not markedly decrease, while the charge efficiency went up to values approaching the theoretical limit of one
(unity). This meant that the energy consumption significantly decreased (being inversely proportional to charge efficiency), even without considering energy recovery, something that is possible with
positive discharge voltages.
In the same study it was also found that with a non-zero discharge voltage, it becomes easier in CDI to achieve a stable effluent concentration by using constant-current operation; something that
before this study was thought to be possible only for membrane-assisted CDI. As senior author prof.dr. J. Yoon remarks: "This was a very insightful study that clearly showed the potential of tuning
the operational conditions of CDI to enhance the performance of a CDI cell. It was remarkable how accurately the porous electrode transport theory, using the Donnan concept to describe salt
adsorption, could describe the data. For design purposes, such a model is indispensable." | {"url":"https://cdi-electrosorption.org/author/p-m-biesheuvel/","timestamp":"2024-11-13T15:58:25Z","content_type":"text/html","content_length":"65559","record_id":"<urn:uuid:baa86af6-8f74-4edb-9dc4-9418f5169f26>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00456.warc.gz"} |
Guide To What Type Of Cell Reference Is $C$19?
When working with spreadsheets, understanding cell references is essential for efficiently performing calculations and data analysis. In a nutshell, cell references are used to identify and
manipulate data in a spreadsheet. There are three main types of cell references in spreadsheets: relative, absolute, and mixed. In this blog, we will delve into the specifics of absolute cell
references and answer the burning question: What type of cell reference is $C$19?
Key Takeaways
• Understanding the different types of cell references in spreadsheets is essential for efficient data analysis and calculations.
• Absolute cell references, such as $C$19, are useful for maintaining the fixed reference to a specific cell when copying and pasting formulas.
• Using absolute cell references can help avoid issues with cell references changing unintentionally in formulas.
• It is important to be aware of the potential drawbacks of using absolute cell references, such as difficulty in adapting formulas when the referenced cell needs to change.
• Best practices for using $C$19 and other absolute cell references include incorporating relative and mixed references as needed for dynamic calculations.
Understanding $C$19 as a cell reference
When working with formulas and references in spreadsheets, it's important to understand the different types of cell references. One common type of cell reference is $C$19, which uses dollar signs to
lock the column and row. Let's take a closer look at what this means and how it impacts your spreadsheet.
A. Explanation of the dollar signs in the cell reference
When you see a cell reference like $C$19, the dollar signs play a crucial role. In Excel or Google Sheets, the dollar sign before the column letter (in this case, C) and before the row number (19)
indicates that the reference is absolute. This means that no matter where you copy or drag the formula, the reference will always point to that specific cell.
B. Discussing the significance of the dollar signs in relation to the column and row
Understanding the significance of the dollar signs in relation to the column and row is essential for working with cell references. When the column letter (C) has a dollar sign, it means that the
reference will always point to column C, regardless of where the formula is located. Similarly, when the row number (19) has a dollar sign, it indicates that the reference will always point to row
So, in the case of $C$19, both the column and the row are locked, making it an absolute cell reference. This can be particularly useful when working with fixed values or constants that need to be
referenced consistently across a spreadsheet.
Advantages of using $C$19 in formulas
When working with formulas in spreadsheets, using an absolute cell reference like $C$19 can provide several advantages that make your work easier and more efficient.
• A. How using an absolute cell reference can make formulas easier to copy and paste
When you use an absolute cell reference, such as $C$19, in a formula, it means that the formula will always refer to that specific cell, regardless of where it is copied or pasted within the
spreadsheet. This can be extremely useful when you have a formula that you need to apply to multiple cells, as it ensures that the references remain consistent and accurate.
• B. Avoiding issues with cell references changing when copying formulas
One common issue that can arise when working with formulas in spreadsheets is that cell references can change when formulas are copied or moved to different locations. By using an absolute cell
reference like $C$19, you can avoid this problem altogether, as the reference will remain fixed and unchanged no matter where the formula is used in the spreadsheet.
Guide to What type of cell reference is $c$19?
When working with formulas in spreadsheets, it's important to understand the different types of cell references and how they can be used. One common type of cell reference is $C$19, which is known as
an absolute cell reference. Let's take a look at some common scenarios for using this type of cell reference in formulas.
A. Calculating totals for specific items in a budget spreadsheet
One common scenario for using $C$19 in a formula is when calculating totals for specific items in a budget spreadsheet. For example, if cell C19 contains the total cost of groceries for the month,
you may want to use this value in a formula to calculate the total cost of all expenses for the month. By using $C$19 as an absolute cell reference in the formula, you can ensure that the total cost
of groceries is included in the calculation, regardless of where the formula is copied or moved within the spreadsheet.
B. Applying a fixed tax rate to a range of values
Another common scenario for using $C$19 in a formula is when applying a fixed tax rate to a range of values. For example, if cell C19 contains the fixed tax rate, you may want to use this value in a
formula to calculate the total tax amount for a range of expenses. By using $C$19 as an absolute cell reference in the formula, you can ensure that the same tax rate is applied to each expense in the
range, regardless of where the formula is copied or moved within the spreadsheet.
Potential drawbacks of using $C$19 in formulas
When using $C$19 as a cell reference in formulas, there are a few potential drawbacks to consider:
• Difficulty in adapting formulas when the referenced cell needs to change
Using an absolute cell reference like $C$19 can make it challenging to adapt formulas when the referenced cell needs to change. If the formula is copied to another location or the referenced cell
is moved, the formula will still point to the original cell, potentially leading to errors or incorrect calculations.
• The risk of errors if the wrong type of cell reference is used in a formula
Another drawback of using $C$19 in formulas is the risk of errors if the wrong type of cell reference is used. For example, if a mixed cell reference (such as $C19 or C$19) is intended but a
simple reference ($C$19) is used instead, it can cause unexpected results in the formula.
Best practices for using $C$19 in spreadsheets
When working with spreadsheets, it's important to understand the different types of cell references and how to effectively use them. The $C$19 cell reference is an absolute reference, and it's
crucial to know when and how to use it in your calculations.
A. Using absolute cell references for static values that should not change
• 1. Understanding absolute cell references
Absolute cell references, denoted by the dollar signs before the column and row identifiers (e.g. $C$19), lock the reference to a specific cell. This means that no matter where you copy or move
the formula, the reference will always point to the same cell.
• 2. Best practices for using absolute cell references
Use absolute cell references for constant values such as tax rates, conversion factors, or fixed data that should not change. This ensures that the calculations remain accurate and consistent,
regardless of any changes to the layout or structure of the spreadsheet.
B. Incorporating relative and mixed cell references as needed for dynamic calculations
• 1. Understanding relative and mixed cell references
Relative cell references adjust when copied to different cells, based on their position relative to the original cell. Mixed cell references, where either the row or column is locked with a
dollar sign, provide a combination of absolute and relative references.
• 2. Best practices for incorporating relative and mixed cell references
Use relative and mixed cell references for dynamic calculations that involve changing values, such as monthly sales figures, inventory levels, or employee salaries. This allows the formulas to
adapt to new data without needing manual adjustments.
Understanding the benefits of using the $C$19 cell reference in formulas can greatly improve your efficiency and accuracy in spreadsheet applications. By using an absolute cell reference, you can
ensure that specific cells are always included in your calculations, regardless of where the formula is copied or moved within the spreadsheet. This can save time and reduce errors in your work.
I encourage readers to experiment with different types of cell references in their own spreadsheets. By familiarizing yourself with relative and mixed cell references, you can gain greater control
and flexibility in your formulas, leading to more efficient and accurate data analysis.
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