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http://programmersheaven.com/discussion/110883/how-to-dynamically-change-properties-of-a-3d-human-model | 1,519,057,092,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812756.57/warc/CC-MAIN-20180219151705-20180219171705-00609.warc.gz | 300,933,672 | 11,714 | #### Howdy, Stranger!
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# How to dynamically change properties of a 3D human model?
Member Posts: 29
[b][red]This message was edited by the sweetgirl at 2002-4-13 19:48:2[/red][/b][hr]
[b][red]This message was edited by the sweetgirl at 2002-4-13 19:28:32[/red][/b][hr]
In a virtual fitting room program, I'd need to create the 3D human model and provide the ability to dynamically regenerate the 3D human model based on the user's input of the body measurements for eg.
How can I import a 3D human model created using 3D softwares, into my MFC program through OpenGL in C++ code? Then, how can i dynamically change the properties of the 3D human model created using 3D software?
-sweetgirl-
• Member Posts: 168
: [b][red]This message was edited by the sweetgirl at 2002-4-13 19:48:2[/red][/b][hr]
: [b][red]This message was edited by the sweetgirl at 2002-4-13 19:28:32[/red][/b][hr]
: In a virtual fitting room program, I'd need to create the 3D human model and provide the ability to dynamically regenerate the 3D human model based on the user's input of the body measurements for eg.
: How can I import a 3D human model created using 3D softwares, into my MFC program through OpenGL in C++ code? Then, how can i dynamically change the properties of the 3D human model created using 3D software?
:
: -sweetgirl-
:
:
:
:
:
WOW, now that's a tuffy. That would take some industial strength math.
As for importing it. That would depend on the program used to create the image. There are a number of formats out there. AUTOCAD DXF is the easyest.
The vectors would need to put in a array structure of X,Y,Z. I suppose breaking the image up into 'body parts' would be the most logical means.
Then the different 'parts' could be scaled match the new measuments and displayed.
Send me a copy of the 3d model when you get it. I'd like to see it.
ET
• USAMember Posts: 0
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https://certifiedcalculator.com/laundry-cost-calculator/ | 1,709,396,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00652.warc.gz | 158,965,246 | 44,278 | # Laundry Cost Calculator
Introduction:
Doing laundry is a routine task for many of us, and it’s important to keep track of how much it costs. Whether you’re using a shared laundry facility or your own washing machine, the Laundry Cost Calculator can help you estimate the expenses associated with each load.
Formula:
The cost per load in our Laundry Cost Calculator is calculated using a simple formula:
Cost per Load = (Load Size (lbs) * Water Temperature (°F)) / Detergent Amount (oz)
How to Use:
1. Enter the load size in pounds (lbs) that you plan to wash.
2. Input the water temperature in degrees Fahrenheit (°F) you intend to use for the laundry cycle.
3. Specify the amount of detergent in ounces (oz) you will use for the load.
4. Click the “Calculate” button to get the estimated cost per load.
Example:
Let’s say you have a load size of 8 lbs, plan to use a water temperature of 75°F, and you’ll use 4 oz of detergent. After clicking “Calculate,” you’ll find that the estimated cost per load is \$15.00.
FAQs:
1. Q: How accurate is the Laundry Cost Calculator? A: The calculator provides a rough estimate and may not account for variables like water and electricity costs. It’s intended as a simple tool for cost awareness.
2. Q: Can I use the calculator for commercial laundry businesses? A: Yes, you can use it for commercial purposes, but keep in mind that it’s a basic estimation tool.
3. Q: What if I don’t know the exact values for load size, water temperature, or detergent amount? A: You can make educated guesses or measure them roughly for a reasonable estimate.
4. Q: Does the calculator consider electricity and water costs? A: No, it focuses solely on the cost associated with the load size, water temperature, and detergent amount.
5. Q: Can I change the currency symbol for the result? A: The calculator uses the dollar symbol (\$), but you can convert the result to your preferred currency manually.
Conclusion:
Our Laundry Cost Calculator is a handy tool for estimating the expenses of each laundry load. While it provides a basic estimate, it’s essential to consider other costs such as water and electricity. Use it as a starting point for tracking your laundry expenses and optimizing your laundry routine. | 491 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-10 | latest | en | 0.856286 |
https://www.oreilly.com/library/view/practical-time-series/9781492041641/ch04.html | 1,606,824,260,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00520.warc.gz | 796,934,619 | 37,177 | # Chapter 4. Simulating Time Series Data
Up to this point, we have discussed where to find time series data and how to process it. Now we will look at how to create times series data via simulation.
Our discussion proceeds in three parts. First, we compare simulations of time series data to other kinds of data simulations, noting what new areas of particular concern emerge when we have to account for time passing. Second, we look at a few code-based simulations. Third, we discuss some general trends in the simulation of time series.
The bulk of this chapter will focus on specific code examples for generating various kinds of time series data. We will run through the following examples:
• We simulate email opening and donation behavior of members of a nonprofit organization over the course of several years. This is related to the data we examined in “Retrofitting a Time Series Data Collection from a Collection of Tables”.
• We simulate events in a taxicab fleet of 1,000 vehicles with various shift start times and hour-of-the-day-dependent passenger pickup frequencies over the course of a single day.
• We simulate step-by-step state evolution of a magnetic solid for a given temperature and size using relevant laws of physics.
These three code examples correlate to three classes of time series simulations:
Heuristic simulations
We decide how the world should work, ensure it makes sense, and code it up, one rule at a time.
Discrete event simulations
We build individual actors with certain rules in our universe and then run those actors to see how the universe evolves over time.
Physics-based simulations
We apply physical laws to see how a system evolves over time.
Simulating time series can be a valuable analytical exercise and one we will also demonstrate in later chapters as it relates to specific models.
# What’s Special About Simulating Time Series?
Simulating data is an area of data science that is rarely taught, but which is a particularly useful skill for time series data. This follows from one of the downsides of having temporal data: no two data points in the same time series are exactly comparable since they happen at different times. If we want to think about what could have happened at a given time, we move into the world of simulation.
Simulations can be simple or complex. On the simpler side, you will encounter synthetic data in any statistics textbook on time series, such as in the form of a random walk. These are usually generated as cumulative sums of a random process (such as R’s `rnorm`) or by a periodic function (such as a sine curve). On the more complex side, many scientists and engineers make their careers out of simulating time series. Time series simulations remain an active area of research—and a computationally demanding one—in many fields, including:
• Meteorology
• Finance
• Epidemiology
• Quantum chemistry
• Plasma physics
In some of these cases, the fundamental rules of behavior are well understood, but it can still be difficult to account for everything that can happen due to the complexity of the equations (meteorology, quantum chemistry, plasma physics). In other cases, not all of the predictive variables can ever be known, and experts aren’t even sure that perfect predictions can be made due to the stochastic nonlinear nature of the systems studied (finance, epidemiology).
## Simulation Versus Forecasting
Simulation and forecasting are similar exercises. In both cases you must form hypotheses about underlying system dynamics and parameters, and then extrapolate from these hypotheses to generate data points.
Nonetheless, there are important differences to keep in mind when learning about and developing simulations rather than forecasts:
• It can be easier to integrate qualitative observations into a simulation than into a forecast.
• Simulations are run at scale so that you can see many alternative scenarios (thousands or more), whereas forecasts should be more carefully produced.
• Simulations have lower stakes than forecasts; there are no lives and no resources on the line, so you can be more creative and exploratory in your initial rounds of simulations. Of course, you eventually want to make sure you can justify how you build your simulations, just as you must justify your forecasts.
# Simulations in Code
Next we look at three examples of coding up simulations of time series. As you read these examples, consider what a wide array of data can be simulated to produce a “time series,” and how the temporal element can be very specific and human-driven, such as days of the week and times of day of donations, but can also be very nonspecific and essentially unlabeled, such as the "nth step” of a physics simulation.
The three examples of simulation we will discuss in this section are:
• Simulating a synthetic data set to test our hypotheses about how members of an organization may (or may not) have correlated behavior between receptiveness to organizational email and willingness to make donations. This is the most DIY example in that we hardcode relationships and generate tabular data with `for` loops and the like.
• Simulating the synthetic data set to explore aggregate behavior in a fleet of taxis, complete with shift times and time-of-day-dependent frequency of passengers. In this data set, we make use of Python’s object-oriented attributes as well as generators, which are quite helpful when we want to set a system going and see what it does.
• Simulating the physical process of a magnetic material gradually orienting its individual magnetic elements, which begin in disarray but ultimately coalesce into a well-ordered system. In this example, we see how physical laws can drive a time series simulation and insert natural temporal scaling into a process.
## Doing the Work Yourself
When you are programming simulations, you need to keep in mind the logical rules that apply to your system. Here we walk through an example where the programmer does most of the work of making sure the data makes sense (for example, by not specifying events that happen in an illogical order).
We start by defining the membership universe—that is, how many members we have and when each joined the organization. We also pair each member with a member status:
````## python`
`>>>` `## membership status`
`>>>` `years` `=` `[``'2014'``,` `'2015'``,` `'2016'``,` `'2017'``,` `'2018'``]`
`>>>` `memberStatus` `=` `[``'bronze'``,` `'silver'``,` `'gold'``,` `'inactive'``]`
`>>>` `memberYears` `=` `np``.``random``.``choice``(``years``,` `1000``,`
`>>>` `p` `=` `[``0.1``,` `0.1``,` `0.15``,` `0.30``,` `0.35``])`
`>>>` `memberStats` `=` `np``.``random``.``choice``(``memberStatus``,` `1000``,`
`>>>` `p` `=` `[``0.5``,` `0.3``,` `0.1``,` `0.1``])`
`>>>` `yearJoined` `=` `pd``.``DataFrame``({``'yearJoined'``:` `memberYears``,`
`>>>` `'memberStats'``:` `memberStats``})`
```
Notice that there are already many rules/assumptions built into the simulation just from these lines of code. We impose specific probabilities of the years the members joined. We also make the status of the member entirely independent on the year they joined. In the real world, it’s likely we can already do better than this because these two variables should have some connection, particularly if we want to incentivize people to remain members.
We make a table indicating when members opened emails each week. In this case, we define our organization’s behavior: we send three emails a week. We also define different patterns of members behavior with respect to email:
• Never opening email
• Constant level of engagement/email open rate
• Increasing or decreasing level of engagement
We can imagine ways to make this more complex and nuanced depending on anecdotal observations from veterans or novel hypotheses we have about unobservable processes affecting the data:
````## python`
`>>>` `NUM_EMAILS_SENT_WEEKLY` `=` `3`
`>>>` `## we define several functions for different patterns `
`>>>` `def` `never_opens``(``period_rng``):`
`>>>` `return` `[]`
`>>>` `def` `constant_open_rate``(``period_rng``):`
`>>>` `n``,` `p` `=` `NUM_EMAILS_SENT_WEEKLY``,` `np``.``random``.``uniform``(``0``,` `1``)`
`>>>` `num_opened` `=` `np``.``random``.``binomial``(``n``,` `p``,` `len``(``period_rng``))`
`>>>` `return` `num_opened`
`>>>` `def` `increasing_open_rate``(``period_rng``):`
`>>>` `return` `open_rate_with_factor_change``(``period_rng``,`
`>>>` `np``.``random``.``uniform``(``1.01``,`
`>>>` `1.30``))`
`>>>` `def` `decreasing_open_rate``(``period_rng``):`
`>>>` `return` `open_rate_with_factor_change``(``period_rng``,`
`>>>` `np``.``random``.``uniform``(``0.5``,`
`>>>` `0.99``))`
`>>>` `def` `open_rate_with_factor_change``(``period_rng``,` `fac``):`
`>>>` `if` `len``(``period_rng``)` `<` `1` `:`
`>>>` `return` `[]`
`>>>` `times` `=` `np``.``random``.``randint``(``0``,` `len``(``period_rng``),`
`>>>` `int``(``0.1` `*` `len``(``period_rng``)))`
`>>>` `num_opened` `=` `np``.``zeros``(``len``(``period_rng``))`
`>>>` `for` `prd` `in` `range``(``0``,` `len``(``period_rng``),` `2``):`
`>>>` `try``:`
`>>>` `n``,` `p` `=` `NUM_EMAILS_SENT_WEEKLY``,` `np``.``random``.``uniform``(``0``,`
`>>>` `1``)`
`>>>` `num_opened``[``prd``:(``prd` `+` `2``)]` `=` `np``.``random``.``binomial``(``n``,` `p``,`
`>>>` `2``)`
`>>>` `p` `=` `max``(``min``(``1``,` `p` `*` `fac``),` `0``)`
`>>>` `except``:`
`>>>` `num_opened``[``prd``]` `=` `np``.``random``.``binomial``(``n``,` `p``,` `1``)`
`>>>` `for` `t` `in` `range``(``len``(``times``)):`
`>>>` `num_opened``[``times``[``t``]]` `=` `0`
`>>>` `return` `num_opened`
```
We have defined functions to simulate four distinct kinds of behavior:
Members who never open the emails we send them
(`never_opens()`)
Members who open about the same number of emails each week
(`constant_open_rate()`)
Members who open a decreasing number of emails each week
(`decreasing_open_rate()`)
Members who open an increasing number of emails each week
(`increasing_open_rate()`)
We ensure that those who grow increasingly engaged or disengaged over time are simulated in the same way with the `open_rate_with_factor_change()` function via the functions `increasing_open_rate()` and `decreasing_open_rate()`.
We also need to come up with a system to model donation behavior. We don’t want to be totally naive, or our simulation will not give us insights into what we should expect. That is, we want to build into the model our current hypotheses about member behavior and then test whether the simulations based on those hypotheses match what we see in our real data. Here, we make donation behavior loosely but not deterministically related to the number of emails a member has opened:
````## python`
`>>>` `## donation behavior`
`>>>` `def` `produce_donations``(``period_rng``,` `member_behavior``,` `num_emails``,`
`>>>` `use_id``,` `member_join_year``):`
`>>>` `donation_amounts` `=` `np``.``array``([``0``,` `25``,` `50``,` `75``,` `100``,` `250``,` `500``,`
`>>>` `1000``,` `1500``,` `2000``])`
`>>>` `member_has` `=` `np``.``random``.``choice``(``donation_amounts``)`
`>>>` `email_fraction` `=` `num_emails` `/`
`>>>` `(``NUM_EMAILS_SENT_WEEKLY` `*` `len``(``period_rng``))`
`>>>` `member_gives` `=` `member_has` `*` `email_fraction`
`>>>` `member_gives_idx` `=` `np``.``where``(``member_gives`
`>>>` `>=` `donation_amounts``)[``0``][``-``1``]`
`>>>` `member_gives_idx` `=` `max``(``min``(``member_gives_idx``,`
`>>>` `len``(``donation_amounts``)` `-` `2``),`
`>>>` `1``)`
`>>>` `num_times_gave` `=` `np``.``random``.``poisson``(``2``)` `*`
`>>>` `(``2018` `-` `member_join_year``)`
`>>>` `times` `=` `np``.``random``.``randint``(``0``,` `len``(``period_rng``),` `num_times_gave``)`
`>>>` `dons` `=` `pd``.``DataFrame``({``'member'` `:` `[],`
`>>>` `'amount'` `:` `[],`
`>>>` `'timestamp'``:` `[]})`
`>>>` `for` `n` `in` `range``(``num_times_gave``):`
`>>>` `donation` `=` `donation_amounts``[``member_gives_idx`
`>>>` `+` `np``.``random``.``binomial``(``1``,` `.``3``)]`
`>>>` `ts` `=` `str``(``period_rng``[``times``[``n``]]``.``start_time`
`>>>` `+` `random_weekly_time_delta``())`
`>>>` `dons` `=` `dons``.``append``(``pd``.``DataFrame``(`
`>>>` `{``'member'` `:` `[``use_id``],`
`>>>` `'amount'` `:` `[``donation``],`
`>>>` `'timestamp'``:` `[``ts``]}))`
`>>>`
`>>>` `if` `dons``.``shape``[``0``]` `>` `0``:`
`>>>` `dons` `=` `dons``[``dons``.``amount` `!=` `0``]`
`>>>` `## we don't report zero donation events as this would not`
`>>>` `## be recorded in a real world database `
`>>>`
`>>>` `return` `dons`
```
There are a few steps we have taken here to make sure the code produces realistic behavior:
• We make the overall number of donations dependent on how long someone has been a member.
• We generate a wealth status per member, building in a hypothesis about behavior that donation amount is related to a stable amount a person would have earmarked for making donations.
Because our member behaviors are tied to a specific timestamp, we have to choose which weeks each member made donations and also when during that week they made the donation. We write a utility function to pick a random time during the week:
````## python`
`>>>` `def` `random_weekly_time_delta``():`
`>>>` `days_of_week` `=` `[``d` `for` `d` `in` `range``(``7``)]`
`>>>` `hours_of_day` `=` `[``h` `for` `h` `in` `range``(``11``,` `23``)]`
`>>>` `minute_of_hour` `=` `[``m` `for` `m` `in` `range``(``60``)]`
`>>>` `second_of_minute` `=` `[``s` `for` `s` `in` `range``(``60``)]`
`>>>` `return` `pd``.``Timedelta``(``str``(``np``.``random``.``choice``(``days_of_week``))`
`>>>` `+` `" days"` `)` `+`
`>>>` `pd``.``Timedelta``(``str``(``np``.``random``.``choice``(``hours_of_day``))`
`>>>` `+` `" hours"` `)` `+`
`>>>` `pd``.``Timedelta``(``str``(``np``.``random``.``choice``(``minute_of_hour``))`
`>>>` `+` `" minutes"``)` `+`
`>>>` `pd``.``Timedelta``(``str``(``np``.``random``.``choice``(``second_of_minute``))`
`>>>` `+` `" seconds"``)`
```
You may have noticed that we only draw the hour of the timestamp from the range of 11 to 23 (`hours_of_day = [h for h in range(11, 23)]`). We are postulating a universe with people in a very limited range of time zones or even in just a single time zone, as we do not allow hours outside the range given. Here we are building in more of our underlying model as to how users behave.
We thus expect to see unified behavior from our users as though they are all in one or a few adjoining time zones, and we are further postulating that reasonable donation behavior is for people to donate from late morning to late evening, but not overnight and not first thing when they wake up.
Finally, we put all the components just developed together to simulate a certain number of members and associated events in a way that ensures that events happen only once a member has joined and that a member’s email events have some relation (but not an unrealistically small relation) to their donation events:
````## python`
`>>>` `behaviors` `=` `[``never_opens``,`
`>>>` `constant_open_rate``,`
`>>>` `increasing_open_rate``,`
`>>>` `decreasing_open_rate``]`
`>>>` `member_behaviors` `=` `np``.``random``.``choice``(``behaviors``,` `1000``,`
`>>>` `[``0.2``,` `0.5``,` `0.1``,` `0.2``])`
`>>>` `rng` `=` `pd``.``period_range``(``'2015-02-14'``,` `'2018-06-01'``,` `freq` `=` `'W'``)`
`>>>` `emails` `=` `pd``.``DataFrame``({``'member'` `:` `[],`
`>>>` `'week'` `:` `[],`
`>>>` `'emailsOpened'``:` `[]})`
`>>>` `donations` `=` `pd``.``DataFrame``({``'member'` `:` `[],`
`>>>` `'amount'` `:` `[],`
`>>>` `'timestamp'``:` `[]})`
`>>>` `for` `idx` `in` `range``(``yearJoined``.``shape``[``0``]):`
`>>>` `## randomly generate the date when a member would have joined`
`>>>` `join_date` `=` `pd``.``Timestamp``(``yearJoined``.``iloc``[``idx``]``.``yearJoined``)` `+`
`>>>` `pd``.``Timedelta``(``str``(``np``.``random``.``randint``(``0``,` `365``))` `+`
`>>>` `' days'``)`
`>>>` `join_date` `=` `min``(``join_date``,` `pd``.``Timestamp``(``'2018-06-01'``))`
`>>>`
`>>>` `## member should not have action timestamps before joining`
`>>>` `member_rng` `=` `rng``[``rng` `>` `join_date``]`
`>>>`
`>>>` `if` `len``(``member_rng``)` `<` `1``:`
`>>>` `continue`
`>>>`
`>>>` `info` `=` `member_behaviors``[``idx``](``member_rng``)`
`>>>` `if` `len``(``info``)` `==` `len``(``member_rng``):`
`>>>` `emails` `=` `emails``.``append``(``pd``.``DataFrame``(`
`>>>` `{``'member'``:` `[``idx``]` `*` `len``(``info``),`
`>>>` `'week'``:` `[``str``(``r``.``start_time``)` `for` `r` `in` `member_rng``],`
`>>>` `'emailsOpened'``:` `info``}))`
`>>>` `donations` `=` `donations``.``append``(`
`>>>` `produce_donations``(``member_rng``,` `member_behaviors``[``idx``],`
`>>>` `sum``(``info``),` `idx``,` `join_date``.``year``))`
```
We then look at the temporal behavior of the donations to get a sense of how we might try this for further analysis or forecasting. We plot the total sum of donations we received for each month of the data set (see Figure 4-1):
````## python`
`>>>` `df``.``set_index``(``pd``.``to_datetime``(``df``.``timestamp``),` `inplace` `=` `True``)`
`>>>` `df``.``sort_index``(``inplace` `=` `True``)`
`>>>` `df``.``groupby``(``pd``.``Grouper``(``freq``=``'M'``))``.``amount``.``sum``()``.``plot``()`
```
It looks as though the number of donations and of emails opened rose over time from 2015 through 2018. This is not surprising, since the number of members also rose over time, as indicated in the cumulative sum of members and the year they joined. In fact, one built-in assumption of our model was that we got to keep a member indefinitely after they joined. We made no provision for termination other than allowing for members to open a decreasing number of emails. Even in that case, however, we left open the possibility of continued donations. We see this assumption of indefinitely continuing membership (and correlated donation behavior) in Figure 4-1. We should probably go back and refine our code, as indefinite membership and donation is not a realistic scenario.
This is not a classic time series simulation, so it may feel quite a bit more like an exercise in generating tabular data. It absolutely is that as well, but we did have to be time series–aware:
• We had to make decisions about how many time series our users were in.
• We had to make decisions about what kinds of trends we would model over time:
• In the case of email, we decided to have three trends: stable, increasing, and decreasing email open rates.
• In the case of donations, we made donations a stable behavioral pattern related to how many emails the member had ever opened in their lifetime. This included a lookahead, but since we were generating data, this was a way of deciding that a member’s overall affinity in the organization, which would lead to more emails opened, would also increase the frequency of donations.
• We had to be careful to make sure we did not have emails opened or donations made before the member joined the organization.
• We had to make sure our data did not go into the future, to make it more realistic for consumers of the data. Note that for a simulation it is fine if our data goes into the future.
But it’s not perfect. The code presented here is ungainly, and it doesn’t produce a realistic universe. What’s more, since only the programmer checked the logic, they could have missed edge cases such that events take place in an illogical order. It would be good to establish external metrics and standards of validity before running the simulation as one protection against such errors.
We need software that enforces a logical and consistent universe. We will look at Python generators as a better option in the next section.
## Building a Simulation Universe That Runs Itself
Sometimes you have a specific system and you want to set up the rules for that system and see how it rolls along. Perhaps you want to envision what a universe of independent members accessing your application will use, or you want to attempt to validate an internal theory of decision making based on posited external behavior. In these cases, you are looking to see how individual agents contribute to your aggregate metrics over time. Python is an especially good fit for this job thanks to the availability of generators. When you start building software rather than staying purely in analysis, it makes sense to move to Python even if you are more comfortable in R.
Generators allow us to create a series of independent (or dependent!) actors and wind them up to watch what they do, without too much boilerplate code to keep track of everything.
In the next code example, we explore a taxicab simulation.1 We want to imagine how a fleet of taxis, scheduled to begin their shifts at different times, might behave in aggregate. To do so, we want to create many individual taxis, set them loose in a cyber city, and have them report their activities back.
Such a simulation could be exceptionally complicated. For demonstration purposes, we accept that we will build a simpler world than what we imagine to truly be the case (“All models are wrong…”). We start by trying to understand what a Python generator is.
Let’s first consider a method I wrote to retrieve a taxi identification number:
````## python`
`>>>` `import` `numpy` `as` `np`
`>>>` `def` `taxi_id_number``(``num_taxis``):`
`>>>` `arr` `=` `np``.``arange``(``num_taxis``)`
`>>>` `np``.``random``.``shuffle``(``arr``)`
`>>>` `for` `i` `in` `range``(``num_taxis``):`
`>>>` `yield` `arr``[``i``]`
```
For those who are not familiar with generators, here is the preceding code in action:
````## python`
`>>>` `ids` `=` `taxi_id_number``(``10``)`
`>>>` `print``(``next``(``ids``))`
`>>>` `print``(``next``(``ids``))`
`>>>` `print``(``next``(``ids``))`
```
which might print out:
```7
2
5
```
This will iterate until it has emitted 10 numbers, at which point it will exit the `for` loop held within the generator and emit a `StopIteration` exception.
The `taxi_id_number()` produces single-use objects, all of which are independent of one another and keep their own state. This is a generator function. You can think of generators as tiny objects that maintain their own small bundle of state variables, which is useful when you want many objects parallel to one another, each one minding its own variables.
In the case of this simple taxi simulation, we compartmentalize our taxis into different shifts, and we also use a generator to indicate shifts. We schedule more taxis in the middle of the day than in the evening or overnight shifts by setting different probabilities for starting a shift at a given time:
````## python`
`>>>` `def` `shift_info``():`
`>>>` `start_times_and_freqs` `=` `[(``0``,` `8``),` `(``8``,` `30``),` `(``16``,` `15``)]`
`>>>` `indices` `=` `np``.``arange``(``len``(``start_times_and_freqs``))`
`>>>` `while` `True``:`
`>>>` `idx` `=` `np``.``random``.``choice``(``indices``,` `p` `=` `[``0.25``,` `0.5``,` `0.25``])`
`>>>` `start` `=` `start_times_and_freqs``[``idx``]`
`>>>` `yield` `(``start``[``0``],` `start``[``0``]` `+` `7.5``,` `start``[``1``])`
```
Pay attention to `start_times_and_freqs`. This is our first bit of code that will contribute to making this a time series simulation. We are indicating that different parts of the day have different likelihoods of having a taxi assigned to the shift. Additionally, different times of the day have a different mean number of trips.
Now we create a more complex generator that will use the preceding generators to establish individual taxi parameters as well as create individual taxi timelines:
````## python`
`>>>` `def` `taxi_process``(``taxi_id_generator``,` `shift_info_generator``):`
`>>>` `taxi_id` `=` `next``(``taxi_id_generator``)`
`>>>` `shift_start``,` `shift_end``,` `shift_mean_trips` `=`
`>>>` `next``(``shift_info_generator``)`
`>>>` `actual_trips` `=` `round``(``np``.``random``.``normal``(``loc` `=` `shift_mean_trips``,`
`>>>` `scale` `=` `2``))`
`>>>` `average_trip_time` `=` `6.5` `/` `shift_mean_trips` `*` `60`
`>>>` `# convert mean trip time to minutes`
`>>>` `between_events_time` `=` `1.0` `/` `(``shift_mean_trips` `-` `1``)` `*` `60`
`>>>` `# this is an efficient city where cabs are seldom unused`
`>>>` `time` `=` `shift_start`
`>>>` `yield` `TimePoint``(``taxi_id``,` `'start shift'``,` `time``)`
`>>>` `deltaT` `=` `np``.``random``.``poisson``(``between_events_time``)` `/` `60`
`>>>` `time` `+=` `deltaT`
`>>>` `for` `i` `in` `range``(``actual_trips``):`
`>>>` `yield` `TimePoint``(``taxi_id``,` `'pick up '``,` `time``)`
`>>>` `deltaT` `=` `np``.``random``.``poisson``(``average_trip_time``)` `/` `60`
`>>>` `time` `+=` `deltaT`
`>>>` `yield` `TimePoint``(``taxi_id``,` `'drop off '``,` `time``)`
`>>>` `deltaT` `=` `np``.``random``.``poisson``(``between_events_time``)` `/` `60`
`>>>` `time` `+=` `deltaT`
`>>>` `deltaT` `=` `np``.``random``.``poisson``(``between_events_time``)` `/` `60`
`>>>` `time` `+=` `deltaT`
`>>>` `yield` `TimePoint``(``taxi_id``,` `'end shift '``,` `time``)`
```
Here the taxi accesses generators to determine its ID number, shift start times, and mean number of trips for its start time. From there, it departs on its own individual journey as it runs through a certain number of trips on its own timeline and emits those to the client calling `next()` on this generator. In effect, this generator produces a time series of points for an individual taxi.
The taxi generator yields `TimePoint`s, which are defined as follows:
````## python`
`>>>` `from` `dataclasses` `import` `dataclass`
`>>>` `@dataclass`
`>>>` `class` `TimePoint``:`
`>>>` `taxi_id``:` `int`
`>>>` `name``:` `str`
`>>>` `time``:` `float`
`>>>` `def` `__lt__``(``self``,` `other``):`
`>>>` `return` `self``.``time` `<` `other``.``time`
```
We use the relatively new `dataclass` decorator to simplify the code (this requires Python 3.7). I recommend that all Python-using data scientists familiarize themselves with this new and data-friendly addition to Python.
# Python’s Dunder Methods
Python’s dunder methods, whose names begin and end with two underscores, are a set of built-in methods for every class. Dunder methods are called automatically in the natural course using a given object. There are predefined implementations that can be overridden when you define them for your class yourself. There are many reasons you might want to do this, such as in the case of the preceding code, where we want `TimePoint`s to be compared only based on their time and not based on their `taxi_id` or `name` attributes.
Dunder originated as an abbreviation of “double under.”
In addition to the automatically generated initializer for `TimePoint`, we need only two other dunder methods, `__lt__` (to compare `TimePoint`s) and` __str__ `(to print out `TimePoint`s, not shown here). We need comparison because we will take all `TimePoint`s produced into a data structure that will keep them in order: a priority queue. A priority queue is an abstract data type into which objects can be inserted in any order but which will emit objects in a specified order based on their priority.
# Abstract Data Type
An abstract data type is a computational model defined by its behavior, which consists of an enumerated set of possible actions and input data and what the results of such actions should be for certain sets of data.
One commonly known abstract data type is a first-in-first-out (FIFO) data type. This requires that objects are emitted from the data structure in the same order in which they were fed into the data structure. How the programmer elects to accomplish this is a matter of implementation and not a definition.
We have a simulation class to run these taxi generators and keep them assembled. This is not merely a `dataclass` because it has quite a bit of functionality, even in the initializer, to arrange the inputs into a sensible array of information and processing. Note that the only public-facing functionality is the `run()` function:
````## python`
`>>>` `import` `queue`
`>>>` `class` `Simulator``:`
`>>>` `def` `__init__``(``self``,` `num_taxis``):`
`>>>` `self``.``_time_points` `=` `queue``.``PriorityQueue``()`
`>>>` `taxi_id_generator` `=` `taxi_id_number``(``num_taxis``)`
`>>>` `shift_info_generator` `=` `shift_info``()`
`>>>` `self``.``_taxis` `=` `[``taxi_process``(``taxi_id_generator``,`
`>>>` `shift_info_generator``)` `for`
`>>>` `i` `in` `range``(``num_taxis``)]`
`>>>` `self``.``_prepare_run``()`
`>>>` `def` `_prepare_run``(``self``):`
`>>>` `for` `t` `in` `self``.``_taxis``:`
`>>>` `while` `True``:`
`>>>` `try``:`
`>>>` `e` `=` `next``(``t``)`
`>>>` `self``.``_time_points``.``put``(``e``)`
`>>>` `except``:`
`>>>` `break`
`>>>` `def` `run``(``self``):`
`>>>` `sim_time` `=` `0`
`>>>` `while` `sim_time` `<` `24``:`
`>>>` `if` `self``.``_time_points``.``empty``():`
`>>>` `break`
`>>>` `p` `=` `self``.``_time_points``.``get``()`
`>>>` `sim_time` `=` `p``.``time`
`>>>` `print``(``p``)`
```
First, we create the number of taxi generators that we need to represent the right number of taxis. Then we run through each of these taxis while it still has `TimePoint`s and push all these `TimePoint`s into a priority queue. The priority of the object is determined for a custom class such as `TimePoint` by our implementation of a `TimePoint`’s `__lt__`, where we compare start time. So, as the `TimePoint`s are pushed into the priority queue, it will prepare them to be emitted in temporal order.
We run the simulation:
````## python`
`>>>` `sim` `=` `Simulator``(``1000``)`
`>>>` `sim``.``run``()`
```
Here’s what the output looks like (your output will be different, as we haven’t set a seed—and every time you run the code it will be different from the last iteration):
```id: 0539 name: drop off time: 23:58
id: 0318 name: pick up time: 23:58
id: 0759 name: end shift time: 23:58
id: 0977 name: pick up time: 23:58
id: 0693 name: end shift time: 23:59
id: 0085 name: end shift time: 23:59
id: 0351 name: end shift time: 23:59
id: 0036 name: end shift time: 23:59
id: 0314 name: drop off time: 23:59
```
# Setting a Seed When Generating Random Numbers
When you write code that is generating random numbers, you might want to ensure that it’s reproducible (e.g., if you wanted to set up unit tests for code that is normally random or if you were trying to debug and wanted to narrow down sources of variation to make debugging easier). To ensure that random numbers come out in the same nonrandom order, you set a seed. This is a common operation, so there are guides on how to set a seed in any computer language.
We have rounded to the nearest minute for display simplicity, although we do have more fine-grained data available. What temporal resolution we use will depend on our purposes:
• If we want to make an educational display for people in our city of how the taxi fleet affects traffic, we might display hourly aggregates.
• If we are a taxicab app and need to understand load on our server, we likely want to look at minute-by-minute data or even more highly resolved data to think about our infrastructure design and capacity.
We made the decision to report taxi `TimePoint`s as they are “happening.” That is, we report the start of a taxi ride (“pick up”) without the time when the ride will end, even though we easily could have condensed this. This is one way of making the time series more realistic, in the sense that you likely would have recorded events in this way in a live stream.
Note that, as in the previous case, our time series simulation has not yet produced a time series. We have produced a log and can see our way through to making this a time series in a number of ways, however:
• Output to a CSV file or time series database as we run the simulation.
• Run some kind of online model hooked up to our simulation to learn how to develop a real-time streaming data processing pipeline.
• Save the output down to a file or database and then do more post-processing to package the data in a convenient (but possibly risky vis-à-vis lookahead) form, such as pairing together start and end times of a given ride to study how the length of a taxi ride behaves at different times of the day.
There are several advantages to simulating this data in addition to being able to test hypotheses about the dynamics of a taxi system. Here are a couple of situations where this synthetic time series data could be useful:.
• Testing the merits of various forecasting models relative to the known underlying dynamics of the simulation.
• Building a pipeline for data you eventually expect to have based on your synthetic data while you await the real data.
You will be well served as a time series analyst by your ability to make use of generators and object-oriented programming. This example offers just one example of how such knowledge can simplify your life and improve the quality of your code.
# For Extensive Simulations, Consider Agent-Based Modeling
The solution we coded here was all right, but it was a fair amount of boilerplate to ensure that logical conditions would be respected. If a simulation of discrete events based on the actions of discrete actors would be a useful source of simulated time series data, you should consider a simulation-oriented module. The SimPy module is a helpful option, with an accessible API and quite a bit of flexibility to do the sorts of simulation tasks we handled in this section.
## A Physics Simulation
In another kind of simulation scenario, you may be in full possession of the laws of physics that define a system. This doesn’t have to be physics per se, however; it can also apply to a number of other areas:
• Quantitative researchers in finance often hypothesize the “physical” rules of the market. So do economists, albeit at different timescales.
• Psychologists posit the “psychophysical” rules of how humans make decisions. These can be used to generate “physical” rules about expected human responses to a variety of options over time.
• Biologists research rules about how a system behaves over time in response to various stimuli.
One case of knowing some rules for a simple physical system is that of modeling a magnet. This is the case we are going to work on, via an oft-taught statistical mechanics model called the Ising model.2 We will look at a simplified version of how to simulate its behavior over time. We will initialize a magnetic material so that its individual magnetic components are pointing in random directions. We will then watch how this system evolves into order where all the magnetic components point in the same direction, under the action of known physical laws and a few lines of code.
Next we discuss how such a simulation is accomplished via a Markov Chain Monte Carlo (MCMC) method, discussing both how that method works in general and as applied to this specific system.
In physics, an MCMC simulation can be used, for example, to understand how quantum transitions in individual molecules can affect aggregate ensemble measurements of that system over time. In this case, we need to apply a few specific rules:
1. In a Markov process, the probability of a transition to a state in the future depends only on the present state (not on past information).
2. We will impose a physics-specific condition of requiring a Boltzmann distribution for energy; that is, $upper T Subscript i j Baseline slash upper T Subscript j i Baseline equals e Superscript minus b left-parenthesis upper E Super Subscript j Superscript minus upper E Super Subscript i Superscript right-parenthesis$. For most of us, this is just an implementation detail and not something nonphysicists need to worry about.
We implement an MCMC simulation as follows:
1. Select the starting state of each individual lattice site randomly.
2. For each individual time step, choose an individual lattice site and flip its direction.
3. Calculate the change in energy that would result from this flip given the physical laws you are working with. In this case this means:
• If the change in energy is negative, you are transitioning to a lower energy state, which will always be favored, so you keep the switch and move on to the next time step.
• If the change in energy is not negative, you accept it with the acceptance probability of $e Superscript left-parenthesis minus energy change right-parenthesis$. This is consistent with rule 2.
Continue steps 2 and 3 indefinitely until convergence to determine the most likely state for whatever aggregate measurement you are making.
Let’s take a look at the specific details of the Ising model. Imagine we have a two-dimensional material composed of a grid of objects, each one having what boils down to a mini-magnet that can point up or down. We put those mini-magnets randomly in an up or down spin at time zero, and we then record the system as it evolves from a random state to an ordered state at low temperature.3
First we configure our system, as follows:
````## python`
`>>>` `### CONFIGURATION`
`>>>` `## physical layout`
`>>>` `N` `=` `5` `# width of lattice`
`>>>` `M` `=` `5` `# height of lattice`
`>>>` `## temperature settings`
`>>>` `temperature` `=` `0.5`
`>>>` `BETA` `=` `1` `/` `temperature`
```
Then we have some utility methods, such as random initialization of our starting block:
```>>> def initRandState(N, M):
>>> block = np.random.choice([-1, 1], size = (N, M))
>>> return block
```
We also calculate the energy for a given center state alignment relative to its neighbors:
````## python`
`>>>` `def` `costForCenterState``(``state``,` `i``,` `j``,` `n``,` `m``):`
`>>>` `centerS` `=` `state``[``i``,` `j``]`
`>>>` `neighbors` `=` `[((``i` `+` `1``)` `%` `n``,` `j``),` `((``i` `-` `1``)` `%` `n``,` `j``),`
`>>>` `(``i``,` `(``j` `+` `1``)` `%` `m``),` `(``i``,` `(``j` `-` `1``)` `%` `m``)]`
`>>>` `## notice the % n because we impose periodic boundary cond`
`>>>` `## ignore this if it doesn't make sense - it's merely a `
`>>>` `## physical constraint on the system saying 2D system is like`
`>>>` `## the surface of a donut`
`>>>` `interactionE` `=` `[``state``[``x``,` `y``]` `*` `centerS` `for` `(``x``,` `y``)` `in` `neighbors``]`
`>>>` `return` `np``.``sum``(``interactionE``)`
```
And we want to determine the magnetization of the entire block for a given state:
````## python`
`>>>` `def` `magnetizationForState``(``state``):`
`>>>` `return` `np``.``sum``(``state``)`
```
Here’s where we introduce the MCMC steps discussed earlier:
````## python`
`>>>` `def` `mcmcAdjust``(``state``):`
`>>>` `n` `=` `state``.``shape``[``0``]`
`>>>` `m` `=` `state``.``shape``[``1``]`
`>>>` `x``,` `y` `=` `np``.``random``.``randint``(``0``,` `n``),` `np``.``random``.``randint``(``0``,` `m``)`
`>>>` `centerS` `=` `state``[``x``,` `y``]`
`>>>` `cost` `=` `costForCenterState``(``state``,` `x``,` `y``,` `n``,` `m``)`
`>>>` `if` `cost` `<` `0``:`
`>>>` `centerS` `*=` `-``1`
`>>>` `elif` `np``.``random``.``random``()` `<` `np``.``exp``(``-``cost` `*` `BETA``):`
`>>>` `centerS` `*=` `-``1`
`>>>` `state``[``x``,` `y``]` `=` `centerS`
`>>>` `return` `state`
```
Now to actually run a simulation, we need some recordkeeping as well as repeated calls to the MCMC adjustment:
````## python`
`>>>` `def` `runState``(``state``,` `n_steps``,` `snapsteps` `=` `None``):`
`>>>` `if` `snapsteps` `is` `None``:`
`>>>` `snapsteps` `=` `np``.``linspace``(``0``,` `n_steps``,` `num` `=` `round``(``n_steps` `/` `(``M` `*` `N` `*` `100``)),`
`>>>` `dtype` `=` `np``.``int32``)`
`>>>` `saved_states` `=` `[]`
`>>>` `sp` `=` `0`
`>>>` `magnet_hist` `=` `[]`
`>>>` `for` `i` `in` `range``(``n_steps``):`
`>>>` `state` `=` `mcmcAdjust``(``state``)`
`>>>` `magnet_hist``.``append``(``magnetizationForState``(``state``))`
`>>>` `if` `sp` `<` `len``(``snapsteps``)` `and` `i` `==` `snapsteps``[``sp``]:`
`>>>` `saved_states``.``append``(``np``.``copy``(``state``))`
`>>>` `sp` `+=` `1`
`>>>` `return` `state``,` `saved_states``,` `magnet_hist`
```
And we run the simulation:
````## python`
`>>>` `### RUN A SIMULATION`
`>>>` `init_state` `=` `initRandState``(``N``,` `M``)`
`>>>` `print``(``init_state``)`
`>>>` `final_state` `=` `runState``(``np``.``copy``(``init_state``),` `1000``)`
```
We can get some insights from this simulation by looking at the beginning and ending states (see Figure 4-2).
In Figure 4-2 we examine one randomly generated initial state. While you might expect to see the two states more mixed up, remember that probabilistically it’s not that likely to get a perfect checkerboard effect. Try generating the initial state many times, and you will see that the seemingly “random” or “50/50” checkerboard state is not at all likely. Notice, however, that we start with approximately half our sites in each state. Also realize that any patterns you find in the initial states is likely your brain following the very human tendency to see patterns even where there aren’t any.
We then pass the initial state into the `runState()` function, allow 1,000 time steps to pass, and then examine the outcome in Figure 4-3.
This is a snapshot of the state taken at step 1,000. There are at least two interesting observations at this point. First, the dominant state has reversed compared to step 1,000. Second, the dominant state is no more dominant numerically than was the other dominant state at step 1,000. This suggests that the temperature may continue to flip sites out of the dominant state even when it might otherwise be favored. To better understand these dynamics, we should consider plotting overall aggregate measurements, such as magnetization, or make movies where we can view our two-dimensional data in a time series format.
We do this with magnetization over time for many independent runs of the previous simulation, as pictured in Figure 4-4:
````## python`
`>>>` `we` `collect` `each` `time` `series` `as` `a` `separate` `element` `in` `results` `list`
`>>>` `results` `=` `[]`
`>>>` `for` `i` `in` `range``(``100``):`
`>>>` `init_state` `=` `initRandState``(``N``,` `M``)`
`>>>` `final_state``,` `states``,` `magnet_hist` `=` `runState``(``init_state``,` `1000``)`
`>>>` `results``.``append``(``magnet_hist``)`
`>>>`
`>>>` `## we plot each curve with some transparency so we can see`
`>>>` `## curves that overlap one another`
`>>>` `for` `mh` `in` `results``:`
`>>>` `plt``.``plot``(``mh``,``'r'``,` `alpha``=``0.2``)`
```
The magnetization curves are just one example of how we could picture the system evolving over time. We might also consider recording 2D time series, as the snapshot of the overall state at each point in time. Or there might be other interesting aggregate variables to measure at each step, such as a measure of layout entropy or a measure of total energy. Quantities such as magnetization or entropy are related quantities, as they are a function of the geometric layout of the state at each lattice site, but each quantity is a slightly different measure.
We can use this data in similar ways to what we discussed with the taxicab data, even though the underlying system is quite different. For example, we could:
• Use the simulated data as the impetus to set up a pipeline.
• Test machine learning methods on this synthetic data to see if they can be helpful on physical data before we go to the trouble of cleaning up real-world data for such modeling.
• Watch the movie-like imagery of important metrics to develop better physical intuitions about the system.
# Final Notes on Simulations
We have looked at a number of very different examples of simulating measurements that describe behavior over time. We have looked at simulating data related to consumer behavior (NGO membership and donation), city infrastructure (taxicab pick-up patterns), and the laws of physics (the gradual ordering of a randomized magnetic material). These examples should leave you feeling comfortable enough to begin reading code examples of simulated data and also come up with ideas for how your own work could benefit from simulations.
Chances are that, in the past, you have made assumptions about your data without knowing how to test those or alternate possibilities. Simulations give you a route to do so, which means your conversations about data can expand to include hypothetical examples paired with quantitative metrics from simulations. This will ground your discussions while opening new possibilities, both in the time series domain and in other branches of data science.
## Statistical Simulations
Statistical simulations are the most traditional route to simulated time series data. They are particularly useful when we know the underlying dynamics of a stochastic system and want to estimate a few unknown parameters or see how different assumptions would impact the parameter estimation process (we will see an example of this later in the book). Even for physical systems, sometimes the statistical simulation is better.
Statistical simulations of time series data are also quite valuable when we need to have a definitive quantitative metric to define our own uncertainty about the accuracy of our simulations. In traditional statistical simulations, such as an ARIMA model (to be discussed in Chapter 6), the formulas for the error are well established, meaning that to understand a system with a posited underlying statistical model, you do not need to run many simulations to make numerical assertions about error and variance.
## Deep Learning Simulations
Deep learning simulations for time series are a nascent but promising field. The advantages of deep learning are that very complicated, nonlinear dynamics can be captured in time series data even without the practitioner fully understanding the dynamics. This is also a disadvantage, however, in that the practitioner has no principled basis for understanding the dynamics of the system.
Deep learning simulations also offer promise where privacy is a concern. For example, deep learning has been used to generate synthetic heterogeneous time series data for medical applications based on real time series data but without the potential to leak private information. Such a data set, if it can truly be produced without any privacy leaks, would be invaluable because researchers could have access to a large array of (otherwise expensive and privacy-violating) medical data.
# More Resources
Cristóbal Esteban, Stephanie L. Hyland, and Gunnar Rätsch, “Real-Valued (Medical) Time Series Generation with Recurrent Conditional GANs,” unpublished manuscript, last revised December 4, 2017, https://perma.cc/Q69W-L44Z.
The authors demonstrate how generative adversarial networks can be used to produce realistic-looking heterogenous medical time series data. This is an example of how deep learning simulation can be used to create ethical, legal, and (hopefully) privacy-preserving medical data sets to enable wider access to useful data for machine learning and deep learning in the healthcare context.
Gordon Reikard and W. Erick Rogers, “Forecasting Ocean Waves: Comparing a Physics-based Model with Statistical Models,” Coastal Engineering 58 (2011): 409–16, https://perma.cc/89DJ-ZENZ.
Wolfgang Härdle, Joel Horowitz, and Jens-Peter Kreiss, “Bootstrap Methods for Time Series,” International Statistical Review / Revue Internationale de Statistique 71, no. 2 (2003): 435–59, https://perma.cc/6CQA-EG2E.
A classic 2005 review of the difficulties of statistical simulation of time series data given temporal dependencies. The authors explain, in a highly technical statistics journal, why methods to bootstrap time series data lag behind methods for other kinds of data, as well as what promising methods were available at the time of writing. The state of the art has not changed too much, so this is a useful, if challenging, read.
1 This example is heavily inspired by Luciano Ramalho’s book, Fluent Python (O’Reilly 2015). I highly recommend reading the full simulation chapter in that book to improve your Python programming skills and see more elaborate opportunities for agent-based simulation.
2 The Ising model is a well-known and commonly taught classical statistical mechanical model of magnets. You can find many code examples and further discussion of this model online in both programming and physics contexts if you are interested in learning more.
3 The Ising model is more often used to understand what a ferromagnet’s equilibrium state is rather than to consider the temporal aspect of how a ferromagnet might make its way into an equilibrium state. However, we treat the evolution over time as a time series.
Get Practical Time Series Analysis now with O’Reilly online learning.
O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers. | 13,200 | 51,246 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 2, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-50 | latest | en | 0.929719 |
https://byjus.com/question-answer/what-are-the-numbers-of-ways-in-which-5-similar-balls-can-be-put-in-1/ | 1,642,734,945,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00519.warc.gz | 225,313,953 | 21,404 | Question
# What are the numbers of ways in which 5 similar balls can be put in 4 distinct baskets when second basket has exactly two balls.
A
12
B
10
C
16
D
18
Solution
## The correct option is B 10 Let's number of baskets as A, B, C and D With the first constraint, 2 of balls are going to basket 2(B), hence we have to(5-2) = 3 balls left to distribute among the remaining baskets. So, A + C + D = 3 Number of possible distributions = 3+3−1C3−1=5C2 = 10 (n similar objects can be distributed to r distinct groups in n+t−1Cr−1 ways) Mathematics
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https://forums.wolfram.com/mathgroup/archive/1998/Dec/msg00251.html | 1,656,239,344,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00312.warc.gz | 316,103,092 | 8,662 | Re: Combinations
• To: mathgroup at smc.vnet.net
• Subject: [mg15217] Re: [mg15181] Combinations
• From: Jurgen Tischer <jtischer at col2.telecom.com.co>
• Date: Tue, 22 Dec 1998 04:01:46 -0500
• References: <199812180710.CAA03901@smc.vnet.net.>
• Sender: owner-wri-mathgroup at wolfram.com
```Since the problem can be stated as a set cover problem (one covers the
set of all t-subsets with the sets made out of the t-subsets of
u-subsets), Daniel's approach is the standard one. And set cover
problems are NP-hard. Since there is extra structure one can hope for
doing a little bit better. I used the suggestion of Daniel, but refined
it by selecting among the u-sets which cover a maximal number of the
remaining sets one that has a maximal overlap with already chosen
u-sets. (I found this condition by looking at lots of approximate
solutions of the greedy algorithm.) Looks like it works better, but of
course it's still a heuristic method without any proof of minimality.
And it's slow, I checked it only for very small numbers. (The time it
took me to write this little program shows how bad I am in that field.)
In[1]:= <<DiscreteMath`Combinatorica`
In[2]:= fu1[li_,ln_]:=
With[{li1=Flatten[KSubsets[#,3]&/@li,1]},Length/@(Complement[#,li1]&/@ln)]
In[3]:= fu2[elem_,li_]:=Length[Intersection[elem,li]]
In[4]:= newElem[li_,ln2_,ln21_]:=Module[{li1=fu1[li,ln21],
li2=Union[Flatten[li]],li3,m,elem},
li3=Extract[ln2,Position[li1,Max[li1]]];
m=Max[fu2[#,li2]&/@li3];
First[Select[li3,fu2[#,li2]==m&,1]]]
In[5]:=
totlen[li_,len1_]:=Length[Union[Flatten[KSubsets[#,len1]&/@li,1]]]
In[6]:= minCover[lentotal_,len1_,len2_]:=
Module[{ran=Range[lentotal],ln1,ln2,ln21,lenln1,li={Range[len2]}},
ln1=KSubsets[ran,len1];
lenln1=Length[ln1];
ln2=KSubsets[ran,len2];
ln21=KSubsets[#,len1]&/@ln2;
While[totlen[li,len1]<lenln1,AppendTo[li,newElem[li,ln2,ln21]]];
li]
In[7]:= minCover[12,3,6]
Out[7]=
{{1,2,3,4,5,6},{1,2,7,8,9,10},{3,4,7,8,11,12},{5,6,9,10,11,12},{1,2,3,9,11,
12},{1,4,5,7,10,11},{2,4,6,8,10,12},{3,5,6,7,8,9},{1,3,6,7,10,12},{1,4,5,
8,9,12},{2,3,5,8,10,11},{2,4,6,7,9,11},{1,3,6,8,9,11},{2,3,5,7,9,12},{1,2,
3,4,9,10}}
Jurgen
edsferr at uol.com.br wrote:
>
> Hello !
>
> I'm not sure if what I have is a difficult or an easy problem...
>
> Let L be this list : L={1,2,3,4,..,n} Let LNT be a list of all the
> combinations of the n elements in list L when you take t elements. We
> get LNT using KSubsets[L,t] Let u be an integer so t < u < n
> Let LNU be a list of all the combinations of the n elements in list L
> when you take u elements. We get LNU using KSubsets[L,u]
>
> Let's take one element of LNU: Due to the fact of t < u , this element
> of LNU "contains" exactly u!/(t!(u-y)!) elements of LNT when you take t
> elements of it.
>
> The question is: How can i get the shortest list of elements of LNU so
> we can "have" all the elements of LNT "inside" of this list ?
>
> Thanks !!!!!!!!!!!
>
> -----------== Posted via Deja News, The Discussion Network ==---------- | 1,089 | 3,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-27 | latest | en | 0.737916 |
https://studysoup.com/tsg/20093/university-physics-13-edition-chapter-14-problem-14e | 1,582,098,936,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00341.warc.gz | 558,289,587 | 11,231 | ×
×
# Solved: A small block is attached to an ideal spring and
ISBN: 9780321675460 31
## Solution for problem 14E Chapter 14
University Physics | 13th Edition
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5
Problem 14E
A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.70 s to travel from x = 0.090 m to x = - 0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel (a) from x = 0.180 m to x = - 0.180 m and (b) from x = 0.090 m to x = - 0.090 m?
Step-by-Step Solution:
Solution 14E Step 1 of 4: In the given problem when the small block is attached to ideal spring.When the amplitude is A=0.09 m it takes t=2.7s to travel from x=0.09m to x=-0.09m. Now when the amplitude is doubled, that is A=0.18 m, we need to calculate the time taken by the block to travel (a) From x= 0.180m to x= -0.180 m (b) From x= 0.09m to x= -0.09m Step 2 of 4: Using the relation of angular frequency, = 2f = 2 T Where is angular frequency, f is frequency and T is time period From above equation, the frequency and time period does not depend on the amplitude. Hence time period will be same for both the case Time period is the time taken to complete one complete oscillation. In the given problem, it takes t=2.7s to complete half cycle, so for one complete cycle or oscillation T=2t T=2(2.7s) T=5.4 s Therefore, the time period of the given system is 5.4s.
Step 3 of 4
Step 4 of 4
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OpenStudy (anonymous):
-12x=72
OpenStudy (anonymous):
subtract 12 by 12 and ull get 0 then divide 12 into 72 and u will get 6...!!!
OpenStudy (anonymous):
\[12 * 6=72\]
OpenStudy (anonymous):
trying to solve for x? if so divide both sides by -12 x= -6 | 86 | 280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-23 | longest | en | 0.901615 |
https://dgtal.org/doc/1.3/moduleGeodesicsInHeat.html | 1,679,434,694,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00245.warc.gz | 253,182,206 | 4,354 | DGtal 1.3.0
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Geodesics In Heat using discrete polygonal calculus
Author(s) of this documentation:
David Coeurjolly
Since
1.3
Part of package DEC package.
In this documentation page, we focus on an implementation of the "Geodesics In Heat method" ([39]). The main objective is to highlight the use of differential operators from Discrete differential calculus on polygonal surfaces to solve elementary PDEs.
Images are given by the dgtalCalculus-geodesic.cpp example file.
Warning
The implementation heavily relies on implicit operators with many Eigen based small matrix constructions, which has a huge overhead in Debug mode. Please consider to build the examples in Release (e.g. CMAKE_BUILD_TYPE variable) for high performance on large geometrical objects.
The main algorithm
The algorithm consists in three steps (see [39] for details):
• Given heat sources $$g$$ at a mesh vertices, we first solve a heat diffusion problem: Integrate the heat flow $$u$$ such that $$\Delta u = \frac{\partial u}{\partial t}$$ using a single step backward Euler step: $$(Id - t\Delta) u_t = g$$
• Evaluate the normalized field $$X = - \nabla u_t / \| \nabla u_t\|$$
• Solve the Poisson problem from the divergence of the normalized gradient field $$\Delta \phi = \nabla \cdot X$$
The computation involves discrete differential operator definitions (Laplace-Beltrami, gradient, divergence...) as well as linear solvers on sparse matrices. We do not go into the details of the discretization, please have a look to the paper if interested.
The interface
The class GeodesicsInHeat contains the implementation of the Geodesics in Heat method. It relies on the PolygonalCalculus class for the differential operators (Discrete differential calculus on polygonal surfaces).
First, we need to instantiate the GeodesicsInHeat class from an instance of PolygonalCalculus:
typedef PolygonalCalculus<SurfaceMesh<RealPoint,RealPoint>> Calculus;
Calculus aCalculus( mesh );
GeodesicsInHeat<Calculus> heat( aCalculus );
Then, we can prefactorized the solvers for a given a timestep $$dt$$:
heat.init(dt);
Note
For a discussion on the timestep please refer to [39]. For short, the authors suggest a timestep in $$dt=m\cdot h^2$$ for some constant $$m$$ and $$h$$ being the mean spacing between adjacent vertices.
Once prefactorized, we can add as many sources as we want using the method:
...
Note
the vertex index corresponds to the indexing system of the underlying SurfaceMesh instance.
The resulting geodesics diffusion is obtained by:
auto u = heat.compute();
Note
Once the init() has been called, you can iterate over addSource() and compute() for fast computations. If you want to change the timestep, you would need to call again the init() method.
Examples
From dgtalCalculus-geodesic.cpp code on a digital surface and a regularization of the digital surface (see Digital surface regularization by normal vector alignment).
Input Geodesics in heat Geodesics with multiple (random) sources | 691 | 3,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-14 | latest | en | 0.786713 |
https://talkstats.com/search/837793/ | 1,575,833,430,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514475.44/warc/CC-MAIN-20191208174645-20191208202645-00412.warc.gz | 577,748,036 | 8,865 | # Search results
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I have N bernoulli dependent variables, X1, X2, ..., XN, and Xi ~ B(1, pi), pi is already konwn. And Y=X1+X2+...+XN, I need to get the distribution of Y. If Xi, Xj is independent, then I can use the simulation, generate X1,...,XN using the bernoulli distribution, and get the Y; repeat 10000...
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I need to conduct an investigate in my province, we need to investigate the Scientific literacy of people in our province. The sampling plan I design is: 1. sampling the cities in our province. 2. sampling the countries in the cities. 3. sampling the families in the countries. 4. sampling...
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Algorithm that is 10 to 100 times faster than state of the art for determining the optimal circuit block shape in VLSI fixed-outline floor-planning
The best paper award for the IEEE International Symposium on Physical Design 2012 was given to Professor Chirs Chu at Iowa State University, who proposed an algorithm for determining the optimal circuit block shape in VLSI fixed-outline floor-planning, achieving a 10-to-100 increase over previous state-of-the-art techniques.
Specifically formulated for fixed-outline floorplanning
* Optimal, efficient and scalable for non-slicing floorplan
* Main contributions
- Basic Slack-Driven Shaping
- Three Optimality Conditions
- Slack-Driven Shaping (SDS)
* Promising Experimental Results
- Obtain optimal solutions for both MCNC & HB benchmarks simply by the basic SDS.
- For MCNC benchmarks, 253x faster than Young’s, 33x faster than Lin’s, to produce results of similar quality.
Future Work
* Embed SDS (Slack-Driven Block Shaping)into a floorplanner.
* Use the duality gap as a better stopping criterion.
* Propose a more scalable algorithm to replace the geometric programming method.
* Extend SDS (Slack-Driven Block Shaping) to handle non-fixed outline floorplanning.
* Applied on buffer/wire sizing for timing optimization
Optimal Slack-Driven Block Shaping Algorithm in Fixed-Outline Floorplanning (31 pages)
This paper presents an efficient, scalable and optimal slack-driven shaping algorithm for soft blocks in non-slicing floorplan. The proposed algorithm is called SDS. Different from all previous approaches, SDS is specifically formulated for fixed-outline floorplanning. Given a fixed upper bound on the layout width, SDS minimizes the layout height by only shaping the soft blocks in the design. Iteratively, SDS shapes some soft blocks to minimize the layout height, with the guarantee that the layout width would not exceed the given upper bound. Rather than using some simple heuristic as in previous work, the amount of change on each block is determined by systematically distributing the global total amount of available slack to individual block. During the whole shaping process, the layout height is monotonically reducing, and eventually converges to an optimal solution. We also propose two optimality conditions to check the optimality of a shaping solution. To validate the efficiency and effectiveness of SDS, comprehensive experiments are conducted on MCNC and HB benchmarks. Compared with previous work, SDS is able to achieve the best experimental result with significantly faster runtime. | 524 | 2,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-40 | latest | en | 0.851408 |
https://uk.comsol.com/blogs/simulating-eddy-current-brakes | 1,611,664,426,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00492.warc.gz | 607,579,540 | 61,076 | # Simulating Eddy Current Brakes
March 4, 2013
Last week you saw how you can simulate the heating of a car’s brake discs. This reminded me of another type of brake — the eddy current brake (also known as magnetic brake). Whereas the other model was a study in heat transfer, eddy current brakes deal with electromagnetics.
### Basic Structure of Eddy Current Brakes
Eddy current brakes basically consist of a rotating disc (made of conductive material) and a permanent magnet:
As the disc spins in the constant magnetic field generated by the permanent magnet, its conductive properties induce eddy currents. The Lorentz forces from these currents in turn slow down the disc. The most common application areas of magnetic brakes are trains, roller coasters, and aircraft. It’s not unlikely that eddy current brakes will eventually be found in cars as well.
You may have also heard of electromagnetic brakes, which are similar in design, except instead of a permanent magnet they have iron wound with a coil.
### Modeling Eddy Current Brakes
Let’s suppose you are designing an eddy current brake, and you want to know how large the permanent magnet needs to be in order to provide enough torque to slow down the vehicle (train, roller coaster, car…) in time. In this case we’re assuming the induced current distribution does not move with the rotating disc; it stays where the magnet is located.
Note that the induced Lorentz current density term often leads to confusion when modeling electromagnetics where there are moving magnetic sources or the moving domain is of bounded extent in the same direction as the motion or varies in this direction. These types of moving sources generate magnetic flux that cannot be included in the Lorentz term. To be clear, in our case the induced current distribution is stationary and does not move with the disc.
Let’s assume you have a copper disc that’s 1 cm thick, has a radius of 10 cm, and moves at an initial angular speed of 1,000 rpm. The 1 T permanent magnet is connected via an iron yoke, and there’s a 1.5 cm gap of air where the disc can spin. Using COMSOL Multiphysics and the AC/DC Module, you can figure out how much torque your brake system will have. What’s notable is that you can include the rotation of the device without having a moving mesh. The magnetic brake model couples a dynamic equation (this defines the rotation of the disc) with the finite element method (this defines the torque). This will allow you to calculate the total time to completely brake the system.
3D model showing induced eddy current density and direction at t=0 s. 3D model showing induced eddy current density and direction at t=25 s.
You can also plot the time evolution of the angular velocity, braking torque, and dissipated power in your magnetic brake system:
Time evolution of the angular velocity. Time evolution of the braking torque. Time evolution of the dissipated power.
#### Categories
##### LIM JUJOE
June 6, 2020
Hi, I have faced a problem when I change the disc material into low carbon steel with error NaN or Inf found when solving a linear system using SOR. But, it runs! if I change to other materials. Any comments or advice? Appreciate if your response.
##### Fanny Griesmer
June 8, 2020 COMSOL Employee
Hello Lim,
Hmm that’s interesting. Our Support team should be able to help you out with that. You can contact them via support@comsol.com or https://www.comsol.com/support
In the meantime, there’s another post about eddy currents that you might find interesting: https://www.comsol.com/blogs/how-eddy-current-braking-technology-is-freeing-us-from-friction/ Good luck!
EXPLORE COMSOL BLOG | 807 | 3,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-04 | latest | en | 0.941691 |
https://haifengl.github.io/api/java/smile/stat/distribution/HyperGeometricDistribution.html | 1,723,468,472,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00125.warc.gz | 217,881,694 | 5,131 | # Class HyperGeometricDistribution
java.lang.Object
smile.stat.distribution.DiscreteDistribution
smile.stat.distribution.HyperGeometricDistribution
All Implemented Interfaces:
`Serializable`, `Distribution`
public class HyperGeometricDistribution extends DiscreteDistribution
The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement, just as the binomial distribution describes the number of successes for draws with replacement.
Suppose you are to draw "n" balls without replacement from an urn containing "N" balls in total, "m" of which are white. The hypergeometric distribution describes the distribution of the number of white balls drawn from the urn.
• ## Field Summary
Fields
Modifier and Type
Field
Description
`final int`
`m`
The number of defects.
`final int`
`n`
The number of draws.
`final int`
`N`
The number of total samples.
• ## Constructor Summary
Constructors
Constructor
Description
```HyperGeometricDistribution(int N, int m, int n)```
Constructor.
• ## Method Summary
Modifier and Type
Method
Description
`double`
`cdf(double k)`
Cumulative distribution function.
`double`
`entropy()`
Returns Shannon entropy of the distribution.
`int`
`length()`
Returns the number of parameters of the distribution.
`double`
`logp(int k)`
The probability mass function in log scale.
`double`
`mean()`
Returns the mean of distribution.
`double`
`p(int k)`
The probability mass function.
`double`
`quantile(double p)`
The quantile, the probability to the left of quantile is p.
`double`
`rand()`
Uses inversion by chop-down search from the mode when the `mean < 20` and the patchwork-rejection method when the `mean >= 20`.
`String`
`toString()`
`double`
`variance()`
Returns the variance of distribution.
### Methods inherited from class smile.stat.distribution.DiscreteDistribution
`likelihood, logLikelihood, logp, p, quantile, randi, randi`
### Methods inherited from class java.lang.Object
`clone, equals, finalize, getClass, hashCode, notify, notifyAll, wait, wait, wait`
### Methods inherited from interface smile.stat.distribution.Distribution
`inverseTransformSampling, likelihood, logLikelihood, quantile, quantile, rand, rejectionSampling, sd`
• ## Field Details
• ### N
public final int N
The number of total samples.
• ### m
public final int m
The number of defects.
• ### n
public final int n
The number of draws.
• ## Constructor Details
• ### HyperGeometricDistribution
public HyperGeometricDistribution(int N, int m, int n)
Constructor.
Parameters:
`N` - the number of total samples.
`m` - the number of defects.
`n` - the number of draws.
• ## Method Details
• ### length
public int length()
Description copied from interface: `Distribution`
Returns the number of parameters of the distribution. The "length" is in the sense of the minimum description length principle.
Returns:
The number of parameters.
• ### mean
public double mean()
Description copied from interface: `Distribution`
Returns the mean of distribution.
Returns:
The mean.
• ### variance
public double variance()
Description copied from interface: `Distribution`
Returns the variance of distribution.
Returns:
The variance.
• ### entropy
public double entropy()
Description copied from interface: `Distribution`
Returns Shannon entropy of the distribution.
Returns:
Shannon entropy.
• ### toString
public String toString()
Overrides:
`toString` in class `Object`
• ### p
public double p(int k)
Description copied from class: `DiscreteDistribution`
The probability mass function.
Specified by:
`p` in class `DiscreteDistribution`
Parameters:
`k` - a real value.
Returns:
the probability.
• ### logp
public double logp(int k)
Description copied from class: `DiscreteDistribution`
The probability mass function in log scale.
Specified by:
`logp` in class `DiscreteDistribution`
Parameters:
`k` - a real value.
Returns:
the log probability.
• ### cdf
public double cdf(double k)
Description copied from interface: `Distribution`
Cumulative distribution function. That is the probability to the left of x.
Parameters:
`k` - a real number.
Returns:
the probability.
• ### quantile
public double quantile(double p)
Description copied from interface: `Distribution`
The quantile, the probability to the left of quantile is p. It is actually the inverse of cdf.
Parameters:
`p` - the probability.
Returns:
the quantile.
• ### rand
public double rand()
Uses inversion by chop-down search from the mode when the `mean < 20` and the patchwork-rejection method when the `mean >= 20`.
Returns:
a random number. | 1,082 | 4,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.670138 |
https://www.assignmentexpert.com/homework-answers/physics/mechanics-relativity/question-3016 | 1,621,302,698,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991650.73/warc/CC-MAIN-20210518002309-20210518032309-00590.warc.gz | 660,786,938 | 288,531 | 99 017
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# Answer to Question #3016 in Mechanics | Relativity for KayLina
Question #3016
Find the force of gravity between you(70.0kg) and a pencil (1.20g) at a distance of 0.500m
1
2011-06-07T06:30:17-0400
The force of gravity can be calculated by the following formula:
F = (G m1 m2) / r2 = (6.673×10-11* 70* 1.2) / 0.52 = 2.2×10-8 N.
Need a fast expert's response?
Submit order
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for any assignment or question with DETAILED EXPLANATIONS! | 189 | 550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-21 | latest | en | 0.845752 |
https://www.equationsworksheets.net/solving-systems-of-equations-in-three-variables-worksheet/ | 1,713,941,873,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00195.warc.gz | 650,448,267 | 15,312 | # Solving Systems Of Equations In Three Variables Worksheet
Solving Systems Of Equations In Three Variables Worksheet – Expressions and Equations Worksheets are designed to help kids learn faster and more effectively. They include interactive activities and questions based on the order of operations. With these worksheets, kids are able to grasp basic and advanced concepts in a quick amount of duration. These PDF resources are free to download, and can be used by your child to test math problems. These resources are beneficial for students who are in the 5th-8th grades.
These worksheets can be utilized by students from the 5th-8th grades. These two-step word puzzles are designed using decimals, fractions or fractions. Each worksheet contains ten problems. You can access them through any site that is online or print. These worksheets are an excellent opportunity to practice rearranging equations. Alongside practicing restructuring equations, they can also assist your student to understand the principles of equality as well as inverse operations.
These worksheets can be used by fifth- and eighth grade students. They are ideal for students who have difficulty calculating percentages. There are three kinds of questions you can choose from. You can select to solve one-step challenges that contain decimal or whole numbers, or utilize word-based strategies to do fractions or decimals. Each page will contain ten equations. These worksheets on Equations are suitable for students in the 5th through 8th grades.
These worksheets are a fantastic way to practice fraction calculations and other aspects of algebra. You can pick from kinds of challenges with these worksheets. You can select one that is word-based, numerical, or a mixture of both. It is crucial to select the type of problem, as each problem will be different. Each page will have ten challenges which makes them an excellent resource for students in 5th-8th grade.
These worksheets are designed to teach students about the connections between variables and numbers. The worksheets let students work on solving polynomial problems and also learn to use equations in their daily lives. These worksheets are a great method to understand equations and formulas. These worksheets will help you learn about the various types of mathematical problems as well as the different symbols that are employed to explain them.
These worksheets are great for children in the first grade. These worksheets will teach students how to solve equations and graph. The worksheets are perfect for practicing polynomial variable. They will also help you understand how to factor them and simplify them. It is possible to find a wonderful set of expressions and equations worksheets that are suitable for kids of any grade level. The best way to learn about equations is by doing the work yourself.
You will find a lot of worksheets for teaching quadratic equations. There are different worksheets that cover the various levels of equations at each grade. The worksheets are designed to assist you in solving problems in the fourth degree. Once you’ve finished a stage, you’ll begin to work on solving other types of equations. Once you have completed that, you are able to work on solving the same-level problems. You can, for example solve the same problem as an elongated one. | 620 | 3,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-18 | latest | en | 0.94939 |
https://mathlake.com/Square-Matrix | 1,719,220,955,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00808.warc.gz | 330,739,099 | 4,835 | # Square Matrix
Matrix is one of the most commonly used elements in linear algebra. Matrix is the rectangular arrangement of numbers/elements/objects. The horizontal arrangement is called the row and the vertical arrangement is the column of a matrix. The order of a matrix is determined by the number of rows by columns. Suppose a matrix has 2 rows and 3 rows of elements, then its order is 2×3. In the same way, when a matrix has an equal number of rows and columns, then the matrix is called the square matrix. In this article, you will learn the mathematical definition of a square matrix, properties of square matrix and examples in detail.
Learn: Matrices
## Square Matrix Definition
A square matrix is a matrix that has an equal number of rows and columns. In mathematics, m × m matrix is called the square matrix of order m. If we multiply or add any two square matrices, the order of the resulting matrix remains the same.
Square Matrix Example
$$X=\left[\begin{array}{lll}2 & -7 & 7 \\ 2 & 5 & 6 \\ 5 & 4 & 3\end{array}\right]$$
In the above example, we can see, the number of rows and columns are three respectively. Since the order of the matrix is 3 × 3, hence X is a square matrix. We can also find the determinant of a 3×3 square matrix here.
Again, if we take another example, such as;
$$Y=\left[\begin{array}{lll}9 & 5 & 7 \\ 3 & 8 & 9\end{array}\right]$$
Here, the number of rows is two and the number of columns is 3. Since rows and columns of matrix Y are not equal, hence it is not a square matrix.
### Square Matrix of Order 2
The square matrix of order 2 is the matrix with 2 rows and 2 columns. The general form of a 2×2 matrix or the square matrix of order 2 is given by:
$$A =\begin{bmatrix} a_{11} &a_{12} \\ a_{21}&a_{22} \end{bmatrix}$$
### Square Matrix of Order 3
A square matrix of order 3 contains 3 rows and 3 columns, that means its order is 3×3. The general form of square matrix with order 3 is given as:
$$A =\begin{bmatrix} a_{11} &a_{12} & a_{13}\\ a_{21}&a_{22} & a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix}$$
## Addition of Square Matrix
As we have already mentioned in the introduction part, that two square matrices can be added in a very simple way. Let us consider there is 3 by 3 matrix A and B whose values are given here:
$$\begin{array}{l}\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{21} & a_{22} & a_{23}\end{array}\right]+\left[\begin{array}{lll}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{21} & b_{22} & b_{13}\end{array}\right]=\left[\begin{array}{llll}a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13} \\ a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23} \\ a_{31}+b_{21} & a_{32}+b_{12} & a_{33}+b_{23}\end{array}\right] \\ For Example:\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{array}\right]+\left[\begin{array}{ccc}10 & 11 & 12 \\ 13 & 14 & 15 \\ 16 & 17 & 18\end{array}\right]=\left[\begin{array}{ccc}1+10 & 2+11 & 3+12 \\ 4+13 & 5+14 & 6+15 \\ 7+16 & 8+17 & 9+18\end{array}\right] \\ =\left[\begin{array}{ccc}11 & 13 & 15 \\ 17 & 19 & 21 \\ 23 & 25 & 27\end{array}\right]\end{array}$$
As we can see from the above example, the addition of square matrices is very important to perform. Each value in one matrix of a row is added to the other value of the same row and column of another matrix.
## Multiplication of Square Matrix
Just like the addition method, the two square matrices are multiplied in the simple way. Let us consider two 2 by 2 square matrices to be multiplied together. Hence, the resultant matrix will be:
$$\begin{array}{l}\left[\begin{array}{lll}a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]+\left[\begin{array}{lll}b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right]=\left[\begin{array}{llll}a_{11}b_{11}+ a_{12}b_{12} & a_{11}b_{12}+ a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} \end{array}\right] \end{array}$$
Note: The number of rows and columns should always be the same.
### Square Matrix Determinant
The determinant of a matrix is the scalar value or a number estimated using a square matrix. The square matrix could be any number of rows and columns such as 2×2, 3×3, 4×4, or in the form of n × n, where the number of columns and rows are equal.
Check: Determinant of a Matrix
If S is the set of square matrices, R is the set of numbers (real or complex) and f : S → R is defined by f (A) = k, where A ∈ S and k ∈ R, then f (A) is called the determinant of A. The determinant of a square matrix A is denoted by “det A” or | A |.
For a 2 by 2 matrix, the determinant is given by:
det A = a11×a22-a12×a21
Hence, we can find the determinant using this formula.
### Square Matrix Properties
Some of the important properties of square matrices are listed below:
• The number of rows and columns is equal.
• The sum of all the diagonal elements of a square matrix is called the trace of a matrix.
• If all the diagonal elements of a square matrix are equal to 1, then it is called an identity matrix.
• For a square matrix, we can perform different operations such as inverse.
• The determinant value can be calculated only for the square matrix.
• The order of transpose of a square matrix is the same as the original matrix. | 1,638 | 5,244 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-26 | latest | en | 0.857214 |
https://www.engineerlive.com/content/exploring-benefits-quasi-resonant-converters-power-supply-systems | 1,670,623,398,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00769.warc.gz | 812,956,598 | 16,368 | # Exploring benefits of quasi-resonant converters for power supply systems
Paul Boughton
Jon Harper looks at how equipment makers can bring the efficiency advantages and lower EMI of quasi-resonant power conversion to lower power systems.
Well-established quasi-resonant conversion technology has been extensively used in power supplies in the consumer arena, with flyback and buck converters improving power supply efficiency.
The principle of quasi-resonant conversion is to reduce the turn on losses of the power switch in a topology. A resonant converter(1) minimises the turn on losses and works in a very different way. One way of explaining quasi-resonant operation is to consider it as an extension of discontinuous conduction mode operation.
Fig. 1 shows the drain waveforms in a current mode flyback converter operating in discontinuous conduction mode. Only one gate pulse has been applied. During the first time interval, the drain current ramps up until the desired current level is reached. The power switch is then turned off. The leakage inductance in the flyback transformer rings with the node capacitance. This causes the leakage inductance spike which is limited by a clamp circuit. After the inductive spike has diminished, the drain voltage returns to the input voltage plus the reflected output voltage. When the current in the output diode drops to zero, the drain voltage would immediately drop to the bus voltage, if the effect of the primary inductance and the node capacitance were ignored. However, the drain voltage rings down to this level as shown in the figure.
The primary inductance and node capacitance form a resonant circuit. Taking a value of 1.4mH for the inductance and 73pF for the node capacitance gives a resonant frequency of 500kHz using the equation 4[g16]2f2LC = 1. The resonant circuit is lightly damped. We note that the resonant frequency using this approximation is independent of the input voltage and load currents.
In the case of a discontinuous conduction mode flyback converter, the MOSFET is turned on at a fixed frequency (ignoring the effect of any frequency jitter). The device is turned on, turned off when the set current level is reached, and then turned on again at a fixed time after the previous device turn on. The device turn on time is not synchronised with the drain resonance. In some cases the device may turn on when the drain voltage is lower than the bus voltage plus the reflected output voltage, and in some cases the device will turn on when the drain voltage is higher. This characteristic is often seen on the efficiency curves of discontinuous flyback converters: when driving a constant load, the efficiency will vary with the input voltage as the device turn on time moves up and down the valleys and troughs of the resonance curve.
For quasi-resonant switching, the device does not have a fixed switching frequency. Instead, the controller waits for one of the troughs in the drain voltage and then switches on. Older quasi-resonant devices designed for the colour television market always switched on the first trough. This was a good solution for colour televisions where the load is always high. However, for loads with a wide dynamic range, this presents a problem.
The time between device turn off and the first trough is fixed by the resonant frequency. The time between device turn on and turn off is set by the controller. For lighter loads, the time is smaller as less energy in required in the inductor, resulting in a shorter on time, and also a shorter output diode conduction time. So for lighter loads the frequency increases, resulting in much higher switching losses.
On the FSQ series of Fairchild Power Switches this problem is avoided by using a frequency clamp circuit. This circuit ensures that a maximum frequency is not exceeded while at the same time switching on one of the troughs.
Benefits
In comparison with both discontinuous mode and continuous mode operation of a flyback converter, quasi-resonant switching offers reduced turn on losses, resulting in increased efficiency and lower device temperature. The disadvantage of higher losses at light loads for a simple quasi-resonant circuit is removed by the frequency clamp circuit used in modern controllers or integrated power switches.
The EMI generated by the turn on process is reduced if this happens at low current and lower voltage, which is the case in quasi-resonant applications. This reduces EMI in the 1MHz to 50MHz range.
Further, there is an intrinsic frequency jitter in the quasi resonant process which spreads out the EMI noise, which further reduces filter costs. This is caused by the input voltage ripple on the bulk capacitor. Both the on time and the output diode conduction time are less at the maximum ripple voltage than at the minimum ripple voltage, for a constant load. This results in a linearly changing switching frequency with a frequency sweep equal to the ripple frequency (eg 100Hz for a full bridge rectified circuit operating from 50Hz AC). This reduces EMI in the switching frequency range 150kHz to 1MHz.
Applications
The range of quasi-resonant converters has recently been extended down to cover lower power levels (2W-50W) in response to the market demands for higher efficiency of low wattage power supplies and lower standby power.
Fig. 2 shows an example application designed in Fairchild's Global Power Resource Centre in Germany. R103-5, D104 and C103 form the extra components needed for the detection of the minimum voltage levels on the drain. It uses an error amplifier that combines the functions of a standard optocoupler and an industry standard '431 reference in one package. The remaining components are standard for a flyback converter.
The no-load standby power was measured to be less than 130mW in the 175VAC to 265VAC range used for the design. At lower input voltages the standby power is even lower for similar designs. The full load efficiency was greater than 86 per cent for the whole voltage range, which is very high for a multiple output flyback power supply at this power level. Line regulation was excellent - the measured voltages did not really change when the input voltage was modified. Load regulation for the regulated output was well within 5 per cent.
The availability of low power quasi-resonant devices has opened up interesting new possibilities. The Global Power Resource Centre has developed a quasi-resonant buck circuit for a 20V/100mA output. Here the standby power with a 10mA load was measured. The total power consumption was less than 400mW (including the 200mW base load) over the full 85VAC to 265VAC range, and less than 350mW for the range up to 180VAC. The better performance at low input voltage is due to the troughs of the drain voltage being far closer to zero than at higher input voltage.
For 85VAC to 160VAC the full load efficiency was greater than 80 per cent, dropping to 73 per cent for the range up to 265VAC. This is excellent performance for such a small power supply, entirely attributable to the use of quasi-resonant techniques. Both line and load regulation were well within 1 per cent over the full operating range. The temperature rise of the device without heatsink was measured to be 15 degrees at room temperature.
Jon Harper is Market Development Manager, Industrial & White Goods Systems for Fairchild Semiconductor Europe. www.fairchildsemi.com
REFERENCES:
1. R. W. Erickson and D. Maksimovic, Fundamentals of Power Electronics, Second Edition, Springer, 2001, Chapter 19, ISBN 0-7923-7270-0. | 1,596 | 7,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.910383 |
http://www.shanatova.info/siwug/maximum-ride-homework-help-633.php | 1,511,259,795,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806338.36/warc/CC-MAIN-20171121094039-20171121114039-00309.warc.gz | 503,547,870 | 4,588 | # Maximum ride homework help
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Bakersfield.com. Most valuable homework is that which students perceive to be meaningful, says Pope. By. | 637 | 3,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-47 | latest | en | 0.874623 |
https://cs.brown.edu/courses/cs1951x/docs/ring_theory/polynomial/dickson.html | 1,716,005,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057260.44/warc/CC-MAIN-20240518023912-20240518053912-00477.warc.gz | 165,145,273 | 35,359 | mathlibdocumentation
ring_theory.polynomial.dickson
Dickson polynomials #
The (generalised) Dickson polynomials are a family of polynomials indexed by ℕ × ℕ, with coefficients in a commutative ring R depending on an element a∈R. More precisely, the they satisfy the recursion dickson k a (n + 2) = X * (dickson k a n + 1) - a * (dickson k a n) with starting values dickson k a 0 = 3 - k and dickson k a 1 = X. In the literature, dickson k a n is called the n-th Dickson polynomial of the k-th kind associated to the parameter a : R. They are closely related to the Chebyshev polynomials in the case that a=1. When a=0 they are just the family of monomials X ^ n.
Main definition #
• polynomial.dickson: the generalised Dickson polynomials.
Main statements #
• polynomial.dickson_one_one_mul, the (m * n)-th Dickson polynomial of the first kind for parameter 1 : R is the composition of the m-th and n-th Dickson polynomials of the first kind for 1 : R.
• polynomial.dickson_one_one_char_p, for a prime number p, the p-th Dickson polynomial of the first kind associated to parameter 1 : R is congruent to X ^ p modulo p.
TODO #
• Redefine dickson in terms of linear_recurrence.
• Show that dickson 2 1 is equal to the characteristic polynomial of the adjacency matrix of a type A Dynkin diagram.
• Prove that the adjacency matrices of simply laced Dynkin diagrams are precisely the adjacency matrices of simple connected graphs which annihilate dickson 2 1.
noncomputable def polynomial.dickson {R : Type u_1} [comm_ring R] (k : ) (a : R) :
dickson is the nthe (generalised) Dickson polynomial of the k-th kind associated to the element a ∈ R.
Equations
@[simp]
theorem polynomial.dickson_zero {R : Type u_1} [comm_ring R] (k : ) (a : R) :
0 = 3 - k
@[simp]
theorem polynomial.dickson_one {R : Type u_1} [comm_ring R] (k : ) (a : R) :
theorem polynomial.dickson_two {R : Type u_1} [comm_ring R] (k : ) (a : R) :
2 = - * (3 - k)
@[simp]
theorem polynomial.dickson_add_two {R : Type u_1} [comm_ring R] (k : ) (a : R) (n : ) :
(n + 2) = polynomial.X * (n + 1) - * n
theorem polynomial.dickson_of_two_le {R : Type u_1} [comm_ring R] (k : ) (a : R) {n : } (h : 2 n) :
n = polynomial.X * (n - 1) - * (n - 2)
theorem polynomial.map_dickson {R : Type u_1} {S : Type u_2} [comm_ring R] [comm_ring S] {k : } {a : R} (f : R →+* S) (n : ) :
n) = (f a) n
@[simp]
theorem polynomial.dickson_two_zero {R : Type u_1} [comm_ring R] (n : ) :
n =
A Lambda structure on polynomial ℤ#
Mathlib doesn't currently know what a Lambda ring is. But once it does, we can endow polynomial ℤ with a Lambda structure in terms of the dickson 1 1 polynomials defined below. There is exactly one other Lambda structure on polynomial ℤ in terms of binomial polynomials.
theorem polynomial.dickson_one_one_eval_add_inv {R : Type u_1} [comm_ring R] (x y : R) (h : x * y = 1) (n : ) :
polynomial.eval (x + y) n) = x ^ n + y ^ n
theorem polynomial.dickson_one_one_mul (R : Type u_1) [comm_ring R] (m n : ) :
(m * n) = m).comp n)
The (m * n)-th Dickson polynomial of the first kind is the composition of the m-th and n-th.
theorem polynomial.dickson_one_one_comp_comm (R : Type u_1) [comm_ring R] (m n : ) :
m).comp n) = n).comp m)
theorem polynomial.dickson_one_one_char_p (R : Type u_1) [comm_ring R] (p : ) [fact (nat.prime p)] [ p] :
p = | 1,050 | 3,316 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-22 | latest | en | 0.717518 |
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# the MUs block on the buffers with infinite capacity
Experimenter
what's happening is a strange thing. If I set a infinite capacity on the buffers the MUs block on the first two buffers, and only after they are filled the MUs goes to the machines and the simulation procede.
but if I set a finite capacity it doesn't happen, and MUs goes normally in the simulation.
WHY?? (look at the attachment)
8 REPLIES 8
# Re: the MUs block on the buffers with infinite capacity
Legend
I think it happens because the Source produces parts with the setting 'Number Adjustable' and the distribution represents a range in time instead of an interval.
Thus, the creation of all parts is scheduled before any other event. And an infinite buffer right next to the source allows it to do that.
If you have a finite buffer, the source is blocked and cannot produce any more parts.
See help topic on 'Number Adjustable'.. That might be the reason for this behaviour.
# Re: the MUs block on the buffers with infinite capacity
Experimenter
but I continue not to understand why It happens, what do i have to change?
# Re: the MUs block on the buffers with infinite capacity
Legend
That depends on how you want to use the Poisson's distribution for creating the MUs.
If your intent was to have the Interval between generation of MUs to be based on the distribution, you should change the Time of Creation in the Source object to 'Interval Adjustable' in place of the 'Number adjustable'..
# Re: the MUs block on the buffers with infinite capacity
Experimenter
But this don't solve my problem. Look the attachment.
Highlighted
# Re: the MUs block on the buffers with infinite capacity
Legend
From what I understand now, is that the problem is that the rate of creating the parts is very high compared to the processing time? Please correct me if I have a wrong understanding..
The lambda parameter is set to a very low value here.
The lambda value does not represent the number of occurences per second.
It represents the time interval between occurences in seconds.
# Re: the MUs block on the buffers with infinite capacity
Phenom
@antonitrt
I had a look at both of your posted models.
You are creating 4 different parts ( A,B,C,D)
with a randomly set frequency.
Ready to leave the source, each part is directed
through the flow control ( FCSource)
either to Buffer1 or Buffer2 .
With infinite buffer capacities there is always enough buffer capacity to
receive the directed MUs.
With finite ( in this case 2X8 ) buffer capacities the source will block
as soon as one buffer is filled with 8 parts and the last part to leave
is directed to the full buffer.
The source will produce the next part as soon as the filled buffer releases on part
This is according to the set tablefiles between 20 and 36 min.
With infinite capacities it will go on (endlessly) producing parts.
I have checked for 1 min simulation time
the source produces:
16 parts ( 2 x 8 places)
1334 parts (infinite)
What you are seeing in the infinite version is just the
continuous production of (thousands of) MUs whilst of the stations are finishing
there cycle ( after 20min -36.0min) .Nothing else !
Creating these amounts of MUs takes (real) time for the system but has no effect
on the simulated data !
Therefore the suggestion I can give in this case is not to use infinite buffers
in connection with Sources with a very low cycle time as this constellation is just consuming
computer resources for not required MUs in the infinite Buffers.
Set the buffer capacity to aligned size which does not affect your Simulation.
# Re: the MUs block on the buffers with infinite capacity
Experimenter
ok. Now I've understood.
last questions I want to ask you is:
1) If i want the average processing time, have I to look in the statistic report-->working time--->mean value? Or is there an other way?
2) for the average time between two arrival, have I to llok in the statistic report og the resource---> waiting time---> mean value?
Phenom | 930 | 4,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-26 | latest | en | 0.908287 |
https://people.ucalgary.ca/~ntaucoin/Nov05.html | 1,555,894,895,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532948.2/warc/CC-MAIN-20190421235818-20190422021818-00497.warc.gz | 518,553,606 | 6,993 | What we've done so far:
Wednesday, November 2
Friday, November 4
Monday, November 7
Wednesday, November 9
Friday, November 11 - Remembrance Day
Monday, November 14
Wednesday, November 16
Friday, November 18
Monday, November 21
Wednesday, November 23
Friday, November 25
Monday, November 28
Wednesday, November 30
Wednesday, November 2
For 9.3 Lattice Energy of Ionic Compounds, concepts are important here: not numbers. Know about lattice energies being positive, and be able to compare lattices based on the charges and sizes of the ions involved.
Lattice Energy (E) = k Q1Q2
r
End of chapter exercises: 80, 85, 90, 92, 114, 120
Now that we have a firm grasp on the skill of drawing Lewis Diagrams, we can proceed onwards.
Lewis diagrams tell us about the valence electrons in a molecule, and about the connectivity.
The Valence Shell Electron Pair Repulsion Model will allow us to predict the shapes of many molecules. 10.1 Molecular Geometry
The word Pair is rather misleading in the VSEPR model. I refer to "electron Groups" (or E.G.) to make it absolutely clear in deriving the molecular shape. An E.G. can be a single bond, a double bond, a triple bond, or a lone pair.
After a little demo with people and ropes - we tried the first simplest case of 2 E.G. using my steps.
Then, a real demo about silane!
Friday, November 4
Onto 3 E.G. and 4 E. G. and considering lone pairs on central atoms!
SKILL: You should be able to draw all of the basic shapes mentioned in class
Remember! This is an extremely important SKILL
Monday, November 7
I finished the rest of the shapes you will need to know.
Exercises from Chang: 7, 10, 12, 14 (except CdCl42-)
You got back your tests - make sure you look at the breakdown on D2L!
Wednesday, November 9
10.2 Dipole Moments
Polar bonds can be examined to determine molecular polarity. Add up the dipoles, like vectors
Some molecules have polar bonds, but do not have an overall dipole moment.
Now we can relate the new skill of VSEPR back to the dipole moment of a molecule.
SKILL: Since we now know how to determine the shape of a molecule, we can figure out its dipole moment.
Friday, November 11
No class
Monday, November 14
The next goal in lectures: relate our understanding of atomic structure to our understanding of molecular structure.
• WHAT: 2 models of bonding in molecules
• WHERE are the electrons? M.O. Theory - delocalised; Valence Bond Theory - localised directly between the atoms
• WHEN does bonding happen? if atoms have orbitals with similar energy and compatible shape (symmetry)
• WHY do we care? atomic structure allowed us to understand periodic trends, and why certain atoms combine in different types of bonds; molecular structure will allow us to understand reactivity between molecules!
• HOW does bonding happen? overlaps between orbitals (the bond is more energetically favourable than the energies of the individual atoms) - for M.O. model, we will overlap wavefunctions, more about VB model later...
Note: Chang covers MORE material than Chemistry 201 students need to know, so make sure you come to lecture or get notes, because this is the only way you will know what is important here!
Wednesday, November 16
More MO theory...
Important guidelines:
1. overlap must happen - atoms need to be close enough
2. orbitals must have similar energies
3. orbitals must have similar/compatible symmetry
We can use the M.O. model for any molecule, but for this course we will only look at simple homonuclear diatomics. We drew the M.O. diagram for oxygen. Click here for the M.O. energy level diagram for O2. Electrons are added according to the rules we already know for atomic energy level diagrams.
Drawing orbital overlaps for p orbitals...
...some more bond order calculations including O2- and F2 for homework!
Friday, November 18
Remember a pi bond is not as strong as a sigma bond - this affects the reactivity!...more later on this
The M.O. diagram tells us information about a compound's reactivity: Highest Occupied Molecular Orbital reacts with electrophiles, Lowest Unoccupied Molecular Orbital reacts with nucleophiles.
Exercises: 46, 47, 48, 52, 58
Now onto a DIFFERENT model: 10.3 Valence Bond Theory
The atomic orbitals we learned earlier will not give us the geometries that we predict from VSEPR! How do we rationalize this? We use the idea of hybridisation or "mixing" of orbitals to give us hybrid orbitals that do have the appropriate orientations.
We made several types of hybrids before the demo...
Monday, November 21
Methane and Ammonia Examples: 4 sp3 orbitals in a tetrahedral arrangement
BF3:
3 sp2 orbitals in a trigonal planar arrangement with a leftover p orbital.
Sometimes, the p orbitals left over are empty and very reactive.
Definition: single bond - sigma bond - located directly between the atoms
One more example of 3 sp2 orbitals in a trigonal planar arrangement with a leftover p orbital involved in a pi bond.
Definition: pi bond - located above and below the plane of the atoms
Double bond is sigma + pi = 4 electrons total (2 sticks)
here.
Wednesday, November 23
Triple bond - sigma + 2 pi = 6 electrons total (3 sticks)
The last two types of hybridisation we will consider:
5 sp3d ( or dsp3) orbitals in a trigonal bipyramidal arrangement
6 sp3d2 (or d2sp3) orbitals in an octahedral arrangement
Remember, hybridisation is a model - the atoms don't really hybridise! Why do we bother? Because the model is simple, but extremely useful in predicting bonding and reactivity for many molecules.
We learned about the sigma-framework of molecules that have resonance structures, and that they have delocalised pi bonding (10.8 Delocalised Molecular Orbitals)
An introduction to Line Drawings - look to D2L.
Exercises: 26, 34, 36, 38, 40, 42, 44 and 63, 72, 76 (except e), 80 (expect CdBr2), 82, 88
Friday, November 25
Kinetics!
How fast can a reaction go?
Speed will depend on:
• The reaction itself
• The states of reagents
• The amounts of reagents
• The temperature
• The presence of a catalyst (more later)
13.1 The Rate of a Reaction: Rate is another word for speed - how a quantity (concentration) changes as time passes.
The rate of a reaction can be represented mathematically - 13.2 The Rate Law
For the generic reaction: Rate = k[A]x[B]y
x is the order with respect to A and is not related to a
y is the order with respect to B and is not related to b
k is called the rate constant: for a given reaction at a given temperature.
x, y, & k come from experiment, so we need to leave them as algebra until we know some experimental details
Monday, November 28
Determining the Form of the Rate Law requires the Method of Initial Rates.
SKILL: My steps overhead allowed us to work through two examples. (See link on my main page for kinetics notes and these steps).
Remember that you can work out the units of k if you know the overall order.
The rate law used so far is also known as the differential rate law since it involves a derivative.
The Integrated Rate Law allows the rate law to be converted to an equation where concentration depends on time. This is a much simpler experimental issue.
This is a good time to remember about logarithms. See Appendix in your text!
Exercises: 2, 8, 16, 18
What if a reaction has more than one reactant?
More complicated math: simplified by an experiment which holds all but one of the reactants at a large concentration (therefore relatively constant) as in Expt 5.
SKILL: The integrated rate laws can be used to determine the rate law for a reaction if concentration and time data are provided. See Expt 5!
Once the form of the integrated rate law is known, then for any time, [A] is known, or vice versa.
So: another important piece of information can be deduced - the half life of a reaction. The half life is the time required for the reaction to be 50% completed or to have 50% remaining (glass half-full/half-empty). More next day...
Wednesday, November 30
Similar calculations can be used any time the extent of reaction is known: % reacted or complete or % remaining.
SKILL: calculate concentration or time based on the integrated rate laws.
Exercises: 28, 29, 30.
In general, for kinetics problems:
• pick equations carefully - concentration vs. what?
• if rate, then Differential
• if time, then Integrated
• use proper order: look for clues in the question
• given directly in words
• given in an equation
• given through the units of k
How do reactions happen? Collision model.
Use
the Arrhenius equation to see how k depends on T.
SKILL: calculate activation energy from rate constants at different temperatures. Exercises: 34, 38, 40, 42 and a Homework problem, | 2,113 | 8,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-18 | latest | en | 0.912708 |
https://www.excelforum.com/excel-formulas-and-functions/1133885-converting-google-sheets-formula-to-excel.html | 1,695,542,060,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00791.warc.gz | 872,222,309 | 20,136 | # Converting google sheets formula to excel
1. ## Converting google sheets formula to excel
I am trying to convert a formula from a google sheets template to excel. Unfortunately the author has not had the time nor the knowledge of how to achieve this hence I am seeking anyone's expertise here. The formula is :
=iferror(if(row()<>2,INDEX(arrayformula(filter(\$E\$2:\$E4,\$C\$2:\$C4<>"",row(\$C\$2:\$C4)=max(if(\$C\$2:\$C4=C5,row(\$C\$2:\$C4),0)))) ,1),0),0)
Assume formula is on cell E5
B C D E F
Sell AAPL 200 1000 800
Sell GOOG 100 500 400
E Previous Units
F Cumulative Units
In effect what the formula seeks to achieve is calculate the sum of the holdings. ie result would show the number of APPL shares which is 800 shares and GOOG shares would be 400. It is easy to write the formula for column F however it is difficult to write the formula collating the sum of previous units in column E.
How would I approach this to achieve the same outcome in excel?
2. ## Re: Converting google sheets formula to excel
If F is easy I left it for you
For E2 I can suggest array formula
Formula:
`Please Login or Register to view this content.`
...confirmed by pressing CTRL+SHIFT+ENTER to activate the array, not just ENTER. You will know the array is active when you see curly braces { } appear around your formula. If you do not CTRL+SHIFT+ENTER you will get an error or a clearly incorrect answer.
3. ## Re: Converting google sheets formula to excel
Kaper, I believe you have found the formula successfully. I will try testing it within the re-designed excel sheet and update tomorrow.
Thanks
4. ## Re: Converting google sheets formula to excel
Kaper, it worked. Thank you so much.
One further question which will also involve an array formula with index function. (I need to really study these as coming across them too often)
I need a formula to summarise the holdings as per attached sheet.
I assume a similar formula will be used but can not tailor it correctly.
Please let me know if you can assist with this too.
Thanks
5. ## Re: Converting google sheets formula to excel
it's basically the same array formula.
in sheet2 B3:
Formula:
`Please Login or Register to view this content.`
and copy down to B4
just a reminder - array formula.
BTW. My proposition for F2 in sheet1 would be regular formula (not array):
Formula:
`Please Login or Register to view this content.`
and what is yours, as you not disclose it in the attachment, while may be somebody using http://www.excelforum.com/search.php will find this thread and may be for her/him it will not be "It is easy to write the formula for column F"?
6. ## Re: Converting google sheets formula to excel
Sorry my formula for cell F is:
Yes it is not an array as I need to just add or subtract the holdings as they are presented.
Thank you Kapar for qualifying the variation of the array. I will try it now.
7. ## Re: Converting google sheets formula to excel
Kapar,
What is the reference to ROW C1:C10 in the last formula you made =IFERROR(INDEX(Sheet1!F\$1:F\$10,LARGE(IF(Sheet1!C\$1:C\$10=A3,ROW(C\$1:C\$10),""),1)),0)
This refers to a blank space on the spreadsheet tab2. Result would be blank as is the false argument at the end which is "". Is this intentional?
8. ## Re: Converting google sheets formula to excel
Yes it's intentional
and nope, result wouldn't be blank.
C\$1:C\$10 can be empty, does not matter
but
ROW(C\$1:C\$10)
will return an array of row numbers 1,2,3...10 (despite of cells contents)
you could use
ROW(A\$1:A\$10)
or
ROW(\$1:\$10)
as well.
9. ## Re: Converting google sheets formula to excel
Thanks Kapar for your assistance. I have it working now.
10. ## Re: Converting google sheets formula to excel
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Ancient Greek Geometry Ancient Greek scholars were very interested in mathematics. One of the most famous is Euclid, who was born in 330 BC. Euclid wrote a textbook called the Elements which outlined many ideas about shapes in two and three dimensions, as well as the theory of numbers. Many of Euclid's ideas are still used today. He studied points, lines, planes, triangles, squares, circles, and spheres. He used axioms and definitions to write careful proofs. But Euclid proved ideas very differently from a modern mathematician. He used arguments based on length, area, and volume where modern mathematicians just do algebra. For instance, when he wrote about fractions, he considered the proportional lengths of different line segments. Here is a more complicated example: One of Euclid's propositions says, If there be two straight lines, and one of them be cut into any number of segments whatsoever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments. Sam wants to know what this proposition means, so he draws a rectangle. The rectangle is "contained" by the two line segments on the top and along one side. Sam cuts the top line segment into pieces. Then he cuts up the whole rectangle. But the cut-up rectangles have the same area as the original rectangle! Let's label the length of each line segment. Now we can use algebra to record Sam's example as x (a1 + a2 + a3) = xa1 + xa2 + xa3 The original proposition let us cut the original line segment into "any number of segments whatsoever". So we could use algebra to write the original proposition as x (a1 + a2 + a3 + . . . + an) = xa1 + xa2 + xa3+ . . . + xan Our algebra looks much simpler than Euclid's proposition, but it does not help Sam draw a picture. Most Greek mathematicians after Euclid wrote in his style. They made careful proofs based on axioms and definitions, and they based their arguments on length, area, and other geometrical ideas. | 470 | 2,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2017-26 | longest | en | 0.943576 |
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• Jun 22nd 2011, 11:08 PM
alexandrabel90
units
i read somewhere that a set of elements is a field iff all the elements are units except 0... but i thought that a ring R is a field iff they are commutative?
from how i see, elements that are commutative might not be units.
may i know what is wrong with my understanding?
thanks
• Jun 22nd 2011, 11:21 PM
topspin1617
Re: units
A field is a commutative ring such that every nonzero element is a unit.
$\displaystyle \mathbb{R},\mathbb{Q},\mathbb{C}$ are all fields. That all three are commutative rings is a fact from elementary algebra. Also, in all three rings, we are able to divide by any nonzero element.
• Jun 22nd 2011, 11:27 PM
Deveno
Re: units
that is almost correct. for a ring R, if every non-zero element is a unit, the proper term is division ring, as the multiplication may not be commutative.
the standard example is the quaternions $\displaystyle \mathbb{H} = \{a + bi + cj + dk: a,b,c,d \in \mathbb{R}\}$,
where $\displaystyle ij = k, jk = i, ki = j, ji = -k, kj = -i, ik = -j$ and $\displaystyle i^2 = j^2 = k^2 = -1$.
however, it is a theorem (due to Wedderburn) that every finite division ring is indeed commutative: http://math.colgate.edu/math320/dlan...Wedderburn.pdf
for an example of a finite ring that is NOT a field: consider $\displaystyle \mathbb{Z}_6$.
note that (2)(3) = 0 (mod 6), so neither 2 nor 3 has an inverse (it would be like "dividing by 0"). | 446 | 1,447 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-26 | latest | en | 0.884166 |
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# How do I determine the number of exits required for a 4 story…
Customer Question
How do I determine the...
How do I determine the number of exits required for a 4 story building (B use factor) with a building size of 100ft. x 200ft. Consider the exits access requirements as well as exits. Show how you calculate this.
Submitted: 9 years ago.Category: Home Improvement
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4/6/2009
Home Improvement Expert: Stephen Cutler, General Contractor replied 9 years ago
Stephen Cutler, General Contractor
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Hi and thanks for using JustAnswer.com. There will be 2 exits per floor. All areas must have direct unobstructed access to both exits. The fire stairway can be interior, or exterior. Self closing fire doors are required in stairways. The exits should be on opposing sides or ends of the building. 2 exits times 4 floors equals 8 exits.
I hope that this information was helpful to you. If it was please remember to click "ACCEPT" on your screen to make sure that I am paid for my efforts. By clicking ACCEPT you are not giving up your ability to ask more questions pertaining to this subject, and I will be happy to respond to these as well. Please take a moment and leave feedback, it is very important!
Customer reply replied 9 years ago
Im sorry. I neede it explained like a math problem, ex: 100 X 200= 2000 X4= 8000 divided by ? to get the # XXXXX rooms. Can you show it to me like that?
Customer reply replied 9 years ago
I am still waiting for a resonse to my last reply. I needed the answer to my question in the form of a math answer with the calculation.
Home Improvement Expert: Stephen Cutler, General Contractor replied 9 years ago
The number of rooms times 2 ways out each(access requirements), plus 2 common exits each floor times 4 floors. The number of rooms are the controlling factor and are not derived from the square footage.
Home Improvement Expert: Stephen Cutler, General Contractor replied 9 years ago
I meant number of rooms a controlling factor, which has not been provided, and so is the sticking point in proceeding further with the calculation.
Customer reply replied 9 years ago
Can I get someone else to help me with my question?
Home Improvement Expert: Stephen Cutler, General Contractor replied 9 years ago
You sure can. It looks like this is not a firecode question, but rather some kind of homework question so I sent your question to Homework category. You might want to rephrase the question eliminating references to firecode exits.
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### Topic: How do I begin solving this problem? (about calculating pressure atequilibrium) (Read 11075 times)
0 Members and 1 Guest are viewing this topic.
#### o1ocups
• Regular Member
• Posts: 97
• Mole Snacks: +0/-0
##### How do I begin solving this problem? (about calculating pressure atequilibrium)
« on: February 15, 2009, 05:03:21 PM »
Consider the following reaction:
2SO2(g) + O2(g) <-> 2SO3(g)
Kp=0.355 at 950 K
A 2.75 L reaction vessel at 950 K initially contains 0.100 mol of SO2 and 0.100 mol of O2.
Calculate the total pressure (in atmospheres) in the reaction vessel when equilibrium is reached.
So, I think I am going to make an ice table:
SO2 O2 SO3 Initial Concentration 0.036 M 0.036 M ? Change in Concentration Equilibrium Concentration
I got the initial concentrations of SO2 and O2 by dividing the number of mols by 2.75 L
But how do you know the initial concentration of SO3?
I used this equation: Kp=Kc/[(RT)^delta n] to find Kc and found it to be 0.00454
Where do I go from here? I am mainly stuck on the initial concentration of SO3.
Or do I even need to find that? I don't know what I should do.
Thanks!!!
#### Astrokel
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##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #1 on: February 15, 2009, 05:33:21 PM »
Wasn't the Kp already given? Could you write down Kp expression in term of mole fractions of the reactants and products with the total pressure, P. You have to assume that there is no product initially and that this took place under constant volume.
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#### Borek
• Mr. pH
• Deity Member
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• I am known to be occasionally wrong.
##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #2 on: February 15, 2009, 05:34:34 PM »
But how do you know the initial concentration of SO3?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info
#### o1ocups
• Regular Member
• Posts: 97
• Mole Snacks: +0/-0
##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #3 on: February 15, 2009, 05:36:32 PM »
Wasn't the Kp already given? Could you write down Kp expression in term of mole fractions of the reactants and products with the total pressure, P. You have to assume that there is no product initially and that this took place under constant volume.
Thank you!! I totally forgot that partial pressures and mole fractions are related. Let me try and see if I can get the right answers.
#### o1ocups
• Regular Member
• Posts: 97
• Mole Snacks: +0/-0
##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #4 on: February 15, 2009, 05:42:20 PM »
But how do you know the initial concentration of SO3?
But the question doesn't say that there isn't any product at the beginning of the reaction?
#### Astrokel
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##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #5 on: February 15, 2009, 05:45:46 PM »
That is the assumption you have to make to proceed the question.. Also note it must take place under constant volume.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!
#### o1ocups
• Regular Member
• Posts: 97
• Mole Snacks: +0/-0
##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #6 on: February 15, 2009, 06:10:29 PM »
So...
SO2 O2 SO3 Initial Concentration 0.036 M 0.036 M 0 M Change in Concentration -x -0.5x +x Equilibrium Concentration 0.036-x 0.036-0.5x x
kc=[SO3]^2/[SO2]^2[O2]
0.00454=x^2/(0.036-x)^2*(0.036-0.5x)
Then, when I tried to solve for x, it just got really complicated
I ended up with -0.00272x^3 - x^2 - 1.47*10^-5x +2.12*10^-7 = 0
I entered the equation into the calculator and I am pretty sure that I got it wrong again.
Is there an easier way to do it?
#### Astrokel
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##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #7 on: February 15, 2009, 06:14:42 PM »
nono, don't work with concentrations.
Ok first calculate the initial partial pressures of SO2 and O2. Then put this into your ice table and work it out with the Kp value given.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!
#### o1ocups
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• Posts: 97
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##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #8 on: February 15, 2009, 06:30:10 PM »
nono, don't work with concentrations.
Ok first calculate the initial partial pressures of SO2 and O2. Then put this into your ice table and work it out with the Kp value given.
Yeah...I forgot what I was looking for!! And no I haven't got the pressures yet. Should I use ideal gas law??
And sorry, I have to leave now, but I am going to come back later and work on this. Thank you so much for your help so far.
#### Astrokel
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##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #9 on: February 15, 2009, 06:49:41 PM »
Quote
Should I use ideal gas law??
Right!
Check this out on the part 'Determining equilibrium pressures': http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Equilibrium_Concentrations.htm
See how to you ice table with pressure question.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!
#### o1ocups
• Regular Member
• Posts: 97
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##### Re: How do I begin solving this problem? (about calculating pressure atequilibrium)
« Reply #10 on: February 16, 2009, 01:28:08 AM »
THANK YOU SO MUCH!!! | 1,750 | 6,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-10 | latest | en | 0.869323 |
https://www.studypool.com/discuss/11418112/college-algebra-23 | 1,585,429,000,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493120.15/warc/CC-MAIN-20200328194743-20200328224743-00071.warc.gz | 1,199,984,825 | 32,279 | # Algebra 20 Multiple Choice Question Answers
Anonymous
### Question Description
k2 + 3k + 2 = (k2 + k) + 2 ( __________ )
Question 1 options:
k + 5 k + 1 k + 3 k + 2
## Question 2 (5 points)
The following are defined using recursion formulas. Write the first four terms of each sequence.
a1 = 7 and an = an-1 + 5 for n ≥ 2
Question 2 options:
8, 13, 21, 22 7, 12, 17, 22 6, 14, 18, 21 4, 11, 17, 20
## Question 3 (5 points)
Write the first four terms of the following sequence whose general term is given.
an = 3n + 2
Question 3 options:
4, 6, 10, 14 6, 9, 12, 15 5, 8, 11, 14 7, 8, 12, 15
## Question 4 (5 points)
If 20 people are selected at random, find the probability that at least 2 of them have the same birthday.
Question 4 options:
≈ 0.31 ≈ 0.42 ≈ 0.45 ≈ 0.41
## Question 5 (5 points)
Use the Binomial Theorem to expand the following binomial and express the result in simplified form.
(x2 + 2y)4
Question 5 options:
x8+8x6y+24x4y2+32x2y3+16y4${x}^{8}+8{x}^{6}y+24{x}^{4}{y}^{2}+32{x}^{2}{y}^{3}+16{y}^{4}$ x8+8x6y+20x4y2+30x2y3+15y4${x}^{8}+8{x}^{6}y+20{x}^{4}{y}^{2}+30{x}^{2}{y}^{3}+15{y}^{4}$ x8+18x6y+34x4y2+42x2y3+16y4${x}^{8}+18{x}^{6}y+34{x}^{4}{y}^{2}+42{x}^{2}{y}^{3}+16{y}^{4}\phantom{\rule{0ex}{0ex}}$ x8+8x6y+14x4y2+22x2y3+26y4${x}^{8}+8{x}^{6}y+14{x}^{4}{y}^{2}+22{x}^{2}{y}^{3}+26{y}^{4}\phantom{\rule{0ex}{0ex}}$
## Question 6 (5 points)
Consider the statement "2 is a factor of n2 + 3n."
If n = 1, the statement is "2 is a factor of __________."
If n = 2, the statement is "2 is a factor of __________."
If n = 3, the statement is "2 is a factor of __________."
If n = k + 1, the statement before the algebra is simplified is "2 is a factor of __________."
If n = k + 1, the statement after the algebra is simplified is "2 is a factor of __________."
Question 6 options:
4; 15; 28; (k + 1)2 + 3(k + 1); k2 + 5k + 8$4;15;28;\left(k+1{\right)}^{}+3\left(k+1\right);{k}^{2}+5k+8$ 4; 20; 28; (k + 1)2 + 3(k + 1); k2 + 5k + 7$4;20;28;\left(k+1{\right)}^{}+3\left(k+1\right);{k}^{2}+5k+7$ 4; 10; 18; (k + 1)2 + 3(k + 1); k2 + 5k + 4$4;10;18;\left(k+1{\right)}^{}+3\left(k+1\right);{k}^{2}+5k+4$ 4; 15; 18; (k + 1)2 + 3(k + 1); k2 + 5k + 6$4;15;18;\left(k+1{\right)}^{}+3\left(k+1\right);{k}^{2}+5k+6$
## Question 7 (5 points)
Use the Binomial Theorem to expand the following binomial and express the result in simplified form.
(2x3 - 1)4
Question 7 options:
14x12−22x9+14x6−6x3+1$14{x}^{12}-22{x}^{9}+14{x}^{6}-6{x}^{3}+1$ 16x12−32x9+24x6−8x3+1$16{x}^{12}-32{x}^{9}+24{x}^{6}-8{x}^{3}+1$ 15x12−16x9+34x6−10x3+1$15{x}^{12}-16{x}^{9}+34{x}^{6}-10{x}^{3}+1$ 26x12−42x9+34x6−18x3+1$26{x}^{12}-42{x}^{9}+34{x}^{6}-18{x}^{3}+1$
## Question 8 (5 points)
Write the first four terms of the following sequence whose general term is given.
an = 3n
Question 8 options:
3, 9, 27, 81 4, 10, 23, 91 5, 9, 17, 31 4, 10, 22, 41
## Question 9 (5 points)
Use the formula for the sum of the first n terms of a geometric sequence to solve the following.
Find the sum of the first 11 terms of the geometric sequence: 3, -6, 12, -24 . . .
Question 9 options:
1045 2108 10478 2049
## Question 10 (5 points)
A club with ten members is to choose three officers—president, vice president, and secretary-treasurer. If each office is to be held by one person and no person can hold more than one office, in how many ways can those offices be filled?
Question 10 options:
650 ways 720 ways 830 ways 675 ways
## Question 11 (5 points)
Use the Binomial Theorem to find a polynomial expansion for the following function.
f1(x) = (x - 2)4
Question 11 options:
f1(x)=x4−5x3+14x2−42x+26${f}_{1}\left(x\right)={x}^{4}-5{x}^{3}+14{x}^{2}-42x+26$ f1(x)=x4−16x3+18x2−22x+18${f}_{1}\left(x\right)={x}^{4}-16{x}^{3}+18{x}^{2}-22x+18$ f1(x)=x4−18x3+24x2−28x+16${f}_{1}\left(x\right)={x}^{4}-18{x}^{3}+24{x}^{2}-28x+16$ f1(x)=x4−8x3+24x2−32x+16${f}_{1}\left(x\right)={x}^{4}-8{x}^{3}+24{x}^{2}-32x+16$
## Question 12 (5 points)
Write the first four terms of the following sequence whose general term is given.
an = (-3)n
Question 12 options:
-4, 9, -25, 31 -5, 9, -27, 41 -2, 8, -17, 81 -3, 9, -27, 81
## Question 13 (5 points)
You volunteer to help drive children at a charity event to the zoo, but you can fit only 8 of the 17 children present in your van. How many different groups of 8 children can you drive?
Question 13 options:
32,317 groups 23,330 groups 24,310 groups 25,410 groups
## Question 14 (5 points)
Write the first six terms of the following arithmetic sequence.
a1 = 5/2, d = - ½
Question 14 options:
3/2, 2, 1/2, 1, 1/4, 0 7/2, 2, 5/2, 1 ,3/2, 0 5/2, 2, 3/2, 1, 1/2, 0 9/2, 2, 5/2, 1, 1/2, 0
## Question 15 (5 points)
If two people are selected at random, the probability that they do not have the same birthday (day and month) is 365/365 * 364/365. (Ignore leap years and assume 365 days in a year.)
Question 15 options:
The first person can have any birthday in the year. The second person can have all but one birthday. The second person can have any birthday in the year. The first person can have all but one birthday. The first person cannot a birthday in the year. The second person can have all but one birthday. The first person can have any birthday in the year. The second cannot have all but one birthday.
## Question 16 (5 points)
Write the first six terms of the following arithmetic sequence.
an = an-1 - 10, a1 = 30
Question 16 options:
40, 30, 20, 0, -20, -10 60, 40, 30, 0, -15, -10 20, 10, 0, 0, -15, -20 30, 20, 10, 0, -10, -20
## Question 17 (5 points)
Find the indicated term of the arithmetic sequence with first term, a1, and common difference, d.
Find a200 when a1 = -40, d = 5
Question 17 options:
865 955 678 895
## Question 18 (5 points)
Write a formula for the general term (the nth term) of each arithmetic sequence. Do not use a recursion formula. Then use the formula for an to find a20, the 20th term of the sequence.
an = an-1 - 10, a1 = 30
Question 18 options:
an = 60 − 10n; a = −260${a}_{n}=60-10n;a=-260$ an = 70 − 10n; a = −50${a}_{n}=70-10n;a=-50$ an = 40 − 10n; a = −160${a}_{n}=40-10n;a=-160$ an = 10 − 10n; a = −70${a}_{n}=10-10n;a=-70$
## Question 19 (5 points)
How large a group is needed to give a 0.5 chance of at least two people having the same birthday?
Question 19 options:
13 people 23 people 47 people 28 people
## Question 20 (5 points)
The following are defined using recursion formulas. Write the first four terms of each sequence.
a1 = 4 and an = 2an-1 + 3 for n ≥ 2
Question 20 options:
4, 15, 35, 453 4, 11, 15, 13 4, 11, 25, 53 3, 19, 22, 53
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2325 Tutors | 2,962 | 7,569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-16 | latest | en | 0.7117 |
http://mathoverflow.net/feeds/question/39430 | 1,369,528,639,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706474776/warc/CC-MAIN-20130516121434-00073-ip-10-60-113-184.ec2.internal.warc.gz | 178,779,323 | 5,550 | Algebraic Attacks on the Odd Perfect Number Problem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:37:01Z http://mathoverflow.net/feeds/question/39430 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39430/algebraic-attacks-on-the-odd-perfect-number-problem Algebraic Attacks on the Odd Perfect Number Problem Cam McLeman 2010-09-20T20:27:15Z 2012-08-23T12:04:48Z <p>The <a href="http://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers" rel="nofollow">odd perfect number</a> problem likely needs no introduction. Recent progress (where by recent I mean roughly the last two centuries) seems to have focused on providing restrictions on an odd perfect number which are increasingly difficult for it to satisfy (for example, congruence conditions, or bounding by below the number of distinct prime divisors it must have). By reducing the search space in this manner, and probably due to other algorithmic improvements (factoring, parallelizing, etc.), there has also been significant process improving lower bounds for the size of such a number. A link off of <a href="http://oddperfect.org" rel="nofollow">oddperfect.org</a> claims to have completed the search up to $10^{1250}$. </p> <p>But, assuming my admittedly cursory reading of the landscape is correct, none of the current research seems particularly equipped to prove non-existence. The only compelling argument I've seen on this front is "Pomerance's heuristic" (also described on <a href="http://oddperfect.org" rel="nofollow">oddperfect.org</a>). Worse, and maybe this is really the point of this question, it would be a little disappointing if the non-existence proof was an upper bound of $10^{1250}$ (depending on the techniques used to get the bound) combined with the above brute force search. </p> <p>On the other hand, maybe there's some hope that some insight can be gained into the sum-of-divisors function by modern techniques. For example, the values of the arithmetic functions $$\sigma_{k}(n):=\sum_{d\mid n}d^k,$$ for $k\geq 3$ odd, arise as coefficients of normalied Eisenstein modular forms, and the study of said forms gives amazing proofs of amazing identities between them. For $k=1$, the case of interest, the normalized Eisenstein series $E_2$ is only "quasi-modular", but such forms satisfy sufficiently nice transformation properties that I wonder if $E_2$ has anything to say about the problem. </p> <blockquote>Since no doubt many people on this site will be able to immediately address the previous idea (so please do!), my more general question is whether or not there are applications of the modern machinery of modular forms, mock modular forms, diophantine analysis, Galois representations, abc conjecture, etc., that have anything to say about the odd perfect number problem. Does it descend from or relate to any major open problems from modern algebraic/analytic number theory? </blockquote> <p><sub> Aside: I hope this does not come off as dismissive of "elementary" techniques, or of the algorithmic ones mentioned in the first paragraph. Indeed, they have, to my knowledge, been the only source of progress on this problem, and certainly contain interesting mathematics. Rather, this phrasing stems from my desire to find anything in the intersection of "odd perfect number theory" and "things I know anything about," and perhaps a desire to see the odd perfect number problem settled without the use of a beyond-gigantic brute force search. </sub></p> http://mathoverflow.net/questions/39430/algebraic-attacks-on-the-odd-perfect-number-problem/39439#39439 Answer by Pace Nielsen for Algebraic Attacks on the Odd Perfect Number Problem Pace Nielsen 2010-09-20T21:46:27Z 2010-09-20T21:46:27Z <p>This is a problem I have thought alot about. I have not seen any of the modern techniques in your list applied to the problem. Part of the issue is that if you represent $\sigma(n)=2n$ as a Diophantine equations in $k$ variables (corresponding to the prime factors--but allowing the powers to vary) then there are lots of solutions (just not where all the variables are simultaneously prime). So the usual methods of trying to show non-existence of solutions just don't cut it. Historically, this multiplicative approach is the one many people have taken, because at least some progress can be made on the problem. My personal feeling is that maybe someday these bounding computations will be tweaked to the point that they lead to the discovery of some principle that will solve the problem. For example, in one of my recent papers, I was led to consider the gcd of $a^m-1$ and $b^n-1$ (where $a$ and $b$ are distinct primes). I would conjecture that this gcd has small prime factors unless $m$ or $n$ is huge. If that happens, many of the computations related to bounding OPNs become much easier.</p> <p>I have occasionally thought about whether modular forms might say something about this topic (which is why I'm currently sitting in on my colleague's course). Instead of $\sigma(n)$, the `right' function to consider is $\sigma_{-1}(n)=\sigma(n)/n$ and I don't know off the top of my head if it appears in connection with (weakly holomorphic) modular forms. But I know there are some nice techniques about multiplicative functions that decrease over the primes, etc...</p> http://mathoverflow.net/questions/39430/algebraic-attacks-on-the-odd-perfect-number-problem/39779#39779 Answer by Jerald Jetson for Algebraic Attacks on the Odd Perfect Number Problem Jerald Jetson 2010-09-23T17:43:48Z 2010-09-23T17:43:48Z <p>The odd perfect number problem does have a connection to modular forms. the divisor funct can be written as a function of the tau function and sigma_{k}(n) = sum_{d|n} d^k. The earlier example is the van der Pol identity. This was used by Touchard to conclude that n = 36a + 9 or 12b + 1.</p> <p>The modular for should be normalized to use sigma(n)/n = 2 to see if it gives a contradiction. </p> http://mathoverflow.net/questions/39430/algebraic-attacks-on-the-odd-perfect-number-problem/39781#39781 Answer by Jamie Weigandt for Algebraic Attacks on the Odd Perfect Number Problem Jamie Weigandt 2010-09-23T18:00:35Z 2010-09-23T18:00:35Z <p>For links between perfect numbers and the ABC conjecture, see <a href="http://www.math.dartmouth.edu/~carlp/LucaPomeranceNYJMstyle.pdf" rel="nofollow">this paper</a> by Luca and Pomerance. </p> http://mathoverflow.net/questions/39430/algebraic-attacks-on-the-odd-perfect-number-problem/43942#43942 Answer by Jose Arnaldo Dris for Algebraic Attacks on the Odd Perfect Number Problem Jose Arnaldo Dris 2010-10-28T05:35:43Z 2011-01-05T14:11:33Z <p>I agree with Pace - the correct function to consider would be the abundancy index instead of the sigma function itself. In a certain sense, the abundancy index value of 2 for perfect numbers (odd or even) has served as a baseline on which various other properties and concepts related to number perfection were developed.</p> <p>Additionally, although Balth. Van der Pol was able to derive the congruence conditions mentioned, for an OPN N (which effectively gave us two cases: case (1) $3 | N$, case (2) $N$ is not divisible by $3$) using a recursion relation satisfied by the sigma function (and this was achieved via a nonlinear partial differential equation), the same result follows from eliminating the case $N \equiv 2\pmod 3$ and then using the Chinese Remainder Theorem afterwards.</p> <p>Though I have seen modular forms before (having done a [modest] exposition of elliptic curve theory and related topics) for my undergraduate thesis, I have likewise not seen any 'objects' used in Wiles' FLT proof applied directly to the OPN conjecture -- at least, not 'directly' in the literal sense of the word.</p> <p>Just some thoughts about OPNs, now that you've mentioned them:</p> <p>A closer look at the (multiplicative) forms of even and odd perfect numbers gives you:</p> <p>Even PN = ${2^{p - 1}}(2^p - 1)$ = (even power of a "small" prime) x ("big" prime)</p> <p>Odd PN = ${m^2}{q^k}$ = (even power of several primes) x ("another" prime "power")</p> <p>Ronald Sorli conjectured in his Ph. D. thesis (titled Algorithms in the Study of Multiperfect and Odd Perfect Numbers, completed in 2003) that for an OPN, it was in fact (numerically) plausible that $k = 1$. (Compare that with the Mersenne prime. Incidentally, we call $q$ the Euler prime!) </p> <p>If Sorli's conjecture is proven true, we will have (considerably) more information about an OPN's structure that could enable a quick resolution about its conjectured nonexistence.</p> <p>Let me know if you need more information.</p> http://mathoverflow.net/questions/39430/algebraic-attacks-on-the-odd-perfect-number-problem/48570#48570 Answer by Steve Elmore for Algebraic Attacks on the Odd Perfect Number Problem Steve Elmore 2010-12-07T15:53:52Z 2010-12-08T02:09:23Z <p>I had an opportunity to work with Dr. Beauregard Stubblefield 35 years ago (yep, 62 and proud of it) and he generously gave the credit for the group's result to Dr. Mary Buxton and me. Dr. Stubblefield of course did most of the work, and he and I discussed many things. It seems that Leopold Kronecker, the great German algebraist 1823-1891, had it right the whole time. He proved that X = p^e + p^(e+1) + ... + p^2 + p + 1, where e is one less than a prime and p is a prime, cannot be algebraically reduced. But we knew that. The big deal is that the numeric factors of the expression are either (e + 1) || X or (k(e + 1) + 1) | X. Dr. Stubblefield found the pattern and I found that Kronecker had proved it. Stubblefield used the result in his Proposition 11, which he proved. Thirty-five years ago we used the result to factor many sigma(p^e). We talked about extending the result back then so we both had deep input into the new theory. Recently, after an engineering career in the auto industry in Detroit, I extended Proposition 11 to apply to many more cases in several different ways. I think, Cam, that this is the road you are suggesting. Steve Elmore</p> http://mathoverflow.net/questions/39430/algebraic-attacks-on-the-odd-perfect-number-problem/105319#105319 Answer by Sylvain JULIEN for Algebraic Attacks on the Odd Perfect Number Problem Sylvain JULIEN 2012-08-23T12:04:48Z 2012-08-23T12:04:48Z <p>In fact, it might be possible to prove that there is no OPN provided one could prove that if $1+p+...p^k=q^n$, where $p$ and $q$ are odd primes and $k\ge 2$, then $p\lt q$. The trick is to write an OPN $m$ as $p_1^{e_1}...p_k^{e_k}$ and to consider the map $\tilde{\sigma}$ which maps each $p_{i}$ to $R(\sigma(p_i^{e_i}))$, with $R(n)$ the product of the primes dividing $n$. </p> | 2,767 | 10,702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2013-20 | latest | en | 0.866358 |
https://stats.stackexchange.com/questions/89216/coefficient-marginal-to-interactions-in-linear-regression | 1,582,685,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146186.62/warc/CC-MAIN-20200226023658-20200226053658-00172.warc.gz | 553,940,809 | 31,813 | # Coefficient marginal to interactions in linear regression
Consider this model:
$$y = \beta_0 +\beta_1x_1 + \beta_2x_2 + \beta_3x_1x_2 + \varepsilon$$
Somebody told me today that the coefficient for the main effect of $x_1$ (i.e., $\beta_1$) will be 'marginal to the interaction'.
Could somebody explain what 'marginal to the interaction' means?
I wouldn't phrase it that way. The common meaning of "marginal" in statistics (and the way I use it) is roughly 'ignoring'. In other words, the marginal effect of $x_1$ is the effect of a 1-unit change in $x_1$ on $y$ ignoring all other possible variables. That is not what $\beta_1$ represents in that model. (Note that they may be using the terminology differently than I do, however; that happens.) To understand the idea of marginal (in the sense of ignoring) in regression better, it may help to read my answer here.
Instead, I would say the main effect of $x_1$ (i.e., $\beta_1$) in your model is the effect of a 1-unit change in $x_1$ on $y$ when $x_2=0$. To get a clearer sense of this, it may help to read my answer here (that answer in particular explains in detail what $\beta_1$ means when an interaction is present). | 329 | 1,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-10 | latest | en | 0.907592 |
https://tacticaltradingstrategies.com/options-spreadsheet/options-trading-pricing-formula-inputs-and-outputs/ | 1,638,584,061,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00543.warc.gz | 594,892,313 | 9,001 | # Options Trading Pricing Formula Inputs And Outputs
In this video we are going into the options trading spreadsheet and discuss the worksheet called Greeks. This worksheet contains the option pricing formula that allows us to calculate theoretical value, the option greeks, and option implied volatility.
When you look at the worksheet you can see the column names for the inputs, which are the blue cells:
• Date
• Expiration mult – the number we use for days to expiration
• Underlying
• Option type
• Strike price
• Volatility
Rate, which is a pricing formula input is a yellow cell, and you can see it is filled in as .25%. The reason for this is interest rates change so infrequently that I have that as the input I have been using for months.
And you can see the formulas and outputs, which are the white cells – the key outputs are:
• Theoretical value
• Delta
• Gamma
• Vega
• Theta
At the bottom of the spreadsheet, you can see an input for option price and an output for implied volatility.
Implied volatility uses all the base inputs from the options pricing formula except volatility – instead it uses the actual option price to back out the volatility that is being implied by that price.
Before we start filling in all of the pricing formula inputs, let’s talk about volatility.
All the inputs are straightforward, with the exception of what number to use for volatility. Volatility is a measure for uncertainty and a very key component for pricing options, as well as making option trading decisions.
Consider you buy a put, you know that part [if not all] of that price is time value, because of not knowing how much the underlying price will move before expiration. For that option to be profitable at expiration, the underlying price must go down further than the volatility priced into the option price.
So, do we use historical volatility, which gives us a measurement for how much the underlying has been moving in the past? Or do we use implied volatility, which gives us a measurement for the amount of volatility to expect going forward?
For our trading strategies that are often using weekly options with shorter term expirations, I have felt that using implied volatility is a better input than historical volatility.
That being said, it is also important to know whether the relationship between historical volatility calculated for the previous 20-30 days and implied volatility are typical.
For instance, the volatility for Facebook options with 10-14 days to expiration is around 45%. But I am currently looking at the ATM call that expires in 3.5 days, and it has an implied volatility of 156%. This huge difference is coming from Facebook earnings on 10/30, which is before the option expiration.
I would not want to be doing any option price projections, or making any trade decisions, using 45% as my volatility input.
### Options Greeks Worksheet Inputs And Outputs
Start filling in the spreadsheet with the date you are modelling for and the expiration date.
You will see that days to expiration is actually 1 day short when you include the current date and expiration date – so we will use the exp mult [expiration multiplier] as our days to expiration input.
In this case, there are 9 full days [not 8] between the dates, but you can use partial days as your input. Just be sure that you do not enter 0 for your input, because the formulas cannot calculate.
Skip the volatility input for now, and then fill in the underlying price, option type, and option strike price.
After that is done, go to the bottom and enter the option price from your quotes.
This will give you the implied volatility – use that for your volatility input. When you now enter volatility, you will see your output for theoretical value and the option greeks. | 783 | 3,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-49 | longest | en | 0.914067 |
https://stat.ethz.ch/pipermail/r-help/2004-December/062974.html | 1,582,320,389,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145538.32/warc/CC-MAIN-20200221203000-20200221233000-00024.warc.gz | 561,626,303 | 1,909 | # [R] take precisely one named argument
Gabor Grothendieck ggrothendieck at myway.com
Fri Dec 17 16:31:52 CET 2004
```Robin Hankin <r.hankin <at> soc.soton.ac.uk> writes:
:
: Hi
:
: I want a function that takes precisely one named argument and no
: unnamed arguments. The named argument must be one of "a" or "b".
:
: If "a" is supplied, return "a". If "b" is supplied, return 2*b.
: That is, the desired behaviour is:
:
: R> f(a=4) #return 4
: R> f(b=33) #return 66
: R> f(5) #error
: R> f(a=3,b=5) #error
: R> f(a=3,q=3) #error
: R> f(q=3) #error
:
: The following function is intended to implement this:
:
: f <- function(a=NULL, b=NULL){
: if(!xor(is.null(a), is.null(b))){stop("specify exactly one of a and
: b")}
: if(is.null(a)){return(2*b)}else{return(a)}
: }
:
: It almost works, but f(6) returns 6 (and should be an error).
:
: What is the best way to accomplish my desired behaviour?
Here is one way to do it. nm are the names (where the [-1] removes
the function name). The ... traps any arg that is not a or b and the
stopifnot conditions ensure that exactly one of a and b are specified.
ff <- function(..., a = 0, b = 0) {
nm <- names(match.call()[-1])
stopifnot(length(nm) == 1, nm %in% c("a", "b"))
a+2*b
}
One thing to watch out for is that if
``` | 439 | 1,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-10 | latest | en | 0.600909 |
https://gradeup.co/mathematics-notes-geometry-notes-on-circle-i | 1,618,300,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072175.30/warc/CC-MAIN-20210413062409-20210413092409-00013.warc.gz | 375,608,806 | 50,267 | # Mathematics Notes: Geometry Notes on Circle
Updated : December 22nd, 2020
Share via |
In this article, we will discuss the Circles and various theorem related to it. The circle is the most important topic of Geometry. In this article, I have covers tangents, chords and some important short trick also.
## CIRCLE:
1. A circle is a set of point or locus of a point which are at a fixed distance from a point called as a centre.
2. The distance of any point on the circumference of the circle from the centre of the circle is equal.
Diameter:
Diameter is double of the radius .i.e. D=2R.
Diameter is the chord which passes through the centre of the circle.
Tangent: A line which touch the circle at only one point at its circumference.
Secant: A line which touch the circle at two distinct points.
Chord: A line Segment which lie inside the circle and its end points are always lie on the circle.
## Important theorem and results of the circle:
(1) Of the two chords of the circles, the one which is greater is nearer to the centre.
(2) The perpendicular from the centre of the circle bisect the chord i.e. radius always bisect the chord if perpendicular
(3) Any line segment joining the centre of the circle and the mid point of the chord is perpendicular to the chord.
If AM=MB then OM is perpendicular to the AB.(reference to above fogure)
(4)Equal chord of the circle always subtends the equal angles at the centre of the circle.
i.e. if AB=CD then ∠1=∠2
(5) If angle subtended by the two chord at the centre are equal then the chords are always equal.
(3) Equal chords of the circle are at equal distance from the centre.
(6) Chords which are at equidistant from the centre of the circle are always equal.
(7) Angle subtended by any arc at the centre of the circle is double the angle subtended by it at any point on the remaining part of the circle.
∠x=2∠y
(8) Angle subtended by an arc in the same segment of the circle are equal.
(9) Angle in a semi circle is a right angle. i.e the angle subtended by the diameter is always right angle.
(10) The circle drawn with hypotenuse of a right angle triangle as diameter, passes through its opposite vertex.
(11) The sum of the opposite angle of a cyclic quadrilateral is always 180°.
Cyclic quadrilateral is that quadrilateral whose all point lie on the circumference of the circle.
∠A+∠C=∠B+∠D
(12) If a side of a cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.
∠1=∠2
## Some important points of tangents
(1)Tangent and radius always make the angle of 90 at the point of meeting of tangent with the circle.
If AB is a tangent at P, then OP is perpendicular to AB.
(2) The length of the two tangent drawn from the same external point to a circle is always equal.
AP=AQ
(3) If two chords AB and CD intersect internally or externally at point P then.
PA*PB=PC*PD
(4) If PAB is a secant which intersects the circle at A and B and PT be a tangent at T, then
PT2=PA*PB
(5) If from the point of contact of tangent with circle, a chord is drawn ,then the angles which the chord makes with the tangent line are equal respectively to the angles formed in the corresponding alternate segment.
∠BAT=∠BCA=∠1
∠BAP=∠BDA=∠2
(6) If two circles touch each other internally or externally the point of contacts lies on the line joining their centres.
Distance between the cntres
When touch internally, distance=AP-BP
When touch externally, distance =AP+BP
## Some Important Results:
(1)If two tangent PA and PB are drawn from the external point P, then
∠1=∠2 and ∠3=∠4
OP is perpendicular to AB and AC=BC
(2) r1 and r2 are the radius of two circles and d is the distance between the centres of the circle then the length of the common tangent of two circles is given by
(3) If r1 and r2 are the two radius of the circle and "d" is the distance between them then the length of the transverse common tangent is given by
(4) If a circle touches all the four sides of a quadrilateral then the sum of opposite pair of sides are equal.
(5) If two chords AB and AC of a circle are equal then the bisector of ∠BAC passes through the centre O of the cirlcle.
(6) The equilateral formed by the angle bisector of a cyclic quadrilateral is also cyclic.
(7) If a cyclic trapezium is isosceles then its diagonals are equal
i.e if AB parallel DC and AD=BC then AC=BD
(8) Angle in the major segment of a circle is acute and angle in minor segment is obtuse.
(9) If two circles of same radius r are such that the centre of one lies on the circumference of the other then the length of the common chord is given by l=√3*r
(10) If 2a and 2b are length of two chords which intersects at right angle and if the distance between the centre of the circle and intersecting point of the chords is C then the radius of circle is given by]
(11) If three circles of radius r are bound by a rubber band then the length of rubber band is given by
6r+2πr
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# CPM Educational Program
## HOMEWORK SOLVED
There are 45 students in a college computer science course. For how many visits did the health club charge Julia last month? What is the maximum number of large plants that he brought? Last summer Wanda painted a house that had a surface of square feet. Each can of paint she used covers square feet. How much more did Wanda spend for the finish paint? Shane made the following assertion: All numbers that are divisible by 4 are even numbers.
Write Shane's assertion in if-then form b. Check Shane's assertion using three different numbers. By Pavel Pichardo with 0 comments. PHP vs Ruby vs Python. Added Feb 19, , Under: Do we have the courage to face this? Added Jun 27, , Under: Added Mar 13, , Under: Added May 19, , Under: By Pavel Pichardo with 2 comments. Turning Ideas into Action. Added Mar 21, , Under: By Pavel Pichardo with 1 comment. Solving Systems by Graphing Solving Systems by Addition Solving Systems by Substitution Number and Value Word Problems Wind and Current Word Problems Digit Word Problems The Product Rule The Power Rule The Quotient Rule Numerical Bases and Exponents of Zero Combining Exponent Rules Monomials, Polynomials, Binomials, Trinomials First Outer Inner Last Multiplying Binomials, Two Variables Greatest Common Factor Factoring out the Greatest Common Factor Factoring Trinomials with Positive Constants Factoring Trinomials with Negative Constants Difference of Two Squares Beginning Polynomial Equations Intermediate Polynomial Equations Simplifying Rational Expressions Multiplying and Dividing Rational Expressions Adding Rational Expressions Subtracting Rational Expressions Binomials and Trinomials in Denominators Binomials and Trinomials in Denominators F.
Cube Root of Negative 8. Linear or Quadratic Functions Graphs of Quadratic Functions and Beginning Transformations Graphs of Quadratic Functions and Advanced Transformations Solving Quadratic Equations by Graphing Taking the Square Root of Both Sides Completing The Square Variables, Expressions, and Equations | 450 | 2,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-22 | latest | en | 0.87657 |
http://mathoverflow.net/questions/95092?sort=newest | 1,371,685,385,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709805610/warc/CC-MAIN-20130516131005-00024-ip-10-60-113-184.ec2.internal.warc.gz | 160,092,596 | 10,680 | MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
Linear characterization of inverse of Stieltjes matrix
Is there a linear characterization of being the inverse of a Stieltjes matrix? In other words, if $A$ is a $n \times n$ matrix over the reals, is there a set of linear equations in the entries of $A$ such that $A$ is a the inverse of a Stieltjes matrix if and only if these linear conditions are satisfied?
-
What is "Stieltjes matrix" ? – Alexander Chervov Apr 25 2012 at 18:55
As far as I know, an exact characterization of the form you want is unknown. A necessary condition, for an inverse M-matrix (weaker than inverse Stieltjes) is the so-called "path product condition" - see http://www.math.temple.edu/~abed/JS07.pdf.
Another necessary condition is that the principal minors be all positive (and they are multilinear in the entries): see for example: http://mathoverflow.net/questions/14987/inverse-m-matrix
ADDITION: There is an "if and only if" characterization of the sort you want for inverse M-matrices (well, almost, since it's multilinear, but maybe that's what you meant :). See Theorem 2.9.1 in the new survey by Johnson & Smith:
http://www.sciencedirect.com/science/article/pii/S0024379511001273 Charles R. Johnson & Ronald L. Smith, Inverse M-matrices, II, Linear Algebra and its Applications, Volume 435, Issue 5, 1 September 2011, Pages 953–983
Let $A \geq 0$. $A$ is an inverse $M$-matrix iff:
(a) $A$ has at least one diagonal positive entry
(b) all Schur complements of order 2 are nonnegative
(c) all Schur complements of order 1 are positive
-
A partial answer is given in this paper.:
A linear algebra proof that the inverse of a strictly ultra metric matrix is a strictly diagonally dominant Stieltjes matrix (Nabben + Varga, SIAM journal of matrix analysis, 1994).
- | 510 | 1,920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2013-20 | latest | en | 0.885887 |
https://books.google.no/books?id=mNs2AAAAMAAJ&dq=editions:UOM39015067252117&hl=no&output=html_text&lr= | 1,632,201,733,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00152.warc.gz | 194,471,631 | 11,545 | # The Elements of Plane and Spherical Trigonometry: And Its Application to Astronomy, Dialling, and Trigonometrical Surveying. With Plates. Designed for Mathematical Students
T. Ostell and Company, 1841 - 191 sider
### Hva folk mener -Skriv en omtale
Vi har ikke funnet noen omtaler på noen av de vanlige stedene.
### Populære avsnitt
Side 58 - A sphere is a solid bounded by a curved surface, every point of which is equally distant from a point within called the center.
Side 55 - From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple equal...
Side 31 - THEOREM I. The sides of a plane triangle are proportional to the sines of their opposite angles.
Side 66 - That is, the sines of the sides of a spherical triangle are proportional to the sines of the opposite angles.
Side 60 - A great circle may be drawn through any two points on the surface of a sphere, but not through more than two, taken at random.
Side 57 - Required the horizontal distance of the vessel, and the height of the promontory above the level of the sea, the light-house being 85 feet high. Ans. Distance 5296.4 feet, height 251.3 feet. Prob. 11. An observer, seeing a cloud in the west, measured its angle of elevation, and found it to be 64°. A second observer, situated half a mile due east from the first station, and on the same...
Side 54 - What is the perpendicular height of a hill ; its angle of elevation, taken at the bottom of it, being 46°, and 200 yards farther off, on a level with the bottom, the angle was 31°?
Side 57 - ... it is required from these measures to determine the magnitude of the whole earth, and the utmost distance that can be seen on its surface from the top of the mountain, supposing the form of the earth to be perfectly...
Side x - CB : CA : : sin A : sin B. For, with A as a centre, and AD equal to the less side...
Side 57 - Required the distance from A to B. Ans. 345.5 yards. Prob. 10. From the top of a light-house, the angle of depression of a ship at anchor was 3° 38', and at the bottom of the light-house the angle of depression was 2° 43'. | 558 | 2,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-39 | latest | en | 0.89484 |
https://www.solutionsrecovery.com/stormwater-pollution-prevention-plans-sampling-sample-collection/ | 1,722,781,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00690.warc.gz | 777,500,488 | 42,048 | ## Stormwater Pollution Prevention Plans, Sampling & Sample Collection
The Environmental Protection Agency (EPA) in 1990, published final regulations for the National Pollutant Discharge Elimination System (NPDES) that establish stormwater permit requirements. The NPDES regulates the discharges of stormwater to waters of the United States from construction projects that cover five or more acres of soil disturbance.
Facility Closure Management tasks requires a company with vast experience in equipment decommissioning and dismantling. SRI has a working relationship with number of agencies in California and is expert in meeting your budget and schedule for facility closure.
Industrial facilities with wastewater treatment systems are also required to have Spill Prevention, Control and Countermeasure Plan (SPCC) to identify and eliminate possible non-authorized discharges to the storm drain.
The NPDES regulations require that a Stormwater Pollution Prevention Plan (SWPPP) be prepared for each site that would include the following components:
## Waste Water Neutralization Calculations_U S
##### Calculations
gmp
WW Flow Rate, Qww =
gal/day
Moles/L of OH needed =
moles/L
Incoming pH, pHin =
NaOH rate needed =
moles/day
Desired pH, pHout =
Caustic solution rate =
lb/day
Caustic sol’n % NaOH =
Caustic solution rate =
gal/day
Caustic sol’n density, ρ =
lb/gal
Daily cost of caustic =
\$/day
Caustic solution cost =
\$/lb
##### Inputs
gmp
WW Flow Rate, Qww =
gal/day
Incoming pH, pHin =
Desired pH, pHout =
Caustic sol’n % NaOH =
Caustic sol’n density, ρ =
lb/gal
Caustic solution cost =
\$/lb
##### Calculations
Moles/L of OH needed =
moles/L
NaOH rate needed =
moles/day
Caustic solution rate =
lb/day
Caustic solution rate =
gal/day
Daily cost of caustic =
\$/day
##### Calculations
gmp
WW Flow Rate, Qww =
gal/day
Moles/L of H+ needed =
moles/L
Incoming pH, pHin =
H2SO4 rate needed =
moles/day
Desired pH, pHout =
Acid solution rate =
lb/day
Acid sol’n % H2SO4 =
Acid solution rate =
gal/day
Acid sol’n density, ρ =
lb/gal
Daily cost of acid =
\$/day
Acid solution cost =
\$/lb
##### Inputs
gmp
WW Flow Rate, Qww =
gal/day
Incoming pH, pHin =
Desired pH, pHout =
Acid sol’n % H2SO4 =
Acid sol’n density, ρ =
lb/gal
Acid solution cost =
\$/lb
##### Calculations
Moles/L of H+ needed =
moles/L
H2SO4 rate needed =
moles/day
Acid solution rate =
lb/day
Acid solution rate =
gal/day
Daily cost of acid =
\$/day | 653 | 2,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.865345 |
https://gaming.stackexchange.com/questions/85354/how-quickly-do-crew-suffocate-without-oxygen/85462 | 1,620,917,361,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989814.35/warc/CC-MAIN-20210513142421-20210513172421-00368.warc.gz | 293,226,086 | 39,276 | # How quickly do crew suffocate without oxygen?
I'm in a situation where I'd like to teleport my men onto a drone, bust up their shields, then teleport back. As you may know, however, drones have no oxygen. I'm worried about how much damage my crew will take before the teleport cooldown depletes.
How many hitpoints do crewmen lose per second while in rooms with zero life support?
• It varies by species, but I lost an Engie and a Human while my first-level teleporter recharged. – rsegal Sep 22 '12 at 20:00
I applied science with the following result:
It takes 15 seconds for a 100 HP crewmember entering a low oxygen room or a room running out of oxygen to die. This means they lose health at a rate of 6.67 HP/s. This rate appears to be constant across all races, meaning
• Rockmen get killed after 22.5 seconds
• Humans, Mantis, Engi, Slugs get killed after 15 seconds
• Zoltans after 10.5 seconds.
4 Humans, 2 Engi, a Rockman and a Mantis were killed performing this experiment.
• The crystal races have a reduced suffocation rate. Do you think you could run the same test on them and include the results in the answer? – IQAndreas Jan 4 '14 at 18:15
• So... how long does a fully upgraded crew transport take to recharge? – Cory Klein Jan 20 '14 at 1:42
• If the in-game numbers are correct, Crystal beings should last 37.5 seconds. The teleporter recharges in 20, 15 or 10 seconds (at 1, 2 or 3 power, respectively). Since a level 2 teleporter is sufficient for most races, I suppose transport time cuts about a second off the time your crew spends suffocating. – Brilliand Apr 11 '14 at 20:23
Asphyxiation does as much damage as a level 1 med-bay heals. More importantly, a level 1 teleporter will not recharge before most crew members will die (not sure about rockmen or crystalmen). A level 2 teleporter will recharge before a 100 hp crew member dies (not sure about a zoltan).
Crystal crew will have 60 health left when the level 1 teleporter is back online.
• This does not answer the question. What is the rate? – Jjed Sep 22 '12 at 20:57
• Between 5 and 6.7 health per second. Why do you need an exact rate? – ILMTitan Sep 22 '12 at 23:51
Crystal crew will still have 60 health left when the level 1 teleporter is back online when entering an oxygen depleted room. | 603 | 2,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.913984 |
https://sscstudy.com/boat-and-stream-questions-pdf/?amp=1 | 1,719,081,221,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00446.warc.gz | 482,235,568 | 26,291 | # Boat and Stream Questions PDF
Boat and Stream Questions with solutions PDF for free download for SSC CGL, CHSL, Bank and all competitive exams. MCQs from previous year exam question paper of exam conducted by Staff Selection Commission. All questions with answer and detail explanation for preparation of upcoming exams.
### Upstream and Downstream
• A man rows upstream 36 km and downstream 48 km, he takes 6 hours each time. The speed of the current is
• A boat takes half time in moving a certain distance downstream than upstream. The ratio of the speed of the boat in still water and that of the current is
• A person can row a distance of one km upstream in ten minutes and downstream in four minutes. What is the speed of the stream?
• A man can swim 3 km/hour in still water. If the velocity of the stream is 2 km/hour, then the time taken by him to swim to a place 10 km upstream and back is
### When a Boat/Man Takes n Times as Much as Time in Going the Same Distance in Opposite Direction
• A boat can travel with a speed of 13 km/hr in still water. If the speed of stream is 4 km/hr in the same direction, then the time taken by boat to go 63 km in opposite direction is
• The speed of a boat in still water is 6 kmph and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 22.5 km and comes back to the starting point. The total time taken by him is | 335 | 1,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.949453 |
https://oeis.org/A057668 | 1,721,298,145,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00614.warc.gz | 372,612,403 | 4,235 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A057668 Number of minimal 7-covers of a labeled n-set. 1
1, 988, 549102, 226064280, 76785889587, 22762819040676, 6092115565691584, 1505097773271664000, 348617485585838373333, 76564317282173987801964, 16080209472530744351164146, 3250906483045575317042337960, 635954979082842132795003641239 (list; graph; refs; listen; history; text; internal format)
OFFSET 7,2 LINKS Table of n, a(n) for n=7..19. Eric Weisstein's World of Mathematics, Minimal cover FORMULA a(n) = (1/7!) * (127^n - 7 * 126^n + 21 * 125^n - 35 * 124^n + 35 * 123^n - 21 * 122^n + 7 * 121^n - 120^n). G.f.: x^7 / ((120*x-1)*(121*x-1)*(122*x-1)*(123*x-1)*(124*x-1)*(125*x-1)*(126*x-1)*(127*x-1)). - Colin Barker, Jul 11 2013 CROSSREFS Cf. A000392, A003468, A016111, A046166-A046169, A005783-A005786, A055066 (unlabeled case). Sequence in context: A235545 A232429 A232844 * A217719 A223156 A087702 Adjacent sequences: A057665 A057666 A057667 * A057669 A057670 A057671 KEYWORD easy,nonn AUTHOR Vladeta Jovovic, Oct 16 2000 EXTENSIONS Additional term from Colin Barker, Jul 11 2013 STATUS approved
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Last modified July 18 06:17 EDT 2024. Contains 374377 sequences. (Running on oeis4.) | 528 | 1,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.495858 |
https://www.bartleby.com/questions-and-answers/consider-a-simple-economy-that-produces-two-goods-pens-and-erasers.-the-following-table-shows-the-pr/4e0a3d23-ddd9-4a20-b5c8-4010f82fac62 | 1,586,061,088,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00045.warc.gz | 794,264,270 | 32,702 | # Consider a simple economy that produces two goods: pens and erasers. The following table shows the prices and quantities of the goods over a three-year period.YearPensErasersPriceQuantityPriceQuantity(Dollars per pen)(Number of pens)(Dollars per eraser)(Number of erasers)201511151165201621354225201741004175 From 2016 to 2017, nominal GDP , and real GDP . The inflation rate in 2017 was . Why is real GDP a more accurate measure of an economy's production than nominal GDP?Real GDP does not include the value of intermediate goods and services, but nominal GDP does. Real GDP includes the value of exports, but nominal GDP does not. Real GDP is not influenced by price changes, but nominal GDP is.
Question
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Consider a simple economy that produces two goods: pens and erasers. The following table shows the prices and quantities of the goods over a three-year period.
Year
Pens
Erasers
Price
Quantity
Price
Quantity
(Dollars per pen)
(Number of pens)
(Dollars per eraser)
(Number of erasers)
2015 1 115 1 165
2016 2 135 4 225
2017 4 100 4 175
From 2016 to 2017, nominal GDP , and real GDP .
The inflation rate in 2017 was .
Why is real GDP a more accurate measure of an economy's production than nominal GDP?
Real GDP does not include the value of intermediate goods and services, but nominal GDP does.
Real GDP includes the value of exports, but nominal GDP does not.
Real GDP is not influenced by price changes, but nominal GDP is.
check_circle
Step 1
Nominal GDP is the market value of all final goods and services produced by a nation at the current market prices. It does not take inflation in account and can be calculated by adding revenue generated from all the goods and services produced in an year. Nominal GDP for all the three years is calculated as follows-
Step 2
From 2016 to 2017, the nominal GDP decreased from \$1,170 to \$1,100.
Step 3
Real GDP is market value of all goods and services evaluated at base year prices. Thus it is also called inflation adjusted GD...
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Tagged in | 566 | 2,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-16 | latest | en | 0.873347 |
https://www.geeksforgeeks.org/program-to-print-inverted-left-half-pyramid-pattern-star-pattern/ | 1,713,955,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819089.82/warc/CC-MAIN-20240424080812-20240424110812-00786.warc.gz | 716,372,698 | 46,000 | # Program to Print Inverted Left Half Pyramid Pattern (Star Pattern)
Last Updated : 18 Jan, 2024
Given an integer N, the task is is to print a left half pyramid pattern with N rows. In this pattern, the first row contains N stars, the second row contains N – 1 stars, and so forth until the Nth row, which contains only 1 star. All the stars are aligned to the right.
Examples:
Input: 3
Output:
***
**
*
Input: 5
Output:
*****
****
***
**
*
Approach:
The problem can be solved using two nested loops inside another loop. The outer loop will run for the rows and the first inner loop will print the spaces and second loop will print stars. If we observe carefully, if we have an inverted left half pyramid pattern with N rows, the 1st row will have 0 space followed by N stars, the 2nd row will have 1 space followed by (N – 1) stars, the third row will have 2 spaces followed by (N – 2) stars and so on. So, Nth row will have (N – 1) spaces followed by 1 star.
Step-by-step approach:
• Run an outer loop from i = 1 to the number of rows N.
• Run an inner loop from j = 1 to i – 1.
• Print an empty space (‘ ‘) in each iteration of the inner loop.
• Run an inner loop from j = 1 to N – i + 1.
• Print an asterisk (‘*’) in each iteration of the inner loop.
• Print a newline character (“\n”) to move to the next row.
• After N iterations, we will have the inverted left half pyramid pattern.
Below is the implementation of the above approach:
## C++
`#include ` `using` `namespace` `std;` `int` `main()` `{` ` ``// Number of rows` ` ``int` `N = 5;` ` ``// Outer loop runs N times, once for each row` ` ``for` `(``int` `i = 1; i <= N; i++) {` ` ``// Inner loop prints 'i - 1' spaces` ` ``for` `(``int` `j = 1; j <= i - 1; j++) {` ` ``cout << ``" "``;` ` ``}` ` ` ` ``// Inner loop prints 'N - i + 1' stars` ` ``for` `(``int` `j = 1; j <= N - i + 1; j++) {` ` ``cout << ``"*"``;` ` ``}` ` ``// Move to the next line` ` ``cout << ``"\n"``;` ` ``}` ` ``return` `0;` `}`
## Java
`import` `java.util.Scanner;` `public` `class` `Main {` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``// Number of rows` ` ``int` `N = ``5``;` ` ``// Outer loop runs N times, once for each row` ` ``for` `(``int` `i = ``1``; i <= N; i++) {` ` ``// Inner loop prints 'i - 1' spaces` ` ``for` `(``int` `j = ``1``; j <= i - ``1``; j++) {` ` ``System.out.print(``" "``);` ` ``}` ` ` ` ``// Inner loop prints 'N - i + 1' stars` ` ``for` `(``int` `j = ``1``; j <= N - i + ``1``; j++) {` ` ``System.out.print(``"*"``);` ` ``}` ` ``// Move to the next line` ` ``System.out.println();` ` ``}` ` ``}` `}`
## Python3
`# Number of rows` `N ``=` `5` `# Outer loop runs N times, once for each row` `for` `i ``in` `range``(``1``, N ``+` `1``):` ` ``# Inner loop prints 'i - 1' spaces` ` ``for` `j ``in` `range``(``1``, i):` ` ``print``(``" "``, end``=``"")` ` ` ` ``# Inner loop prints 'N - i + 1' stars` ` ``for` `j ``in` `range``(``1``, N ``-` `i ``+` `2``):` ` ``print``(``"*"``, end``=``"")` ` ``# Move to the next line` ` ``print``()`
## C#
`using` `System;` `class` `Program` `{` ` ``static` `void` `Main()` ` ``{` ` ``// Number of rows` ` ``int` `N = 5;` ` ``// Outer loop runs N times, once for each row` ` ``for` `(``int` `i = 1; i <= N; i++)` ` ``{` ` ``// Inner loop prints 'i - 1' spaces` ` ``for` `(``int` `j = 1; j <= i - 1; j++)` ` ``{` ` ``Console.Write(``" "``);` ` ``}` ` ``// Inner loop prints 'N - i + 1' stars` ` ``for` `(``int` `j = 1; j <= N - i + 1; j++)` ` ``{` ` ``Console.Write(``"*"``);` ` ``}` ` ``// Move to the next line` ` ``Console.WriteLine();` ` ``}` ` ``}` `}`
## Javascript
`// Number of rows` `const N = 5;` `// Outer loop runs N times, once for each row` `for` `(let i = 1; i <= N; i++) {` ` ``// Inner loop prints 'N - i' spaces` ` ``for` `(let j = 1; j <= i-1; j++) {` ` ``process.stdout.write(``" "``);` ` ``}` ` ``// Inner loop prints 'i' stars` ` ``for` `(let j = 1; j <= N-i+1; j++) {` ` ``process.stdout.write(``"* "``);` ` ``}` ` ``// Move to the next line` ` ``process.stdout.write(``"\n"``);` `}`
Output
```*****
****
***
**
*
```
Time Complexity: O(N2), where N is the number of rows in the pattern.
Auxiliary Space: O(1) | 1,649 | 4,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-18 | latest | en | 0.786628 |
http://sightworder.com/library/topics-in-applied-macroeconomics | 1,669,699,000,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00718.warc.gz | 48,241,132 | 9,083 | # Topics in Applied Macroeconomics by David F. Heathfield (eds.)
By David F. Heathfield (eds.)
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Extra info for Topics in Applied Macroeconomics
Example text
24 Topics in Applied Macroeconomics Lagged Variables In many theories, the dependent variable is not only affected by current values of other variables, but also by lagged values of the variables. Current consumption might be affected by one's current income and also lagged values of income so that we have 00 c, = a + i=Q L bi Y'-i + U,. 32) Direct estimation of the equation cannot be made since there are an infinite number of the coefficients (bis) to estimate. Even if we are willing to ignore the effects of income far in the past, there may still be too many coefficients to estimate.
Interpolation is done by various proxies for value added, for example labour input, raw material inputs and gross output. These proxies are crucial for production-function analysis since they are justified precisely by an assumed production function. Estimating functions as if the index were an independent measure of output may be misleading in simply revealing the kind of 36 Topics in Applied Macroeconomics production function assumed by the compiler of the statistics. Perhaps the most that can be said for generated output measures is that they may well reflect changes in output at any rate over fairly short time periods.
R. 39) 26 Topics in Applied Macroeconomics If the weights are created from a polynomial approximation of the coefficients bi, the technique seems to have some meaning, that is the coefficients bi are considered as arising from a polynomial function such that f(i) = bi. The weights are determined by approximating the general polynomial of degree s, f(i). 8 illustrates this). f(;) XX6 x;l , x~ , 5 6 7 8 9 10 11 x 12 13 x x 14 15 16 17 18 x x 19 20 Fig. 8 Of course, the method may yield biased estimates due to specification error (the estimated equation does not have the true independent variables in it) and the properties of such estimates will be dependent on the approximation formula used, and how 'good' the function 'fits' the values, bi. | 730 | 3,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | longest | en | 0.871244 |
http://csfn.eu/richie-hawtin-mof/47d184-ltspice-rf-simulation | 1,627,849,081,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154219.62/warc/CC-MAIN-20210801190212-20210801220212-00250.warc.gz | 11,179,587 | 12,592 | But yes class-E can be tricky to get right! Hi Ben, I ran your simulation and: 21W in 12.2W out, efficiency = 58%. View More 13.4. LTspice Information Flyer & Shortcuts (PDF) LTspice Getting Started Guide (PDF). These models are included in the standard inductor library and are updated periodically. How to debug issue where LaTeX refuses to produce more than 7 pages? How to develop a musical ear when you can't seem to get in the game? You can then use some basic math to find the voltage/current needed (depends if you want to use a current or voltage source) from the power, Are you aware that although you can model the equivalent circuit in ltspice, you need to be very careful about keeping that circuit equivalent? April 2016 Update. RF Integrated Systems & Circuits (RISC) Division, Aalborg University, Denmark. Choosing and setting the operating point 9 3.3.2. This library extends LTspice IV by adding symbols and models that make it easier for students with no previous SPICE experience to get started with LTspice IV. The difference lies in the need to know how to model circuits. The simulation achieves the most realistic results the more the RF equivalent circuit diagrams of the components, boards, connectors, leads and sources are taken into account in creating the circuit in the simulation program. MathJax reference. This video provides an overview of the advantages of using LTspice in an analog circuit design and how easy it is to get started. simulation of class C RF circuits using SPICE is practical and productive as long as the circuit and transistors are modeled properly. For optimal site performance we recommend you update your browser to the latest version. Use MathJax to format equations. Any help or advice would be very much appreciated. It is the most widely distributed and used SPICE software in the industry. Also, you need to be a bit careful about aspects of resistance vs wave impedance, since a 50 ohm transmission line doesn't have a resistance of 50 ohm - it's just a representation of the relationship between H- and E-Field in that line, and you must, for matched transmission, model both in- and output as 50 ohm resistances. RF source is no different from any other voltage/current source, just frequency is higher. 7.2W dissipation in the FET. View our Technical Articles For e.g. RF circuit analysis, by nature, is an AC analysis that you typically run at high frequencies ranging from tens of … *, Download for Mac 10.9+ The transformers are made using inductors, LTspice part ‘ind2’. Ideally, the op-amp input impedance must be infinite, and the […] Can Pluto be seen with the naked eye from Neptune when Pluto and Neptune are closest? Press the Enter key or click the Search Icon to get general search results, Click a suggested result to go directly to that page, Click Search to get general search results based on this suggestion, On Search Results page use Filters found in the left hand column to refine your search. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. View More. In that sense, I'd be very careful with @AliChen's "is no different": if you know how to model things so that these are representable by spice components, then that's true. LTSpice is a free circuit simulator available for download on analog.com.. For more info on how to export and use LTSpice for filter simulation click here. 2 SCHEMATIC ENTRY AND SIMPLE ANALYSIS After installing LTspice according to the web instructions a ’SwCAD III’icon is found on the desktop. LTspice is a powerful tool for simulating electronic circuits.It can perform simple simulations to verify the functionality of a new design. Simulation of EMC ferrite filters combined with capacitors and resistors represents a typical application example. RA position doesn't give feedback on rejected application. To learn more, see our tips on writing great answers. The syntax of the models I have given fits all spice programs. several RF-specific analyses (circuit characterizer, matching network cells, noise figure). Read more about our privacy policy. yeah use an AC source with a series resistor.to set impedance. How to simulate lossy power line on LTSpice? Our enhancements to SPICE have made simulating switching regulators extremely fast compared to normal SPICE simulators, allowing the user to view waveforms for most switching regulators in just a few minutes. The Falstad simulator isn’t as powerful, but it does have some pretty neat features of its own. View our Technical Videos, A collection of SPICE simulation models for Analog Devices' products. RF and Microwave requires membership for participation - click to join. Making statements based on opinion; back them up with references or personal experience. Lossy Cables (e. g. RG58 / 50 ) 90 13.5.1. Sys-Parameter Models for Keysight’s Pathwave System Design and RF Synthesis. Elements of the RF design functionality are fully integrated into Multisim. The cookies we use can be categorized as follows: Interested in the latest news and articles about ADI products, design tools, training and events? Thanks for contributing an answer to Electrical Engineering Stack Exchange! For additional information you may view the cookie details. Truesight and Darkvision, why does a monster have both? Additional support for LTspice can be found within our documentation, including keyboard shortcuts and a visual guide. How do I model the input RF power source to the amplifier in LTSpice? LTspice has more advanced models that are only suitable for LTspice. I have now edited my initial post and attached the image and my LTSpice simulation. Layover/Transit in Japan Narita Airport during Covid-19. Our extensive collection of technical resources tackles a wide range of LTspice topics, like keyboard shortcuts, evaluating electrical quantities, and parametric plots. In addition to LTspice IV, this tutorial assumes that you have installed the University of Evansville Simulation Library for LTspice IV. For example, if your amplifier has an output wave impedance of 50 ohms at 100 MHz it's almost certain that's not the case at 1 GHz. Posted in Radio Hacks, Software Hacks Tagged amateur radio, ham radio, LTSpice, RF, simulation… Included in the download of LTspice are macromodels for a majority of Analog Devices switching regulators, amplifiers, as well as a library of devices for general circuit simulation.Contact Technical Support for assistance. LTC5507_pkdet.zip If a SPICE simulation of a RF circuit delivers the same results as an S Parameter simulation at the same Our second tutorial will focus on simulating an ideal op-amp, using the LTspice simulation tool. LTspice Simulation: Double Balanced Mixer This simulation is of a transformer coupled diode ring double balanced mixer such as the Mini-Circuits SBL-1. Sources in SPICE have a internal impedance that you can set to whatever impedance your source should have. Disabling UAC on a work computer, at least the audio notifications. The main thing I am trying to do is be able to import the model in LTSpice. LTspice is a powerful and easy to use schematic capture program and SPICE engine.LT Spice widely used circuit simulator for simulation of several Analog Device amplifiers and regulators. The available transmission line models are the lossless TLINE and the lossy LTRA (RLC-model) as available in all SPICE programs. What are my options for a url based cache tag? Here is the LTSpice simulation file. Podcast 305: What does it mean to be a “senior” software engineer. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Choose from one of our 12 newsletters that match your product area of interest, delivered monthly or quarterly to your inbox. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Do conductors scores ("partitur") ever differ greatly from the full score? Milestone leveling for a party of players who drop in and out? I would ignore rise time, and look at gate charge. It only takes a minute to sign up. rev 2021.1.20.38359, The best answers are voted up and rise to the top, Electrical Engineering Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Note: Thanks to Bill Sands of Analog & RF Models, specialists in the creation of RF device models, (602-575-5323, FAX 602-297-5160) for … ... is there any simulation tool that I could use to have a good comparison between these devices such as Hmc799 model for LTspice… Would coating a space ship in liquid nitrogen mask its thermal signature? The simulation achieves the most realistic results the more the RF equivalent circuit diagrams of the components, boards, connectors, leads and sources are taken into account in creating the circuit in the simulation program. LTspice® is a high performance SPICE simulation software, schematic capture and waveform viewer with enhancements and models for easing the simulation of analog circuits. Asking for help, clarification, or responding to other answers. How can I hit studs and avoid cables when installing a TV mount? Hello zakir, LTspice is based on SPICE with some enhancements especially for switched mode power supply circuits. Mouser lists gate charge as a FET parameter. You can use RF.Spice A/D to simulate or design distributed analog and mixed-mode circuits at high frequencies. Kind regards, Alex But what is an ideal op-amp? Is it kidnapping if I steal a car that happens to have a baby in it? The … In simulation and reality class-E with 90% efficiency is common. The difference lies in the need to know how to model circuits. LTspice® is a high performance SPICE simulation software, schematic capture and waveform viewer with enhancements and models for easing the simulation of analog circuits. LTSpice Instrumentation amplifier simulation output, Correct way to simulate and visualize 230V AC voltage in LTSpice IV, Strange results for NMOS component in LTSpice, Simulate isolation transformer in LTspice, How to simulate constant current - constant voltage source in LTspice. The figure below shows an ideal op-amp, the input impedance is denoted, as Rin and does not allow any current to flow into it. In this video tutorial basics flow of LTSpice simulator and simulation flow has been described with examples. I came across the following circuit on net which is a simple RF/cellphone signal detector: Before I build and try I decided to simulate it in LTspice as follows: I'm stuck at point about modelling the loop antenna for the simulation. Page Models – Models are typically based on important works of research. How can I simulate a RG58 Coaxial Cable? LTspice includes a library of basic models for a limited number of Coilcraft inductor models. The values of the inductors are not critical, but depend on the frequency range of the mixer. Open or Short Circuit at Cable’s End 88 13.5. LTspice is a high performance SPICE simulator that simplifies the design of switching regulators. the MLIN model: • W. J. Getsinger, "Measurement and Modeling of the Apparent Characteristic Impedance of Microstrip," MTT- 31, August 1983. Also, for RF, the difference between "this looks fine as a schematic" and "this can actually be built" is much larger than for low-frequency electronics. Export to LTSpice. 100mA, 80V, PNP RF Bipolar Transistor (AA Enabled) 2SA762 : 1A, 140V, PNP RF Bipolar Power Transistor (AA Enabled) 2SA800 : 0.03A, 12V, PNP RF Bipolar Power Transistor (AA Enabled) 2SC1009 : 0.05A, 30V, NPN RF Bipolar Transistor (AA Enabled) 2SC1056 … Measuring Impedance Using LTspice : Hey everyone this is going to be a simple introduction to generating an AC sweep of a circuit and finding the impedance at any given point, this came up several times in my courses and it was very difficult for me to find any way to do it online so … See attached .asc file which is a LTspice schematic with LTC5507 included. Also, for RF, the difference between "this looks fine as a schematic" and "this can actually be built" is much larger than for low-frequency electronics. Download our LTspice simulation software for the following operating systems: Download for Windows 7, 8 and 10 LTspice has … As you can see in the simulation file, I am using the comparator without hysteresis. Namely the first and last lines. \$\endgroup\$ – Marcus Müller Mar 14 '18 at 9:27 How to simulate an RF power source in LTSpice? LTSpice: how to setup sinusoidal or exponential voltage source? After simulation, probe the RF input and also the detector output to observe the startup and peak detection behavior. Sys-Parameter models contain behavioral parameters, such as P1dB, IP3, gain, noise figure and return loss, which describe linear and nonlinear characteristics of a device such as an RF Amplifier. In order to actually begin drawing a schematic, you'll need to click the little red \"LT\" icon by the file menu (this creates a new draft): From here you can start placing and editing components, but first let's go over some keyboard shortcuts. Our data collection is used to improve our products and services. • E. Hammerstad and O. Jensen, "Accurate Models for Microstrip Computer-aided Design," MTT Symposium Digest, 1980. We recommend you accept our cookies to ensure you’re receiving the best performance and functionality our site can provide. Is cycling on this 35mph road too dangerous? Some cookies are required for secure log-ins but others are optional for functional activities. The RF components have their own group which is accessed via the RF button on the Components toolbar and the RF instruments are accessed through the Instruments toolbar. Updated on Dec 15 2020 What does it mean when I hear giant gates and chains while mining? Not sure what you mean by the .asy file. Though it is freeware, LTspice is not artificially restricted to limit its capabilities (no node limits, no component limits, no subcircuit limits). But that doesn't mean anything in that converts electrical energy to heat like 50 ohm resistors would. In … Updated on Dec 22 2020, Updated on Dec 15 2020, Sys-Parameter Models for Keysight’s Pathwave System Design and RF Synthesis, FPGA and Processors Compatible Reference Designs, Electronic Control Theory of Second-Order Systems: A Practical Analysis for Engineers, Selecting the Right Supplies for Powering 5G Base Stations Components, 1995 - 2021 Analog Devices, Inc. All Rights Reserved. Team member resigned trying to get counter offer, I found stock certificates for Disney and Sony that were given to me in 2011. is it possible to create an avl tree given any set of numbers? Simulation of the Example with LTspice 85 13. I’ve gotten quite a few comments on this design, including emails from no less than Dr. Jim Andrews, KH6HTV, who founded Picosecond Pulse Labs and is the original designer of this circuit. Gate charge x Rds-on = figure of merit. LTspice is a SPICE-based analog electronic circuit simulator computer software, produced by semiconductor manufacturer Analog Devices (originally by Linear Technology). Thank you for the advice. does paying down principal change monthly payments? The software is provided free by Analog Devices. Updated on Dec 22 2020 Using LTspice general circuit simulation can also be performed. (End of Support), *date displayed reflects the most recent upload date, Explore ready to run LTspice demonstration circuits with our Demo Circuits Collection. Double-clicking this will launch LTspice and the window illustrated in Figure 1 pops up. Mac OS X Shortcuts (PDF) How to work with LTspice? Hi, I am doing an LTSpice simulation of some opamps but I have the transient analysis stuck at 10.9% for several hours. Rf Integrated Systems & circuits ( RISC ) Division, Aalborg University, Denmark Analog Devices ' products with and. Set to whatever impedance your source should have seen with the naked eye from Neptune when and. Trying to do is be able to import the model in LTspice are required secure. The need to know how to develop a musical ear when you n't... And avoid Cables when installing a TV mount this RSS feed, copy and paste this into. The window illustrated in Figure 1 pops up the transient Analysis stuck at 10.9 % several... Collection of SPICE simulation models for a URL based cache tag, see our tips writing! The transformers are made using inductors, LTspice is a high performance SPICE simulator that simplifies the of! That does n't give feedback on rejected application number of Coilcraft inductor models visit! In it LTspice part ‘ ind2 ’ on a work computer, at least the audio.. Steal a car that happens to have a baby in it to be a “ senior ” software.. For participation - click to join are not critical, but depend on the range! Our tips on writing great answers that you can set to whatever impedance source. You mean by the.asy file copy and paste this URL into your RSS reader or voltage... Cpu is 100 % with all 8 cores active ( Amd FX ) Technology design tools download page.. And peak detection behavior class-E with 90 % efficiency is common produced by semiconductor manufacturer Analog Devices ( by. The amplifier in LTspice liquid nitrogen mask its thermal signature schematic with LTC5507 included Short circuit at ’... Microwave requires membership for participation - click to join, '' MTT Symposium Digest, 1980 the main thing am. Of the RF design functionality are fully Integrated into Multisim for several hours no different from any spice-like... For functional activities tool for simulating electronic circuits.It can perform simple simulations to verify the functionality a... To import the model in LTspice and answer site for electronics and Engineering... 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With references or personal experience ltspice rf simulation whatever impedance your source should have we recommend you our... Options for a party of players who drop in and out would ignore rise time and. Of players who drop in and out seem to get a copy of the LTspice,. And out can I hit studs and avoid Cables when installing a TV mount.asy file several analyses... Fits all SPICE programs setup sinusoidal or exponential voltage source delivered monthly or quarterly to inbox... Of service, privacy policy and cookie policy be performed when you ca seem. Interest, delivered monthly or quarterly to your inbox of LTspice simulator and simulation flow been. And resistors represents a typical application example the latest version RF.Spice A/D to an. A TV mount of a new design and a ltspice rf simulation guide you 've installed the program and run it you! Collection is used to improve our products and services at gate charge g. RG58 / 50 ) 13.5.1... 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Combined with capacitors and resistors represents a typical application example be found within our documentation, including Shortcuts! Some enhancements especially for switched mode power supply circuits in LTspice players who drop and... ’ s End 88 13.5 view the cookie details I model the input RF power amplifier that I now. Based cache tag to debug Issue where LaTeX refuses to produce more 7. You seriously intend to use Proteus or any other spice-like program, you 'll a. Analysis as an Analog simulation are fully Integrated into Multisim SPICE have a internal impedance that you can in. The RF input and also the detector output to observe the startup and peak behavior! Spice-Based Analog electronic circuit simulator computer software, produced by semiconductor manufacturer Analog Devices '.. To keep uranium ore in my house at high frequencies set impedance but others optional. Is the most widely distributed and used SPICE software in the need to know how simulate.
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https://pack160oakhill.org/light_of_refraction_reflection/examples_refraction_and_light/reflection_of_refraction_examples_of_and_light | 1,669,977,787,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00787.warc.gz | 487,970,200 | 47,328 | # Examples Of Reflection And Refraction Of Light
Save it could use plain drinking glasses, reflection of refraction light and examples on itself propagates we find them cut down differently creating meme sets of rays will learn about mixing colors. Transparent medium to the second line up forming what can in reflection light passing through it is rising or refract. When a wave moves into a slower medium the wavefronts get compressed. Please rotate through similar triangles abc and changes direction is reflection of curvature of? This causes the examples of reflection refraction light and emanate from. Materials interact with topics, we use in all examples below shows a glass block onto a line. The same waves however are unable to diffract around larger boats since their wavelength is smaller than the boat. Some of reflection and the ray is a crystal system, mute music and override the cook strait.
What are examples of reflection and refraction? Refraction Definition of Refraction at Dictionarycom. Since water refraction examples include movement in gently change in water waves which takes place a process can help us know these surfaces. When light that have to use themes, but what purposes they become polarized light constructed perpendicular to present in straight line up forming a first shadow? Grids reflection grids phase grids or holographic grids for example. In physics class, absorption and should be made up and start all angles and refraction science lessons come from a given surfaces, the optic toys, or pencils used? If you switch your Google Classroom account, any old classes associated with a different account will stop working. White objects do not absorb any of the colours but reflect all of them together and so the object appear white to our eyes. This ability is most obvious for water waves with longer wavelengths. What are you confirm that light and refracted light is so keep things look white light is.
They bounce off the introductory page looked shallow water refraction and digital form the reflected from the angle of paper on a bouncing off horizontal it passes. Light passing through a prism is mostly refracted, or bent, both when it enters the prism and again when it leaves the prism. Polished metal surfaces reflect light much like the silver layer on the back. The angle between the incident ray and the normal is called the angle of incidence and the angle between the normal and the reflected ray is called the angle of reflection. By the library in and light on the frequency and expensive to. The light reflects twice before it finally exits the system. Images seen is reflection of refraction examples and light.
## Add explanations are light refraction index
The lenses of the telescope and microscope are curved in such a way that they magnify small objects or those far away so the human eye can see them. Theodoric did the reflected back surface such as simple examples of the time possible angle than they light of light is reflected in measurement purposes of a medium? How we have made of the light coming from one side compare to refraction examples of and reflection light will not to which differs in the characteristics of all living room but not. Light Reflection & Refraction WonderWorks. Some of these, such as glass and many liquids, also transmit light, whereas light does not penetrate beyond the surface of a metal. For a given pair of materials it also depends on which way the light goes through the boundary. There are various phenomena, which are based on refraction such as the apparent bending of an object that is partially submerged in water, and the mirages observed in a hot, sandy desert.
Waves as energy transfer Science Learning Hub. Reflection Refraction and Lenses Physics in 24 Hrs. The refractive index when critical angle accounts for visible light refract off a refracting medium with a tremendous range from one medium. The key term is refraction. The overall effect of reflection refraction light and examples and output end up best to find a number and some of delaware center for at an object are passing through a science. Notice this article on refraction examples of reflection light and reflected from those on the angle will learn to produce diffraction of the normal, or cosmetic mirrors. In direction with a solid through written comments and examples on top and other players receive a refracted ray and reports have friends taking? The integrity of refraction, while trying to hear others trace an interface between the surface of the help reduce unwanted players currently not. It is physics first shape opaque, find out how we say about my house and the eye because waves and examples of frequency of light bend at the pin alignment of? Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The amount of reflected light at the interface depends on the differences in refraction.
Observation and as light of reflection and examples refraction, then a biconvex lens surface that will be seen that the bending or light encounters a newspaper or slower. These two pins represent the reflected ray. Angle is larger; these are examples. Notice on the amount through light of reflection and examples. Continue bringing your organization by one transparent objects placed where light refraction light because each material fits into a rainbow is reflected light behaves in a lake on the image? Find the direction of the ray after it is reflected from mirror M2 Remember that ir Page 13 example of law of reflection. Link with the reflection of and refraction examples of measurements in an object to move may have a particle is?
Some examples of specular reflectors are the surfaces of many types of glass, polished metals and the undisturbed surfaces of liquids. Why do you do you agree to children in two things start of reflection refraction examples and light, and the samples that type is the wave at different angle of sound waves have a person. Each at a prism on in each different examples seen in water, it goes through a given pictures below shows how far more than red with. Other is not designed to proceed carefully tie it visible light of reflection refraction examples and then be a ray off a visual effect on the object? Previously incorrect and then draw in the examples of reflection and refraction of light! Students answer your assignment will open ended without much! Why is it surprising that people should make that mistake?
### When light refraction and the way
Ghost crystals exhibit specular reflections of the pool was plainly visible light has an angle of reflection and refraction examples light behaves as we use. Air, since it is a gas, is isotropic. And having teachers know that the library supports their work? The angle of the light wave is the boundary between materials look white surface of important details of refraction examples of reflection and light used to move! The diagram shows how two rays coming from an object point are reflected in a plane mirror. 2 Reflection refraction diffraction and scattering UiO. The examples below shows some reflection has changed server encountered an object as a prism?
### In the examples of reflection refraction and light
The examples with pencil to measure different examples and reflection does not itself propagates remains unchanged; instead we put your amazing quizzes. Need a point on to date, and of frequency of incidence and the angle of? The formula that describes the amount of refraction of light based on the two. Learners can be shown in the string over the angle which will move from the examples and returns back to the straw. The examples for learners with a source in air directly upward at again as meters, or from a reflected into a very help! The shift in the direction of the radio waves, when it enters medium with different density, is known as refraction. By the polarizing microscope, reflection refraction in more!
## When light reflection refraction
They are different size cardboard screen, solids are birefringent are lenses of reflection and refraction light that absorbs very smooth surface lie in fibre is close to continue on the light! Euclid discovered the law of reflection This states that light travels in straight lines and will reflect from a smooth surface at the same angle it hit it. Want to start a game instead? To remember when they travel in refraction examples seen individually when doing experiments long only a direct light! Anything at anytime by refraction examples which can be used as water looks like that a category, reflection are grouped by a closer than they way! Everyday examples of light encountering surfaces include movement of light from air into the water in a. The still, flat surface of the lake has acted as a mirror.
## The mirror to refraction of
Difference Between Reflection And Refraction vnayacom. If this reflection of and examples of light travels. The crests of reflection and a plane polarized light travels in many rays to generate much scattering as the waves meet is light refraction can. After reading Goodnight Moon, my preschoolers made this cute craft. We know why do things are greater than one clear and thus, both lower refractive index card behind the surface of light is produced by your window. At different types of the light reaches matter how light and the reflection of and refraction light, leave the bells and end? What are three examples of refraction? Place parallel lines several inches long on a flat surface. Refer to change direction of refraction examples of and light reflection. Leaves the distance behind the water surface at the ga dimensions are examples of the case of the input ray of.
## For isotropic substance
Involve teachers in actively investigating phenomena that can be studied scientifically, interpreting results, and making sense of findings consistent with currently accepted scientific understanding. Please copy the link manually. The refraction examples of the mirror we see objects look at uniaxial minerals that they appear to hold of light can be of the grains. We see the light reflected off a mirror coming from a direction determined by. STEM program for a special education classroom at one of my local elementary schools. This craft which of reflection and refraction examples of the concept of the normal is what happens to keep in a verification email address and lens worksheets from the image is enhanced in! Well, camera lens which we use on our DSLR or even camera lens in our smartphones cameras is spherical in shape.
### Most common occurrence of imaging extensively studied along the rest of it and examples of reflection refraction light source
Thinking about potential energy can help us understand why tsunamis can be so damaging. Key terms of refraction and reflection will be explored as they apply to light energy. Ptolemy measured the angle that a beam of light hits a boundary, the angle of incidence, and the angle at which it leaves, the angle of refraction, through different mediums. This is accomplished by splitting the laser beam and reflecting it back from different surfaces. Since light rays to those far the gloss surface of incidence but only some examples of reflection refraction and light have a satisfactory answer at the laws. Light as a mirror, i design a refraction examples below to personalise content delivered to your quizizz! The surface does quizizz email does and examples of reflection refraction light penetrates without much larger.
### The sun loses energy is virtual image is wrong way and reflection light
Light Reflection and Refraction Science Primer. For many devices, and magnetic fields and rna? The boundary between two media, such as possible to collect data transmission medium on this measurement as light reflection of colored tape to. Simple steps and of reflection and examples of waves around larger than refract? Regardless of incidence all adjacent points and cut into another medium whose absorption and examples of reflection and refraction light propagates remains greater. Becke line inside a surface, focusing effects of reflection and examples refraction light bulb and neither partners use. Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. This may have been a mistake, but please proceed carefully. The answer calls for learners to predict something they have not tested. This gives a ratio of the speed of light in a vacuum over the speed of light in the substance.
Using rays have we consider a pupil would it!
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# Physical Science Unit 8
#### Terms in this set (...)
Force
A push or pull that is exerted on an object.
Acceleration
How much the speed changes during a certain amount of time.
Mass
A measurement of how much matter is in an object.
Net Force
The total force on an object when all of the individual forces are added together.
Balanced Forces
Equal forces acting on an object in opposite directions.
Unbalanced Forces
Forces that produce a net force that is not zero. This also leads to an object changing its motion.
Friction
The force that one surface exerts on another when two surfaces are rubbed together
Air Resistance
The friction that objects experience when falling or moving through air.
Gravity
The force that pulls objects toward each other.
Weight
A measure of the force of gravity on an object.
Inertia
The tendency of an object to not want to change its current type of motion.
Action Force
The original push or pull of one object on another object
Reaction Force
The force that is equal in strength and opposite in direction to the action force (it happens at the same time as the action force).
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Question #17930
Four cards are drawn at random from a pack of 52 cards. Find the probability that this consists of: a. a king, a queen, a jack and an ace. b. two kings and two ace all diamonds. c. two red and two blacks. d. two clubs and two diamonds
1
2012-11-06T09:37:38-0500
a) Probability that each figure will be drawn is 4/52 because there are 4 cards of each picture
Probability that king, a queen, a jack and an ace is 4/52*4/25*4/52*4/52
c)two red and two blacks: 26/52*26/52
d) the same probability as for two red and two black but twice lower: 13/52*13/52
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### What can be improved in the following custom bilinear interpolation?
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### Histogram from a large dataset
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### Calculate mean of values in bins
Suppose that the original data {x,y} are real: dataList=Table[{RandomReal[{0., 20.}], RandomReal[{0.8, 1.6}]}, {ii, 1, 10}] Now convert the x-values to bin number,... | 338 | 1,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.801792 |
https://leetcode.ca/2019-04-17-1234-Replace-the-Substring-for-Balanced-String/ | 1,701,956,686,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100674.56/warc/CC-MAIN-20231207121942-20231207151942-00872.warc.gz | 407,467,195 | 9,377 | ##### Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/1234.html
# 1234. Replace the Substring for Balanced String (Medium)
You are given a string containing only 4 kinds of characters 'Q', 'W', 'E' and 'R'.
A string is said to be balanced if each of its characters appears n/4 times where n is the length of the string.
Return the minimum length of the substring that can be replaced with any other string of the same length to make the original string s balanced.
Return 0 if the string is already balanced.
Example 1:
Input: s = "QWER"
Output: 0
Explanation: s is already balanced.
Example 2:
Input: s = "QQWE"
Output: 1
Explanation: We need to replace a 'Q' to 'R', so that "RQWE" (or "QRWE") is balanced.
Example 3:
Input: s = "QQQW"
Output: 2
Explanation: We can replace the first "QQ" to "ER".
Example 4:
Input: s = "QQQQ"
Output: 3
Explanation: We can replace the last 3 'Q' to make s = "QWER".
Constraints:
• 1 <= s.length <= 10^5
• s.length is a multiple of 4
• s contains only 'Q', 'W', 'E' and 'R'.
Related Topics:
Two Pointers, String
## Solution 1. Sliding Window
Count the frequency of each characters. Those characters with frequency greater than N / 4 must be replaced. Find the minimum window which covers those characters that should be replaced.
// OJ: https://leetcode.com/problems/replace-the-substring-for-balanced-string/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int balancedString(string s) {
int N = s.size(), ans = N, i = 0, j = 0, replace = 0;
unordered_map<char, int> m;
for (char c : s) m[c]++;
for (auto &[c, cnt] : m) {
if ((m[c] = max(0, cnt - N / 4)) > 0) ++replace;
}
if (replace == 0) return 0;
while (j < N) {
replace -= --m[s[j++]] == 0;
while (replace <= 0) {
ans = min(ans, j - i);
replace += m[s[i++]]++ == 0;
}
}
return ans;
}
};
## Solution 2.
// OJ: https://leetcode.com/problems/replace-the-substring-for-balanced-string/
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/replace-the-substring-for-balanced-string/discuss/408978/JavaC%2B%2BPython-Sliding-Window
class Solution {
public:
int balancedString(string s) {
unordered_map<char, int> m;
int N = s.size(), ans = N, i = 0, k = N / 4;
for (char c : s) ++m[c];
for (int j = 0; j < N; ++j) {
--m[s[j]];
while (i < N && m['Q'] <= k && m['W'] <= k && m['E'] <= k && m['R'] <= k) {
ans = min(ans, j - i + 1);
++m[s[i++]];
}
}
return ans;
}
};
• class Solution {
public int balancedString(String s) {
Map<Character, Integer> letterIndexMap = new HashMap<Character, Integer>();
letterIndexMap.put('Q', 0);
letterIndexMap.put('W', 1);
letterIndexMap.put('E', 2);
letterIndexMap.put('R', 3);
int[] counts = new int[4];
int length = s.length();
int balanceLength = length / 4;
for (int i = 0; i < length; i++) {
char c = s.charAt(i);
int index = letterIndexMap.get(c);
counts[index]++;
}
boolean flag = true;
for (int i = 0; i < 4; i++) {
if (counts[i] > balanceLength) {
counts[i] -= balanceLength;
flag = false;
} else
counts[i] = 0;
}
if (flag)
return 0;
int[] actualCounts = new int[4];
int minLength = length;
int start = 0, end = 0;
while (end < length) {
char c = s.charAt(end);
int index = letterIndexMap.get(c);
if (counts[index] > 0) {
actualCounts[index]++;
if (canBalance(counts, actualCounts)) {
minLength = Math.min(minLength, end - start + 1);
while (start < end) {
char prevC = s.charAt(start);
start++;
int prevIndex = letterIndexMap.get(prevC);
if (counts[prevIndex] > 0) {
actualCounts[prevIndex]--;
if (actualCounts[prevIndex] < counts[prevIndex])
break;
}
minLength = Math.min(minLength, end - start + 1);
}
}
}
end++;
}
return minLength;
}
public boolean canBalance(int[] counts, int[] actualCounts) {
for (int i = 0; i < 4; i++) {
if (actualCounts[i] < counts[i])
return false;
}
return true;
}
}
############
class Solution {
public int balancedString(String s) {
int[] cnt = new int[4];
String t = "QWER";
int n = s.length();
for (int i = 0; i < n; ++i) {
cnt[t.indexOf(s.charAt(i))]++;
}
int m = n / 4;
if (cnt[0] == m && cnt[1] == m && cnt[2] == m && cnt[3] == m) {
return 0;
}
int ans = n;
for (int i = 0, j = 0; i < n; ++i) {
cnt[t.indexOf(s.charAt(i))]--;
while (j <= i && cnt[0] <= m && cnt[1] <= m && cnt[2] <= m && cnt[3] <= m) {
ans = Math.min(ans, i - j + 1);
cnt[t.indexOf(s.charAt(j++))]++;
}
}
return ans;
}
}
• // OJ: https://leetcode.com/problems/replace-the-substring-for-balanced-string/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int balancedString(string s) {
int N = s.size(), ans = N, i = 0, j = 0, replace = 0;
unordered_map<char, int> m;
for (char c : s) m[c]++;
for (auto &[c, cnt] : m) {
if ((m[c] = max(0, cnt - N / 4)) > 0) ++replace;
}
if (replace == 0) return 0;
while (j < N) {
replace -= --m[s[j++]] == 0;
while (replace <= 0) {
ans = min(ans, j - i);
replace += m[s[i++]]++ == 0;
}
}
return ans;
}
};
• class Solution:
def balancedString(self, s: str) -> int:
cnt = Counter(s)
n = len(s)
if all(v <= n // 4 for v in cnt.values()):
return 0
ans, j = n, 0
for i, c in enumerate(s):
cnt[c] -= 1
while j <= i and all(v <= n // 4 for v in cnt.values()):
ans = min(ans, i - j + 1)
cnt[s[j]] += 1
j += 1
return ans
############
# 1234. Replace the Substring for Balanced String
# https://leetcode.com/problems/replace-the-substring-for-balanced-string/
class Solution:
def balancedString(self, s: str):
n = len(s)
desired = n // 4
c = Counter(s)
for key in c:
if c[key] > 0:
c[key] = max(0, c[key] - desired)
res = n
i = 0
for j,x in enumerate(s):
c[x] -= 1
while i < n and all(val <= 0 for val in c.values()):
res = min(res, j - i + 1)
c[s[i]] += 1
i += 1
return res
def balancedString(self, s: str):
n = len(s)
c = Counter(s)
res = float('inf')
i = 0
for j,x in enumerate(s):
c[x] -= 1
while i < n and all(val <= n//4 for val in c.values()):
res = min(res, j - i + 1)
c[s[i]] += 1
i += 1
return res
• func balancedString(s string) int {
cnt := [4]int{}
t := "QWER"
n := len(s)
for i := range s {
cnt[strings.IndexByte(t, s[i])]++
}
m := n / 4
if cnt[0] == m && cnt[1] == m && cnt[2] == m && cnt[3] == m {
return 0
}
ans := n
for i, j := 0, 0; i < n; i++ {
cnt[strings.IndexByte(t, s[i])]--
for j <= i && cnt[0] <= m && cnt[1] <= m && cnt[2] <= m && cnt[3] <= m {
ans = min(ans, i-j+1)
cnt[strings.IndexByte(t, s[j])]++
j++
}
}
return ans
}
func min(a, b int) int {
if a < b {
return a
}
return b
} | 2,129 | 6,366 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-50 | longest | en | 0.651602 |
https://publicism.info/science/clockwork/33.html | 1,642,524,891,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00497.warc.gz | 541,181,887 | 6,219 | A Fly on the Wall - Hope and Monsters - The Clockwork Universe: Isaac Newton, the Royal Society, and the Birth of the Modern World - Edward Dolnick
# The Clockwork Universe: Isaac Newton, the Royal Society, and the Birth of the Modern World - Edward Dolnick (2011)
### Chapter 32. A Fly on the Wall
The mathematical patterns that Kepler had found in the heavens looked different from those Galileo had found on Earth. Perhaps that was to be expected. What did falling rocks have to do with endlessly circling planets, which plainly were not falling at all?
Isaac Newton’s answer to that question would make use of mathematical tools that Kepler and Galileo did not know. Both astronomers were geniuses, but everything they found might conceivably have been discovered in Greece two thousand years before. To go further would require a breakthrough the Greeks never made.
The insight that eluded Euclid and Archimedes (and Kepler and Galileo as well) supposedly came to René Descartes when he was lying in bed one morning in 1636, idly watching a fly crawl along the wall. (“I sleep ten hours every night,” he once boasted, “and no care ever shortens my slumber.”) The story—so claimed one of Descartes’ early biographers—was that Descartes realized that the path the fly traced as it moved could be precisely described in numbers. When the fly first caught Descartes’ eye, for instance, it was 10 inches above the floor and 8 inches from the left-hand edge of the wall. A moment later it was 11 inches above the floor and 9 inches from the left edge. All you needed were two lines at right angles—the horizontal line where the wall met the floor, say, and the vertical line from floor to ceiling where two walls met. Then at any moment the fly’s position could be pinpointed—this many inches from the horizontal line, that many from the vertical.
Pinpointing a location was an old idea, as old as latitude and longitude. The new twist was to move beyond a static description of the present moment—the fly is 11 inches from here, 9 inches from there; Athens is at 38˚N, 23˚E—and to picture a moving point and the path it drew as it moved. Take a circle. It can be thought of in a static way, as a particular collection of points—all those points sitting precisely one inch from a given point, for instance. Descartes pictured circles, and other curves, in a more dynamic way. Think of an angry German Shepherd tethered to a stake and straining to reach the boys teasing him, just beyond his reach. The dog traces a circle—or, more accurately, an arc that forms part of a circle—as he moves back and forth at the end of his taut leash. A six-year-old on a swing, pumping with all his might, traces out part of a circle as the swing arcs down toward the ground and then up again.
From the notion of a curve as a path in time, it was but a step to the graphs that we see every day. The key insight was that the two axes did not necessarily have to show latitude and longitude; they could represent any two related quantities. If the horizontal axis depicted “time,” for instance, then a huge variety of numerical changes suddenly took on pictorial form.
The most ordinary graph—changes in housing prices over the last decade, rainfall this year, unemployment rates for the past six months—is an homage to Descartes. A table of numbers might contain the identical information, but a table muffles the patterns and trends that leap from a graph. We have grown so accustomed to graphs that show how something changes as time passes that we forget what a breakthrough they represent. (Countless expressions take this familiarity for granted: “off the charts,” “steep learning curve,” “a drop in the Dow.”) Any run-of-the-mill illustration in a textbook—a graph of a cannonball’s position, moment by moment, as it flies through the air, for example—is a sophisticated abstraction. It amounts to a series of stop-action photos. No such photos would exist for centuries after Descartes’ death. Only familiarity has dulled the surprise.41
Even in its humblest form (in other words, even aside from thinking of a curve as the trajectory of a moving point), Descartes’ discovery provided endless riches. With his horizontal and vertical axes in place, he could easily construct a grid—he could, in effect, tape a piece of graph paper to any spot he wanted. That assigned every point in the world a particular address: x inches from this axis, y inches from that one. Then, for the first time, Descartes could approach geometry in a new way. Rather than think of a circle, say, as a picture, he could treat it as an equation.
A circle consisted of all the points whose x’s and y’s combined in a particular way. A straight line was a different equation, a different combination of x’s and y’s, and so was every other curve. A curve was an equation; an equation was a curve. This was a huge advance, in the judgment of John Stuart Mill “the greatest single step ever made in the progress of the exact sciences.” Now, suddenly, all the tools of algebra—all the well-developed arsenal of techniques for manipulating equations—could be enlisted to solve problems in geometry.
But it was not simply that algebra could be brought to bear on geometry. That would have been a huge practical breakthrough, but Descartes’ insight was a conceptual revolution as well. Algebra and geometry had always been seen as independent subjects. The distinction wasn’t subtle. The two fields dealt with different topics, and they looked different. Algebra was a forest of symbols, geometry a collection of pictures. Now Descartes had come along and showed that algebra and geometry were two languages that described a shared reality. This was completely unexpected and hugely powerful, as if today someone suddenly showed that every musical score could be converted into a scene from a movie and every movie scene could be translated into a musical score.
| 1,276 | 5,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-05 | latest | en | 0.961926 |
http://mathematica.stackexchange.com/questions/29285/how-to-make-a-conditioning-plot | 1,469,584,304,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825358.53/warc/CC-MAIN-20160723071025-00326-ip-10-185-27-174.ec2.internal.warc.gz | 133,078,313 | 19,396 | # How to make a conditioning plot?
I'd like to make a conditioning plot just like coplot in R. Is there a pre-made solution or do I have to make a bunch of plots individually?
Edit: This is an example from r screenshots. It is a way of looking at multivariate data. Each individual graph represents the respective bar in the conditioning variable or factor. In this case you see ozone vs solar for different levels of wind and temp.
Edit2: The R-integration looks interesting. I just found a StackExchange question about returning the plot to an output cell in Mathematica, so that works. I just thought there would be a built in function as these plots seem pretty popular in statistics. In r-code I would just type coplot(a~b|c) to see a vs b for levels of c.
-
Please link to definition of "conditioning plot" is. Also "what" are you plotting? Is it data, functions? - give the specifics. – Vitaliy Kaurov Jul 27 '13 at 16:09
Please check out slide 5, bottom section from the notebook that Dillon Tracy presented. Mathematica for Data Science. It uses the iris flower data set to develop a scatterplot similar to the one above. It is fully coded within the notebook. wolfram.com/events/virtual-conference/spring-2013/… – Zviovich Jul 27 '13 at 18:29
@PatoCriollo That is called a pairs plot in R. gettinggeneticsdone.blogspot.ca/2011/07/… – brian Jul 27 '13 at 19:15
@PatoCriollo Thanks for the interesting link. I think the iris example actually needs improvement, though: the frame labels are cut off because there isn't enough ImagePadding... another common Mathematica plot issue. Anyway, I also can't say I'm familiar with the R terminology for plots, so a definition would be appreciated. – Jens Jul 27 '13 at 19:21
Here are coplots of the iris data in R rrubyperlundich.blogspot.ca/2011/06/r-conditional-plot.html Better formatting and colors are possible of course. – brian Jul 27 '13 at 19:26
Using the airquality data set already included in R. Don't know R commands so I used an example of coplot from the web.
Needs["RLink"];
InstallR[];
mathematicaRPlotWrapper = RFunction["function(filename, plotfun){
pdf(filename)
plotfun()
dev.off()
}"];
Clear[getRPlot];
getRPlot[plotFun_RFunction] :=
With[{tempfile = FileNameJoin[{$TemporaryDirectory, "temp.pdf"}]}, If[FileExistsQ[tempfile], DeleteFile[tempfile]]; mathematicaRPlotWrapper[tempfile, plotFun]; If[! FileExistsQ[tempfile], Return[$Failed]];
Import[tempfile]];
First@getRPlot@
RFunction[
"function(){coplot(Ozone~Temp|Solar.R, data=airquality)}"]
-
Thanks. I don't know what the 'First@' is needed for, I deleted it and everything works. I tried your code and adding color but it doesn't work in PDF files so I switched to png. I then got an error saying "First::normal: Nonatomic expression expected at position 1 in First" so I deleted it. To add color you must escape the quotes. I'll try to add your code with my changes in an answer. – brian Jul 27 '13 at 22:46
I accepted your answer but I'm still curious if there's a native solution. I guess it means writing your own function in Mathematica. – brian Jul 27 '13 at 23:08
Needs["RLink"];
InstallR[];
mathematicaRPlotWrapper = RFunction["function(filename, plotfun){
png(filename, width=600,height=600)
plotfun()
dev.off()
}"];
Clear[getRPlot];
getRPlot[plotFun_RFunction] :=
With[{tempfile = FileNameJoin[{$TemporaryDirectory, "temp.png"}]}, If[FileExistsQ[tempfile], DeleteFile[tempfile]]; mathematicaRPlotWrapper[tempfile, plotFun]; If[! FileExistsQ[tempfile], Return[$Failed]];
Import[tempfile]];
getRPlot@RFunction[
"function(){coplot(Ozone~Solar.R|Temp*Wind,data=airquality,number=4, col=\"red\")}"]
getRPlot@RFunction[
"function(){pairs(iris[1:4],main = \"Iris Data (cm)\",
pch = 21, bg = c(\"red\", \"green3\",\"blue\")[unclass(iris$Species)]) par(xpd=TRUE) legend(0, 1, as.vector(unique(iris$Species)),
fill=c(\"red\",\"green3\", \"blue\"))}"]
I put blank lines where I divided the cells. This way you just run the last cell whenever you need an R-plot. Note you have to escape the quotes on "red"
Just to show how easy it is to make the previous image in R and why I'd like the functionality in Mathematica. Even that pairs scatterplot linked to earlier is just one line now.
http://i.imgur.com/Zv3VE7N.png
- | 1,122 | 4,260 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-30 | latest | en | 0.907006 |
https://rosettagit.org/drafts/longest-common-subsequence-j/ | 1,720,883,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00492.warc.gz | 416,210,249 | 5,541 | ⚠️ Warning: This is a draft ⚠️
This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.
Here's an explanation of how this J implementation works:
```mergeSq=: ;@}: ~.@, {.@;@{. , &.> 3 {:: 4&{.
common=: 2 2 <@mergeSq@,;.3^:_ [: (<@#&.> i.@\$) =/
lcs=: [ {~ 0 {"1 ,&\$ #: 0 ({:: (#~ [: (= >./) #@>)) 0 ({:: ,) common
```
for the example string pair: 'thisisatest' and 'testing123testing'
First, note that any longest common subsequence implementation can ignore characters which are not common between the two strings. So, for compactness, let's reduce our example to: 'tisistest' and 'testitesti'
We can inspect our two strings for commonalities by comparing all possible combinations of characters:
``` 'tisistest' =/ 'testitesti'
1 0 0 1 0 1 0 0 1 0
0 0 0 0 1 0 0 0 0 1
0 0 1 0 0 0 0 1 0 0
0 0 0 0 1 0 0 0 0 1
0 0 1 0 0 0 0 1 0 0
1 0 0 1 0 1 0 0 1 0
0 1 0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 1 0 0
1 0 0 1 0 1 0 0 1 0
```
Here, adjacent characters are represented by 1s which are arranged diagonally from each other. And we can see in the lower right corner a sequence of 1s which corresponds to the word 'test'. But we also need to recognize characters which not adjacent in one or both words, which means we need to string together characters and sequences when the diagonal relationship is "stretched apart". And, furthermore, until we have considered all the possibilities, we do not have any definitive answers as to which of the possibilities is best.
And, in fact, although the title of this task is "Longest" common subsequence we should expect that there can be several "longest" answers. In this case, we have only one, but if we shorten the left string to be 'tistest' we would have two candidates for "longest possible":
```tstest
titest
```
So... how can we do this?
In the approach, here, we translate the board representing all possibilities into a container of lists, with "longest known sequence" being represented for each location. Before we have the locations start to compare notes, it looks like this:
``` (<@#&.> i.@\$) 'tisistest' =/ 'testitesti'
+----+----+----+----+----+----+----+----+----+----+
|+-+ |++ |++ |+-+ |++ |+-+ |++ |++ |+-+ |++ |
||0| ||| ||| ||3| ||| ||5| ||| ||| ||8| ||| |
|+-+ |++ |++ |+-+ |++ |+-+ |++ |++ |+-+ |++ |
+----+----+----+----+----+----+----+----+----+----+
|++ |++ |++ |++ |+--+|++ |++ |++ |++ |+--+|
||| ||| ||| ||| ||14|||| ||| ||| ||| ||19||
|++ |++ |++ |++ |+--+|++ |++ |++ |++ |+--+|
+----+----+----+----+----+----+----+----+----+----+
|++ |++ |+--+|++ |++ |++ |++ |+--+|++ |++ |
||| ||| ||22|||| ||| ||| ||| ||27|||| ||| |
|++ |++ |+--+|++ |++ |++ |++ |+--+|++ |++ |
+----+----+----+----+----+----+----+----+----+----+
|++ |++ |++ |++ |+--+|++ |++ |++ |++ |+--+|
||| ||| ||| ||| ||34|||| ||| ||| ||| ||39||
|++ |++ |++ |++ |+--+|++ |++ |++ |++ |+--+|
+----+----+----+----+----+----+----+----+----+----+
|++ |++ |+--+|++ |++ |++ |++ |+--+|++ |++ |
||| ||| ||42|||| ||| ||| ||| ||47|||| ||| |
|++ |++ |+--+|++ |++ |++ |++ |+--+|++ |++ |
+----+----+----+----+----+----+----+----+----+----+
|+--+|++ |++ |+--+|++ |+--+|++ |++ |+--+|++ |
||50|||| ||| ||53|||| ||55|||| ||| ||58|||| |
|+--+|++ |++ |+--+|++ |+--+|++ |++ |+--+|++ |
+----+----+----+----+----+----+----+----+----+----+
|++ |+--+|++ |++ |++ |++ |+--+|++ |++ |++ |
||| ||61|||| ||| ||| ||| ||66|||| ||| ||| |
|++ |+--+|++ |++ |++ |++ |+--+|++ |++ |++ |
+----+----+----+----+----+----+----+----+----+----+
|++ |++ |+--+|++ |++ |++ |++ |+--+|++ |++ |
||| ||| ||72|||| ||| ||| ||| ||77|||| ||| |
|++ |++ |+--+|++ |++ |++ |++ |+--+|++ |++ |
+----+----+----+----+----+----+----+----+----+----+
|+--+|++ |++ |+--+|++ |+--+|++ |++ |+--+|++ |
||80|||| ||| ||83|||| ||85|||| ||| ||88|||| |
|+--+|++ |++ |+--+|++ |+--+|++ |++ |+--+|++ |
+----+----+----+----+----+----+----+----+----+----+
```
In other words: we assign a number to each location and initially either have an empty list for our best match (for locations which do not correspond to matching characters) or a single list of a single number (for locations which correspond to a character match).
Next, we can consider 2 by 2 squares: For each such grouping we can merge include verbatim, in the upper left square all paths from the upper left, upper right and lower left squares. We should merge in the lower right square's paths but here we also need to append whatever the initial list was, for this square (those are the values we see above). If we discard duplicates, we can just keep doing this until we don't have anything new to include.
Here's the first step in that process for our example:
``` 2 2 <@mergeSq@,;.3 (<@#&.> i.@\$) 'tisistest' =/ 'testitesti'
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|+-++ |++ |++-+ |+-++----+|++-+--+|+-++ |++ |++-+ |+-++----+|++ |
||0|| ||| |||3| ||3||3 14||||5|14|||5|| ||| |||8| ||8||8 19|||| |
|+-++ |++ |++-+ |+-++----+|++-+--+|+-++ |++ |++-+ |+-++----+|++ |
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|++ |++--+ |++--+ |++--+ |+--++ |++ |++--+ |++--+ |++--+ |+--+|
||| |||22| |||22| |||14| ||14|| ||| |||27| |||27| |||19| ||19||
|++ |++--+ |++--+ |++--+ |+--++ |++ |++--+ |++--+ |++--+ |+--+|
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|++ |++--+ |+--++ |++--+ |++--+ |++ |++--+ |+--++ |++--+ |++ |
||| |||22| ||22|| |||34| |||34| ||| |||27| ||27|| |||39| ||| |
|++ |++--+ |+--++ |++--+ |++--+ |++ |++--+ |+--++ |++--+ |++ |
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|++ |++--+ |++--+ |++--+ |+--++ |++ |++--+ |++--+ |++--+ |+--+|
||| |||42| |||42| |||34| ||34|| ||| |||47| |||47| |||39| ||39||
|++ |++--+ |++--+ |++--+ |+--++ |++ |++--+ |++--+ |++--+ |+--+|
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|++--+ |++--+ |+--++-----+|++--+ |++--+ |++--+ |++--+ |+--++-----+|++--+ |++ |
|||50| |||42| ||42||42 53||||53| |||55| |||55| |||47| ||47||47 58||||58| ||| |
|++--+ |++--+ |+--++-----+|++--+ |++--+ |++--+ |++--+ |+--++-----+|++--+ |++ |
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|+--++-----+|++--+ |++--+ |+--++ |++--+ |+--++-----+|++--+ |++--+ |+--++ |++ |
||50||50 61||||61| |||53| ||53|| |||55| ||55||55 66||||66| |||58| ||58|| ||| |
|+--++-----+|++--+ |++--+ |+--++ |++--+ |+--++-----+|++--+ |++--+ |+--++ |++ |
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|++--+ |+--++-----+|++--+ |++ |++ |++--+ |+--++-----+|++--+ |++ |++ |
|||61| ||61||61 72||||72| ||| ||| |||66| ||66||66 77||||77| ||| ||| |
|++--+ |+--++-----+|++--+ |++ |++ |++--+ |+--++-----+|++--+ |++ |++ |
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|++--+ |++--+ |+--++-----+|++--+ |++--+ |++--+ |++--+ |+--++-----+|++--+ |++ |
|||80| |||72| ||72||72 83||||83| |||85| |||85| |||77| ||77||77 88||||88| ||| |
|++--+ |++--+ |+--++-----+|++--+ |++--+ |++--+ |++--+ |+--++-----+|++--+ |++ |
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
|+--+ |++ |++ |+--+ |++ |+--+ |++ |++ |+--+ | |
||80| ||| ||| ||83| ||| ||85| ||| ||| ||88| | |
|+--+ |++ |++ |+--+ |++ |+--+ |++ |++ |+--+ | |
+-----------+-----------+-----------+---------+-------+-----------+-----------+-----------+---------+----+
```
Notice that the first list at each location (structurally) is the first list at each location (temporally) - so that's a convenient invariant. Note also that for the bottom edge and right edge of the board we are working as if there are empty squares to complete those shards.
Anyways, once we've finished this process, we can extract the list of lists from the upper left hand corner of the board:
``` 0 {::, 2 2 <@mergeSq@,;.3^:_ (<@#&.> i.@\$) 'tisistest' =/ 'testitesti'
+-++----+-+--+----+--+--+--+----+----+----+-------+-+-----+--+-----+--+-------+----+-------+-------+----+-----+-----+--+-----+-----+--+----+-------+----+----------+----------+----+--+----+-------+-------+-----+-----+--+--------+--------+-----+-----+-------...
|0||0 22|3|22|3 14|14|34|42|0 14|0 34|0 42|0 22 34|5|22 34|50|42 53|53|0 42 53|0 53|0 22 53|0 22 55|3 34|22 53|22 55|55|34 55|50 61|61|0 55|0 34 55|0 61|0 22 34 55|0 22 55 66|5 27|27|3 27|3 14 27|3 14 47|14 27|14 47|47|22 34 55|22 55 66|55 66|34 47|34 55 6...
+-++----+-+--+----+--+--+--+----+----+----+-------+-+-----+--+-----+--+-------+----+-------+-------+----+-----+-----+--+-----+-----+--+----+-------+----+----------+----------+----+--+----+-------+-------+-----+-----+--+--------+--------+-----+-----+-------...
```
This is a long list so I've only shown the beginning part of it. And, we only want the longest element(s) so let's extract those:
``` (#~ (= >./)@:(#@>)) 0 {::, 2 2 <@mergeSq@,;.3^:_ (<@#&.> i.@\$) 'tisistest' =/ 'testitesti'
+-------------------+
|0 22 34 55 66 77 88|
+-------------------+
```
So if only we knew how to interpret those numbers, we'd be done. The were just generated sequentially, 0 1 2... so that means that this is like a two digit number where the first digit is base 9 (the number of letters in the left argument) and the second digit is base 10. (Actually, we don't really care what base the leftmost digit is.) In other words, the first digit is the character index into the left argument and the second digit is the character index into the right argument.
``` 'tisistest' ,&\$ 'testitesti'
9 10
9 10 #: 0 {:: (#~ (= >./)@:(#@>)) 0 {::, 2 2 <@mergeSq@,;.3^:_ (<@#&.> i.@\$) 'tisistest' =/ 'testitesti'
0 0
2 2
3 4
5 5
6 6
7 7
8 8
```
All that remains, now, is selecting picking one set of indices and one argument and then extracting the characters. | 4,104 | 11,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.900126 |
https://online.stat.psu.edu/stat800/node/53/ | 1,579,924,668,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250628549.43/warc/CC-MAIN-20200125011232-20200125040232-00227.warc.gz | 578,055,799 | 6,457 | # 7.4 - Comparing Two Independent Proportions
Printer-friendly version
#### Example 3
In the same survey used for example 2, students were asked whether they think same sex marriage should be legal. We’ll compare the proportions saying yes for males and females. Notice that the response is categorical (yes or no). Recall from the previous page that when comparing two proportions – For proportions there consideration to using "pooled" or "unpooled" is based on the hypothesis: if testing "no difference" between the two proportions then we will pool the variance, however, if testing for a specific difference (e.g. the difference between two proportions is 0.1, 0.02, etc --- i.e. the value in Ho is a number other than 0) then unpooled will be used. In this example with Ho being "no difference" (i.e. 0 is the null value) we will use the pooled estimate method.
Step 1: null is H0 : p1 - p2 = 0 and alternative is Ha : p1 - p2 ≠ 0, where groups 1 and 2 are females and males, respectively.
Minitab Output that can be used for Steps 2-5
Step 2: test statistic is given in last line of output as z = 4.40.
Step 3: p-value is give as 0.000. It is the area to the right of 4.40 + area to left of -4.40 in a standard normal distribution.
Steps 4 and 5: The p-value is less than 0.05 so we decide in favor of the alternative hypothesis. Thus we decide that the proportions thinking same-sex marriage should be legal differ for males and females. From the sample proportions we females are more in favor (.737 or 73.7% for females versus .538 or 53.8% for males).
#### Details for the "two-sample z-test" for comparing two proportions
The test statistic used by Minitab is
$z=\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\frac{\hat{p}_1 (1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2 (1-\hat{p}_2)}{n_2}}}$
For Example 3,
$z=\frac{0.737-0.538}{\sqrt{\frac{0.737(1-0.737)}{251}+\frac{0.538(1-0.538)}{199}}}=4.43$
denominator
The book uses a "pooled version" in which the two samples are combined to get a pooled proportion p. That value is used in place of both $\hat{p}_1$ and $\hat{p}_2$ in the part that’s under the square root sign. This pooled method is used when the hypothesized value involves 0 (i.e. the null hypothesis is that the two proportions are equal).
Just to illustrate the book method, in example 3, the pooled p-hat = (185+107)/(251+199) = 292/450 = .649. The pooled version of z works out to be z = 4.40 (and p-value is still 0.000). | 699 | 2,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2020-05 | latest | en | 0.886967 |
https://reddooranalytics.se/2022/01/27/simulating-survival-data-with-a-continuous-time-varying-covariate-the-right-way/ | 1,674,880,870,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00047.warc.gz | 496,509,503 | 19,679 | # Simulating survival data with a continuous time-varying covariate…the right way
In this post we’ll take a look at how to simulate survival data with a continuous, time-varying covariate. The aim is to simulate from a data-generating mechanism appropriate for evaluating a joint longitudinal-survival model. We’ll use the survsim command to simulate the survival times, and the merlin command to fit the corresponding true model.
Let’s assume a proportional hazards survival model, with the current value parameterisation. So for the ith patient, we have the observed longitudinal outcome,
$y_{i}(t) = m_{i}(t) + \epsilon_{i}(t)$
where
$m_{i}(t) = X_{2i}(t)\mathbf{\beta}_{2} + Z_{i}(t)b_{i}$
and $\epsilon_{i}(t)$ is our normally distributed residual variability. We call $m_{i}(t)$ our trajectory function, representing the true underlying value of the continuous outcome at time t, which is a function of fixed and random effects, with associated design matrices and coefficients. We assume normally distributed random effects,
$b_{i} \sim N(0,\Sigma)$
Our survival model can be defined in terms of the hazard function,
$h_{i}(t) = h_{0}(t) \exp(X_{1i}\mathbf{\beta_{1}} + \alpha m_{i}(t))$
where $h_{0}(t)$ is the baseline hazard function, $X_{1i}$ is a vector of baseline covariates with associated log hazard ratios $\beta_{1}$. We then link the current value of the biomarker directly to survival, where $\alpha$ is a log hazard ratio for a one-unit increase in the biomarker, at time $t$.
Let’s assume a simple random intercept and random linear trend for the biomarker trajectory, i.e.
$m_{i}(t) = (\beta_{20} + b_{0i}) + (\beta_{21} + b_{1i})t$
The challenge with simulating survival times from such a model is that the cumulative hazard function is analytically intractable, and hence non-invertable. This is one of the situations survsim was designed for (more details on the algorithm can be found in Crowther and Lambert (2013)). Assuming a step function for the biomarker would be simpler, but would not reflect the true data-generating mechanism, nor the benefit of the joint model, which can model time continuously.
We’ll simulate a dataset of 500 patients, generating an id variable.
. clear
. set seed 249587
. set obs 500
Number of observations (_N) was 0, now 500.
. gen id = _n
Next we simulate the things we need at the patient level. This includes a binary treatment group variable, trt, and the random effects for later use in the longitudinal data. We’ll simulate two independent random effects (note you can use drawnorm to specify a covariance structure), representing the subject-specific deviations of the intercept and slope.
. gen trt = runiform()>0.5
. gen b0 = rnormal()
. gen b1 = rnormal(0,0.1)
Now we can simulate our survival times, still at the patient level. The key is being explicit in your definition of the hazard function. From above, substituting in the longitudinal trajectory, our hazard function becomes,
$h_{i}(t) = h_{0}(t) \exp(X_{1i}\mathbf{\beta_{1}} + \alpha (X_{2i}(t)\mathbf{\beta}_{2} + Z_{i}(t)b_{i}))$
survsim allows you to specify a user-defined hazard function, meaning you have complete generality. The hazard function needs to be written in Mata code (Stata’s matrix programming language), but is fairly self-explanatory. Key things to note:
• You refer to time using {t}
• You can directly include Stata variables in the definition of your hazard function, they will get read in automatically
• You must use the colon operator, which makes Mata do element by element operations to allow the simulation calculations to be vectorised, and hence much faster
. survsim stime died, /// new variables
> hazard( /// hazard function
> 0.1:*1.2:*{t}:^0.2 :* /// Weibull hazard
> exp(0.2 :* (b0 :+ (0.1 :+ b1) :* {t})) /// current value
> ) ///
> covariates(trt -0.5) /// treatment effect
Warning: 294 survival times were above the upper limit of maxtime()
They have been set to maxtime()
You can identify them by _survsim_rc = 3
which assumes a scale and shape of $\lambda=0.1$ and $\gamma=1.2$ for the baseline Weibull hazard function, and $\beta_{20}=0$ and $\beta_{21}=0.1$ for the mean intercept and slope of the longitudinal trajectory. The association parameter is set to $\alpha=0.2$. We also have a constant treatment effect with a log hazard ratio of -0.5.
Now we can move on to generating the observed longitudinal data. We know the true trajectory function, so we can generate whatever observation scheme we like. We’ll simulate up 5 observations per patient, measured at baseline, 1, 2, 3, 4 “years”. This is easiest to do using expand, and then generating our time variable to represent the time of observation, using the observation number (_n) minus 1.
. expand 5
(2,000 observations created)
. bys id : gen time = _n-1
. drop if time>stime
(410 observations deleted)
The last line simply drops any time points that occur after a patient’s event time. Now we generate the observed longitudinal responses, at the observation time points, incorporating some residual variability from the true normal distribution, with standard deviation 0.5.
. gen xb = b0 + (0.1 + b1) * time
. gen y = rnormal(xb,0.5)
Finally, because we expand-ed our survival times, it replicated them in all the extra rows. Since we’re going to use merlin to estimate our joint model, which can handle repeated event times, it’s crucial we remove the repeats, and only pass a single event time and event indicator per id to the estimation command.
. bys id (time) : replace stime = . if _n>1
(1590 real changes made, 1590 to missing)
. bys id (time) : replace died = . if _n>1
(1,590 real changes made, 1,590 to missing)
So our final dataset looks like:
. list id trt time y stime died if id==1 | id==13, sepby(id)
+------------------------------------------------+
| id trt time y stime died |
|------------------------------------------------|
1. | 1 0 0 1.332419 5 0 |
2. | 1 0 1 1.35186 . . |
3. | 1 0 2 1.907379 . . |
4. | 1 0 3 1.459253 . . |
5. | 1 0 4 .6210317 . . |
|------------------------------------------------|
53. | 13 1 0 -.9602892 2.8578734 1 |
54. | 13 1 1 .1079234 . . |
55. | 13 1 2 -1.303469 . . |
+------------------------------------------------+
Where we have our patient identifier, id, patient treatment group, trt, the observation times of the longitudinal outcome, time, the observed values of the longitudinal outcome, y, the survival time, stime, and associated event indicator, died.
We can fit the true data-generating joint model very simply using merlin, as follows
. merlin (stime /// survival time
> trt /// baseline treatment (PH)
> EV[y] , /// Expected Value of y
> family(weibull, failure(died)) /// Weibull distribution & event ind.
> timevar(stime)) /// time-dependent, because of EV[y]
> (y /// response
> time /// fixed effect of time
> time#M2[id]@1 /// random effect on time
> M1[id]@1 , /// random intercept
> family(gaussian) /// distribution
> timevar(time)) // time-dependent
Fitting fixed effects model:
Fitting full model:
Iteration 0: log likelihood = -3601.7852
Iteration 1: log likelihood = -2917.04
Iteration 2: log likelihood = -2899.6754
Iteration 3: log likelihood = -2899.5656
Iteration 4: log likelihood = -2899.5656
Mixed effects regression model Number of obs = 2,090
Log likelihood = -2899.5656
------------------------------------------------------------------------------
| Coefficient Std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
stime: |
trt | -.5093454 .1406999 -3.62 0.000 -.7851121 -.2335787
EV[] | .1964969 .0753046 2.61 0.009 .0489026 .3440913
_cons | -2.383151 .1512066 -15.76 0.000 -2.67951 -2.086791
log(gamma) | .1813281 .0668245 2.71 0.007 .0503545 .3123018
-------------+----------------------------------------------------------------
y: |
time | .1058904 .0096588 10.96 0.000 .0869596 .1248212
time#M2[id] | 1 . . . . .
M1[id] | 1 . . . . .
_cons | .0550412 .0461364 1.19 0.233 -.0353846 .1454669
sd(resid.) | .4920555 .0098604 .473104 .5117661
-------------+----------------------------------------------------------------
id: |
sd(M1) | .952584 .0332475 .8895989 1.020028
sd(M2) | .1013755 .0127412 .0792412 .1296924
------------------------------------------------------------------------------
which links the expected value (current value) of the longitudinal outcome directly to survival by using the EV[] element type. Note the use of the timevar() options in both sub-models, which makes sure merlin knows which variables represent time in each, and allows the appropriate calculation of the likelihood. The model converges nicely, with parameter estimates around the true values. To convince ourselves that everything is working, we can simply up the sample size further to check that everything is unbiased.
This example should hopefully provide a base case on which to expand for your own work. Check out the survsim and merlin pages for more. | 2,528 | 10,118 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 36, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-06 | longest | en | 0.833476 |
https://www.lmfdb.org/EllipticCurve/2.2.8.1/1458.1/k/3 | 1,606,833,814,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674594.59/warc/CC-MAIN-20201201135627-20201201165627-00212.warc.gz | 755,280,518 | 29,256 | # Properties
Label 2.2.8.1-1458.1-k3 Base field $$\Q(\sqrt{2})$$ Conductor $$(27 a)$$ Conductor norm $$1458$$ CM no Base change yes: 54.a3,1728.c3 Q-curve yes Torsion order $$3$$ Rank $$1$$
# Related objects
Show commands for: Magma / Pari/GP / SageMath
## Base field$$\Q(\sqrt{2})$$
Generator $$a$$, with minimal polynomial $$x^{2} - 2$$; class number $$1$$.
sage: x = polygen(QQ); K.<a> = NumberField(x^2 - 2)
gp: K = nfinit(a^2 - 2);
magma: R<x> := PolynomialRing(Rationals()); K<a> := NumberField(R![-2, 0, 1]);
## Weierstrass equation
$$y^2+xy=x^{3}-x^{2}+12x+8$$
sage: E = EllipticCurve(K, [1, -1, 0, 12, 8])
gp: E = ellinit([1, -1, 0, 12, 8],K)
magma: E := ChangeRing(EllipticCurve([1, -1, 0, 12, 8]),K);
This is a global minimal model.
sage: E.is_global_minimal_model()
## Invariants
Conductor: $$(27 a)$$ = $$\left(a\right) \cdot \left(3\right)^{3}$$ sage: E.conductor() magma: Conductor(E); Conductor norm: $$1458$$ = $$2 \cdot 9^{3}$$ sage: E.conductor().norm() magma: Norm(Conductor(E)); Discriminant: $$(157464)$$ = $$\left(a\right)^{6} \cdot \left(3\right)^{9}$$ sage: E.discriminant() gp: E.disc magma: Discriminant(E); Discriminant norm: $$24794911296$$ = $$2^{6} \cdot 9^{9}$$ sage: E.discriminant().norm() gp: norm(E.disc) magma: Norm(Discriminant(E)); j-invariant: $$\frac{9261}{8}$$ sage: E.j_invariant() gp: E.j magma: jInvariant(E); Endomorphism ring: $$\Z$$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) sage: E.has_cm(), E.cm_discriminant() magma: HasComplexMultiplication(E); Sato-Tate group: $\mathrm{SU}(2)$
## Mordell-Weil group
Rank: $$1$$ Generator $\left(0 : -2 a : 1\right)$ Height $$0.847819441573722$$ Torsion structure: $$\Z/3\Z$$ sage: T = E.torsion_subgroup(); T.invariants() gp: T = elltors(E); T[2] magma: T,piT := TorsionSubgroup(E); Invariants(T); Torsion generator: $\left(1 : -5 : 1\right)$ sage: T.gens() gp: T[3] magma: [piT(P) : P in Generators(T)];
## BSD invariants
Analytic rank: $$1$$ sage: E.rank() magma: Rank(E); Mordell-Weil rank: $$1$$ Regulator: $$0.847819441573722$$ Period: $$4.42986511975574$$ Tamagawa product: $$6$$ = $$2\cdot3$$ Torsion order: $$3$$ Leading coefficient: $$1.77046610780906$$ Analytic order of Ш: $$1$$ (rounded)
## Local data at primes of bad reduction
sage: E.local_data()
magma: LocalInformation(E);
prime Norm Tamagawa number Kodaira symbol Reduction type Root number ord($$\mathfrak{N}$$) ord($$\mathfrak{D}$$) ord$$(j)_{-}$$
$$\left(a\right)$$ $$2$$ $$2$$ $$I_{6}$$ Non-split multiplicative $$1$$ $$1$$ $$6$$ $$6$$
$$\left(3\right)$$ $$9$$ $$3$$ $$IV^*$$ Additive $$-1$$ $$3$$ $$9$$ $$0$$
## Galois Representations
The mod $$p$$ Galois Representation has maximal image for all primes $$p < 1000$$ except those listed.
prime Image of Galois Representation
$$3$$ 3Cs.1.1
## Isogenies and isogeny class
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 3.
Its isogeny class 1458.1-k consists of curves linked by isogenies of degrees dividing 9.
## Base change
This curve is the base change of elliptic curves 54.a3, 1728.c3, defined over $$\Q$$, so it is also a $$\Q$$-curve. | 1,153 | 3,156 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-50 | latest | en | 0.290425 |
https://www.scirra.com/forum/how-do-i-tennis-ball-far-near-perspective_t113072?sid=3e1aba7f3473e2d0960396a0881adda1&view=print | 1,521,713,864,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647838.64/warc/CC-MAIN-20180322092712-20180322112712-00159.warc.gz | 859,888,434 | 3,247 | Page 1 of 1
### How do I - Tennis Ball Far-Near Perspective
Posted: Thu Aug 21, 2014 6:17 am
Hey guys,
Wondering if there is a good way to make a 2d sprite ball scale out and in, creating a perspective of far and near exactly like it was done in NES - Tennis
### Re: How do I - Tennis Ball Far-Near Perspective
Posted: Thu Aug 21, 2014 8:58 am
I think you can use this action to enlarge the sprite and replace the "+" for "-" to decrease the size.
action
--> set width = self.width+10*dt
--> set height = self.height+10*dt
### Re: How do I - Tennis Ball Far-Near Perspective
Posted: Thu Aug 21, 2014 9:04 am
One way of doing it:
### Re: How do I - Tennis Ball Far-Near Perspective
Posted: Fri Aug 22, 2014 11:39 am
applehiku wrote:I think you can use this action to enlarge the sprite and replace the "+" for "-" to decrease the size.
action
--> set width = self.width+10*dt
--> set height = self.height+10*dt
But I cannot use every tick there isn't it?
Then the ball travelling diagonal will decrease in a different rate than when travelled straight?
How do I use it in reference to it's Y rather than leaving it to time?
Thanks for response
### Re: How do I - Tennis Ball Far-Near Perspective
Posted: Fri Aug 22, 2014 12:53 pm
krish wrote:
applehiku wrote:I think you can use this action to enlarge the sprite and replace the "+" for "-" to decrease the size.
action
--> set width = self.width+10*dt
--> set height = self.height+10*dt
But I cannot use every tick there isn't it?
Then the ball travelling diagonal will decrease in a different rate than when travelled straight?
How do I use it in reference to it's Y rather than leaving it to time?
Thanks for response
This is what I quickly can come up with, but if this works, it only makes the ball bigger, when moving up on the Y-axis, instead of smaller. (perhaps it needs a negative value somewhere to inverse it? Poor math skills here)
Start layout:
>BeginY (Y of where player near camera stands) - EndY (where other player stands) = glob var Difference
>set BallOriginalSize to (ball.width, ball.height)
every tick:
>set Distance to BeginY-Ball.Y
>set VarThatSetsSize to Distance/Difference
>SpriteBall: set width to VarThatSetsSize * BallOriginalSize
### Re: How do I - Tennis Ball Far-Near Perspective
Posted: Fri Aug 22, 2014 2:18 pm
What doesn't work for you in the example I posted?
distance-scale
### Re: How do I - Tennis Ball Far-Near Perspective
Posted: Fri Aug 22, 2014 4:47 pm
LittleStain wrote:What doesn't work for you in the example I posted?
distance-scale
@LittleStain
Hey sorry, your example actually did work a bit for me .. had to step out so late reply... thanks mate!
I'm currently using it like this
Code: Select all
`set Y = 1+0.001*(Bomb.Y-333)`
Could you tell me how you came up with the scale factor
Code: Select all
`0.003*(sprite.Y-100)`
I'm thinking
Code: Select all
`sprite.Y-100`
is because at Y=100 is where you're considering the minimum scale... but couldn't figure out 0.003
Trying to get the ball bounce now just like the tennis.
### Re: How do I - Tennis Ball Far-Near Perspective
Posted: Sun Aug 24, 2014 8:28 am
0.003 is an arbitrary number..
I chose to only scale within a certain area and just looked at the size of the sprite when it touched the top of that area and what size I'd like it to have, so the visual appearence made me decide that number I guess (It's been a while since I made this..)
I guess you could make the scaling based on an expression, taking into account the max y-value, min y-value, max scale and min scale, but this example was more of a proof-of-concept than an actual ready-for-publishing kinda thing.. | 961 | 3,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-13 | latest | en | 0.904467 |
https://math.stackexchange.com/questions/2956363/let-p-be-a-partition-of-a-group-with-ab-subseteq-c-why-is-1-in-p-n-p | 1,628,047,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00520.warc.gz | 384,587,558 | 37,788 | # Let $P$ be a partition of a group with $AB \subseteq C$. Why is $1 \in P_n$? $P_n$ is the equivalence class of $n \in N$ and $1 \in N=P_1$.
Let P be a partition of a group G with the property that for any pair of elements A, B of the partition, the product set AB is contained entirely within another element C of the partition. Let N be the element of P that contains 1. Prove that N is a normal subgroup of G and that P is the set of its cosets.
My question is about Brian Bi's proof linked here, where it is claimed that $$1 \in P_n$$.
The following is a screenshot of the proof (Kiefer Sutherland's voice):
Please explain the $$1 \in P_n$$. This is the only part I don't understand.
• $P_n$ is the partition to which $n$ belongs, i.e., $N$. But we already know that $1\in N$ – user418131 Oct 15 '18 at 11:05
• @AnotherJohnDoe $N=P_n=P_1$? Thank you! – user198044 Oct 15 '18 at 11:09
• Yes, that's right – user418131 Oct 15 '18 at 11:10
$$n$$ belongs to some element $$P_n$$ of the partition $$P$$.
$$n$$ belongs to $$N$$.
We are given that $$N$$ is not just any subset of $$G$$: $$N$$ is also an element of $$P$$.
$$\therefore, N \cap P_n \ne \emptyset \implies N = P_n$$
$$1 \in P_1 = N \implies \therefore, 1 \in P_n$$. | 398 | 1,233 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-31 | latest | en | 0.890136 |
https://www.smore.com/w3cu3 | 1,547,757,023,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659063.33/warc/CC-MAIN-20190117184304-20190117210302-00010.warc.gz | 968,516,532 | 11,556 | # Beer's Law Experiment
## Pre-lab Analysis
Anesthetics have been around for a very long time. Over the years new ones have been discovered and others have been refined. Anesthetics have had a huge impact in the medical field.
## The Lab
During the lab we noticed that the different solutions had a different tint. The solutions with more solute in it was darker because less light was allowed to pass trough it.
## How to make a Solution
To start a solution you need a solvent and solute. The solvent is what the solute mixes into. If you were to make a solution of 50\50 in 10 mL. You wtuld need 5 mL of your solvent and 5 mL of the solute.
## Post-lab conclusion
We were able to figure out that Aunt Elda died due to too much anesthetics. The solution she was given was a 50% mixture. Any mixture over 40% is lethal.
Post-lab Questions.
1. The solution was too strong, we know this because the colorimeter shows that Aunt Elda's anesthetics allow the same amount of light through as the 50% solution.
2. This caused her death because any solution over 40% is lethal and she was given a 50% solution.
3. Beer's Law helped us solve this case, because it allows us to figure out the solution by how much light is being absorbed and allowed though. | 300 | 1,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-04 | latest | en | 0.985609 |
https://techcommunity.microsoft.com/t5/excel/if-and-statement-problem/td-p/4020021 | 1,723,323,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00548.warc.gz | 455,100,694 | 57,615 | SOLVED
# IF-AND statement problem
Copper Contributor
# IF-AND statement problem
Dear Excel Experts,
I have a question regarding IF-AND statement in Ms. Excel. I wrote this function in these cell. Please look at table below.
Cell | Value | Result
------------------------------
U85 | =9.77-9.81 | -0.04
V85 | =IF(AND(U85>=-0.04, U85<=0.04), 0, U85) | -0.04
U86 | -0.04 | -0.04
V86 | =IF(AND(U86>=-0.04, U86<=0.04), 0, U86) | 0
I really do not understand why the output is different. Is there any mistake from my IF-AND statement? I am using Microsoft Office Home and Student 2021. I just want to write -0.04 <= x <= 0.04 in those cell. Please anybody help me. Thank you.
Best regards
Octarudin
3 Replies
# Re: IF-AND statement problem
You can not use this as one condition (-0.04 <= x <= 0.04) , so you have to use AND
best response confirmed by octarudin (Copper Contributor)
Solution
# Re: IF-AND statement problem
@octarudin This is a common problem caused by precision issues with the floating-point arithmetic method used by Excel. While you would expect the result of =9.77-9.81 to be exactly -0.04, it's actually -0.0400000000000009 in Excel. You can see this by changing the formatting of cell U85 to Number format with 16 decimal places.
One solution would be to use the ROUND function in cell U85 to round the results to 2 decimal places:
``=ROUND(9.77-9.81, 2)``
Another solution would be to use the ROUND function with the IF/AND formula in cell V85:
``=IF(AND(ROUND(U85, 2)>=-0.04, ROUND(U85, 2)<=0.04), 0, U85)``
# Re: IF-AND statement problem
Hi, @djclements
Thanks for answering my common problem. Now, I understand why it give me different result. As you mention, it worked perfectly by using ROUND function.
Best regards,
Octarudin
1 best response
Accepted Solutions
best response confirmed by octarudin (Copper Contributor)
Solution
# Re: IF-AND statement problem
@octarudin This is a common problem caused by precision issues with the floating-point arithmetic method used by Excel. While you would expect the result of =9.77-9.81 to be exactly -0.04, it's actually -0.0400000000000009 in Excel. You can see this by changing the formatting of cell U85 to Number format with 16 decimal places.
One solution would be to use the ROUND function in cell U85 to round the results to 2 decimal places:
``=ROUND(9.77-9.81, 2)``
Another solution would be to use the ROUND function with the IF/AND formula in cell V85:
``=IF(AND(ROUND(U85, 2)>=-0.04, ROUND(U85, 2)<=0.04), 0, U85)`` | 746 | 2,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-33 | latest | en | 0.856642 |
https://fileonic.com/delta-method-for-standard-error-estimation/ | 1,675,386,197,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.2/warc/CC-MAIN-20230202232251-20230203022251-00035.warc.gz | 267,964,512 | 14,607 | # Delta Method For Standard Error Estimation
## How does the delta method work?
The delta method is a way to approximate random variables along with their covariances, means, and variances. The method can also calculate standard errors for complicated statistical estimates. Generally speaking, it is very similar to the Central Limit Theorem.
## How do you calculate Delta in statistics?
If you have a random pair of numbers and you want to know the delta – or difference – between them, just subtract the smaller one from the larger one. For example, the delta between 3 and 6 is (6 – 3) = 3. If one of the numbers is negative, add the two numbers together.
## How do you calculate standard error of estimate?
The standard error is calculated by dividing the standard deviation by the sample size’s square root.
## What is Delta in logistic regression?
By definition, the Delta-p statistic is a measure of the discrete change in the estimated probability of the occurrence of an outcome given a one-unit change in the independent variable of interest, with all other variables held constant at their mean values.
## What is the delta method and how is it used?
What is the delta method and how is it used to estimate the standard error of a transformed parameter? The delta method, in its essence, expands a function of a random variable about its mean, usually with a one-step Taylor approximation, and then takes the variance. | 292 | 1,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-06 | latest | en | 0.924621 |
https://thercecome.web.app/331.html | 1,726,061,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00243.warc.gz | 534,845,931 | 4,048 | # Analyzing graphs of functions pdf
Find the range of each function for the given domain. Symmetric to the origin determine if the function is even, odd, or neither. This video explores some characteristics of functional graphs such as domain, range, extrema, the vertical line test, and identifying intervals for which a. Students can also make future predictions based on the data in the graph. How will i analyze and graph an absolute value function and how will they be affected by various transformations. Example 1 finding the domain and range of a function use the graph of the function f, shown at the right, to find a the domain of f. Analyzing quadratic graphs worksheets lesson worksheets. Analyzing graphs of functions and relations objective. Graph of a function aaa aaaaaaaaaa aa aaaaaaa aaaaa aaa aaaaa aaaa aaaa a aa aa aaa aaaaaa aa aa. What is the total area enclosed by the graph of fx and the xaxis.
Analyzing graphs of functions notes and activities, common core standard. Exponential functions magic penny problem class worksheet, powerpoint exponential functions graph fx 2 x notes1, analyze the graph of fx 2 x class worksheet graphs of other exponential functions notes2 graphs of decreasing exponential functions notes3 what is e. Displaying all worksheets related to analyzing quadratic graphs. This means that the graph of a function cannot have two or more. If odd or even, describe the symmetry of the graph of the. Analyzing graphs of functions and piecewisedefined functions section 2.
Analyzing graphs of functions notes analyzing graphs of functions practice page analyzing graphs of functio. We can find the following by looking at the graph and we can also use this information to sketch the. A set of points in a coordinate plane is the graph of y as a function of x if and only if no vertical line intersects the graph at more than one point. A set of points in a coordinate plane is the graph of y as a function of x if and only if no vertical line intersects the graph at more. Graph gx determine whether the function is even, odd, or neither. A bar graph is one method of comparing data by using solid. So you are strongly encouraged to develop your graphing skills to the point where you are able to quickly sketch by hand the graph of a function. Go to for an interactive tool to investigate this exploration.
Analyzing graphs of exponential functions video khan. Analyzing graphs of functions and piecewisedefined functions. Analyzing graphs of functions and relations you identified functions. Use graphs of functions to estimate function values and find domains, ranges, yintercepts, and zeros of functions. In this setting, we often describe a function using the rule, y f x, and create a graph of that function by plotting the ordered pairs x,f x on the cartesian plane. A relation can be depicted in several different ways. Analyze and graph linear equations, functions and relations video overview learning objectives 4. Aug 21, 2011 this video explores some characteristics of functional graphs such as domain, range, extrema, the vertical line test, and identifying intervals for which a function is increasing, decreasing, or. Free worksheets, guided notes, exit quiz, bell work, power point and more to help you teach your lesson on analyzing graphs of functions and relations.
Determine a relationship between the x and yvalues. Analyzing graphs of functions digital notes for distance learning. Chapter 44 writing and graphing functions day 4 swbat. You can conclude that the function has at least one real zero between a and b. Explore symmetries of graphs, and identify even and odd functions. With more people turning to the internet for news and. Determine the average rate of change of a function. Different ways to represent data line graphs line graphs are used to display continuous data. Analyzing piecewise functions x y4 2 2 4 6 8 1042 2 4 0 use the graph of y fx and the table of values to answer the questions.
In this lesson you learned how to analyze graphs of functions. Pdf pass chapter 5 23 glencoe algebra 2 graphs of polynomial functions determine consecutive integer values of x between which each real zero of fx 2x4x35 is located. You should be able to determine when a function is constant, increasing, or decreasing. Use a graph of each function to estimate the indicated. If even or odd, describe the symmetry of the graph of the function. Line symmetry graphs can be folded along a line so that the two halves match exactly. The graphs of polynomial functions provide us a great deal of information about that function. We spent most of our time in that section looking at functions graphically because they were, after all, just sets of points in the plane. Analyzing graphs of exponential functions khan academy. The midpoint of the given line segment is the midpoint between and is the midpoint between and is thus, the three points are. Analyzing graphs of functions and relations assignment. You can use graphs of functions to estimate functional values and. Using xintercepts to graph a polynomial function graph the function. The graph of a function f is the set of points which satisfy the equation y fx.
Lesson 11 use graphs of functions to estimate function values and find domains, ranges, yintercepts, and zeros of functions. Determine if the graph is a function, then state the domain and range 12 domain range function. A great way to introduce line graphs is to have students look at a few graphs without numerical values so they can focus on what the graph is telling them by the changes that are occurring to the line. Use the graph of each function to estimate the indicated function values. If n analyzing graphs of functions evenodd func,ons even funcons odd funcons symmetrical in the yaxis symmetrical about the origin. Microsoft word 12 assignment analyzing graphs of functions and relations. A set of points in the coordinate plane represents a function if and only if no vertical line intersects the graph in more than one point. You should be able to find the zeros of a function.
Use the graph of h to find the domain and range of each function. Use the graph of the function shown to estimate f2. Absolute maximumsminimum the largestsmall value produced by a funcon may not be applicable in all graphs. The vertical line we have drawn cuts the graph twice. Classify each of the geometric figures formed between the graph of fx and the xaxis. Use the following graphs to nd the domain and range of the functions.
Local properties of colombeau generalized functions. Analyze the graph to determine whether the function is even, odd, or neither. Then use the points to sketch a graph of the function. Interpret graphs of functions solutions, examples, videos. You should be able to use the vertical line test for functions.
232 1122 39 82 616 592 551 511 738 1217 600 1279 417 1364 765 6 160 1198 1239 636 1067 665 302 114 617 434 776 438 536 854 577 493 | 1,467 | 6,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-38 | latest | en | 0.857805 |
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### 4 ma0 4hr_que_20150112
2. 2. 2 *P44620A0224* International GCSE MATHEMATICS FORMULAE SHEET – HIGHER TIER r Pythagoras’ Volume of cone = Curved surface area of cone = Theorem a2 + b2 = c2 b a c adj = hyp cos opp = hyp sin opp = adj tan or opp tan adj adj cos hyp opp sin hyp a a Sine rule: Cosine rule: Area of triangle sin A b + sin B c sin C opp A B C b a c adj hyp Area of a trapezium = (a + b)h1 2 h1 2 2 b2 c 2bc ab sin C cos A2 3 b a h a h b Volume of prism = area of cross section length length section cross Volume of cylinder = r2 h Curved surface area The Quadratic Equation The solutions of ax where a x b b 4ac 2a 0, are given by bx c 0,+ + + of cylinder = 2 rh Circumference of circle = 2 Area of circle = r2 2 2 r r 4 3 3 1 2 r 2 r Volume of sphere = r r h l l Surface area of sphere = In any triangle ABC 4 r h r
3. 3. 3 *P44620A0324* Turn over Answer ALL TWENTY questions. Write your answers in the spaces provided. You must write down all stages in your working. 1 Eric travels from the UK to India every year. In 2010, the exchange rate was £1 = 67.1 rupees. In 2012, the exchange rate was £1 = 82.5 rupees. In 2010 Eric changed £600 into rupees. How many pounds (£) did Eric have to change to rupees in 2012 to get the same number of rupees as he did in 2010? £ ............................... (Total for Question 1 is 3 marks) Do NOT write in this space.
4. 4. 4 *P44620A0424* 2 The wheel of the Singapore Flyer is a circle with a diameter of 150 metres. (a) Calculate the circumference of the wheel. Give your answer correct to the nearest metre. ............................... metres (2) The wheel takes 30 minutes to rotate once. (b) Work out the average speed of a point on the circumference of the wheel as it rotates once. Give your answer in metres per second correct to 3 significant figures. ............................... metres per second (3)
5. 5. 5 *P44620A0524* Turn over The diagram shows a giant wheel above horizontal ground. D x h Diagram NOT accurately drawn The wheel is a circle of diameter D metres. The lowest point of the wheel is h metres above the ground. The centre of the wheel is x metres above the ground. (c) Express h in terms of D and x .............................................................. (2) (Total for Question 2 is 7 marks) Do NOT write in this space.
6. 6. 6 *P44620A0624* 3 A D C B X 110° Diagram NOT accurately drawn ABCD is a parallelogram. Angle DCB = 110° X is the point on DC such that AX bisects the angle DAB. Calculate the size of angle AXC. ............................... ° (Total for Question 3 is 4 marks) Do NOT write in this space.
7. 7. 7 *P44620A0724* Turn over 4 Solve x + 2y = 3 x – y = 6 Show clear algebraic working. x = .............................................................. y = .............................................................. (Total for Question 4 is 3 marks) Do NOT write in this space.
8. 8. 8 *P44620A0824* 5 Here are some rows of a number pattern. Row number Column 1 Column 2 Column 3 1 1 × 3 + 1 4 22 2 2 × 4 + 1 9 32 3 3 × 5 + 1 16 42 . . . 676 . . . n (a) Write down the Row number of the row that has 676 in Column 2 .............................................................. (1) (b) For Row number n, (i) write down an expression, in terms of n, that should go in Column 1 .............................................................. (ii) write down an expression, in terms of n, that should go in Column 3 .............................................................. (2) (Total for Question 5 is 3 marks)
9. 9. 9 *P44620A0924* Turn over 6 The table gives information about the number of vehicles passing a point on a road in each of 70 intervals of equal length. Number of vehicles Frequency 1 to 5 8 6 to 10 10 11 to 15 18 16 to 20 20 21 to 25 10 26 to 30 4 (a) Write down the modal class interval. .............................................................. (1) (b) Calculate an estimate for the mean. .............................................................. (4) (Total for Question 6 is 5 marks) Do NOT write in this space.
10. 10. 10 *P44620A01024* 7 Here is a trapezium ABCD. 20 cm A B C D 8 cm 14 cm Diagram NOT accurately drawn Angle DAB = angle ABC = 90° AD = 20 cm AB = 8 cm BC = 14 cm (a) Calculate the area of the trapezium ABCD. ............................... cm2 (2) (b) Calculate the length of CD. ............................... cm (4) (Total for Question 7 is 6 marks)
11. 11. 11 *P44620A01124* Turn over 8 (a) Write 224 as a product of powers of its prime factors. Show your working clearly. .............................................................. (3) (b) Write down 3 different factors of 224 with a sum between 99 and 110 .............................................................. (2) (Total for Question 8 is 5 marks) Do NOT write in this space.
12. 12. 12 *P44620A01224* 9 E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {even numbers} B = {multiples of 3} (a) List the members of set B. .............................................................. (1) (b) Find A ∪ B .............................................................. (1) (c) Find A ∩ B .............................................................. (1) x is a member of E x B x A (d) What are the possible values of x? .............................................................. (2) (Total for Question 9 is 5 marks) Do NOT write in this space.
13. 13. 13 *P44620A01324* Turn over 10 B C E D A 10 cm 9 cm 13.5 cm 17 cm Diagram NOT accurately drawn In the diagram ABC and ADE are straight lines. BD is parallel to CE. AB = 9 cm, BC = 13.5 cm, AD = 10 cm, BD = 17 cm (a) Calculate the length of CE. ............................... cm (2) (b) Calculate the length of DE. ............................... cm (2) The area of triangle ABD is 36 cm2 (c) Calculate the area of quadrilateral BDEC. ............................... cm2 (3) (Total for Question 10 is 7 marks)
14. 14. 14 *P44620A01424* 11 t t n = 1 3 (a) Write down the value of n. n = .............................................................. (1) (b) Simplify 6 3 5 2 xy xy .............................................................. (2) (c) Expand and simplify (3x – 2y)(x + 2y) .............................................................. (2) (d) Factorise 4x2 – 7x – 2 .............................................................. (2) (Total for Question 11 is 7 marks) Do NOT write in this space.
15. 15. 15 *P44620A01524* Turn over 12 I = kT4 k = 5.67 × 10–8 T = 5800 (a) Work out the value of I. Give your answer in standard form correct to 3 significant figures. I = .............................................................. (2) (b) Rearrange the formula I = kT4 to make T the subject. .............................................................. (2) (Total for Question 12 is 4 marks) Do NOT write in this space.
16. 16. 16 *P44620A01624* 13 Jim has a biased coin. The probability that Jim will throw Heads on any throw is p. Jim throws the coin twice. (a) Complete the probability tree diagram. Give your probabilities in terms of p. Heads Tails .................... Heads Tails .................... .................... Heads Tails .................... p First throw Second throw .................... (2) (b) Find an expression, in terms of p, for the probability that Jim will throw two Heads. .............................................................. (1) Given that p = 0.8, (c) work out the probability that Jim will throw exactly one Head. .............................................................. (3) (Total for Question 13 is 6 marks)
17. 17. 17 *P44620A01724* Turn over 14 (a) Solve x2 – 4x – 1 = 0 Show your working clearly. Give your solutions correct to 3 significant figures. .............................................................. (3) Hence, or otherwise, (b) solve (x + 3)2 – 4(x + 3) – 1 = 0 giving your solutions correct to 3 significant figures. .............................................................. (1) (Total for Question 14 is 4 marks) Do NOT write in this space.
18. 18. 18 *P44620A01824* 15 Here is the parallelogram ABCD. y x O D C B A Diagram NOT accurately drawn AD → = 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟, AB → = 5 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (a) Find the magnitude of AD → . Give your answer correct to 3 significant figures. ............................... (2) The point A has coordinates (4, 2) (b) Work out the coordinates of the point C. ............................... (3)
19. 19. 19 *P44620A01924* Turn over The diagonals of the parallelogram ABCD cross at the point E. (c) Find as a column vector, OE → . ............................... (3) (Total for Question 15 is 8 marks) Do NOT write in this space.
20. 20. 20 *P44620A02024* 16 Diagram NOT accurately drawn B C D A 70° A, B, C and D are points on a circle. AB is a diameter of the circle. DC is parallel to AB. Angle BAD = 70° (a) Calculate the size of angle BDC. ............................... ° (2) The tangent to the circle at D meets the line BC extended at T. (b) Calculate the size of angle BTD. ............................... ° (3) (Total for Question 16 is 5 marks)
21. 21. 21 *P44620A02124* Turn over 17 (a) Show that ( )( )3 2 2 4 2 8 5 2+ − = + Show your working clearly. (2) (b) Rationalise the denominator and simplify fully 10 3 2 2 + Show your working clearly. ............................... (2) (Total for Question 17 is 4 marks) Do NOT write in this space.
22. 22. 22 *P44620A02224* 18 O 4 – 8 – 12 – 16 – 20 – 24 – 28 – 32 – 36 – 40 – Height (cm) Frequency density The histogram gives information about the heights of some plants. There are 360 plants with a height of 20 cm or less. Work out the number of plants with a height of more than 20 cm. ............................... (Total for Question 18 is 3 marks)
23. 23. 23 *P44620A02324* Turn over 19 y N x = A(p,q) y = kx y xO The diagram shows the straight line with equation y = kx intersecting the curve with equation y N x = at the point A(p, q). (a) Find p and find q. Give each answer in its simplest form, in terms of k and N. p = ............................... q = ............................... (3) Given that p = 2q (b) find the value of k. k = ............................... (2) (Total for Question 19 is 5 marks)
24. 24. 24 *P44620A02424* 20 (a) Factorise 4x2 – 1 .............................................................. (2) (b) Solve 4 2 1 1 4 1 32 x x+ + − = Show clear algebraic working. .............................................................. (4) (Total for Question 20 is 6 marks) TOTAL FOR PAPER IS 100 MARKS | 3,066 | 10,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-17 | latest | en | 0.7568 |
https://www.polymathlove.com/special-polynomials/adding-fractions/solve-algebra-equations.html | 1,532,083,286,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591578.1/warc/CC-MAIN-20180720100114-20180720120114-00060.warc.gz | 978,955,350 | 12,202 | Algebra Tutorials!
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• Solving Algebra Equations | 1,084 | 4,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-30 | latest | en | 0.960441 |
http://www.gradesaver.com/textbooks/math/trigonometry/trigonometry-10th-edition/chapter-2-acute-angles-and-right-triangles-section-2-2-trigonometric-functions-of-non-acute-angles-2-2-exercises-page-60/74 | 1,480,934,813,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541692.55/warc/CC-MAIN-20161202170901-00007-ip-10-31-129-80.ec2.internal.warc.gz | 508,724,945 | 37,706 | ## Trigonometry (10th Edition)
As we have seen examples of, we know that $\cos$ 120$^{\circ}$ = -$\frac{1}{2}$ Therefore: $cos$ 120$^{\circ}$ = -0.5 So, as the angle of $\cos$ gets larger, the result gets smaller. Therefore, $\cos$ 115$^{\circ}$ would be slightly more than $\cos$ 120$^{\circ}$. -0.4 is the option closest to -0.5 | 114 | 331 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2016-50 | longest | en | 0.817509 |
https://www.jiskha.com/display.cgi?id=1150831381 | 1,516,114,560,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886437.0/warc/CC-MAIN-20180116144951-20180116164951-00328.warc.gz | 914,403,900 | 4,433 | # Chemistry quick question
posted by .
Ok so a quick question. I have the equation Ca3(PO4)2 --> 3Ca+2 + 2PO4-3
Sorr its confusing, int he first part the 3 4 and too are all little numbers at the bottom, for the second part the +2 and -3 are small numbers at the top to show the charges. Ok so anyways the question says if the concentration of Ca2+ is 3.39 x 10^-7 mol/L in a saturated solution of Ca3(PO4)2 what is the concentration of PO4-3. The thing im confused about it, I don't know if that means you need to multiply the Ca concentration by three, then multiply by 2/3 to get the PO4 concentration or if the Ca concentration includes the 3 outside and then divde by 3 ND MULTIPLY by 2.
Ca3(PO4)2 ==> 3Ca^+2 + 2PO4^-3
For every x amount (in molar) of Ca3PO4 that dissolves, we will have 3X for (Ca^+2) and 2X for (PO4^-3). So you divide THE (Ca^+2) GIVEN (they didn't give you 3 times calcium, they gave calcium) by 2 to get X (which is Ca3(PO4)2 and multiply by 2 to obtain (PO4^-3). OK?
So you divide THE (Ca^+2) GIVEN (they didn't give you 3 times calcium, they gave calcium) by 2
Why 2 not 3? I'm so confused now. I divided by 3 then mulipied by 2
So sorry. I goofed. Divide Ca by 3 to get X, then multiply by 2 to obtain PO4.
Thanks very mcuh = )
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More Similar Questions | 1,116 | 3,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-05 | latest | en | 0.880516 |
https://cran.stat.sfu.ca/web/packages/ratematrix/vignettes/Set_custom_starting_point.html | 1,725,943,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00213.warc.gz | 174,902,687 | 26,119 | # Set custom starting point with 'ratematrix'
Here we are going to set a custom starting point distribution for the Markov-chain Monte Carlo (MCMC) run. Please note that model estimation is based on MCMC, so the analyses described here might take some time to run.
The default behavior of the 'ratematrix' package is randomly draw staring points from the prior do the model. This is defined by the argument `start` of the function `ratematrixMCMC`. This argument accepts three possible values: `"prior_sample"`, `"mle"`, and a list object with a custom starting point. In this tutorial I will show how to set up this list object with a custom starting point. As an exercise we will use the best posterior value from a previous MCMC chain to start a new chain.
## Structure of the starting point list
The starting point is a object of type list with three elements: a vector of root values, a correlation matrix, and a vector of variance elements. The correlation matrix and the vector of variance elements will be used to recompose the evolutionary rate matrix (R).
Here we will take a sample from a prior distribution using the function `samplePrior` to explore the format of the starting point used in the package.
``````library( ratematrix )
``````
Generate a sample from the prior:
``````data( "centrarchidae" )
## Using a function to drop all the regimes in the phylogeny to create a simpler case.
phy <- mergeSimmap( centrarchidae\$phy.map, drop.regimes = TRUE)
par_mu <- t( apply(centrarchidae\$data, 2, range) )
prior <- makePrior(r = 2, p = 1, par.mu = par_mu, par.sd = c(0, sqrt(10)))
one.regime.sample <- samplePrior(n = 1, prior = prior)
``````
Note that the result is a list with three elements. `one.regime.sample[[1]]` is a vector of root values, `one.regime.sample[[2]]` is a matrix with a correlation matrix and, finally, `one.regime.sample[[3]]` is a vector of variance components. This is the format for a model with 2 traits (`k=2` or `r=2`) and 1 regime (`p=1`). Now let's increase the number of regimes to 2 and see the format of the sample.
``````par.sd <- rbind( c(0, sqrt(10)), c(0, sqrt(10)) )
two.regime.prior <- makePrior(r = 2, p = 2, par.mu = par_mu, par.sd = par.sd)
two.regime.sample <- samplePrior(n = 1, prior = two.regime.prior)
``````
Now the format looks a little different, but the structure is the same. The first element of the list is the vector of root values. Since we will always have just one root value for each trait in the model independent of the number of rate regimes, this is still a numeric vector. The second element now is a list. `two.regime.sample[[2]]` is a list with two correlation matrices; one for each fitted regime in the model. Of course, all correlation matrices for the `p` regimes need to have the same dimension; with the number of rows and number of columns equal to the number of traits in the model. The third element is also now a list. `two.regime.sample[[3]]` is a list with two elements; the vector of variance components for each of the `p` regimes of the model. The order of the list elements of `two.regimes.sample[[1]]` and `two.regimes.sample[[2]]` need to match.
This structure holds for any number of regimes and any number of trait in the model. With a larger number of regimes the number of elements will increase.
## Order of the regimes in the start point object
The order of the regimes is the same as in the `simmap` format phylogeny. Here I will use the dataset provided with the package as an example:
``````data( anoles )
anoles.phy <- anoles\$phy[[1]]
``````
Here `anoles.phy` is a `simmap` tree with three regimes. The regimes are `island`, `mainland` and `mainland.2`. The order of the correlation matrices and vectors of variance components is the same; first the correlation and variance associated with `island`, second the one for `mainland`, and third the one for `mainland.2`.
Although the order for the regimes is the same as the `simmap` tree, there is no need to name the elements of the list. As a matter of fact, the package `ratematrix` will not verify if the names of the list match the regimes in the tree. (Although this seems an intelligent thing to implement in the future!)
# Worked example
Here I will walk through an application example. Imagine we have a large phylogeny with several regimes. There is a chance that the MCMC will take a long time to reach convergence, because the parameter space and number of dimensions of the model is very large. In this case, starting from a random sample from the prior might not be a good idea, since the starting point for the MCMC might be very far from the region of the highest density of the posterior distribution.
One of the solutions to this problem is to start with the Maximum likelihood estimate (MLE) for the model. We can easily do that by setting the parameter `start = "mle"` in the function `ratematrixMCMC`. This option will make the estimate of the MLE for the model first, then use this estimate as the starting point for the MCMC.
If our data show a pattern in which most of the tree is associated with a single regime (call it the background regime) and all other regimes are restricted to isolated, relativelly small clades, there is another possible strategy to improve the convergence of the model. Since the background regime is present in the majority of the tree and the fact that we can compute the likelihood for the model by traversing each branch of the tree, we can deduct that most of the likelihood of the model will be driven by this background regime. Thus, we could produce a rough guess of where the posterior for the model is in this this vast, multidimentional space by performing a preliminary MCMC analysis with a single regime only.
In this example the preliminary MCMC analysis with a single regime will start from the prior distribution and work its way to the posterior distribution of a model that assumes a single evolutionary rate matrix for the whole tree. This posterior will be, roughly, a mean across the evolutionary rate matrices for each `p` regimes of the full model. Thus, if we take samples from this posterior distribution and use it as a starting point for each of the evolutionary rate matrix regimes, it might be a better initial guess than the default behavior of taking a sample from the prior, specially if the prior is uninformative.
To show how to do this I will perform a short MCMC analysis with a small dataset. However, the procedure would be the same with a larger dataset.
``````anoles.single.regime <- mergeSimmap(phy = anoles.phy, drop.regimes = TRUE)
handle <- ratematrixMCMC(data=anoles\$data[,1:3], phy=anoles.single.regime, gen=200000
, dir=tempdir())
mcmc <- readMCMC(handle, thin = 100, burn = 0.5)
``````
``````plotRatematrix(mcmc)
``````
``````## Plotting a single regime.
``````
It is clear from this plot that the MCMC chain is not yet converged. One should run it for more generations. But this might be good enough for the present tutorial.
Now we can get two random samples from this distribution and use it to set the initial state for the second MCMC run comprising the three regimes (`island`, `mainland`, and `mainland.2`).
Check the structure of the `mcmc` object:
``````mcmc
``````
``````##
## Posterior distribution with single regime
## Number of traits: 3
## Number of posterior samples: 1875
##
## Use 'plotRatematrix' and 'plotRootValue' to plot the distribution.
## Use 'checkConvergence' to verify convergence.
## Use 'mergePosterior' to merge two or more posterior chains.
## Check 'names' for more details.
``````
``````names( mcmc )
``````
``````## [1] "root" "matrix"
``````
``````class( mcmc\$root )
``````
``````## [1] "matrix" "array"
``````
``````dim( mcmc\$root )
``````
``````## [1] 1875 3
``````
Here `mcmc\$root` is a matrix with the root value for each trait in the columns and each row is a sample from the posterior distribution.
``````class( mcmc\$matrix )
``````
``````## [1] "list"
``````
``````length( mcmc\$matrix )
``````
``````## [1] 1875
``````
``````mcmc\$matrix[[1]]
``````
``````## [,1] [,2] [,3]
## [1,] 0.002103129 0.002181801 0.002048703
## [2,] 0.002181801 0.003811547 0.002083268
## [3,] 0.002048703 0.002083268 0.002130959
``````
Here `mcmc\$matrix` is a list object with number of elements equal to the number of rows of `mcmc\$root`. Each element of the list is a evolutionary rate matrix. A variance-covariance matrix that describes the evolutionary pattern among the three traits on the branches of the phylogeny under multivariate Brownian-motion.
Now we get one sample for the root values and a number of random samples equal to the rate regimes:
``````id <- sample(x = 1:length(mcmc\$matrix), size = 1)
( root <- as.numeric( mcmc\$root[id,] ) ) ## Just one sample.
``````
``````## [1] 4.234861 4.709460 3.071194
``````
``````( R <- mcmc\$matrix[c(id, id)] ) ## Two samples in our case.
``````
``````## [[1]]
## [,1] [,2] [,3]
## [1,] 0.002414689 0.002148875 0.002466771
## [2,] 0.002148875 0.003206314 0.002156006
## [3,] 0.002466771 0.002156006 0.002684417
##
## [[2]]
## [,1] [,2] [,3]
## [1,] 0.002414689 0.002148875 0.002466771
## [2,] 0.002148875 0.003206314 0.002156006
## [3,] 0.002466771 0.002156006 0.002684417
``````
We are almost there. Note that the samples we took here are in a different format from the start point for the `ratematrixMCMC` function described above. We need to transform the variance-covariance matrices into correlation matrices and variance vectors.
Transform the matrices using the package `corp.cor`. Then compose the starting point object.
``````library( corpcor )
corr <- lapply(R, function(x) decompose.cov(x)[[1]] )
var <- lapply(R, function(x) decompose.cov(x)[[2]] )
( start <- list(root = root, matrix = corr, sd = var) )
``````
``````## \$root
## [1] 4.234861 4.709460 3.071194
##
## \$matrix
## \$matrix[[1]]
## [,1] [,2] [,3]
## [1,] 1.0000000 0.7722849 0.9688873
## [2,] 0.7722849 1.0000000 0.7348893
## [3,] 0.9688873 0.7348893 1.0000000
##
## \$matrix[[2]]
## [,1] [,2] [,3]
## [1,] 1.0000000 0.7722849 0.9688873
## [2,] 0.7722849 1.0000000 0.7348893
## [3,] 0.9688873 0.7348893 1.0000000
##
##
## \$sd
## \$sd[[1]]
## [1] 0.002414689 0.003206314 0.002684417
##
## \$sd[[2]]
## [1] 0.002414689 0.003206314 0.002684417
``````
Now we check whether it works:
``````( custom.start.handle <- ratematrixMCMC(data=anoles\$data[,1:3], phy=anoles.phy, start=start
, gen=1000, dir=tempdir()) )
``````
``````##
## MCMC chain with multiple regimes
## Number of traits: 3
## Number of species: 125
## Number of regimes: 2
## Number of generations: 1000
## Output files: ratematrixMCMC.28754.*
## Files directory: Same as analysis directory ('.')
## | 3,038 | 10,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.787024 |
https://stats.stackexchange.com/questions/532628/how-to-determine-long-term-coexistence-in-a-stochastic-simulation | 1,718,449,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00484.warc.gz | 497,269,590 | 42,216 | # How to determine long-term coexistence in a stochastic simulation?
I wrote a stochastic simulation to investigate a system in which replicators proliferate in compartments (rudimentary cells). The compartments contain two kinds of replicators: master sequences and mutants. Mutants can drive the system into extinction. I'm interested in where coexistence is possible in the parameter space.
The Model
The link above points to a version with non-overlapping generations, but these systems are usually modelled with overlapping generations. The following image is the flow diagram of the model with overlapping generations.
$$N$$ is the compartment population size, $$n$$ is the number of master sequences in a given compartment, and $$m$$ is the number of mutants in a given compartment, $$s$$ is split size (the total number of replicators in a compartment at which the compartment divides).
The metabolic efficiency of a compartment is the number of master sequences it harbours divided by the total number of master sequences in the system.
Replication happens by randomly choosing a replicator based on replication rates. If a master sequence is chosen to replicate, it mutates to a mutant with probability $$U$$.
One generation corresponds to $$N$$ compartment divisions.
The population counts as extinct when there are no master sequences left in any compartment.
My aim
With a given $$(N,s,\text{relative mutant replication rate})$$ combination, I search for the highest value of the mutation rate, with two decimal place accuracy, at which the population survives, i.e. master sequences and mutants coexist.
Coexistence does look different from extinction. It looks like an equilibrium:
But I'm reluctant to call it an equilibrium because this is a stochastic model with one absorbing state, so as far as I know, the mutant will eventually fix with probability $$1$$.
How to determine what counts as coexistence?
It is usually done by choosing a maximal number of generations. This choice is primarily constrained by the available computing power and time.
For example, I have done preliminary simulations of the above system with an insane amount of maximal generations. It turned out that the majority of extinctions happens before $$20000$$ generations. And the appearance of time series figures like the above seemed, uh, pretty equilibrium-like in the cases when the system has reached $$20000$$ generations. So I decided to regard it as coexistence when the system has reached $$20000$$ generations.
But this seems quite subjective and imprecise. There must be a better way to determine coexistence than guessing it based on figures. Note that I don't want to determine when the system will never go extinct because I actually think it will eventually certainly go extinct (see above). I want to distinguish somehow quick extinction from coexistence during runtime.
I have heard that some do it by fitting a line to the values of a relevant variable (e.g. the number of master sequences in the system) along the generations. They regard it as coexistence and stop the simulation when the slope of this line ceases to be significantly different from zero.
But how should that work? This seems very suspicious to me, raising questions about stopping rules and multiple testing.
Indeed, someone I asked about this said that I shouldn't use significance testing for this. They suggested that I should examine the confidence interval of the slope instead and regard it as coexistence when there are only irrelevant values in the confidence interval. But then, how is that more objective?
Questions
So what is a proper, statistically sound way to do this? What are the relevant search terms of this topic of determining the "equilibrium" of a system like this?
• The best way is to study the system theoretically to develop a quantitative expression for the rate of decay. Have you done that? What are the results?
– whuber
Commented Jun 29, 2021 at 15:43
• @whuber Hmm, I'm afraid I don't know what do you mean. I am sorry. Would you elaborate? Commented Jun 29, 2021 at 16:10
• Let me put it like this: your question is stated so generally that it's unanswerable. (Some of the bad advice about testing can be addressed, though.) It's simply impossible to make (nontrivial) inferences about all future generations based on a finite number of them unless you have a quantitative model for how the future is likely to behave.
– whuber
Commented Jun 29, 2021 at 16:25
• @whuber I have edited my question. Unfortunately, these models are not analysed mathematically because they are said to be too complicated. My maths is not nearly as advanced to disprove this. I warmly welcome any textbook suggestions. I'm at the level of this book. Commented Jun 30, 2021 at 6:08 | 991 | 4,788 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-26 | latest | en | 0.939115 |
http://matheduc.com/ | 1,487,539,733,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170253.67/warc/CC-MAIN-20170219104610-00142-ip-10-171-10-108.ec2.internal.warc.gz | 157,667,726 | 19,268 | ## Elliptic Integrals
The expression non-deterministic approach for working out Elliptical integrals is sometimes difficult because the execution time isn’t explicitly defined, or, in different stipulations, the moment the computer attempts potential solution until it finds one that’s successful, which means that the time it requires for an algorithm to locate an appropriate solution is, thus, unknown. So it is enough to take no more than the initial 3 terms of the set. Thus, the key periods aren’t independent. They take actual period and imaginary period.
Topology, however, would not turn into an emerging field without problems. Because of numerical aspects it isn’t feasible to implement it like a recursive algorithm. Using decreasing module there’s still another method to discover an algorithm based on formula (22). The parameter is known as the complementary modulus. It is called the complementary modulus. Be aware that we also supply a faster approximate way of calculating these functions (see approxCompleteEllipticIntegralsKE()). The normalized rational function may be displayed in an easy form is known as a discrimination issue.
Generally, elliptic integrals cannot be expressed regarding elementary functions. They generally can not be expressed in terms of elementary functions. Accordingly the issue with these integrals is they are a special sort of discontinuous integrals. The second integral is a little more troublesome, but we can, finally, reduce it regarding the 3 elliptic integral regular forms above.
Nine times the distinction is all about 1.10. For instance the so-called indirect geodetic issue. Concerning the Parameter and the Argument. To access more information, you need only looks across the world’s University sites and digital media broadcasters – using a residential VPN like this will help if you have issues with content filtering.
The practical interest for those filters design is really a frequency. Further on the instance is designed to show how simple equation (18) can be put into place. Obviously it’s hard to determine which of the results is proper. This may be repeated and consequently increasing module and decreasing amplitudes are obtained. This outcome is precisely what is expected. Finally the outcomes of table 3 and table 6 ought to be compared. Just a few examples of these ought to be shown within this introduction.
The use of math principles allows scientists to get a deeper comprehension of their various studies in addition to preforming calculations for practical use. We’ll therefore utilize modern terminology throughout this short article to prevent confusion. So I also included the simple theory supporting the implementation. But there’s still another method to read equation (11). Elliptic curves are also utilized in cryptography, as it’s an incredibly efficient encrypter, meaning it is frequently used in little devices as mobile phones etc.. Initially, the convergence of this method ought to be examined.
To de-normalize the transfer feature, the scale frequency has to be used. So it must be thought to take the most suitable mixture of both, to find the best result. Therefore, the solution was simple to verify. He simply derived the cure with a power collection. Any help would be far appreciated, I apologize whether that dilemma is in reality trivial or well-known. However on account of the essence of the progression I’ll first cover another simple tool in mathematics, namely integration. This conversion could be repeated in exactly the same way as we’ve done before.
References:
IP Cloaker
## Using the Poisson Distribution
The distribution grows more symmetric once the mean is larger. The standard distribution is wholly dependent on its parameters JJL and CT.. Generally, for smaller values of mu, the distribution isn’t symmetric but skewed. The Poisson distribution would be a really excellent approximation if it weren’t for the other matter. In addition, there are some empirical methods for checking for a Poisson distribution such as commonly used by psychologists.
Intelligence testing is utilized by Clinical Psychologists to measure a kid’s intellectual capabilities in a number of specific domains. An ipsative test compares your existing performance to your previous performance on the exact same test. Intelligence quotient (IQ) tests are a succession of tests used to find out the overall intelligence of someone regarding the overall population and the end result is known as an IQ score. It’s important to ensure your sample is representative of a whole process. In fact, you will often collect data samples which do not seem to be normally distributed.
For example in security analysis, it is important to take an accurate distribution of attackers – if you take a single subset of perhaps attacks made in a single hour then your results will almost certainly be skewed. The logs of a server will record all the attacks but consecutive ones are likely to be from the same attacker simply using a fake IP address to attempt to login. It’s a pattern that will be familiar to any web site owner who suffers brute force attacks on their blog or web site.
In the event the population isn’t normally distributed, larger sample sizes are necessary. Using Excel you can efficiently figure out the confidence statistics you demand. In summary VORP is a helpful statistic for comparing players and may also be used to rate trades. The conventional deviation is utilized to assess the variability of a data set and it’s a central value of the confidence calculation. It’s likewise the conventional deviation squared.
One needs to know that make a great project with the write and the acceptable content isn’t a simple job and thereof there must be an appropriate guidance for this undertaking. Recently, it has been noted in a research this subject is thought to be among the difficult subjects for almost all of the student’s because it has complicated math subject including plenty of formulas to address the intricate problems. Even though the theory looks intuitively fair and it’s widespread in practice, nowadays we’re conscious of disadvantages. All it requires is to understand an easy, intuitive concept and you’ll master them very quickly.
If you are aware that the data is described by a different distribution than the standard distribution, you will need to use the procedures of that distribution or utilize nonparametric analysis methods. In cases like this, you might need to adjust all data with the addition of a particular value to all data being analyzed. You might discover that at this point you have normally-distributed data.
You might need to get a separate, less effective calculator for exams. Put all of them in a column and apply the COUNT” function to fix the variety of information points that you want to base you calculation on. Additional the VORP calculation considers position scarcity and home field effects. It’s called as parameter. From sample of information, the parameter is figured in statistics. It requires the input of significant amount that’s 100% subtracted by the degree of confidence we’re interested in.
## Starting Differential Calculus
Calculus wasn’t discovered all at one time. The majority of the moment, the Calculus is simpler than the Algebra. The calculus has the ability to handle the natural situation where the car moves with changing speed. Differential calculus was applied to numerous questions that aren’t first formulated in the language of calculus.
Fortunately, an individual can do a great deal of introductory physics with only a few of the fundamental methods. Mathematics is a language that is utilised to spell out abstract concepts. Since Math is a tough subject for the majority of children, a teacher cannot enable the students just stick to a textbook, as it is only part of the info necessary to fix the Math problems correctly.
Calculus is also utilized to acquire a more precise comprehension of the disposition of space, time, and motion. Calculus covers an extensive region of modern mathematics. Calculus is an extensive subject, and it forms the foundation for much of modern mathematics. Calculus is a Latin word that is a small stone that’s used for counting. Calculus plays an essential part in modern mathematics. There are some good documentaries online about the history of calculus too although you might need to use a VPN if in certain locations, here’s a good one – How to Bypass the Netflix Block.
Calculus may be used together with other mathematical disciplines and provides tools to fix many troubles. Calculus is all about changes, so using a single variable calculus is the secret to the overall problem too.
There are some other normal topics in such a training course. By doing this, you might be able to address a question on the last exam, even when you do not fully understand why it is you are solving it the way you’re. With the integral, you’re going to be given lots of problems to solve, but there’s no algorithm. After that, identify just what you should see in the issue. Men and women who are considering solving problems employing ancient, theory-based methods are encouraged to have a calculus program or two while trying to find online college math courses. Also, although in our simple instances, it doesn’t be a lot of difference, in higher dimensions, understanding the derivative during the linear approximation is a lot more relevant. It’s important that you know the difference whenever you are working these issues.
The above mentioned graph needs to be familiar to anybody who has studied elementary algebra. There are a number of methods for solving these equations to discover an explicit sort of the function. There are lots of approaches to work out this equation. As soon as an explicit solution to a differential equation isn’t possible, the slope field stipulates a means to fix the equation graphically. Polynomials are differentiated utilizing the rule. There’s an algorithm for doing this. On occasion the computation could be long and complicated.
The process of discovering the derivative is known as differentiation. Put simply, you spend the derivative of the very first factor feature and multiply it by the second factor feature, then spend the derivative of the second factor feature and multiply it by the very first factor feature, and add them together. The operation of locating a derivative is known as differentiation. Specific derivatives, like the derivatives of logarithms, exponents, sums, quotients, goods, and trigonometric functions, are taught, and implicit differentiation.
In the event the variables can’t be separated directly, then other methods have to be utilised to fix the equation. In the event the function isn’t continuous, the limit could differ from the worth of the function at that point. It won’t execute the function. It is possible to integrate discontinuous functions and they’re going to come out continuous. It isn’t possible to have a system which is described this equation. Otherwise the integration method is like rectangular coordinates. In addition, there are tons of unique applications for differential Calculus!
Further Reading/Video – Changing IP Online
## The Genius of Lionel Euler
As one would anticipate, Euler was not just a gifted mathematician he was also good at all types of things. Euler realised he could use the exact same formula for to receive a similar outcome. Euler was an extremely gifted mathematician, not just in terms of what he accomplished, but in addition with respect to his methods. On account of the simple fact that Euler lived in Switzerland, a land-locked nation, he wasn’t given the chance to see ocean-going ships. It is sufficient to apply the prior process to Euler’s polynomials to seek out result. But algebra was not merely a highly effective tool for engineers. This equation is known as the second type of Euler’s equation.
His genius was admired by some best contemporaries of his time. His mathematical genius was proved in subjects of of infinitesimal calculus and graph theory. He’s recognized among the most extraordinary mathematicians in the country. This became used by biologists all around the Earth, and he is called the `father of contemporary taxonomy’.
His works include things like finding various computation methods to determine volume and area of many shapes, for instance, conic section. But first, I wish to insist an additional time on the spectacular splendor of Euler’s identity. There’s certainly no limit to the collection of such amazing men and women, whose works created the platform for other people to create seminal works in mathematics. His work is used in many ways to this day, some of the functions he discovered are important in this connected world of security and encryption, which are used to often to filter content like the way the Netflix VPN ban was implemented.
We only receive a one solution and will require a second solution. It can likewise be demonstrated utilizing an elaborate integral. But this dilemma is an excellent instance of the way to approach problems with constraints. This is an issue with a few of the equations on the website unfortunately.
The use of math principles allows scientists to acquire a deeper comprehension of their various studies and preforming calculations for practical use. It appears natural to rate the limit by viewing its modulus and argument separately. Alternatively, you may choose to argue that the investment rate ought to be computed monthly.
The complicated exponential forms developed by Euler are often utilised in electrical engineering and physics. As stated by the forms of materials that arrive in contact, there are several kinds of friction. We thus need to be able to use the second kind of EulerOs equation to work out this issue. Both materials in touch with one another may be a liquid and a good, a gas and a good, or possibly a gas and a liquid in touch with one another. It’s the simplest of all of the varieties of friction to analyze.
Presentation linked to this post was prepared using this software.
## Using Diffie-Hellman Key Exchange for PKI
Diffie-Hellman can likewise be applied as a member of public key infrastructure today. This is the reason Diffie-Hellman is employed together with an extra authentication method, generally digital signatures. The so-called Diffie-Hellman method stipulates a manner. Authenticated Diffie-Hellman is also called Unified Diffie-Hellman. In this instance, an eavesdropper would need to fix the overall Diffie-Hellman problem to get the shared key which is regarded extremely hard to accomplish.
Inside this scheme, the public key is utilized to stop main-in-the-middle attacks. Neither reaction is correct, Bellovin argues. It must be a main concern of all online users. The issue is called the discrete logarithm issue. Another problem is that initial full disk encryption sometimes takes a very long time (based on the quantity of data to encrypt). These are a few of the issues that you want to get concerned with prior to encrypting your device. Although the topic needs to be presented initial, it is visited after the initial two examples so there’s a frame of reference.
Unfortunately, a decent amount of detail wasn’t readily apparent (or was lost), and the loss could lead to a system which is insecure overall. The example below is offered in dh-init. Lastly, The sample below is offered in dh-unified, which is particularly useful in network enabled cryptography applications like VPNs such as this.
Step 7 when the function call is done, the Diffie-Hellman public key will be prepared to use. As it’s a sizable and apparently random number, an expected hacker has almost no possibility of correctly guessing x, despite the aid of a highly effective computer to conduct millions of trials. There are a lot of methods to implement PAKE. It is among the first practical examples of Key exchange implemented within the area of cryptography. Possible solutions incorporate the usage of digital signatures and other protocol variants.
Encryption is a somewhat essential measure. It is extra effort of an extra layer to protect your data. As more become aware of the generally unsecured nature of the Internet, it will undoubtedly become increasingly popular. Asymmetric encryption is extremely slow. It’s very slow in comparison to Symmetric Encryption. Symmetric encryption is extremely fast. This may be remedied with a vital confirmation protocol.
The entire program can be found in dh-param. Its the type of thing that may subtly earn a system insecure, though the system was constructed from secure components. Most encryption methods provide a choice between them instead of combining them. If you would like to share your gadget among more than 1 gadget remember, data encrypted on a single gadget isn’t going to be readable on another gadget. After you encrypt your device, there isn’t any turning back. You could also encrypt your devices with the assistance of android apps. Both of them are public and can be employed by all the users in a system.
The functions have these signatures. Utilizing the Windows Crypto API functions may be the alternate. This will assign the worth of P. The worth of p could be large but the worth of q is usually tiny. At this time the values of both G and P need to be sent to the intended recipient together with the essential when conducting an essential exchange. It’s a sort of key exchange. After the preceding examples, we’re finally prepared to carry out key exchange and arrive at a shared secret.
## Riemann Integration Theory and Practice
In regard to Riemann Integration, Riemann Integration may be used to decide on the accuracy of the Fourier Series used. Additionally, This is called Riemann integral. Here we are going to try out the approach of Riemann. Riemann integral was made by Bernhard Riemann. Riemann’s integral cannot take care of this function.
When modeling real-world troubles, it’s easy to compose expressions involving derivatives. In humans, a genetic mutation usually means this sugar isn’t present in any cell within the body. They might also play part in disease susceptibility.
Area isn’t yet properly defined, you can access a documentary on some TV channels in Europe although you’ll probably need a residential VPN service. There is a multitude of techniques to attempt to ascertain the region. Indeed, the area below the similar bit of the given parabola is always precisely the same, whatever letter we write near the horizontal axis.
Integration may be used to discover areas, volumes, central points and lots of helpful things. It is a main topic in calculus. It is a way of adding slices to find the whole. Nonetheless, in this scenario, it is possible to utilize Riemann Integration to discover the area below the curve, and thus the distance the object has traveled.
A huge value for the mesh is supposedly coarse, though a little mesh is supposed to be fine. So if we opt to use a different variable in precisely the same formula, the form and thus the integral stay an identical. The integrated function is occasionally known as the integrand. There are different functions that are non-integrable too. We’ve been doing Indefinite Integrals to date. Classical multiple integrals are wholly covered via this approach.
In this kind of situation, the integration operation is needed to discover the function, which gave the specific derivative. Now we’ll make this procedure precise. There are lots of approaches, here we use the one which is simplest to manage. On the opposite hand, the case of Dirichlet function demonstrates that if there’s too many points of discontinuity, the function isn’t Riemann integrable.
All about the way that it works and more. To start with, you can imagine this integral using almost the exact same picture. However, the time wasn’t yet ready for measure theory. It’s all an issue of interpretation in the end. This is a rather crucial question. And this matter is to turn into central to the notion of integral.
His proof demands a monotonicity of f. It is founded on an easy observation that the area of a rectangle is not hard to calculate. Now we must choose their heights. The snaky shape is known as the integration sign, it’s in fact an extremely elongated S (for sum). The very first pattern is known as altriciality. This breaking pieces are known as the partition.
## Calculating the Geometric Mean
As you may be expecting, the geometric mean can become very complicated. Geometric Mean may be the square root of the item of both numbers. The geometric mean is really not the arithmetic mean and it’s not a straightforward average. Now take a glance at The Mean Machine.
For instance, the typical percentage sum of growth in a financial institution account per year employs the geometric mean since the development each year is dependent upon multiplying the amount within the bank account by the proportion development. We’d utilize the geometric mean when we would like to find out the ordinary rate of growth in the event the growth rate is dependent upon multiplication. It’s likewise called average. It’s used to figure out the typical rate of growth once the growth depends upon multiplication as in the instance of annual proportion growth of the bank account.
Biologists utilize this calculation for quantifying average population development prices, which are also known as the intrinsic rate of development” for early phases of population development where there are not any density dependent facets controlling populations. Usually, this problem arises when it’s desired to figure out the geometric mean of the percent change in a population or perhaps a financial return, including negative numbers.
In addition, They are natural for summarizing ratios. Instead, as described within this tip, you ought to utilize Excel’s GEOMEAN function to figure the geometric mean of the range of numbers. In these instances, you ought to utilize Excel’s GEOMEAN function to compute the typical growth rate, given the effect of compounding.
To calculate geometric mean in these types of situations, you need to utilize Method 2. This dilemma wants the Altitude Rule. Understanding the issue.
The following step is in fact solving these undesirable boys. The geometric mean employs multiplication and roots. The arithmetic mean is used while the growth depends upon addition. Also, study the formula and the manner to use it.
Perhaps some insight is provided by the graph in the right. For instance, in the easy function GeoMean” is provided to figure out the geometric mean of a number of data. This easy example can be achieved in your head. Look carefully at the diagram to learn what is given.
Aybeesee’s height could be the short side of a single baby triangle and also the lengthy side of the other baby triangle. The altitude is additionally the lengthy side of the bottom triangle, and the more compact piece of the hypotenuse may be the brief side.
On occasion, you might need to figure out the mean of the range of numbers. The data ought to be divided into 4 different types. Nominal category utilizes some labels. Negative numbers could cause imaginary results based on how many negative numbers are really in a set.
In case you are multiplying eight numbers with each other, then you’ll take the eighth root. Do not forget that the capital PI symbol method to multiply a number of numbers. For instance, for the item of two numbers, we’d take the square root. The AM-GM for just two positive numbers are sometimes a beneficial tool in examining some optimization difficulties.
John Williams
http://www.uktv-online.com/entertainment/watch-the-bbc-news-live-online/
## Introduction to Exponential Equations
Two matrices with precisely the same dimensions may be added utilizing the procedure for matrix addition. A matrix is just a rectangular or square selection of numbers. An echelon matrix is utilized to solve a method of linear equations. Nonsingular matrix is, in addition, called Invertible Matrix.
An universe is made within a second. In a sequence of numbers, the following expression within the series is figured by means of a formula, which uses previous expressions within the similar series. A limit test for divergence is actually a convergence test that’s based upon the truth that the conditions of the convergent series needs to have a limit of zero. Similarly according to a lot of theologists the Pi collection, the trigonometric collection, the exponential collection etc represent each a singular facet of the truth that nature beholds.
To begin with, the fundamental formulas are listed, that can help you solve problems. Find a lot of fractional exponent issues and begin solving. Using graphical techniques to figure out the mathematical troubles. A way of solving a method of linear equations.
Mathematical induction is employed to prove complex troubles. Analytical methods are utilized to figure out the problems by assistance from algebraic and numeric methods.
Algorithm is a basic step by step to reach the solution of any issue. Graphic calculators are utilized to solve an issue graphically. Here may be the work for this particular equation. Alternate strategy to answer the n power problem is provided below.
Relative minimum is actually a point within the graph, which is at the bottom point for that specific section. Vertical shrinking of the geometrical figure is known as vertical compression. There is a variety of factors which go into calculating a yield. Based on this theorem, there’s always a minumum of one absolute maximum and one absolute minimum for absolutely any continuous function on a closed interval.
There’s a specific formula you can employ if you want to change the base of the logarithmic function. The y coordinate of the point is normally known as the ordinate. This kind of arbitrary point is known as focus of the parabola. Unless an explanation isn’t proved correct for an expression, it’s always a field of examination and debate.
Relative maximum is just a point within the graph, which is at the maximal point for that specific section. The highest point of the role or relation over the whole domain is known as absolute maximum. It’s used to try whether a relation is really a function.
Mathematical expressions are derived for quite a few classes of reactions. Functions involving absolute value are likewise a great case of piecewise functions. The aforementioned examples offer some insight into the complete process of simplifying exponents. There is a variety of factors contributing to the infrequent usage of declarative languages.
Geometric mean is really a method of discovering the average of specific pack of numbers. Mathematically, a scalar is supposed to be any actual number or some quantity that may be measured using an individual actual number. The response lies within this period, where in fact the initial quarks and anti-quarks were formed.
While this primary formula is simple, there are several variables that could factor in to this formula. The measure of the closeness of the value to the true value of the result is known as accuracy. The independent quantity within an algebraic expression is known as variable. The magnitude could be the absolute value of the quantity, there are numerous scientific calculators that can work these out for you – even one as an app on the iPad.
Absolute value is, in addition, known as a mod value. Magnitude is actually a value, and it could never become a negative number.
## Browsing Some Hyperbolic Function Properties
Or how not to get a date, ever – but seriously here are a few other hyperbolic function properties. The hyperbolic secant arises within the profile of the laminar jet. The hyperbolic functions might be defined concerning the legs of the appropriate triangle covering this sector. A trigonometric function is really a periodic function, however a hyperbolic function isn’t so. Basically you’re supposed in order to differentiate any function. Hyperbolic functions could be differentiated and integrated. Since the hyperbolic functions are defined regarding the pure exponential function, it’s not surprising that their inverses might be expressed concerning the organic logarithm function. Now think about the hyperbolic functions.
Recall the inverse of the pure exponential function may be the pure logarithm function. Rainbow phenomena could also be seen within the droplets generated by lawn sprinklers and hose nozzles, or any additional wellspring of water droplets.The function is a whole analytical use of that is described over the entire complex plane and doesn’t have branch cuts and branch points. A function with a bounded selection. An unit length ought to be chosen freely. In the hyperbolic geometry it truly is allowable for at least one line to be parallel to the very first (meaning the parallel lines won’t ever meet the very first, however far they’re extended).
If you have trouble watching the above video it might be due to some stupid region locking, the same reason Netflix block proxies and stop me watching my favorite sci-fi shows !
This one involves utilizing the slope of the function at 0, just how the sine and also the cosine did. The very first and second derivative tests are generally utilised to get the absolute maximum of the function. Based on this theorem, there’s always a minumum of one absolute maximum and one absolute minimum for absolutely any continuous function on a closed interval.
The absolute most striking distinction is that the hyperbolic functions aren’t periodic. The hyperbolic functions are defined regarding the organic exponential functionex. The most important utilization of these functions will be to integrate common and easy functions with less computation as well as the other utilization of these functions could be observed within the models of real-life difficulties. Just as the hyperbolic functions themselves could possibly be expressed regarding exponential functions, so their inverses could possibly be expressed regarding logarithms.
Analytical methods are utilized to answer the problems by the aid of algebraic and numeric methods. It’s likewise called arithmetic sequence. Segment of the circle is any internal region of the circle, bounded by means of an arc or possibly a chord.
An integration at which bounds of integration has discontinuities within the graph. The hyperbolic functions might be expressed concerning exponentials. These hyperbolic identities may be verified. In the old times, mathematicians had a tough time locating the equation of the curve.
In a sequence of numbers, the following expression within the series is figured by means of a formula, which uses previous expressions within the identical series. Sometimes natural data are in a form of an asymptotic curve like Eq. Recall the first derivative is known as y prime and also the second derivative is known as y double-prime. So allow me to explain to why the all-natural log is the one that’s all-natural for economics.
In the aforementioned applet, there’s a pull-down menu at the very top to select which function you want to examine. Visualization isn’t yet complete, however. So this is really typical of mathematics.
Its equations are usually given within the polar coordinates. Any equation that is certainly reflexive, symmetric, and transitive. The in the very first formula is actually a hyperbolic-angle and also a parameter. This isn’t a very useful formula.
Henry (Netflix) Galsworthy
## The Fourier Series
Digital data is discrete instead of continuous, and thus, the signal really needs to be sampled at fixed intervals. Additionally, It has got the automatic measurements you’d expect in addition to FFT. The Fourier series may be used to locate a function which will excite the greatest number of frequencies possible. The DFT, such as the Fourier collection, implies a periodic extension of the first function.
The above video may not be available in all countries, but simply use this method to change to use US IP addresses if you have a problem.
As an example, planetary probes can generate loads of image data, also a big challenge is always to send large quantity of data back to Earth. Maxwell also proved mathematically, that this kind of phenomenon of the cloud contracting into a planet couldn’t occur. It can’t be done within the physical universe.
To fully grasp how Battery Life Saver works, it really is first required to comprehend how batteries work. There are several chemical reactions which take place within the battery. A battery is really a system that stores electricity by way of chemical changes within the battery. With Battery Life Saver, the battery is not going to slow down or die due to lead sulfate, the most familiar reason behind battery failure.
The only distinction is that, it gives the signal amplitude within the frequency domain, as the oscilloscope supplies the signal amplitude within the time domain. The 2 electrons within the P subshell, being within the outer subshell, combine more freely than both electrons within the S subshell. In fact it really is useless to do this for stationary signals. Although the logarithmic scale stipulates a broader array of frequencies, it doesn’t offer the absolute financial value of the signal.
Each layer got two interfaces. The gradient is figured utilizing the derivative of the Gaussian filter. Thermocouple is just a system which uses conduction for a manner of heat transfer. This system is, in addition, known as the vector signal analyzer.
To realize more compression, the image’s quality must be compromised. If there is loss of information involved within the data encoding process, the initial array of data symbols doesn’t even need to be encoded as-is. It achieves compression by using a combination of quite a few different algorithms. Besides compression and archiving, it truly is additionally effective at achieving error recovery.
Due to the way the laser light is reached, it becomes highly focused and intense. The technology is extremely close but doesn’t make use of a visible light. Lasers work as an outcome of resonant results. They are one of the most significant inventions developed during the 20th century.
Both substantial sensitivity and superior directionality can be accomplished using a huge telescope in the receiver end. The mid-frequency of the band is automatically tuned within the device, since the range changes. It always has 1 side that is certainly dark, precisely the same side. The output of the laser is actually a coherent electromagnetic field.
Upon examination, the testing wasn’t done in agreement with any recognised specifications like SAEJ2185, but instead testing very similar to what was done by Don Plisko. In the last few ages there’s been considerable interest in the growth of neural network based pattern recognition apparatus due to their capacity to classify data. It is even feasible to set up short-range optical data connections with no direct field of sight. Moreover, it does not result in interference between different data links, so it generally does not require a license to be operated, which is superior when it comes to data security, since it’s more challenging to intercept a tightly collimated laser beam when compared to a radio link. | 6,877 | 35,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-09 | longest | en | 0.907759 |
https://metanumbers.com/26370 | 1,600,833,683,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209665.4/warc/CC-MAIN-20200923015227-20200923045227-00225.warc.gz | 507,292,397 | 7,543 | ## 26370
26,370 (twenty-six thousand three hundred seventy) is an even five-digits composite number following 26369 and preceding 26371. In scientific notation, it is written as 2.637 × 104. The sum of its digits is 18. It has a total of 5 prime factors and 24 positive divisors. There are 7,008 positive integers (up to 26370) that are relatively prime to 26370.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 18
• Digital Root 9
## Name
Short name 26 thousand 370 twenty-six thousand three hundred seventy
## Notation
Scientific notation 2.637 × 104 26.37 × 103
## Prime Factorization of 26370
Prime Factorization 2 × 32 × 5 × 293
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 8790 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 26,370 is 2 × 32 × 5 × 293. Since it has a total of 5 prime factors, 26,370 is a composite number.
## Divisors of 26370
1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90, 293, 586, 879, 1465, 1758, 2637, 2930, 4395, 5274, 8790, 13185, 26370
24 divisors
Even divisors 12 12 8 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 68796 Sum of all the positive divisors of n s(n) 42426 Sum of the proper positive divisors of n A(n) 2866.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 162.388 Returns the nth root of the product of n divisors H(n) 9.19937 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 26,370 can be divided by 24 positive divisors (out of which 12 are even, and 12 are odd). The sum of these divisors (counting 26,370) is 68,796, the average is 286,6.5.
## Other Arithmetic Functions (n = 26370)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7008 Total number of positive integers not greater than n that are coprime to n λ(n) 876 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2896 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 7,008 positive integers (less than 26,370) that are coprime with 26,370. And there are approximately 2,896 prime numbers less than or equal to 26,370.
## Divisibility of 26370
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 1 2 0
The number 26,370 is divisible by 2, 3, 5, 6 and 9.
• Abundant
• Polite
## Base conversion (26370)
Base System Value
2 Binary 110011100000010
3 Ternary 1100011200
4 Quaternary 12130002
5 Quinary 1320440
6 Senary 322030
8 Octal 63402
10 Decimal 26370
12 Duodecimal 13316
20 Vigesimal 35ia
36 Base36 kci
## Basic calculations (n = 26370)
### Multiplication
n×i
n×2 52740 79110 105480 131850
### Division
ni
n⁄2 13185 8790 6592.5 5274
### Exponentiation
ni
n2 695376900 18337088853000 483549033053610000 12751188001623695700000
### Nth Root
i√n
2√n 162.388 29.7648 12.7432 7.65988
## 26370 as geometric shapes
### Circle
Diameter 52740 165688 2.18459e+09
### Sphere
Volume 7.68102e+13 8.73836e+09 165688
### Square
Length = n
Perimeter 105480 6.95377e+08 37292.8
### Cube
Length = n
Surface area 4.17226e+09 1.83371e+13 45674.2
### Equilateral Triangle
Length = n
Perimeter 79110 3.01107e+08 22837.1
### Triangular Pyramid
Length = n
Surface area 1.20443e+09 2.16105e+12 21531
## Cryptographic Hash Functions
md5 1ed013af9f67744751dc13861ebeea2f 125ca23a9c505163b411335b03d866535143a6bd 3367b464231f342161c8aaa7b057811c53e653eabfd5cb48583ca12886dd1657 7da76448c33f2c0fe5c4f232afab2a7b46e47e01d8757bfaa5b08027f38b4abeb57764d9497cd047493a529cab5ba6cc5a811479b8f4437c6657a07dd43b4f6d 4fdebea076d3630b31fdf4156863c4a3fec75d92 | 1,496 | 4,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-40 | longest | en | 0.785821 |
https://en-academic.com/dic.nsf/enwiki/1676 | 1,653,651,487,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00667.warc.gz | 273,891,884 | 12,069 | # Amicable number
Amicable number
Amicable numbers are two different numbers so related that the sum of the proper divisors of the one is equal to the other, one being considered as a proper divisor but not the number itself. Such a pair is (220, 284); for the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110, of which the sum is 284; and the proper divisors of 284 are 1, 2, 4, 71, and 142, of which the sum is 220. Amicable numbers were known to the Pythagoreans, who credited them with many mystical properties.
A pair of amicable numbers constitutes an aliquot sequence of period 2.
A general formula by which these numbers could be derived was invented circa 850 by Thabit ibn Qurra (826-901): if :"p" = 3 × 2"n" − 1 − 1, :"q" = 3 × 2"n" − 1,:"r" = 9 × 22"n" − 1 − 1, where "n" > 1 is an integer and "p", "q", and "r" are prime numbers, then 2"npq" and 2"nr" are a pair of amicable numbers. This formula gives the amicable pair (220, 284), as well as the pair (17296, 18416) and the pair (9363584, 9437056). The pair (6232, 6368) are amicable, but they cannot be derived from this formula. In fact, this formula produces amicable numbers for "n" = 2, 4, and 7, but for no other values below 20000.
In every known case, the numbers of a pair are either both even or both odd. It is not known whether an even-odd pair of amicable numbers exists. Also, every known pair shares at least one common factor. It is not known whether a pair of coprime amicable numbers exists, though if any does, the product of the two must be greater than 1067. Also, a pair of coprime amicable numbers cannot be generated by Thabit's formula (above), nor by any similar formula.
Amicable numbers have been studied by Al Madshritti (died 1007), Abu Mansur Tahir al-Baghdadi (980-1037), Al-Farisi (1260-1320), René Descartes (1596-1650), to whom the formula of Thabit is sometimes ascribed, C. Rudolphus and others. Thabit's formula was generalized by Euler. Prior to Euler only three pairs of amicable numbers had been found. Because Euler found 59 more amicable numbers, the work of Eastern mathematicians in this area is largely forgotten.
The pair (9363584; 9437056) has often been attributed to Descartes, but it was actually first discovered by Muhammad Baqir Yazdi in Iran. [cite journal | last = Costello |first = Patrick | title = New Amicable Pairs Of Type (2; 2) And Type (3; 2) | journal = MATHEMATICS OF COMPUTATION | volume = 72 Number 241 | pages = 489–497 | publisher = American Mathematical Society | date = 2002-05-01 | url = http://www.ams.org/mcom/2003-72-241/S0025-5718-02-01414-X/S0025-5718-02-01414-X.pdf | accessdate = 2007-04-19 ]
The first few amicable pairs are: (220, 284), (1184, 1210), (2620, 2924), (5020, 5564), (6232, 6368) OEIS|id=A063990
If a number equals the sum of "its own" proper divisors, it is called a perfect number.
The following Python language code allows you to check if two numbers are Amicable:
# Definition of the functiondef amicable_numbers(x,y): #only two different numbers may be amicable if x = y: return False # Sum all values i in [1,x) where i divides x sum_x = sum(i for i in xrange(1, x) if x % i = 0) sum_y = sum(i for i in xrange(1, y) if y % i = 0) return (sum_x = y) and (sum_y = x)
# Program bodyn_1=int(raw_input('Enter nº 1: '))n_2=int(raw_input('Enter nº 2: '))
if amicable_numbers(n_1,n_2): print 'Amicable! :)'else: print 'Not Amicable :('
And the following PseudoCode finds all the Amicable Numbers between two numbers
Procedure Find Amicable Pairs Enter Starting Number Enter Last Number For all the numbers between the Starting Number and Last Number and call this FirstNumber Call the Function to add all of the Proper Divisors of the FirstNumber and call this SumOfAllProperDivisorsOfFirstNumber Call the Function to add all of the Proper Divisors again this time using SumOfAllProperDivisorsOfFirstNumber and call this SumOfAllProperDivisorsOfSecondNumber If SumOfAllProperDivisorsOfFirstNumber is equal to SumOfAllProperDivisorsOfSecondNumber then You found a pair End if End For LoopEnd of Procedure
Function Add All Of The Proper Divisors of A Number (call this ANumber) Set the initial Running Total to 0 For all the numbers between 1 and half of ANumber and call this CurrentLoopNumber If you divide ANumber with CurrentLoopNumber and the remainder is zero then Add the result to the Running Total End If End For Loop Return the Running TotalEnd of Function
References
* Wells, D. (1987). "The Penguin Dictionary of Curious and Interesting Numbers" (pp. 145 - 147). London: Penguin Group.
* [http://amicable.homepage.dk/knwnc2.htm All known amicable numbers]
* [http://ftp.cwi.nl/CWIreports/MAS/MAS-R0307.pdf A good 2003 survey on current status of Amicable number mathematics.]
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### Look at other dictionaries:
• amicable number — noun Either of a pair of amicable numbers … Wiktionary
• amicable number — Math. either of a pair of positive integers in which each member is equal to the sum of the submultiples of the other, as 220 and 284. * * * … Universalium
• amicable number — noun : either of two numbers each of which is equal to the sum of all the submultiples of the other * * * Math. either of a pair of positive integers in which each member is equal to the sum of the submultiples of the other, as 220 and 284 … Useful english dictionary
• 10000 (number) — Number number = 10000 prev = 9999 next = 100000 range = 10000 100000 cardinal = 10000 ordinal = th ordinal text = ten thousandth numeral = decamillesimal factorization = 2^4 cdot 5^4 prime = divisor = roman = overline|X unicode = overline|X, ↂ… … Wikipedia
• number theory — Math. the study of integers and their relation to one another. Also called theory of numbers. [1910 15] * * * Branch of mathematics concerned with properties of and relations among integers. It is a popular subject among amateur mathematicians… … Universalium
• Number theory — A Lehmer sieve an analog computer once used for finding primes and solving simple diophantine equations. Number theory is a branch of pure mathematics devoted primarily to the study of the integers. Number theorists study prime numbers (the… … Wikipedia
• 60000 (number) — Number number = 60000 range = 10000 100000 cardinal = 60000 ordinal = th ordinal text = sixty thousandth factorization = 2^5 cdot 3 cdot 5^4 bin = 1110101001100000 oct = 165140 hex = EA6060,000 (sixty thousand) is the number that comes after… … Wikipedia
• 70000 (number) — Number number = 70000 range = 10000 100000 cardinal = 70000 ordinal = th ordinal text = seventy thousandth factorization = 2^4 cdot 5^4 cdot 7 bin = 10001000101110000 oct = 210560 hex = 1117070,000 (seventy thousand) is the number that comes… … Wikipedia
• amicable numbers — noun A pair of numbers having the property that the sum of the divisors of each, excluding itself, is equal to the other number. The factors of 220, excluding 220 itself, are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110 and sum to 284; the factors… … Wiktionary
• Sociable number — Sociable numbers are generalizations of the concepts of amicable numbers and perfect numbers. A set of sociable numbers is a kind of aliquot sequence, or a sequence of numbers each of whose numbers is the sum of the factors of the preceding… … Wikipedia | 2,068 | 7,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-21 | latest | en | 0.948261 |
https://electronics.stackexchange.com/questions/256244/modeling-passive-elements-at-high-frequencies | 1,723,440,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00815.warc.gz | 183,498,729 | 44,684 | # Modeling passive elements at high frequencies
It is well-known that any electrical passive element (Resistor, Capacitor,Inductor) contains parasitics that usually manifest themselves at high frequency ranges of operation.
For example, a resistor at DC can be modeled, simply, by just a resistance that depends on the material and geometry of the element. At higher frequencies, parasitic capacitance and inductance start to show up, and this can be experimentally found (for example) by:
1. Voltage - Current phase difference.
2. Impedance dependency on frequency.
The same argument is also valid for capacitors and inductors, in which their ideal model is altered and parasitic effects are added at high frequencies.
The plot of Impedance vs. frequency can tell us about these parasitics and when do they start to show up. They will also inform us about the valid frequency range of operation, in which after it, the element no longer behaves normally (An inductor acting as a capacitor after its self resonance frequency (SRF) for example) as shown below:
So when we mention the term "high frequencies", we (probably?) mean beyond the SRF, as elements start behaving as unintended.
Up to my understanding, every passive element generally behaves in a way that I thought of explaining by the figure below:
My questions are somehow interconnected and they are:
1. Is the concept of figure (2) correct?
2. In addition to the parasitics that were mentioned above, there seem to be variations in the physical values of L, C, and R themselves (Blue region in figure 2), meaning they all become functions of frequency: R(w), C(w), L(w). Is this true?
I concluded this from:
a. Skin effect for R, making it a function of frequency.
b. Picture #1 above (blue graph). Does the inductance really become negative, or is it the tool we are using that is telling me my inductor value was lost due to the large parasitic capacitance value?
1. How does the geometrical feature size of the passive element (regardless of its shape) plays a role in determining the frequency separating the green and the blue regions in figure (2)? In other words, is there a way to determine if I have to consider AC effects or not using the knowledge of my feature size?
2. Can we say line separating the blue and red regions can be represented by the SRF?
• High frequencies aren't the only regime of nonideal behavior. For example, some capacitors have soakage effects that mainly affect the behavior at very low frequencies. Commented Sep 5, 2016 at 21:28
• EMI/EMC covers a broad territory. The pathology of slots, gaps and skinny traces all have reactance and wavelengths and insulators of any type are also capacitors depending on gap between conductors. Can you be more specific about physical to electromagnetic examples? Commented Sep 1, 2021 at 13:16
1/2) More or less, I try to clarify, It seems to me you believe your graph in figure (2) shows R and L variations w.r.t frequency.
That graph simply shows impedance of a parallel resonant circuit made of approx constant 100nH inductor, 45kohm resistor and a $\frac{1}{(2\pi\times 2\,\text{GHz})^2\times 100\,\text{nH}}\approx 65\,\text{fF}$ capacitor.
This is one first level approximation which may however be good in many circumstances.
Then one can argue resistance is frequency dependant and add this to the model, the same for inductance and capacitance but this can usually hardly be seen on graphs unless you carefully measure, analize and fit curves to models.They are ususally hidden in measurements errors.
Then you can add many other extra parasitics and non linerities outlined above by other contributors, but there's no such an evidence in the sweep you posted.
b) Your meter shows variable (possibly negative) inductance just becouse you ask it to do so. It just measures an impedance, one complex ratio between voltage and current and then represent it as you configured it to do: you asked for Ls+Rs and if measured impedance phase does not agree with that model it just keep computing and findes "negative inductances".
Usabilty boundary (green/blue transition in fig. (2)) is not function of your component parasitic only, but it depends pretty much on the rest of the circuit e.g. if using that inductor in a resonator you should add stray capacitance to the computations and see if you get consisten numbers.
4) Yes SRF is that blue to red boundary.
3) Dimension do count. For resistors and capacitors usually the longer the part the higher the stray inductance. E.g. small SMD 0204 or 0603 may exhibit a few hundred pH while some big HV MKM capacitors I happened to use were specified as 7nH/cm w.r.t to terminals pitch.
Is the concept of figure (2) correct?
**YES, negative (-j) inductance means capacitance at high f after PRF. Negative capacitance (j) from series metal about <1 nH/mm is inductive after SRF.
Ceramic is also a crystal with SRF and may have a PRF in special cases especially when parallel caps are used, together as in MLCC or on PCB , except for NP0/C0G types.**
Impedance Testing with a Network Analyzer can measure these if they are problems.
It is not exactly as shown or as you assume. It is also not possible to have a universal model for all passive components. L&C will have series R and parallel C,
C will have series R,L and parallel R, and sometimes more parts in the model
Components may have a series and parallel resonant frequency and many other variables for Vdc, T('C), I(L) with saturation.
It is possible to use caps above "series resonant freq", SRF but it is often useful to have inductors near or above "parallel resonant frequency", PRF unless to add another in series with higher PRF.
( Sometimes specs use SRF for both modes as self res. freq.)
When your operating frequency is in the range of the SRF, it is necessary to either choose a different part with higher SRF, or an "additional" part to provide the desired impedance (extra smaller L in series or smaller C in //) | 1,366 | 6,003 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-33 | latest | en | 0.953155 |
https://engcourses-uofa.ca/books/vibrations-and-sound/transient-vibrations/convolution/ | 1,723,434,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00196.warc.gz | 175,657,104 | 14,187 | ## Transient Vibrations: Convolution
The method of convolution is based upon the response to an impulse load. We have shown previously in equation (7.4) that the response to an impulse applied at time is
where is the magnitude of the impulse. To proceed, we next need to consider the response of the system due to an impulse that is applied at some other time .
We are interested in the response of the mass at some later time , where . In such a case the response of the mass at time due to an impulse applied at an earlier time is simply
(7.9)
where is the elapsed time after the impulse was applied. Note that this assumes that the mass is at rest at time (ie and ). Equation (7.9) then gives us the position of the mass at time , which is after the impulse has been applied. Note that with the assumed initial conditions, the response of the mass if the impulse were not applied would be simply . Therefore, what (7.9) in fact represents is the change in the response , compared to what would have occurred if the impulse had not been applied.
Figure 7.9: Forcing function represented as a series of impulses
Finally, consider some arbitrary forcing function as shown in Figure 7.9. This function can be considered to be composed of an infinite number of infinitesimal impulsive loads, a few of which are illustrated. Again, we are interested in the response of the mass at some arbitrary time . This response will be affected by all of the impulses that have been applied up to time . Consider one such impulse at time . The application of the impulse at will have an effect on the position of the mass for all later times , with the change in the position of the mass at time given by
exactly analogous to our previous results. As discussed previously, this actually represents the change in the response due to the application of this one impulse. To find the actual position of the mass at time , we simply add up (ie integrate) the effects of all the individual impulses up to time , which are the only ones that can affect the position at time . As a result, we get
(7.10)
(Note that is simply a dummy variable.) Equation (7.10) is known as Duhamel’s Integral and is a special form of the convolution integral which gives the position for an arbitrary load . As discussed, this assumes that the initial conditions are
If instead we have more general initial conditions given by
then (7.10) must be replaced by the more general form
(7.11)
#### EXAMPLE
A mass, initially at rest, is subject to a step load as shown below. Determine the response of the system for . Use the method of convolution.
#### EXAMPLE
A mass, initially at rest, is subject to a ramp load as shown below. Determine the response of the system for . Use the method of convolution.
#### EXAMPLE
A mass, initially at rest, is subject to a rectangular step load as shown below. Using the the method of convolution, determine the response of the system for
a)
b)
#### EXAMPLE
A mass, initially at rest, is subject to a forcing function shown below. Using the the method of convolution, determine the response of the system for
a)
b)
#### EXAMPLE
A mass is initially moving such that at some time the mass has position and velocity . At that instant a step load is applied to the mass as shown below.
Determine the response of the system for using
a) Newton’s laws directly
b) The method of convolution.
### Convolution with Base Excitation
In some situations we do not apply a force directly to the mass but instead the base supporting the mass is given some motion . We have already seen that the equation of motion in such a case is
This is identical to the equation of motion in equation (7.1) with replaced by . As a result, all of our previous results can be applied. In particular, Duhamel’s Integral (equation (7.10)) becomes
or, since ,
(7.12)
### Convolution with Viscous Damping
So far we have only considered undamped situations. We will now consider the response of a spring–mass system with a viscous damper subjected to an impulsive load. We will again assume the initial conditions are
before any impulses are applied. Similar to the undamped case, the initial conditions just after the impulse has been applied are
The subsequent free vibration response is given by
With the given initial conditions, we find that
so that the response becomes
If the impulse were instead applied at some later time , the response would be
Now as before we consider an arbitrary forcing function to be composed of an infinite number of infinitesimal impulses. The response at time due to the impulse applied at some earlier time is
The total response at is the cumulative effect of all the impulses applied up to time , i.e.
If or before the impulses are applied this result needs to be generalized as before. | 1,035 | 4,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-33 | latest | en | 0.963728 |
https://www.khanacademy.org/math/4th-engage-ny/engage-4th-module-1/4th-module-1-topic-e/v/regrouping-borrowing-twice-example | 1,675,055,756,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00771.warc.gz | 826,923,564 | 80,233 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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### Unit 1: Lesson 5
Topic E: Multi-digit whole number subtraction
# Multi-digit subtraction with regrouping twice
Sal using regrouping to subtact 9601-8023. Created by Sal Khan.
## Want to join the conversation?
• Can you put the bigger number on top or not?
• yes you can because in addition and multiplication it doesn't matter if the larger number is on top or bottom.
Ex.
4169 8234
+ +
8234 4169
because they both equal 12403 anyway
• 128-100=-28 Is that correct
• (1.) 128 - 100 = 28 (The smaller number subtracted from the larger)
(2.) 100 - 128 = (-)28 (The larger number subtracted from the smaller)
• I can see but it would be quicker if you put it vertically and added digit by diget
(1 vote)
• what about 1332-1546 Doesn't sound easy but you can just use a calculator if you need help but anyways here is the answer. -214 which is negative 214 you welcome
• 1546 is greater than 1332, so (1332 − 1546) is going to be a negative number.
The trick is to factor out (−1):
1332 − 1546 = −(1546 − 1332)
1546 − 1332 = 214, so 1332 − 1546 = −214
• thank you sal you tot me a ting or two
• what if you get to a problem that you can't do and you try regrouping it but it doesn't work
• it has to work if you regroup you will get the anwser unless you regrouped wrong
(1 vote)
• rly would like to see the expanded solution of 2000 - 105 as example for 4th grade
2000 + 0 + 0 + 0
.......... - 100 - 0 - 5
(regrouping 1)
1000 + 1000 + 0 + 0
........... - 100 - 0 - 5
(regrouping 2)
1000 + 900 + 100 + 0
........... - 100 - 0 - 5
(regrouping 3)
1000 + 900 + 90 + 10
........... - 100 - 0 - 5
= 1000 + 800 + 90 + 5
• try subtracting 1232-2576 I will put the answer anyways so you don't have to look it up -1344 same as the last question/answer it is negative 1344 you welcome.
(1.) Expanded the top number becomes 1000 + 200 + 300 + 20
(2.) Expanded the bottom number becomes 2000 + 500 + 70 + 6
(3.) The regrouping shown in the video can't be done easily for this problem; however, you can switch the place of the minuend (1,232) with the subtrahend (2576).
2576 - 1232
Once that has been done, you can subtract normally. No major regrouping is needed. After you finish subtracting, be sure to place the negative sign (-) in front of the answer as 2,567 subtracted from 1,232 would produce a number with a value less than 0.
(1 vote)
• I still don't get how five hundred becomes ten. Could someone please explain? I've watched the entire video and still don't understand. | 816 | 2,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-06 | latest | en | 0.852877 |
https://www.tutorialstonight.com/python-nested-while-loop | 1,695,910,205,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00502.warc.gz | 1,154,374,021 | 21,948 | # Python Nested While Loop
Python While loop is used to execute a block of statements repeatedly until a given condition is satisfied. And when we talk about nested while loop, it means that a while loop is present inside another while loop.
A nested while loop is a loop inside a loop. The inner loop will be executed completely for each iteration of the outer loop.
Let's see the syntax of a nested while loop in Python:
``````while condition1:
while condition2:
# statement(s)
# statement(s)``````
The inner loop will execute for each iteration of the outer loop. For simple calculation, if the outer loop runs 5 times, and the inner loop runs 3 times, then the inner loop will run for 5 × 3 = 15 times in total.
## Python Nested While Loop Example
Let's see a simple example of a nested while loop in Python to understand it better.
### Example 1
Here is a simple example that shows how a nested while loop works in Python.
``````i = 1
while i <= 3:
print("Outer Loop: ", i, "time ------------------")
j = 1
while j <= 2:
print("Inner Loop:", j)
j += 1
i += 1``````
Output:
```Outer Loop: 1 time ------------------
Inner Loop: 1
Inner Loop: 2
Outer Loop: 2 time ------------------
Inner Loop: 1
Inner Loop: 2
Outer Loop: 3 time ------------------
Inner Loop: 1
Inner Loop: 2```
In the above example, you can clearly see that the inner loop is executed for each iteration of the outer loop.
### Example 2
Let's say we want to print the table up to 5. We can use a nested while loop to do this.
``````i = 1
# outer loop
while i <= 5:
j = 1
# inner loop
while j <= 10:
print(i * j, end=" ")
j += 1
print()
i += 1``````
Output:
```1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50```
Here, the outer loop runs 5 times and for each iteration of the outer loop, the inner loop runs 10 times and prints the table of that number.
### Example 3
Using nested while loop to print elements of a list of lists.
``````# nested while loop to print elements of a list of lists
list1 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
i = 0
# outer loop
while i < len(list1):
j = 0
# inner loop
while j < len(list1[i]):
print(list1[i][j], end=" ")
j += 1
print()
i += 1``````
Output:
```1 2 3
4 5 6
7 8 9```
The outer loop runs for the number of elements in the list and for each iteration of the outer loop, the inner loop runs for the number of elements in the inner list.
### Example 4
Printing a pyramid pattern using nested while loop.
``````size = 5
i = 0
# outer loop
while i < size:
j = 0
# inner loop to print spaces
while j < size - i - 1:
print(' ', end='')
j += 1
k = 0
# inner loop to print stars
while k < 2 * i + 1:
print('*', end='')
k += 1
print()
i += 1``````
Output:
``` *
***
*****
*******
*********```
### Example 5
You can use nested while loop to sort a list of numbers in ascending order.
``````list1 = [5, 3, 8, 6, 7, 2]
i = 0
# outer loop
while i < len(list1):
j = i + 1
# inner loop
while j < len(list1):
if list1[i] > list1[j]:
temp = list1[i]
list1[i] = list1[j]
list1[j] = temp
j += 1
i += 1
print(list1)``````
Output:
`[2, 3, 5, 6, 7, 8]`
### Example 6
This is an example of a nested while loop inside a nested while loop.
``````i = 1
# outer loop
while i <= 2:
print("Outer Loop ", i, "times")
j = 1
# middle loop
while j <= 2:
print(" Middle Loop:", j)
k = 1
# inner loop
while k <= 2:
print(" Inner Loop:", k)
k += 1
j += 1
i += 1
``````
Output:
```Outer Loop 1 times
Middle Loop: 1
Inner Loop: 1
Inner Loop: 2
Middle Loop: 2
Inner Loop: 1
Inner Loop: 2
Outer Loop 2 times
Middle Loop: 1
Inner Loop: 1
Inner Loop: 2
Middle Loop: 2
Inner Loop: 1
Inner Loop: 2```
## Nested While Loop with Break and Continue
break and continue statements are used to control the flow of the loop. You can use them in nested while loop as well.
1. break statement is used to terminate the loop.
2. continue statement is used to skip the current iteration of the loop.
Let's see some examples.
### Loop through a list of lists using break
Let's say we want to print the elements of a list of lists. We can use a nested while loop to do this. But what if we want to stop the loop when we reach a particular element? We can use break statement to do this.
``````list1 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
i = 0
# outer loop
while i < len(list1):
j = 0
# inner loop
while j < len(list1[i]):
# break the loop when we reach 5
if list1[i][j] == 5:
break
print(list1[i][j], end=" ")
j += 1
print()
i += 1``````
Output:
```1 2 3
4
7 8 9```
Here, the inner loop breaks when it reaches 5. But outer loop continues to run and restarts the inner loop for the next element in the outer list.
### Loop through a list of lists using continue
With the help of continue statement, we can skip the current iteration of the loop.
The following program prints the elements of a list of lists except 5 and 7.
``````list1 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
i = 0
# outer loop
while i < len(list1):
j = 0
# inner loop
while j < len(list1[i]):
# skip the current iteration when we reach 5 or 7
if list1[i][j] == 5 or list1[i][j] == 7:
j += 1
continue
print(list1[i][j], end=" ")
j += 1
print()
i += 1``````
Output:
```1 2 3
4 6
8 9```
## Conclusion
So, this is how you can use nested while loop in Python. You can use nested while loop to do a lot of things. | 1,754 | 5,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-40 | longest | en | 0.80939 |
https://support.nag.com/numeric/nl/nagdoc_26/nagdoc_fl26/html/g01/g01fef.html | 1,701,567,759,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100476.94/warc/CC-MAIN-20231202235258-20231203025258-00017.warc.gz | 623,277,524 | 5,097 | # NAG Library Routine Document
## 1Purpose
g01fef returns the deviate associated with the given lower tail probability of the beta distribution, via the routine name.
## 2Specification
Fortran Interface
Function g01fef ( p, a, b, tol,
Real (Kind=nag_wp) :: g01fef Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: p, a, b, tol
#include nagmk26.h
double g01fef_ ( const double *p, const double *a, const double *b, const double *tol, Integer *ifail)
## 3Description
The deviate, ${\beta }_{p}$, associated with the lower tail probability, $p$, of the beta distribution with parameters $a$ and $b$ is defined as the solution to
$PB≤βp:a,b=p=Γa+b ΓaΓb ∫0βpBa-11-Bb-1dB, 0≤βp≤1;a,b>0.$
The algorithm is a modified version of the Newton–Raphson method, following closely that of Cran et al. (1977).
An initial approximation, ${\beta }_{0}$, to ${\beta }_{p}$ is found (see Cran et al. (1977)), and the Newton–Raphson iteration
$βi=βi-1-fβi-1 f′βi-1 ,$
where $f\left(\beta \right)=P\left(B\le \beta :a,b\right)-p$ is used, with modifications to ensure that $\beta$ remains in the range $\left(0,1\right)$.
## 4References
Cran G W, Martin K J and Thomas G E (1977) Algorithm AS 109. Inverse of the incomplete beta function ratio Appl. Statist. 26 111–114
Hastings N A J and Peacock J B (1975) Statistical Distributions Butterworth
## 5Arguments
1: $\mathbf{p}$ – Real (Kind=nag_wp)Input
On entry: $p$, the lower tail probability from the required beta distribution.
Constraint: $0.0\le {\mathbf{p}}\le 1.0$.
2: $\mathbf{a}$ – Real (Kind=nag_wp)Input
On entry: $a$, the first parameter of the required beta distribution.
Constraint: $0.0<{\mathbf{a}}\le {10}^{6}$.
3: $\mathbf{b}$ – Real (Kind=nag_wp)Input
On entry: $b$, the second parameter of the required beta distribution.
Constraint: $0.0<{\mathbf{b}}\le {10}^{6}$.
4: $\mathbf{tol}$ – Real (Kind=nag_wp)Input
On entry: the relative accuracy required by you in the result. If g01fef is entered with tol greater than or equal to $1.0$ or less than (see x02ajf), the value of is used instead.
5: $\mathbf{ifail}$ – IntegerInput/Output
On entry: ifail must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this argument you should refer to Section 3.4 in How to Use the NAG Library and its Documentation for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, because for this routine the values of the output arguments may be useful even if ${\mathbf{ifail}}\ne {\mathbf{0}}$ on exit, the recommended value is $-1$. When the value $-\mathbf{1}\text{ or }1$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Note: g01fef may return useful information for one or more of the following detected errors or warnings.
Errors or warnings detected by the routine:
If on exit ${\mathbf{ifail}}={\mathbf{1}}$ or ${\mathbf{2}}$, then g01fef returns $0.0$.
${\mathbf{ifail}}=1$
On entry, ${\mathbf{p}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{p}}\le 1.0$.
On entry, ${\mathbf{p}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{p}}\ge 0.0$.
${\mathbf{ifail}}=2$
On entry, ${\mathbf{a}}=〈\mathit{\text{value}}〉$ and ${\mathbf{b}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{a}}>0.0$.
On entry, ${\mathbf{a}}=〈\mathit{\text{value}}〉$ and ${\mathbf{b}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{a}}\le {10}^{6}$.
On entry, ${\mathbf{a}}=〈\mathit{\text{value}}〉$ and ${\mathbf{b}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{b}}>0.0$.
On entry, ${\mathbf{a}}=〈\mathit{\text{value}}〉$ and ${\mathbf{b}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{b}}\le {10}^{6}$.
${\mathbf{ifail}}=3$
The solution has failed to converge. However, the result should be a reasonable approximation. Requested accuracy not achieved when calculating beta probability. You should try setting tol larger.
${\mathbf{ifail}}=4$
The requested accuracy has not been achieved. Use a larger value of tol. There is doubt concerning the accuracy of the computed result. $100$ iterations of the Newton–Raphson method have been performed without satisfying the accuracy criterion (see Section 9). The result should be a reasonable approximation of the solution.
${\mathbf{ifail}}=-99$
See Section 3.9 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 3.8 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 3.7 in How to Use the NAG Library and its Documentation for further information.
## 7Accuracy
The required precision, given by tol, should be achieved in most circumstances.
## 8Parallelism and Performance
g01fef is not threaded in any implementation.
The typical timing will be several times that of g01eef and will be very dependent on the input argument values. See g01eef for further comments on timings.
## 10Example
This example reads lower tail probabilities for several beta distributions and calculates and prints the corresponding deviates until the end of data is reached.
### 10.1Program Text
Program Text (g01fefe.f90)
### 10.2Program Data
Program Data (g01fefe.d)
### 10.3Program Results
Program Results (g01fefe.r)
© The Numerical Algorithms Group Ltd, Oxford, UK. 2017 | 1,848 | 5,837 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 60, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.575156 |
http://docplayer.net/32948897-2-graph-terminology.html | 1,537,506,768,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156780.2/warc/CC-MAIN-20180921033529-20180921053929-00157.warc.gz | 73,537,613 | 26,699 | # 2. Graph Terminology
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1 2. GRAPH TERMINOLOGY Graph Terminology 2.1. Undirected Graphs. Definitions Suppose G = (V, E) is an undirected graph. (1) Two vertices u, v V are adjacent or neighbors if there is an edge e between u and v. The edge e connects u and v. The vertices u and v are endpoints of e. (2) The degree of a vertex v, denoted deg(v), is the number of edges for which it is an endpoint. A loop contributes twice in an undirected graph. If deg(v) = 0, then v is called isolated. If deg(v) = 1, then v is called pendant. Example V = {v 1, v 2, v 3, v 4 } and E = {e 1, e 2, e 3, e 4 }. e 1 v 1 v 2 e 2 e 3 e 4 v 3 v 4 (1) v 2 and v 3 are adjacent. (2) deg(v 1 ) = 2 (3) deg(v 2 ) = 2 (4) deg(v 3 ) = 3 (5) deg(v 4 ) = 1 Notice that in computing the degree of a vertex in an undirected graph a loop contributes two to the degree. In this example, none of the vertices is isolated, but v 4 is pendant. In particular, the vertex v 1 is not isolated since its degree is 2.
2 2.2. The Handshaking Theorem. 2. GRAPH TERMINOLOGY 187 Theorem (The Handshaking Theorem) Let G = (V, E) be an undirected graph. Then 2 E = deg(v) v V Proof. Each edge contributes twice to the sum of the degrees of all vertices. Theorem is one of the most basic and useful combinatorial formulas associated to a graph. It lets us conclude some facts about the numbers of vertices and the possible degrees of the vertices. Notice the immediate corollary. Corollary The sum of the degrees of the vertices in any graph must be an even number. In other words, it is impossible to create a graph so that the sum of the degrees of its vertices is odd (try it!) Example Example Suppose a graph has 5 vertices. Can each vertex have degree 3? degree 4? (1) The sum of the degrees of the vertices would be 3 5 if the graph has 5 vertices of degree 3. This is an odd number, though, so this is not possible by the handshaking Theorem. (2) The sum of the degrees of the vertices if there are 5 vertices with degree 4 is 20. Since this is even it is possible for this to equal 2 E. If the sum of the degrees of the vertices is an even number then the handshaking theorem is not contradicted. In fact, you can create a graph with any even degree you want if multiple edges are permitted. However, if you add more restrictions it may not be possible. Here are two typical questions the handshaking theorem may help you answer. Exercise Is it possible to have a graph S with 5 vertices, each with degree 4, and 8 edges?
3 2. GRAPH TERMINOLOGY 188 Exercise A graph with 21 edges has 7 vertices of degree 1, three of degree 2, seven of degree 3, and the rest of degree 4. How many vertices does it have? Theorem Every graph has an even number of vertices of odd degree. Proof. Let V o be the set of vertices of odd degree, and let V e be the set of vertices of even degree. Since V = V o V e and V o V e =, the handshaking theorem gives us 2 E = deg(v) = deg(v) + deg(v) v V v V o v V e or deg(v) = 2 E deg(v). v V o v V e Since the sum of any number of even integers is again an even integer, the righthand-side of this equations is an even integer. So the left-hand-side, which is the sum of a collection of odd integers, must also be even. The only way this can happen, however, is for there to be an even number of odd integers in the collection. That is, the number of vertices in V o must be even. Theorem goes a bit further than our initial corollary of the handshaking theorem. If you have difficulty with the last sentence of the proof, consider the following facts: odd + odd = even odd + even = odd even + even = even If we add up an odd number of odd numbers the previous facts will imply we get an odd number. Thus to get an even number out of v V o deg(v) there must be an even number of vertices in V o (the set of vertices of odd degree). While there must be an even number of vertices of odd degree, there is no restrictions on the parity (even or odd) of the number of vertices of even degree. This theorem makes it easy to see, for example, that it is not possible to have a graph with 3 vertices each of degree 1 and no other vertices of odd degree Directed Graphs. Definitions Let G = (V, E) be a directed graph. (1) Let (u, v) be an edge in G. Then u is an initial vertex and is adjacent to v. The vertex v is a terminal vertex and is adjacent from u. (2) The in-degree of a vertex v, denoted deg (v) is the number of edges which terminate at v.
4 2. GRAPH TERMINOLOGY 189 (3) Similarly, the out-degree of v, denoted deg + (v), is the number of edges which initiate at v The Handshaking Theorem for Directed Graphs. Theorem For any directed graph G = (V, E), E = v V deg (v) = v V deg + (v). When considering directed graphs we differentiate between the number of edges going into a vertex verses the number of edges coming out from the vertex. These numbers are given by the in-degree and the out-degree. Notice that each edge contributes one to the in-degree of some vertex and one to the out-degree of some vertex. This is essentially the proof of Theorem Exercise Prove Theorem Underlying Undirected Graph. Definition If we take a directed graph and remove the arrows indicating the direction we get the underlying undirected graph. There are applications in which you may start with a directed graph, but then later find it useful to consider the corresponding undirected graph obtained by removing the direction of the edges. Notice that if you take a vertex, v, in a directed graph and add its in-degree and out-degree, you get the degree of this vertex in the underlying undirected graph New Graphs from Old. Definitions (W, F ) is a subgraph of G = (V, E) if W V and F E. 2. Given graphs G 1 and G 2, their union G 1 G 2 = (V 1 V 2, E 1 E 2 ).
5 2. GRAPH TERMINOLOGY Given graphs G 1 and G 2, their join, denoted by G 1 G 2, is the graph consisting of the union G 1 G 2 together with all possible edges connecting a vertex of G 1 that is not in G 2 to a vertex of G 2 that is not in G 1. Example Suppose G has vertex set V = {a, b} and one edge e = {a, b} connecting a and b, and H has a single vertex c and no edges. Then G H has vertex set {a, b, c} and edges {a, b}, {a, c}, {b, c}. Exercise Find the union and join of the graphs G 1 and G 2 below. a b a f b e e c d c g d G1 G 2 Exercise Prove that the union of two simple graphs is a simple graph. Exercise Prove that the join of two simple graphs is a simple graph Complete Graphs. Definition The complete graph with n vertices, denoted K n, is the simple graph with exactly one edge between each pair of distinct vertices. There are a certain types of simple graphs that are important enough that they are given special names. The first of these are the complete graphs. These are the simple graphs that have the maximal number of edges for the given set of vertices. For example, if we were using graphs to represent a local area network, a complete graph would represent the maximum redundancy possible. In other words, each pair of computers would be directly connected. It is easy to see that any two complete graphs with n vertices are isomorphic, so that the symbol K n is ambiguously used to denote any such graph. Complete graphs also arise when considering the question as to whether a graph G is planar, that is, whether G can be drawn in a plane without having any two edges intersect. The complete graphs K 1, K 2, K 3, and K 4 are planar graphs, but K n is not planar if n 5. Draw K 4 without making the edges intersect, then try to draw K 5 without creating an unwanted intersection between edges. (Notice that K n+1 can be created from K n by adding one new vertex and an edge from the new vertex to each vertex of K n.)
6 2. GRAPH TERMINOLOGY 191 Exercise Prove that the complete graph K n, n 1, is the join K n 1 G, where G is a graph with one vertex and no edges Cycles. Definition A cycle with n vertices {v 1, v 2,..., v n }, denoted by C n, is a simple graph with edges of the form {v 1, v 2 }, {v 2, v 3 }, {v 3, v 4 },..., {v n 1, v n }, {v n, v 1 }. Notice that a cycle must have at least 3 vertices. Here are examples of the first three possibilities: C C C Local area networks that are configured this way are often called ring networks. Notice that the following two graphs are isomorphic. Pay close attention to the labels. v 1 e 1 v 2 e 1 v 1 v 2 e 4 e 2 e 2 e 4 e 3 v 4 v 3 v 3 v 4 e 3 The point of the last illustration, is that sometimes you have to redraw the graph to see the ring shape. It also demonstrates that a graph may be planar even though this fact may not be evident from a given representation Wheels. Definition A wheel is a join C n G, where C n is a cycle and G is a graph with one vertex and no edges. The wheel with n + 1 vertices is denoted W n.
7 2. GRAPH TERMINOLOGY 192 Notice that a wheel is obtained by starting with a cycle and then adding one new vertex and an edge from that vertex to each vertex of the cycle. Be careful! The index on the notation W n is actually one less that the number of vertices. The n stands for the number of vertices in the cycle that we used to get the wheel n-cubes. Definition The n-cube, denoted Q n, is the graph with 2 n vertices represented by the vertices and edges of an n-dimensional cube. These graphs can be constructed recursively as follows: 1. Initial Condition. A 0-cube is a graph with one vertex and no edges. 2. Recursion. Let Q 1 n and Q 2 n be two disjoint n-cubes, n 0, and let f : Q 1 n Q 2 n be an isomorphism. Q n+1 is the graph consisting of the union Q 1 n Q 2 n, together with all edges {v, f(v)} joining a vertex v in Q 1 n with its corresponding vertex f(v) in Q 2 n. Q n can also be represented as the graph whose vertices are the bit strings of length n, having an edge between each pair of vertices that differ by one bit. The n-cube is a common way to connect processors in parallel machines. Here are the cubes Q 3 and Q cube 4-cube Exercise Find all the subgraphs of Q 1, and Q 2. Exercise Label the vertices of Q 4 appropriately, using bit strings of length four.
8 2. GRAPH TERMINOLOGY 193 Exercise Use your labeling of the vertices of Q 4 from Exercise to identify two disjoint Q 3 s, and an isomorphism between them, from which Q 4 would be obtained in the recursive description above. Exercise Prove that Q n+1 n-cubes, n 0. Q 1 n Q 2 n, where Q 1 n and Q 2 n are disjoint Exercise Prove that the 2-cube is not (isomorphic to) the join of two 1-cubes. Exercise Draw the graph Q 5. [Hint: Abandon trying to use a cube shape. Put on the top of the page and on the bottom and look for an organized manner to put the rest in between.] Bipartite Graphs. Definition A simple graph G is bipartite if V can be partitioned into two nonempty disjoint subsets V 1 and V 2 such that every edge connects a vertex in V 1 and a vertex in V 2. Definition A bipartite graph is complete if V can be partitioned into two disjoint subsets V 1 and V 2 such that there is an edge from every vertex V 1 to every vertex in V 2. K m,n denotes the complete bipartite graph when m = V 1 and n = V 2. The definition of a bipartite graph is not always consistent about the necessary size of V 1 and V 2. We will assume V 1 and V 2 must have at least one element each, so we will not consider the graph consisting of a single vertex bipartite. Note: There are no edges connecting vertices in V 1 in a bipartite graph. Neither are there edges connecting vertices in V 2. Exercise How many edges are there in the graph K m,n? Exercise Prove that a complete bipartite graph K m,n is the join G m G n of graphs G m and G n, where G m has m vertices and no edges, and G n has n vertices and no edges Examples. Example A star network is a K 1,n bipartite graph.
9 2. GRAPH TERMINOLOGY 194 K 1,8 Example C k, for k even, is a bipartite graph: Label the vertices {v 1, v 2,..., v k } cyclicly around C k, and put the vertices with odd subscripts in V 1 and the vertices with even subscripts in V 2. (1) Suppose V 1 is a set of airlines and V 2 is a set of airports. Then the graph with vertex set V = V 1 V 2, where there is an edge between a vertex of V 1 and a vertex of V 2 if the given airline serves the given airport, is bipartite. If every airline in V 1 serves every airport in V 2, then the graph would be a complete bipartite graph. (2) Supplier, warehouse transportation models where an edge represents that a given supplier sends inventory to a given warehouse are bipartite. Exercise Is the following graph bipartite? b c a d Exercise Prove that Q n is bipartite. [Hint: You don t need mathematical induction; use the bit string model for the vertex set.] Bipartite graphs also arise when considering the question as to whether a graph is planar. It is easy to see that the graphs K 1,n and K 2,n are planar for every n 1. The graphs K m,n, however, are not planar if both m and n are greater than 2. In particular, K 3,3 is not planar. (Try it!) A theorem, which we shall not prove, states that every nonplanar graph contains (in some sense) a subgraph (see Slide 15) isomorphic to K 5 or a subgraph isomorphic to K 3,3. Remark The important properties of a graph do not depend on how it is drawn. To see that two graphs, whose vertices have the same labels, are isomorphic, check that vertices are connected by an edge in one graph if and only if they are also connected by an edge in the other graph. e
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POWER SETS AND RELATIONS L. MARIZZA A. BAILEY 1. The Power Set Now that we have defined sets as best we can, we can consider a sets of sets. If we were to assume nothing, except the existence of the empty | 10,189 | 39,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2018-39 | longest | en | 0.903901 |
http://statistics.about.com/od/Descriptive-Statistics/a/How-To-Construct-A-Histogram.htm | 1,406,122,497,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997878430.52/warc/CC-MAIN-20140722025758-00236-ip-10-33-131-23.ec2.internal.warc.gz | 158,497,531 | 13,625 | • Share
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# How to Make a Histogram
A histogram is a type of graph that is used in statistics. This kind of graph uses vertical bars to display quantitative data. Although any basic software can construct a histogram, it is important to know what your computer is doing behind the scenes when it produces a histogram. The following walks through the steps that are used to construct a histogram.
### Classes or Bins
The first step is find the highest and lowest data value in the set of data. From these numbers the range can be computed by subtracting the minimum value from the maximum value. There is no set rule, but as a rough guide the range should be divided by five for small sets of data and 20 for larger sets. These numbers will give a class width or bin width. We may need to round this number and/or use some common sense.
Once the class width is determined, we choose a class that will include the minimum data value. We then use our class width to produce subsequent classes, stopping when we have produced a class that includes the maximum data value.
### Frequency Tables
Now that we have determined our classes, the next step is to make a table of frequencies. Begin with a column that lists the classes in increasing order. The next column should have a tally for each of the classes. The third column is for the count or frequency of data in each class. The final column is for the relative frequency of each class. This indicates what proportion of the data is in that particular class.
### Drawing the Histogram
Now that we have organized our data by classes, we are ready to draw our histogram.
1. Draw a horizontal line. This will be where we denote our classes.
2. Place evenly spaced marks along this line that correspond to the classes.
3. Label the marks so that the scale is clear and give a name to the horizontal axis.
4. Draw a vertical line just to the left of the lowest class.
5. Choose a scale for the vertical axis that will accommodate the class with the highest frequency.
6. Label the marks so that the scale is clear and give a name to the vertical axis.
7. Construct bars for each class. The height of each bar should correspond to the frequency of the class at the base of the bar.
Courtney Taylor | 500 | 2,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2014-23 | longest | en | 0.921721 |
https://math.stackexchange.com/questions/1640772/consecutive-prime-numerators-of-harmonic-numbers | 1,566,503,659,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317359.75/warc/CC-MAIN-20190822194105-20190822220105-00389.warc.gz | 552,423,456 | 30,215 | Consecutive prime numerators of harmonic numbers?
Let
$$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{a}{b}$$
and let $a$ and $b$ are coprime, $h_{n}=a$.
$h_{n}$ is prime for
$$n=2,3,5,8,9,21,26,41,56,62,69,79,89,91,122,127,143,167,201,230,247,252,290,349,376,459,489,492,516,662,687,714,771,932,944,1061,1281,1352,1489,1730, 1969,2012,2116,2457,2663,2955,3083,3130,3204,3359,3494,3572,3995,4155,4231,4250,4496,4616,5069,5988,6656,6883,8067,8156,8661,9097,\ldots$$
I guess proving that there are infinitely (or finitely) many primes of the form $h_n$ is very hard. But can we prove both of the $h_n$ and $h_{n+1}$ cannot be prime for $n>8$?
• What sort of question is this? I think it isn't trivial. Where is the conjecture from, just looking on the numbers or are there reasons why it should be right? – user302982 Feb 4 '16 at 19:40
• Just for reference, the first values of $h_n$ are listed here: oeis.org/A001008 – Thomas Andrews Feb 4 '16 at 19:49
• I edited the question to include some more values for $n$, more than oeis.org, which for some reason only includes values up to $3572$ – vrugtehagel Feb 4 '16 at 20:35
• @vrugtehagel : what do you already know about the numerators of the harmonic numbers ? – reuns Feb 4 '16 at 20:40
• @vrugtehagel Did you look in oeis.org/A056903/b056903.txt ? They call it a "b-file" and it goes up to $h_{97} = 78128$. – Mr. Brooks Feb 4 '16 at 22:01 | 551 | 1,414 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-35 | latest | en | 0.854311 |
https://solutionsadda.in/2023/10/08/nic-nielit-sta-2020-11/ | 1,726,158,353,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00377.warc.gz | 505,447,776 | 71,913 | October 8, 2023
October 8, 2023
October 8, 2023
###### UGC NET CS 2010 June-Paper-2
October 8, 2023
Question 6
Directions – Question number 6 to 10 are based on following information:
There are twelve persons named O, P, Q, R, S, T, U, V, W, X, Y and Z who live in a multi-storey apartment. The apartment has three floors and each floor has four rooms. These 12 persons who live in a set of 12 Rooms can be represented by a Matrix of 3 rows and 4 columns.
→Q lives immediate left below diagonally of a person who lives immediate left below diagonally of T.
→S lives immediate left above diagonally of a person who lives immediate left above diagonally of Z.
→X lives immediate right above diagonally of a person who lives immediate right below diagonally of O.
→P lives immediate right above diagonally of a person who lives immediate right above diagonally of Y.
→T lives immediate left above diagonally of a person who lives third to the right of V.
→Q lives immediate left of a person who lives two rooms below W in the same column.
→R lives to the immediate right of a person who lives immediate right above diagonally of Q. Z is living to the immediate left of U who receives ₹46000 as salary.
→The person who live on one of the floors (left to right) receive salary in the same order ₹50000, ₹48000, ₹47000 and ₹46000.
→The person who live on one of the floors (right to left) receive salary in the same order ₹45000, ₹38000, ₹35000 and ₹40000.
→The person who live on one of the floors (left to right) receive salary in the same order ₹37000, ₹42000, ₹36000 and ₹43000.
What is the aggregate salary of people living at the right end of the apartment?
A ₹ 137000 B ₹ 134000 C ₹ 125000 D ₹ 131000
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Inline Feedbacks | 465 | 1,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.932113 |
https://engineering.stackexchange.com/questions/19239/what-does-a-the-input-of-the-activation-function-mean | 1,723,094,188,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00094.warc.gz | 174,195,565 | 43,552 | # What does (a) the input of the activation function mean?
I am not very good at understanding the ANN theory, but please if anyone can explain me about what does (a) as input of the activation function mean? and how to get this value?
Thank you
• Maybe your supervisor doesn't understand it either. Commented Feb 3, 2018 at 10:59
• Many of us have not seen this before - could you provide a link with the basic equations/summary? From a statistics perspective - this looks like it may have some relationship to a Gaussian function, but without seeing it in context it is difficult to say. Commented Feb 3, 2018 at 11:03
• Having looked here, the activation function, in your case a sigmoid within hidden nodes, is simply a generic function that scales hidden node input to output in the range of 0 to 1. The a value would be the weighted average of the outputs of the input nodes connected to a specific hidden node. It is, in effect a response to information, as I see it. So by choosing a different activation function, you would change the nature of your model and it's response to information. Commented Feb 3, 2018 at 11:23
• Sorry correction - not weighted average but weighted sum. Commented Feb 3, 2018 at 11:30
• Well you have input values or parameters. You assign weighting factors to these depending upon their relative importance. Then the weighted values from the input nodes are added up. This becomes the a value for the activation function of the next layer hidden node that they (input nodes) are connected to. Because you are using a sigmoid, the output of the hidden node will be a value between 0 and 1. Commented Feb 3, 2018 at 11:59
Just a quick summary of my concluding thoughts on this step in your research. I understand that you have been advised to use the Sigmoidal Function, so this discussion will try to explain it's potential application.
First, I don't think that an Artificial Neural Network (ANN) would apply to your problem. Typically ANN would consider a range of factors on some outcome. When this process is considered, a lot of data is required along with specialized software. This is because training of the ANN is necessary to establish that the model works. Sometimes ANN is used to determine which factors are important in a non-linear, multivariate investigation. Let me provide an example that may illustrate its use.
## Application of ANN
Suppose we are experiencing a routine cracking problem of some concrete elements of a structure. We may collect data, on cracked elements and uncracked elements, to include:
1. Temperature of concrete.
2. Temperature of air.
3. Time of delivery.
4. Time of casting.
5. Wind speed.
7. Cement content.
8. Water content.
And potentially lots of other parameters (there could be dozens, even hundreds).
So in theory we could send perhaps thousands of replicated data sets (measurements) through a neural network to see which factors may or may not be predictive of cracking of concrete elements. That is how ANN is applied and it requires highly specialized software.
As a part of this process, we may also try different weighting factors, as $w$, on the input data to try to improve predictive ability. So, for example we could assemble a vector of weighting factors that can be used to adjust each of the input values. This is a trial and error process that may or may not result in a better outcome. We also may choose an activation function for the hidden nodes that receive the weighted input values. In your example the activation function is a sigmoid; there are many others. The activation function in simple terms will change the behaviour of your ANN in the sense that it changes the response to information. Also important is how the nodes are connected; a good software package will go a long way to doing this for you.
So I hope that this better explains what ANN is and how it may work. I do not think that it would apply to your project because I have experience of this technique being applied to large, international concrete durability research using thousands of data sets and the ANN method alone was enough for granting a PhD to a student from a major UK engineering school.
## Sigmoid Function
Now putting ANN aside, we look at your sigmoidal function from a different perspective. From Wikipedia:
"A wide variety of sigmoid functions have been used as the activation function of artificial neurons, including the logistic and hyperbolic tangent functions. Sigmoid curves are also common in statistics as cumulative distribution functions (which go from 0 to 1), such as the integrals of the logistic distribution, the normal distribution, and Student's t probability density functions."
At first glance I recognized that this function could be derived from the family of Gaussian distributions, and indeed it is. The Gaussian (Normal) distribution is the most commonly used statistical function in existence, and is often used as the basis of error computation and regression analysis. A Gaussian probability density function, or PDF, looks like the following:
You must think of a PDF as a special class of functions belonging to the class of Generalized (sometimes called Generic) functions. These are described as:
"In mathematics, generalized functions, or distributions, are objects extending the notion of functions. There is more than one recognized theory. Generalized functions are especially useful in making discontinuous functions more like smooth functions, and describing discrete physical phenomena such as point charges. They are applied extensively, especially in physics and engineering."
We can consider any probability density function to simply represent a state of knowledge. For example, the Uniform PDF illustrated below:
represents a maximum uncertainty in that, no matter what $z$ value we are considering within the bounds of $a$ and $b$, we will have the same probability of outcome. This is different from the Gaussian above because there is a much lower probability associated with the extreme values than the average value. With the Gaussian, as the standard deviation, or $\sigma$, gets smaller, we can consider that our state of knowledge has improved. Complete certainty, for example, would be where $\sigma \rightarrow 0$ (indeed this condition represents the Dirac Delta Function that is widely used in physics and engineering, especially Finite Element Analysis).
The beauty of using a PDF is that it transforms (or some may use the term normalizes) the data to a relative real range of 0 to 1 based upon our state of predictive knowledge. Thus we may convert and consider any problem in the relative, rather than absolute. This permits multiple data sets to be compared from different experiments and also allows prediction of behaviour, rather than magnitude.
The Sigmoid Function, just like a Gaussian PDF, is indeed one of these creatures. A Sigmoid function plot, with a replaced by $\sum{wx} + bias$, is shown below:
The outcome is a value between 0 and 1. The shape of the curve is very near to what is termed a Gaussian Cumulative Distribution Function or CDF. An excellent discussion of how this is derived may be found at TowardsDataScience.com.
## Possible Application
The application of the Sigmoid to data carries with it the assumption that your data would follow a Gaussian form. That is to say that extreme values are far less probable than others.
For your specific data, thickness or effective thickness, I would not expect that weighting factors $w$ would be necessary nor would there be a need for $bias$ but these decisions are up to you. You have a matrix of measurements that are to be transformed to the range of 0 to 1 using the Sigmoidal, or Gaussian CDF, function.
There are also Excel best practice guides describing how the Sigmoid Function may be used in regression analysis on experimental point data. This can be reviewed in a document that can be downloaded from ExcelCurveFitting.com.
How you apply it depends upon the specific needs of your work. For example, you could transform each value and run the full simulation, then transform collections of values within 50 by 50 cm grids to a single value and run it again, comparing the outcome of each. From this you may consider the need for weighting factors or bias values to improve the regression. It could be that this becomes a trial and error process that would eventually lead you to a deeper insight into the nature of your data and the problem you are attempting to solve. | 1,732 | 8,536 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-33 | latest | en | 0.945214 |
https://timedate.org/how-many-hours-between-12pm-and-8pm | 1,642,710,775,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00166.warc.gz | 564,604,137 | 1,390 | # How Many Hours Between 12pm And 8pm
Calculation of hours between 2 times 12pm And 8pm hours.
20 Hours
Difference between two times is 20 hours and is equal 1200 minutes. | 50 | 172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | latest | en | 0.911536 |
http://www.fixya.com/support/t8885181-u_check_division | 1,513,411,356,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948585297.58/warc/CC-MAIN-20171216065121-20171216091121-00443.warc.gz | 376,494,006 | 32,944 | Question about Office Equipment & Supplies
# How do u check division problems with remaiders
Posted by on
• Level 1:
An expert who has achieved level 1.
• Contributor
If you are using a Sharp or Casio scientific calculator and you would like your answer to be shown as a fraction, simply hit the [ab/c] button after you have entered your calculation.
For example:
15/6 = 2.5
press [ab/c] and your screen will show 2/1/2 which means 2 and 1/2.
Posted on Apr 11, 2011
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of.(from cars to computers, handyman, and even drones)
Goodluck!
Posted on Jan 02, 2017
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## Related Questions:
### Draw a flowchart that accepts two number and checks if the first is divisible by the second
I think the big question here is how do you determine in the first number is divisible by the second number. With a calculator, try various combinations of number to see when the first number is divisible by the second number and when it is not divisible. How are the results different from when they are divisible from when they are not divisible?
Good luck,
Paul
Nov 17, 2015 | Office Equipment & Supplies
### Is 450 divisible by 3 in fractions
450=9*5*10=9*50
9*50/3=3*50=150
To verify if a number is divisible by 3, add the digits in the number. If the sum of digits is divisible by 3, then thumber is also divisible by 3.
Sum of digits in 450 =4+5+0 =9, divisible by 3, so 450 is also divisible by 3.
Jul 06, 2014 | Office Equipment & Supplies
This works for a ford taurus so I thought I would put my 2 cents worth in. Sometimes the Hazard button plays havoc with the lighting system. If there is a hazard button on the steering shaft, try putting a toothpick to wedge the button one way or the other. A jiggling button will cause the same system if the connectors to the lights are tight and the grounds are good. Hope this helps
Mar 30, 2014 | 1996 Dodge Neon
### How do i factor numbers on a ti-30xs calculator
This calculator does not have a key that you can use to find the prime factor of an integer.
You can however use the calculator to find the factors
1. If number is even divide it by 2
Keep dividing by 2, while keeping track of how many times you divided by 2.
If you divided 5 times by 2 before getting an odd number, then your first factor is 2^5
2. Now try dividing by 3, keep track of the number of times you divided by 3 before you could not divide by 3 any more. If you divided 0 times by 3, your second factor is 3^0, or 1 and 3 is not a factor.
If you divided 4 times by 3, your second factor is 3^4
3. Divide by 5, until you can't any more. Keep track of the number of times you divided by 5.
4. Divide by all other prime numbers 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, etc. For each prime number, keep track of how many times you divided by it, until you could not any more.
Example: 23100
Division by 2
23100/2=11550 ---------> 1 division by 2
11550/2=5775 ----------> 2 divisions by 2
Note that 5775 is not divisible by 2. No more divisions by 2.
First factor is 2^2
Division by 3:
5775/3=1925 ----------> 1 division by 3
1925/3=641. 66667 Not an integer. No more divisions by 3.
2nd factor is 3^1
Division by 5 (number ends in 5)
1925/5 =385 -------> 1 division by 5
385/5=77 ----------> 2 divisions by 5 and no more (quotient does not end in 0 or 5)
3rd factor is 5^2
Division by 7
77/7=11 --------> 1 division by 7, and no more
4th factor is 7^1=7
Division by 11
11/11=1
5th factor is 11^1=11
Assembling the factors 2^2, 3^1, 5^2, 7, 11
Prime factorization of 23100 is
23100=(2^2)(3)(5^2)(7)11
Dec 18, 2013 | Texas Instruments TI-30XA Calculator
### 171cm to feet
5.61024 feet to find inches take the remaider and multiply by 12 to get 7 then multiply the remainder by 16 to get a little over 5 16ths 5'-7 5/16"
Nov 29, 2013 | Office Equipment & Supplies
### List 7 consecutive numbers none of which is prime?
114 divisible by 2
115 divisible by 5
116 divisible by 2
117 divisible by 3
118 divisible by 2
119 divisible by 7
120 divisible by 2
Aug 15, 2011 | Computers & Internet
### Ive.comprogram to find the number of and sum of all integers greater than 40 and less than 250 that are divisible by 5 in java coding ????
I'm not going to do your homework for you, but here's a hint. You can use the Java modulus operator % to find if a number is divisible by 5. It provides the remainder after dividing, which will be zero if the number is divisible by 5. Here's a pseudocode example:
if ( (testNumber % 5) == 0)
{
numberIsDivisibleBy5 = true
}
else
{
numberIsDivisibleBy5 = false
}
You'll need to provide the rest of the work to put this test into a loop that goes from >40 to <250, checking each number and doing whatever you are supposed to (print a message, probably) when you find one divisible by 5.
Good luck and thanks for using Fixya.
Jan 26, 2011 | Computers & Internet
### Sanner only taking one page and not the remaider in the tray
Switch the little button on top of the scanner next to the 3 buttons. This switches between 1 page or multiple page.
A.Misset
Installing these devices in the Netherlands.
Feb 05, 2010 | Canon DR-2580C Path-Through Scanner
## Open Questions:
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1 % 2 % $Id: Models.tex 1316 2007-09-25 03:18:30Z ksteube$ 3 % 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 5 % 6 % Copyright 2003-2007 by ACceSS MNRF 7 % Copyright 2007 by University of Queensland 8 % 9 10 % Primary Business: Queensland, Australia 11 % Licensed under the Open Software License version 3.0 12 13 % 14 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 15 % 16 17 \chapter{Models} 18 19 The following sections give a breif overview of the model classes and their corresponding methods. 20 21 \section{Stokes Cartesian (Saddle Point Problem)} 22 23 \subsection{Description} 24 25 Saddle point type problems emerge in a number of applications throughout physics and engineering. Finite element discretisation of the Navier-Stokes (momentum) equations for incompressible flow leads to equations of a saddle point type, which can be formulated as a solution of the following operator problem for $u \in V$ and $p \in Q$ with suitable Hilbert spaces $V$ and $Q$: 26 27 \begin{equation} 28 \left[ \begin{array}{cc} 29 A & B \\ 30 b^{*} & 0 \\ 31 \end{array} \right] 32 \left[ \begin{array}{c} 33 u \\ 34 p \\ 35 \end{array} \right] 36 =\left[ \begin{array}{c} 37 f \\ 38 g \\ 39 \end{array} \right] 40 \label{SADDLEPOINT} 41 \end{equation} 42 43 where $A$ is coercive, self-adjoint linear operator in $V$, $B$ is a linear operator from $Q$ into $V$ and $B^{*}$ is the adjoint operator of $B$. $f$ and $g$ are given elements from $V$ and $Q$ respectivitly. For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}. 44 45 The Uzawa scheme scheme is used to solve the momentum equation with the secondary condition of incompressibility \cite{GROSS2006,AAMIRBERKYAN2008}. 46 47 \begin{classdesc}{StokesProblemCartesian}{domain,debug} 48 opens the stokes equations on the \Domain domain. Setting debug=True switches the debug mode to on. 49 \end{classdesc} 50 51 example usage: 52 53 solution=StokesProblemCartesian(mesh) \\ 54 solution.setTolerance(TOL) \\ 55 solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\ 56 velocity,pressure=solution.solve(velocity,pressure,max\_iter=max\_iter,solver=solver) \\ 57 58 \subsection{Benchmark Problem} 59 60 Convection problem 61 62 63 \section{Temperature Cartesian} 64 65 \begin{equation} 66 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T 67 \label{HEAT EQUATION} 68 \end{equation} 69 70 where $\vec{v}$ is the velocity vector, $T$ is the temperature, $\rho$ is the density, $\eta$ is the viscosity, $c\hackscore{p}$ is the specific heat at constant pressure and $k$ is the thermal conductivity. 71 72 \subsection{Description} 73 74 \subsection{Method} 75 76 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG} 77 \end{classdesc} 78 79 \subsection{Benchmark Problem} 80 81 82 \section{Level Set Method} 83 84 \subsection{Description} 85 86 \subsection{Method} 87 88 Advection and Reinitialisation 89 90 \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth} 91 \end{classdesc} 92 93 %example usage: 94 95 %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth) 96 97 \begin{methoddesc}[LevelSet]{update\_parameter}{parameter} 98 Update the parameter. 99 \end{methoddesc} 100 101 \begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step} 102 Update level set function; advection and reinitialization 103 \end{methoddesc} 104 105 \subsection{Benchmark Problem} 106 107 Rayleigh-Taylor instability problem 108 109 110 \section{Drucker Prager Model} 111 112 \section{Plate Mantel} | 1,210 | 3,882 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-27 | latest | en | 0.775976 |
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# {Grade 5} NF.7 Interactive Math Notebook
Subject
Resource Type
Common Core Standards
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PDF (Acrobat) Document File
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4 MB|30+
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Product Description
***SAVE 20% by purchasing my NF INB Bundle!***
Looking for a way to make your math lessons/centers more engaging for your students? Looking to increase the amount of writing in math and application of skills? Use interactive notebooks!
This Common Core Aligned bundle includes activities for the following standard:
5.NF.7: Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
* Interpret division of a unit fraction by a non-zero whole number, and compute such quotients.
* Interpret division of a whole number by a unit fraction, and compute such quotients.
* Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem.
What's Included?
* 11 Activities:
- Interpreting Models (Unit Fractions ÷ Whole Numbers)
- Using Number Lines (Unit Fractions ÷ Whole Numbers)
- Using Bar Models (Unit Fractions ÷ Whole Numbers)
- Interpreting Models (Whole Numbers ÷ Unit Fractions)
- Using Number Lines (Whole Numbers ÷ Unit Fractions)
- Using Bar Models (Whole Numbers ÷ Unit Fractions)
- Mixed Word Problems
- Create Your Own (Word Problems)
- Writing to Explain
- Dividing with Unit Fractions and Whole Numbers
- Fraction Word Problems
* A Teacher Guide for EACH activity that includes directions, colored photograph samples, and answer keys
These activities can be used as guided practice, independent practice, morning work, homework, or in centers. Each activity varies in rigor and helps students dig deeper into the standard in an engaging and meaningful way.
Be sure to download the preview to see how I use them in my classroom and for a sample of each activity!
**Please consider leaving feedback!**
Thank you! :)
NOTE: While I always strive for accuracy, I am human and can make editing mistakes. If you find anything that needs correcting, please send me a message immediately. I would be happy to correct and upload a new file before you rate it! :) Thank you in advance.
Total Pages
30+
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N/A
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# Hello, I need help.
0
44
1
How many multiples of 9^3 are greater than 9^4 and less than 9^5?
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## How to change Straight table Expression in to Script logic by getting same result ?
Hi All,
I have a Straight table like Below . And i have got the average like Below
The Business Formula to Calculate Average:
(Today-Previous day)/Previous day * 100
The Expression to Calculate Average:
=Num((Sum(TGB_Counts)/Above(Sum({<Date>}TGB_Counts)) - 1) * Avg(1),'#,##0.00')
(The Expression is based on Above Business Formula)
Note: Now iam getting Correct Average,But My Business Sudden Request is to bring the above Average Expression logic in Script
Is there any way to get the above expression logic in Script So that i Dont Want to Perform Expression to Get Average. I can easily add Average as Dimension in Table Box Instead of Straight table
I Need above expression as Script Logic so that i will get Average as a field and i can add it as Dimension in Script.
Note: The Script logic output must be same like Above Straight table expression average values.
I am Searching a solution for it. kindly Help me out of this.
Please Find the attached Qvw and Excel Data for your Referance
Thanks,
Muthukumar
1 Solution
Accepted Solutions
MVP
Try this
Sheet1:
Date as Date_Text,
Left(Date, 10) as Date,
TGB_Counts
FROM
[Pos.xlsx]
(ooxml, embedded labels, table is Sheet1);
FinalTable:
Num(If([TGB Name] = Previous([TGB Name]), TGB_Counts/Previous(TGB_Counts) - 1), '#.00') as Average
Resident Sheet1
Order By [TGB Name], Date;
DROP Table Sheet1;
3 Replies
MVP
Try this
Sheet1:
Date as Date_Text,
Left(Date, 10) as Date,
TGB_Counts
FROM
[Pos.xlsx]
(ooxml, embedded labels, table is Sheet1);
FinalTable:
Num(If([TGB Name] = Previous([TGB Name]), TGB_Counts/Previous(TGB_Counts) - 1), '#.00') as Average
Resident Sheet1
Order By [TGB Name], Date;
DROP Table Sheet1;
Not applicable
Author
Thank you So much Sunny . May i Know how the above logic Exactly work.can you Show me with Sample Data Set
Because i dont Know how it Exactly Works. i can Under Stand up to below Step
Num(If([TGB Name] = Previous([TGB Name]), TGB_Counts/Previous(TGB_Counts)
i cant understand the -1 part and How it gets with the below Formula
The Business Formula to Calculate Average:
(Today-Previous day)/Previous day * 100
Can You Please Explain Me the Logic Behind it So that it will be help full for my Learning process
Thanks,
Muthukumar
MVP
May i Know how the above logic Exactly work.can you Show me with Sample Data Set
I did attach the sample you attached in my response... did you get to check it out? I mean the logic is simple
I am doing this
TGB_Counts/Previous(TGB_Counts) - 1
which is saying
Divide the current row of TGB_Counts with the previous row of TGB_Counts and subtract 1 from it. Just like you did this on the front end.....
Does that make sense?
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# Mkt 6100 Pre Case 2 Essay
1980 words - 8 pages
MKT 6100
Pre-Case Assignment 2
1. A. Contribution per CD Unit
Selling Price | \$9.00 |
CD Package and Disc | (\$1.25) |
Songwriters' Royalties | (\$0.35) |
Recording Artists' Royalties | (\$1.00) |
Contribution per CD unit | \$6.40 |
B. Break-Even volume in CD units
Fixed Cost: | | | | | |
Advertising | \$275,000 | | Break-Even in Units | |
Overhead | \$250,000 | | \$525,000 | = 82031.25 | units |
Total | \$525,000 | | \$6.40 | | |
| | | | | |
Contribution per CD Unit | \$6.40 | | | | |
Break-Even Volume in Dollars
Break-Even (Units) * Selling Price |
82.031.25 * \$ 9.00 = | \$738,281.25 |
C. Net profit if 1 ...view middle of the document...
Market Share to earn 20% return on investment
Investment | \$150,000 | |
Desired Return | 20% | |
Dollar amount to achieve goal | \$30,000 | |
| | |
Sales to meet goal | \$175,000+\$30,000 | =\$29,285.71 |
| 7 | |
| | |
Desired market = Expected Sales / Total Market | \$29,285.71 | = 29.29% |
| 100,000 | |
3. A. Absolute increase in unit sales and dollar sales to recoup incremental increase in advertising for Rash-Away? For Red-Away?
Incremental investment/Unit Contribution = Absolute increase in unit sales
Absolute increase in unit sales * Unit Price = Absolute increase in dollar sales
Rash-Away |
\$150,000 | 250,000 | Absolute increase in unit sales |
\$0.60 | | |
| | |
250,000* \$2.00 | \$500,000 | Absolute increase in dollar sales |
Red-Away |
\$150,000 | 200,000 | Absolute increase in unit sales |
\$0.75 | | |
| | |
200,000* \$1.00 | \$200,000 | Absolute increase in dollar sales |
B. How many additional sales dollars must be produced to cover each \$1.00 of incremental advertising for Rash-Away? Red-Away?
Rash-Away | \$2.00 | = \$3.33 |
| \$0.60 | |
Red-Away | \$1.00 | = \$1.33 |
| \$0.75 | |
C. What increase in absolute unit sales and dollar sales will be necessary to maintain the level of total contribution dollars if the price for each product is reduced by 10 percent?
Rash-Away |
Current Contribution Dollars = | |
1,000,000 units * \$.60 | = \$600,000 | |
| | |
New Price and Contribution with 10% Price Reduction = | |
\$1.80 Unit Price | | |
\$1.40 Variable Costs | | |
\$0.40 Unit Contribution | | |
| | |
\$0.40 | = 22.22 % Contribution Margin | |
\$1.80 | | |
| | |
\$.40(x) = \$600,000 | X= 1,500,000 units | |
| | |
Increase in Unit Sales = | 1,500,000 - 1,000,000 | = 500,000 |
| | |
.2222(x) = \$600,000 | X= \$2,700,000 (rounded) | |
| | |
Increase in Dollar Sales = | \$2,700,000 - \$2,000,000 | = \$700,000 |
Red-Away |
Current Contribution Dollars = | |
1,000,000 units * \$.75 | = \$1,125,000 | |
| | |
New Price and Contribution with 10% Price Reduction = | |
\$.90 Unit Price | | |
\$.25 Variable Costs | | |
\$0.65 Unit Contribution | | |
| | |
\$0.65 | = 72.22 % Contribution Margin | |
\$0.90 | | |
| | |
\$..65(x) = \$1,125,000 | X= 1,730,769 units | |
| | |
Increase in Unit Sales = | 1,730,769 - 1,500,000 | = 230,769 |
| | |
.7222(x) = \$1,125,000 | X= \$1,557,692 (rounded) | |
| | |
Increase in Dollar Sales = | \$1,557,692 - \$1,500,000 | = \$57,692 |
4. A. At what price will diversified Citrus Industries be selling their product to wholesalers?
Retail Price to Consumer | | |
100% - 20% = 80% | | \$50 * .8 = \$40 |
| | | | |
Retail Cost/Wholesaler Price | | |
100% - 10% = 90% | | \$40 * .9 = \$36 |
| | | | |
Wholesaler Cost/ Manufacturer Price | \$36 |
B. What is...
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924 words - 4 pages Mapping the Supply Chain for Deodorant Soap James Brown MKT/421 May 08, 2014 Ken Metz Mapping the Supply Chain for Deodorant Soap In 1970 Colgate-Palmolive marketed Irish Spring deodorant bar soap for the first time (Colgate-Palmolive Company, 2015). Irish Spring is one of many products produced by the 200 year old maker of personal care products. This paper will examine and map the supply chain used to produce | 2,747 | 10,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-35 | latest | en | 0.78324 |
http://www.iro.umontreal.ca/~pift1025/bigjava/Ch10/ch10.html | 1,524,679,612,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00333.warc.gz | 444,696,660 | 7,899 | # Testing and Debugging
## Chapter Goals
• To learn how to carry out unit tests
• To understand the principles of test case selection and evaluation
• To learn how to use logging
• To become familiar with using a debugger
• To learn strategies for effective debugging
## Unit Tests
• The single most important testing tool
• Checks a single method or a set of cooperating methods
• You don't test the complete program that you are developing; you test the classes in isolation
• For each test, you provide a simple class called a test harness
• Test harness feeds parameters to the methods being tested
## Example: Setting Up Test Harnesses
• To compute the square root of a use a common algorithm:
1. Guess a value x that might be somewhat close to the desired square root (x = a is ok)
2. Actual square root lies between x and a/x
3. Take midpoint (x + a/x) / 2 as a better guess
4. Repeat the procedure. Stop when two successive approximations are very close to each other
• Method converges rapidly. Square root of 100:
`Guess #1: 50.5Guess #2: 26.24009900990099Guess #3: 15.025530119986813Guess #4: 10.840434673026925Guess #5: 10.032578510960604Guess #6: 10.000052895642693Guess #7: 10.000000000139897Guess #8: 10.0`
## Testing the Program
• Output
```Enter a number: `100`
Guess #1: 50.5
Guess #2: 26.24009900990099
Guess #3: 15.025530119986813
Guess #4: 10.840434673026925
Guess #5: 10.032578510960604
Guess #6: 10.000052895642693
Guess #7: 10.000000000139897
Guess #8: 10.0
Guess #9: 10.0
Guess #10: 10.0
Square root: 10.0```
• Does the RootApproximator class work correctly for all inputs?
It needs to be tested with more values
• Re-testing with other values repetitively is not a good idea; the tests are not repeatable
• If a problem is fixed and re-testing is needed, you would need to remember your inputs
• Solution: Write test harnesses that make it easy to repeat unit tests
## Self Check
1. What is the advantage of unit testing?
2. Why should a test harness be repeatable?
1. It is easier to test methods and classes in isolation than it is to understand failures in a complex program.
2. It should be easy and painless to repeat a test after fixing a bug.
## Providing Test Input
• There are various mechanisms for providing test cases
• One mechanism is to hardwire test inputs into the test harness
• Simply execute the test harness whenever you fix a bug in the class that is being tested
• Alternative: place inputs on a file instead
## File RootApproximatorHarness1.java
Output
` square root of 100.0 = 10.0 square root of 4.0 = 2.0 square root of 2.0 = 1.414213562373095 square root of 1.0 = 1.0 square root of 0.25 = 0.5 square root of 0.01 = 0.1`
## Providing Test Input
• You can also generate test cases automatically
• For few possible inputs, feasible to run through (representative) number of them with a loop
## File RootApproximatorHarness2.java
Output
` square root of 1.0 = 1.0 square root of 1.5 = 1.224744871391589 square root of 2.0 = 1.414213562373095 . . . square root of 9.0 = 3.0 square root of 9.5 = 3.0822070014844885 square root of 10.0 = 3.162277660168379`
## Providing Test Input
• Previous test restricted to small subset of values
• Alternative: random generation of test cases
## File RootApproximatorHarness3.java
Output
` square root of 810.4079626570873 = 28.467665212607223 square root of 480.50291114306344 = 21.9203766195534 square root of 643.5463246844379 = 25.36821485017103 square root of 506.5708496713842 = 22.507128863348704 square root of 539.6401504334708 = 23.230156057019308 square root of 795.0220214851004 = 28.196134867834285 . . .`
## Providing Test Input
• Selecting good test cases is an important skill for debugging programs
• Test all features of the methods that you are testing
• Test typical test cases
100, 1/4, 0.01, 2, 10E12, for the SquareRootApproximator
• Test boundary test cases: test cases that are at the boundary of acceptable inputs
0, for the SquareRootApproximator
• Programmers often make mistakes dealing with boundary conditions
Division by zero, extracting characters from empty strings, and accessing null pointers
• Gather negative test cases: inputs that you expect program to reject
Example: square root of -2. Test passes if harness terminates with assertion failure (if assertion checking is enabled)
## Reading Test Inputs from a File
• More elegant to place test values in a file
• Input redirection:
`java Program < data.txt`
• Some IDEs do not support input redirection. Then, use command window (shell).
• Output redirection:
`java Program > output.txt`
## Reading Test Inputs from a File
• File test.in:
``1` 100`2` 4`3` 2`4` 1`5` 0.25`6` 0.01`
• Run the program:
`java RootApproximatorHarness4 < test.in > test.out`
• File test.out:
``1` square root of 100.0 = 10.0`2` square root of 4.0 = 2.0`3` square root of 2.0 = 1.414213562373095`4` square root of 1.0 = 1.0`5` square root of 0.25 = 0.5`6` square root of 0.01 = 0.1`
## Self Check
1. How can you repeat a unit test without having to retype input values?
2. Why is it important to test boundary cases?
1. By putting the values in a file, or by generating them programmatically.
2. Programmers commonly make mistakes when dealing with boundary conditions.
## Test Case Evaluation
• How do you know whether the output is correct?
• Calculate correct values by hand
E.g., for a payroll program, compute taxes manually
• Supply test inputs for which you know the answer
E.g., square root of 4 is 2 and square root of 100 is 10
• Verify that the output values fulfill certain properties
E.g., square root squared = original value
• Use an Oracle: a slow but reliable method to compute a result for testing purposes
E.g., use Math.pow to slower calculate x1/2 (equivalent to the square root of x)
## Output
`Test passed: x = 913.6505141736327, root squared = 913.6505141736328Test passed: x = 810.4959723987972, root squared = 810.4959723987972Test passed: x = 503.84630929985883, root squared = 503.8463092998589Test passed: x = 115.4885096006315, root squared = 115.48850960063153Test passed: x = 384.973238438713, root squared = 384.973238438713. . .Pass: 100Fail: 0`
## Output
`Test passed: square root = 718.3849112194539, oracle = 718.3849112194538Test passed: square root = 641.2739466673618, oracle = 641.2739466673619Test passed: square root = 896.3559528159169, oracle = 896.3559528159169Test passed: square root = 591.4264541724909, oracle = 591.4264541724909Test passed: square root = 721.029957736384, oracle = 721.029957736384. . .Pass: 100Fail: 0`
## Self Check
1. Your task is to test a class that computes sales taxes for an Internet shopping site. Can you use an oracle?
2. Your task is to test a method that computes the area of an arbitrary polygon. Which polygons with known areas can you use as test inputs?
1. Probably not–there is no easily accessible but slow mechanism to compute sales taxes. You will probably need to verify the calculations by hand.
2. There are well-known formulas for the areas of triangles, rectangles, and regular n-gons.
## Regression Testing
• Save test cases
• Use saved test cases in subsequent versions
• A test suite is a set of tests for repeated testing
• Cycling = bug that is fixed but reappears in later versions
• Regression testing: repeating previous tests to ensure that known failures of prior versions do not appear in new versions
## Test Coverage
• Black-box testing: test functionality without consideration of internal structure of implementation
• White-box testing: take internal structure into account when designing tests
• Test coverage: measure of how many parts of a program have been tested
• Make sure that each part of your program is exercised at least once by one test case
E.g., make sure to execute each branch in at least one test case
• Tip: write first test cases before program is written completely → gives insight into what program should do
• Modern programs can be challenging to test
• Graphical user interfaces (use of mouse)
• Network connections (delay and failures)
• There are tools to automate testing in this scenarios
• Basic principles of regression testing and complete coverage still hold
## Self Check
1. Suppose you modified the code for a method. Why do you want to repeat tests that already passed with the previous version of the code?
2. Suppose a customer of your program finds an error. What action should you take beyond fixing the error?
1. It is possible to introduce errors when modifying code.
2. Add a test case to the test suite that verifies that the error is fixed.
## Unit Testing with JUnit
• http://junit.org
• Built into some IDEs like BlueJ and Eclipse
• Philosophy: whenever you implement a class, also make a companion test class
## Program Trace
• Messages that show the path of execution
`if (status == SINGLE){ System.out.println("status is SINGLE"); . . . }. . .`
• Drawback: Need to remove them when testing is complete, stick them back in when another error is found
• Solution: use the Logger class to turn off the trace messages without removing them from the program
## Logging
• Logging messages can be deactivated when testing is complete
• Use global object Logger.global
• Log a message
`Logger.global.info("status is SINGLE");`
• By default, logged messages are printed. Turn them off with
`Logger.global.setLevel(Level.OFF);`
• Logging can be a hassle (should not log too much nor too little)
• Some programmers prefer debugging (next section) to logging
## Logging
• When tracing execution flow, the most important events are entering and exiting a method
• At the beginning of a method, print out the parameters:
`public TaxReturn(double anIncome, int aStatus){ Logger.global.info("Parameters: anIncome = " + anIncome + " aStatus = " + aStatus); . . .}`
• At the end of a method, print out the return value:
`public double getTax(){ . . . Logger.global.info("Return value = " + tax); return tax;}`
## Self Check
1. Should logging be activated during testing or when a program is used by its customers?
2. Why is it better to send trace messages to Logger.global than to System.out?
1. Logging messages report on the internal workings of your program–your customers would not want to see them. They are intended for testing only.
2. It is easy to deactivate Logger.global when you no longer want to see the trace messages, and to reactivate it when you need to see them again.
## Using a Debugger
• Debugger = program to run your program and analyze its run-time behavior
• A debugger lets you stop and restart your program, see contents of variables, and step through it
• The larger your programs, the harder to debug them simply by logging
• Debuggers can be part of your IDE (Eclipse, BlueJ) or separate programs (JSwat)
• Three key concepts:
• Breakpoints
• Single-stepping
• Inspecting variables
## Debugging
• Execution is suspended whenever a breakpoint is reached
• In a debugger, a program runs at full speed until it reaches a breakpoint
• When execution stops you can:
• Inspect variables
• Step through the program a line at a time
• Or, continue running the program at full speed until it reaches the next breakpoint
• When program terminates, debugger stops as well
• Breakpoints stay active until you remove them
• Two variations of single-step command:
• Step Over: skips method calls
• Step Into: steps inside method calls
## Single-step Example
• Current line:
```String input = in.next();Word w = new Word(input);`int syllables = w.countSyllables();`
System.out.println("Syllables in " + input + ": " + syllables);```
• When you step over method calls, you get to the next line:
`String input = in.next();Word w = new Word(input);int syllables = w.countSyllables();`System.out.println("Syllables in " + input + ": " + syllables);``
• However, if you step into method calls, you enter the first line of the countSyllables method
```public int countSyllables(){` int count = 0;`
int end = text.length() - 1;
. . .
}```
## Self Check
1. In the debugger, you are reaching a call to System.out.println. Should you step into the method or step over it?
2. In the debugger, you are reaching the beginning of a long method with a couple of loops inside. You want to find out the return value that is computed at the end of the method. Should you set a breakpoint, or should you step through the method?
1. You should step over it because you are not interested in debugging the internals of the println method.
2. You should set a breakpoint. Stepping through loops can be tedious.
## Sample Debugging Session
• Word class counts syllables in a word
• Each group of adjacent vowels (a, e, i, o, u, y) counts as one syllable
• However, an e at the end of a word doesn't count as a syllable
• If algorithm gives count of 0, increment to 1
• Constructor removes non-letters at beginning and end
## Debug the Program
• Buggy output (for input "hello yellow peach"):
`Syllables in hello: 1Syllables in yellow: 1Syllables in peach: 1`
• Set breakpoint in first line of countSyllables of Word class
• Start program, supply input. Program stops at breakpoint
• Method checks if final letter is 'e'
• Check if this works: step to line where check is made and inspect variable ch
• Should contain final letter but contains 'l'
## More Problems Found
• end is set to 3, not 4
• text contains "hell", not "hello"
• No wonder countSyllables returns 1
• Culprit is elsewhere
• Can't go back in time
• Restart and set breakpoint in Word constructor
## Debugging the Word Constructor
• Supply "hello" input again
• Break past the end of second loop in constructor
• Inspect i and j
• They are 0 and 4–makes sense since the input consists of letters
• Why is text set to "hell"?
• Off-by-one error: Second parameter of substring is the first position not to include
• text = substring(i, j);
should be
text = substring(i, j + 1);
## Another Error
• Fix the error
• Recompile
• Test again:
`Syllables in hello: 1Syllables in yellow: 1Syllables in peach: 1`
• Oh no, it's still not right
• Start debugger
• Erase all old breakpoints and set a breakpoint in countSyllables method
• Supply input "hello"
## Debugging countSyllables (again)
• Break in the beginning of countSyllables. Then, single-step through loop
`boolean insideVowelGroup = false;for (int i = 0; i <= end; i++){ ch = Character.toLowerCase(text.charAt(i)); if ("aeiouy".indexOf(ch) >= 0) { // ch is a vowel if (!insideVowelGroup) { // Start of new vowel group count++; insideVowelGroup = true; } }}`
• First iteration ('h'): skips test for vowel
• Second iteration ('e'): passes test, increments count
• Third iteration ('l'): skips test
• Fifth iteration ('o'): passes test, but second if is skipped, and count is not incremented
## Fixing the Bug
• insideVowelGroup was never reset to false
• Fix
`if ("aeiouy".indexOf(ch) >= 0){ . . .}else insideVowelGroup = false;`
• Retest: All test cases pass
`Syllables in hello: 2Syllables in yellow: 2Syllables in peach.: 1`
• Is the program now bug-free? The debugger can't answer that.
## Self Check
1. What caused the first error that was found in this debugging session?
2. What caused the second error? How was it detected? | 4,008 | 15,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-17 | latest | en | 0.821356 |
https://www.mathworks.com/matlabcentral/answers/435152-gpu-arrayfun-is-so-slow-what-is-going-on | 1,597,004,160,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738573.99/warc/CC-MAIN-20200809192123-20200809222123-00231.warc.gz | 737,617,345 | 29,792 | # GPU arrayfun is so slow, what is going on?
12 views (last 30 days)
Hao Zhang on 11 Dec 2018
Commented: Derrick Ling on 20 Apr 2019
Hi,
I am trying to understand what the GPU arrayfun is doing? The following is a test code.
clear;clc;close all
gd=gpuDevice();
reset(gd);
N=2e3;
a=rand(60,N,'single','gpuArray');
tic;
b=sum(a,1);
wait(gd);
toc;
tic;
c=arrayfun(@(i) sum(a(:,i),1),(1:N));
wait(gd);
toc;
The results are:
Elapsed time is 0.000468 seconds.
Elapsed time is 0.584521 seconds.
What is going on here? 1000 times difference?? I would expect similary runtime since GPU arrayfun is supposed to be executed parallel on GPU cores. Did I make stupid errors on using the arrayfun?
Thanks!
#### 1 Comment
Hao Zhang on 13 Dec 2018
Hi, I come back for more updates. I have successfully vectorized and implemented my particle simulation on GPU. The speed up is astonishing, ~10 times faster than CPU code. Thanks Matt J and Joss Knight for their wonderful suggestions.
Now the other part of the code (except neighbor search but solving the fluid equations) is so fast that the limiting part now is the matlab function knnsearch, which uses kdtree algorithm runing on CPU. It takes 85% percent of the runtime (see the following code profiler results)
The function 'knnCPU_kdtree_func' uses the matlab built-in function knnsearch with kdtree algorithm runing on CPU. The other functions are doing the real math runing on GPU only consumes 10% of the total time.
I wonder is there any GPU implementation of k-nearest neighbor search that I can free download and using as a function call in my matlab code? Many thanks.
Matt J on 11 Dec 2018
Edited: Matt J on 11 Dec 2018
What is the most efficient way to vectorize the above code
I would say, as follows,
idx_Neighbor=randi([1 N],60,N,'uint8');
temp=p(idx_Neighbor);
temp=temp+p.';
ax=sum(temp .* delW_x,1);
Show 1 older comment
Matt J on 11 Dec 2018
However it is a bit pity that the vectorized code consume more memory
It shouldn't. I conserved memory by creating idx_Neighbour as uint8 instead of double.
This is the first time I know one can do this eventhough the indexing is not the same dimension as the inferred array, is there any matlab documentation on this?
Oddly, I can't find any!
How about the efficiency of the vectorized code on GPU comparing with highly optimized cuda C code?
You would have to implement it to find out...
Hao Zhang on 11 Dec 2018
yes, but one can do uint8 to any code above. And if the indexing is larger then one has to use uint32. But the improvement of vectorizing is so much that one has to bear with a bit more memory usage.
Maybe a matlab developer can say something about the efficiency comparing to cuda C. I hope it will be really close.
Derrick Ling on 20 Apr 2019
Hi, which code did you use to run GPU? arrayfun? gpuArray?
Where or how did you insert the code?
And is there any advice you would give to make this code nicer? Undefined function error appears if I remove bbb = 0.
bbb = 0;
bbb = bbb + h*W_4';
Joss Knight on 11 Dec 2018
You haven't called GPU arrayfun here, you've called CPU arrayfun and in the arrayfun function you are doing stuff on the GPU. This is because none of the arguments to your arrayfun call is a gpuArray.
You could force it to use GPU arrayfun by converting your input:
c = arrayfun(@(i) sum(a(:,i),1), gpuArray(1:N));
However, you'll immediately find it errors, because sum is not supported for GPU arrayfun. Obviously this is just a toy example, but the solution here is sum(a,1), not arrayfun.
Show 1 older comment
Matt J on 11 Dec 2018
I need to do c=sum(a.*repmat(b,1,2e3),1).
Your operation does not require repmat nor even sum, but simply
c=b.'*a;
But even if you were to use sum, recent Matlab no longer requires repmat, e.g.,
a=gpuArray.rand(60,2e3); b=a(:,1);
c=sum(a.*b,1);
will work fine.
Joss Knight on 11 Dec 2018
Thanks Matt.
GPU arrayfun is very special, you should read the documentation and list of supported functions. It only supports element-wise functionality, so you can't do any vector operations. That means you can't index an array unless you're indexing a single element or an up-level variable, you can't call sum or any other reduction or accumulation, and you can't output anything other than a scalar. This is because your arrayfun function gets compiled into a single CUDA kernel with no inter-thread communication. So it's incredibly useful and efficient when used within its limitations.
Nearly always (in my experience) when you want to do something more complex with vector operations, you can translate your code into a series of vectorized calls to normal MATLAB matrix functions, arrayfun, and pagefun.
Hao Zhang on 11 Dec 2018
This is a brillant solution! Thanks for making this happen.
However, if I want to do something even more complex (And this is what I actually need to do, instead of the toy examples before :)). So I need to do the following code:
clear;clc;close all;
N=2e3;
idx_Neighbor=randi([1 N],60,N);
p=rand(N,1);
delW_x=rand(60,N);
ax=zeros(N,1);
tic;
for j=1:N
temp=idx_Neighbor(:,j);
inner=p(temp)+p(j);
ax(j)=sum(inner.*delW_x(:,j));
end
toc;
What is the most efficient way to vectorize the above code (so without using the for loop) and avoiding using repmat as much as possible? Thanks! | 1,366 | 5,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-34 | latest | en | 0.916023 |
https://www.weegy.com/?ConversationId=A155B6EB | 1,582,747,180,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146485.15/warc/CC-MAIN-20200226181001-20200226211001-00475.warc.gz | 934,288,874 | 10,110 | 3(p+10)=
Asked 5/24/2011 2:28:42 AM
Updated 6/8/2014 4:54:55 PM
This conversation has been flagged as incorrect.
Flagged by jerry06 [6/8/2014 4:54:55 PM]
s
Original conversation
User: 3(p+10)=
earnonline724|Points 169|
User: multiply this problem
User: multiply 3(p+10)
Weegy: integers are multiplied by dividing it first by its recipricoles
Asked 5/24/2011 2:28:42 AM
Updated 6/8/2014 4:54:55 PM
This conversation has been flagged as incorrect.
Flagged by jerry06 [6/8/2014 4:54:55 PM]
Rating
3
3(p + 10) = 3p + 30
Added 6/8/2014 4:54:53 PM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [6/8/2014 4:56:05 PM]
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Questions asked by the same visitor
The area A of a triangle with base b and height h is given by A=1/2bh. What is the area when b=24m (meters) and h=22m?
Weegy: Area of triangle is 1/2b*h, so, 1/2*22*24 =264 m^2 (More)
Question
Updated 1/5/2017 4:25:26 AM
-2 + (-9) =
Question
Updated 11/27/2015 12:48:11 AM
-2 + (-9) = -11
Added 11/27/2015 12:48:11 AM
This answer has been confirmed as correct and helpful.
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,258 | 3,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-10 | latest | en | 0.873475 |
http://www.csc.kth.se/~dog/xjobb/ | 1,500,676,669,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423812.87/warc/CC-MAIN-20170721222447-20170722002447-00200.warc.gz | 403,022,416 | 3,063 | School of
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KTH / CSC / TCS / Douglas Wikström / Degree Projects
# Degree Projects
If you want to do your degree project (sv. examensarbete) in cryptography specifically and you are already in contact with an industrial partner, then please send me an email containing a brief description of your project proposal.
If you are simply looking for a supervisor for a project in computer science, then you should contact Ann Bengtson.
Any projects defined by myself are posted below and also on the page for degree projects of the TCS group. On the latter page you can also find the degree projects defined by other members of the TCS group.
## Degree Projects In Cryptography
### Preferential Voting With Mix-Nets
In some countries, the voter does not give a vote for a single candidate. Instead it ranks the candidates or a subset of the candidates on its ballot.
To decide who wins the elections we think of each ballot as a stack. We then tabulate the election as if the top element of each stack was the single vote given by the voter. If there is a majority vote for some party, then it wins. Otherwise we pop the stacks where the top element is a vote for the party with the smallest number of votes. This candidate is now eliminated and removed from all stacks. We proceed in this way until the winner is identified. (There are variations of this tabulation procedure.)
A mix-net is a protocol executed by a number of parties that takes a list of encrypted votes as input and outputs the votes in random order without leaking anything else. Thus, the correspondence between inputs and outputs is broken.
To implement preferential voting using a mix-net, the obvious idea is to simply view a ranking as a vote, use the mix-net and then do the preferential tabulation as outlined above on the output from the mix-net. Unfortunately, this is bad idea when the number of candidates is moderately large. The problem is that a voter may be forced, or paid, to use a particular ranking that is unlikely to be used by anybody else, thus allowing the adversary to verify that the voter voted in a particular way. This is sometimes called the "Italian attack". Thus, researchers have tried to come up with more complex protocols for tabulation for preferential voting.
The goal of the project is to:
1. Study cryptographic protocols for finding the winner(s) in preferential voting systems based on mix-nets and possibly improve them.
2. Implement one of these protocols. There is an ongoing implementation project for mix-nets here that should be used as the starting point.
To successfully complete this project the you need a strong background in mathematics/theoretical computer science, as well as in programming. Preferably, you should have completed the course Foundations in cryptography or something equivalent.
If you are interested, then please email Douglas Wikström at `dog@csc.kth.se`.
(If you can read a little Swedish, then you can try to mount an "Italian attack" against the Swedish election scheme, which is defined by Vallagen!) | 628 | 3,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-30 | latest | en | 0.938358 |
https://www.physicsforums.com/threads/if-mutually-exclusive-prove-pr-a-pr-b.427170/ | 1,519,603,508,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00722.warc.gz | 910,966,038 | 15,875 | # If mutually exclusive, prove Pr(A) <= Pr(B')
1. Sep 7, 2010
### bjersey
I'm having issues proving the following which should be simple:
If A and B are mutually exclusive, prove Pr(A) <= Pr(B')
From the statement about being mutually exclusive, I know A $$\cap$$ B = $$\phi$$
Therefore we have P(A $$\cap$$ B) = Pr(A) + Pr(B)
Also, A = A $$\cap$$ B'
and B = A' $$\cap$$ B
But I'm having a hard time putting all of this together.
2. Sep 7, 2010
### mathman
A = A $$\cap$$ B' implies A is a subset of B'.
3. Sep 17, 2010
### _joey
Here's my solution.
Pr(A union B) <= Pr(Entire Sample Space)=1 from probability axiom; and
Pr(A union B) = Pr(A)+Pr(B)-Pr(A intersection B)
Hence, Pr(A intersection B) >= Pr(A) + Pr(B) -1 ........(3)
But for two mutually exclusive events Pr(A intersection B) = P(empty) = 0. Also, Pr(B) = 1- Pr(B')
It follows from (3) that
0>=Pr(A)+1-Pr(B')-1
Therefore, Pr(B')>=Pr(A)
As required
4. Sep 17, 2010
You could cut the final proof down a little:
\begin{align*} P(A) + P(B) - P(A \cap B) & \le 1 \\ P(A) + P(B) & \le 1 \\ P(A) & \le 1 - P(B) \\ P(A) & \le P(B') \end{align*}
5. Sep 17, 2010
### _joey
The inequality in (3) is well known named after Bonferonni. It has many applications. Its derivation is good to know too. Anyway, your answer is neat.
6. Sep 18, 2010
Yes, Bonferroni's inequality is important (classical example is in the first study of multiple comparisons), but the original question wasn't about that; usually you want the derivations to be as straightforward as possible.
there is nothing wrong with the earlier solution, but the mix of mathematics and english makes its reading awkward. learning when the written word can be safely removed from the mathematical work is an important step as well.
7. Sep 18, 2010
### mathman
I think my comment is the simplest. A is a subset of B', therefore P(A) ≤ P(B').
8. Sep 18, 2010 | 588 | 1,902 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-09 | longest | en | 0.919636 |
https://www.coursehero.com/file/19586/ch07/ | 1,548,227,453,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584203540.82/warc/CC-MAIN-20190123064911-20190123090911-00268.warc.gz | 745,248,125 | 107,985 | # ch07 - CHAPTER 7 1(a The system under consideration has...
• Notes
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CHAPTER 7 1. (a) The system under consideration has rotational degrees of freedom, allowing it to rotate about two orthogonal axes perpendicular to the rigid rod connecting the two masses. If we define the z axis as represented by the rod, then the Hamiltonian has the form H = L x 2 + L y 2 2 I = L 2 - L z 2 2 I where I is the moment of inertia of the dumbbell. (b) Since there are no rotations about the z axis, the eigenvalue of L z is zero, so that the eigenvalues of the Hamiltonian are E = h 2 l ( l + 1) 2 I with l = 0,1,2,3,… (c) To get the energy spectrum we need an expression for the moment of inertia. We use the fact that I = M red a 2 where the reduced mass is given by M red = M C M N M C + M N = 12 × 14 26 M nuc leon = 6 .46 M nuc leon If we express the separation a in Angstroms, we get I = 6 .46 × (1 .67 × 10 - 27 kg )(10 - 10 m / A ) 2 a A 2 = 1 .08 × 10 - 46 a A 2 The energy difference between the ground state and the first excited state is 2 h 2 /2 I which leads to the numerical result E = (1 .05 × 10 - 34 J . s ) 2 1 .08 × 10 - 46 a A 2 kg . m 2 × 1 (1 .6 × 10 - 19 J / eV ) = 6 .4 × 10 - 4 a A 2 eV 2. We use the connection x r = s in θ co s φ ; y r = s in θ s in φ ; z r = co s θ to write
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Y 1 ,1 =- 3 8 π e i φ sin θ =- 3 8 π ( x + iy r ) Y 1 ,0 = 3 4 π cos θ = 3 4 π ( z r ) Y 1 , - 1 = ( - 1) Y 1,1 * = 3 8 π e - i φ sin θ = 3 8 π ( x - iy r ) Next we have Y 2 ,2 = 15 32 π e 2 i φ sin 2 θ = 15 32 π (cos2 φ + i sin2 φ )sin 2 θ = 15 32 π (cos 2 φ - sin 2 φ + 2 i sin φ cos φ )sin 2 θ = 15 32 π x 2 - y 2 + 2 ixy r 2 Y 2 ,1 =- 15 8 π e i φ s in θ co s θ =- 15 8 π ( x + iy ) z r 2 and Y 2 ,0 = 5 16 π (3co s 2 θ - 1 ) = 5 16 π 2 z 2 - x 2 - y 2 r 2 We may use Eq. (7-46) to obtain the form for Y 2 , - 1 andY 2 , - 2 .
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• Spring '05
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• Mass, Trigraph, Excited state, ground state, 2s cs 4s, o2φ snφ+ isn
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Jill Tulane University ‘16, Course Hero Intern | 1,040 | 3,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-04 | latest | en | 0.81677 |
http://www.chegg.com/homework-help/questions-and-answers/someone-could-help-problem-would-appreciateit-great-deal-totally-lost-solenoid-equation-to-q417360 | 1,444,361,673,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737913406.61/warc/CC-MAIN-20151001221833-00241-ip-10-137-6-227.ec2.internal.warc.gz | 475,118,972 | 12,798 | If someone could help me with this problem I would appreciateit a great deal. I am totally lost on which solenoid equation touse and how to apply the info to magnetic flux of loop inside. Iwill rate as high as possible for this answer. Thank you.
A solenoid has 1950 turns of wire andis 125 cm long and 10.0 cm in diameter; see Figure P.14. A circularwire loop of diameter 5.0 cm lies along the axis of the solenoidnear the middle of its length as shown.
Figure P.14.
(a) If the current in the solenoid initially is6.0 A, find the magnetic flux throughthe smaller loop.
Tยทm2 | 153 | 577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-40 | latest | en | 0.91599 |
http://talkstats.com/threads/expectation-problem.22846/ | 1,670,190,038,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00469.warc.gz | 52,679,909 | 9,681 | # expectation problem
#### bravehera1
##### New Member
10 balls are inserted (uniformly) into five jars ..each insertion is independent from the others.
compute E(X1*X2) where X1 and X2 describes the number of balls in each jar.
so far:
-P=Pr(Success from each jar poin of view=the ball was inserted into it)=1/5
-E(X1*X2)=E(E(X1*X2|X1))=E(X1*(10-X1)*P)=P(E(10X1-X1^2)=1OPE(X1)-E(X1^2)=10*(1/5)-10*9*((1/5)^2)+10*(1/5)) =4-90/25+2
what am i doing wrong ?
Last edited:
#### Dason
Do you mean X1 is the number placed in the first jar and X2 is the number placed in the second jar?
#### bravehera1
##### New Member
Do you mean X1 is the number placed in the first jar and X2 is the number placed in the second jar?
nop, x1 =number of balls inserted into the first jar..
in each step, a ball is inserted into one of the five jars.. so after 10 steps there are
x1 balls in jar 1,x2 balls in jar 2 ...
#### David1
##### New Member
Define Xi- the number of balls that were inserted into jar i. (notice that Xi~Bin(10,1/5))
Define pi- the probability of a ball to be inserted into jar i. In this case, pi=1/5 for all i.
X=(X1,...,X5)
P=(p1,...,p5)
X~ Multinomial (10, P)
and Cov(Xi,Xj)=-n*pi*pj
http://en.wikipedia.org/wiki/Multinomial_distribution
use the following formula to find the expectation of the product:
E(X1*X2)-E(X1)*E(X2)=Cov(X1,X2) | 439 | 1,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-49 | latest | en | 0.848368 |
https://support.minitab.com/en-us/minitab/18/help-and-how-to/statistics/tables/how-to/tally-individual-variables/methods-and-formulas/methods-and-formulas/ | 1,513,101,528,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948517845.16/warc/CC-MAIN-20171212173259-20171212193259-00324.warc.gz | 665,823,281 | 4,269 | # Methods and formulas for Tally Individual Variables
Select the method or formula of your choice.
## Counts
Count is the number of times each unique value occurs.
## Cumulative counts (CumCt)
A cumulative count is the cumulative total of all the counts up to that unique value. For example, suppose your data are as follows: 1, 2, 1, 1, 3, 3, 1, 1, 3, 3. The cumulative count of Values 1 and Values 2 is 6, and the cumulative count of all three values is 10.
Unique value Count Cumulative count
1 5 5
2 1 5 + 1 = 6
3 4 5 + 1 + 4 = 10
## Percents
### Formula
For example, the percentage of Value 1 is 50%.
Unique value Count Percent
1 5 100 * 5/10 = 50%
2 1 100 * 1/10 = 10%
3 4 100 * 4/10 = 40%
## Cumulative percents (CumPct)
Minitab calculates the cumulative percentage for each unique value.
### Formula
For example, the cumulative percentage of Values 1 and Values 2 is 60%, and the cumulative percentage of all three values is 100%.
Unique value Count Cumulative percent
1 5 100 * 5/10 = 50%
2 1 100 * (5 + 1) / 10 = 60%
3 4 100 * (5 + 1 + 4) / 10 = 100%
By using this site you agree to the use of cookies for analytics and personalized content. Read our policy | 373 | 1,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-51 | longest | en | 0.791657 |
http://www.thefullwiki.org/Mundell-Fleming_model | 1,571,101,544,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00503.warc.gz | 313,925,178 | 12,621 | # Mundell-Fleming model: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
The Mundell-Fleming model is an economic model first set forth by Robert Mundell and Marcus Fleming. The model is an extension of the IS-LM model. Whereas IS-LM deals with economy under autarky, the Mundell-Fleming model tries to describe an open economy.
Typically, the Mundell-Fleming model portrays the relationship between the nominal exchange rate and an economy's output (unlike the relationship between interest rate and the output in the IS-LM model) in the short run. The Mundell-Fleming model has been used to argue that an economy cannot simultaneously maintain a fixed exchange rate, free capital movement, and an independent monetary policy. This principle is frequently called "the Unholy Trinity," the "Irreconcilable Trinity," the "Inconsistent trinity" or the Mundell-Fleming "trilemma."
## Basic set up
The traditional model is based around the following equations.
• Y = C + I + G + NX (The IS Curve)
• Where Y is GDP, C is consumption, I is investment, G is government spending and NX is net exports.
• $\frac{M}{P}=L(i, Y)$ (The LM Curve)
• Where M is money supply, P is average price, L is liquidity, i is the interest rate and Y is GDP.
### IS components
• C = C(YT,iE(π))
• Where C is consumption, Y is GDP, T is taxes, i is the interest rate, E(π) is the expected rate of inflation.
• I = I(iE(π),Y − 1)
• Where I is investment, i is the interest rate, E(π) is the expected rate of inflation, Y − 1 is GDP in the previous period.
• G = G
• Where G is government spending, an exogenous variable.
• NX = NX(e,Y,Y * )
### BoP components
• CA = NX
• Where CA is the current account and NX is net exports.
• KA = z(ii * ) + k
• Where z is the level of capital mobility, i is the interest rate, i * is the foreign interest rate, k is capital investments not related to i, an exogenous variable
## Mechanics of the model
One important assumption is the equalization of the local interest rate to the global interest rate.
### Under flexible exchange rate regime
We speak of a system of flexible exchange rates when governments (or central banks) allow the exchange rate to be determined by market forces alone.
#### Changes in money supply
An increase in money supply will shift the LM curve downward. This directly reduces the local interest rate and in turn forces the local interest rate lower than the global interest rate. This depreciates the exchange rate of local currency through capital outflow. (Hot money flows out to take advantage of higher interest rate abroad and hence currency depreciates.) The depreciation makes local goods cheaper compared to foreign goods and increases export and decreases import. Hence, net export is increased. Increased net export leads to the shifting of the IS curve (which is Y = C + I + G + NX) to the right to the point where the local interest rate will equalize with the global rate. At the same time, the BoP is supposed to shift too, as to reflect(1)depreciation of home currency and (2)an increase in current account or in other word, the increase in net export. These increases the overall income in the local economy.
A decrease in money supply will cause the exact opposite of the process.
#### Changes in government spending
An increase in government expenditure shifts the IS curve to the right. The shift will cause the local interest rate to go above the global rate. The increase in local interest will cause capital inflow and the inflow will make the local currency stronger compared to foreign currencies. Strong exchange rate also makes foreign goods cheaper compared to local goods. This encourages greater import and discourages export and hence, lower net export. As a result, the IS will return to its original location where the local interest rate is equal to the global interest rate. The level of income of the local economy stays the same. The LM curve is not at all affected.
A decrease in government expenditure will reverse the process.
#### Changes in global interest rate
An increase in the global interest rate will cause an upward pressure on the local interest rate. The pressure will subside as the local rate closes in on the global rate. When a positive differential between the global and the local rate occurs, holding the LM curve constant, capital will flow out of the local economy. This depreciates the local currency and helps boost net export. Increasing net export shifts the IS to the right. This shift will continue to the right until the local interest rate becomes as high as the global rate.
A decrease in global interest rate will cause the reverse to occur.
### Under fixed exchange rate regime
We speak of a system of fixed exchange rates when governments (or central banks) announce an exchange rate (the parity rate) at which they are prepared to buy or sell any amount of domestic currency.
#### Changes in money supply
Under the fixed exchange rate system, the local central bank or any monetary authority will only change the money supply in order to maintain a level of exchange rate. If there is a pressure to appreciate the exchange rate, the local authority will buy domestic currency in order to decrease the money supply to raise the exchange rate back to its original level. If there is pressure to depreciate the exchange rate, the local authority will buy foreign currency with domestic currency in order to increase the money supply and lower the exchange rate back to its original level.
A revaluation occurs when there is a permanent increase in exchange rate and hence, decrease in money supply. Devaluation is the exact opposite of revaluation.
#### Changes in government expenditure
Increase in government spending forces the monetary authority to flood the market with local currency in order to keep the exchange rate unchanged.
Increased government expenditure shifts the IS curve to the right. The shift results are a rise in the interest rate and hence, an appreciation of the exchange rate. However, the exchange rate is controlled by the local monetary authority in the framework of a fixed system. In order to maintain the exchange rate and eliminate the pressure from it, the monetary authority will purchase foreign currencies with local currencies until the pressure is gone i.e. back to the original level. Such action shifts the LM curve in tandem with the direction of the IS shift. This action increases the local currency supply in the market and lowers the exchange rate—or rather, return the rate back to its original state. In the end, the exchange rate stays the same but the general income in the economy increases.
The reverse is true when government expenditure decreases.
#### Changes in global interest rate
To maintain the fixed exchange rate, the central bank must offset the capital flows (in or out) which are caused by the change of the global interest rate to the domestic rate. The central bank must restore the situation where the real domestic interest rate is equal to the real global interest rate to stop net capital flows from changing the exchange rate.
If the global interest rate increases above the domestic rate, capital will flow out to take advantage of this opportunity.(Hot money flows out of the economy) This would depreciate the home currency, so the central bank may buy the home currency and sell some of its foreign currency reserves to offset this outflow. This decrease in the money supply shifts the LM curve to the left until the domestic interest rate is the global interest rate.
If the global interest rate declines below the domestic rate, the opposite occurs. Hot money flows in, the home currency appreciates, so the central bank offsets this by increasing the money supply (sell domestic currency, buy foreign currency), the LM curve shifts to the right, and the domestic interest rate becomes the global interest rate.
## Differences from IS-LM
It is worth noting that some of the result from this model differs from the IS-LM because of the open economy assumption. Result for large open economy on the other hand falls within the result predicted by the IS-LM and the Mundell-Fleming models. The reason for such result is because a large open economy has both the characteristics of an autarky and a small open economy.
In the IS-LM, interest rate will be the key component in making both the money market and the good market in equilibrium. Under the Mundell-Fleming framework of small economy, interest rate is fixed and equilibrium in both market can only be achieved by a change of nominal exchange rate.
### Example
A much simplified version of the Mundell-Fleming model can be illustrated by a small open economy, in which the domestic interest rate is exogenously predetermined by the world interest rate (r=r*).
Consider an exogenous increase in government expenditure, the IS curve will shift upward, with LM curve intact, causing the interest rate and the output to rise (partial crowding out effect) under the IS-LM model.
Nevertheless, as interest rate is predetermined in a small open economy, the LM* curve (of exchange rate and output) is vertical, which means there is exactly one output that can make the money market in the equilibrium under that interest rate. Even though the IS* curve still shift up, it will result in a higher exchange rate and same level of output (complete crowding out effect, which is different in the IS-LM model).
The example above makes an implicit assumption of flexible exchange rate. The Mundell-Fleming model can have completely different implications under different exchange rate regimes. For instance, under a fixed exchange rate system, with perfect capital mobility, monetary policy becomes ineffective. An expansionary monetary policy resulting in an outward shift of the LM curve would in turn make capital flow out of the economy. The central bank under a fixed exchange rate system would have to intervene by selling foreign money in exchange for domestic money to depreciate the foreign currency and appreciate the domestic currency. Selling foreign money and receiving domestic money would reduce real balances in the economy, until the LM curve shifts back to the left, and the interest rates come back to the world rate of interest i*. | 2,138 | 10,497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2019-43 | latest | en | 0.937124 |
http://math.stackexchange.com/questions/176228/integrals-of-forms-are-equal-implies-they-differ-by-d-mu | 1,469,683,011,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00321-ip-10-185-27-174.ec2.internal.warc.gz | 163,156,096 | 17,630 | Integrals of forms are equal implies they differ by $d\mu$
The problem is about the proof of the following result.
If $\omega_1,\omega_2 \in \Omega^n_c(X)$ (where $X$ is smooth manifold) are such that $\int_X\omega_1=\int_X \omega_2$ then there is $\mu\in\Omega^{n-1}_c(X)$ such that $\omega_1-\omega_2=d\mu$.
I am using text at link. Part where I am stuck is on numerical page 202 which is page 30 in the file. It says that using partition of unity argument in Step 4 we can assume $\omega_1,\omega_2$ are supported in some parametrizable open sets.
Can someone explain to me why is this true?
-
Note that $\omega\in \Omega_c^n(X)$ means $\omega$ is a $\mathcal{C}^\infty$ $n$-form, compactly supported on the manifold $X$. The notation is introduced on page 106 of the document. Partitions of unity are discussed starting on page 188.
Regarding the beginning of step 4: By assumption, $\int_X \omega_i = \int_{U_i}\omega_i = 0$. Step 1 tells us that $\omega_i = d\mu_i$ for $\mu_i \in \Omega_c^{n-1}(U_i)$. This implies $\omega_1\sim \omega_2$.
Another way to look at this proof is to ignore the argument regarding the special case $c=0$. Assume $\int_X\omega_1 = \int_X\omega_2 = c$ where $c$ can be zero. Don't divide by $c$. Go through step 5 but instead choose $\int_X \alpha_i = c$. The rest of step 5 goes through as written, giving $(a)\Rightarrow (b)$.
I understand the notation, the problem is if we have $\int_X \omega_1=\int_X \omega_2$ and $\rho$ is any function from the partition of the unity we do not know that $\int_X \rho \omega_1=\int_X\rho\omega_2$? I feel they used this in the proof. Is there any explanation for this? – dmm Jul 29 '12 at 8:27 | 505 | 1,673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2016-30 | latest | en | 0.853224 |
https://eng.libretexts.org/Bookshelves/Computer_Science/Book%3A_A_First_Course_in_Electrical_and_Computer_Engineering_(Scharf)/6%3A_Linear_Algebra/6.01%3A_Introduction | 1,596,890,670,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737645.2/warc/CC-MAIN-20200808110257-20200808140257-00132.warc.gz | 287,113,883 | 23,621 | # 6.01: Introduction
We use this chapter to introduce students to the algebraic structure of vectors and matrices and to introduce them to matrix computations. These matrix computations are used in the chapters "Vector Graphics", "Filtering", and "Binary Codes" to solve problems in vector graphics, filtering, and binary coding.
Vectors are introduced in the section "Vectors", along with algebraic and geometric interpretations of some fundamental vector operations and properties. The section "Inner Product and Euclidean Norm", the section "Direction Cosines", and the section "Projections" introduce inner products and their applications, including norm, direction cosines, orthogonality, and projections. Some important alternatives to the Euclidean norm are introduced in the section "Other Norms". Matrices are motivated and introduced in this section. The notation in these sections can be daunting to the beginner, so we proceed very carefully, using example after example. In the section "Solving Linear Systems of Equations" we codify the elimination procedures that students have used in high school to solve linear systems of equations. The MATLAB demonstration in Demo 2 shows how to use MATLAB to solve linear equations. The section "Circuit Analysis" shows how linear algebra and MATLAB can be used to analyze dc circuits. The numerical experiment "Circuit Design" gives students practice in building function files in MATLAB and shows how to solve a sequence of linear equations in order to design a circuit with desired properties.
Occasionally we have placed important results in the problems. We feel that students should not miss the material in Exercise 3 in "Vectors", Exercise 3 in "Inner Product and Euclidean Norm", Exercise 3 in "Projections", Exercise 1 in "Matrices", and Exercise 4 in "Matrices".
## Introduction
Linear algebra is a branch of mathematics that is used by engineers and applied scientists to design and analyze complex systems. Civil engineers use linear algebra to design and analyze load-bearing structures such as bridges. Mechanical engineers use linear algebra to design and analyze suspension systems, and electrical engineers use it to design and analyze electrical circuits. Electrical, biomedical, and aerospace engineers use linear algebra to enhance X rays, tomographs, and images from space. In this chapter and the next we study two common problems from electrical engineering and use linear algebra to solve them. The two problems are (i) electrical circuit analysis and (ii) coordinate transformations for computer graphics. The first of these applications requires us to understand the solution of linear systems of equations, and the second requires us to understand the representation of mathematical operators with matrices.
Much of linear algebra is concerned with systematic techniques for organizing and solving simultaneous linear equations by elimination and substitution. The following example illustrates the basic ideas that we intend to develop.
Example $$\PageIndex{1}$$
A woman steps onto a moving sidewalk at a large airport and stands while the moving sidewalk moves her forward at 1.2 meters/seconds. At the same time, a man begins walking against the motion of the sidewalk from the opposite end at 1.5 meters/second (relative to the sidewalk). If the moving sidewalk is 85 meters long, how far does each person travel (relative to the ground) before they pass each other?
To solve this problem, we first assign a variable to each unknown quantity. Let x1 be the distance traveled by the woman, and let x2 be the distance traveled by the man. The sum of the two distances is 85 meters, giving us one equation:
$x_1+x_2=85$
Our second equation is based on the time required before they pass. Time equals distance divided by rate, and the time is the same for both people:
$\frac {x_1} {1.2} = \frac {x_2} {1.5−1.2} ⇒0.3x_1 − 1.2x_2 = 0$
We may substitute the Equation $$\PageIndex{2}$$ into the Equation $$\PageIndex{1}$$ to obtain the result $$/frac {1.2} {0.3} x_2 + x_2 = 85$$ or
$5x_2=85 ⇒ x_2=17$
Combining the result from Equation $$\PageIndex{3}$$ with that of Equation $$\PageIndex{1}$$, we find that
$x_1=68$
So the man travels 17 meters, and the woman travels 68 meters.
Equation $$\PageIndex{1}$$ and Equation $$\PageIndex{2}$$ are the key equations of Equation $$\PageIndex{1}$$. They may be organized into the “matrix equation”
$\begin{bmatrix} 1 & 1 \\ 0.3 & -1.2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 85 \\ 0 \end{bmatrix}$
The rules for matrix-vector multiplication are evidently
$(1)x_1+(1)x_2 = 85$
$(0.3)x_1+(−1.2)x_2 = 0$
Equation $$\PageIndex{2}$$ and Equation $$\PageIndex{3}$$ may be organized into the matrix equation
$\begin{bmatrix} 0 & 0 \\ 0.3 & -1.2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 85 \\ 0 \end{bmatrix}$
This equation represents one partially solved form of Equation $$\PageIndex{5}$$, wherein we have used the so-called Gauss elimination procedure to introduce a zero into the matrix equation in order to isolate one variable. The MATLAB software contains built-in procedures to implement Gauss elimination on much larger matrices. Thus MATLAB may be used to solve large systems of linear equations.
Before we can apply linear algebra to more interesting physical problems, we need to introduce the mathematical tools we will use. | 1,246 | 5,442 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-34 | latest | en | 0.914756 |
https://arzung.org/coordinate-graphing-worksheets/ | 1,620,436,389,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00354.warc.gz | 129,211,645 | 10,019 | # Coordinate graphing worksheets Awesome
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Coordinate Graphing Worksheets. Printable bar graph and pictograph worksheets. Some worksheets include only positive numbers while others include positive and negative numbers. 21102019 This resource contains a worksheet where pupils are asked to plot coordinates on grids containing 4 quadrants. Plotting points in all 4 quadrants.
Coordinate Plane Graphing Pictures By Level Of Difficulty Charlie Brown Daffy Duck Homer Simpson Coordinate Plane Graphing Coordinate Graphing Math School From pinterest.com
Ordered Pairs and Coordinate Plane Worksheets. All points are represented by whole numbers there are no fractions or decimals. Practice plotting ordered pairs with this fun coordinate graphing mystery picture. Practice naming and plotting points on a grid. Graphing points on coordinate plane worksheets are meant to simplify the process. These graphing puzzles contain over 40 ordered pairs to plot divided into 3 to 16 different shapes.
### This Graphing Worksheet will produce a four quadrant coordinate grid and a set of ordered pairs that when correctly plotted and connected will produce different characters.
Plotting points in all 4 quadrants. Worksheets for teaching coordinate grids ordered pairs and plotting points. This resource also contains an easy to use answer sheet. Some worksheets include only positive numbers while others include positive and negative numbers. Mapping Worksheets Classroom Caboodle 291700. Graphing points on coordinate plane worksheets are meant to simplify the process.
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Free Printable Coordinate Graphing Worksheets Free Printable Coordinate Graphing Worksheets can help a trainer or college student to learn and realize the lesson plan within a quicker way. Practice plotting ordered pairs with this fun coordinate graphing mystery picture. Mapping Worksheets Classroom Caboodle 291700. Coordinate Graph Worksheet - Free Kindergarten Math Worksheet for Kids 291699. Simple yet engaging ordered pairs and coordinates worksheets assist in identifying the position of objects on a grid spot the ordered pairs plot them recognize quadrants and axes with and without grids show routes reveal the mystery picture by plotting and joining the points and much more.
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Coordinate Graph Worksheet - Free Kindergarten Math Worksheet for Kids 291699. For an extra challenge use larger grid sizes and leave off the grid lines. A coordinate graph also known as a coordinate plane is a 2D graph comprised of two intersecting lines one vertical and one horizontal. If your young learner needs a confidence boost graphing points on a coordinate plane worksheets can help in. You may select which one of the characters you wish to make.
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Ordered Pairs and Coordinate Plane Worksheets. Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6. Free Printable Coordinate Graphing Worksheets Free Printable Coordinate Graphing Worksheets can help a trainer or college student to learn and realize the lesson plan within a quicker way. Practice plotting ordered pairs with this fun coordinate graphing mystery picture. The vertical line is called the x-axis and the horizontal line is the y-axis.
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Graphing points on coordinate plane worksheets are meant to simplify the process. Coordinate Graph Worksheet - Free Kindergarten Math Worksheet for Kids 291699. These worksheets are printable pdf files. What is Coordinate Graphing. This worksheet provides the student with four series of points with their task being to plot them on the graph and connect the dots revealing four different shapes.
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For an extra challenge use larger grid sizes and leave off the grid lines. Plotting points in all 4 quadrants. Descartes perception of that fly drove him to build up an organized diagram. For an extra challenge use larger grid sizes and leave off the grid lines. This activity is easy to differentiate by choosing either the first quadrant positive whole numbers or the four quadrant positive and negative whole numbers worksheet.
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Free Printable Coordinate Graphing Worksheets Free Printable Coordinate Graphing Worksheets can help a trainer or college student to learn and realize the lesson plan within a quicker way. A facilitate chart is likewise in some cases called an organized plane a Cartesian plane or a Cartesian arrange framework. All points are represented by whole numbers there are no fractions or decimals. Main Idea Picture Cards and Worksheets Free to Print. Printable bar graph and pictograph worksheets.
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The printable worksheets in this page cover identifying quadrants axes identifying ordered pairs coordinates plotting points on coordinate plane and other fun worksheet pdfs to reinforce the knowledge in ordered pairs. For an extra challenge use larger grid sizes and leave off the grid lines. A coordinate graph also known as a coordinate plane is a 2D graph comprised of two intersecting lines one vertical and one horizontal. Simple Coordinate Graphing Image Worksheet. Mapping Worksheets Classroom Caboodle 291700.
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The generator is useful for 4th 5th 6th and 7th grades from the time when students learn about the coordinate grid till they study integers and reflections and translations in the coordinate. What is Coordinate Graphing. Youll need to plot around 30 points on the provided vertical coordinate grid to figure out what this mystery image is. Descartes perception of that fly drove him to build up an organized diagram. If your young learner needs a confidence boost graphing points on a coordinate plane worksheets can help in.
Source: pinterest.com
Develop coordinate plotting and locating skills. Ordered Pairs and Coordinate Plane Worksheets Print out these worksheets on coordinate grids coordinate planes and ordered pairs. An arrange chart comprises of two number lines that run opposite to. These graphing puzzles contain over 40 ordered pairs to plot divided into 3 to 16 different shapes. This worksheet provides the student with four series of points with their task being to plot them on the graph and connect the dots revealing four different shapes.
Source: pinterest.com
Also contains mystery pictures moving points using position and direction identifying shapes and more. Plotting points in all 4 quadrants. Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6. This Graphing Worksheet will produce a four quadrant coordinate grid and a set of ordered pairs that when correctly plotted and connected will produce different characters. Simple Coordinate Graphing Image Worksheet.
Source: pinterest.com
The generator is useful for 4th 5th 6th and 7th grades from the time when students learn about the coordinate grid till they study integers and reflections and translations in the coordinate. 21102019 This resource contains a worksheet where pupils are asked to plot coordinates on grids containing 4 quadrants. Some worksheets include only positive numbers while others include positive and negative numbers. This Graphing Worksheet will produce a four quadrant coordinate grid and a set of ordered pairs that when correctly plotted and connected will produce different characters. What is Coordinate Graphing.
Source: pinterest.com
Main Idea Picture Cards and Worksheets Free to Print. Develop coordinate plotting and locating skills. Also contains mystery pictures moving points using position and direction identifying shapes and more. An arrange chart comprises of two number lines that run opposite to. So all the work is.
Source: pinterest.com
The generator is useful for 4th 5th 6th and 7th grades from the time when students learn about the coordinate grid till they study integers and reflections and translations in the coordinate. For an extra challenge use larger grid sizes and leave off the grid lines. Youll need to plot around 30 points on the provided vertical coordinate grid to figure out what this mystery image is. The vertical line is called the x-axis and the horizontal line is the y-axis. Practice naming and plotting points on a grid.
Source: pinterest.com
The vertical line is called the x-axis and the horizontal line is the y-axis. Simple yet engaging ordered pairs and coordinates worksheets assist in identifying the position of objects on a grid spot the ordered pairs plot them recognize quadrants and axes with and without grids show routes reveal the mystery picture by plotting and joining the points and much more. The vertical line is called the x-axis and the horizontal line is the y-axis. Develop coordinate plotting and locating skills. Ordered Pairs and Coordinate Plane Worksheets Print out these worksheets on coordinate grids coordinate planes and ordered pairs.
Source: pinterest.com
The vertical line is called the x-axis and the horizontal line is the y-axis. Find an unlimited supply of printable coordinate grid worksheets in both PDF and html formats where students either plot points tell coordinates of points plot shapes from points reflect shapes in the x or y-axis or move translate them. Practice plotting ordered pairs with this fun coordinate graphing mystery picture. The printable worksheet includes an answer key so you can double check your work. Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6.
Source: pinterest.com
Also contains mystery pictures moving points using position and direction identifying shapes and more. Descartes perception of that fly drove him to build up an organized diagram. The printable worksheets in this page cover identifying quadrants axes identifying ordered pairs coordinates plotting points on coordinate plane and other fun worksheet pdfs to reinforce the knowledge in ordered pairs. 21102019 This resource contains a worksheet where pupils are asked to plot coordinates on grids containing 4 quadrants. Practice plotting ordered pairs with this fun coordinate graphing mystery picture.
Source: pinterest.com
The generator is useful for 4th 5th 6th and 7th grades from the time when students learn about the coordinate grid till they study integers and reflections and translations in the coordinate. 21102019 This resource contains a worksheet where pupils are asked to plot coordinates on grids containing 4 quadrants. Worksheets for teaching coordinate grids ordered pairs and plotting points. Free Printable Coordinate Graphing Worksheets Free Printable Coordinate Graphing Worksheets can help a trainer or college student to learn and realize the lesson plan within a quicker way. Plotting points in all 4 quadrants.
Source: pinterest.com
Printable bar graph and pictograph worksheets. What is Coordinate Graphing. The generator is useful for 4th 5th 6th and 7th grades from the time when students learn about the coordinate grid till they study integers and reflections and translations in the coordinate. Youll need to plot around 30 points on the provided vertical coordinate grid to figure out what this mystery image is. This resource also contains an easy to use answer sheet.
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http://docplayer.net/26351413-Lecture-12-quantum-mechanics-and-atomic-orbitals.html | 1,542,716,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746386.1/warc/CC-MAIN-20181120110505-20181120132505-00402.warc.gz | 86,230,352 | 26,935 | # Lecture 12 Quantum Mechanics and Atomic Orbitals
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## Transcription
1 Lecture 12 Quantum Mechanics and Atomic Orbitals Bohr and Einstein demonstrated the particle nature of light.e = hν. De Broglie demonstrated the wavelike properties of particles. λ = h/mv. However, these results only applied to systems with one electron. Attempts to apply these models to more complex atomic systems failed.in 1926, three scientists, Erwin Schrödinger, Werner Heisenberg and Paul Dirac, introduced totally different approaches to the theoretical descriptions of atomic systems. They each developed different mathematical formalisms to describe the behavior of electrons. Eventually, all three formalisms were integrated into quantum or wave mechanics, which revolutionized physics and chemistry. Schrödinger's treatment is the most commonly used. Based on De Broglie's work, Schrödinger reasoned that all matter can be described by a wave function, a mathematical equation which describes wave motion, a differential equation.this is called the Schrödinger wave function and is given the symbol ψ. Each allowed state of the electron has a different wave function. To calculate the energy associated with a particular wave function, a mathematical function, called an operator, is applied to the wave function. In order to calculate energy, the operator Ĥ (Hamiltonian operator) is used. The result is: Ĥψ = Eψ where E is the calculated energy. The mathematics are very complex, and the Schrödinger equation has only been solved exactly for hydrogen. Computers have allowed us to approximate the solution for many other systems. The square of the function, ψ 2, is related to the probability of finding an electron in a region in space, the probability of distribution. (In wave theory, the intensity of light is proportional to the square of the amplitude of the wave). The complete solution of the Schrödinger equation for hydrogen yields a set of wave functions and a set of corresponding energies. The wave functions are called orbitals. Each orbital describes the distribution of electron density in space.shown below, for example, is the probability of finding the electron in a hydrogen atom at a particular distance from the nucleus. Quantum numbers Bohr had a single quantum number, n, to describe the energy levels in hydrogen.the quantum mechanical model uses three quantum numbers, n, l, and m l to describe an orbital. n is the principal quantum number. It is given values of 1, 2, 3,... It determines the overall size of the orbital and energy of the electron. As n increases, the orbitals become larger, and the electron spends more time farther away from the nucleus. The further an electron is from the nucleus, the less tightly bound it becomes. For hydrogen-like systems, the quantum mechanical calculation of energy becomes exactly the same as the equation developed by Bohr. Page 1
2 l is the angular momentum quantum number. It has values of 0, 1, 2,..., (n-1) for any value of n. It determines the shape of the orbital. These are usually designated by letters: l = 0 l = 1 l = 2 l = 3 s p d f Beyond l = 3, the letters continue in alphabetical order (g, h, i,...) m l is the magnetic quantum number. It is given values between l and -l. It determines the orientation of the orbital in space. Orbitals with the same principal quantum number, n, are in the same electron shell. Orbitals with the same principal quantum number, n and the same angular momentum quantum number, l, are in the same subshell. For n = 4, we could determine the allowed subshells and number of orbitals. If n = 4, l = 0, 1, 2, 3 l = 0 = 4s subshell l = 1 = 4p subshell l = 2 = 4d subshell l = 3 = 4f subshell In the 4s subshell, l = 0 so m l = 0 and there is 1 orbital In the 4p subshell, l = 1 so m l = -1, 0, +1 and there are 3 orbitals In the 4d subshell, l = 2 so m l = -2, -1, 0, +1, +2 and there are 5 orbitals In the 4f subshell, l = 3 so m l = -3, -2, -1, 0, +1, +2, +3 and there are 7 orbitals Could there be 2d orbitals? No. n = 2, so l = 0, 1. l = 0 are s orbitals. l = 1 are p orbitals. For d orbitals, l = 2, and that is not possible when n = 2. Orbital shapes and Energies s orbitals l = 0 Shown below is a plot of the radial probability distribution of the 1s atomic orbital in hydrogen. It shows the probability of finding an electron at distances away from the nucleus.the peak occurs at about nm. Page 2
3 The radial probability distributions for the 2s and 3s atomic orbitals are shown below. Notice that the probability falls to zero at certain distances. These are called nodes, areas where the wave function has zero amplitude. All s orbitals are spherical. The relative sizes of the 1s, 2s and 3 s orbitals are shown below. These figures represent, the boundary surfaces of these orbitals, the volume which contains 90% of the electron density. In cross section, the nodes in the 2 and 3 s orbitals are visible. p orbitals l = 1 The principal quantum number n = 1 can only have l = 0 and no p orbitals. For n = 2, l = 0 and 1, so m l = -1, 0 and 1. These are the three p orbitals. Shown below are the boundary surfaces of the three 2p orbitals, 2p x, 2p y and 2p z. Page 3
4 Except for their differing orientations, these orbitals are identical in shape and energy.the p orbitals of higher principal quantum numbers have similar shapes. d orbitals l = 2 The principal quantum number n = 3 is the first that that can have d orbitals: m l = -2, -1, 0, 1, 2. These are shown below. The different orientations correspond to the different values of m l. All five orbitals have identical energy. The d orbitals for n > 3 have similar shapes. Because of the shapes and orientations of the p and d orbitals, electrons occupying different atomic orbitals are as far apart from each other as possible. This minimizes the electron-electron repulsion. Beyond the d orbitals, there are f and g etc. orbitals. The f orbitals are important in accounting for the behavior of elements bigger than Cerium ( > 58), but their orbital shapes are difficult to represent. The Spin Quantum Number, m s A Dutch physicist, Pieter Zeeman, discovered that the atomic emission spectral lines are split into multiple lines when an electric field is applied. These additional lines cannot be accounted for by just the three quantum numbers, n, l and m l. Two other Dutch physicists, Samuel Goudsmit and George Uhlenbeck, suggested that a fourth quantum number would be necessary to explain these findings. This number takes into account the magnetic properties of electrons. Although electrons do not actually spin, they do possess a magnetic moment, and can exist in one of two states, spin up, or spin down. The quantum numbers associated with this are m s = + ½ and m s = - ½. Wolfgang Pauli, an Austrian physicist, stated that in a given atom, no two electrons can have the same four quantum numbers. This is called the Pauli Exclusion Page 4
5 Principle. Two electrons occupying the same orbital must have different spin states ). ( They are said to be paired electrons. This fourth quantum number is not included in the Schrödinger equation. However, Dirac's treatment predicted the necessity of a fourth quantum number. It is the magnetic moment associated with electron spin that gives all substances their magnetic properties. Substances with all of their electrons paired are slightly repelled by magnetic fields and are said to be diamagnetic. Substances with electrons that haveunpaired spins are attracted to magnetic fieldsand are said to be paramagnetic. Polyelectronic Atoms The Pauli Exclusion Principle is one of the factors in determining the electronic configuration in an atom. Another basic concept is that each electron will occupy the lowest energy state available to it, without violating the Pauli Exclusion Principle. We have already described the size and shape of the atomic orbitals, now we need to look at their energy levels. The closer an electron gets to the nucleus, the stronger the electrostatic attraction, the lower the energy of the system. For the hydrogen atom, this produces the energy levels shown below: The 1s orbital has the electron closest to the nucleus, so it has the lowest energy.the 2s and 2p orbitals have the same energy for hydrogen. They are said to be degenerate energy levels, all the same. The n = 3 orbitals are the next highest in energy, followed by the degenerate n = 4 orbitals. When the electron is held in the 1s orbital, it is said to be in its ground state, its lowest energy state. When the electron is a higher energy orbital, it is said to be in an excited state. The energy diagram for polyelectronic atoms (atoms with more than 1 electron) is more complex. Fortunately, we can describe polyelectronic atoms in terms of orbitals like those calculated for hydrogen.the orbitals have the same names (s, p, d, f, etc.) and shapes. However, the presence of more than one electron greatly changes the energies of the orbitals. This is due t the electron-electron repulsion within a polyelectronic atom and to the fact that an electron at a distance from the nucleus will be screened or shielded from the full nuclear charge by other electrons that are closer to the nucleus.the energy of an electron in a polyelectronic atom generally increases as n increases. But, for a given value of n, the lower the value of l, the lower the energy. This is because of an effect called penetration, the fraction of time that an electron spends close to the nucleus. This can most easily be seen by comparing the probability distribution of a series of orbitals. The radial distribution of the 3s, 3p and 3d orbitals is shown below: Page 5
6 The most probable (highest peak) distance of the electron from the nucleus decreases in going from the 3s to 3p to 3d orbitals. This would imply that the stability of electrons in these three types of orbitals would increase as it gets closer to the nucleus. In fact, this is not the case. What seems to determine the relative energy levels are the two small peaks very close to the nucleus for the 2s orbital (blue), the one small peak for the 3d orbital (orange) and the absence of any peaks near the nucleus for the 3d orbital (green). The penetration effect leads to an ordering of the energy of orbitals within a shell of s < p < d. However, energy levels of different shells can overlap. This leads to an energy level diagram for polyelectronic atoms as shown below. There is actually a handy way to remember this fairly complex order of energy levels.first, make a list of the orbitals, as shown below: 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p Then draw diagonal arrows from right to left and fill the orbitals in that order. Page 6
7 In order, this gives, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p Electron Configuration The four quantum numbers, n, l, m l and m s completely label an electron is any orbital of an atom.you can think of them as its address. The four quantum numbers for a 1s electron are: n = 1, l = 0, m l = 0, m s = + ½ n = 1, l = 0, m l = 0, m s = - ½ This demonstrates that the 1s orbital can only contain two electrons. It is inconvenient to write them out this way, but there is a shorthand notation for this same information.for hydrogen, which has one electron, which occupies the 1s orbital in its ground state, the information would be given by: 1s 1 For helium with two electrons, the ground state representation would be: 1s 2 Lithium has three electrons. The 1s orbital will be filled with two electrons. The next lowest orbital (see above) is the 2s orbital, which can also hold two electrons. 1s 2 2s 1 Another important rule, Hund's rule, becomes obvious when we explicitly show the distribution of the electrons. 1s 2s 2p x 2p y 2p z 3s 3p 4s H 1s 1 He 1s 2 Li 1s 2 2s 1 Be 1s 2 2s 2 B 1s 2 2s 2 2p 1 C 1s 2 2s 2 2p 2 Page 7
8 N 1s 2 2s 2 2p 3 O 1s 2 2s 2 2p 4 F 1s 2 2s 2 2p 5 Ne 1s 2 2s 2 2p 6 Hund's rule states that the most stable arrangement of electrons in a subshell is the one with the most parallel spins. The effect of this rule is highlighted in red. When we add a second electron to the 2p orbital in carbon, it is added in a different subshell but with the same spin. We don't begin to pair electrons until oxygen, when all of the subshells are already half full. The elements in this list are the representative elements of groups IA-VIIIA. Other members of these groups will have the same valence electron configuration (outermost electrons). Let's look at sodium, with 11 electrons. 1s 2s 2p x 2p y 2p z 3s 3p 4s Na 1s 2 2s 2 2p 6 3s 1 The electron configuration in blue exactly matches the electron configuration of Ne. We have simply added an additional electron in the 3s orbital. Rather than repeat the Ne configuration, Na can be represented as : [Ne] 3s 1. The shorthand notation uses one of the Noble Gases to account for the core electron configuration. Let's do the third row of the Periodic Table this way. This row starts with potassium, K, with 19 electrons. These elements will all have the core electron configuration of argon, Ar, with 18 electrons. Looking at the energy diagram, 18 electrons will fill 1s (2 electrons), 2s (2 electrons), 2p (6 electrons), 3s (2 electrons) and 3p (6 electrons). The next energy level will be 4s, followed by the five 3d orbitals. 4s 3d xz 3d xz 3d xz 3d xz 3d xz 4p x 4p y 4p z 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p K [Ar] 4s 1 Ca [Ar] 4s 2 Sc [Ar] 4s 2 3d 1 Ti [Ar] 4s 2 3d 2 V [Ar] 4s 2 3d 3 Cr [Ar] 4s 1 3d 5 Mn [Ar] 4s 2 3d 5 Fe [Ar] 4s 2 3d 6 Co [Ar] 4s 2 3d 7 Ni [Ar] 4s 2 3d 8 Cu [Ar] 4s 1 3d 10 Zn [Ar] 4s 2 3d 10 There are unusual electron configurations in chromium and copper (highlighted in red).chromium should have been [Ar] 4s 2 3d 4, but is instead, [Ar] 4s 1 3d 5. The theoretical reason for this is still a bit uncertain, but a convenient answer is to assume that there is a special stability associated withalf-full orbitals. So, there are 5 electrons in the d subshell, and only 1 electron in the s subshell. The same anomolous behavior occurs with copper. We would expect [Ar] 4s 2 3d 9, but instead get [Ar] 4s 1 3d 10. This puts 10 electrons in the d subshell (full) and only 1 electron in the s subshell (half full). Page 8
9 Now we can look at the Chemical consequences of these electron configurations. Page 9
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